Page 1

ELEMENT DESIGN TO SHAPE A STRUCTURE SABAH SHAWKAT

II


In memory of our dear colleague Prof. Miloslav Mudrončík


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Reviewer: Cover Design: Software Support: Printing/ Binding:

Prof. Dipl. Ing. Ján Hudák, PhD. Richard Schlesinger asc. Applied Software Consultants, s.r.o., Bratislava, Slovakia Tribun EU, s.r.o., Brno, Czech Republic

All rights reserved. No part of this book may be reprinted, or reproduced or utilized in any form or by any electronic, mechanical or other means, including photocopying, without permission in writing from the author.

Element Design to Shape a Structure II. ©

Assoc. Prof. Dipl. Ing. Sabah Shawkat, MSc, PhD. 1. Edition, Tribun EU, s.r.o. Brno 2017 ISBN xxxxxxxxxxx


He teach students of architecture several structural engineering subjects. Moreover, he regularly organize workshops

for students and exhibitions of their projects and construction models. He also actively practise in projecting and building constructions as well as reconstructions and modernizations of buildings.

Sabah Shawkat is also a passionate

expert in reinforced concrete, prestressed concrete structures and structural design. He has published numerous articles in professional journals and wrote several books.

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Department of Architecture Sabah Shawkat Zahawi is the Head of Engineering Room at the Academy of Fine Arts and Design in Bratislava, Slovakia.


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Introduction Example 4.2-14 Verify the stresses in concrete and in steel 4. Limitation of Stress

1

4. 1 Verification at the Serviceability Limit States

22

of a T-beam

1

Example 4.2-15 Verify the position of the neutral axis of a T-beam 24

4.1.1 Permissible Stresses

1

Example 4.2-16 Design the flexural reinforcement for the I-beam 25

Example 4.1.1-1 Verification of the stresses in concrete and in steel

2

Example 4.2-17 Verify the stresses in concrete

3

Example 4.2-18 Verify the stresses in concrete

26

and in steel of a T-beam

reinforcement 4.2 Service Load Stress-Straight-Line Theory 4.2.1 Analysis for Stresses-Section cracked and elastic Example 4.2.1 Compute the stresses in concrete and steel

and in steel of a T-beam

3 4

Example 4.2-19 Verify the stresses of concrete

29

and steel of a T-column

reinforcement for a given bending moment Example 4.2-2 Compute the specified load stress in concrete

27

5

Example 4.2-20 Verify the stresses in concrete

30

and in steel of a T-beam

and steel Example 4.2-3 Compute the stress in concrete and steel

7

Example 4.2-21 Design the flexural reinforcement for the T beam 32

Example 4.2-4 Check the stresses of concrete and steel

8

Example 4.2-22 Assessment of tension parts of PPRCB according 33 to limit state of crack width

for a given bending moment and axial load Example 4.2-5 Verify the stresses in concrete and in steel

9

Example 4.2-23 Assessment the risk of longitudinal cracks due to 39

11

Example 4.2-24 Limitation of concrete stress due to

increased pressure stress concrete

reinforcement Example 4.2-6 Verify the stresses in concrete and in steel

an increased creep

reinforcement Example 4.2-7 Check the stresses in concrete and in

11 5. Deformation Behaviour of Reinforced Concrete Beams

steel reinforcement Example 4.2-8 Check the stresses in concrete and reinforcement

13

5.1 Deformation Behaviour of Reinforcement Concrete Beams

44 49

for I, T and rectangular sections

caused by a bending moment Ms and external normal tension load

5.1.1 Specimen and Material Details

50 50 52

Example 4.2-9 Verify the stresses in concrete and in steel

14

5.1.2 Loading and Instrumentation Details

Example 4.2-10 Verify the stresses in concrete and in steel,

16

5.1.3 Methods 5.1.4 Discussion and Analysis of the Result

the cross- section caused by a bending moment

5.2 Determination of Strain Energy on Reinforced Concrete Beams

and normal compression force Example 4.2-11 check the stresses in concrete and in steel

41

17

5.2.1 Methods 5.3 Crack Development and The Strain Energy in Reinforced Concrete Beams

reinforcement Example 4.2-12 Determine the entire area of steel reinforcement

19

5.3.1 Formation, Development and Width of Cracks

Example 4.2-13 Compute the steel and concrete stresses

20

5.3.2 Evaluation of Cracks

54 55 55 58 59 59


Example 6.2.1-3 Static scheme of rafter with overhangs from left 117

shearing crack of reinforced concrete beam Example 5-2 To assess the crack width perpendicular

Example 6.2.1-4 Rafter with overhanging ends of the

65 66

perpendicular cracks Example 5-4 The calculation of the stress in the reinforcement

69

after full cracking Example 5-5 Calculation of shear crack widths on a reinforced

71

73 Example 5.4-1 Calculation of deflections for rectangular

Example 6.2.1-7 Calculation of the thickness t and stress

119

Example 6.2.1-8 Design dimensions of the timber elements

119

Example 6.2.1-9 Design dimensions of the timber elements

120 121

Example: 6.3-1 Calculation of carrying capacity of the column

124

Example 6.3-2 Calculate the required width of the support

125

of beam on column 125

dimensions of the elements

75

Example 6.3-4 Assessment of wooden column

127

75

bending moments Example 5.4-4 The calculation of the stress in the reinforcement

119

Example 6.3-3 Top chord of truss beam be designed

for rectangular cross-section Example 5.4-3 Evaluation of deflections due to shear forces and

118

Example 6.2.1-6 Buckling calculation

73

reinforced concrete beam Example 5.4-2 Detailed calculation of the coefficient χ

Example 6.2.1-5 Rafter as a continuous beam

6.3 Concentric compression members

concrete beam according to CEB - FIP 5.4 Methods

117

right and of the left

to the centreline of the reinforced concrete slab Example 5-3 Assessment according to limit state the widths of

116

7. Profiled steel sheeting

128

7.1 Structural steel

128

76

7.1-1 Profiled steel sheeting

after cracking and crack width determination

128

84

7.2 The permanent load

129

Example 5.4-6 The calculation of loads using trapezoidal rules

90

7.3 Methods of analysis and design

130

Example 5.4-7 Calculation of ideal load

93

7.3-1 Ultimate limit state

Example 5.4-8 Calculation of the deflection by the ideal of loads

94

7.3-2 Bending failure and bending strength

131

7.3-3 Anchorage failure, anchorage fatigue resistance and

131

Example 5.4-5 Theorem of reciprocity of virtual work

6. Behaviour and Conception of Timber Structures Example 6-1 : Compute the stress in Rafter for a given bending

anchorage strength

95 96

moment and external force at the section Example 6-2 : Compute the stress in Rafter for a given bending

96

moment and external force at the section

8. Masonry 6.1 How Structural Systems Carry The Load - Timber Engineering 6.2 Structural Design

97

7.3-4 Shear failure

132

7.3-5 Design of the slab for point and line load

133

7.3-6 Reinforcement of the support

133

7.3-7 Typical structural details of composite slabs

134

7.3-8 Serviceability limit state for composite slabs 152

137

Example 7.1 Composite steel and concrete joists

8.1 Unreinforced masonry walls subjected112 to vertical loading

6.2.1 Proposal for a family house roof using steel elements

130

113

8.1-1 Verification masonry Example 6.2.1-1 Dimensioning reinforcementofin unreinforced piles 115

walls

8.1-2 Characteristic compressive, flexural and shear

152 152 152

strength of masonry wall Example 8.1

157

Example 8.2

159

Example 8.3

160

142

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Example 6.2.1-2 Static scheme rafters as a simple beam

63

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Example 5-1 Calculate the distance of the first an inclined


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8. Masonry

152

8.1 Unreinforced masonry walls subjected to vertical loading

152

8.1-1 Verification of unreinforced masonry walls

152

8.1-2 Characteristic compressive, flexural and shear

152

strength of masonry wall Example 8.1

157

Example 8.2

159

Example 8.3

160

9. Terminology

161

10. Realisation Projects

169


Timber structures can be highly durable when properly treated, detailed and built. Examples of

timber and profiled steel sheeting that are of interest to design engineers and architects. Several

this are seen in many historic buildings all around the world. Timber structures can easily be

step-by-step worked examples are provided to illustrate the design methods explained in this

reshaped or altered, and if damaged they can be repaired.

book. In chapter 4 gives an overview of verification of stress on reinforced concrete cross section subjected to bending and bending moment with axial load in serviceability limit state

In chapter 7 gives an overview of metal cladding systems provide an efficient, attractive and reliable solution to the building envelope needs of single storey buildings.

can easily be performed. Calculations are demonstrated by examples. In order to check the sufficient bearing capacity of reinforced concrete rectangular column in centric compression for axial force taking into account the existing reinforcement. In chapter 5 the experiments, the results of which are presented in this book were designed and realised with the aim of determining the deflections and strain energy, as well as the work of external load. A method of measuring the deformations was applied, making possible to separate the bending induced deformations from those induced by the shear force. At every loading level were the characteristics of cracking and the deformations of the basic fictitious truss system measured by means of mechanical strain deformometers. In the case of a reinforced beam, a certain amount of the strain energy is dissipated in crack formation and propagation, and in other irreversible deformations. However, there are still not enough experimental results available for the calculation of strain energy. It is suitable if the system of the measured values is proposed in advance enabling the determination of the strain energy as exactly as possible. In chapter 6 gives an overview of the properties of timber are very sensitive to environmental conditions; for example moisture content, which has a direct effect on the strength and stiffness, swelling or shrinkage of timber. A proper understanding of the physical

In chapter 8 is devoted to the design of structural masonry is written primarily for structural engineers. The book will also provide an invaluable reference source for practising engineers in many building, civil and architectural design. In chapter 9 provide Architecture & Construction defines more terms in architecture and building construction. Because there have been significant changes, advances, and new developments in building materials and services, construction techniques, and engineering practices The design of structures/elements is explained and illustrated using numerous detailed, relevant and practical worked examples. These design examples are presented in a format typical of that used in design office practice in order to encourage students to adopt a methodical and rational approach when preparing structural calculations.

I kindly ask readers of this textbook who have questions, suggestions for improvements, or who find errors, to write to me. I thank you in advance for taking the time and interest to do so. Finally, I am grateful to my colleague Richard Schlesinger for his help, constructive criticism, patience and encouragement that have made this project possible.

characteristics of timber enables the building of safe and durable timber structures. Timber from well-managed forests is one of the most sustainable resources available and it is one of the oldest known materials used in construction. It has a very high strength to weight ratio, is capable of transferring both tension and compression forces, and is naturally suitable as a flexural member. Timber is a material that is used for a variety of structural forms such as beams, columns, trusses, girders, and is also used in building systems such as piles, deck members, railway sleepers and in formwork for concrete.

Bratislava 2017

Sabah Shawkat

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This text book provides a brief description of the engineering properties of concrete,

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Introduction


4. Limitation of Stress

4.1.1 Permissible Stresses

Mechanical properties of concrete are governed by the properties of its Components

In reinforced concrete structures, longitudinal cracks parallel to the reinforcing bars

and the interaction between them. In the simplest case concrete can be regarded as a three

may occur if the stress level in the concrete under the rare combination of actions

phase material consisting of a continuous phase (mortar) and a Particle phase (coarse

Gk  j  P k  Q k  1   0  i Q k  i

aggregates). The described phases interact through a third phase, namely the interface. The

durability. In detail EC 2 requires the verification of stresses, associated with the following

strength, the stiffness and the cracking of the normal weight, normal strength concrete is

permissible values, for the following reasons:

basically governed by the properties of the mortar and the ability of the interface to mobilize

1) For preventing the development of longitudinal cracks, the concrete compressive stresses

the strength and stiffness of the aggregates. Properties of the mortar and the interface are in

under rare combinations of actions are limited to 0.6 f ck in absence of other methods (for

turn governed by the water/cement ratio; the lower w/c ratio the stronger the mortar and interface stage. Consequently, decreased w/c ratio improves the interaction between the mortar and the coarse (gross) aggregate phases. Introduction of superplasticizers in concrete technology has made it possible to produce concrete with Extremely low w/c ratio, resulting in the production of concretes with very strong mortar phase and interface. In differ to the normal strength concrete, the crack trajectory does not avoid the coarse aggregates, but it may pass through them.

exceeds a critical value. Such cracking may lead to a reduction in

example, increasing the concrete cover of confinement of the compression zone). c  0.6 f ck

2) For preventing excessive creep, the concrete compressive stresses under quasi - permanent combinations of actions Gk  j  P k   2  i Q k  i are limited to 0.45 f ck , c  0.45 f ck

3) for reinforcing steel subjected to loads and restraints, s  0.8 f yk

4) for reinforcing steel subjected to restraints only,

4. 1 Verification at the Serviceability Limit States

s  f yk

Limitation of Stress under Serviceability Conditions The provisions at the ultimate limit states in Eurocode 2 may lead to excessive stresses in

5) for prestress tendon  p  0.75  f pk

concrete, reinforcing steel. These stresses may, as a consequence, adversely affect the appearance and performance in service conditions and the durability of concrete structures.

 c'

- non-linear creep of concrete due to excessive compressive stresses,

40

- increased permeability of the concrete surface due to micro - cracking of concrete around the

35

reinforcing bars, - yielding of steel in service condition leading to cracks with unacceptable width, - cracks parallel to the reinforcing bars,

The common serviceability limit states are:

2

As

, 

100  b  d

bd c

´c    0.6  fck.cyl overall w idth of a cross-section [m] effective depth of a c ross -section [m]

30

b d

25

M s bending moment [MNm]

As area of reinforcement [cm2 ]

15

d

Fc

Ms

As 10

 s'

b 5

- crack control

0 0

0.25

0.5

0.75

1

1.25

1.5

1.75

Figure: 4.1.1-1

Limitation of Stress

0,6 f ck,cyl

2

z= .d

20

- stress limitation - deflection control

Ms

c<

- excessive deflection of concrete members caused by high stresses, cracking and creep of the members,

Verification of stress at the serviceability limit states.

x= .d

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1

2.25

2.5

Fs

2.75

3


cross-section figure 4.1.1.1-1. Verify the stresses in concrete and in steel reinforcement. The section properties: Characteristic value of concrete-cylinder compressive strength (MPa):

f ck

25 MPa

Characteristic yield stress of reinforcement (MPa):

f yk

410 MPa

Cross-section ( m ):

b

0.4 m

d

0.55 m

Bending moment at the section :

Ms

Distance of the centroid tension and compression reinforcement

d1

0.05 m

from the extreme fibre of the cross-section (m):

d2

0.05 m

Area of compression reinforcing longitudinal bars ( m ):

A1

0.0008m 

Area of tension reinforcing longitudinal bars ( m2 ):

A2

0.0026m 

2

0.15 MN m

2

Reinforced concrete sections are usually transformed into equivalent concrete sections. Note that the transformed section consists of the gross concrete section plus n times the area of steel minus the concrete area displaced by the embedded bars A 1  A 2 . n

Where n

15

Es , Ec

Es

Figure: 4.1.1.1-1

2

Moment of inertia of the transformed cracked cross-section about the neutral axis ( m4 ) 3

b x

I cr

3

2

I cr

M s

c

x

c

I cr

5.70 MPa

 c  0.6 f ck

Allowable concrete stress:

0.6 f ck

is the modulus of elasticity of steel, or concrete respectively

A knowledge of the modulus of elasticity of concrete is necessary for all computations of deformations as well as for design of sections by the working stress design procedure. The term Young’s modulus of elasticity has relevance only in the linear elastic part of a stress-

4

0.006 m

Bending concrete stress at a distance x from the neutral axis (MPa)

is known as the modular ratio

Ec

2

 n  A 2 ( d  x)  n  A 1 d 1  x

The assumption was correct.

15 MPa

from which the steel stress is (MPa)  s1

nMs 

 s2

nMs 

 x  d1  Icr

66.76MPa

strain curve. To determine the location of the neutral axis, the moment of the tension area about the axis is set equal to the moment of the compression area, which gives: b

2

x

2

 n  A 2 ( d  x)  n  A 1 x  d 1

0

b x

x

 n  A 2  n  A 1  x  n  A 2 d  n  A 1 d 1

2 0.228 m

As above.

120.59MPa

 s2  0.8f yk

Using the design aid in figure B3-B3.3

out

or

0.228 m 2

Icr

From the graf figure B3-B3.3 we found

Depth of the compression zone (m): x

 d1  x

 0

Ms b d

2

´c

A2 b d

 100

1.18

5.27 MPa

´s2

98.86 MPa

1.23967

The compressive stress in the concrete at the top fibres of the section (MPa): c

´ bc  

6.53 MPa

Limitation of Stress

whichever is lesser

 c  f ck

OK

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Example 4.1.1-1: Pure bending with tension and compression reinforcement in a rectangular

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2


4.2 Service Load Stress-Straight-Line Theory Stress in tension reinforcement (MPa):  s2

´ s2  

whichever is lesser

122.55 MPa

 s2  0.8  f yk

4.2.1 Analysis for Stresses-Section cracked and elastic

OK

Sections designed for strength (Ultimate Limit States) under factored loads must be checked for serviceability (deflection, crack width, etc.) under the specified loads. Under service loads, flexural members are generally in the cracked phase with linear distribution of strains and stresses. The computation of stresses using the straight-line theory (elastic theory) for cracked section is discussed below. Figure 4.2.1.1a shows a beam of rectangular section, subjected to a specified load moment, Ms. For this beam, the corresponding transformed section, neglecting the concrete in the tension side of the neutral axis, is shown in figure 4.2.1.1b. The centroid of

Figure: 4.1.1.1-2

the transformed section locates the neutral axis.

To compute stresses, and strains if desired, the device of the transformed section can still be used. One need only take account of the fact that all the concrete which is stressed in tension is assumed cracked, and therefore effectively absent. As shown in figure, the transformed section then consists of the concrete in compression on one side of the axis and n times the steel area on the other. The distance to the neutral axis, in this stage, is conventionally expressed as x. (Once the concrete is cracked, any material located below the steel is ineffective, which is why d is effective depth of the beam.)

Knowing the neutral axis location and moment of inertia, the stresses in the concrete (and steel) in this composite section may be computed from the flexure formula bc

 s'

Verification of stress at the serviceability limit states.

6000

s´   0.8 fyk Ms

The internal resultant forces in concrete and steel will then be:

5000

b d

M

b d c

d

3000

Ms

As

2000

 s

b

'

Icr Fs

bc

0 0,25

0,5

0,75

1

1,25

1,5

1,75

2

2,25

A2  s2

d

x 3

2,5

2,75

Fc  z

M

or

Fs z

M

bx

3

3

 n  As  ( d  x )

2

from which

< 0,8.fyk

1000

0

Fs

2

bx 

The moment of inertia of the section, Icr is given by:

Fc z= .d

4000

bc

bx 

Equating the external applied moment M and the moment of resistance,

As

2

.

of internal forces and external applied moment.

z s

x Icr

The inner lever arm will be:

8000 7000

Ms

The same results could be obtained more simply and directly considering the static equilibrium

Fc

x=.d

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3

3

M

x Icr

Figure: 4.1.1.1-3

Limitation of Stress

and

s

M A2  z

bc 2

z

or

M

A2  s z


Using the design aid in diagram

dimensions shown in fig. 4.2.1-1. Compute the stresses in concrete and steel reinforcement for

Ratio of tension reinforcement (-):

a given bending moment. Assumptions:

´ bc

25  MPa

Characteristic yield stress of the steel (MPa): fyk

0.6  fck

call

h

´ s

247.47 MPa

0.80708

The allowable concrete stress (MPa): call

15  MPa

The compressive stress in the concrete at the top fibres of the section (MPa): 0.90  m

Distance of the tension reinforcement centroid from the extreme fibre of the cross-section (m): d2

2

bd

250 kN  m

0.4  m

and

Ms

410 MPa

Cross-section ( m): b

0.45

7.27681 MPa

Bending moment at the section ( MN  m): Ms

2

 10

bd

From the diagram we obtain

Characteristic value of concrete cylinder compressive strength (MPa): fck

As

bc

  ´ bc

5.87  MPa

bc

whichever is lesser

bc  call

allowable

The steel stresses (MPa):

0.02  m

Effective depth of a cross-section ( m): d

h  d2

d

n

20  mm

2

As

n   

2

4

246 MPa

5

The entire area of tension reinforcing bars ( cm ): 

sall

Stress in tension reinforcement (MPa):

Diameter, number of bars: 

0.6  fyk

sall

0.88  m

As

15.70  cm

2

s

  ´ s

s

199.72 MPa

whichever is lesser s  sall

allowable

Reinforced concrete sections are usually transformed into equivalent concrete sections. Note that the transformed section consists of the gross concrete section plus n times the area of steel minus the concrete area displaced by the embedded bars As . Modular ratio Es

n

Ec

n  15

To locate the neutral axis (centroid) of the transformed section, we have to equate the moments of areas about the neutral axis. b

x

2

2

 n  As  ( d  x )

0

Figure: 4.2.1-1

Limitation of Stress

Solving

x

0.268

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Example 4.2-1: A reinforced concrete beam of rectangular section has the cross-section

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4


Example 4.2-2: The cross-section of a beam with compression reinforcement are shown in

Using the design aid in diagram

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5

0.309

x

d

x

figure 4.2.2-1a. The beam is subjected to a specified load moment. Compute the specified

0.272  m

load stress in concrete and steel. The material properties are

The moment of inertia of the section, Icr is given by: Icr

bx

3

3

 n  As  ( d  x )

2

Icr

0.01139 m

Assumed:

4

Characteristic value of concrete cylinder compressive strength (MPa):

Equating the applied moment to the moment of resisting forces (here taken as the moment of compressive forces in concrete about tension reinforcement):

Ms  b  x 

bc

2

 

d 

x

Solving

0

3

bc

fyk

as before

5.87  MPa

n  bc 

30  MPa

410 MPa

Bending moment at the section ( MN  m): Ms

0.185  MN  m

Cross-section ( m):

Tension steel stress: s

fck

Characteristic yield stress of the steel (MPa):

b

dx

s

x

as before

199.34 MPa

With the distribution of concrete and steel stresses shown in fig. 4.2.1.1c the internal resultant forces in concrete and steel forming the internal resisting couple are Fc and Fs and the lever arm of this couple is z, where:

0.3  m

h

0.60  m

Distance of the centroid of the tension reinforcement from the extreme fibre of the crosssection (m): d1

0.05  m

d2

0.05  m

Effective depth of a cross-section ( m): Fc z

bx  d

bc

2

Fc

316.71 kN

z

0.78936 m

x 3

Fs

A s  s

Fs

306.69 kN

d

h  d2

d

0.55  m

Compression reinforcement: Diameter, number of bars:

The allowable service load moment for the beam will then be (kN m): M

1 2

 

 call  MPa  b  x   d 

x

 3

c M

16  mm

nc

2

643.92 kN  m

2

The entire area of compression reinforcing bars ( m ):

Compressive stresses in the concrete shall be limited in order to avoid longitudinal cracks, micro-cracks or high levels of creep, where they could result in unacceptable

A1

nc   

c 4

2

4

 10

A1

effects on the function of the structure.

Tension reinforcement:

Stresses in the reinforcement shall be limited in order to avoid inelastic strain, to avoid

Diameter, number of bars:

unacceptable cracking or deformation. t

25  mm

Limitation of Stress

nt

4

0.0004 m

2


A2

t

nt   

2

4

 10

4

A2

0.00196 m

c

2

M s

x

as above see bc

9.77MPa

c

Icr

The specified load stresses are computed by the straight line theory applied to the cracked

Using the design aid in diagram B3-B3.3

transformed section, shown in figure 4.2.2-1b.

Ratio of tension reinforcement (-):

The modular ratio

Es

n

n  15

Ec

about the neutral axis (see figure 4.2.2-1b): b

x

2

 n  A1  x  d1  n  A2  ( d  x )

Solving,

0

x

bc

 2 

d 

x

3

 n  A1 

bc  x  d1

x

 d  d1

x

 x  d1

and

5.279 MPa

Ms

2

bd

call

Solving,

´ s

2.03857

0.6  fck

call

18  MPa

The compressive stress in the concrete at the top fibres of the section (MPa):

b2

bc

  ´ bc

bc

10.76 MPa

The allowable steel stresses (MPa):

7.68  MPa 4

Moment of inertia of the transformed cracked cross-section about neutral axis ( m ) Icr

bx 3

3

2

 n  A2  ( d  x )  n  A1  d1  x

2

Icr

0.00443 m

sall

0.6  fyk

sall

246 MPa

4

Stress in tension reinforcement (MPa):

Compression and tension steel stress (MPa): s s1

n  M s

s2

n  M s

x  d1 Icr dx Icr

98.86 MPa

9.77  MPa

bc

reinforcement of compressive level, c2 (MPa) is: bc

´ bc

The allowable concrete stress (MPa):

Referring to the stress distribution in figure 4.2.2-1c, the concrete stress at compression

b2

1.19

form (-):

0.23385 m

compressive forces Fc in concrete and Fs in steel, about tension reinforcement):

bx 

To apply design diagram the transformed action Ms have to be brought into a dimensionless

Equating the applied moment to the moment of resisting forces (here taken as the moment of

Ms

2

 10

From diagram we obtain

To locate the neutral axis (centroid) of the transformed section, equate the moments of areas 2

A2 bd

s1

115.25 MPa

or

s1

n  b2

s2

198.19 MPa

or

s2

n  bc 

s1

dx x

s2

115.25 MPa

198.19 MPa

  ´ s

s

whichever is lesser s  sall

allowable

Limitation of Stress

201.53 MPa

whichever is lesser

bc  call

allowable

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Using the flexure formula,

2

The entire area of tension reinforcing bars ( m ):

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Effective depth of a cross-section ( m): d

h  d2

d

0.65  m

Combined actions have to be transformed in relation to the centre of gravity of the tension reinforcement ( MN  m): M

 Ms

Ns 

 Ns

 

 d 

h 



M

2 

0.295  MN  m

Limiting values for the related neutral axis depth:

2

b  d  sall

Example 4.2-3: The member shown in figure 4.2.3-1 is subjected to bending moment and external normal load. Compute the stress in concrete and steel.

1

3

0.40822

Lever arm of the internal forces (m):   z d1   3 

z

0.56155 m

As

0.00153 m

2

Assumptions: Characteristic value of concrete cylinder compressive strength (MPa): fck

90  M

Figure: 4.2.2-1

25  MPa

The entire required steel area ( m ) is:  M  N  1 As  s z  sall

Characteristic yield stress of reinforcement (MPa): fyk

410 MPa

Allowable stress of reinforcement (MPa): sall

0.6  fyk

sall

246 MPa

Bending moment at the section ( MN  m): Ms

0.25  MN  m

Axial load at the section ( MN): Ns

0.15  MN

Cross-section ( m): b

0.35  m

h

Figure: 4.2.3-1

0.70  m

Distance of the tension reinforcement from the extreme fibre of the cross-section (m): d2

0.05  m

Diameter, number of bars: nt

4

Limitation of Stress

t

25  mm

2


A2

nt   

t

2

A2

4

0.001963 m

the stresses of concrete and steel for a given bending moment and axial load. Assumptions:

2

Characteristic value of concrete cylinder compressive strength (MPa):  fck

The allowable concrete stress (MPa): 0.6  fck

call

15  MPa

call

Characteristic yield stress of reinforcement (MPa): fyk

Modular ratio: n

Es

n

Ec

bc

0.001963 m sall

n

sall

2

1

11.31 MPa

bc

whichever is lesser

bc  call

allowable

A2 bh

 100

0.80122

Ms 2

bd

5.96  MPa

d

´ bc  

bc

Ms ´ s

143.36 MPa

whichever is lesser

10.08 MPa

  ´ s

whichever is lesser

0.75  m

0.05  m

h  d2

d

0.7  m

s

242.38 MPa

s  sall

0.15  MN  m

Axial external load at the section ( MN): Ns

bc  0.6  fck

allowable

0.60  MN

Combined actions have to be transformed in relation to the centre of gravity of the tension reinforcement ( MN  m):

Stress in tension reinforcement (MPa): s

h

Bending moment at the section( MN  m):

The compressive stress in the concrete at the top fibres of the section (MPa): bc

0.35  m

Effective depth of a cross-section ( m):

1.69062

From the diagram we obtain ´ bc

246 MPa

sall

Distance of the tension reinforcement from the extreme fibre of the cross-section (m): d2

Using the design aid in diagram 

0.6  fyk

Cross-section ( m): b

410 MPa

Allowable stress of reinforcement (MPa):

15

The compressive stress in the concrete at the top fibres of the section (MPa): A2

30  MPa

M

 Ms

Ns 

 Ns

 

 d 

h 



M

2 

0.345  MN  m

allowable

Limiting values for the related neutral axis depth: 

90  M 2

b  d  sall

1 3

Limitation of Stress

0.40957

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Example 4.2-4: Compute the steel area to the cross-section shown in figure 4.2.4-1 and check

2

The entire provided steel area ( m ) will then be:

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Example 4.2-5: Pure bending with tension and compression reinforcement in a rectangular

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cross-section figure 4.2.5-1. Verify the stresses in concrete and in steel reinforcement. The section properties: Characteristic value of concrete-cylinder compressive strength (MPa): fck

25  MPa

Characteristic yield stress of reinforcement (MPa): fyk

410 MPa

Cross-section ( m):

b

0.4  m

d

0.55  m

Bending moment at the section ( MN  m): Ms

Figure: 4.2.4-1

Distance of the centroid tension and compression reinforcement from the extreme fibre of the

Lever arm of the sectional forces (m):

z

 

d1 

cross-section (m): d2



 3

0.15  MN  m

z

0.05  m

d1

0.05  m

0.60443 m

2

Area of compression reinforcing longitudinal bars ( m ): A1

2

The entire steel area ( cm ) is:

0.0008 m

2

2

Area of tension reinforcing longitudinal bars ( m ): As

 M  N   1  104  s z  sall

As

1.187 cm

A2

2

0.0026 m

2

Reinforced concrete sections are usually transformed into equivalent concrete sections. Note Since the sign of the amount of reinforcement is negative, the reinforcement is not necessary

that the transformed section consists of the gross concrete section plus n times the area of steel

and the concrete alone resists the bending moment and external axial load.

minus the concrete area displaced by the embedded bars A1  A2 .

Depth of the compression zone (m):

A knowledge of the modulus of elasticity of concrete is necessary for all computions of deformations as well as for design of sections by the working stress design procedure.

x

 h Ms     3  2 Ns 

x

0.375 m

x h

Modular ratio:

The compressive stress in the concrete at the top fibres of the section (MPa):

n

Es Ec

n

15

The term youngs modulus of elasticity has relevance only in the linear elastic part of a stressbc

2  Ns bx

bc

9.14  MPa

whichever is lesser

bc  0.6  fck

strain curve.

Limitation of Stress


s1

n  M s

x  d1

s2  0.6  fyk

66.76 MPa

s1

Icr

s2

n  Ms

dx Icr

s2

120.59 MPa

allowable

Using the design aid in diagram

A2

1.18

 100

bd

From diagram, follows

Fig: 4.2.5-1 To determine the location of the neutral axis, the moment of the tension area about the axis is set equal to the moment of the compression area, which gives: b

x

2

2

 n  A2  ( d  x )  n  A1  x  d1

´ bc

0

b

0.228 m x

2

2

´ s2

Ms

2

bd

Depth of the compression zone (m):

x

5.27  MPa

98.86  MPa

1.23967

The compressive stress in the concrete at the top fibres of the section (MPa):

or

bc

 n  A2  n  A1  x  n  A2  d  n  A1  d1

x

0

´ bc  

bc

6.53  MPa

whichever is lesser

bc  0.6  fck

allowable

s2  0.6  fyk

allowable

As above

0.228 m

Stress in tension reinforcement (MPa): 4

Moment of inertia of the transformed cracked cross-section about the neutral axis ( m ) Icr

bx

3

2

 n  A2  ( d  x )  n  A1  d1  x

3

2

Icr

0.006  m

s2

4

bc

M s

Icr

122.55 MPa

whichever is lesser

still be used. One need only take account of the fact that all the concrete which is stressed in tension is assumed cracked, and therefore effectively absent. As shown in figure 4.2.5-1b, the

bc

Allowable concrete stress:

0.6  fck

s2

To compute stresses, and strains if desired, the device of the transformed section can

Bending concrete stress at a distance x from the neutral axis (MPa)

x

´ s2  

5.70  MPa

bc  0.6  fck

The assumption was correct.

transformed section then consists of the concrete in compression on one side of the axis and n times the steel area on the other. The distance to the neutral axis, in this stage, is conventionally expressed as x. (Once the concrete is cracked, any material located below the steel is ineffective, which is why d is effective depth of the beam.)

15  MPa

Limitation of Stress

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From which the streel stress is (MPa)

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Having obtained x by solving this quadratic equation, one can determine the moment of

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inertia and other properties of the transformed section as in the preceding case. Alternatively, one can proceed from basic principles by accounting directly for the forces which act in the cross section these are shown in figure 4.2.6-1b. The concrete stress, with maximum value bc at the outer edge, is distributed linearly as shown. The entire steel area A2 is subjected to the stress s2 . Icr

bx

3

 n  A2  ( d  x )

3

2

Icr

0.00559 m

2

Figure 4.2.5-2: Isometric view of reinforced concrete beam

Example 4.2-6:

Use the section properties from example 4.2.5, the beam is without

compression reinforcement shown in figure 4.2.6-1. Verify the stresses in concrete and in steel

Bending concrete stress at a distance x from the neutral axis (MPa) bc

reinforcement. To determine the location of the neutral axis, the moment of the tension area about the axis is set equal to the moment of the compression area, which gives section modulus 3

of the transformed cracked of cross-section about the neutral axis ( m ).

M s

x

bc

Icr

6.55  MPa

bc  0.6  fck

allowable

The assumption was correct.

Stress in tension reinforcement (MPa): s2

n  M s

dx Icr

s2

123.11 MPa

s2  sall

Example 4.2-7: A rectangular member has the dimensions figure 4.2.7-1 subjected to a bending moment and external normal force. The entire section is in compression, and symmetrically reinforced, check the stresses in concrete and in steel reinforcement. The section properties Characteristic value of concrete cylinder compressive strength (MPa): fck

30  MPa

Characteristic yield stress of the steel (MPa): fyk

410 MPa

Cross-section: h

Figure: 4.2.6-1 bx

 n  A2  x  n  A2  d

b

x

0

x

0.2442 m

 n  A2  ( d  x )

0

0.30  m

d2

or

0.05  m

d1

0.05  m

Effective depth of cross-section (m):

2

2

b

Distance of the tension and compression reinforcement from the extreme fibre of the crosssection (m):

2

2

0.60  m

x

0.2442 m

As above.

d

h  d2

Limitation of Stress

d

0.55  m


A1

10.50 cm

2

A2

If

A1

Ms Ns

Bending moment at the section (MN m): Ms

Igg´

The cross-section is whole section in compression

Bo  n   A1  A2   v2

0.115  MN  m

4

Moment of inertia of the transformed uncracked section ( m )

Axial compression load at the section (MN): Ns

3

1.20  MN

I

bh 3

Reinforced concrete sections are usually transformed into equivalent concrete sections. Note that the transformed section consists of the gross concrete section plus n times the area of steel minus the concrete area displaced by the embedded bars A1  A2 .

b

Igg´

3

Ms

Ns

2

  v1  v2 3

2

2   Bo  v1

I

3

2 2    n  A1   v1  d1  A2   d  v1  

Igg´

0.09583 m

Ns Ms

 n   A2  d  A1  d1

Bo  n   A1  A2   v2

Igg´

Igg´

0.00737 m

0.00737 m

0.10108 m

The assumption was correct.

Bo  n   A1  A2   v2

The compressive stress in the concrete at the top fibres of the section (MPa): b1

Ns Bo

Ms v1

b1

I

10.35 MPa

Allowable concrete stress:

b1  0.6  fck

0.6  fck

allowable

18  MPa

The compressive stress in the concrete at the bottom fibres of the section (MPa): Figure: 4.2.7-1

b2

Reinforced concrete sections are usually transformed into equivalent concrete sections. Note that the transformed section consists of the gross concrete section plus n times the area of steel minus the concrete area displaced by the embedded bars A1  A2 . 2

b  h  n  A2  A1

Bo

0.2115 m

2

The distance of the extreme fibre from the neutral axis (centroid) (m): 2

bh v1 v2

2

 n  A2  d  A1  d1

Bo h  v1

 v1

0.3  m

v2

Ms h  v2

b2

I

0.99  MPa

b2  0

Stress in compression reinforcement (MPa): s2

Area of the transformed uncracked section ( m ): Bo

Ns Bo

 Ns

n

 Bo

Ms d  v1



I

 

s2

26.58 MPa

s2  0.6  fyk

allowable

s1  0.6  fyk

allowable

Stress in compression reinforcement (MPa):

s1

 Ns

n

 Bo

Ms v2  d1 I

0.3  m

Limitation of Stress

  

s1

143.63 MPa

4

4

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Area of reinforcing longitudinal bars:

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13 Example 4.2-8: A rectangular column has the dimensions shown in figure 4.2.8-1. Check the

Reinforced concrete sections are usually transformed into equivalent concrete sections. Note

stresses in concrete and reinforcement caused by a bending moment Ms and external normal

that the transformed section consists of the gross concrete section plus n times the area of steel

tension load Ns.

minus the concrete area displaced by the embedded bars A1  A2 .

Section properties: Characteristic value of concrete cylinder compressive strength (MPa): fck

30  MPa

Characteristic yield stress of the steel (MPa): fyk

0.30  m

h

0.05  m

d1

0.0  m

Effective depth of cross-section (m): d

h  d2

d

2

q

2  c  90 

2

3

A1 b A1 b

A2

2  90

 c  d1  90   c  d1

b

 ( d  c)

A2 b

 ( d  c)

0

2

y

0.82381 m 2

p

2.21916 m

q

1.16833 m

3

0.65053 m

cy

x

0.173 m 3

Section modulus of the transformed cracked cross-section about the neutral axis ( m ):

2 2

Area of tension reinforcing longitudinal bars ( cm ): A2

3  c  90 

x

210 kN

0  cm

p

c

2

The depth of the compression zone (m):

110 kN  m

Area of compression reinforcing longitudinal bars ( cm ): A1

e

3

Axial tension load at the section (kN): Ns

0.52381 m

h

c

y  py  q

0.55  m

Bending moment at the section (kN m): Ms

e

Ns

0.60  m

Distance of the tension and compression reinforcement from the extreme fibre of the crosssection (m): d2

Ms

We find point e on the outside of the section, thus the cross-section is partially compressed

410 MPa

Cross-section (m): b

e

22.30 cm

2

S

K

bx

2

2

 n  A1  x  d1  A2  ( d  x )

Ns

S

S

0.0081 m

K

25935.122

3

In case that the external load is tension force then we have to substitute a negative sign for the tension normal force Ns, otherwise we substitute a positive sign for the compression normal force Ns. The compressive stress in the concrete at the top fibres of the section (MPa): bc

K x

Allowable concrete stress: Figure: 4.2.8-1

0.6  fck

18  MPa

Limitation of Stress

bc

4.49  MPa

bc  0.6  fck

allowable


s1

n  K  x  d1

s1

2

Area of tension reinforcing longitudinal bars ( cm ):

67.411 MPa

A2

Stress in tension reinforcement (MPa): s2

n  K  ( d  x)

s2

40.21 cm

2

Reinforced concrete sections are usually transformed into equivalent concrete sections. Note

146.553 MPa

that the transformed section consists of the gross concrete section plus n times the area of steel minus the concrete area displaced by the embedded bars A1  A2 .

Example 4.2-9: A rectangular column has the dimensions shown in figure 4.2.9-1. Verify the stresses in concrete and in steel, the cross-section caused by a bending moment Ms and normal

e

Ms

e

Ns

1.102 m

compression force Ns. The section properties: Characteristic value of concrete cylinder compressive strength (MPa): fck

30  MPa

Characteristic yield stress of the steel (MPa): fyk

410 MPa

Cross-section (m): b

0.60  m

h

1.1  m

Distance of the tension and compression reinforcement from the extreme fibre of the crosssection (m): d2

0.07  m

d1

0.055 m

Effective depth of the cross-section (m): d

h  d2

d

Figure: 4.2.9-1

1.03  m

We find point e on the outside of the cross-section, thus the cross-section is partially compressed Bending moment at the section (kN m): Ms

c

430 kN  m

h

c

2

0.55256 m

We substitute for c a negative sign in case that the position of c is situated outside of the crosssection, in order to determine the values of p, q and y. Then we can compute the depth of the

Axial compression load at the section (kN): Ns

e

compression zone of concrete.

390 kN

2

2

Area of compression reinforcing longitudinal bars ( cm ): A1

15.71 cm

2

Area of the transformed uncracked section ( m ): Bo

b  h  n  A2  A1

Limitation of Stress

Bo

0.74388 m

2

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Stress in compression reinforcement (MPa):

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This transformed concrete area is seen to consist of the actual concrete area plus n times the

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15 Depth of the compression zone (m):

area of the reinforcement. x

The distance of the extreme fibre from the neutral axis (m):

y  c

x

0.47  m

Section modulus of the transformed cracked cross-section about the neutral axis: 2

bh 2

v1 v2

 n  A2  d  A1  d1

S v1

Bo

0.57  m K

h  v1

v2

bx

2

2 Ns

 n  A1  x  d1  A2  ( d  x )

S

S

0.04284 m

K

9104.38

3

0.53  m

In case that the external load is tension force then we have to substitute a negative sign for the

If Ms Ns

tension normal force Ns, otherwise we substitute a positive sign for the compression normal

Igg´

Then the cross-section is partially compressed.

Bo  n   A1  A2   v2

The compressive stress in the concrete at the top fibres of the section (MPa):

4

Moment of inertia of the transformed uncracked section ( m ):

b

3 3 2 2   v1  v2   n  A1   v1  d1   A2   d  v1   3 

Igg´ Ms

Igg´

0.08582 m

K x

bc

0.19682 m

Bo  n   A1  A2   v2

Igg´

2

p

3  c  90 

q

2  c  90 

3

A1 b A1 b

3

0

bc

 n  A2  ( d  x )  n  A1  d1  x

A2

2  90

b

 ( d  c)

A2 b

 ( d  c)

2

2

p

0.18172 m

q

1.26015 m

s1

n  K  x  d1

4.29  MPa

 0.6  fck

18  MPa

s2

n  K  ( d  x)

y

s1

56.88 MPa

s2

76.26 MPa

3

Compute the value of y (m): y  py  q

allowable

Stress in compression and tension reinforcement (MPa):

 c  d1  90 

 c  d1

bx

Allowable concrete stress:

The assumption was correct.

bc  0.6  fck

Ns x 2

2

Bo  n  A1  A2   v2

4.29  MPa

Other way of solution is checking the compressive stress in concrete as follows (MPa):

bc

Igg´

Ns

bc 4

1.10256 m

Ns

Ms

force Ns.

1.0241

Limitation of Stress


the stresses in concrete and in steel, the cross-section caused by a bending moment Ms and normal compression force Ns. The section properties: Characteristic value of concrete cylinder compressive strength (MPa): fck

25  MPa

Characteristic yield stress of the steel (MPa): fyk

410 MPa

Cross-section (m): b

0.50  m

h

1.0  m

d1

0.05  m

Distance of the tension and compression reinforcement from the extreme fibre of the crosssection (m): d2

0.05  m

Figure: 4.2.10-1

effective depth of the cross-section (m): d

h  d2

d

Reinforced concrete sections are usually transformed into equivalent concrete sections.

0.95  m

2

Area of the transformed uncracked section ( m ):

Bending moment at the section (kN m): Ms

90  kN  m

Bo

Axial compression load at the section (kN): Ns

410 kN

15.71 cm

e

38.21 cm Ms

2

bh

h 2

e

c

0.219 m

0.28  m

2

v1

The location of e is inside of the section, thus the section is partially compressed c

0.58088 m

2

The distance of the extreme fibre from the neutral axis (m):

2

e

Ns

Bo

area of the reinforcement.

2

Area of tension reinforcing longitudinal bars: A2

This transformed concrete area is seen to consist of the actual concrete area plus n times the

Area of compression reinforcing longitudinal bars: A1

b  h  n  A2  A1

v2

 n  A2  d  A1  d1

Bo h  v1

 v1

0.52615 m

v2

0.47385 m

We keep the positive sign, since c is situated inside of the section. If Ms Ns

Igg´

Bo  n  A1  A2   v2

Limitation of Stress

Then the cross-section is partially compressed.

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Example 4.2-10: A rectangular member has the dimensions as shown in figure 4.2.10-1. Verify

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Other way of solution is the compressive stress in concrete as follows (MPa):

4

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17

Moment of inertia of the transformed uncracked section ( m ):

b

Igg´

3

Ms

Ns

p

q

3

3

 n  A1  v1  d1

2  A2  d  v1 2

Igg´

2

3  c  90 

2  c  90 

A1 b

A1 b

 c  d1

4

A2 b

2  90

b

 ( d  c)

p

2

q

0.15928 m

0.36745 m

0.6  fck

15  MPa

s1

n  K  x  d1

s2

n  K  ( d  x)

s1

23.143 MPa

s2

0.718 MPa

s1  0.6  fyk

allowable

s1  0.6  fyk

allowable

3

entire section is in compression, and unsymmetrically reinforced, check the stresses in concrete

0.642 m

and in steel reinforcement.

yc

x

The section properties:

0.92  m

Characteristic value of concrete cylinder compressive strength (MPa): 3

Section modulus of the transformed cracked section about the neutral axis ( m ) bx

2

2

S

 n  A1  x  d1  A2  ( d  x )

Ns

0.23  m

K

S

3

fck

25  MPa

Characteristic yield stress of the steel (MPa): fyk

1767.57

410 MPa

Cross-section (m):

In case that the external load is tension force then we have to substitute a negative sign for the tension normal force Ns, otherwise we substitute a positive sign for the compression normal force Ns.

b

0.30  m

h

0.60  m

Distance of the compression reinforcement from the extreme fibre of the cross-section (m): d2

0.05  m

d1

0.05  m

Effective depth of the cross-section (m): The compressive stress in the concrete at the top fibres of the section (MPa): bc

1.63  MPa

2

Depth of the compression zone ( m):

K

bc

 n  A2  ( d  x )  n  A1  d1  x

Stresses in compression and tension reinforcements (MPa):

0 y

S

bx

Example 4.2-11: A rectangular member has the dimensions as shown in figure 4.2.11-1. The

3

y  py  q

x

Ns x 2

Allowable concrete stress:

 ( d  c)

A2

bc

2

The assumption was correct.

 c  d1  90 

0.05765 m

0.18  m

Bo  n   A1  A2   v2

Bo  n  A1  A2   v2

3

Igg´

Igg´

0.21  m

Ns

Ms

  v1  v2

K x

bc

1.63  MPa

bc  0.6  fck

allowable

d

h  d2

d

0.55  m

Bending moment at the section (kN m): Ms

70  kN  m

Limitation of Stress


Ns

v2

1700 kN

h  v1

v2

0.31  m

4

Moment of inertia of the transformed uncracked section ( m ): 2

Area of compression reinforcing longitudinal bars ( cm ): A1

10.40 cm

2

b

Igg´

3

2

  v1  v2 3

3

2 2    n  A1   v1  d1  A2   d  v1  

Igg´

0.00677 m

4

Area of tension reinforcing longitudinal bars ( cm ): A2

4.50  cm

2

If the location of c is inside of the section, then the cross-section is partially compressed.

Reinforced concrete sections are usually transformed into equivalent concrete sections. e

Ms

e

Ns

h

0.041 m

6

0.1  m

e

h 6

The entire section is in

c

h

 v1

2

c

0.01093 m

When the position of c is inside of the cross-section, then the sign remains positive for the

compression.

determination of p,q and y values in order to compute the depth of the compression zone x. MG

Ms  Ns c

MG

51.41216 kN  m

If MG Ns

Igg´

Then the cross-section is entire in compression.

Bo  n  A1  A2   v2

4

Moment of inertia of the transformed uncracked section ( m ): 3

bh

I

3

Figure: 4.2.11-1

b

Igg´

2

3

Area of the transformed uncracked section ( m ): Bo

b  h  n  A2  A1

Bo

0.20235 m

MG

2

 n   A2  d  A1  d1 2

  v1  v2 3

2

  Bo  v1

2

I

3

2 2    n  A1   v1  d1  A2   d  v1  

0.00677 m

or

4

Igg´

0.03024 m

Ns

Igg´

0.00677

Bo  n   A1  A2   v2

0.09694 m

This transformed concrete area is seen to consist of the actual concrete area plus n times the MG

area of the reinforcement.

Ns

The distance of the extreme fibre from the neutral axis (m): 2

bh v1

2

 n  A2  d  A1  d1

Bo

Igg´

The assumption was correct.

Bo  n   A1  A2   v2

The compressive stress in the concrete at the top fibres of the section (MPa):

 v1

0.29  m

b1

Ns Bo

Ms v1 I

Limitation of Stress

b1

11.38 MPa

b1  0.6  fck

allowable

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Axial compression load at the section (kN):

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The compressive stress in the concrete at the bottom fibres of the section (MPa):

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19

Ns

b2

Bo

Ms h  v2

b2

I

5.41  MPa

b2  0.6  fck

allowable

Distance of the centroid tension reinforcement from the extreme fibre of the cross-section (m): d1

0.04  m

d2

0.06  m

Effective depth of the cross-section (m): Stresses in reinforcement (MPa):  Ns Ms  v2  d1 n   I  Bo 

s1

 Ns Ms  d  v1  n   I  Bo 

s2

s1

s2

d

166.47 MPa

85.56 MPa

s1  0.6  fyk

allowable

s2  0.6  fyk

allowable

and check the stresses of concrete and steel for a known bending moment and external tension axial load.

Characteristic value of concrete cylinder compressive strength (MPa): 30  MPa

Characteristic yield stress of the steel (MPa): 410 MPa

Allowable stress of steel (MPa): sall

0.6  fyk

sall

Ms Ns

a1

0.385 m

zs

h  d1  d2

a2

0.115 m

A1

0.54

e

0.125 m

a1

zs

0.5  m

a2

246 MPa

Ns a2

Provided:

sall  z s

Numbers of bars: A2

The section properties:

fyk

d

0.30  m

h

n1

2

Ns a1

A1

3.74  cm

z s  a1

2

Provided:

sall  z s

Numbers of bars:

n2

4

A2

12.52  cm

As1

n1   

16  mm

1

2

Diameter:  2

2

As2

n2   

As1

4

4  20

4.02  cm

s1 s1

Ns a2

2

2

4

As2

12.56  cm

Ns a1

d  d1  As1

s2

d  d1  As2

228.78 MPa

s2

245.09 MPa

s1  0.6  fyk

allowable

s2  0.6  fyk

sall

allowable

0.60  m

0.05  MN  m

0.40  MN

Figure: 4.2.12-1

Limitation of Stress

0.6  fyk

2

20  mm

Stresses in tension reinforcement (MPa):

External axial tension load at the section (MN): Ns

 e  d1

Diameter:  1

Bending moment at the section (MN m): Ms

2

2  16

Cross-section (m): b

h

The entire area of tension reinforcing longitudinal bars:

Example 4.2-12: Determine the entire area of steel reinforcement as shown in figure 4.2.12-1

fck

e

h  d2

sall

246 MPa

2


Example 4.2-13: A T-beam having the cross-sectional dimensions shown in figure 4.2.13-1

Characteristic value of concrete cylinder compressive strength (MPa):

is subjected to a specified load bending moment Ms. Compute the steel and concrete stresses. The section properties:

fck  25  MPa

Characteristic value of concrete cylinder compressive strength (MPa):

Characteristic yield stress of the steel (MPa):

fck

25  MPa

fyk  410 MPa

Characteristic yield strength of reinforcing steel (MPa):

Cross-Section (m): b  0.30  m

fyk

400 MPa

h  0.50  m

The concrete dimensions:

Bending moment at the section (MN m):

Total depth of the T-beam (m):

M  0.03  MN  m

h

0.60  m

External axial compression load at the section (MN): Width of the web (m):

N  0.30  MN

Distance of the tension and compression reinforcement from the extreme fibre of the crosssection (m): d2  0.06  m

d1  0.04  m

C 

C  0.1 m

N

ea 

h 2

 C  d2

hf

ea  0.09 m

2

A1  2.74  cm

d2

N  ea

d  d1  A1

s1  234.6194MPa s1  s

Allowable steel stress:

0.10  m

Effective depth to the steel centroid (m):

2

d

Stresses in tension reinforcement (MPa) at SLS: s1 

0.10  m

Distance of the tension reinforcement from the extreme fibre of the cross-section (m):

Area of reinforcement in the tension and compression zone: A2  9.45  cm

1.00  m

Flange depth (m):

d  d2  0.4 m

M

0.25  m

Flange width (m): b

Effective depth of cross-section (m): d  h  d1

bw

s2 

d

0.5  m 2

  d  d1  A2

Area of reinforcing longitudinal bars ( cm ):

N  d  d1  ea

A2

42.00 cm

2

The bending moment at the section (kN m):

s2  249.43311MPa

Ms

250 kN  m

The modular ratio is:

s2  s s  0.6  fyk

h  d2

s  246 MPa

n

Es

Ec

n

15

Limitation of Stress

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Section properties:

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is the modulus of elasticity of steel, or concrete respectively

b

x

2

 n  A2  ( d  x )

2

Solving

0

x

0.196  m

The reinforcement area is transformed to concrete. The section is considered cracked and the concrete area on the tension side of the neutral axis is ignored. For elastic stress distribution

Since this is greater than hf = 100 mm, the assumption is not correct and the neutral axis

under specified load, the neutral axis passes through the centroid of the effective transformed

passes through the web. Taking moments of areas for the case in figure 4.2.13-1c

section. The neutral axis may lie in the flange as shown in Fig. 4.2.13-1b (that is x < hf ) or it may pass through the web as in Fig. 4.2.13-1c ( x > hf ).

bw 

x

2

 

 b  bw  hf   x 

2

hf 

  n  A2  ( d  x )

0

2

Solving

x

0.21  m

The stresses may be found using the stress distribution or by using the flexure formula. Considering the stress distribution shown in figure 4.2.13-1d, the moment is given by the moment of the compressive forces about the centroid of steel. Taking the compressive area as the difference between two rectangles bx

b  bw  x  hf 

and

The moment equation gives: Ms

bx 

bc 2

 

d 

x

   b  bw   x  hf  

 x  hf    x    d  h  x  hf  f 3  2 

bc  

3

Solving

6.64  MPa

bc

Allowable concrete stress: 0.6  fck

15  MPa

The assumption was correct.

bc  0.6  fck

From the stress distribution diagram, s2

n  bc 

dx

s2

x

133.25 MPa

Allowable steel stress (MPa): 0.6  fyk

The assumption was correct.

s2  0.6  fyk

Alternatively, to use the flexure formula,

Figure: 4.2.13-1 To locate the neutral axis, first assuming that it lies in the flange, equating the moments of areas about neutral axis figure 4.2.13-1b gives the equation:

240 MPa

bc

M x Icr

4

Moment of inertia of the transformed cracked section ( m ): Icr

b

x

3

3

Limitation of Stress

b  bw  x  hf  3

3

 n  A2  ( d  x )

2

Icr

0.00805 m

4


Ms x

bc

bc

Icr

6.64  MPa

For very small eccentricities e, the distance to the neutral axis x may exceed the depth h of the section. Then the entire cross-section area will be in compression, so that the stress in

as before

the steel A2 will also be compressive, though mostly of magnitude smaller than sall .

Stress in tension reinforcement (MPa): and

( d  x)

n  M s

s2

s2

Icr

For very small eccentricities e, the distance to the neutral axis x may exceed the depth h of the

133.25 MPa

section. Then the entire cross-section area will be in compression, so that the stress in the steel

as before

A2

Evidently, for reasons of equilibrium, the external load must act along the same line as the

Check the location of the neutral axis 0.6  fyk

sall

sall

If

240 MPa

resultant of all internal forces, i.e., must pass through the plastic centroid. In a symmetrical section the center and the plastic centroid coincide.

Mt  Ms

Then the neutral axis will then be in the web, otherwise the neutral axis will lie in the flange hf  2  b  hf   d    sall 3  M 93.33  kN  m M M Mt

30  d  hf

will also be compressive, though mostly of magnitude smaller than sall .

t

t

s

The assumption was correct.

Example 4.2-14: Verify the stresses in concrete and in steel of a T-beam figure 4.2.14-1 caused by bending moment Ms and external compression force Ns. The section properties:

The neutral axis of a T beam may be either in the flange or in the web, depending upon the proportions of the cross-section, the amount of tensile steel, and the strengths of materials. If the calculated depth to the neutral axis is less than or equal to the slab thickness hf , the beam

Characteristic value of concrete cylinder compressive strength (MPa): fck

25  MPa

Characteristic yield strength of reinforcing steel (MPa):

can be analyzed as if it were a rectangular beam of width equal to b, the effective flange width. The reason for this is illustrated in figure 4.2.13-1b, which shows a T beam with neutral axis in

M s  Mt

if

M s  Mt

if

neutral axis is inside of the flange neutral axis is outside of flange or is inside of the web

The entire cross-section area will be in compression if: Ms Ns

Bo  v2

Where Ms is the bending moment at the section Ns is

Bo

the external axial load at the section

is the moment of inertia of the transformed uncracked section is the area of the transformed uncracked section

400 MPa

Total depth of the T-beam (m): h

0.65  m

Width of the web (m): bw

Igg´

Igg´

fyk

The concrete dimensions

the flange. The compressive area is indicated by the shaded portion of the figure.

0.25  m

Flange width (m): b

0.70  m

Flange depth (m): hf

0.10  m

Distance of the tension and compression reinforcement, from the extreme fibre of the crosssection (m):

Limitation of Stress

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The compressive stress in the concrete at the top fibres of the section (MPa):

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22


d2

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0.03  m

d1

The center of gravity of the concrete section from the upper edge:     

0.04  m

Effective depth to the steel centroid (m): d

h  d1

d

2

A2

3.39  cm

2 3

  bd  bw  hd1    hd1

2

9.42  cm

2

Area of reinforcing longitudinal bars ( cm ): A1

hd   1     bd  bw hd  H  2   0.5  bh  bw hh1  3 hh1  hh    hs  hh  0.5 

0.5   bw h  bh  bw hh

0.61

agc 

2

2

Ac

agc  0.26536 m

The bending moment at the section (kN m): Ms

62  kN  m

Ic 

Axial external load at the section (kN): Ns

1300 kN

1 12

3

3

 bw  h   bh  bw   hh   bh  bw  

hh1

2

3

3

3

  bd  bw   hd   bd  bw  

hd1

3

3

2

 bw  h   0.5

hd   1   agc   bh  bw  hh1    agc  hh    bd  bw   hd   h  2 2     2 1  1    bd  bw  hd1    h  hd   hd1  agc 2  3 

 

  bh  bw  hh   agc 

n = 15 where n = Es / Ec is known as the modular ratio. Es, Ec is the modulus of elasticity of steel, or concrete respectively

hh  2

4

Ic  0.00884 m

This transformed concrete area is seen to consist of the actual concrete area plus n times the area of the reinforcement. The distance of the extreme fibre from the neutral axis ( m):

v2

Figure 4.2.14-1: Entire T beam section is in compression hd  0

hd  0

Ac  bw  h 

B

hh1  0

o



2

h  v1

b

 bw hf 2

v2

2

 n A

1

d1  A

2

d



 

v1

0.25905 m

0.39095 m

or agi 

Sectional area of the concrete

 b  h 2 w

1

v1

Ac  agc   e   A2  d  A1  d1  Ap  ap Ai

agi  0.25905 m

bd  0

 bh  bw hh   bd  bw hd  0.5  bh  bw  hh1  0.5  bd  bw  hd1 2

Ac  0.2075 m

h  agi  0.39095 m 4

Moment of inertia of the transformed uncracked section ( m ):

2

Area of the transformed uncracked section ( m ): Bo

or

bw  h  b  bw  hf  n  A1  A2

Ai  Ac   e   A1  A2  Ap 

Bo

0.22672 m

2

2

Ai  0.22672 m

bw   v1  v2 3

Igg´

Igg´

3 0.00973 m

Limitation of Stress

4

3

2  2   b  b  h   hf   v  hf    n  A  v  d 2  A  d  v 2  w f  12 1 1 2  1   1 1 2    


Ii  Ic  Ac   agi  agc   e  A1   agi  d1  A2   h  d2  agi   Ap   ap  agi  2

2

2

2

Example 4.2-15: Verify the position of the neutral axis of a T-beam section figure 4.2.15-1 caused by the bending moment Ms.The section properties and material characteristic of concrete and steel are as follows:

4

Ii  0.01019 m

The section properties: Characteristic value of concrete cylinder compressive strength (MPa):

Check the location of the neutral axis: If

fck

Ms

Ns Ms Ns

o

Igg´

The entire cross-section area will be in compression.

B o  v2

fyk

Igg´ Bo  v2

Ns

o

Bo

Total depth of the T-beam (m): 0.109  m

The assumption was correct.

K

5.73  MPa

h

o  K  v1

Ns Ai

b1

K

Igg´

6367.781

Ms

b

hf

o  K  v2

 c2 

Ns Ai

b2

Ms

 Ii     h  agi 

1.1  m

Flange depth (m):

section (m):

or

3.24  MPa

0.15 m

Distance of the tension and compression reinforcement, from the extreme fibre of the cross-

The compressive stress in the concrete at the bottom fibres of the section (MPa): b2

0.25  m

Flange width (m):

 c1  7.31095 MPa

 Ii     agi 

bw

or

7.38  MPa

0.65  m

Width of the web (m):

Ms

The compressive stress in the concrete at the top fibres of the section (MPa):

 c1 

410 MPa

The concrete dimensions:

0.047  m

b1

20  MPa

Characteristic yield strength of reinforcing steel (MPa):

d2

 c2  3.35423 MPa

0.06  m

d1

0.00  m

Effective depth to the steel centroid (m): d

h  d2

d

0.59  m

Bending moment at the section (kN m):

Stresses in compression reinforcement (MPa): 0.6  fyk s1

call

240 MPa

n  o  K  v1  d1 

s2  0.6  fyk

allowable

0.6  fck

s1

106.93 MPa

s2

n  o  K  d  v1

s1  0.6  fyk



s2

2

52.48  MPa

 

b  hf   d 

allowable Mt

call

12  MPa

sall

0.6  fyk

sall

246 MPa

hf 

  sall

3

30  d  hf

Limitation of Stress

Mt

249.07 kN  m

Mt  Ms

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The moment of inertia of ideal cross‐section 

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Example 4.2-16: Design the flexural reinforcement for the I-beam section figure 4.2.16-1

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caused by the bending moment Ms. The section properties and material characteristic of concrete and steel are as follows: Characteristic value of concrete cylinder compressive strength (MPa): fck

20  MPa

Characteristic yield strength of reinforcing steel (MPa): fyk

450 MPa

Cross-section (m):

Figure: 4.2.15-1 2

Area of compression reinforcing bars ( m ): A1

0m

b1

2

0.0016 m

b2

0.60  m

h

bw

0.9  m

0.20  m

d2

0.10  m

d1

Es, Ec is the modulus of elasticity of steel, or concrete respectively

The centroid of the steel bars from the top of the beam is

Depth of the compression zone (m):

2

2

d

 n  A2  n  A1  b  bw  hf  x  n  A2  d  n  A1  d1  b  bw 

hf

2

0

x

0.140  m

4

3

3

 b  bw 

x  hf 3 3

Ms

2

Icr

0.00587

x

bc

Icr

call

5.50  MPa

0.6  fck

12  MPa

bc  0.6  fck

allowable

sall

Stress in tension reinforcement (MPa):

n  M s

dx Icr

s2

1300 kN  m

264.44 MPa

0.6  fck

call

12  MPa

Allowable stress in steel (MPa):

s2

0.8  m

Allowable stress in concrete (MPa): 

The compressive stress in the concrete at the top fibres of the section (MPa): M s

d

Es, Ec is the modulus of elasticity of steel, or concrete respectively 2

 n  A2  ( d  x )  n  A1  d1  x

bc

h  d2

n = 15 where n = Es / Ec is known as the modular ratio.

bx

0.25  m

Bending moment at the section (kN m)

2

Moment of inertia of the transformed cracked section ( m ): Icr

h2

0.05  m

x

0.10  m

from the extreme fibre of the cross-section (m):

2

n = 15 where n = Es / Ec is known as the modular ratio.

bw 

h1

Distance of the tension and compression reinforcement

2

Area of tension reinforcing bars ( m ): A2

1.10  m

0.6  fyk

sall

270 MPa

k1

sall

0.6  fyk

246 MPa

s2  0.6  fyk

allowable

Depth of the compression zone (m): x

1  d

Limitation of Stress

x

0.36

x  h1

n

k1

18

1

n n  k1

1

0.45455


Example 4.2-17: Verify the stresses in concrete and in steel of a T-beam section figure 4.2.17-1 caused by the bending moment Ms and external compression force Ns.. The

Fc

h1  call   bw  x   b1  bw  h1   2    x  2  

Fc

1367.86 kN

section properties Characteristic value of concrete cylinder compressive strength (MPa):

Inner lever arm of internal forces (m):

fck

20  MPa

z

d

h1 2

Characteristic yield strength of reinforcing steel (MPa):

 2  2  x  h1 2 2 6  b1  x   b1  bw   x  h1   3

b1  h1  bw  x  h1

z

0.73  m

fyk

450 MPa

Allowable stress of concrete (MPa): call

Stresses in compression reinforcement (MPa): s1

n  x  d1

x

 call

s1

0.6  fck

call

12

Allowable stress of steel (MPa): sall

155.25 MPa

0.6  fyk

sall

270 MPa

Bending moment at the section 2

Area of reinforcement in the tension and compression zone ( m ):

Ms

0.550  MN  m

Total depth of the T-beam (m): A1 A1

M s  Fc  z

d  d1  sall 0.00593 m

2

A2 A2

Fc  A1  s1

h

1.10  m

Flange depth (m):

sall 0.00593 m

hf

2

0.15  m

Width of the web (m): bw

0.30  m

Flange width (m): b

0.70  m

Distance of the tension and compression reinforcement, from the extreme fibre of the crosssection (m): d2

0.08  m

d1

0.0  m

The centroid of the steel bars from the top of the beam is (m): d

h  d2

d

1.02  m

Position of the neutral axis:    if  Mt  Ms

  

Figure: 4.2.16-1

2

 

b  hf   d  Mt

The neutral axis will then be in the web of the T-beam.

hf 

  sall

3

30  d  hf

Limitation of Stress

Mt

158.04 kN  m

Mt  Ms

The assumption was correct.

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Compression force acting in the concrete (kN):

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Example 4.2-18: Verify the stresses in concrete and in steel of a T-beam section figure 4.2.18-

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1 caused by bending moment Ms and external compression force Ns. The section properties Characteristic value of concrete cylinder compressive strength (MPa): fck

20  MPa

Characteristic yield strength of reinforcing steel (MPa): fyk

410 MPa

Total depth of the T-beam (m):

hf

2

Width of the web (m):

2

bw

2

Area of tension reinforcing bars ( m ): A2

0.002198 m

15  cm

Area of compression reinforcing bars ( m ): 0m

90  cm

Flange depth (m):

Figure: 4.2.17-1

A1

h

25  cm

Flange width (m): b

2

110 cm

2

Area of reinforcement in the tension and compression zone ( cm ):

Depth of the compression zone (m): bw 

x

2

2

A2

 n  A2  n  A1  b  bw  hf  x  n  A2  d  n  A1  d1  b  bw 

hf

2

2

0

x

d2

4

Moment of inertia of the transformed cracked section ( m ): bx

3

3

 b  bw 

x  hf 3 3

2

 n  A2  ( d  x )  n  A1  d1  x

2

Icr

0.02288 m

The compressive stress in the concrete at the top fibres of the section (MPa): bc

M s

x

bc

Icr

6.77  MPa

0.6  fck

12  MPa

bc  0.6  fck

n  M s

dx Icr

s2

266.10 MPa

15.70  cm

4

5.0  cm

d1

5.0  cm

The centroid of the steel bars from the top of the T-beam section ( cm) is: d

75  cm

Bending moment at the section (kN m): allowable

Ms

450 MN  m

Axial compression external load at the section (kN):

Stress in tension reinforcement (MPa):

s2

A1

2

Distance of the tension and compression reinforcement, from the extreme fibre of the crosssection ( cm ):

0.28  m

Icr

15.70  cm

2

Ns

0.6  fyk

270 MPa

s2  0.6  fyk

allowable

850 kN

The depth of the neutral axis (centroid) (cm): 2

yG

bw  h  b  bw  hf

2

2  bw  h  b  bw  hf

Limitation of Stress

yG

31.43617 m


e

Ms Ns

 100

Section modulus of a transformed T-beam section c

e  yG

c

21.50501 m

e

52.94  m S

Since the external compression force applied a significant distance outside the cross-

bx

2

2

b  bw  x  hf 2 2

 n  A1  x  d1  A2  ( d  x )

S

26955.19

section, this causes that the section tension governs. Position of the neutral axis:

K

If the expressions H2 and H3 have the same sign then the neutral axis will then be in the web of the section. Otherwise the position of the neutral axis will be in a flange.

K

100 S

(For the tension force we substitute a minus sign).

0.315

The compressive stress in the concrete at the top fibres of the section (MPa):

H2

b  bw  3  c  2  hf   hf 2  90 A1  c  d1  d1  A2  ( d  c)  d

H2

11742646.95

H3

hf  3  c  b  hf 2  90 A1  c  d1  d1  hf   A2  ( d  c)  d  hf 

H3

6194870.01

Ns 1000

bc

K x

bc

9.12  MPa

Other way of solution is checking the compressive stress in concrete as follows:

Since H2 and H3 have the same sign the neutral axis is situated in the web of the T-section. p

q

3bc

2

 bw

bw 2bc

3

bw

3

y  py  q

b

 3

 1  c  hf

2 

90  A1 bw

 c  d1 

90  A2 bw

 ( d  c)

 b  1  c  h 3  90  A1  c  d 2  90  A2  ( d  c ) 2  2  1  f bw bw  bw  0

y

p

12515.52657

yc

x

Ns 10  x bw  x

q

b  bw  2  x  hf   hf

2

2

2

bc

 n  A1  d1  x  n  A2  ( d  x )

724139.69811 bc  0.6  fck

48.65  m

allowable

Depth of the compression zone (m): x

bc

Stress in tension reinforcement (MPa):

28.94  m

s2

n  K  ( d  x)

s2

n  bc 

dx x

Figure 4.2.18-1: The location and distribution of the internal and external forces.

Limitation of Stress

s2

217.87 MPa

s2

217.87 MPa

or

9.12  MPa

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Eccentricity (cm):

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Example 4.2-19: Verify the stresses of concrete and steel of a T-column figure 4.2.19-1 caused

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29 Allowable stress of concrete (MPa):

by the bending moment Ms and tension force Ns. 0.6  fck

call

call

12

Allowable stress of steel (MPa): 0.6  fyk

sall

sall

270 MPa

Distance of the tension and compression reinforcement from the extreme fibre of the crosssection (m): d2

0.05  m

d1

0.05  m

The centroid of the steel bars from the top of the T-beam (m) is: d

Figure: 4.2.19-1 Since the point c is inside of the section or between the reinforcement As1 and As2 and the axial

h  d2

d

0.8  m

The depth of the neutral axis (centroid) (m):

load is a tension load entire section will then be in tension. 2

The section properties: Characteristic value of concrete cylinder compressive strength (MPa): fck

bw  h  b  bw  hf

yG

2

yG

2  bw  h  b  bw  hf

0.29  m

20  MPa

Bending moment at the section (kN m): Characteristic yield strength of reinforcing steel (MPa): fyk

Ms

83  kN  m

450 MPa

Axial external tension load at the section (kN): Total depth of the T-beam (m): h

0.85  m

Flange depth (m): hf

Ns

370 kN

Eccentricity (m): e

Ms

e

Ns

0.224  m

0.15  m

2

Area of compression reinforcing bars ( m ): Width of the web (m): bw

0.25  m

Flange width (m): b

1.10  m

A1

  d  d1  sall

Ns d  y G  e

A1

0.00052 m

As1

0.0006 m

Provided: 3  16

Limitation of Stress

2

2


Ns

A2

sall

1 caused by bending moment Ms and external compression force Ns.  A1

A2

0.0008552 m

2

The section properties Characteristic value of concrete cylinder compressive strength (MPa):

Provided:

fck

3  20

As2

0.00094 m

Characteristic yield strength of reinforcing steel (MPa): fyk

Modular ratio            Es /Ec = n  

n

sall

bw  h  b  bw  hf  n  As1  As2

0.6  fyk

sall

270 MPa

Total depth of the T-beam (m): Bo

h

0.36317

This transformed concrete area is seen to consist of the actual concrete area plus n times the area of the reinforcement. The distance of the extreme fibre from the neutral axis

1.05

Width of the web (m): bw

0.30  m

Flange depth (m): hs = hf hf

0.15  m

Flange width (m):

v1



 bw  h2  b  bw  hf 2    n   As1  d1  As2  d 2 Bo  2 

v1

0.30738 m

v2

h  v1

v2

0.542  m

1

450 MPa

Allowable steel stress (MPa):

15

Area of the transformed uncracked section Bo

30  MPa

2

b

0.85  m

Stress in tension reinforcement (MPa):

s1

  d  d1  As1

Ns d  v1  e

s1

219.5045 MPa

Stress in tension reinforcement (MPa):

s2

  d  d1  As2

Ns v1  d1  e

s2

252.27 MPa



Figure: 4.2.20-1

H2

H3

hf  3  c  b  hf 2  90 A1  c  d1  d1  hf   A2  (d  c)  d  hf 

2

b  bw  3  c  2  hf  hf  90  A1  c  d1  d1  A2  ( d  c )  d

Limitation of Stress

H2

1.43438

H3

0.93143

STRUCTURAL ENGINEERING ROOM

Example 4.2-20: Verify the stresses in concrete and in steel of a T-beam section figure 4.2.20-

2

Area of tension reinforcing bars ( m ):

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30


STRUCTURAL ENGINEERING ROOM

Department of Architecture

31 If the expressions H2 and H3 have the same sign the neutral axis will then be in the web of the section. Otherwise the position of the neutral axis will be in a flange. Since H2 and H3 have the same sign the neutral axis is situated in the web of the T-section.

p

q

3bc

2

2bc

b

 3

 bw

bw

3

b

 2

 bw

bw

3

y  py  q

Figure 4.2.20-2: The location and distribution of the internal and external forces.

d2

0.05  m

h  d2

0.900  MN  m

0.250  MN 2

Area of reinforcement in the tension and compression zone ( m ): The depth of the neutral axis (centroid) (m):

30  d  hf

yG

bc

0.4316

e

Ns

e

2 

 d1  c

90  A2 bw

 (d  c)

2

20.38778

q

28.00861

3.52843 m m

0.36  m

Mt

Ns x

bw  x

2

b  bw  2  x  hf   hf 2

bc

s2

which is lesser than allowable stress

c

yG  e

c

3.1684 m

n  bc 

s2 which sall

dx x

s2

bc  0.6  fck

256.40 MPa

is lesser than allowable stress sall of reinforcement

270 MPa

Limitation of Stress

9.61  MPa

 n  A1  d1  x  n  A 2  ( d  x )

Stress in tension reinforcement (MPa):

3.6  m

If Mt  Ms

0.175  MN  m

Eccentricity (m): Ms

bw

p

The assumption was correct

2

2

2  bw  h  b  bw  hf

90  A1

 ( d  c)

Other way of solution is checking the position of the neutral axis as follows:

bc

3 

bw

The compressive stress in the concrete at the top fibres of the section (MPa):

0.0035

x

Mt  Ms

2

0

yc

x

Axial compression external load at the section ( MN):

bw  h  b  bw  hf

90  A2

 d1  c 

y

t

yG

 1  c  hf

Compute Mt (MN m): hf  2  b  hf   d    sall 3  M

A2

bw

Then the neutral axis will then be in the web of T-beam.

Bending moment at the section ( MN  m):

Ns

90  A1

Ms

2 

Depth of the compression zone (m):

The centroid of the steel bars from the top of the T-beam is (m): d

Distance of the tension reinforcement from the extreme fibre of the cross-section (m):

 1  c  hf

sall

0.6  fyk

allowable


Section properties: Characteristic value of concrete cylinder compressive strength (MPa): fck

Mt

0.85 

1.5

fcd

1.15

282.60 MPa

z

0.35  m

1.20  m

h  d2

d1

0.05  m

d

0.59  m

0.55  MN  m

0.720  MN

Nsd  0

The eccentricity (m): e

Ms Ns

2

from the upper edge (MN). Ffl

1.445  MN

hf

z

2

0.515  m

Mfl

0.74  MN  m

Msd  Nsd  d  y G

Mext

0.80  MN  m

Moment subjected to the web will then be ( MN  m): Msds

Mext  Mfl

Msds

Msds

2

0.061  MN  m

0.045

bw  d  fcd

From the diagram we obtain:

Axial tension load at the section (MN): Nsd

hf

Ffl  z

Mext

Bending moment at the section ( MN  m): Msd

0.235  m

External bending moment calculated to the centroid of the steel bars ( MN  m):

The centroid of the steel bars from the top of T beam (m) is: d

d

Mfl

0.15  m

0.06  m

yG

The ultimate moment of flange refers to the tension reinforcement (MN m):

Distance of the tension and compression reinforcement, from the extreme fibre of the crosssection (m): d2

Thus the lever arm of the inner forces is known (m):

0.65  m

Flange width (m): b

Mt  Msd

2

b  bw  hf  fcd

Ffl

Flange depth (m): hf

2  bw  h  b  bw  hf

distance of

Web width (m): bw

0.312  MN  m

The resultant of the concrete compressive force in the flange my be assumed to act in a

Total depth of T beam (m): h

2

fyd

Mt

bw  h  b  bw  hf

yG

Design yield stress of reinforcement (MPa): fyd

30  d  hf

325 MPa

fyk

  fyd

3

The depth of the neutral axis (centroid) (m):

11.33  MPa

Characteristic yield strength of reinforcing steel (MPa): fyk

hf 

The assumption was correct

Design value of concrete cylinder compressive strength (MPa): fcd

 

2

b  hf   d 

20  MPa

fck

Position of the neutral axis: If Then the neutral axis will then be in the web of T-beam. Mt  Ms

2

The required tension reinforcement A2 ( cm ) can be calculated from: A2

e

0.764

0.01475

  bw  d  fcd  100 

Ns fyd

Limitation of Stress

4

 10

A2

28.92  cm

2

STRUCTURAL ENGINEERING ROOM

Example 4.2-21: Design the flextural reinforcement for the T beam figure 4.2.21-1 caused by bending moment Msand axial tension load Ns .

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32


STRUCTURAL ENGINEERING ROOM

Department of Architecture

33 Provided:

Example 4.2-22: We assess tension parts of PPRCB according to limit state of crack width.

Diameter, Number of bars:

Beam of (C25 / 30 concrete and reinforced 10335), for which satisfies w lim

As2

25  mm

t

6  25

nt   

t

nt

main bearing structure reinforcement As It was determined according to ULS. Operational

6

2

2

As2

4

0.3  mm. The

0.00295 m

load torque from the unique combination of load As

2

combinations M

kva

Which is greater than A2

2

and the quasi-permanent load

.

To locate the neutral axis, first assuming that it lies in the flange, equating the moments of areas about neutral axis gives the equation: b

x

2

 n  As2  ( d  x )

2

Solving

0

x

0.175  m

Since this is greater than hf = 150 mm, the assumption is not correct and the neutral axis passes through the web. Taking moments of areas for the case in:

bw 

x

2

2

 

 b  bw  hf   x 

hf 

  n  As2  ( d  x )

2

0

Solving

x

0.176  m

Figure: 4.2.22-1 Cracks do not, per se, indicate a lack of serviceability or durability, in reinforced concrete structures, cracking may be inevitable due to tension, bending, shear, torsion (resulting from either direct loading or restraint of imposed deformations), without necessarily impairing serviceability or durability. Bending Moment Md  520  kN m

L  15  m

Mkva  465  kN m

L H  15

H  1m

Where L is the span of the beam. H is the depth of the beam

Figure: 4.2.21-1

Material characteristics: Concrete 30/37: fck  30  MPa

Limitation of Stress

fctm  2.9  MPa

Ecm  32.0  GPa

fcd  0.85 

fck 1.5

fcd  17  MPa


Cross-sectional dimension:

fyk  412  MPa Es  200  GPa  2  25  mm

A2 

fyk

fyd 

 2

 2

A2  0.0004909 m

n2  4 Ast1  n2  A2

A1 

 

 1

2

A1  0.0000785 m

4

n1  6

Ast2  n1  A1

2

Es

b  0.0  m

zb  0  m

zb  0.25  m

zb  0.65  m zb  0.65  m

zb  1.0  m

zb  1.0  m

Ast

Ac  d  fcd

2

1  12  mm

Ast  0.0024347 m

2

  b  d  fcd

Mu  Md

  0.175

Ac  0.29  m

 

Ast Ac  fcd

Mu    Ac  d  fcd

ok

 100  MPa   0.0493861

n1  4

Mu  819.6125 m kN

hh1  0  m

hs  0.4  m

hd1  0  m

as1  0.05  m

as2  0.05  m

bs  0.2  m

hh  hd  hh1  hd1  hs  1 m d  H  as2

bd  0.35  m hd  0.35  m

d  0.95 m

s1

0 0.0875

A

s2

A

A a    A c

c

e

gi

p

s2

d  A A

s1

a

s1

 A a

p p

i

The moment of inertia of ideal cross-section: i

I

by

c 

A a

gi

a

2 2 2 2    A   a  a   A   a  d  A   a  a   c s1 s2 gi p gi p  e  s1 gi

2

As  n1   

1

As  n2   

2 3 4

2

4

4

As  n5   

2

4

3

As  n4   

2

4

2

As  n3   

2

4

1

5

2

4

5

4

5

2

As

1

As

2

zs  0.05  m

 0.00045 m

1

2

 0.0001571 m

2

As

3

As

4

As

5

 0.0003142 m

2

 0.0001571 m

2

 0.0019635 m

2 2 zb  zb  zb   zb   ( k 1) k k   ( k 1) 0

m

zs  0.20  m 2

zs  0.70  m 3

zs  0.95  m 4

zs  0.95  m 5

0.0625

zbk 1  zbk 0 m

2

m

0.25

0.1875

0.5

0.08

0.6475

0.9

-0.0975

1.2675

1.3

0.1225

2.0725

1.65

0.35

3

2

 

j

0.0001571

 

As  zs 2   j j 

As  0.0004524

I

3

0.0375

Distance of center of gravity of ideal cross-section of the upper edge: a

n5  4

b( k 1) zbk  bk zb( k 1)

Ideal sectional area: e

2

6

c

n4  2

5  25  mm

bh  0.35  m hh  0.25  m

A  A

n3  4

4  10  mm

Cross-sectional dimension:

i

n2  2

3  10  mm

A

zb  0.25  m

b  0.35  m

5

7

2  10  mm 

1

4

Area and location of reinforcement:

Msd

3

j  1  5

Ast  Ast1  Ast2

Ecm

b  0.35  m

6

2

Ast2  0.0004712 m  e 

b  0.2  m

2

7

2

Ast1  0.0019635 m

 1  10  mm

b  0.35  m b  0.35  m b  0.2  m 1

2

4

k  1  7

fyd  358.261  MPa

1.15

2

m

0.0000011 0.0000063

As j  zs j   4

m

0.0000226 0.0000314

0.0003142

0.0001539

0.0002199

0.0001571

0.0001418

0.0001492

0.0019635

0.0017721

0.0018653

Limitation of Stress

3

m

STRUCTURAL ENGINEERING ROOM

Steel - Reinforcement:

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34


STRUCTURAL ENGINEERING ROOM

Department of Architecture

35 Description of full-sectional acting

Ideal characteristics of a fully-acting sectional:

Area of the concrete section

Area ideal cross-section:

Ac 

1  2

7

k 1

Ac  bs  H 

2

Ac  0.29 m

b( k 1) zbk  bk zb( k 1)

or

Ai 

1  2

 bh  bs  hh   bd  bs  hd  0.5  bh  bs  hh1  0.5  bd  bs  hd1

7

b( k 1) zbk  bk zb( k 1)  E  cm

k 1

Ai1  Ac   e  As  As  As  As  As

2

Ac  0.29 m

5

Es

1

2

3

4

5

j 1

As

2

Ai  0.3090263 m

j

2

Ai1  0.3090263 m

Static moment: The static moment of area of the concrete section to the upper edge:

Si 

7  1  3 Sc    b  zb  b   zb   zb  zb  Sc  0.148 m      ( k  1 ) k k k  1 k  1 k    6  k  1 

agc  0.5103448 m

agc 

Ac

agc 

2   bd  bs   hd1    hd1 3

hd    1    bd  bs   hd H  2   0.5  bh  bs  hh1  3 hh1  hh    hs  hh  0.5  2

agi 

1

Si Ai

agi1 

agi  0.525208 m

or

Ac  agc   e  As  zs  As  zs  As  zs  As  zs  As  zs 1

1

2

2

3

3

4

4

5

5

Ai1

agi1  0.525208 m

The moment of inertia to the upper edge:

j 1

Distance of center of gravity from the top of the cross-section:

Ac

7  1    12 k  

5

 As j zs j 

3

agc  0.5103448 m

I c0 

k 1

Es

b( k 1) zbk  bk  zbk 1  zbk 1  zbk  E  cm

Si  0.1623031 m

or

2

7

Sc

0.5  bs  H   bh  bs   hh

1  6

b  ( k

Ic0  0.1037167 m

4

1)

 zb

k

 b

k

The moment of inertia of ideal cross-section:

 zb

(k 1) 

 Sc    Ac 

Ic  Ic0  A c  

2



  zb

k 1

2

 zb

(k 1)

 zb

Ic  0.0281856 m

k

 zb k  2    

Ii  0

4

 1  7  2 2 b   zb  b  zb   zb  zb  zb   zb   ( k 1) k k  12   ( k 1) k k ( k 1)  k 1  k  1   Es 5 2    As  zs  Ecm j  1  j j  

 

4

Ii  0.1166865 m 0

Ii  Ii  Ai   agi  0

Limitation of Stress

2

4

Ii  0.0314436 m

      



or


3 3 1  b H3  b  b h 3  b  b  hh1  b  b h 3  b  b  hd1  Ic          s h s h h s d s d d s 12 3 3  2 2 h h  bs H 0.5 H  agc 2   bh  bs  hh agc  h    bd  bs  hd  H  d  agc  2  2     2 2  1  1 1  1    bh  bs  hh1  agc  hh  hh1   bd  bs  hd1   H  hd  hd1  agc 2 3 2 3     

        

4

Ic  0.0362294 m

s

may be Taken equal to fyk if adequate anchorage is secured; a lower value may, however,

be needed to satisfy the crack width limit. f

ct eff

is the upper fractal of the concrete strength in tension at the moment when the first

crack is expected to appear. f K

ct eff

3  MPa

is the factor correcting the value A, as found by technical theory, against the real Act

values taking into account non-linear stress distributions (self –equilibrating effects); the

following rules may be applied.

When the cross-section subjected to Pure bending then the depth of tension zone from the bottom will be:

Restraint of extrinsic imposed deformations K=1 Restraint of intrinsic imposed deformations of rectangular sections. K= 0.8 for < 300 mm

H  agi  0.474792 m

K= 0.5 for >800 mm (linear interpolation is possible)

The effective concrete area in tension Ac,ef accounts for the non-uniform normal stress distribution by bond forces into the concrete cross-section at the end of the transmission length. Entire area of the bottom flange is in tension. Protruding part of the lower flange without main reinforcement have area according to figure A Act1   bd  bs   hd

Kc accounts for the scheme of tensile stress distribution. Kc= 1 for pure tension Kc= 0.4 under flexural conditions without axial compressive force. Kc= 0.4-1.0 for a combination of pure tension and flexure.

ct1

In prestress members, the minimum reinforcement for crack control is not necessary in

2

Act1  0.0525 m

areas where, under the rare combination of loads and the characteristics value of prestress or For pure bending it is kct  0.4

normal force, the concrete remains in compression.

Where:

Otherwise, the required minimum area may be calculated by means of equation, with the

 s  fyk

fcteff  3.0  MPa

fctm  2.9  MPa

following values for kc.

fcteff  fctm

For the combination of pure tension and flexure, in the absence of a more rigorous

method, the following simplified procedure may be applied for the calculation of the required area of minimum reinforcement within the tensioned concrete zone. A

s

k k f c

A

ct eff 

For box sections Kc= 0.45 for the webs Kc= 0.9 for the tension chord

ct

For rectangular section

s

Kc= 0.45 under flexure conditions without axial compressive force.

A  denotes the area of the concrete tension zone just before the formation of cracks, ct

calculated with the technical theory in the uncracked stage.

Kc= 0 when concrete remain in compression, or the depth of the tension zone, calculated in the basis of a cracked section under the loading conditions leading to formation of the first crack, does not exceed the lesser of h/4 or 0.3m.

Limitation of Stress

STRUCTURAL ENGINEERING ROOM

or

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36


Linear interpolation between both values is possible.

STRUCTURAL ENGINEERING ROOM

Department of Architecture

37

Prestress tendon may be taken into account as minimum reinforcement within a 300 mm

The calculation of the Stress at the top and bottom of cross-section

square surrounding the tendon, provided the different bond behaviour of the tendons and

The value of the coefficient k We can specify the event of eccentric tension, if we know the c

stress of concrete on both edges of the ideal cross-section  and  before cracking c max c min then: 0.70  0.3 

c

Ii Md

 c2 

Ii

  agi 

 c1  8.6856518  MPa  s1  0

  H  agi 

 c2  7.8518935  MPa  s2  0

The tension part must reinforced to avoid the cracks. The cross-section is relieved of compressive force.

c max

Pure bending

kc  0.4 As

fcteff  3  MPa

k  1.0 2

3

 0.0003142 m

Asct1  As

3

 As

As

2

Act1  0.0525 m 2

4

 0.0001571 m

As  As 3

 s  412  MPa

Asct1  0.0004712 m

Asct1´  kc  k  fcteff 

As  As  Assct1´ 3 4

Act1 s

xr 

2

Asct1´  0.0001529 m

bt 

5

 As

4

 1  

2 b e

As 1 as1  As 5  As 4 d  2  As  As 5 4 

xr  0.2318656 m

the moment of inertia of the weakened cross-section:

Area considered the main tensile reinforcement. 5

b

 As

The depth of compression zone xr - weakened cross section ag = xr. So we can directly determine

The inner part of the flange A the main bearing reinforcement ct2

4

e

ok

 As

Act2  0.0525 m

Cross-section with cracks:

2

 0.0004712 m

4

2

Act2   bd  bs   hd

kc  0.4

4

2

Ast2  As

 c1  fctm  c2  fctm

c min

k

Md

 c1 

reinforcement are taken into account.

I 

2

 Asct1

Ast2  0.0016493 m

bd  bs  bs   H  agi  hd

hd   H  agi  hd

H  agi  hd  0.124792 m

1 3

 b  xr   e  As   xr  as1   As  As 3

2

1

5

4

 d  xr 2

The stress in the reinforcement at cracking the exceptional load combinations:

bt  0.2 m

Ii d  xr

 sr  fctm  

I H  agi

 e

 sr  102.8293179  MPa

Area of the longitudinal reinforcement:

Stresses in reinforcement for long-term and short-term component of load: Asmin 

0.6  bt   H  as2  fyk

 MPa

Asmin  0.0015  bt   H  as2 

2

Asmin  0.0002767 m

 slt 

or

2

 sst 

Asmin  0.000285 m

0.6  Mkva I

  d  xr    e

0.4  Mkva

Limitation of Stress

I

  d  xr    e

 slt  149.3806061  MPa

 sst  99.5870708  MPa

4

I  0.0083829 m


2.5  ( H  d)  0.125 m

 s   slt   sst

 s  248.9676769  MPa

 1  1

for long-acting, or multiple repeated load

 lt2  0.5

 sst2  1  2 

 slt   sst

Aceff  2.5  ( H  d)  bs

5  0.025 m

 r 

2

Ast2  0.0016493 m

The average final crack length:

for a one-time short

 slt   lt2   sst   sst2

bs  0.2 m

srm  50  mm  0.25  k1  k2   2  0.7

5

Ast2

2

Aceff  0.025 m

 r  0.0659734

Aceff

srm  0.087894 m

r

The following inequality should be observed Wk<Wlim

The mean value of average strain reinforcement at the appropriate load combinations determined taking into account the effect of tension zone of concrete – (tension stiffening)

Wk denotes the characteristic crack width. Wlim denotes the nominal limit value of crack width which is specified for cases of

between cracks. In a cracked cross-section all tensile forces are balanced by the steel only. However, between adjacent cracks, tensile forces are transmitted from the steel to the surrounding

expected functional consequences of cracking, or for some particular cases related to durability problems. Computing crack widths w

concrete by bond forces. The contribution of the concrete may be considered to increase the stiffness of the tensile reinforcement. Therefore, this effect is called the Tension Stiffening

the cracks of direct load, h  800  mm

Effect. During the state of cracks formation one crack after the other occurs thus decreasing the stiffness of the member. When cracks appear, single cracks play an important role. In this state some part of the area between the cracks remain in state I (s =c). when the crack pattern has stabilized, the distance between the cracks, is equal to or less than twice the transmission length

 sm

2    sr     1   1  2    Es   s  

 smlt 

 slt

Es

  

  sr    s 

 1   1   lt2  

 sst

Es

 

wklt     smlt  srm

k2  1

pure tension for reinforcement with great bond

wkst  0.0000617 m

0.5 

h  xr xr

compression with eccentricity

and equal in length

2.5  ( h  d)

  

 smlt  0.0006832

2

  

wklt  0.00010208 m wklt  wkst  0.0001638 m

 smst  0.000413

wlim  0.3  mm

wkst    srm   smst

wlim  0.3  mm

Stress of concrete and reinforcement from the effect of quasi-permanent load combination will

pure bending Effective cross-sectional area of the tensile concrete that surrounds the tensile reinforcement k2  0.5

2

  sr    s 

 1   1   sst2  

smooth surface

k2

wk  0.0001638 m

The cross-section is satisfied according to limit state of crack width:

k1  1.6

k1  0.8

wk    srm   sm

  1.7

wlim  0.3  mm

 smst 

 sm  0.0010962

for cross sections h  300  mm

(length over which slip between steel and concrete occurs). s

  1.3

k

be:  c1  

Mkva I

 xr

Limitation of Stress

 c1  12.8615598  MPa

STRUCTURAL ENGINEERING ROOM

The total stress in the reinforcement:

Department of Architecture

38


STRUCTURAL ENGINEERING ROOM

Department of Architecture

39 Example 4.2-23

The stress in the lower and upper concrete reinforcement  s2 

 e  Mkva

I

  d  xr 

 s1 

 s2  248.9677  MPa

As  kc  k  fcteff 

 e  Mkva

I

Partial prestressed beams according to figure, the proposed concrete C35 / 45 f ck, f ctm ,

  as1  xr 

Ecm . Reinforcement As1, As2, Es , Tendons Ap, Ep . For service state is a section for stress loading in exceptional combination

 s1  63.0504107  MPa

Act1

2

2

As  0.000253 m

Ast2  0.0016493 m

 G k j   P  Qk 1   0 1 Qk 1

Md , normal force Nd (tension) in the distance a c (See figure), and the normal force Np (Pressure) from

the

Pm t

P0  P c  P t ( t )  P

The main tensile reinforcement satisfied according to limit state of crack width when Ast2> As

increased pressure stress concrete.

 s2

Inner forces - Bending moment

mean

bias

force

for

pretensioned

Pm t

element

 ( x)  . It should be asses the risk of longitudinal cracks due to

Tension part of the web A and Surface longitudinal reinforcement ct3 Act3  bs   H  agi   hd kc  0.4

2

Act3  0.0249584 m  cteff  3.0  MPa

k  0.5

As  kc  k  fcteff 

Act3 s

For pure bending  s  248.9676769  MPa

2

As  0.0000601 m

We choose both sides J6   A 0.15 m   s  6  mm

ns  6

As  ns   

s

2

4

2

As  0.0001696 m

The function of the structure should not be harmed by the cracks formed. The durability of the structure during its intended lifetime should not be harmed by the cracks formed. Tensile stress in reinforcement should be limited with an appropriate safety margin below the Figure 4.2.23-1: Values of ideal cross-section and stress distribution

yielding strength, preventing uncontrolled cracking. In calculating the stress, account shall be taken of whether the section is expected to crack under service loads and also of the effects of creep, shrinkage and relaxation of prestressing steel. Other indirect actions which could influence the stress, such as temperature, should also be considered. Tensile stress in concrete Depending on the stress limit chosen different limit states may be required, but the LS of decompression is considered to be the most significant.

Material characteristics

fck  35MPa

fctm  3.2MPa 2

Ecm  33.5GPa 2

As1  8.042cm

As2  12.56cm

Es  200GPa

Ep  200GPa

Md  2.35MN m

Nd  1.20MN

Np1  4.20MN

Limitation of Stress

,

eh  0.5m

2

Ap  33.98cm

e



Es Ecm


The moment of inertia of ideal cross-section

ap  0.85m

as1  0.05m

as2  0.05m

bs  0.15m

bh  0.4m

bd  0.3m

H  1m

hh  0.2m

hs  0.45m

hh1  0.1m

hd1  0.05m

hd  0.2m

d  H  as2

H  hh  hh1  hs  hd1  hd

2 2 2 2 Ii1I  IcI  AcI  agi1I  agcI   e As1  agi1I  as1  As2  H  as2  agi1I  Ap  ap  agi1I 

Ii1I  0.0347682m

H  1m

AcI  bs H   bh  bs  hh   bd  bs  hd   0.5 bh  bs  hh1  0.5 bd  bs  hd1 2

etotiI 

Sectional area ideal Ai1I  AcI   e  As1  As2  Ap 

Ai1I  0.2788363m

  bd  bs  hd1  hd1  hs  hh  0.5 2

agcI 

 c1

hd   2   0.5 bh  bs  hh1  1 hh1  hh     bh  bs  hh    bd  bs  hd  H 

3

2

3

etotiI  0.2849709m



Nd  Np1  agi1I   1  Ai1I etotiI  Ai1I I i1I  

 c1

 22.8478319MPa

 c2

 1.7410728MPa

The lower edge

 c2

AcI

agcI  0.4567682m



Nd  Np1  H  agi1I   1  Ai1I etotiI  Ai1I I i1I  

Control of cracking in the upper edge of the cross section

The center of ideal cross-section

agi1I 

Nd  Np1

The top edge

The center of gravity of the concrete section from the upper edge 2

 H  a   N  a  a gi1I p1  p gi1I 2 

Md  Nd 

Calculation of concrete stress in the extreme fiber cross-section

2

0.5 bs H

4

Distance resultant horizontal forces of gravity ideal cross-section

Sectional area of the concrete

AcI  0.24625m

When calculating the stress it is whether under the action of service load expected or not expected cracking. Weakening cracks should be considered in cases where exceptional load

AcI agcI   e  As2 d  As1 as1  Ap ap

 G k j   P  Qk 1   0 1 Qk 1

combinations in accordance

agi1I  0.4916376m

Ai1I

or

 G k j   P  Qk 1

valid for the largest cross-section of the tension without cracking :  c  f ctm , where f ctm the

The moment of inertia of the concrete cross section

mean value of the tensile strength of concrete. The cross-section without crack assumes the full effect of the concrete section and the elastic

IcI 

3

3

hh1 hd1 3 3 3 2 bs H   bh  bs  hh   bh  bs     bd  bs  hd   bd  bs    bs H 0.5H  agcI  3 3 12 2 2 2 hh  hd    1 1    bd  bs  hd  H    bh  bs  hh  agcI   agcI    bh  bs  hh1   agcI  hh  hh1   1

2

1 1   bd  bs  hd1   H  hd  hd1  agcI  2

IcI  0.0293517m

4

3

2

2

2

3

behaviour of concrete and reinforcement in tension and compression. Therefore, it is necessary to begin by calculating the limit state stress limitation of certain parameters of ideal cross-section. Because EC2 is considering a uniform value of modulus of elasticity for concrete and tendons Es single-value ratio  e

Limitation of Stress

Es Ecm

.

Ep We can when calculating the ideal sections to a

STRUCTURAL ENGINEERING ROOM

cross-sectional dimension

Department of Architecture

40


The cross-section without crack assumes the full effect of the concrete section and the elastic

STRUCTURAL ENGINEERING ROOM

Department of Architecture

41 Limitation of concrete compressive stress due to increased creep

behaviour of concrete and reinforcement in tension and compression.

When you prestressed beams should be an assessment especially at the stage of brought in by pre-stress concrete. It assessed the fulfillment of the conditions of reliability for

Therefore, it is necessary to begin by calculating the limit state stress limitation of certain parameters of ideal cross-section. Because EC2 is considering a uniform value of modulus of elasticity for concrete and tendons Es single-value ratio  e  c1

Ep We can when calculating the ideal sections to a

G k j

 P   2 1 Qk 1

Es Ecm

 22.8478319MPa

Example 4.2-24: Limitation of concrete stress due to an increased creep

fctm  3.2MPa

 c1

 fctm

compressive strength of concrete at the stage of prestressing where the moment of its self-

weight element Md, and the mean value of normal force initial prestressing force is Np, Initial

Control of cracking at the bottom of the cross section

 1.7410728MPa

Nonprestressed partial beam of the image is proposed concrete C35/45 f ck,  f ctm ,  Ecm.  Concrete reinforcement    As1,  As2,  Es  , prestress reinforcement   Ap,  Ep.   is required to assess

Thus, the cracks do not arise

 c2

compressive stress in the concrete at the quasi-permanent load combinations

prestressing force it is Np. Because the quasi-permanent load exceeds 50% of the total load

fctm  3.2MPa

element is assumed to be the ratio of the modulus of elasticity of the material  e

Thus, the cracks do not arise. Limitation of concrete compressive stress for evidence of longitudinal cracks Limit state limit pressure stress in the concrete terms of the threat of longitudinal cracks control of assistance terms of reliability: c

 0.6f ck

Satisfies the compressive stress c in absolute terms this condition or not  fck It is characteristic concrete compressive strength, the greatest tension concrete compressive

load at an exceptional combination  G k j   P  Qk 1   0 1 Qk 1 or 

 G k j   P  Qk 1, resp.  G k j  P  0.9Q k 1  0.6fck

 21MPa

 c1

 22.8478319MPa

The reliability condition is not satisfied, it is necessary to increase the grade of concrete. Formation longitudinal cracks reduces the durability of the element. Failure Conditions c

 0.6f ck we can strengthen the transverse reinforcement element or increase the thickness

of the concrete cover.

Limitation of Stress

Figure: 4.2.24-1

15.


As2  0.001256m

bs  0.4m

Np  2.8MN

Md  0.14MN m

H  ap  0.15m

2

2

Ap  25.49cm H  1 m

Variable cross-section with cracks

Calculation of the lower (compression) edge cross-section Md  Np  H  ap

e 

Sectional area ideal Ai  bs H   e  As1  As2  Ap 

Ai  0.48184m

The center of gravity of the cross section of the concrete from the top of the cross-section

12

agi  0.4693363m

Ii  0.0453753m

4

2

Position resultant of horizontal load

etoti 



Md  Np agi   H  ap 

etoti  0.3693363m

Np

 c2

 c2



Np  agi   1  Ai etoti  Ai  Ii  Np  H  agi   1  Ai etoti  Ai  Ii 

6  e

As1  e  as1  As2 ( e  d )  Ap  e  ap  x  bs 

0

As1 as1  e  as1  As2 d ( e  d )  Ap ap  e  ap  bs  x  0.7677839m

The area of an ideal cross section with cracks A  bs x   e  As1  As2  Ap 

A  0.3889535m

2

The center of gravity of the concrete cross section of the upper extremity with cracks

ag 

The calculation of stress at the edges of the concrete cross section at the lower surface

 c1

6  e

x  Find( x)

 H  a    A  a  a 2  A  H  a  a 2  A  a  H  a 2 gi s1  s2  s2 gi p  p  e  s1  gi  gi   2 

 bs H

2

x  3 e x 

The moment of inertia of ideal cross-section

3 1

x  0 m

Given 3

Ii  bs H 

e  0.2 m

Np

The height of compression zone of cross-section

2

bs H0.5H   e As1 as1  As2  H  as2  Ap  H  ap  agi  Ai

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2

As1  8.02cm

 c1

 16.5076358MPa

 c1

 fctm

bs x0.5x   e As1 as1  As2  H  as2  Ap  H  ap 

The moment of inertia of ideal cross-section with cracks

3 1

I  bs x 

12

 fctm

 c2

 6.283225MPa

 fctm be assumed with weakened cross-section (compression with high eccentricity).

4

2

 x  a    A  a  a 2  A  H  a  a 2  A  a  H  a 2 s1 s2  s2 g p  p  e  s1  g g    2 g

 bs x

I  0.0251826m  c2

ag  0.3703358m

A

The greatest concrete compressive stress  c1



Np  ag  1  A  ag  e A  I 

Limitation of Stress

 c1

 14.2127021MPa

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Checking terms of reliability

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43

 c1

 14.2127021MPa

0.45fck

 15.75MPa

The result shows that the prestress should be brought in only at the moment when the concrete reaches 96% 28- daily strength. Limiting tensile stress reinforcement

EC2 prescribes the conditions in terms of usability limit stress in reinforcement and so as not to wide, permanently opining cracks. Achieving this limit state control conditions reliability for the largest tensile stress in exceptional combination of load: for concrete reinforcement s

 0.8fyk

fyk where the yield strength of concrete reinforcement e corresponding to a standard strength concrete reinforcement fyk according to EC- for prestress steel (tendon): p

 0.75fpk

where fpk is the characteristic strength of the tendon professional standards comparable tensile strength according to EC

Limitation of Stress


Since the tensile stress in concrete is below the modulus of rupture, the assumption that the section is uncracked is correct. Both concrete and steel stress fall within the intial linear ranges of the stress-strain relations and hence the assumption of the linear elastic behaviour is justified. As the applied moment is increased, the stress also increase and cracking of concrete on the tension side will begin as the maximum tensile stress reaches the tensile strength in flexure. Because the concrete is considered ineffective in tension, the effective depth of the member is taken as the distance from the extreme compression fibre to the centroid of the tensile reinforcement.

were measured by means of deformometers, while the lengths of the bases were 140mm for the upper and lower chords, 203mm for the diagonals and 140mm for the verticals. From the ratio of the geometrical dimensions L/h = 5.6 (where L is the span, h is the depth of the test beams) it is evident, that the influence of the shearing deformation on the total deflection will be significant. It was observed that the average strains in the pure bending zone follow a linear distribution along the depth of the cross section under short-term loading. That is, the plane sections remain plane under short-term loading. This observation serves a basis for the assumptions made in the analytical model described below.

The deflection model includes both bending and shear effects. Twelve specimens were precracked with a diagonal crack and tested under stationary loading for short-term. It was found that the contribution of shear effect to short-term deflection is significant. The behaviour of reinforced concrete structures under service conditions can be considerably different from linearly elastic behaviour due to the cracking of concrete which, even though it arises only in a limited number of cross sections, modifies the stiffness of the structural elements and therefore leads to non-negligible redistribution of the stress resultants. Reinforced concrete elements subjected to bending are characterized when the bending moment is higher than the first cracking moment, by the formation of cracks which lead to transferring the tensile loads to the reinforcing bars while the concrete comprised between two consecutive cracks is still reacting. This effect is usually referred

Figure 5-1: Distribution scheme of interconnected measurment basis for deformation and distribution of deflection - meters: a - measurment basis, b - deflection meters

to as tension stiffening. The evaluation of the deflection of reinforced concrete beams is strongly affected by the cracking of concrete. In order to develop a correct calculation of deflection, it is necessary to define an adequate tension-stiffening model and the exact prediction of the values of shear and bending stiffening. Theoretical and experimental analysis of the influence of the short term stationary load on deformation properties of reinforced concrete rectangular beams at different load levels is given. Deflections due to bending moment and shear force were calculated from deformations of the fictitious truss system, using a method based on Williot-Mohr translocation polygons. Strains of the upper and lower choards and the diagonals of the truss analogy system

Figure 5-2: Unit force acting on tested reinforced concrete beam For an uncraked section, the elasticity method has been commonly used to establish the curvature formula for calculation of deflection. In reality, most RC beams have cracks under

Deformation Behaviour of Reinforced Concrete Beams

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5. Deformation Behaviour of Reinforced Concrete Beams

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service loads. To calculate the curvature of cracked section in this study, it is assumed that (1)

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45 The first member of relationship represents deflection due to the effects of bending moments,

plane sections remain plane under short-term loading, (2) tensile strength of concrete can be

and the second member represents deflection due to the effects of shearing forces on the total

ignored.

deflection the effect of second member on the deflection is very small and can be ignored in

The loading procedure is shown in (Figure 5-1) where the relationships of the mid-span

most cases.

deflection am, av, atot versus the corresponding loading force F (Figure 5-1) The moment -

The deflections am, av, atot in the middle of the span of the beams for the selected load level

deflection relationship may be approximated by a bilinear relation which represents in some way

are given and were calculated on the basis of the deformations of the bases of the fictitious truss

the overall effect of the moment-curvature relationships described previously.

system. The deflection am corresponds to the effects of the rotation, deflection av to the effects

The primary aim is to find such coefficients that would describe the state in which the effect of load and the consequent development of cracks causes reduction of beam stiffness and an increase of deflection. These coefficients are supposed to facilitate an easy and simple determination of the influence of bending moments and shearing forces on deflection.

of shearing deformation, and atot are the total deflections. Morever we determined the relative value of the influence of shear forces (av/atot) and bending moments (am/atot) on the beam deflections due to the load. This influence was 17% for The series I, 21% for the series II, 18% for the series III and 23% for the series IV. This indicates

The virtual work principle may be applied to relate a system of forces in equilibrium to a system of compatible displacements. For example, if a structure in equilibrium is given a set

that the beams with higher amount of shear reinforcement show smaller growth of deflection due to shear strain in comparison to the beams with lower amount of shear reinforcement.

of small compatible displacement, then the work done by the external loads on these external displacements is equal to the work done by the internal forces on the internal deformation. In plastic analysis, internal deformations are assumed to be concentrated at plastic hinges. Deflection according to virtual work is usually calculated by means of the following expression . a

     M M  VV ds    ds   Ec Jc  G Ac  

  where M , V are the values of the inner force from the unit force in the cross-section in which the deflection is calculated. M Ec Ic

1

 el

I

is the curvature (the flexural stiffness EcIc being taken into calculation is that of the transformed section)



V G Ac

 el

is shear deformation, the ratio of the shear force to the value of shear stiffness GAc of the transformed cross-section

The shape coefficient  has a value of 1.2 for the rectangular section.

Figure 5-3: Deflection of beam due to shearing deformation and rotation The coefficient  depends on the shape of the cross-section. Its value in the linear state is 1,2 for the rectangular section.

Deformation Behaviour of Reinforced Concrete Beams


bh

h

bd

d

b

bs

s

b

h

b

hh h

d

hd h

s

Coefficient of the cross-section shape  hs h

Relative area of cross-section

a

Relative integrals for the calculation of the  coefficient d

 i1

 h  h   s  s   d  d

A

b h a

60

60

2



  s   d   s  2   d  d

1

h

h

h   h

d

J

1 3

2

d   d

3

bh j

zd

 h  zh

3

3

wd

h

 i4

  hh  h  s h  dd   d   s d  3

J zd

zh

3

hh

4

4

2

3

 wh

J zh

5

4

2

3

5

    6  5 h      8      2  5 h s h 

Relative Moment of Inertia j

60

2  a

d

3

60

a 2

j

 

 

 

   

     h  15   h   h  10   h   h  3   h   s       h   h   s   30   h 2   h 3  10   h 2   h 2   h 3      1

 i3

  h   s  h

2

    6  5 d      8      2  5 d s d 

and internal flange edges

h

5

    d  15   d   d  10   d   d  3   d    s       d   d   s   30   d 2   d 3  10   d 2   d 2   d 3      1

 i2

Relative position of central axis of cross-section regarded to ultimate cross-section fibres

2

5

 8   d  3   d  10   d   d  15   d   d

5

5

2

3

4

 8   h  3   h  10   h   h  15   h   h

 

   

  i1   i2   i3   i4

The curvatures

1

i

were determined by means of the deformation of the upper and lower

chord of the truss analogy system. Deflection due to the bending moment in the middle of the beams was calculated as well and indicated by the index "exp" as we departed from the measured values. n

amexp

i 1

1

i

 Mis

From the horizontal and vertical displacements of the truss analogy along the beams, shear strains were determined as well as deflections av,exp due to shear forces along the beams at all loading levels.

Figure: 5-4 Dimension of the cross-section for the evaluation of coefficient 

n

avexp

  iVis

i 1 where i are shear strains caused by loading force,

Deformation Behaviour of Reinforced Concrete Beams

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Relative dimensions of the cross-section with regard to the dimensions of h, b

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s - length of measurement base,

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47

n - number of measurement bases along the beam. Coefficients for the average bending and shear stiffness were derived from the given deflections amexp, avexp and amtheor, avtheor, at each loading level

 cr

amexp amtheor

avexp

 cr

avtheor

These coefficients express the state after crack development, while its values increase with an increasing loading level, i.e., cracks development. The advantage of these coefficients is that they permit the direct separation of the effect of shear forces on the deflections of reinforced concrete beams from the effect of bending moments. With these coefficients, the deflection due to bending moments and shear forces in state II will be:

a

Figure 5-5: Diagram bending moment versus curvature V

 GAc 

      M M V V   s  d     ds   cr cr Ec  Ic G Ac    

 pl

 GAc 

( - inclination)

The stiffness after cracking was determined from the diagrams "shear force - shear deformation" obtained from the test results. From the flexural and shear stiffness the coefficients of the degradation related to

For the calculation of the reduction of flexural rigidities also another approach was used

the load level were determined as follows:

based on the calculation of the flexural rigidities in linear state I and the flexural rigidities in state II after the development of cracks. The bending stiffness in state II was evaluated from the diagrams bending moment versus the curvature relationship. M EIc

1

 II

 pl

EIc

( - inclination)

The evaluation of the deformations of the bases of the fictitious truss system enables to

 

determinate the flexural stiffness EJc of the test beams in the whole range of the loading, using the diagrams "moment-curvature" as a basis. A similar method was applied for the calculation of the reduction of the shear stiffness, the shear stiffness of the transformed cross-section and the

shear stiffness in the whole range of the loading force GAc were applied.

Figure 5-6: Diagram shear force versus shear strain

Deformation Behaviour of Reinforced Concrete Beams


n

then

Ec Ic

 cr

 cri L

i 1

n

s

 crc

i 1

 cri L1 

s

BEAM IA

40 MPa 35 MPa

1

 crII´

1

 crII´c

 cr

 crII

 crc

 crII´   crII´c

30 MPa 25 MPa

2

20 MPa

M sd 8

2 b. d

16 MPa

6 4

level load

12 MPa

n

 GAc

 cr.i

then

G Ac

 cr

i 1

 cri L

n

s

 crc

i 1

 cri L1 

2

s

 crII´

1

 crII´c

 cr

 crII

 crc

1

atot(deflection)[mm] 6

1

5

 crII´   crII´c

4

2

1

0

1

2

3

5

6

2 2

SHAWKAT

level load

4

trus s m ethod

deflection will be presented in a new form. We can write:

atot(deflection)[mm]

0

SBETA

2

After the arrangement of the calculated deflections according to virtual work, the

gs=1

gs=1

6

1

fc k,c y l=25,25 MPa was calculated

2

fc k,c y l=22,17 MPa was obtained from fck ,cube

m

8

Diagram of m versus deflection atot .

a

Figure: 5-7

      V V M M d s    ds   Ec Ic  crII  G Ac  crII  

BEAM IC

40 MPa

By means of the coefficients  m ,  m , which express the increase of the relative average of curvature and shear strain of the reinforced concrete tests beams in state II it is possible to calculate the deflection after the full development of cracks.

35 MPa

30 MPa 25 MPa

M sd 8 2 b. d

20 MPa 16 MPa

6 4

level load

12 MPa

n

k

  si   ci Ec Ic hMiL

i 1

  si   ci Ec Ic

n

 ck

hMiL1

i 1

m

n

s

 k   ck 2

k

i 1

 iG Ac  ViL

s

2

s

 ck

i 1

m

 iG Ac  ViL1

s

Determination of deflections at the mid span of beams evaluated by the SBETA (Computer

5

4

2

1

0

1

2

3

5

6

2 2

SHAWKAT

4

trus s m ethod

1

fc k,c y l=19,57 MPa was calculated

2

fc k,c y l=21,50 MPa was obtained from fck ,cube

6 8

m

Diagram of m versus deflection atot .

Program for Nonlinear Finite Element Analysis of Reinforced Concrete Structures in Plane Stress State), by the fictitious truss analogy method and by the author´s

atot (deflection)[mm]

0

SBETA

 k   ck 2

1

atot(deflection)[mm] 6

n

gs=1

Figure: 5-8

own method of calculation. (Figure 5-7 - Figure 5-8).

Deformation Behaviour of Reinforced Concrete Beams

level load

gs=1

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 EJc

 cr.i

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49 5.1 Deformation Behaviour of Reinforcement Concrete Beams for I, T and rectangular

Simplified method for the calculation of short-term deflection

sections

0,50

The evaluation of the deflection of reinforced concrete beams is strongly affected by the

0,40

cracking of concrete. In order to develop a correct calculation of deflection, it is necessary to define an adequate tension-stiffening model and the exact prediction of the values of shear and

0,30

k s5

k s4

0,20

bending stiffening. The evaluation of experimental results of concrete beams with different reinforcement types subjected to stationary and moving load are given.

0,10 0,00 1,50

1,00

0,50

0,00

0,50

1,00

1,50

0,10 0,20 0,30

k s6

0,40 0,50

ks

-concrete, beams, reinforcement, cracking, stationary and moving loading, deflection, bending

12 MPa

Msd

2

a

2

b d fcd

rigidities, shear rigidities, curvature, shear strain, degradation of shear and bending rigidity.

16 MPa

L

ksi  kbid d

20 MPa

The behaviour of reinforced concrete structures under service conditions can be

25 MPa

a Deflection d Effective depth of a cross-section (m b Overall width of a cross-section (m) L Span of the beam (m) Msd Bending design moment (kN m)

30 MPa

considerably different from linearly elastic behaviour due to the cracking of concrete which,

35 MPa

even though it arises only in a limited number of cross sections, modifies the stiffness of the

40 MPa

Key Words

structural elements and therefore leads to non-negligible redistribution of the stress resultants.

Figure: 5-9

Reinforced concrete elements subjected to bending are characterized when the bending moment is higher than the first cracking moment, by the formation of cracks which lead to transferring

Simplified method for the calculation of short-term deflection

the tensile loads to the reinforcing bars while the concrete comprised between two consecutive

cracks is still reacting: this effect is usually referred to as tension stiffening.

0,50

The study of the influence of shear forces on total deflection of reinforced concrete

0,40

beams is dependent on applying a suitable evaluation method. Such method permits by the

0,30

0,20

k s2

Ms 2

b d fcd

means of average strain to determine separately the deflection due to bending moments and the

2

a

L ksi  kbid d k s1

deflection due to shear forces. These condition fulfill the truss analogy method (Hájek, Hanečka, Nurnbergová, 1985).

0,10

1,50

1,00

0,50

0,00 0,00

0,50

0,10

1,00

1,50 12 MPa 16 MPa

0,20

k s3

20 MPa 25 MPa

0,30

30 MPa 35 MPa

0,40

kb 0,50

Figure: 5-10

40 MPa

ks

The present experimental research quantifies the effect of short-term, gradually growing load and moving load on the process of deformation and failure of reinforced, partial prestressed and steel fibre beams with different section shapes: rectangular (Shawkat, Cesnak, Bolha, Bartók, 1994), (Shawkat,1993), (Shawkat, Križma, Cesnak, 1995) I-section (Hanečka, Križma, Ravinger, Shawkat, 1994), (Križma, Hanečka, Shawkat, 1993), (Križma, Shawkat, Bolha, 1995) and T-section [11]. An experimental study aimed at the deformation behaviour of reinforced concrete beams subjected to long-term load is being done at the Institute of Construction and Architecture of the Slovak Academy of Sciences in Bratislava. The results will be presented and discussed in a separate paper. The primary aim of this study was to analyse these problems:

Deformation Behaviour of Reinforced Concrete Beams


on reinforced concrete beams (further indicated as RCB), partially prestressed concrete

5.1.2 Loading and Instrumentation Details

All the elements were simply supported reinforced concrete beams. The rectangular

beams-PPCB, steel fibre reinforced concrete beams-SFRCB.

beams had a clear span of 1120 mm and were tested under two concentrated loads. The I-section

2) To find coefficients cr, cr which would most exactly describe the state after the full

beams RCB (5), PPCB and SFRCB with a clear span of 3600 mm were tested under a midspan

development of cracks which are characterized by a decrease in the stiffness and an increase in deflection. These coefficients are supposed to enable easy and simple determination of the influence of shearing forces and bending moments on deflection of

concentrated load. The beam RCB (6) T section had a clear span of 4000 mm and has also been tested under a midspan concentrated load. the beam SFRCB had a clear span of 3600 mm subjected to a moving load.

reinforced beams. 3) To obtain coefficients cr, cr which together with bending rigidities (E.J) and shear rigidities (/(G.A)) in the linear region permit the replacement of the bending and shear rigidities after the full development of cracks. 5.1.1 Specimen and Material Details

We have chosen eight from the total number of beams tested. The beams marked as RCB(i), i=1,.4 were of rectangular cross-sections, the beam RCB(5) of an I-section and the beam RCB(6) of a T-section. The partial prestressed beam PPCB as well as the beam SFRCB reinforced by steel fibres had an I-section: their description as well as the material and reinforcement characteristics are given, e.g. in (Shawkat, Križma, Bolha, Cesnak,), (Hanečka, Križma, Ravinger, Shawkat,), (Shawkat, Križma, Cesnak,), (Shawkat, Križma, Bolha, Cesnak,). The cross-section of the beams are in (fig. 5.1.1-1). Shear reinforcement was provided by 6 mm diameter steel stirrups located at 70 mm center to center for RCB (1) and RCB(2) rectangular beams, resp. 80 mm for RCB(3), RCB(4) and by 8 mm diameter at 80 mm center to center for all the other beams.

Figure 5.1.1-1: The cross-section of the tested beams

Figure 5.1.2-1: Side view of test beams with the network of bases for mechanical gauges (A B,C,D,E-location of the hydraulic arms)

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1) To determine the deflection am due to bending moments and deflection aq due to shear forces

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The moving load has been simulated through five loading jacks acting in the position A

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to E (fig. 5.1.2-1). The value of the service load was F=230 kN. With regard to the topic of this paper, the process of the crack development will not be discussed in more detail; only some interesting facts will be selected. The final state of the development of the cracks for all the beams RCB(i) is in fig. 5.3.2-1. The differences between PPCB and RCB (6), SFRCB can be explained as the influence of the prestress force Np,exp=5,85=425 kN (the prestress force at the beginning of the test). The differences arise from the number of slop cracks and the angle of the cracks. The ductile material in the webs of the SFRCB beam with the connection of stirrups greatly affected the development of the perpendicular cracks to the level of load =0,2. On the bases of the test results we can say that the moving load increases the number of slope cracks and crossing cracks. The crossing crack have arise in the SFRCB at the position D, load level0.5 The beams (I-section) loaded by stationary load have always been destroyed by the crushing of the upper compression part. With the beam loaded by the moving load, we have noticed even distortion caused by diagonal tension.

Figure 5.1.2-2: Side view of test rectangular reinforced concrete beams

Deformation Behaviour of Reinforced Concrete Beams


Strains of the upper and lower choards and the diagonals of the truss analogy system (fig. 5.1.3-1) were measured by means of deformometers, while the lengths of the bases were

hs

s

h

h

hh

d

h

hd

s

0.7143

A

0.1225

h

0.285

h

d

0

Relative area of cross-section

140 mm for the upper and lower choadrs, 203 mm for the diagonals and 140 mm for the verticals (beams RCB (1) to RCB (4)), resp. 360 mm for the upper and lower choards, 587 mm for the

0.58333

a

A

b  h  a

diagonals, 465 mm for the verticals (I-section beams RCB (5), SFRCB and PPCB). For the Tsection beam RCB (6), were these characters - 363 mm for the upper and lower choards, 484

Relative position of central axis of cross-section with regard to the ultimate cross-section fibers

mm diagonals and 320 mm verticals.

and internal flange edges.

Deflection according to virtual work is usually calculated by means of the following expression

a

    

   M M ds     E J  

 V V

G A

  h   s  h2   s    d   s   2  d  d  

h ds

2 a

h

where M and V are moments and shear forces from the united force acting in the middle of the

 h  h



V

1  h

d

0.60204

h

0.11224

d

 d  d

d

0.60204

j

3

0.05104

Then

is shear strain

G A

d

 h  h3    h   s   h3   d  d3    d   s   d3  

j

is curvature

E J

0.39796

Relative moment of inertia

beam, M

h

J

The first member of relationship (1) represents deflection due to the effects of bending

wh

3

J

b  h  j

J

wh

zh

0.00131

zh wd

0.00943

 h h

J

zh

0.13929

wd

zd

zd

 d h

zd

0.21071

0.00623

moments, and the second member on the total deflection is very small and can be ignored in most cases. The coefficient  depends on the shape of the cross-section. Its value in the linear

Relative integrals for the calculation of  coefficient

state is 1,2 for the rectangular section, 1,86 for the I-section and 1,46 for the T-section. The detailed calculation of the coefficient p for all section shapes is given in the example of the

1

T-section. Relative dimensions of the cross-section with regard to the dimensions of h, b (according to Fig. 1) bh d

0.6b s bd b

0.25h s s

0.25b d bs b

h

b s b bh b

0.6h h d

0.1h d 0.4167

0 h s

0.35 0.4167

h

2

d

60

  8  d  3  d  10  d   d  15  d   d

1 60  g

5

5

2

3

4

  d   15  d  10  d   d  3  d

2 4 2 3 5    2 3     d   d   s  30  d   d  10   d2   d2   d3  6  d5  2 5  8   d   s   d

1

Deformation Behaviour of Reinforced Concrete Beams

1

    

2

0

0.00439

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5.1.3 Methods

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STRUCTURAL ENGINEERING ROOM

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3

60  s

r

4

60

  h   15  h  10  h   h  3  h

2 4 2 3 5    2 3   h   h   s  30  h   h  10   h2   h2   h3  6  h5  2 5  8   h   s   h

  8  h  3  h  10  h   h  15  h   h 5

5

2

3

4

    

3

0.0015

a

4

0.00066

      M M V V ds      cr ds  cr EJ GA    

For the calculation of the reduction of flexural rigidities, we also used another approach based on the calculation of the flexural rigidities in linear state I and the flexural rigidities in state II after the development of cracks. The bending stiffness in state II was evaluated from the

Coefficient of the cross-section shape  a

After arrangement, the calculated deflections according to virtual work will be

4

  i 2 j i  1

diagrams bending moment versus the curvature relationship 1.46849

The curvatures i were determined by means of the deformation of the upper and lower choard

M

1

EJc

EJc

( - inclination)

of the truss analogy system. Deflection due to the bending moment in the middle of the beams was calculated as well and indicated by the index "exp" as we departed from the measured

For the calculation of the reduction of shear strain, we used a similar approach departing from

values

the calculation of the shear rigidities in state I and the shear rigidities in the state II after the n

am.exp

   1 M   i s   i1  i

development of cracks from the diagram shear force versus the shear stain relationship V GAc

GAc

(- inclination)

From the horizontal and vertical displacements of the truss analogy along the beams, shear

From the portion of flexural and shear rigidities in state I and state II at all load levels we found

strains were determined as well as deflections av,exp due to shear forces along the beams at all

out that the coefficients have the values given below

loading levels n

av.exp

i1

  i Vi s   

where i are shear strains caused by loading force,

s - length of measurement base,

cr

EJc

 cr

EJ

GAc GA 

After the arrangement of the calculated deflections according to virtual work, the deflection (6) will be present in a new form. We can write

n - number of measurement bases along the beam. a

Coefficients for the average bending and shear stiffness were derived from the given deflections am, exp, av, exp and am, theor, av, theor, at each loading level

cr

am.exp am.theor

 cr

av.exp av.theor

    

 M M

  ds     E J cr  

 V V G A   cr

ds

After that we discovered the average coefficient of the flexural and shear rigidities, which could be replaced by the coefficient of relative rigidities along the beam as follows  cr

1

n

E J l  

i1

( E J)  cri s 

  cr

1 G A 

Deformation Behaviour of Reinforced Concrete Beams

n

l i  1

 ( G A )    s    cr  i  


The percentage ratio of the effect of the bending moments and shear forces on total deflection and the relative values of the flexural cr and shear rigidities cr, average flexural   cr and

shear rigidities cr .

The largest percentage ratio of the influence of shear forces on the deflection was registered on the beam SFRCB which was under the effect of the moving load and its values were 24% at the service load and 26% at the ultimate load. The smallest values on the T section beam (RCB6) were 8% at the service load and 12% at the ultimate load, the bending slenderness being more than 10. To the contrary, the percentage ratio of the effect of the bending moments on deflection was the smallest on the beam SFRCB - 76% at the service load and 74% at the ultimate load and the largest on the T - section beam - 92% at the service load and 88% at the ultimate load. The maximum average value ofcr was determined for the reinforced concrete rectangular beams subjected to stationary load - 2.2 at the service load and 2.6 at the ultimate load. The same values were obtained on the I - section RCB beam subjected to the stationary load. The average value of cr for the beams PPCB and SFRCB was 2.05 at the ultimate load. The average value of cr for the reinforced concrete rectangular beams was 9.4 at the ultimate load, and for the I - section beam it was 9.9 at the ultimate load. The smallest value of

cr was 6.0 measured on the partially prestressed beam PPCB. The effect of the short - term, gradually growing and moving load on the deformation as well as the failure of the reinforced concrete beams, partially prestressed and steel - fibre concrete beams can be summarized 1) Between the pattern of cracks at stationary and moving loading, we can observe a large difference on the webs of beams at moving load; we can see cracks each other. In the case of a beam of the SFRCB series, the initial perpendicular cracking was Figure 5.1.3-1: Distribution scheme of interconnected measurement and distribution of

influenced by the toughness of the web material and by the influence of the shear

deflection – meters.

reinforcement. 2) Maximum curvature was observed in the region of concentrated loads and maximum shear

Deformation Behaviour of Reinforced Concrete Beams

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5.1.4 Discussion and Analysis of the Result

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in the middle of the shear span.

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55 5.2 Determination of Strain Energy on Reinforced Concrete Beams

The magnitude of the shear strain is highly dependent on the formation of inclined

3) The bending slenderness l/h has a great influence on the proportion of shear forces and the bending moments on the deflection, e.g. if the bending slenderness is larger than 10, the

cracks. If no shear cracks exist (state I in shear), the shear deformations are usually small and

influence of shear forces would be very small.

can in most cases be neglected. After the full development of inclined cracking (state II in

4) The average percentage ratio of the influence of shear forces on deflection was 22% for all beams except the T - section beam. This ratio is a function of the bending slenderness.

shear), shear deformations can be quite large, even larger than deformations due to flexion. The main aim of this study was to analyse these problems:

5) The maximum obtained value of cr was 10 for reinforced concrete beams and 6 for the

- To determine the strain energy using reinforced concrete beams and to find the portion of

partially prestressed beams.

strain energy due to shearing forces on the total internal strain energy or external energy.

6) The maximum value of cr for all the reinforced concrete beams was 2 - 3.

- To find a coefficient which would most exactly describe the state after the full development

7) Rigidity of beams can be defined by means of the diagram moment versus curvature relationship and shear force versus shear strain relationship at the inclination angle  and . 8) Beam rigidities are dependent on the level of initial forces M, V, N. 9) Bending rigidity (EJ)c and shear rigidity (GA)c are neither a multiplication of the modulus of elasticity E and the moment of inertia J of the gross concrete section, nor the shear modulus of concrete G and the cross-sectional area A.

of cracks which is characterized by a decrease of stiffness and an increase of deflection. This coefficient is supposed to facilitate an easy and simple determination of the influence of shearing forces on deflection. Due to the difficulties occurring when measuring and calculating the strain energy through a test on reinforced concrete elements this topic occurs only rarely in the literature. The development of the measuring and evaluating techniques also facilitates investigations aimed at the development of a theory to judge the reliability of concrete structures by the means of strain energy. It would be possible to characterize by the means of strain energy the state of construction, especially its component strains, more suitably than by values of strain or statical effects of loading. This suitability comes out of the fact that strain energy is a scalar value which also enables to express in a simple way also the reserve of construction in the given strain state in the relation to the limit state. 5.2.1 Methods

A certain part of the energy used for the development of cracks and irreversible strains is present in the reinforced concrete elements. Through the means of strain energy we could also express more generally the condition of reliability, e.g. in the form: P Uss  Ust  Prel

where P is the probability that the limit state will occur, Uss is the strain energy needed for reaching the given state of strain, Ust is the strain energy needed for reaching the limit state, Prel is the given probability characterizing the reliability.

Deformation Behaviour of Reinforced Concrete Beams


f j

the measured values is proposed in advance, which enables the determination of the strain energy as exactly as possible. For the determination of strain energy we performed tests on beams using stationary loading. We tested 12 reinforced concrete rectangular beams with a slenderness ratio of l/h=5.6

1

U

2

s

1

 Mj k 1  Mj k   rj 

k 2

where:

rj-1 is the increase of curvature

divided into four series I to IV. The cross-section of series I and III was 150/200 mm the shear

Mj,k is the bending moment of cross-section j at loading level k.

reinforcement 8, the longitudinal reinforcement 28 in the compression zone, and 216 in

Strain energy Uf, due to curvature at loading level s on the whole beam, was then calculated

tensile zone or 120/200 mm shear reinforcement 6. The longitudinal reinforcement as above,

according to this relation: n

for series II and IV. Shear stiffness was 1.55 for series I and IV, or 2.05 for series II and III.

U

Using a large number of measuring points, we could create suitable conditions for indirect evaluation of strain energy.

0.5 s 

f

 Uj 1  Uj

j 2

where:

s is the measuring base (the same along the whole beam length) n is the number of cross-sections, including the terminal one

We calculated the strain energy due to shearing forces in the same way. From shear strain we could determine the value of total energy U necessary for reaching a certain deflection in the middle of the beam span. The strain energy in the cross-section j at the loading level s is: 1

U

v j

2

s

 Vj k 1  Vj k  j

k 2

where: Figure 5.2.1-1: Cross-sections and side view of test beams

j is the increase of shear strain

At each loading level along the whole beam span, we measured deflections as well as length

Vj,k is the shear force in the cross-section j at the loading level k.

changes at the nine bases on the compression and tensile chords and on diagonals, using a

The strain energy Uv due to shear strain at the loading level n on the whole beam was then

system of interconnected measurement bases. At the same time we were investigating the

calculated according to the relation:

process of crack development. The measurements of length changes at each base enable the

n

U

v

determination of curvature and shear strain along the beam. By the means of curvature, we could evaluate the value of total energy U needed for reaching a certain deflection am due to a bending moment in the middle of the beam span.

section j at loading level s (marked as U j) will be:

 Uj 1  Uj

j 2

Then we could determine the value of strain energy due to internal forces from the sum of the strain energy due to curvature and shear strain according to the equation:

The value of the strain energy at the given loading level was determined numerically, first calculating the strain energy of the cross-section. For example, the strain energy in cross-

0.5 s 

U

U U f

v

We determined the internal strain energy on beams in the dependence on the load at each loading level. We calculated the amount of the effect of the strain energy due to shear strain on the total internal strain energy. This amount was within an interval from 16% to 24%. This means that

Deformation Behaviour of Reinforced Concrete Beams

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However, there are still not enough results for such an approach. It is suitable if the system of

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the average amount of energy due to shearing forces is 20% of the internal energy. We also

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 

determined the percentage ratio of the energy due to external forces.

  

For the energy due to shearing forces we took 100%, and this ratio was then in an interval from 94% to 106%. The evaluation of deflections of reinforced concrete beams is strongly affected by the cracking of concrete. Between state I and II in shear, a gradual decline of stiffness (as in bending) exists. In most practical cases where deformations are to be calculated for serviceability checks, shear deformations are largely overestimated using the shear strain of state II. The development of the shear strain is a function of applied shear force. Deflection according to virtual work is usually calculated as follows: a

    

   M M ds     EIc  

 V V ( GA)

2

Sz

2 Aj 

2

dA

A bz

where: G is shear rigidity A is the surface of the cross-section Coefficient  has a linear state value of 1.2 for a rectangular section. Our aim is to find such a coefficient that would describe the state in which is the effect of load and the consequent development of cracks causes reduction of beam stiffness and an increase

ds

of deflection.

where:   M and V are the moments and shear forces from the unite force acting in the middle of the beam.

V/(GA) is shear strain

As we have already mentioned, the deflections calculated along the beam at all loading levels enabled us to calculate the deflections due to the effect of shear forces as experimental deflection (acal).

M/(EI)c is curvature The first member of the relation represents deflection due to the effects of bending moments, and the second member represents deflection due to shear forces. The second member was determined according to the Simsons' law:

i 1

     V  ds 

We obtained from the acal and ath deflections at each loading level the coefficientcr. This coefficient expresses the state after crack development, while its value increase with an

x

 i 1 f ( x) dx  x

acal

h  0.33 y o  4 y 1  2 y 2  4 y 3    4 y   yn

n 1

where:

increased loading level, i.e., crack development. The values of coefficient cr in the depending on the loading level are drawn on the diagrams in Fig.2 for all reinforced concrete beams.

n is an odd number

The shape coefficient  from calculating the deformation due to shear forces V is as follows: ath

     

 V V

( GA)

ds

This equation represent deflection due to shear forces according to the strength of the material and we call it theoretical deflection (ath). Its influence on the total deflection is very small and can in most cases be ignored. Coefficient  depends on the shape of the cross-section:

From the diagrams it is obvious that the value of coefficient cr reaches a maximum value of 10 in most cases. The advantage of this coefficient is the fact that it permits the direct separation of the effect of shear forces on the deflection of reinforced concrete beams from the effect of bending moments. With tis coefficient, the deflection due to shear forces will be:    av      

 V V 1  cr

ds

 G A

However, to verify this statement it is necessary to continue with the experiments on this topic.

Deformation Behaviour of Reinforced Concrete Beams


The maximum width of a diagonal crack is about half way between the support and the load

5.3 Crack Development and The Strain Energy in Reinforced Concrete Beams

For calculating the strain energy caused by external forces we used the results of

point above the axis of the beam.

measurements in the bases of the multiple truss analogy system (Willot-Mohr translocation

The function of stirrups is to prevent the widening of cracks, the weaker the stirrups, the bigger

diagrams). The strain included also the crack widths. It is clear that the crack development

the shear cracks. The bigger the shear span, the wider the shear cracks.

directly influences the calculation of strain energy. We intent to devote our next research to the

Cracks open when the tensile stress in the concrete is larger than the tensile strength of the

direct relation between the crack development and strain energy, focusing on the stain. The

beam.

magnitude of the shear strain is highly dependent on the formation of inclined cracks. If no

The moment - deflection relationship may be approximated by a bilinear relation which

shear cracks exist (state I in shear), the shear deformations are usually small and can in most

represents in some way the overall effect of the moment-curvature relationships described

cases be neglected. After the full development of inclined cracking (state II in shear), shear

previously

deformations can be quite large, even larger than deformations due to flexion.

The mode of diagonal failure has been found by experiment to be primarily dependent upon the

Due to the difficulties occurring by measuring and calculating the strain energy by the test on the reinforce concrete element this topic occur in the literature only rarely.

shear span to depth ratio (a/h). The presence of shear reinforcement also appears to have an effect upon the mode of diagonal failure, it can be considered that the internal stress condition for diagonal failure is reached

To study the process of the formation and the development of bending (perpendicular)

when the shear force (applied load) obtains the critical value Vcr independent of (a/h).

cracks at the level of tension reinforcement as well as the shearing (incline) cracks at the level

According to our results the amount of the effect of the shear forces on deflections of reinforced

of tension reinforcement and at the middle of the beam depth.

concrete beams was 20% on the average.

To determine the strain energy using reinforced concrete beams and to find the portion of strain

The same value of 20% was determined by calculating the portion of strain energy due to shear

energy due to shearing forces on the total internal strain energy resp. external energy.

forces on total internal energy. We tested 12 reinforced concrete rectangular beams with the slenderness ratio l/h=5,6. The beams were divided into four series I to IV, the cross section of series I and III was 150/200 mm, resp. 120/200 mm for series I and IV, shear reinforcement 8, longitudinal reinforcement 2 V8 in compression zone, as above, for series II and IV. Shear stiffness was 1,55 for series I and IV, resp. 2,05 for series II and III.

Figure 5.2.2-1: The portion of strain energy due to shear forces on total internal energy

Figure 5.3-1: Cross-sections and side view of test beams

Deformation Behaviour of Reinforced Concrete Beams

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The results presented in this paper allow us to draw the following conclusions:

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5.3.1 Formation, Development and Width of Cracks

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The bending cracks first appear in the region of the maximum moment, then follows the development of cracks in the shearing forces region. These cracks slightly incline in the

On the basis of all diagrams we can state that immediately after the crack formation was the process of beam failure concentrated in to the regions influenced by the shear forces, especially their middle parts.

direction of principle stress compression trajectories and finally a shear crack (from a new or bending crack) is formed which prolongs to the compression region and to the tension beam end. In this cracks appears a failure which is called critical crack.

5.3.2 Evaluation of Cracks

Considering that the most calculation methods for widths of bending and shear cracks used the average crack width to determine the limit crack width, the average crack width at is

Character of shear crack formation in the zone of shear force acting can have two forms:

the most important evaluated parameter.

a) Cracks begin from the tension side of elements as a normal crack in bending region and

The measured values of average bending crack widths at the level of tension reinforcement in

they further develop in the direction of principle stress compression trajectories.

the dependence on the moment were compared to the average values of bending crack widths

b) Shear cracks appear solitary in the middle zone along the element depth.

calculated according to CSN 73 12 01- 86. The values calculated according to the CSN were

We observed the formation, development and widths of cracks at each loading level. by the

higher than the measured once while the values measured on beams in the series II and III were

means of optical device we plotted course of cracks and each crack was labeled with a number

higher than in the series I and IV.

as it was formed. A short perpendicular line indicated the development at the corresponding loading level. The crack widths were measured at three levels:

We also evaluated the dependence of the sum of bending crack widths at the level of

1. At the level of tension edge concrete fiber.

tension reinforcement on the moment, it is obvious that the maximum values were obtained on

2. At the level of the center of gravity of the tension reinforcement.

the beams in series II and III where the distance between the loads was smaller than in the series

3. At the middle of the beam depths.

I and IV.

Experimental beams of series I, II, III and IV were subjected to short term gradual

Further we compared the measured values of average widths of shear cracks at the level of

loading. We plotted the process of the development of bending and shear cracks on the diagrams

tension reinforcement on the left and right side of the beams in the dependence on shear force

both on the level of tension reinforcement and in the middle of the beam depth.

to these value calculated according to CSN 73 12 01 - 86.

Inclinations of shear cracks located closest to the supports were measured on the left

We also evaluated the dependence of the sum of bending crack widths at the level of tension

and right side of the beam. The length of inclination angle was given by the intersection of the

reinforcement on the moment, it is obvious that the maximum values were obtained on the

crack with the longitudinal reinforcement and the end of the crack by the crack failure. After

beams in series II and III where the distance between the loads was smaller than in the series I

the formation of shear cracks an increase of loading caused the growth of the crack widths. In

and IV.

the dependence on the percentual degree of shear reinforcement and the ratio a/h (where a is

Further we compared the measured values of average widths of shear cracks at the level of

the distance of load from the support and h is the section depth) the crack width can gain the

tension reinforcement on the left and right side of the beams in the dependence on shear force

large values, even higher than the limit values. Maximum values of width of the bending and

to these value calculated according to CSN 73 12 01 - 86.

shear cracks wm and wq determined for all beams series at each loading level. According to our investigations for all rectangular beams it is clear that the short-term gradually growing loading caused the formation and development of cracks along the whole beam span. We can also note that the maximum widths of shear cracks were in the dependence on loading growing more intensively than the widths of bending cracks.

Deformation Behaviour of Reinforced Concrete Beams


1

M

V

1

c

1

el

m

[kN]

1

1

pl

2

c

el

V

m

pl

1

3

Mcr

Vcr 3

2

1

1a

1 1 1

1 2a 2

E.I

1

1

1

3

(krivost )

  1 2a 2

1a

3

3

G.A 

3

2

1

2

(skosenie)

Figure 5.3.2-1: Diagram bending moment versus curvature and diagram shear force versus shear strain

Figure 5.3.2-2: Stresses of concrete versus measurment basis due to loading level

Figure: 5.3.2-1

Deformation Behaviour of Reinforced Concrete Beams

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M

[kNm]

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Further we investigated the dependence of the sum of shear crack widths at the

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level of tension reinforcement and in the middle of the beam depth on loading. The final process of crack development along the beam is shown in fig. 5.3.2-1. We summarized the measurements of the dependence of average bending crack widths at and their sum on the moment M at the level of tension reinforcement. We determined the average inclination of shear cracks for all beam series and on the basis of the these results we can state that a higher concrete quality in the compression zone caused an increase of the shear crack inclination. For example in series II on the right side of the beams was the average inclination 45 degrees, in series IV on the left side of the beams the average inclination was 38 and 40 degrees. We compared the moment values by the crack formation Mr calculated according to CSN 73 12 01 -86 to the measured data, the calculated values were smaller than the measured values.

Figure 5.3.2-2: The final process of cracks development along the beam

Deformation Behaviour of Reinforced Concrete Beams


Figure 5.3.2-3: Geometry of beams according to SBETA

Figure: 5.3.2-4

Deformation Behaviour of Reinforced Concrete Beams

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Example 5-1: We calculate the distance of the first an inclined shearing crack of reinforced

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concrete beam on the simple supported beam. The characteristics of the materials, the geometric dimensions of the element and the load are as follow:

Permanent load: Design load: gd

35.88  kN  m

 1

Variable load: gs

gd

1

29.9kNm

gs

1.2

Effective section height: d

h  ast

Shear force: Vd

0.5 m

Vd

Vd1

98.67kN

Qd  2.5

Vd1

3

Figure: 5.1-1

cross-sectional dimension: h

0.5 gd  L

0.3 m

b

Span of the beam: L

5.5 m

material characteristics: Concrete: fctm

1.2 MPa

Ec

32500 MPa

Steel: Es

n

210 GPa

Es Ec

n

6.461

Compression reinforcement at the upper of compression zone of the RCB: 4.03 cm

Asc

Tension reinforcement at the bottom of the tension zone of RCB: 10.08 cm

ast

3.0 cm

2

xi

1 b  h  2  n  Ast  d  Asc  asc  2 b  h  n  Ast  Asc

0.255m

xi

The moment of inertia of ideal cross-section:

2

Ast

Figure: 5.1-2

The depth of ideal cross-section:

2

Ii

 h2

bh

3

 x i  x i  h   n  Ast  d  x i

2  Asc  xi  asc2

Ii

4

0.00356m

Sectional area and the cross-section of an ideal n times the reinforcement area:

Cover: asc

2.5 cm

Ai

b  h  n  Ast  Asc

Ai

2

0.1591m

Deformation Behaviour of Reinforced Concrete Beams

82.225kN


Section modulus:  Ii     h  xi 

w1

w1

3

0.01458m

Shearing force to any service loads: V

g s

2

s

L

Vs

82.22kN

Moment at cracking: Mr

Rctm  w1

Mr

17.436kNm

Mr

Vs Srm  gs

Srm

2

2

Reaction from any service loads: A C

gs

2 30.61kNm

Vs

82.225kN

C

Mr

A

1

14.95kNm

B

Qs

Distance from the first crack of support:

Srm

 2  B  B  4 A  C    2 A  

Srm

0.40m

Deformation Behaviour of Reinforced Concrete Beams

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Example 5-2: To assess the crack width perpendicular to the centreline of the reinforced

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65 The moment of inertia of ideal cross‐section: 

concrete slab. Reinforced concrete slab is made of concrete and reinforcement as follows:

Ii

Concrete:

fcd

11.5 MPa

fctm

fyd

220 MPa

Es

Steel:   

2.45 MPa

Ec

27 GPa

3

210 GPa

M r

f ctm 

h

  n Ast  he  agi2  Asc  agi  asc 2

0.09 m

Ii

Mr

h  a gi

3.47kNm

satisfies

Ms  Mr

1 m

 agi agi  h

Ii

4

0.0000626m

Moment at cracking: 

Cross‐sectional dimension:  b

h 2

b h 

Ms

M slt

p

p

Mu

M slt  M sst

Ms

4kNm

0.416

Calculation of the compression area of the cross section:  xr

5 cm

As 2

1

xr 2  xr b  n 

As  j

j 1

2 b

A st

j

zs j

h  ast

zsm

zs j

1

 n

j 1

 A s  zs   j j

0

xr

0.0178m

Stresses in the reinforcement shall be limited in order to avoid inelastic strain, to avoid

Figure 5.2-1

unacceptable cracking or deformation.

Bending moments Mu  dw  st

6.0 kN m 1.2 for

M slt

2.5 kN m

M sst

Long-term stress in steel reinforcement:

1.5 kN m

bending reinforced concrete members

10 mm

ds

10 mm

dw

n st

ds

5

 zsm

0.01m

A st

n st   

ast  st

20 mm

asc

2

A st

4

2

0.00039m

A sc

The coefficient expresses the relationship of the covering layer: tb

6

ast

 st

 st

h b

2

0m

 1

  zs        j   b  xr   zsm    2 n  As   1   zsm  zs     2 j x j     r    j 1 xr 

1

n

0.00436

Calculation of the cross-section element values:

b  h  n  A st  A sc 2

agi

2

Ai

1 b  h  2 n  A st  h e  A sc  asc  2 b  h  n  A st  A sc

 104.23MPa

w3a

Es

n

Ec

7.77

slt 3   k   tb   0.035   st   E  d w s  

w3a

Area ideal cross-section: Ai

slt

Crack width perpendicular to the centreline of the elements:

Percentage reinforcement cross-section: A st

slt

xr

1.33

tb

h

2 n  M slt  

0 mm

0.09305m

he

agi

h  ast

he

0.07m

Figure 5.2-2

0.0458m

Deformation Behaviour of Reinforced Concrete Beams

0.0001048m


Assessment under Eccentricity

10 425 assessed according to limit state of crack the width of perpendicular crack.

b

350 mm

h

550 mm

ast

material characteristics: Concrete:

17 MPa

fctm

3.15 MPa

Ec

ds

32.5 GPa

asc

35 mm dw

375 MPa

Es

n sc

 sc

14 mm

As

0.00062m

Ai

210 GPa

agi

4

A sc

n sc   

 sc

4

2

A sc

As

A sc

1

Ii

The lower reinforcement n2

20 mm

n 3  

 st

n st

h 6

6

 st

0.01142

b h

tb  1

0.38182

tb

h

dw

b  h  n  A st  A sc

2

As

2

As

3

n 2  

2

4

n2  n3

1 b  h  2 n  A st  d  A sc  asc  2 b  h  n  A st  A sc

h 2

b h 

 agi agi  h 3

 st

4

As

2

2

0.00062832m

n3

5

Mr

Ii fctm  h  agi

0.21069m

agi

0.286m

d

  n Ast  d  agi2  Asc  agi  asc 2

4

0.00588m

70.27kNm

Ii

rt

rt

h  agi Ai

0.106m

2

0.0015708m

Ai

n st   

 st

2

4

A st

2

0.0022m

n

Es Eb

1 n

6.46

Nsst  Nslt

Long-term and short-term components of service values normal force and bending moment

170kN

efmax

Ai

fctm  1

rt

efmax

42.58kN

rt

The condition is not fulfilled, that is to say it is the eccentric

compression. Simplifying the calculation of stress s   

20 mm

h  ast

Ii

Mr

Nsst  Nslt  fctm  A st

2

Ai

Moment at cracking:  2

Nslt

satisfies

moment of inertia ideal cross‐section: 

2

3

efmax 

2

0.00062m

As

ast

tb

A st

 st

2

 st

0.09167m

6

Calculation of values of ideal cross‐section: 

The upper reinforcement

1

h

1.79m

33 mm

Steel: fyd

efmax

Nslt  Nsst

The coefficient the cover layer is expressed by the formula: 

fcd

M slt  M sst

efmax

cross-sectional dimension

120 kN

M slt

225 kN m

Nsst

50 kN

M sst

80 kN m

The long-term components of the load: eslt

M slt Nslt

h 2

 ast   st

eslt

2.095m

Deformation Behaviour of Reinforced Concrete Beams

A bp

2

0 mm

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Example 5-3: Reinforced concrete columns with concrete C30/25 and reinforcement with

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A bp  

from short-term components of the load from the difference between the overall and long-term

Es  A sc

effects:

Ec

b  h  ast   st

0.023

6.462

A st  Es

h  ast   st

4.23

A st  Es

0.90 s

where at is the distance of the center of gravity of the largest reinforced, imposed in extreme tension layer reinforcements drawn from the edge.

xu

h

2

0.01142

d

h fyd  st   d fcd

xu

0.133m

at

 at

es

2.014m

0.26913

d

ast   st

d  xu

Nslt  Nsst  es

A st  h  at

4.069

h  ast   st

 w

s

Ec

w

0.02297

0.89

 295.42MPa

crack width perpendicular to the centreline of element k

2000

w

3  s  slt d w   k     0.035   st     mm Es  

h  ast   st h  xu  ast



Es  A sc

b  h  ast   st

es

0.082

b  h  ast   st  Ec

Determine

 st

Nslt  Nsst

For ratio

eslt

0.08202

b  h  ast   st  Ec

w

M slt  M sst

es

the ratio

A bp 

1.055

w3blim

0.3 mm

w3alim

0.2 mm

w

0.00004637m 

w3b

w3a  w

w3b

0.00021m

satisfies

w3b  w3blim

The eccentrically compressed members the stress s,. Respectively s,lt allow us to determine More accurate solution:

the relationship:

slt



Nslt  eslt

A st  h  at

 w

slt

219.34MPa

Stresses in the reinforcement shall be limited in order to avoid inelastic strain, to avoid unacceptable cracking or deformation

ast1

4 cm

zs 2

0.475m

w3a

2000

w3alim

3  slt d w   k     0.035   st     mm Es  

w3a

0.2 mm

0.00016042m

w3a  w3alim satisfies

asc

zs 2

h  ast  ast1

zs 3

0.515m

zs 3

xr 3  2 agi  efmax xr2  xr  b4  n As 1  agi  efmax  zs 1   

crack width perpendicular to the centreline of the elements: k

zs 1

4 b

    A s 2  agi  efmax  zs 2  A s   agi  efmax  zs  3 3 

      A s   agi  efmax  zs   zs   2 2 2   A s   agi  efmax  zs   zs  3 3 3  

 n  A s   agi  efmax  zs   zs  1 1 1

Deformation Behaviour of Reinforced Concrete Beams

   



h  ast

zs 1

0.033m

0

xr

0.175m


4

3

xr 3  2 agi  efmax xr2  xr  b  n  

4 b

j 1

3

 n

j 1

w

A s j   agi  efmax  zs j     

w3b

A s j   agi  efmax  zs j   zs j    

w3a  w

w3b

0.0002094m

w3blim

0.3 mm

w3b  w3blim

good.

 zsm

zsm

0.0000539m

The reliability condition is satisfied, Conformity approximate and exact calculation is very

2 n  Nslt  

zs 3

w

0

long-term stress in reinforcement:

zsm

3  s  slt d w     mm k     0.035   st   E mm s  

slt

0.515m

xr

 1

    zs j     b  xr  2 n  As  1  j  xr    j 1  3

slt

212.53MPa

Crack width perpendicular to the centreline of the elements: k

2000

w3alim

3  slt d w   k       0.035   st     mm  mm Es  

w3a

w3a

0.2 mm

0.00015544m

w3a  w3alimsatisfies

Effects of short-term components of the load establishing from the difference between the overall and long-term effects:

Figure 5.3-1 In the calculation of stresses, cross sections should be assumed to be cracked, with no contribution from concrete in tension, if the maximum tensile stress in the concrete exceeds fctm under characteristic combination of actions.

efmax

s

M slt  M sst Nslt  Nsst

slt 

Nsst  Nslt Nslt

It should be noted that there are particular risks of large cracks occurring at sections where efmax

there are sudden changes of stress, e.g.

1.79m

- at changes of section - near concentrated loads

s

- sections where bars are curtailed

 301.08MPa

- areas of high bond stress, particularly at the ends of laps

Crack width perpendicular to the centreline of the elements: k

2000

w3alim

0.2 mm

Deformation Behaviour of Reinforced Concrete Beams

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or for expressing we compression areas xr write:

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Example 5-4: To calculate the stress in the reinforcement after cracking and crack width

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determination, reinforced concrete member of concrete C20 and reinforcement R 10 505 is positioned in the current environment of an enclosed building. Concrete column is loaded with axial force Nd and bending moment Md by its own weight.

Crack width perpendicular to the centreline of the elements: k dw

1600 16 mm

1.2

ds

dw

1.2 for tension members

Ms

ef

ef

Ns

0.03837m

Cross-sectional dimension: b

250 mm

h

Percentage reinforcement cross-section:

350 mm

 st

Material characteristics:

A st

b  0.5 h  ef

 st

0.01908

Concrete: fcd

11.5 MPa

fctm

2.45 MPa

Ec

27 GPa

The calculation of values of ideal cross-section:

Es

n

n

Eb

7.77778 Ai

Steel: fyd

As

1

450 MPa

Es

210 GPa

28 mm A sc

As

n sc

16 mm

 sc

As

2

1 b  h  2 n  A st  d  A sc  asc  2 b  h  n  A st  A sc

2

Ai

0.09854m

agi

0.18214m

2

A sc

n sc   

 sc

4

2

A sc

2

0.0004m

Ii

h 2

b h 

3

 agi agi  h

  n Ast  d  agi2  Asc  agi  asc 2

Ms

18 mm

n st

A st

As

0.00102m

4

A st

n st   

 st

4

2

A st

2

asc

zs 2

d

zsm

d

0.00102m

2

2

zs 1

Nsst

110 kN

We determine the eccentricity in tension element with small eccentricity as follow:

Ms

16.5kNm

M slt

16.5 kN m

Ns

Nsst  Nslt

Ns

430kN

 bg

10.5 ef  h 6 ef  h

Deformation Behaviour of Reinforced Concrete Beams

 bg

1.2976

4

0.00113m

Assessment of eccentricity:

 st

M slt

Ii

0.0004m

1

28 mm

320 kN

0.322m

2

Service load: Nslt

d

The moment of inertia of ideal cross-section:

The lower reinforcement:

ast

2

agi

The upper reinforcement:

asc

b  h  n  A st  A sc


ebal

j 1

A s  zs   zs  agi  j j j  ebal

2

j 1

tb

 A s  zs   j j

 etbal

A s  asc  A s  d 1

2

0.1403m

ef

 st

0.038m ef  etbal

w

Small eccentricity - the cross-section does not arise compression area. Calculation of stress  slt:    eflt

M slt

eflt

Nslt 2 1

S1

j 1

A s j   zsm  zs j    

j 1

S1

2

3

0.00011822m

S2

j 1

A s j   zs j  zs 1   

j 1

slt

h

tb

0.48

S3

0.00002276m

A s j   zs j  zs 1   agi  ef  zs j     

S4

0.00003037m

Nslt   zsm  zs  1  S4 S2  S1 S3

0.01908 3  d w  slt  k   0.035   st     mm  mm Es  

w

0.00006291m

w3a

  w 

w3a

0.00009m

w3b

0.00016609m

w3blim

wst

w

w3alim

s  slt slt 0.3 mm

wst w3b

0.00002163m  w3a  w 

0.4 mm

S2

3

0.00029926m

4

A s j   zsm  zs j    agi  ef  zs j     

2

S4

ast

0.05156m

2 1

S3

6

Minimum reinforcement areas:

A s  asc  agi  asc  A s  d  d  agi 1 2

etbal

or

0.1301m

4

Figure 5.4-1 If crack control is required, a minimum amount of bonded reinforcement is required to

slt

205.86MPa 

control cracking in areas where tension is expected. The amount may be estimated from equilibrium between the tensile force in concrete just before cracking and the tensile force in reinforcement at yielding or at a lower stress if necessary to limit the crack width.

Calculation of total stress in steel:

Cracking shall be limited to an extent that will not impair the proper functioning or durability

of the structure or cause its appearance to be unacceptable.

s

Nslt  Nsst   zsm  zs 1 S4

Appropriate limitations, taking into account of the proposed function and nature of the structure s

276.62MPa 

and the costs of limiting cracking, should be established.

S2  S1 S3

Deformation Behaviour of Reinforced Concrete Beams

STRUCTURAL ENGINEERING ROOM

The coefficient the cover layer is the core relationship:

2

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Example 5-5: Calculation of shear crack widths on a reinforced concrete beam according to

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71

CEB - FIP.

hef

Cross-sectional dimension:

bef

475  mm

Acef

420  mm

Nominal section width by height d

186  mm

bmin

hef

0.242m

0.186m

Effective area

Effective section height d

3.5  Ds  7.5  Ds

Effective width

Section depth h

Effective height

bef

bef  hef

0.04501m

Acef

2

Greater shear stress:

bmin

 ss

Asw

0.0023

 ss

bw  s  sin

Material characteristics:

Where Vn is shear force in the considered load combinations Concrete: Compressive strength of concrete determined the cubes of edge 200 mm

34.8  MPa

fcub

fck

0.85  fcub

fck

29.580  MPa

 Rd

0.337  MPa

Steel: Modulus of elasticity of reinforcement Es

210  GPa

k

1.2

for

Ast

s

Acef

0.0337

k1 = 0.4

90  deg

22  mm

Shear reinforcement bars distance measured parallel to the longitudinal axis of the element 120  mm

81.66kN

Vn

 Pmax

Determine the average distance cracks:

Mean shear reinforcement bars placed in circuit elements

UHsin

9

Diameter shear reinforcement bars: Ds (mm)

Shear reinforcement concrete cover

Ds

7

Coefficient for reinforcement with a smooth surface: k2 = 0.25

2

22  mm

Vn

Coefficient for periodic reinforcement profile:

1520  mm

Ast

105  kN

Percentage reinforcement cross-section: s

Sectional area of the tension reinforcement

c

Pmax

Srm

 

2 c 

UHsin  10

Ds

  k1  k2  s 

Srm

0.133m

UHsin shear reinforcement bars distance measured parallel to the longitudinal axis of the element Vcd

2.5   Rd  bmin  d

Vcd

65.81kN

Deformation Behaviour of Reinforced Concrete Beams


72

Vn  Vcd

1

88.21MPa

 ss

bmin  d   ss sin (  )  ( sin (  )  cos (  ) )

 40  MPa

 ss

 as

Determining the average elongation of shear reinforcement:

 as

0.4 

Es

Es   0.4 

2  Vcd     Vn  

 1  

 as

0.00070213

Wm

0.00009349m

0.4 

 ss

Es

0.00038

 ss

Es

The average crack width:

  Vcd  2  1    Es   Vn  

 ss

 ss

 ss

 as

0.00015

 as

0.00017

 as

 0.4 

Wm

 ss

we consider

Es

0.4 

 ss

Srm   as

Crack width at an angle to the longitudinal axis of the element is determined from the

Es

relationship:

The average crack width: Wm

 

S rm   0.4 

 ss

 

Wk

0.00002237m

Wm

Es 

Crack width of the longitudinal axis of the element is determined from the relationship:

Wk

1.7  Wm  k

Wk

0.00004564m

0.12  mm

Wmax

Wmax Wk

Wmax Wk

1.7  Wm  k

Wk

0.00019m

Wmax

0.786

2.63

Smaller shear stress:

 ss

0.0040

Vn

11  Pmax 9

Vn

128.33kN

Stress of shear reinforcement in cross-section with a crack:

 ss

Vn  Vcd

1  bmin  d   ss sin (  )  ( sin (  )  cos (  ) )

 ss

200.06MPa

 ss

 40  MPa

Figure 5.5-1

Deformation Behaviour of Reinforced Concrete Beams

0.15  mm

Wk  Wmax

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 ss

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Determining the average elongation of shear reinforcement:

Reinforcement stress in cracks cross-section (greater shear stress)


5.4 Methods

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Example 5.4-1: Calculation of deflections for rectangular reinforced concrete beam of

First Truss Analogy The study of the influence of shear forces on the total deflection of reinforced concrete beams depends on applying a suitable evaluation method. We found the truss analogy method as most suitable for our purpose.

30

40

10

Y0

Y10

X10

X1i

0

Xi

1i

i

Equation for the calculation of the Williot-Mohr diagram. On the upper chord Xi

X i 1   i

On the lower chord

Yi X1i Y1i

Y1 i 1   X1 i 1   Xi  cot     3i csc   

X1 i 1   1i

X1 i 1  Xi  cot     3i csc     Y1 i 1   X1i  Xi  cot     2i csc   

Wi

Y1n h s

Ui

m s

 Xi

W1i

U1i

atan 

0.809

cot   

m

11

n

6

i

0.952

( 1  n )

01

0.025 mm

02

0.133 mm

03

0.616 mm

04

0.841 mm

05

0.355 mm

06

0.017 mm

11

0.08 mm

12

0.509 mm

13

1.078 mm

14

0.784 mm

15

0.233 mm

16

0.003 mm

b) Lower chords

21

0.881 mm

22

0.870 mm

23

0.218 mm

24

0.06 mm

25

0.138 mm

26

0 mm

31

0.037 mm

32

0.019 mm

33

0.089 mm

34

0.680 mm

35

0.928 mm

36

0.187 mm

Edge conditions

X0 Yo 0 X10 X0 Xi X i 1   i

Y10 X1i

Y1i

Yi

 Xi1   X1i  Xi1  cot     2i csc   

Y0 X1 i 1   1i

X01

01

X00

0 mm

30

0 mm

40

0 mm

X10

0 mm

X11

11

Y00

0 mm

Y10

0 mm

Y1( i)  X1( i)  Xi  cot     3i csc   

Vertical and horizontal displacement on the upper chord

 Y1n    2 i  1  Y1   i  m 

Ui

Y n h s m s

 Xi

Vertical and horizontal displacement on the lower chord W1i

 Y n    2 i  1  Y1   i  m

U1i

1.380

system were measured by the means of deformometers as follows

d) Ascending diagonals

X1i

Second truss analogy Edge condition

Wi

csc   

The values of strains of the upper and lower chords and diagonals of the truss analogy

c) Descending diagonals

Vertical and horizontal displacement on the lower chord  Y1n      2 i  1  Y1i  m 

 hs    s 

s 140 mm h s 147 mm

a) Upper chords

Vertical and horizontal displacement on the upper chord  Y1n      2 i  1  Yi  m 

measurement bases by means of the Williot and Mohr translocation polygon. Length of the bases for the upper and lower chords Length of the bases for the verticals

Edge condition X0

uniform cross-section according to the method of measurements on a system of interconnected

X1i

Deformation Behaviour of Reinforced Concrete Beams

10

0 mm


Edge conditions X00

0 mm

Y00

0 mm

X10

X00

Y10

Y00

Equations for the calculation of the Willot-Mohr diagram on the upper and lower chord X0i

X0( i 1)  0i

Y0i

X1i

X1( i 1)  1i

Y1i

Figure 5.4.1-1: Side view of reinforced concrete beam

First Truss Analogy Equations for the calculation of the Willot-Mohr diagram on the upper and lower chord X0i

X0( i 1)  0i

X1i

Y0i

Y1( i 1)  X1( i 1)  X0i  cot     3i csc   

Y1( i)  X1( i)  X0i  cot     3i csc   

 X0i1  X1( i)  X0i1  cot     2i csc  

XOi [m]

X1i [m]

YOi [m]

Y1i [m]

0.000025 -0.000108 -0.000724 -0.001565 -0.00192 -0.001903

0.00008 0.000589 0.001667 0.002451 0.002684 0.002681

-0.001237857 -0.000986381 0.003633952 0.014168286 0.025090476 0.028674952

-0.001140429 -0.000639286 0.001281429 0.002382667 0.002672238 0.002461905

X1( i 1)  1i

 X1i1  X0i  cot     3i csc    Y1i1  X1( i)  X0i  cot     2i csc   

Y1i

XOi [m]

Y1n

YOi [m]

X1i [m]

Y1i [m]

0.000025

-7.4905E-05

0.00008

-0.001239143

-0.000108 -0.000724

-0.00108633 -0.00049638

0.000589 0.001667

-0.001623952 0.001479714

-0.001565 -0.00192 -0.001903

0.005496857 0.014848857 0.024051

0.002451 0.002684 0.002681

0.009404476 0.019424191 0.028416714

Y0n

W0i W1i

 Y1n 

   ( 2 i  1)  Y0i  m   Y1n   ( 2 i)  Y1   i  m 

U0i U1i

 Y1n h s     X0i  m s 

W0 n

0 mm

W1i

 Y0n   ( 2 i  1)  Y1   i  m 

U1i

X1i

W0i U0i

 Y0n   ( 2 i)  Y0   i  m 

Y0n m s

 h s  X0i

Vertical

Vertical

Horizontal

Horizontal

displacements

displacements

displacements

displacements

on the lower chord on the upper chord on the upper chord on the lower chord

X1i

Vertical Vertical Horizontal Horizontal displacements displacements displacements displacements on the lower chord on the upper chord on the upper chord on the lower chord W1i [m]

WOi [m]

UOi [m]

U1i [m]

0.005612052 0.010369771 0.011639013

0.002261359 0.007645697 0.011428654

0.002320777 0.002187777 0.001571777

0.00008 0.000589 0.001667

0.00808716 0.002440355 0.002186455

0.009808325 0.004829234 0

0.000730777 0.000375777 0.000392777

0.002451 0.002684 0.002681

W1i [m]

WOi [m]

UOi [m]

U1i [m]

0.003747242 0.008459727 0.011752641

0.006451485 0.011413636 0.012006931

0.002762155 0.002629155 0.002013155

0.00008 0.000589 0.001667

0.01586503

0.006686225

0.001172155

0.002451

0.020789087 0.026213048

0.000977662 0.002606814

0.000817155 0.000834155

0.002684 0.002681

Deformation Behaviour of Reinforced Concrete Beams

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Second Truss Analogy

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Example 5.4-2: Detailed calculation of the coefficient  for rectangular cross-section

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75 Relative integrals for the calculation of the  coefficient

Relative dimensions of the cross-section regarded to the dimensions of h, b bh

0.15 m

b

0.15 m

bs

hs

0.20 m

hd

0.0 m

h

h d s

bh b bd b hs

h

1

s

d

1

h

1

s

h

d

0.15 m 0.20 m bs b hh h hd

bs

hh

0.0 m

s

1

h

0

d

h

i1

bd

i2

i3

for the evaluation of coefficient  a A

 h h   s s   d d b  h  a

0.03 m

0.5

2  a

d

 d  d

0.5

h

1

1

4

4

4

2

3

 0  

5

   h  15  h   h  10  h   h  3  h 2

    h   h   s           8   h   s 2  h5 

3

3 30  h   h 5  6  h

5

5

2

2

 

0.5

3

4

 8  h  3  h  10  h   h  15  h   h

J zd wd

3

3

Bottom horizontal displacement node 3:

3

4

b  h  j

0.0001m 

 h  zh J zd

0.1 m 3

0.001 m

3

3

zh wh

 

3

0.08333

 h h

0.1 m

J zh

0.00417

2



0.00417

3

     

0

0.001 m

a

j

2

  i1   i2   i3   i4

Example 5.4-3: Evaluation of shear strain

Upper vertical displacement node 4:

   h  h   h   s   h   d  d   d   s   d

2

Bottom horizontal displacement node 4: 1

     

 10  h   h   h   

Top vertical displacement node 3:

Relative moment of inertia

j

5

Top horizontal displacement node 4:

0.5

 h  h

3

    d   d   s  30  d2  d3  10  d2   d2   d3      6  d5        8   d   s 2  d5  2

60  s

60

2

   d  15  d   d  10  d   d  3  d 2

60  s

h

5

Top horizontal displacement node 3:

  h   s    h 2   s    d   s    2   d   d 1  h

i4

5

Coefficient of the cross-section of shape 

and internal flange edges

d

 8  d  3  d  10  d   d  15  d   d

1

2

Relative position of the central axis of cross-section regarded to ultimate cross-section fibers

h

60

0

Figure 5.4.2-1: Dimension of the cross-section

Relative area of cross-section

d

Bottom Vertical displacement node 3: Bottom Vertical displacement node 4: Length of the bases: Depth of the fictitious truss system of beam:

3

u 3h

1.19710 

u 4h

1.10810 

w 3h

2.26510 

w 4h

3.25110 

u 3d

0.162 10

3

3

3

3

3

u 4d

0.49210 

w 3d

2.32510 

w 4d

3.39410 

L 34

140mm

h

3

3

147mm

Deformation Behaviour of Reinforced Concrete Beams

m

m m m m m m m

1.2


L34h

u4h  u3h

L34h

0.000089m

3

4

3-4

L34h

L34  L34h

L34h

0.13991m

4d-4h 3h-3d

L34d

u4d  u3d

L34d

0.00033m

Example 5.4-4: Evaluation of deflections due to shear forces and bending moments, reduction of the bending stiffness and shear stiffness by means of the measured strains at the fictitious

L34d

L34  L34d

L34d

0.14033m

truss system.

4d-3d 4

3 h3d3h

w3d  w3h

h4d4h

w4d  w4h

h3d3h

0.00006m

h4d4h

0.000143m

h3dh3h

h  h3d3h

h4dh4h

h  h4d4h

h3dh3h

0.14706m

h4dh4h

0.14714m

Shear strain of depth 3-3:  3v

u 3d  u 3h h 3dh3h

Shear strain of depth 4-4:

0.00704

 4v

Shear strain on upper length of the bases 43:  3h

w 4h  w 3h L 34h

Average shear strain:  1

 3v   3h

2

h 4dh4h

0.00419

Shear strain on lower length of the bases 4-3:  4d

0.00705

u 4d  u 4h

w 4d  w 3d L 34d

0.00762

Average shear strain:

0.00704

 2

Total average shear strain of section at midspan of beam:

 p

 4v   4d

0.0059

2

 3v   3h   4v   4d

4

0.00647

Figure 5.4.4-1: Side view of the tested beam Material data: Span of the beam Depth of the fictitious truss system of beam Length of the bases Modulus of elasticity of concrete Shear modulus Width of the cross-section

l

1.12m

hs

0.147m

s

0.14m

Ec G b

30.94GPa 0.435E c 15cm

Deformation Behaviour of Reinforced Concrete Beams

STRUCTURAL ENGINEERING ROOM

Determination of  :

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76


Depth of the cross-section

STRUCTURAL ENGINEERING ROOM

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77

Area of the cross-section Moment of inertia of the cross-section

Ic

Point load

1

Coefficient of the shape of the cross-section

Total span of the reinforced concrete beam

L1

Loading level

s

b h

F s1

0.07 m

M3

0.14 m

M4

0.21 m

M7

0.14 m

M8

0.07 m

M9

0.0 m

M5

0.28 m

   M  s i

3.374mm

i

The vertical displacements at each bases due to shear forces of fictitious truss system at the upper chord

1.2 1.29m

Average strain on the bases of the fictitious truss system

M2

0.21 m

i

51.52 kN

Fs

0.0 m

M6

a mexp

80kN

2

F s1

M1

The evaluation of the deflection due to the bending moment at the middle of the beam span

160kN

F s2 Fs

3

12

F s1

Service load:

at the upper chord

The bending moments along the beam due to unit force

h 20cm A c b h

1.55

Vh 1

0.000 mm

Vh 2

0.017 mm

Vh 3

0.152 mm

Vh 4

0.759 mm

Vh 5

0.853 mm

Vh 6

0.833 mm

Vh 7

0.689 mm

Vh 8

0.184 mm

Vh 9

 0.002 mm

Vh 10

0.00 mm

The vertical displacements at each bases due to shear forces of fictitious truss system at the lower chord

at the lower chord

Vd 11

0.000  mm

Vd 12

0.025  mm

Vd 13

0.678  mm

Vd 14

0.607  mm

Vd 17

0.798  mm

Vd 18

0.686  mm

h

0.057

d

0.1

Vd 15

0.845  mm

Vd 16

0.684  mm

h

0.4

d

3.214

Vd 19

 0.014  mm

Vd 20

0.00  mm

h

1.021

d

1.786

h

1.179

d

2.071

h

1.1

d

1.85

h

1.05

d

1.643

h

0.964

d

3.536

h

0.357

d

2.4

The evaluation of experimental deflections due to shearing forces at the middle of the span

h

0.121

d

0.314

of the fictitious truss system

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

Calculation of the average of shear strain at each basis of the fictitious truss system along the beam Vh

i

Calculation of the curvatures at each bases along the beam i 1 6

 0.00107m-1  2 0.01832m-1 7

0.02459m-1 0.03061m-1

3 8

0.0191m-1 0.01876m-1

4 9

 d  h i i

0.02211m-1  0.00296m-1

h 5

10 3 0.02007m-1

i 1

 Vh i

s

Vd

i 11

 Vd

i 10

s 2

1

0.00002857

4

0.00118571

7

0.00220357

2

0.00293571

5

0.00064643

8

0.00316429

3

0.00191429

6

0.00010714

9

0.00005714

Shearing force due to the unit force along the beam V1

0.50

V2

0.50

V3

0.50

V4

0.50

V6

0.50

V7

0.5

V8

0.50

V9

0.50

Deformation Behaviour of Reinforced Concrete Beams

V5

0


3

  1 V1  4   2 V2  2   3 V3  4   4 V4  2   5 V5  4   6 V6     2    V   4   V     V    7 7 8 8 9 9 

a mexp  a vexp a vexp

0.881mm

0.20713

moment at the middle of the beam span can be expressed by the means

The portion of the experimental and theoretical curvatures by the means of coefficient 

deflection

i

M1

M1

V1 1

80  kN

V1 5

0 kN

1 5

4

M1 M1

2 6

11.2 kN  m M1

M1 M1

5

V1 2

80  kN

V1 6

0 kN

22.4 kN  m

3

M1

7

3

V1 3

80  kN

V1 7

 V13

M1

24.8 kN  m

M1

M1

4 8

V1 4

0 kN

V1 8

 V12

M1

2

V1 9

a mtheor

i

The evaluation of the theoretical deflection at the

 V1

9

M1

The evaluation of the theoretical deflection at the

1

a vtheor a vtheor  a mtheor

6.791

3

2.637

4

2.758

7

4.228

8

5.181

9

0

  1   2   3   4   5   6   7   8   9  s l

3.298

  1   2   3   4   5   6   7   8   9   s

2.503

5

 k  c k

m

2.863

L1

1.140  mm

2

l  t1   m  M 5  2 5

3.08092

3.8722  mm



  G A c  V     s i

i

i

The portion of experimental shear strain to theoretical

 V1i   G A  c 

shear strain is expressed by the means of coefficient 

0.0998 mm

i

1

0.1442

2

14.81681

3

9.66157

4

0

6

0

7

11.12162

8

15.97043

9

0.2884

k ck

 1   2   3   4   5   6   7   8   9  s

5

0

5.3569

l 

m

 1   2   3   4   5   6   7   8   9 s

4.65095

L1 

1.240  mm

0.080

2

2.285

will be:

middle of the beam span due to shearing forces a vtheor  a mtheor

0

6

Deflection due to bending moments through coefficient  m

V1

i

E b J b

1

1

 M 1i   E c J c  M i  s  



i M1i

 ck

middle of the beam span due to bending moments a vtheor

i  t1

i

k

0 kN  m

2.959

a mtheor

i.e. 20,71 % is the effect of shearing force on the total

Bending moment and shearing force along the beam due to the concentrated load Fs2 M1

a mexp

 cr

of coefficient  cr

4.255  mm

a mexp  a vexp

The portion of the experimental and theoretical deflection due to bending

 k  c k 2

5.00392

The computation of shear stiffness according to the diagram "shearing force versus shear strain" at i.e. 8,0 % is the effect of shearing force on the

the level of loading in state II (after the full development of cracks) will be

total theoretical deflection G A i

The portion of the experimental and theoretical deflection due to shearing force at the middle of the beam span can be expressed by the means of coefficient  cr

 cr

a vexp a vtheor

8.827

V1i i

 GA k

V1 i

i

l

GA1

2800 MN

GA2

27.250 MN

GA3

41.791 MN

GA4

0 MN

GA6

0 MN

GA7

36.304 MN

GA8

25.282 MN

GA9

1400 MN

i

s

191.328  MN

GAc k

i

V1 i i

L1

s

Deformation Behaviour of Reinforced Concrete Beams

166.114  MN

GA5

0

STRUCTURAL ENGINEERING ROOM

s

a vexp

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78


Area of the cross-section after the full

STRUCTURAL ENGINEERING ROOM

Department of Architecture

79

GA k

A cr

EJ1

0.56863

G A c 0.83333

EJ5

development of cracks (state II)  GA1  GA2  GA3  GA4  GA5  GA6  GA7  GA8  GA9    G A  0.8333    s 0.56865

A cr

 G A  0.8333    i    GA1  GA2  GA3  GA4  GA5  GA6  GA7  GA8  GA9   s 0.24729 V1

l

V1

  V 1   i

GA el.k

 i   G A  c    s

i

252354.37  kN

l

GA k GA el.k

GA elc.k

 cr1

 crc  crIIc´

cr5

0

 el

i

2

833.59 kN m

 i   G A c     s

219098.37  kN

L1

2

0 kN m

i

i

2

723.74 kN m

 s

L1

i

  i   Ec  Jc     s

i

2

2707.25 kN m

EJelck

EJck

0.307

EJelk

EJ9

  M1

l

EJk

597.17 kN m

M1

i

EJck

EJ8

2

M1

  i   Ec  Jc     s

EJelk

i

i

G A c

cri

cr2

14.816

cr6

0

cr3

9.661

cr7

11.121

 cr1  cr2  cr3  cr4  cr5  cr6  cr7  cr8  cr9 s L 1  0.21262

 s

1121.72 kN m

EJelck

L1

2

2350.48 kN m

0.307

cracks) along the reinforced concrete beam is expressed as coefficient  cr

l 

1

2

EJ4

V 1 

 cr1  cr2  cr3  cr4  cr5  cr6  cr7  cr8  cr9 s

 crc

i

i

EJi E c J c

 crII´ 0.144

731.73 kN m

i

l

crk

cr1

EJ7

2

  M1

expressed as  cr

GAi

1353.73 kN m

1173.06 kN m

M1

i

The portion of shear stiffness in state I to shear stiffness in state II along the beam is

cri

EJ6

2

EJ3

i

0.75817

Shear strain in state I (without cracks)

G A c

2

The portion of the bending stiffness in state II (with cracks) to bending stiffness in state I (without

GAc k

0.75817

GA elc.k

455.56 kN m

M1

  V 1  

i

1235.79 kN m

EJk

2

EJ2 2

l

A cr

2

0 kN m

 crII

 crII´   crIIc´ 2

0.19861

cr4

0

cr8

15.970

0.288

cr9

 crII´ 5.41698

1  cr1

1

The calculation of the bending stiffness according to the "diagram bending moment versus curvature" at the level of loading in state II (with cracks)

0

cr2

0.14724

cr3

0.37914

cr4

0.36255

cr5

0.39942

cr6

0.43753

cr7

0.2365

cr8

0.19301

cr1  cr2  cr3  cr4  cr5  cr6  cr7  cr8  cr9 s l crk

 crII´

L1

 crIIc´

 crck

 crIIc´

 crII´   crIIc´ 2

0.26942

0.26942

cr1  cr2  cr3  cr4  cr5  cr6  cr7  cr8  cr9 s

 crck

 crII

0.23392

EJi

M1i i

EJ2

E c J c EJ3

E c J c EJ4

0

3.71162

1  crck

4.27499

 crII

5.03487

1 crk

1

0.25167

bending stiffness in state II along the beam as follows E c J c

cr9

0.23392

Further we can determine the portion of the bending stiffness in state I to

4.70312

 crII

0.1846

cr1

E c J c EJ5

E c J c EJ6

E c J c EJ7

l

Deformation Behaviour of Reinforced Concrete Beams

E c J c EJ8

 s

3.29825

E c J c EJi

3.97344


the full development of cracks can be expressed as follows

4

J1

0.00001472  m

J5

0.00004  m

4

4

J3

0.00004  m

4

J7

0.00002  m

J2

0.00001  m

J6

0.00004  m

J

M1

i

Ji

4

J4

0.00004  m

4

J8

0.00002  m

i  E c

4 4

4

J9

i

l

L1 Jc Jk

JbII

Jk

Ji

i

Jck

Jc

4

0.00003m 

 s

Jc

4

0.00002339m 

 s

Jck

Jc

Jck

1

3.99331

2

JbII

J2

J3

J4

J5

J6

Jc J7

0.25042

l

Jc J8

Jc  s

Due to shearing force

Total deflection

a vtheor

a vtheor  a mtheor

1

3.29825

J2

Jc J3

Beam Ia

Portion of shearing force on

total deflection

total deflection

0.25042

JbII

J el

i

1 12

 b h

3

Jc J4

Jc J5

Jc J6

Jc

Ji

J7

J8

L1

Average curvature of the beam at span l

1.240 mm

 100

8.05

 s

Due to shearing force

Total deflection

a mexp

a vexp

a mexp  a vexp

0.88 mm

3.29825

 cr

 cr

a vexp a vtheor

a mexp a mtheor

8.82

2.96

Average curvature of the beam at span L1 c k

2.8636

Relative average curvature of the beam

m

3.08092

Average shear strain of the beam at span l

Average shear strain of the beam at span L1

k

Jc

Due to bending moment

total deflection

a vtheor a vtheor  a mtheor

due to bending moment in state II:

2.8636

- experimental:

Portion of bending moment on

91.94

Coefficient  cr expresses the increase of deflection

1.5534

3.37 mm

 100

due to shearing force in state II:

Deflection at the mid span of the beam s

0.0998mm 

Portion of bending moment on

c k

4.65095

Relative average shear strain of the beam

m

5.00392

Average shear stiffness of beam at span

Average shear stiffness of beam at span

l in state II

L1 in state II

GA k

GAc k

k

Loading level:

1.1402mm 

Coefficient  cr expresses the increase of deflection

inertia in state II (with cracks) along the beam can be determined as follows Jc

20.71

4.274

The portion of the moment of inertia in state I (without cracks) to the moment of

Jc

 100

a mtheor

a mtheor

Jc

Jc

a mexp  a vexp

Due to bending moment

a mtheor  a vtheor

(the section is without cracks)

Jc

a vexp

79.28

- theoretical:

3.711

Moment of inertia of the cross-section of the beam in state I

Jc

 100

0 m

i

Jk

a mexp a mexp  a vexp

4.25 mm

5.3569

191328.56503kN 

166114.72313kN 

Portion of shearing force on

Average shear stiffness of beam at span

Average shear stiffness of beam at span

total deflection

l in state I

L1 in state I

Deformation Behaviour of Reinforced Concrete Beams

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The determination of the moment of inertia of the cross-section in state II after

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GAel k

252354.375kN 

GAelc k

219098.37209kN 

Relative average of the moment of inertia of

Relative average of the moment of inertia of

the beam at span l in state I, II respectively

the beam at span L1 in state I, II respectively

Relative average value of coefficient  crII´ ,

Relative average value of coefficient  crIIc´ ,

which expresses the reduction

which expresses the reduction

Jc

of the shear stiffness at span L1 of beam

Jk

of the shear stiffness at span l of beam

Jc

3.71162

Bending stiffness in state I E Jel

i

 crII´

0.1846

 crIIc´

 crII

0.19861

Average bending stiffness of the beam

at span l in state II

at span L1 in state II

Coefficient, which expresses the effect of the increase of initial deformation

EJc k

at span l in state I

at span L1 in state I EJ elck

2

2350.48062kN  m

the beam at span l in state II,I respectively

the beam at span L1 in state II,I respectively EJc k

0.30791

EJ elck

0.44043

 ck

0.30791

Relative average value of coefficient  crII´ ,

Relative average value of coefficient  crIIc´ ,

which expresses the reduction of the bending

which expresses the reduction of the bending

stiffness at span l of beam

stiffness at span L of beam

Reduction of the area of the section of a beam in state II

Uf 1

0 kN m

Uf 2

0.03855kN  m

Uf 3

0.05988kN  m

Uf 4

0.07676kN  m

Uf 5

0.06968kN  m

Uf 6

0.06361kN  m

Uf 7

0.096 kN m

Uf 8

0.02941kN  m

Ufpl

Ufel

i  M 1  s

i

i

Uf 9

0.26942

 crIIc´

  t1 M1  s i

i

i

140.249kN   mm

Uf

Ufpl Ufel

3.093

The evaluation of the strain energy due to shear force along the beam at the

0.23392  crII

0.25167

0 kN m

  mm   1 M1  2 M1  3 M1  4 M1  5 M1     s 433.88419kN  1 2 3 4 5     6 M1  7 M1  8 M1  9 M1  6 7 8 9  

Uv  crII´

 ck A c

Acr

the level of loading by means of diagram moment versus curvature

Relative average of the bending stiffness of

EJ k

 J c

The evaluation of the strain energy due to bending moment along the beam at U f

Relative average of the bending stiffness of

EJ elk

J cr

0.25042

723.744kN  m

Average bending stiffness of the beam

2707.25kNm 

0.85

1

2

Average bending stiffness of the beam

EJ elk

i

G A c

Coefficients, which expresses the reduction of the moment of inertia of the section in state II 

2

V 1 

 el

at loading

Average bending stiffness of the beam

833.598kN  m

Shear strain in state I

E c J c

0.21262

of shear stiffness of the beam in state II

EJ k

4.27499

i

Coefficient  crII represent average value and expresses the reduction

2

Jc k

i

 i  V 1  s i

level of loading by means of diagram shear force versus shear strain

Coefficient  crII represent average value and expresses the reduction

Uv

of bending stiffness of the beam in state II

Uv

1 6

0.00032 kN m

Uv

0 kN m

Uv

2 7

0.03288 kN m

Uv

0.02468 kN m

Uv

3 8

Deformation Behaviour of Reinforced Concrete Beams

0.02144 kN m

Uv

0.03544 kN m

Uv

4 9

0 kN m

Uv

 0.00064 kN m

5

0 kN m


1. By means of curvature i and shear strain  i along the beam in state II U vpl

s 3

    1 V 1  4   2 V 1

  2   3 V 1   4   4 V 1   2   5 V 1   1 2 3 4 5        4   6 V 1   2   7 V 1   4   8 V 1     9 V 1  6 7 8 9     

Strain energy in state I

Uv

U vpl U vel

7.619

U vel

 i

U el

 

  

121.733kN   mm

  V 1i   V     s 15.977 kN mm  G A 1i 

U vel  U vel

U fpl  U vpl

i

156.227kN   mm

i

2. Through the coefficients  cr and  cr , which express the increase of the deflection due to bending moment and shear force of a reinforced concrete beam in state II  M 1 i

  E c  J c  M i

So the portion is 21,91 %

U pl

0.7809

 ´ m

M 1 i

So the portion is 78,09 % The portion of internal energy due to

bending moment in state II to state I

Shear force in state II to state I

U fel

3.093



4.25  mm

E cJ c

 

 i

V1 i G A c

 Vi    s

 

4.28  mm

curvature and shear strain of the reinforced concrete beam in state II

a4

The portion of internal energy due to U fpl

i

i

 M i  s   ´ f     



 M 1i 

´ f

The portion of total strain energy in state II to state I

U vpl U vel

U





 V 1 i 



  k  E c  J c   M    s    k  G A c   V     s i

i

´ m

 V 1 i

  G A c  V      cr   s

4. By means of the coefficients  k ,  k , which express the increase of the relative average of

The portion of internal energy due to bending moment on the total internal energy U fpl

    cr   s  

bending moment and shear force in state II

i

0.2191

i

3. Through the coefficients ´m , ´f , which express the increase of the strain energy due to

555.61752kN   mm

a 3 U pl

4.17  mm

i

i

The portion of internal energy due to shear force on the total internal energy U vpl

i

i

a2

Total strain energy is the sum of energy due to bending moment and shearing force, respectively U pl

   M   s     V   s

a 1

i

4.29  mm

i

5. Through the coefficients  crII´ ,  crII´ , which express the reduction of the relative average of bending stiffness and shear stiffness of reinforced concrete beam in state II

7.619

U pl U el

3.556

a5

 i

M 1   i   M i s   E c  J c   crII´   

 i

V1   i   V  s  G A c   crII´ i   

Summary of The Evaluation of The Deflection at The Mid Span of The Beam in State II

Deformation Behaviour of Reinforced Concrete Beams

4.77  mm

STRUCTURAL ENGINEERING ROOM

Strain energy in state II

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6. By means of the coefficients  , which express the reduction of the moment of inertia of the

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section in state II, and the coefficient  1 , which express the effect of the increase of the initial deformation at loading

a 6

i

M 1   i   M i s   1 E c  J c     

Summary of the Evaluation of the Strain Energy Due to Bending Moment and Shearing Force at the Mid Span of the Beam in State II.

1.By means of curvature i and shear strain  i along the beam in state II 5.35  mm

U 1

    M 1    s      V 1    s i

7. By means of the coefficient  , which expresses the reduction of the moment of inertia of the section in state I

a 7

 i

M 1 i

  M  E cJ c 

i

   s  

i

i

i

548.00  kN  m

i

i

2. Through the coefficients  cr and  cr , which express the increase of the deflection due to bending moment and shear force of a reinforced concrete beam in state II

U2

 M 1i

 V 1i





  E c  J c  M 1    cr   s    G A c  V 1      cr   s i

i

i

i

556.06  kN  mm

3. Through the coefficients ´m , ´f , which express the increase of the strain energy due to

4.55  mm

bending moment and shear force in state II

8. By means of the coefficients  ,  ck , which express the reduction of the moment of inertia and the reduction of the area of the cross-section of the beam in state II

U3



 M 1i 





 V 1i 



  ´ m  E c J c   M 1   s    ´ f   G A c   V 1    s i

i

i

i

555.61 kN  mm

4. By means of the coefficients  k ,  k , which express the increase of the relative average of curvature and shear strain of the reinforced concrete beam in state II

a8

 i

M 1 i

    M s   E cJ c i   

 i

V1

  i   V  s  G  ck  A c i   

4.78  mm

U4

 i

  M 1i   k   E c J c  M   



1i  s 

 i

  V 1i    k  G A c   V 1i    s    

584.16  kN  mm

5. Through the coefficients  crII´ ,  crII´ , which express the reduction of the relative average of bending stiffness and shear stiffness of reinforced concrete beam in state II U5

 i

M 1   i   M 1 s   E c  J c   crII´ i  

 i

V1   i   V  s  G A c   crII´ 1i   

607.10  kN  mm

6. By means of the coefficients  , which express the reduction of the moment of inertia of the section in state II, and the coefficient  1 , which express the effect of the increase of the initial deformation. U6

 

M 1 i

 

   1 E c  J c    M 1    s i

i

658.89  kN  mm

Deformation Behaviour of Reinforced Concrete Beams


section in state II U7

 

Geometrical dimensions and characteristics of the materials reinforced concrete beam:

M 1 i

 

  E c  J c    M 1    s

Beam span:

560.06  kN  mm

i

i

L  3.6  m

8. By means of the coefficients  ,  ck , which express the reduction of the moment of inertia and the reduction of the area of the cross-section of the beam in state II U8

 

M1 i

 

 

V1

 

  E c J c   M 1   s    G  ck A c  V 1     s i

Example 5.4-5: Theorem of reciprocity of virtual work (Betti sentence)

i

i

i

i

596.33 kN mm

L1  4.10  m

The depth of fictitious truss system: h  0.46365  m

Length of base:  s 

0.360  m

Modulus of elasticity of concrete: Ec  40.69  GPa

Shear Modulus: G  0.42  Ec

Cross-sectional area: A  736  cm

2

Sectional moment of inertia: J c  0.002311258  m

4

Figure 5.4.4-2: Diagram bending moment versus curvature Load action on RC beam Fs2

280  kN

 

1.16

The coefficient of cross-sectional shape:  

1.85

The load action Fs2  F

Figure 5.4.4-3: Diagram shear force versus shear strain

J c  0.00231 m

Fs2  280 kN

4

Average strain the base of the lower edge of the fictitious truss system (per thousand) d

1

 0.150

d

2

 0.381

d

3

 0.942

d

4

 0.817

Deformation Behaviour of Reinforced Concrete Beams

d

5

 1.378

STRUCTURAL ENGINEERING ROOM

7. By means of the coefficient  , which expresses the reduction of the moment of inertia of the

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84


d

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85

d

6

 1.158

11

d

7

 1.350

d 

8

1.883

d

9

 1.506

d

10

 0.814

 

 0.244 

Average Strain in the base of the upper edge of a fictitious truss system (per thousand)

h h h

1 6

 0.189

h

 0.961

h

11

2 7

 0.028

h

 0.950

h

3 8

 0.167

h

 1.328

h

4 9

 0.600

h

 0.456

h

5

 0.669

 0.039

10

d

7

 0.086

10

1 4 7



7

 h

7

h d

10

 10

 h

10

h

 0.00008 m

1

 0.00496 m

1

 0.00167 m

d

 

9

3

 10

1

 0.00306 m

10

3

11

 h

9

h



d

 10

 h

11

11

h

2 5

1

9

3

 

 h h

8

 10

3

3

 0.00441 m

1

1

 0.00693 m

11

8

 10

 0.00076 m

8

d

8

1

 0.00034 m

3

6 9

 0.00239 m

1

 0.00457 m

1

 0.00423 m

1

1

Bending moments along the beam from the effect unit force at point B  M  0.3  m 2

M  0  m 1

M  0.5  m 6

M  0.4  m 7

M  0.5  m 3 M  0.3  m 8

M  0.8  m 4

M  0.2  m 9

M  0.6  m 5

M  0.1  m 10

M  0  m 11

Calculation of the effects of experimental deflection due to bending moments at the center of the beam ymexp 

 

 i Mi  s 

ymexp  0.005 m

i

Calculation of the skew on individual bases along the beam Vertical offsets each basic truss system at the top edge

Figure 5.4.5-1 Curvature calculation between individual bases along the beam  

d

3

 

4

3

 h

3

h d

4

 h h

4

 10

3

3

 10

 

d

1

 

6

1

 h

1

h d

6

 h h

6

 10

3

3

 10

 

2

 

5

d

2

 h

2

h d 5  h

h

5

 10

3

3

 10

Vh1  0.000 mm

Vh2  0.040 mm

Vh3  0.484 mm

Vh4  1.109 mm

Vh5  1.903 mm

Vh6  2.242 mm

Vh7  2.797 mm

Vh8  2.761 mm

Vh9  2.119 mm

Vh10  0.685 mm

Vh11  0.062 mm Vh12  0.000 mm

Vertical offsets each basic truss system at lower edge of the beam

Vd11  0.000 mm

Vd12  0.265 mm

Vd13  0.943 mm

Vd14  1.581 mm

Vd15  1.976 mm

Vd16  2.525 mm

Vd17  2.620 mm

Vd18  2.851 mm

Deformation Behaviour of Reinforced Concrete Beams


Vd20  1.634 mm

Vd21  0.340 mm

Vd22  0.0 mm

The calculation of the average individual basic skew fictitious truss system of longitudinal members Vh2  Vh1 1



 s



 s



 s



 s





 s

 s

3

6

 s



Vh9  Vh8  s



Vd14  Vd13

 s

Vd17  Vd16

 10

 s

 s



Vd21  Vd20  s

2

Vd22  Vd21  s

2

1

 0.00031

2

 0.00167

3

 0.00175

4

 0.00165

5

 0.00123

6

 0.0009

7

 0.00027

8

 0.00153

9

 0.00304

 10

 0.00266

Q2  0.7

Q3  0.7

Q4  0.4

Q5  0.3

Q7  0.3

Q8  0.3

Q9  0.3

Q10  0.3

Q11  0.3

Q6  0.3

yqexp    1Q1   2Q2   3Q3   4Q4   5Q5   6Q6   7Q7    s    Q   Q   Q   Q   8 8 9 9 10 10 11 11  yqexp  0.00176 m

yqexp

Vd19  Vd18

Q1  0.7

ymexp  yqexp  0.00676 m

Effect of shear forces on the deflection is 20.06%

 s

Vh11  Vh10

 s

Shear forces from the force unit along the beam at point B

 0.26062

ymexp  yqexp

1

yqexp ymexp  yqexp

 0.73938

Bending moments and shear forces along the beam from that of FS2

2

Vd20  Vd19

 s

2

 s

8

2

Vd18  Vd17

2

 s



Vh7  Vh6

 s

Vh12  Vh11  11

Vh4  Vh3

beam fictitious truss system

Vd13  Vd12 2

Vd15  Vd14

2

 s

 s



Vd16  Vd15

Vh10  Vh9 9

2

2 Vh8  Vh7

7

 s

2 Vh6  Vh5

5

Vh3  Vh2

2 Vh5  Vh4

4

Vd12  Vd11

Experimental calculation of deflection from the effect of shear forces in the center of the

 11

 0.00056

M11  0 kN m

M12  30.2 kN m

M13  60.5 kN m

M14  90 kN m

M15  121 kN m

M16  151.2 kN m

M17  181.4 kN m

M18  211.7 kN m

M19  141.1 kN m

M110  70.6 kN m

M111  0 kN m

Q11  84 kN

Q12  84 kN

Q13  84 kN

Q14  84 kN

Q15  84 kN

Q16  84 kN

Q17  84 kN

Q18  ( 196  84) kN

Q19  196 kN

Q110  196 kN

Q111  196 kN

Calculation of the theoretical deflection of the beam in the middle of the action of bending moments ymteor 

 M1i

  E J Mi  i

c c

 s

ymteor  0.00165 m

Deformation Behaviour of Reinforced Concrete Beams

STRUCTURAL ENGINEERING ROOM

Vd19  2.392 mm

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86


Calculation of the theoretical deflection along the beam from the effect of shear forces

STRUCTURAL ENGINEERING ROOM

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87

yqteor 

 Q1i

  GA Qi 

i

    s  

yqteor  0.00018 m

yqteor  ymteor  0.00183 m

yqteor  ymteor

yqteor  ymteor

 2.84266

1

 2.5718

8

 11

using the coefficient cr(Load level is 1.16)

yqteor

 cr

ymexp ymteor

 

1

0

M1i Ec Jc

2

2

2 t1 2

 2.3709

 10

9

t1 9

 2.82045

 2.2266

GA

 9

Q111

GA 5

GA

9

 

GA

GA

 10

Q19

4

Q13

GA

Q14 GA

2

 

GA

Q12 GA

 10

 

GA

7

Q12

4

 

2



6

Q15

3



3

Q12

5

Q18

2



Q110 GA

 2.52938

2

 13.5125

3

 14.19824

4

 13.36636

0

6

 13.5125

7

 13.5125

8

 9.28282

9

 10.56076

 10

GA1 

 3.02936



 3.07656

9

 9.23584

 11

 1.93678

 11

 1.93678

the level of loading (condition II) of cracked:

The experimentally determined ratio of curvature to the theoretical expressed as:



t1 8



5

GA11 

t1 i

t1 10

9

1

GA6   cr

2

Q11

 11

expressed using the coefficient cr (Load level is 1.16) 

 10

8



Calculation of the shear stiffness on the basis of a working diagram of shear force - skew at

 9.6577

The calculation of the reduction of bending stiffness in the center of the beam can be

 cr

 0.90044

The calculation of the reduction of shear stiffness in the center of the beam can be expressed

yqexp

 

beam.



t1 7



 10

8

The ratio of experimental and theoretical deflection effect of shear forces in the center of the

 cr

8

8

1

 

effect of bending moments on the total deflection 1

6

t1 6

7



7

The ratio of the experimentally determined skew to the theoretical expressed using the :

 0.09956

yqteor



7

effect of shear forces on the total deflection yqteor

6

6

3

3



3 t1 3

 3.71811

4

4



4 t1 4

5



5 t1 5

 3.19355  5  3.43146

Q11 1

Q16 6

GA2 

GA7 

Q12 2

Q17 7

GA3 

GA8 

Q13 3

Q18 8

GA4 

GA9 

Q14 4

Q19 9

GA5 

GA10 

Q15 5

Q110  10

Q111  11

GA1  268800 kN

GA2  50316.13977 kN

GA3  47885.98575 kN

GA4  50866.27418 kN

GA5  68108.10811 kN

GA6  93046.15385 kN

GA9  64379.56204 kN

GA7  310153.84615 kN

GA8  73242.50681 kN

GA10  73615.02347 kN

GA11  351044.77612 kN

Deformation Behaviour of Reinforced Concrete Beams


M110  10

along cr :

M111

EJ11 

 11

2

cr1



cr5



cr9



GA1 GA GA5 GA GA9 G A



cr2





cr6



cr10



GA2 GA GA6 GA





cr3





cr7



GA10 G A



cr11

GA3 GA GA7 GA





cr4





cr8



GA11 G A



cr1

GA4 GA GA8 GA

2

EJ1  0 m kN



EJ2  39666.37394 m kN 2

2

EJ4  29448.48271 m kN EJ5  27406.766 m kN 2

2

EJ7  36567.87391 m kN



EJ8  30568.2669 m kN

2

2

EJ3  25293.80072 m kN 2

EJ6  33083.47334 m kN 2

EJ9  33344.04434 m kN

2

EJ10  42237.01935 m kN EJ11  0 m kN

 0.39535

The bending stiffness of elasticity (as I) without cracks:   

cr2

 0.07401

cr3

 0.07043

cr4

 0.07481

cr5

 0.10017

cr6

 0.13685

cr7

 0.45618

cr8

 0.10773

cr9

 0.09469

cr10

 0.10827



 cr

 0.18745

Coefficient

 cr

1  cr

expresses reduce shear stiffness for a given load level 

EJ5 

M15 5



cr6



cr10

(condition II) of cracked: 

i

cr1

EJ1 Ec Jc

L1

Calculation of the bending stiffness based on the diagram moment‐ curvature at the level of loding 

M1i

EJ1 

EJ6 

M11 1

M16 6

kN

Jc  0.00231 m

4

2

Ec Jc  94045.08802 m kN

individual bases fictitious truss system cr :

 0.51632

 5.33482

EJi 

2

The ratio of flexural rigidity in state II to the bending stiffness over the state I and the

( cr1  cr2  cr3  cr4  cr5  cr6  cr7  cr8  cr9  cr10  cr11 ) 

 cr

1

cr11

Ec  40690000 m

EJ2 

EJ7 

M12 2

M17 7

EJ3 

EJ8 

M13 3

M18 8

EJ4 

EJ9 

M14

EJ6 Ec Jc



EJ2

cr2



cr7



Ec Jc EJ7 Ec Jc

cr3



cr8



EJ3 Ec Jc EJ8 Ec Jc

cr4

EJ4



cr11

Ec Jc



EJ11 Ec Jc

cr5



cr9



EJ5 Ec Jc EJ9 Ec Jc

EJ10 Ec Jc

cr1

0

cr2

 0.42178

cr3

 0.26895

cr4

 0.31313

cr5

 0.29142

cr6

 0.35178

cr7

 0.38883

cr8

 0.32504

cr9

 0.35455

cr10

 0.44911

cr11

0

4

M19 9

( cr2  cr3  cr4  cr5  cr6  cr7  cr8  cr9  cr10  cr11 )  s L1

cr



cr

 0.27787

Deformation Behaviour of Reinforced Concrete Beams

STRUCTURAL ENGINEERING ROOM

EJ10 

The ratio of shear stiffness in the state II to the shear stiffness in state I, expressed as a beam

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Coefficient

1 cr

1

Is the reduction of bending stiffness at the load level

cr

 3.59883

we can also ratio the bending stiffness in the state of I to bending stiffness over the state II 1 expressed by the longitudinal members :  cr

Ec Jc EJ3 Ec Jc EJ6 Ec Jc EJ10

Ec Jc

 3.71811

EJ4 Ec Jc

 2.84266

EJ7

Ec Jc

 3.19355

EJ5 Ec Jc

 2.5718

EJ8

Ec Jc

 3.43146

EJ2

 2.3709

Ec Jc

 3.07656

EJ9

 2.82045

 2.2266

Moment of inertia of the cross-section II of the state of cracks, provided that the modulus of

J6 

J11 

M12

J3 

 2 Ec

M16

J7 

 6 Ec

M13  3 Ec

M17  7 Ec

J4 

J8 

M14

J1 

 4 Ec

M18

M11  1 Ec

J9 

 8 Ec

M19  9 Ec

J5 

J10 

M15  5 Ec

M110  10 Ec

4

J9  0.00082 m

J2  0.00097 m 4

J6  0.00081 m

4

4

J3  0.00062 m

4

J10  0.00104 m

J7  0.0009 m 4

J11  0 m

4

4

J2 Jc J7

 2.5718

Em1  0 m kN

Em2  0.00828 m kN

Em3  0.0521 m kN

Em4   4 M14  s

Em5   5 M15  s

Em6   6 M16  s

Em4  0.09902 m kN

Em5  0.19232 m kN

Em6  0.24877 m kN

Em7   7 M17  s

Em8   8 M18  s

Em9   9 M19  s

Em7  0.32395 m kN

Em8  0.5278 m kN

Em9  0.21495 m kN

Em10   10 M110  s

Em11   11 M111  s

Em10  0.04248 m kN

Em     1 M11   2 M12   3 M13   4 M14   5 M15     s     M1   M1   M1   M1   M1   M1   7 7 8 8 9 9 10 11 11 11    6 6 Em  1.66718 m kN Calculation of internal energy from the effects of shear forces longitudinal members at a

J8  0.00075 m

Jc J3 Jc J8

 3.71811

 3.07656

Jc J4 Jc J9

 3.19355

 2.82045

Jc J5 Jc J10

 2.2266

Jc J6

Eq3   3 Q13  s Eq1  0.00945 m kN

Eq3  0.05305 m kN

Eq4   4 Q14  s

4

Eq5   1 Q15  s

Eq6   6 Q16  s

Eq4  0.04994 m kN

Eq5  0.00945 m kN

Eq6  0.0273 m kN

Eq7   7 Q17  s

Eq8   8 Q18  s

Eq9   9 Q19  s

Eq7  0.00819 m kN

Eq8  0.06166 m kN

Eq9  0.21482 m kN

Eq10   10Q110  s

Eq11   11Q111  s

Eq10  0.18787 m kN

Eq11  0.0394 m kN

4

 3.43146

Eq2   2 Q12  s

Eq2  0.05048 m kN

with cracks on different bases:  2.3709

Em3   3 M13  s

4

J4  0.00072 m

The ratio of moment of inertia in the state I without cracking to the moment inertia in state II

Jc

Em2   2 M12  s

Eq1   1 Q11  s

 11 Ec

J5  0.00067 m

Em1   1 M11  s

given level of work load by shear force diagram - shear strain:

M111

J1  0 m

a given level of loading in the diagram moment - curvature:

Em11  0 m kN

elasticity of the beam is a constant: J2 

Calculation of the internal energy of the effects of bending moments longitudinal members at

 2.84266

Eq 

 s

 1 Q11  4   2 Q12  2   3 Q13  4   4 Q14  2   5 Q15   3    4   6 Q16  2   7Q17  4   8Q18    9Q19    10Q110    11Q111 

Eq  0.4687 m kN

Deformation Behaviour of Reinforced Concrete Beams


Example 5.4-6: This is a simple beam placed on two supports where is loaded in midspan by

action:

single load P. In the middle span is reinforced and thus the moment of inertia of the section

Ecelkove  Em  Eq

with respect to the two-mass inertia, in the quarter span Ii 2Ia figure 5.4.6-1.

Ecelkove  2.13588 m kN

Proportion of internal energy from the effects of shear forces to the total internal energy: Thus, the proportion of energy from the effects of shear forces to the total internal energy is

Ecelkove

beam y m . figure 5.4.6-1b is drawn triangular moment diagram with the greatest ordinate pL

Mmax

21.9% Eq

We are looking for angular rotation in supports a a b and deflection at the center of the

, figure 5.4.6-1c is a diagram

4

on span of secondary beams S

 0.21944

M

L

4

M EI

, figure 5.4.6-1d on the diagram

M EI

as the load

and figure 5.4.6-1. There are four secondary beams with

Proportion of internal energy from the effects of bending moments on the total internal

diagrams

energy:

diagrams can easily determine the ideal loads. As the beam and loads are symmetrical,

Thus, the proportion of energy from the effects of bending moments on the total internal

applies:

energy is 78.0% Em Ecelkove

W0

W4

EI

1

, which act as individual load on each of the secondary beams. Using these

P L L 1    2 8  E  Ia 4 3

2

P L

192 E  Ia

 0.78056

External energy or deformation will be:

 9.232 mm  7.678 mm   1.503 mm   2   2 

W  Fs2 

W  2.15698 m kN

The ratio of external to internal energy will be:

Ecelkove  2.13588 m kN Ecelkove W

 0.99022

W  2.15698 m kN 1

Ecelkove W

 0.00978

loss of energy is minimal: 0.9%

Figure 5.4.6-1: Simple beam with discontinuous variable inertia.

Deformation Behaviour of Reinforced Concrete Beams

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The total internal energy is the sum of energy from the effect of bending and shear energy of

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91

W2

 1 P L L 1 1 P L L 2  2           2 8 E  I 4 3 2 4 E  Ii 4 3 

 P  L2

2 

 192 E  Ii

  48 E  Ii  2

The calculation of loads using trapezoidal rules

2

10 P  L

P L

192 E  Ii

Subsequent relationship is a simplified linearized differential equation crease lines, which

In general, lists any moment of inertia Ic a comparable moment of inertia and further express

solves a double integration. The first integration determines the rotation of the crease lines

the ideal shape for loads

and second crease lines. 2

d y 2

W0

Ic

P L

W1

 192 E  Ic Ia Ic

We put Ic Ii ,

2

Ia

W3

Ic

,

 Ic 2 Ic     Ii  96 E  Ia P l

1,

Ii

2

2 P  L

W1

192  E  Ic

2

10 P  L

W2

Ic

dz

 192 E  Ic Ii

8 P  L

W3

W2

192  E  Ic

2

M ( z) EI ( z )

If we extend the right side of the equation, then we get

Thus, the load get ideal

2

Wo

2

2

10  P  L

2

d y

192  E  Ic

dz

2

Rotation of the cross section in the support beam are equal. The term will write directly to rotate depending on the force in the form of

 M ( z)     E  I( z) 

Ic   2  M ( z)     I E I c ( z)  

Then to calculate the ideal force we get loads we will continue to work with

a

W0  W1 

W2

2

2

1.25

192 E  Ic

4

L

2

 W1 

2

L

P L

4

E  Ic

8 L  2 L  15 L     192 2 192 2 192 4 



2

9 P  L

384 E  Ic

Mm  lava 

Ic Im  lava

 Mm  prava 

Ic Im  prava Ic

3

1.125

P L

48 E  Ic

inertia Ic Ii 2  Ia in span 2   L , we see that the calculation result of the deflection is greater 4

I

triangles load W:

Comparing the equations- moment of inertia constant, and equation – where moment of 1

Ic

, but with diagrams M  , called reduced moment diagram.

Decomposition trapezoid area, which is reduced moment diagrams Mz 

Deflection at the center of the beam

 W0 

and

can write

that the result of the equation of rotation of the support is more about 25%.

2

1 EIc

moments or moments of inertia, respectively. both. Generally, for example, to the point we

1

L

. Now leave out constant

On the end between the sectors may occur due to sudden changes in the discontinuous

constant, and the equation, where the moment of inertia Ic Ii 2  Ia in length 2  L , indicates

a 

Ic I( z)

16 E  Ic

Comparison between the equation in the case of the beam, where the moment of inertia is

ym

 M ( z) 

P L

2

15 P  L

b

M E I

1 E  Ic

Wm

1 E  Ic

Ic Ic 1 1 1 2    Mm  1  prava   S    Mm  lava   S   3 2 3 2 I I m  1  prava m  lava 

  

   Ic Ic 1 2 1 1   Mm  prava   S    Mm  1  lava  S  2 3 2 3  Im  prava Im  1  lava

about 12.5%.

Deformation Behaviour of Reinforced Concrete Beams

I

to get perfect


92

Wm

S

Ic

6  E  Ic Im

 Mm  1  prava  2  Mm  lava 

S

Ic

6  E  Ic Im  1

 2  Mm  prava  Mm  1  lava

If the moment of inertia of each segments of the beam does not change, i,e are constant, then Im  1  prava Im  1  prava

Im  lava Im  lava

a Im

Im  prava

Im  1  lava

Im  prava

Im  1  lava

We reduced this moment M  Im  1

(sectors) ´ : ´m

S

Ic

Ic

leave and move on to further reduced in size sections

I

a

Im

S

´m  1

Ic Im  1

We get : Wm

´m

6  E  Ic

 Mm  1  prava  2  Mm  lava 

´m  1

6  E  Ic

 2  Mm  prava  Mm  1  lava

If moment diagram does not suddenly change, then Mm  1  lava and Mm  1  lava Mm  1  prava .

Mm  1  prava , Mm  lava

Mm  prava

If the moment of inertia is constant, then Im  1

Wm

Im

Im  1

S 6 E  I

I,

Ic

ie we can write

 Mm  1  4  Mm  Mm  1

For the first and last load W we get

W0

Figure 5.4.6-2

Wn

S 6 E  I

S 6 E  I

Ie

I0

  2  M0 

 M1  lava 

Ic I1  lava

Ic 

In 

  Mn  1  prava  2  Mn 

  

In the case where the moment of inertia and moments of the diagram, without sudden changes in:

W0

S0 6 E  I

  2  M 0  M 1

Wn

Sn 6 E  I

 Mn  1  2  M n

Deformation Behaviour of Reinforced Concrete Beams

STRUCTURAL ENGINEERING ROOM

Then we can be written Wm in the form

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Wm

Ic Ic Ic       Mm  1  prava   2  Mm  lava   2  Mm  prava  6  E  Ic  Im  1  prava Im  lava Im  prava  Ic     Mm  1  lava  I  m  1  lava   S


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Example 5.4-7: Calculation of ideal load W Calculation of loads in ideal load cross-section we done according to figure 5.4.7-1. Selected parts diagram M 

Ic I

divided into two trapezoids ( m  1)  c  d  m a m  d  e  ( m  1) on the

remaining two the same area of the parabola with a horizontal base  and vertical cde

f 4

where

is parabola a f is the peak of the whole parabola. For variable parabola is valid the

same relationship as the

ql 8

2

(mid- span moment is reduced to a quarter).

Figure 5.4.7-2: Reduced moment diagram for the beam with I  konst . thus we get: Wm

S 6  E  Ic

Ic

Im  1

  Mm  1 

 4  Mm 

Ic

 Mm  1 

Im

Ic

  

Im  1 

Ic Ic Ic   1 2 S 1 1   2      Mm     Mm  1   Mm  1   2 3 E Ic  4 8  Im Im  1 Im  1  

Then the result of an ideal load W m will shape: Wm

Figure 5.4.7-1

Ic Ic   f Mm     Mm  1   Mm  1   2  Im Im  1 Im  1  Ic

f

1

4

4

 Mm 

1

Ic Im

1 8

Ic

Im  1

  Mm  1 

 Mm  1 

Ic

And from there

 

Im  1 

Ideal load in cross-section consists of two parts: From the contribution of the two trapezoids ( m  1)  c  d  m a m  d  e ( m  1) , Of two equally large areas parabola (since they are symmetrical, then W m since they are symmetrical, then).

  Mm  1 

Ic Im  1

 10 Mm 

Ic Im

 Mm  1 

 In the case of constant inertia term will: Wm

For peak parabola c  d  e will be

S 12 E  Ic S 12 E  I

 Mm  1  10 Mm  Mm  1

Ic

 

(2.8.17)

Im  1 

(2.8.18)

for first ideal load W we get Ic Ic 1  Ic  1 2   1 Ic Ic   S  W0   2  M0   M 1         M 1     M0   M 2    I0 I1  2 3 E Ic  4 I1 8  I0 I0   6  E  Ic  I I I  S c c c W0   3.5 M0   3.0 M1   0.5 M2   I0 I1 I2  12 E  Ic  And similary for the last ideal load W we get ideal Ic Ic Ic   S Wn   0.5 Mn  2   3.0 Mn  1   3.5 Mn   In  2 In  1 In  12 E  Ic 

(2.8.19)

(2.8.20)

Given that the the moment of inertia is constant and moment changes are continuous, we can write (2.8.21) S W0

Wn

12 E  Ic S 12 E  Ic

  3.5 M0  3.0 M1  0.5 M2

 0.5 Mn  2  3.0 Mn  1  3.5 Mn

Deformation Behaviour of Reinforced Concrete Beams

(2.8.22)


In experiments tested at deformation elements located at the top (hi) and lower (di) edge of the beam. under which the curvature is calculated:  hi   di h

curvature

 d1 85

Aid calculated curvature at various levels are obtained ideal values of loads: Wi

1

i

d s

F

 hi

 hn h = 147

1 i

 h1

d i

d si

Ra

Ideálne bremená W1

 dn

ri ra = 560

Wi

Wn

Priebeh momentov od pôsobenia W i = priehyb nosníka od pôsobenia F

f

i

Their summation we get the amount of force Ra (reaction)

Figure 5.4.8-1: 7Reinforced beam loaded by ideal loads n ri center distance of the individual elements of the Ra Wi place in which we calculate the deflection f ra distance forces Ra from the point of determining i 1 the deflection Calculations of the moments on the beams subjected to ideal loads Wi the progress of deflection along the beam due to concentrated force F. n The maximum deflection of the beam at f R a r a  W i r i mid-span and is calculated as follows :

i

1

Element h_d

5_2

8_3

9_26

12_27

13_30

16_31

17_34

20_35

eh ed

1.37E-06

1.71E-05

2.94E-05

1.98E-05

-2.73E-05

-1.19E-04

-1.76E-04

-2.73E-04

1.90E-06

1.55E-05

7.18E-04

8.58E-04

1.03E-03

1.19E-03

1.26E-03

1.32E-03

1/i = curvature

-3.61E-06

1.09E-05

-4.68E-03

-5.70E-03

-7.17E-03

-8.89E-03

-9.75E-03

-1.08E-02

dsi [m]

0.0425

0.0425

0.03875

0.03875

0.03875

0.03875

0.03875

0.03875

Wi

-1.54E-07

4.63E-07

-1.81E-04

-2.21E-04

-2.78E-04

-3.44E-04

-3.78E-04

-4.19E-04

ri [m]

0.62375

0.58125

0.541

0.501875

0.463125

0.424375

0.385625

0.34687

Wi . ri

-9.58E-08

2.69E-07

-9.82E-05

-1.11E-04

-1.29E-04

-1.46E-04

-1.46E-04

-1.45E-04

Element h_d

21_38

24_39

41_54

44_55

45_58

48_59

49_62

52_63

eh ed

-3.50E-04

-4.94E-04

-5.45E-04

-6.43E-04

-6.49E-04

-6.57E-04

-6.61E-04

-6.63E-04

1.37E-03

1.44E-03

1.48E-03

1.50E-03

1.44E-03

1.45E-03

1.45E-03

1.45E-03

1/i = curvature

-1.17E-02

-1.31E-02

-1.38E-02

-1.46E-02

-1.42E-02

-1.43E-02

-1.43E-02

-1.43E-02

dsi [m]

0.03875

0.03875

0.041667

0.041667

0.041667

0.04167

0.041667

0.041667

Wi

-4.52E-04

-5.09E-04

-5.74E-04

-6.07E-04

-5.92E-04

-5.97E-04

-5.97E-04

-5.97E-04

ri [m]

0.308125

0.269375

0.2292

0.1875

0.145733

0.10412

0.0625

0.020833

Wi . ri

-1.39E-04

-1.37E-04

-1.32E-04

-1.14E-04

-8.63E-05

-6.22E-05

-3.73E-05

-1.24E-05

Ra [ - ]

6.35E-03

ra [ m ]

0.56

Total deflection f [m]

2.06E-03

Deformation Behaviour of Reinforced Concrete Beams

STRUCTURAL ENGINEERING ROOM

Example 5.4-8: Calculation of the deflection by the ideal of loads

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6. Behaviour and Conception of Timber Structures

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Timber is a structural material which has similar properties to steel and reinforced

In pitched roofs with insulation at rafter level and in cold flat roofs, a vapour control layer should be included on the warm side, beneath the insulation, to restrict the passage of

concrete in the sense that it can carry both tension and compression with almost equal facility.

moisture and air. Between the insulation and the underlay or sarking, a 50mm minimum

It is therefore capable of resisting bending-type load and may be used for all types of

ventilation space must be provided with vents at eaves and ridge, or at opposite sides of a flat

structural element. It is significantly less strong than either steel or reinforced concrete,

roof. This space, combined with the thickness of the insulation can mean that the depth of the

however, with the result that larger cross-sections are required to carry equivalent amounts of

rafters is determined by thermal rather than structural requirements. Pitched roofs with insulation at ceiling level or between rafters and cold flat roofs

load. There are no specific differences in the roof construction of a timber framed structure compared to other types of construction. It is however, important to ensure that any additional

should incorporate ventilation to reduce the risk of condensation occurring. A vapour control layer at the top floor ceiling will restrict the passage of moisture and can act as an air barrier.

point loads from the roof (from girder trusses, purlins, etc) are adequately supported by additional studs or posts in the timber wall panels. The main design considerations for which flexural members should be examined are 1. Bending (including lateral buckling) 2. Deflection 3. shear 4. Bearing. Generally, bending is critical condition for medium span beams, deflection for long span beams, and shear for heavily loaded short span beams or at notched ends. For designs based on permissible stress philosophy, bending is checked by appling the basic theory of bending principles. In relation to timber design this must also take into account the relevant modification factors for exposure, load duration, load sharing and so on.

Figure 6-1: Construction of purlin roof

From theory of bending we know that M = f . W, where W= bd2/6 for rectangular sections. Knowing the applied loads, the maximum bending moment M may be calculated. Where f

is the permissible stress value of the material. To avoid damage to finishes, ceilings, partitions and so on, the deflection of timber

flextural members when fully loaded should be limited to 0,003 of the span. aadmissible = 0,003* span For flextural member to be adequate in deflection the summation of the actual deflection due to bending am and that due to shear av must not be greater than the permissible value aadmissible am+ av < aadmissible

Roof panels can be used as an alternative to attic roof trusses to provide unobstructed roof spaces for occupation. The panels are similar in construction to timber frame floor cassettes and enable a weatherproof structure to be rapidly achieved on site. Panel roof design relies on four conditions being assessed and calculated by a structural engineer. 1. The floor structure acts in tension to restrain the outward force of the roof panels at the eaves. Any butt or lapped joints in the joists should be designed to withstand tension forces. A loadbearing internal wall or a beam is normally required to reduce the floor joist spans. Openings in the floor framing require special design to ensure continuity of this tension

Behaviour and Conception of Timber Structures


should preferably be located parallel to the joists. 2. The ceiling strut is located where the roof collar is normally situated but, unlike the collar in a simple roof which is a tension member, the ceiling strut is in compression and therefore carries out a quite different function. 3. The panels are sheathed to resist rafter buckling and wind (racking) forces.

Moment of inertia of the cross-section:

Section modulus:

lcr i

Slenderness ratio: 

floors with joists at 600 mm centres provide slightly better sound insulation.

  Nmax  A 1   

1

 N max Mmax     W    A

J

0.0000364m4

0.0519615m

38.9711432

b h

0.000405m3

W

0.5 h

3

12

i

J

W

4. The resistance to horizontal thrust at the eaves relies on effective levelling. Floor joists are normally spaced at either 400 or 600mm centres. Test evidence has shown that

J A

Radius of gyration: i

1

J

2

3100

0.4899194

0.9334677

Which ever lesser  1

5.8618244 MPa

The deflection of the Rafter may be determined as follows:

fmax fmax Figure 6-2: Truss configuration with bolted metal plate connections Example 6-1: Compute the stress in Rafter for a given bending moment and external force at the section: Bending moment at the section: Mmax

1.6 kN m

Exterrnal force at the section

22 kN

Nmax lcr

0.9 L

Modulus of elasticity of the member: E Cross Section: Allowable stress:

b

75 mm

1

h

lcr

2.25 m

2.025m

2 1000 kN cm

1 200

L

0.0023148 m

fallowable

Whichever lesser than fallowable Ok

0.01125 m

Compute the stress in Rafter for a given bending moment and external force at the section:

Exterrnal force at the section

6.3 kN m

Nmax

Modulus of elasticity of the member: E

A

Which ever lesser fallowable

Example 6-2:

Effective length of Ceiling struts: lcr

180 mm

b h

fallowable

fmax

0.0023148 m

13.0 kN

Length of the ceiling struts measured between centres of restraint:

12 MPa

The entire area of cross-section: A

2  M L   5  max  E J  48 

fmax

Bending moment at the section:Mmax

Length of the Rafter measured between centres of restraint: L Effective length of Rafter

2  M L   5  max  E J  48 

0.0135m

2

Cross Section:

b

140 mm

Allowable stress:  1

h

0.9 L

lcr

4.5m

2 1000 kN cm

180 mm

12 MPa

The entire area of cross-section: A

b h

A

0.0252m

Behaviour and Conception of Timber Structures

2

L

5.0 m

STRUCTURAL ENGINEERING ROOM

function. The staircase opening should interrupt the minimum number of floor joists and

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STRUCTURAL ENGINEERING ROOM

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97 1

Moment of inertia of the cross-section: J J A

Radius of gyration: i Slenderness ratio:  Section modulus:

W

  Nmax  A 1   

1

 Nmax Mmax    W    A

i

lcr i

3

12

6.1 How Structural Systems Carry The Load - Timber Engineering

0.000068m4

J

Primary Structural Systems

The number of primary structural systems, their spacing and the positions of supports

0.0519615m

86.6025404

J

W

0.5 h

b h

are governed by the plan layout. The design of the grid depends on the utilisation conditions,

2

3100

2.4193548

e.g. movable partions, lighting. Plan view of primary structural systems

0.000756m3

0.8959933

9.8165373 MPa

Which ever lesser 1

The deflection of the ceiling struts may be determined as follows: fmax

 M L2   5  max   48 EJ 

fdov

1 L 200

fdov

fmax

0.0241127m

Whichever lesser thanfdov

0.025m

Permissible ratio of effective span to effective depth for simplified analysis of deflection

Figure 6.1-1: Plan view of secondary structural systems

limit.

Secondary Structural Systems

Secondary structural systems give form to the roof and also the interior layout. The loadbearing arrangement is determined by the number of the nature of the supports, the number and interconnection of independent loadbearing elements, and the form of the loadbearing members. - Rafter:

h = span / 24

Irequired = 208.3 Mmax (kN.m) L (m) = cm4

- Purlins or joists:

h = span / 16

Irequired = 312.5 Mmax (kN.m) L (m) = cm4 Irequired = 26 q (kN/m) L3 (m) = cm4

- Ceiling tie:

-Columns or Posts (studs): n = 6, EII= 105(kg/cm2) Irequired = (n.F(kg).H2(cm)) / (2. EII) = cm4 -Circle posts

-Square posts

-Rectangular posts

D (H / 18,75)

a  (H / 21,7)

a  (H / 21,7)

a is least lateral dimension of cross-section -Glued laminated timber column: a = H / 22.5 -Glued laminated timber beam: h = span / 16

Figure 6.1-2:

Behaviour and Conception of Timber Structures


Internal walls should not be supported by floating floor decks or on the decking of timber floors unless adequate support is provided. Lateral and vertical movement needs to be considered if internal walls are supported on a floating floor deck. Actual deflections of roofs or floors, particularly long spans, may impose loads onto non loadbearing elements, unless deflections are taken into account. Normally deflected elements are assumed as 0.003 of span (m) under full design load. Deflection of joists which may cause squeaking around fixings must also be considered. If a deflection gap is specified by the structural engineer or truss manufacturer, the head of the wall must still be adequately supported to prevent lateral movement. Figure 6.1-3: Resolve stability of bearing structural members - lateral bracing Stability by means of trusses. Partly enclosed building often include bracing trusses within the walls. Compared to frames, the quantity of material required is considerably less, but the connections are more complicated. The diagonals between posts and beams can be

The external wall consists of two parts: - the loadbearing timber frame wall - the outer cladding. This may be a heavyweight cladding, supported independently by the foundations, or a lightweight cladding attached to the timber frame.

used as a frame-work for the wall construction. The ensuing loads, particularly those perpendicular to the wall, e.g. wind or silo loads, must be included in the design. The first of the frame-works here is capable of achieving equilibrium under the loading shown but is unstable. The insertion of the diagonal element in the second framework renders it capable of achieving stable equilibrium figure 6.1-3. Most structural arrangements require bracing for stability and the devising of bracing systems is an important aspect of structural design. Timber houses have a good reputation for performance in seismic events. This is based on the low weight of timber structures, ductility of joints, clear layout of timber houses and good lateral stability of the house as a whole. As in any kind of building it is usually the inadequate structural design or inadequate supervision during the building process that causes the damages induced by seismic events. For wooden houses vulnerable parts are: the anchorage of the house, the diaphragm action of floors and the first soft storey which sometimes has been left without sufficient lateral bracing (for example crawl spaces, garages). Point loads from beams, eg trimmers carrying joists, should be transmitted directly to the foundations by the use of additional studs, the exact number being determined by

Figure 6.1-4: Lay-out of roof constructions subjected to horizontal load-wind

calculation. Deep beams require pockets to be formed within the wall panels or top hung hangers are specified. Load transfer for beams should be followed through all panels and floor

Behaviour and Conception of Timber Structures

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framing to foundation level when specified by the structural engineer.

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Figure 6.1-6: Plan elevation of primary structural systems and secondary structural systems

Figure 6.1-5: Stability of structures subjected to lateral forces The most common option is to increase the depth of the studs to more than 140mm to allow more insulation to be incorporated. Studs up to 200mm deep have been used. For sections deeper than 200mm, the use of solid timber is unlikely to be economic and designers should consider a timber based structural composite. There are a number of I-section components with solid timber or composite material flanges and webs of various timber based panel products, most commonly oriented strand board or fibreboard. In addition, there are a number of different types of structural timber composites on the market, which have been available for many years.

Figure 6.1-7: Plan elevation of primary structural systems and secondary structural systems

Behaviour and Conception of Timber Structures


Figure 6.1-8: a) The transformation of shear forces, b) The transformation of shear forces to

Figure 6.1-10: Trusses with steel rod

compression respectively. Traction (tension) and bending, c) buckling, d) Tension for bending, e) Compression for bending.

Figure 6.1-9: The three articulated lattice frame

Figure 6.1-11: Trusses with steel rod

Behaviour and Conception of Timber Structures

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Beam structures may be stressed due to a variety of vertical load see figure 6.1-8.

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Carrier systems beam structures shown in sections and methods of supports under the

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effect of vertical load. A support beam structure a- Beam b- Beam structures with rod cFrame

Beam structures stressed mainly by pressure

Beam structures stressed mainly by tension

Beam structures stressed mainly by pressure and tension Beam Beam structures stressed mainly by bending

Beam structures with rod

Frame

Figure: 6.1-13 Expression of different stiffness to the frame structure and transferred the load on the bending moment acting on the horizontal and vertical elements of the frame system.

Beam structures arranged radially

Beam structures merged as surface elements

Figure: 6.1-14 Figure: 6.1-12

Behaviour and Conception of Timber Structures


With hardwood block and steel plates on both sides

Cable-stayed bridges

With members notched to accommodate Isection plus top plate to resist tension

Cable-stayed bridges with single pylon tied back to end support

Cable-stayed bridge with single raking pylon

Cable-stayed pinned beams with triangulated pylon

to tnansverse deck beam

Prop connections- with steel plate let into slits

Prop connections- with T section

Prop connections- with steel plate let into slits

Prop connections- with steel plate let into slits plus end plate

To longitudinal beam via block

Prop connections- with steel plate let into slits

Prop connections- with oblique joint

A-frame pylon as three legged trestle

Cable-stayed bridge tied back to intermediate support

Behaviour and Conception of Timber Structures

To longitudinal beam via hinge pin

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Pinned ridge joints

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102


Collar conections:

Trussed beams

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103 Trussed-beam with tie in timber or steel

with two posts and beam in bending under asymmetric loading

with one posts

Beam- trussing junction: with steel plate on end grain

With hinge pin and reinforced nail plate

Three â&#x20AC;&#x201C; dimensional base details: with cleats- with steel angles let into slitsconnected to steel stanchion with steel shoe

-with oblique dado joint and simplex con -with twin members nailed or dowelled - with oblique dado joint and nailed fish â&#x20AC;&#x201C;plates -with cleat and nailed fishplates

Plane frames

Plane frames

By means of plate welded to nail plate

With steel plate on end grain

Plane frames

Plane frame as solid web or box beam

Plane frame as truss

Behaviour and Conception of Timber Structures

plane frame with rigid corner


System and beam forms: -as purlins of squared timber sections Spacing of purlins depends on roof construction, loading etc. h= l/8 to l/14

Further trusses

Various beam elevations Glued laminated timber

The two-pin frames

h=1/10-1/15

Linear member, frame, arches

Tied triangular frames

h=l/10 to l/20

Mansard roof truss – arch truss – arch truss with raised eaves

Duopitch roof trusses with raised eaves

Various beam elevations

Systems and beam forms As primary loadbearing system nailed, glued flanges or webs h= l/8 to l/14

Duopitch roof trusses with raised eaves

Line of thrust for uniformly distributed load 2nd order parabola

Symetrical trusses- with rising and falling diagonals

With diamond bracing – with posts plus rising and falling diagonals

With clerestory windows on one side – with raised bottom chord and clerestory Windows- with lantern light along length of roof- as monitor roof (for lighting and ventilation) with raised bottom chord Simply supported plane frames - for monopich roof with raise eaves

Cmpression members -Reinforced concrete compression members are usually called columns, timber compression members posts, and steel compression members stanchions. -The vertical loads they support can be concentric or eccentric. If the load is concentric its line of application coincides with the neutral axis (NA) of the member. Such compression members are said to be axially loaded and the stress induced is adirect compressive stress

Behaviour and Conception of Timber Structures

-Stress = (load / area) -Slenderness ratio = (effective length / least radius of gyration) - Raius of gyration = (second moment of area / area)

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Glued laminated timber:

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Three-pin frames- portal frames Of course minimising weight does not necessarily minimise cost. In this instance, overall economy of construction will only be achieved if automated fabrication is adopted

Three-pin plane frames as trusses

Three pin frames- with tie solid crosssections

Three pin plane frames without tie

Troj klbový RAM – s pevným tiahlom

Three pin frames- without tie solid crosssections Troj klbový ram – s pevným tiahlom Statická schéma Change in pitch

Statická schéma

Symmetrical arrangement with raised eaves Three-Pin Frames -Glued laminated timber

Transverse stability is ensured by the frame with its fixed-base column. The tie is placed low to clear the suspended walkway and must thus penetrate the main columns so as to be joined to the cantilevering eaves Three-Pin Frames with Raised Tie (Collar)

Main members with one prop each Main members with two props each

Behaviour and Conception of Timber Structures


The building is braced in the longitudinal direction by the wind girders between the -double T-box and solid-web sections of arches. glued laminated timber -circular arc -asymmetric arch

Arches for Bridges -Driveâ&#x20AC;&#x201C;through -Drive-over

-propped arch -Supported on A- frames with cellular infill -lever arm providing fixity at centre of arch Solid web member with reinforced end and cast â&#x20AC;&#x201C; in side

-cross-sections resolved into individual Important alternative that allows you to

members plus tie.

significantly change the wall thickness of plate girder and that is prestressing. For prestressing the best shape of section I. We bring along the longitudinal compressive

For RC arch bridges the cost of falsework and

stress primarily in a cross-section of the parts

formwork

that are drawn by the action of the load.

comparison

Prestress pressure decreases in a wall of

girder. As a result, arch bridges are economical

mainly tensile stress at a curved prestressing

only under a limited range of topografical and

elements we bring the vertical load exerted

geotechnical conditions. The ratio of arch span

upward.

to rise, l/f, should be chosen between 2:1 and

Glued laminated timber member on elastomeric bearing

for

arch

to

bridges

conventional

is

high

in

cast-in-place

10:1. Arch base detail The mechanism of structural failure

circular arc,asymmetric arch, Propped arch

Solid web member with reinforced end and cast â&#x20AC;&#x201C; in side

Behaviour and Conception of Timber Structures

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System Variation

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107 Plane Frame Systems - Truss frames

-supported on A- frames with cellular infill -lever arm providing fixity at centre of arch -cross-sections resolved into individual members, plus tie

Glued laminated timber member on elastomeric bearing Stress and deformations are normally small when the ratio of span to rise is less than 4:1, regardless

of

the

degree

of

statical

indeterminacy of the arch. As l/f approchase 10:1, it may be necessary to reduce or eliminate redunant

moments

due

to

restrained

deformations by providing hinges at the springing lines and at the crown.

Three â&#x20AC;&#x201C;pin trusses span in the transverse direction. The parallel top and bottom chords are each in two parts, the diagonals and posts just single members.

Behaviour and Conception of Timber Structures


300

2

Load q kg/m

L span(m) a / b (cm) 3 8 / 16 3,20 8 / 16 3,4 8 / 18 3,60 8 / 20 3,8 8 / 20 4,00 8 / 20 4,2 10 / 20 4,40 10 / 22 4,6 10 / 22 4,80 10 / 22 5 12 / 24 5,20 12 / 24 5,4 12 / 26 5,60 12 / 26 5,8 12 / 26 6,00 12 / 26

325 c (cm) 74 62 73 83 71 62 66 77 68 60 81 72 82 73 67 62

a / b (cm) 8 / 16 8 / 18 8 / 18 8 / 20 8 / 20 10 / 20 10 / 22 10 / 22 10 / 22 12 / 24 12 / 24 12 / 24 12 / 26 12 / 26 12 / 26 14 / 26

350 c (cm) 68 80 67 77 66 71 82 71 63 85 77 67 75 67 61 67

a / b (cm) 8 / 16 8 / 18 8 / 20 8 / 20 10 / 20 10 / 20 10 / 22 10 / 22 10 / 24 10 / 24 12 / 26 12 / 26 12 / 26 12 / 26 14 / 26 14 / 26

375 c (cm) 64 74 85 72 74 66 76 66 75 80 70 78 70 62 66 60

a / b (cm) 8 / 16 8 / 18 8 / 20 8 / 20 10 / 20 10 / 22 10 / 22 10 / 22 12 / 24 12 / 24 12 / 26 12 / 26 12 / 26 14 / 26 14 / 26 20 / 26

400 c (cm) 60 69 80 67 71 82 71 61 84 75 83 73 65 69 62 80

a / b (cm) 8 / 18 8 / 18 8 / 20 8 / 20 10 / 20 10 / 22 10 / 22 10 / 24 12 / 24 12 / 24 12 / 26 12 / 26 12 / 26 14 / 26 20 / 26 20 / 26

Xa

425 c (cm) 80 65 75 62 67 76 66 75 78 70 77 69 61 64 84 78

a / b (cm) 8 / 18 8 / 20 8 / 20 10 / 20 10 / 22 10 / 22 10 / 22 12 / 24 12 / 24 12 / 26 12 / 26 12 / 26 14 / 26 14 / 26 20 / 26 20 / 26

c (cm) 75 84 70 74 83 72 62 85 76 83 73 65 67 61 78 73

J

a1

bX

1 3 Ec   12 bc hc  Edr 

1 12

5

Deflection f

c is the distance between the purlins or rafters

fcomposite

L span of the purlins or rafters

116

384

X1

Ec

Ewood

 gs L 

2

I

1

L    b 2 2 S

S

if 

Md

5

b h h  2 4

384

Ntot

n

N16 I

1 12

 b h

Edr J

then the section should be composite

3

Composite member Steel vs concrete

F

Ec Ewood

 n

hc bc

Fwood

Ec    Fwood   Fc   Ewood 

hdr bdr

a Fc 

X

hdr

Ec Ewood

b

2 2

hc 2

 hdr 2

b  Fwood a F

Fc

F

Ec Es

hb bb

Ec

Fo

  Fc  Es  Fo n

stat

bb

a

ho 2

1 m

b

hb 2

wood

gs L

Composite member - Timber vs concrete

Fc

4

cm

4

a  fwood

Jdr

Ntot

2

M X  12MPa J

d fcomposite

Edr Jdr

1 m

b1  Fwood a1

4

A 16 fyd

z

bb

3

Concrete

nsmaller

N16

HX

bdr hdr   Fc 

M Ec  X1  fcd J Ewood

h

f

where a and b is cross-section dimension of purlins or rafters

b1

 ho

Behaviour and Conception of Timber Structures

V b z

V

1 2

ql 2

fdeflection

1.2

q L

8 G A

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Table 6.1-1: Load, Span, Dimensions, and Axis Spacments for Soft timber, Class I, and II

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109

Fc  X

Ec Es

2

Xa

F

bX

Table 6.1 -2: Dimensions, Area, Section Modulus, Second Moment of Area for Timber

2

b  Fo a

b1

X1

b (cm) 8

a1

HX 10

J h

Ec  1 3 Ec 2 2  12 bb hb   Jsteel  Fconcrete   b1  Fo  a1 Es Es   M Ec  X1  fcd J Es

concrete

M X  235MPa J

d

10 12

4

cm

stee

12 12 12 14

Deflection 5  gs L   4

f if

384

 bmax Nred

fcomposite

Es Jsteel Md Jsteel

V b z

a   ocel

5 384

gs L

f

Es J

fcomposite

14 14 14 14 14 16

nsmalle

cross-section need to be composite

Nmax

V = shearing force

0.8N max

4

Nf

2

31.6d

 fcd

1 2

L  bmax  b 2

b = width b=1m

d = is diameter of 10

16 16 16 16 16 18

18 18 18 18 18 20

20 20 20 20 20 22

22 22 22 22 22

h (cm) 8 10 12 14 16 10 12 14 16 18 20 10 12 14 16 18 20 22 24 10 12 14 16 18 20 22 24 26 28 10 12 14 16 18 20 22 24 26 28 10 12 14 16 18 20 22 24 26 28 10 12 14 16 18 20 22 24 26 28 10 12 14 16 18 20 22 24 26 28

F= a x b 64 80 96 112 128 100 120 140 160 180 200 120 144 168 192 216 240 264 288 140 168 196 224 252 280 308 336 364 392 160 192 224 256 288 320 352 384 416 448 180 216 252 288 324 360 396 432 468 504 200 240 280 320 360 400 440 480 520 560 220 264 308 352 396 440 484 528 572 616

W x =(b*h2) / 6 85 133 192 261 341 167 240 327 427 540 667 200 288 392 512 648 800 968 1152 233 336 457 597 756 933 1129 1344 1577 1829 267 384 523 683 864 1067 1291 1536 1803 2091 300 432 588 768 972 1200 1452 1728 2028 2352 333 480 653 853 1080 1333 1613 1920 2253 2613 367 528 719 939 1188 1467 1775 2112 2479 2875

Behaviour and Conception of Timber Structures

Ix=(b*h3 ) / 12 341 667 1152 1829 2731 833 1440 2287 3413 4860 6667 1000 1728 2744 4096 5832 8000 10648 13824 1167 2016 3201 4779 6804 9333 12423 16128 20505 25611 1333 2304 3659 5461 7776 10667 14197 18432 23435 29269 1500 2592 4116 6144 8748 12000 15972 20736 26364 32928 1667 2880 4573 6827 9720 13333 17747 23040 29293 36587 1833 3168 5031 7509 10692 14667 19521 25344 32223 40245

W y=(h*b2) / 6 85 107 128 149 171 167 200 233 267 300 333 240 288 336 384 432 480 528 576 327 392 457 523 588 653 719 784 849 915 427 512 597 683 768 853 939 1024 1109 1195 540 648 756 864 972 1080 1188 1296 1404 1512 667 800 933 1067 1200 1333 1467 1600 1733 1867 807 968 1129 1291 1452 1613 1775 1936 2097 2259

Iy =(h*b3) / 12 341 427 512 597 683 833 1000 1167 1333 1500 1667 1440 1728 2016 2304 2592 2880 3168 3456 2287 2744 3201 3659 4116 4573 5031 5488 5945 6403 3413 4096 4779 5461 6144 6827 7509 8192 8875 9557 4860 5832 6804 7776 8748 9720 10692 11664 12636 13608 6667 8000 9333 10667 12000 13333 14667 16000 17333 18667 8873 10648 12423 14197 15972 17747 19521 21296 23071 24845


Figure 6.1-15: Simple supported beam, rigid frame, rigid frame

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Dimensioning lattice truss beams

Table: 6.1 -3 span

inclination

(m)

(mm)

a

e

d1

d2

6.00

0.40

65x165

65x185

75x130

55x105

0.50

65x165

65x185

75x130

55x105

0.60

65x165

65x185

75x130

55x105

0.80

65x165

65x185

65x130

55x105

0.40

65x165

65x185

75x130

55x105

0.50

65x165

65x185

75x130

55x105

Span

inclination

a

e

d1

d2

m1

m2

(m)

(mm)

6.00

1.00

65x105

65x165

65x130

65x110

55x105

55x105

7.00

0.60

65x105

65x185

65x130

65x130

55x105

55x105

0.80

65x105

65x185

65x130

65x130

55x105

55x105

0.40

65x185

65x185

105x130

105x130

55x105

55x105

0.50

65x185

65x185

105x130

105x130

55x105

55x105

0.60

65x185

65x185

105x130

105x130

55x105

55x105

0.80

65x185

65x185

75x130

75x130

55x105

55x105

0.40

65x165

65x185

65x130

65x130

55x105

55x105

0.50

65x165

65x185

65x130

65x130

55x105

55x105

0.60

65x165

65x185

65x130

65x130

55x105

55x105

7.00

Figure 6.1-16

Table: 6.1-4 Figure 6.1-17

8.00

Figure 6.1-18

9.00

10.00

Figure 6.1-19

11.0

0.80

65x165

65x185

65x130

65x130

55x105

55x105

0.40

65x165

65x185

65x130

65x130

55x105

55x105

0.50

65x165

65x185

65x130

65x130

55x105

55x105

0.60

65x165

65x185

65x130

65x130

55x105

55x105

0.40

65x165

65x185

65x130

65x130

55x105

55x105

Behaviour and Conception of Timber Structures

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Table: 6.1-5 span

inclination

(m)

(mm)

11.00 12.00

14.00

a

e1

e2

d1

d2

d3

m1

m2

m3

0.50

65x165

65x165

65x185

0.60

65x165

65x165

65x185

65x130

65x130

65x130

65x130

65x110

55x105

55x105

55x105

65x110

55x105

55x105

0.40

65x165

65x165

65x185

65x130

55x105

65x130

65x110

55x105

55x105

55x105

0.50

65x165

0.60

65x165

65x165

65x185

65x165

65x185

65x130

65x130

65x110

55x105

55x105

55x105

65x130

65x130

65x110

55x105

55x105

0.40

65x165

75x205

55x105

75x205

75x130

75x130

75x110

75x110

75x110

55x105

0.50

65x165

0.60

65x165

75x205

75x205

75x130

75x130

75x110

75x110

75x110

55x105

75x205

75x205

75x130

75x130

75x110

75x110

75x110

55x105

Table 6.1-6: Transformation at an inclined sectional grades to the inclinations m / m degree

Inclination

degree

m/m

Inclination

degree

m/m

Inclination

degree

m/m

Inclination m/m

1

0.02

24

0.45

47

1.07

70

2.75

2

0.04

25

0.47

48

1.11

71

2.90

3

0.05

26

0.49

49

1.15

72

3.08

4

0.07

27

0.51

50

1.19

73

3.27

5

0.09

28

0.53

51

1.24

74

3.49

6

0.11

29

0.55

52

1.28

75

3.73

7

0.12

30

0.58

53

1.33

76

4.01

8

0.14

31

0.60

54

1.38

77

4.33

9

0.16

32

0.65

55

1.43

78

4.71

10

0.18

33

0.68

56

1.48

79

5.15

11

0.19

34

0.70

57

1.54

80

5.67

12

0.21

35

0.73

58

1.60

81

6.31

13

0.23

36

0.75

59

1.66

82

7.12

14

0.25

37

0.78

60

1.73

83

8.14

15

0.27

38

0.81

61

1.80

84

9.51

16

0.29

39

0.84

62

1.88

85

11.43

17

0.31

40

0.87

63

1.96

86

14.30

18

0.33

41

0.90

64

2.03

87

19.03

19

0.34

42

0.93

65

2.14

88

26.64

20

0.36

43

0.97

66

2.25

89

57.29

21

0.38

44

1.00

67

2.36

90

ď&#x201A;Ľ

Figure 6.1-20: Membrane structures

Figure 6.1-21: The dimensioning of the elements

Behaviour and Conception of Timber Structures


Structural design is the methodical investigation of the stability, strength and rigidity of structures. The basic objective in structural analysis and design is to produce a structure

Structural Design

As with any other type of design, the evolution of the form of a structure is a

capable of resisting all applied loads without failure during its intended life.

creative act which involves the making of a whole network of interrelated decisions. It may be

A structural design project may be divided into three phases, i.e. planning, design and

thought of as consisting of two broad categories of activity: first, the invention of the overall

construction.

form and general arrangement of the structure and, secondly, the detailed specification of the precise geometry and dimensions of all of the individual components of the structure and of

Planning: Planning is the first step of project management philosophy of planning,

the junctions between them.

organizing and controlling the execution of the projects. Project planning and project scheduling is two separate and distinct function of the project management This phase involves consideration of the various requirements and factors affecting the general layout and dimensions of the structure and results in the choice of one or perhaps several alternative types of structure, which offer the best general solution. The primary consideration is the function of the structure. Secondary considerations such as aesthetics,

 Polygonal Frames  Solid-web members, support on different levels  Solid-web members, symmetrical  Lattice beams, symmetrical

sociology, law, economics and the environment may also be taken into account. Project planning is the function in which project and construction managers and their key staff members prepares the master plan. Then this master plan is put into time schedule by scheduling people which is called project scheduling. A project plan is mostly responsible for the success or failure of the project.

 There are floor joist systems where the joists are supported from a ring beam, such as I-section joists on metal hangers and metal-web joists.

Design: This phase involves a detailed consideration of the alternative solutions

defined in the planning phase and results in the determination of the most suitable proportions, dimensions and details of the structural elements and connections for constructing each alternative tructural arrangement being considered. Construction: This phase involves mobilization of personnel; procurement of

materials and equipment, including their transportation to the site, and actual on-site erection. During this phase, some redesign may be required if unforeseen difficulties occur, such as unavailability of specified materials or foundation problems.

 In these systems, lintels may not be required but the ring beam must be continuous over the opening.  The detailed design of a structure is normally carried out by a structural engineer (more probably a team of engineers) but, the overall form of an architectural structure is determined by that of the building which it supports and therefore principally by the architect (or architectural team).

Behaviour and Conception of Timber Structures

Radial arrangements -with tie of the top -with central column

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6.2 Structural Design

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112


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113 Solid-Web Systems

6.2.1 Proposal for a family house roof using steel elements

The process of structural design may be subdivided into two parts: there is a preliminary design stage, when the form and general arrangement of the structure are devised, and a second stage in which structural calculations are performed and the dimensions of the various structural elements are determined. Figure: 6.2-1

Figure: 6.2-2

-Radial-symmetrical arrangements - Symmetrical arrangement -Cumulative arrangement of radial frames Figure: 6.2-3 Proposal roof using steel elements

Behaviour and Conception of Timber Structures


Advantages •

Figure: 6.2-4

good in tension & compression - medium strength - yield stress 10-50MPa (as used)

low weight 600-1000kg/m3 - good strength to weight ratio

easy to cut to length on site - easily joined

reasonably inexpensive - except for special e.g. glulam

may be attractive - use for appearance

Disadvantages •

low Modulus of Elasticity 8000-15000MPa - large deformations - creeps

hard to create rigid joints - pinned and semi-rigid joints

strength time dependent - varies across grain (except for plywood, particleboard)

swells with moisture

small spans (except for glulam)

Figure: 6.2-5

Figure 6.2-6: Proposal roof using steel elements

Figure 6.2-7:

Figure 6.2-8: Proposal roof using steel elements

Behaviour and Conception of Timber Structures

STRUCTURAL ENGINEERING ROOM

What is the advantages and disadvantages of the timber when used as structural materials:

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114


STRUCTURAL ENGINEERING ROOM

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115 Example 6.2.1-1: Support of foundation micro piles, concrete - fcd=11.33 MPa, reinforcement – fyd = 356.52 MPa - the diameter piles:

15  cm

- The characteristic values of resistance of the stem in the

R

250kPa

soil G2 : - Coefficient of action conditions:

- root length:

lk

- Load-bearing capacity of the pile: Load on piles:

0.9

Vd

1.8

  lk R 

N= 550kN

Vd  190.852 kN

550

q

Load which attributable to 1 pilot on a length

2

m

kN m

 0.5

L is appoint an the (m) is simply h in mm and the rafter (a) is appoint an the (mm) Figure 6.2.1.1-2: The proposal dimensions piched roof

of 1 m: q  137.5

kN < m

=> satisfies

V d  190.852 kN

The calculation of slenderness ratio and coefficient of buckling, see table 6.3-2 if

Dimensioning reinforcement in piles: - the force of having one pilot transfer:

N d

- Proposal area needed of reinforcement: Ac As=(N- 0.85 Ac fcd) / 0.85 fyd



137.5  kN

2

4

According to the graph in figure 3.2.1-2 if if

reinforcement

75

1  0.8 

  75

  75

Jrequired

26 q lk

26 q 

3

q

4

cm

kN

l 2 k

cm

3

daN m

4

cm

Calculation of stress, coefficient of buckling  found in table 6.3-2.  II



N max M max   0.85 b h W

Behaviour and Conception of Timber Structures

2

3100

     100 

3100

Jrequired

1 2

1  0.8

Required moment of inertia

Figure: 6.2.1.1-1

1

 

if

sections, I propose only structural As = -5.616 cm2

2

Ac  176.715 cm

=> all the force is transferred concrete

  75

2

E

105 

E

106 

daN 2

cm N

2

cm

2


Example 6.2.1-2: Static scheme rafters as a simple beam, load calculation, dead load, snow

load, wind load q perpend

 g perpend  s perpend  w perpend a

1

kN m

Calculation of deflection:

In this case it must be allowed that the deflection is greater than the calculated deflection.

Calculation of reactions section and respectively. c, and 5

f

384

q l k

the maximum bending moment at mid-span

4

 f max

E II J x

Rac 5

f

84

f

M max l k

5 24

E II J x

 f max

 l k

E II h

qperpend Lk

Mmax

kN

2

2

2

qpepend Lk

Rca

2

kN m

8

The calculation of normal force

 f max

ak

The formula for the calculation of slenderness and radius of gyration for different

  45o

Nca

  45o

Nca

Rcb cos Rcb cos

cross-sectional shapes, for example, square, circle, hexagon, octagon, and is as follows.:

Load and normal force at point c L imin R icircle 4 

irect

0.289 b

i6angular

0.102 a

isquar

0.289 b

i8angular

qII

0.475 R

Nac

 gII  sII a

1

kN m

Nca  qII Lk

For beam, reaction and bending moment Ra

Rb

qo Lx 2

2

Mmax

Figure: 6.2.1.2-1

Behaviour and Conception of Timber Structures

qo Lx 8

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0.85 is a coefficient reductions- represents the ratio of stress in bending to stress in the compression.

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116


Example 6.2.1-3: Static scheme of rafter with overhangs from left

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117 Example 6.2.1-4: Rafter with overhanging ends of the right and of the left: Calculation of the load

Calculation of reactions section and respectively. C

Ra

q perpend L

Rc

kN

2

q perpend

 g perpend  S perpend  w perpend a

When a part of the length Lp of the rafters less than 0.8m

1

kN m

 g II  s II a

q II

The calculation of the maximum bending moment M max

2

q perpend L

Rb

As with beam console - bending moment at point a-

console

reactions

Ma

L

q perpend 

R ca

q perpend 

2

L 2

x

L

Ma L

Na

x

Figure: 6.2.1.3-1

R c x  q perpend 

Rc q perpend

x

2

 q perpend L 1 2

Mb L1 Mb L1

R ab q perpend

 x2   2

R ab x  q perpend  qII L1

Nb

qII L2

 q perpendL p  2

2

Ma

m

In both cases, normal force is allowed Na

 q perpend L 1

M max

2

M max

q perpend L 2

R ab

Ma

2

2

R ba

q perpend L k

R ac

 q perpendL 2

Mb

kN m

8

1

kN m

q II L

Ra

q perpend L p

Rb

q perpend L 2

2

M max

R ba x 2 

Ma

2

R ab

2

q perpend x 2 2

Mb

Behaviour and Conception of Timber Structures

x1

2

 q perpend L 2

R ba

q perpendx 1

R ab x 1 

Mb

Figure: 6.2.1.4-1 M max

2

 q perpend L 1 2

 q perpend L 1 2

Rab qperpen

x2

M a   L1 

  M b L 1   

 M b   L1 

  M a  L 1   Rba qperpen


 L k13  L k23   8 L k  



Md

L k2   M d    q perpend    2    Lk 

R cd

 q perpend L k2

R dc

2

 q perpend L k1

R da

R ad

2

 q perpend L k1 2

N dc N da N ad

Dx

N da  q II L k1

180o  2   90o

kN

 M d   Lk 

kN



 M d   Lk 



kN

L is appoint an the (m), h is in mm and the distance of rafters (a) is appoint an the (mm) Figure 6.2.1.5-2: The proposal dimensions of the roof truss, and the L (m) and a in (mm)

 M d   Lk 



 D perpend sin  sin

 D perpendsin90o sin

N cd  q II L k2 N dc  D II

 M d   Lk 

2

D II

kN



 q perpend L k

D perpend

Figure: 6.2.1.5-1

kN m

Rcd

kN

kN

  45o

N cd

  45o

N cd

Rcd

Ncd

Rcd cos

cos

kN

Where L is appoint an the (m) will then h (mm). Figure 6.2.1.5-3: Proposal dimensions purlin of the truss system with two stationary stands

90o  2 

Behaviour and Conception of Timber Structures

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Example 6.2.1-5: Rafter as a continuous beam

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118


Example 6.2.1-6: Buckling calculation

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119

Example 6.2.1-7:

Figure: 6.2.1.6-1 Calculation and assessment of the strength of the strut N

84.15kN

l1

4.5 m

h

2.8 m

2

l

2

l1  h

l

5.3 m

The axial force in the element h l

sin

S

N

S

sin

159.28kN

Figure: 6.2.1.7-1

cross-sectional dimension

2

bp 22cm hp 22cm Ap bp hp Ap 0.0484m Navrh Slenderness ratio and the value of which can be found  in Table 6.3-2 l 0.28 b p

86.039

Calculation of stress at an angle: t

2.39

 S

 II

Ap

1

 II

0.42

F b  II

 8 t

t

Truly stress at an angle will:

S

N

 II  7.8357MPa

Ap

x

5

 pIId   pIId   pperpendd sin

 p

7.8655

Or in the graph of Figure 6.3-2, calculate the stress as follows 

h

Calculation of the thickness t, respectively. t1:

stress in cross-section in the pressure  II

F 56b

F

cos 2

p

N b  IId

F cos

x

b  II

F cos

 8 t

We see the result (stress in cross-section in the pressure) are almost the same. Example 6.2.1-8:

Calculation of the collet and force in the collet will be: tan

H

h

tan  0.6222

l1

P tan

bhamb

H  52.36kN

22 cm

Ahamb

hhamb

22 cm

hhamb bhamb

hamb = lhamb / 0.289 bhamb

hamb = 78.64

hamb =  H / Ahamb

hamb = 2.152 MPa

lhamb

Ahamb

5 m

0.0484m2

 = 1.99

Or in the graph of figure 3.2.1-2, calculate the stress as follows 1

0.489

 hamb

H

Ahamb

 hamb  2.2123MPa

We see that the result is almost the same. Figure: 6.2.1.8-1

Behaviour and Conception of Timber Structures

t

t1

 p b

t

cos F

70b


t

4 t2

F 70b

 II

kx Fut

h

M Wut

Aeo

Mx

Kx e

M

MA  Mx

 pII   pII   pperpend sin

p

4

MA

0.85

Calculation of stress in an inclined section 

p

2

 pII   pII   pperpend sin

The calculation of the thickness t1, t2 respectively. values of the forces F1, F2, and the

Figure: 6.2.1.8-3

thickness of the an inclined t2

t1

N1

F2 cos

t2sikme

 p b

F1 cos

t2 cos

2

t1sikme

 p b

F1 cos

F2

t2sikme b  p

F1

t1

t1sikme b

1 8

1

g kN m

 L2

Jpot

3

3

26 gL

cm

 pII



Dx A

M Wx

0.85

daN 2

cm

Where Dx is compressive normal force Example 6.2.1-9: Design dimensions of the timber elements that are subjected by bending

cos

F1

F  F2

M

moment as follows: Data

x1

F1 cos 8 II

x2

Kx b II

M

6 kN m

30 deg

sin    0.5

30 deg

b h

10 MPa

cos     0.866

1.4

tan    0.577

Calculates the moment in the x, y Mx

M sin  

Mx  3 kN m

M cos   

My

My  5.196 kN m

The calculation of section modulus and of the section height Mx  c My

Wx

3

Wx  0.00103m

h

3

6 Wx c

I suggest h = 0.22m b

h

b

c

0.157m

I suggest b = 0.16m cst

h b

cst  1.375

The cross-section will have dimensions and sectional module: 150mmx220mm

Wx

2

b h 6

3

Wx  0.00129 m

Figure: 6.2.1.8-2

Behaviour and Conception of Timber Structures

h

0.205

STRUCTURAL ENGINEERING ROOM

h

Department of Architecture

120


6.3 Concentric compression members

Elements of concentric loaded assessed for simple compression or buckling. Calculation of the concentrated elements for buckling appropriate already at low slenderness elements practically in slenderness =l/i > 10.

b a fcod

the width of the rectangular cross-section the dimensions of the square section the design strength in compression wood Six an angular cross-section i min

Table 6.3-1: calculating the necessary area and moment of inertia of the column square section rectangular cross-section circular cross-section

lf

The required cross-sectional area

i l f  21.7a

l f  18.75D A

N f cod D

 0.001l f

2

lf 18.75

a

Eight an angular cross-section

0.102a

i min

0.102R

 75 lf  21.7b

 0.001l f

f cod

A

1.13

D

A

N

2

A

N

A b

c

h

 0.001l f  h

c A

b

21.7

b

Figure 6.3-1: Buckling of compression members

2 1

f cod

And must be lf a

b c

c

Buckling length therefore generally represents a distance of two inflection points at which deviates element in the shape of one half sinusoid, i.e as an element mounted articulated at the end of this length figure 6.3-1.

lf 21.7

Table 6.3-2: Coefficient of buckling  

The moment inertia

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121

l f  18.75D

lf

4

2.13

D i min

J

lf 18.75

D

 75

l f  21.7a Jrequired

D

i

2

lf  fcod 3100 N

4

a 1.86 J And must be lf a 21.7

i min

l f  21.7b

0.289a

4

i

the radius of gyration

lf

the buckling length of the column of figure 6.3-1

D

the diameter of the circular cross-section

h

the height of the rectangular cross-section

4

b

1.86

b i min

J c

lf 21.7 0.289b

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170

Koeficient  0 1 1,01 1,03 1,08 1,14 1,24 1,42 1,64 2,07 2,62 3,22 3,9 4,64 5,45 6,31 7,25 8,25 9,32

1 1 1,01 1,03 1,08 1,15 1,25 1,44 1,67 2,12 2,68 3,29 3,97 4,72 5,53 6,4 7,35 8,35 9,43

2 1 1,01 1,04 1,09 1,16 1,26 1,46 1,7 2,17 2,74 3,36 4,04 4,8 6,62 6,5 7,45 8,47 9,55

3 1 1,01 1,04 1,09 1,17 1,28 1,48 1,74 2,22 2,8 3,42 4,12 4,87 5,7 6,59 7,55 8,51 9,66

4 1 1,01 1,05 1,1 1,18 1,3 1,5 1,78 2,27 2,86 3,48 4,19 4,95 5,79 6,69 7,65 8,67 9,78

Behaviour and Conception of Timber Structures

5 1 1,02 1,03 1,1 1,19 1,32 1,52 1,82 2,33 2,92 3,55 4,26 5,03 5,87 6,78 7,75 8,78 9,88

6 1 1,02 1,06 1,11 1,2 1,34 1,54 1,87 2,39 2,98 3,62 4,33 5,11 5,96 6,88 7,85 8,88

7 1 1,02 1,06 1,11 1,21 1,36 1,56 1,92 2,45 3,04 3,69 4,41 5,2 6,05 6,97 7,95 9

8 1 1,02 1,07 1,12 1,22 1,38 1,58 1,97 2,5 3,1 3,76 4,48 5,28 6,13 7,06 8,05 9,12

9 1 1,02 1,07 1,13 1,23 1,4 1,61 2,02 2,56 3,16 3,83 4,56 5,37 6,22 7,15 8,13 9,22

 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170


slenderness   = lo/i

10 20 30 40 50 60 70 80 90 100 120 140 160 180 200

Lo/b

Lo/D

rectangular crosssection

circular crosssection

2,9 5,8 8,7 11,5 14,4 17,3 20,2 23,1 26,0 28,9 34,6 40,4 46,2 52,0 57,7

2,5 5 7,5 10 12,5 15 17,5 20 22,5 25 30 35 40 45 50

Value 

0,98 0,93 0,85 0,76 0,67 0,58 0,51 0,44 0,38 0,33 0,26 0,20 0,16 0,13 0,11

Figure 6.3-2: Buckling coefficient of wood

Figure: 6.3-3

Behaviour and Conception of Timber Structures

0,99 0,96 0,92 0,86 0,80 0,74 0,67 0,61 0,55 0,50 0,41 0,34 0,28 0,24 0,20

1,0 0,98 0,96 0,93 0,89 0,85 0,80 0,76 0,71 0,67 0,58 0,51 0,44 0,38 0,33

0,93 0,76 0,58 0,44 0,33 0,26 0,20 0,16 0,13 0,11 0,08 0,06 0,05 0,04 0,03

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Table 6.3-3: Value  for different values of slenderness

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123

b  h b

5

1

 7

2

6 Mx 2

 dovoleneh

b  h

Wmax

Mx Wx

My Wy

c

4

1

 7

Wx

c

Wy

3

c

Calculation fixed moment at point B

Jmax

MB

1.4  1.6

 II

3 2

2

R1  R2

2

b h

For rectangular cross-section value c calculated as follows: c

2

2

b h 6

b h 6

h b

a1

N1

2

2

MB

N2

a1  a2 The calculation of the reaction and the bending moment Lx 1 q  q y  s y  w y  kN m

the table below: I 140-I 220

I 240-I 600

U120 to U 160

U180 to U320

8

9 -10

6

7-8

2

Mx qy

qx l 8

My

Wx  l1x    l2x   m  2

a1  a2

2

q

l

R1

R2

Ps

qsneh L1

2

2

kN

Mmax

q

l

kN m

8

Pp

0.5

Ps

sin The minimum moment of inertia of the column:

shape The coefficient c

a2 2

1.2  1.4

Table 6.3-4: The values of coefficients for the different cross-sectional forms are presented in Cross-sectional

N1 a1  N2 a2

qy ls

2

ls

8

qx

l  2 c

Wpot

Mx  My c

 pII

qy  l1x    l2x   m  2

Figure: 6.3-5

2

Jmin Jmin

Figure: 6.3-4

nPs H 2

 EII

4

Jmin

a

12

a

4

the minimum moment of inertia of the column

n

safety factor n = 6

H

total height of the column

F

centric force is in a pillar at the top surface

Behaviour and Conception of Timber Structures

12Jmin

2

MB


0.16m

hs

0.20m

Ls

3.2m

The calculation of slenderness the column with a safety factor 

Ls

  64.879

0.289bs

  75

nehrozi vzper

1.61

Stress of timber in compression parallel to the fibers in the wood is class II 8.5MPa

 pIIdovolene

A

A  0.029m2

bs hs

Calculation of the maximum compressive force in the pillar  pIIdovolene



Fmax

A pIIdovolene

Fmax

A

Fmax  161.053kN

Compressive force applied to the top surface of wooden column F max Figure: 6.3-6

161kN

The calculation of the stress on the lower surface of the column Fmax

 stlpkolmo

Example: 6.3-1

bs hs

 stlpkolmo  5.59MPa

 stlpkolmo   pdovolene

It must be designed pad under the posts of hard wood calculation procedure is as follows Lpodlozka

Fmax

Lpodlozka  0.503m

bs  pdovolene

Pad behaves as a cantilever, the bending moment calculation is M

Fmax  Lpodlozka  bs 

M  6.905kNm

8

Necessary section modulus will M

Wpotrebne

 sdovolene

Wpotrebne  0.0006905m3

Shear stress a

 Lpodlozka  bs 2

a  0.172m

Figure: 6.3.1-1 Calculation of carrying capacity of the column, dimensions and length of the column:

Behaviour and Conception of Timber Structures

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bs

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124


Example 6.3-2: Wooden beams with dimensions bxh, loaded with vertical load laid on

shear force

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125

Fmax a

Tmax

wooden column as shown in the figure below, it is necessary to calculate the required width of

Tmax  54.9kN

Lpodlozka

the support of beam on column.

b

Assessment of shear stress  sIIstlp

3 2

Tmax

 sIIstlp  3.198MPa

bs d

 sIIstlp   sIIdovolene

d

2

Tmax bs  sIIdovolene

d  0.429m

22 cm

 ckolmo

2.5 MPa

F

18 kN

F

2

Apotr  42 cm

 ckolmo

Support width xpotr we find the following b

Unless the thickness of the wooden base is large and it is unreasonable then it can be replaced

h

We calculate the area of the imposition, where F is the concentrated loads in kN Apotr

According Permissible Stress  sIIdovolene the thickness will be: 3

16 cm

16 cm

xpotr

Apotr

xpotr  2.3333 cm

b

We suggest xpotr = 2.5 cm

with a steel washer plate width will correspond to the width of the tie beam and the length of the plates will be as follows: X

Fmax  pdovolenebs

X  0.503m

Calculation of the thickness of the plates have to check the stress  s   sdovolene

140MPa

hplatnicky  8mm

Figure: 6.3.2-1

assessment of stress s

Fmax bs hplatnicky

Example 6.3-3: Top chord of truss beam be designed dimensions of the elements of two parts  s  125.781MPa

 s   sdovolene

on the line which is 1.80 meters subjected to compressive force of 220kN.

Calculation of stress in compression perpendicular to the fibers

Figure: 6.3.3-1 Data: section height h, l is the length of the element which is shown in figure, F is the axial Figure: 6.3.1-2

force in a given bar, preferably h

160mm

l

1800mm

F

220 kN

Behaviour and Conception of Timber Structures

 cII  8.5 MPa

m

2


We calculate the radius of gyration, and slenderness a rod so that we can gain coefficient iy

0.289 h

iy  0.046 m

y

l iy

 y  38.927

 F

2

Apotr  0.029 m

 cII

Apotr

Apotr

h2 h1

The longest segment to buckling will be:

s1

1

b  0.091 m

2 h

i1  0.0289 m

s1max  60 i1 

This is the width of the cross-section as follows: b

0.289 b

1.13

Required area will: Apotr

i1

s1max

s1max

3

s1max  1.734 m

a  3 b

 1  20

i1

60 i1

s1  0.578 m

3 s1

sk

C coefficient expresses the way joints in wooden elements

c=3

I suggest 2x100 / 180 mm, from the true area will then be: b

100 mm

h

180 mm

Askut

2 b h

2

Askut  0.036 m



The clear distance between the elements a

a  0.8 b

80 mm

 y

 w  y  6.906 MPa

 y   cII

Direction x-x

ok

V

3

( 2 b  a) h 12

4

Ix  0.0003293 m

ix

Askut

T ix  0.0956 m

l ix

 w   cII

ok

1.05

 w F

 w  6.417 MPa

Askut

 w F 60

V  3.85 kN

Vs1 2 a

T  13.908 kN

Minimum anchoring nails into the wooden cross section in the realization connection having 8  klinca

Slenderness in the x-x x

w

A wooden base then transferred greatest shear force

radius of gyration Ix

  y

The connection between pad wooden elements waist

The cross-sectional moment of inertia Ix

   20.353

Control the stress

The control stress  F Askut

m 2  x  c   1 2

 klincaLklinca

 klinca

8mm

Where the design with dimensions nail

 x  18.821

For element with a width of 100 mm the radius of gyration will be:

hhlbkaklinca

8  klinca

hhlbkaklinca  0.064 m

Behaviour and Conception of Timber Structures

Lklinca

260 mm

STRUCTURAL ENGINEERING ROOM

m is the number of cross-sections within the rod of direction y-y

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126


Example 6.3-4: Assessment of wooden column, wooden column has a length l, must take the

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127

load F, which may be a column dimensions and provided pure compression

Figure: 6.3.4-1 The data: F is the axial force applied to the upper surface of the column and L is the total length of the column F An

l

60 kN F

4200mm

 cperpend

2 MPa

8.5 MPa

 cII

2

 cperpend

An  300 cm

I suggest column with dimensions a

a

180 mm

180 mm

The actual cross-sectional area Ar

2

a a

Ar  0.0324 m

Control the stress  Dperpend

F Ar

 Dperpend  1.852 MPa

 Dperpend   cperpend

We calculate the radius of gyration and the slenderness of the column as a basis for the expression of buckling factor. Provided on the buckling in the middle column i

0.289 a

i  0.052 m

l i

  80.738

2.12

Control the stress 

 F Ar

   3.926 MPa

    cII

Behaviour and Conception of Timber Structures


7.1 Structural steel

Properties of materials

Standard strength grades for structural steel are given in terms of a nominal yield

The dimensioning of composite slab is based on the instructions laid out in Eurocode 4 (SFS-ENV 1994-1-1). The dimensioning of composite sheets adheres to the instructions laid out in Eurocode 3, part 1.3 (SFS-ENV 1993-1-3:1996). The load specifications and safety coefficients are based on Eurocode 1 (ENV 1991-1:1994).

strength fy and ultimate tensile strength fu, these values may be adopted as characteristic values in calculations. The grade used in worked examples here is S 355, for which fy = 355 MPa

fu = 510 MPa

for elements of all thicknesses up to 40mm.

Concrete

The density of structural steel is assumed to be 7850 kg/m3. its coefficient of linear

Concrete reaches its maximum compressive stress at a strain of between 0.002 and

thermal expansion is given as 12x10-6 per oC. but for simplicity the value 10x10-6 per oC (as for

0.003, and at higher strain it crushes, losing almost all its compressive strength. It is very brittle

reinforcement and normal-density concrete) may be used in the design of composite structures

in tension, having a strain capacity of only about 0.0001(i.e. 0.1mm per metre) before it cracks.

for buildings.

The maximum stress reached by concrete in a beam or column is little more than 80% of its cube strength. Steel yields at strain similar to that given for crushing of concrete, but on further straining the stress in steel continues to increase slowly, until the total strain is at least 40 times the yield strain.

Reinforcing steel Standard strength grades for reinforcing steel will be specified in terms of

The Characteristic compressive strengths at 28 days are fck = 25MPa (cylinder) and fcu = 30MPa

a characteristic yield strength fsk. Values of fsk used in worked examples here are 460MPa, for

(cube). All design formulae use fck, not fcu.

ribbed bars, and 500MPa, for welded steel fabric or mesh. It is assumed here that both types of reinforcement satisfy the specifications for high bond action and high ductility.

Mean tensile strength, fctm = 2.6MPa

The modulus of elasticity for reinforcement, Es, is normally taken as 200GPa, but in composite

With upper and lower 5%: fck0.95 = 3.3MPa

section it may be assumed to have the value for structural steel, Es = 210GPa.

Fck0.05 = 1.8MPa Basic shear strength, Rd = 0.25fck0.05 = 0.30MPa -6

7.1-1 Profiled steel sheeting o

Coefficient of linear thermal expansion, 10x10 per C.

Steel composite sheets are profiled steel sheets designed for use in concrete-steel 3

Normal-density – concrete typically has a density, , of 2400kg/m , it is used for composite columns and web encasement in worked examples here, but the floor slabs are

composite slabs. The composite sheets serve as part of the tension reinforcement in the lower slab surface in the service state and the ultimate limit state. In casting, the sheet functions as a mould.

constructed in lightweight-aggregate concrete with density  = 1900kg/m . The mean secant

The design of Steel sheets results in a smooth and architecturally useful lower surface in slabs, and

modulus of elasticity is given for grade C25/30 concrete as

plastic-coated composite sheets can serve as a finished ceiling without using suspended ceilings.

3

Ecm = 30.5 ( /2400)2 GPa

with in kg/m3 units

The profile of the sheet forms the necessary installation system for the hangers below the slab. A composite slab is a concrete structure formed by concrete and a composite sheet equipped with sufficient bond properties. The composite sheet acts as tension reinforcement, replacing a part or all of the bar reinforcement in the slab. As the concrete reaches sufficient

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7. Profiled steel sheeting

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128


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129 strength after casting, the composite sheet and concrete create a combined effect to anchor the

The serviceability limit state design shows that the deflection of the slab does not exceed the

bond ribs to the concrete.

defined serviceability limits.

This material is available with yield strengths fyp ranging from 235 MPa to at least 460 MPa, in profiles with depths ranging from 45mm to over 20mm, and with a wide range of

7.2 The permanent load (dead load) are the weights of the structures and its finishes. In

shapes. These include both re-entrant and open troughs. There are various methods for

composite members, the structural steel component is usually built first, so a distinction must

achieving composite action with a concrete slab. This material is commonly used as permanent

be made between load resisted by the steel component only, and load applied to the member

formwork for floor slabs in buildings, then known as composite slabs. As it is impracticable to

after the concrete has developed sufficient strength for composite action to be effective. The

weld shear connectors to material that may be less than 1mm thick, shear connection is provided

division of the dead load between these categories depends on the method of construction.

either by pressed or rolled dimples that project in to the concrete, or by giving the steel profile

Composite beams and slabs are classified as propped or un-propped.

a re-entrant shape that prevents separation of the steel from the concrete.

In propped construction, the steel member is supported along its length until the concrete

Sheets are normally between 0.8mm and 1.5mm thick, and are protected from corrosion

has reached a certain proportion, usually three-quarters, of its design strength. The whole of the

by zinc coating about 0.02mm thick on each face. Elastic properties of the material may be

dead load is then assumed to be resisted by the composite member. Where no props are used, it

assumed to be as for structural steel.

is assumed in elastic analysis that the steel member alone resists its own weight and that of the

Prior to the casting of the slab, the composite sheet functions as a mould and installation deck. For construction, the composite sheets must be designed for the following loads:

formwork and the concrete slab. Other dead load such as floor finishes and internal walls are added later, and so are assumed to

- Self-weight of the composite sheet

be carried by the composite member. In ultimate strength methods of analysis it can be assumed

- Construction loads: minimum of 2.0 kN/m line load placed parallels with the propping

that the effect of the method of construction of the resistance of a member is negligible.

beams of the slab. Line loads can be distributed with distances of 1 m. In casting, the composite sheets must be dimensioned for the following loads: - Self-weight of the composite sheet

The principle vertical variable load in a building is uniformly distributed load on each floor. For offices, its characteristics value as

- Self-weight of the concrete

qk = 5.0 kN/m2

- Live point load 1 kN/m2

for checking resistance to point loads a concentrated load

As the spans are determined on the basis of the strength, the concrete weight is considered as a live load. In the load case, the line load, parallel with the propping, has been added to the point where

Qk = 7.0 kN

the combined effect with the self-weight of the concrete is most significant. The calculation

These rather high loads are chosen to allow for a possible change of use of the building. A more

method requires connecting the sheets with end supports.

typical loading qk for an office floor is 3.0kN/m2.

The composite slab is a structural slab in the longitudinal direction of the composite sheet. The slab can be designed as a single-span or a continuous composite slab. When the anchoring strength is sufficient, the cross-section As of the entire composite sheet reaches the yield stress fyd before the breaking mechanisms of the concrete start to become apparent. The design for the ultimate limit state shows that there is sufficient safety against bending failure,

The principle horizontal variable load for a building is wind. They usually consist of pressure or suction on each external surface. Wind load rarely influence the design of composite beams, but can be important in framed structures not braced against sides way and in all tall buildings.

shear failure and anchorage failure.

Profiled Steel Sheeting


lie in the web, the steel flange, or the concrete flange of the member. The theory is not in

The concrete members of the composite slab are designed and produced in accordance

principle any more complex than that used for steel I – beam.

with valid instructions for concrete structures as well as the relevant norms. The cube strength of

Composite slabs can be reinforced with hot-rolled and cold-rolled reinforcement steel

the concrete used must be at least 25 MN/m2. In composite slabs, it is recommended to use concrete

with a yield limit ReL between 400-500 N/mm2. The minimum breaking stretch for hot-rolled steel

with as small water/cement ratio as possible. When using concrete agents, it must be made sure

is 12%, and the minimum total elongation under maximum load is 5%. The minimum total

that the agents do not cause damage to the coatings of the composite sheet. It is not allowed to use

elongation under maximum load for cold-rolled steel is 3.5%.

agents containing chloride salts. The hardened concrete does not cause corrosion problems to the 7.3-1 Ultimate limit state

galvanisation and plastic coating of the sheets. The minimum thickness of the composite slab when using a Steel composite sheet is 100 mm, meaning that the design thickness hc of the concrete slab on top of the composite sheet must be at least 50 mm. The granular size D of the concrete must meet the requirements D  hc/3, D 32 mm. The steel designer will be familiar with the elementary elastic theory of bending, and the simple plastic theory in which the whole cross-section of a member is assumed to be at

The design fatigue resistance of a structure is determined using equations based on the following expression: Rd = R(ad1, ad2,… Xd1, Xd2), where adi = design value of the structure dimensions. In the case in question, the nominal values (anom) of the dimensions are used as the design value.

yield, in either tension or compression. In support areas, continuous slabs are designed as reinforced-concrete structures, with traditional reinforcement on the upper surface of the slab. In this case, the composite sheets do not act as compression reinforcement. It is possible to further reinforce the span areas of the composite slab by adding reinforcing bars to the lower slab surface, which then serve as span reinforcement with the composite sheets. In case of a fire, the reinforcement bars alone act as the span reinforcement. Both theories are used for composite members, the differences being as follows: 1. Concrete in tension is usually neglected in elastic theory, and always neglected in plastic

Xdi = Xki/M = design value of the structure material. In the case of the composite slab, functional reference formulas based on the above expression can be as follows: Rd = R(Xk/M , anom), Rd = Rk/R, (determining the casting spans of the composite sheet), in formulas M = partial safety coefficient of material. - concrete:

theory. 2. In the elastic theory, concrete in compression is transformed to steel by dividing its

- reinforcement:

breadth by the modular ratio Es/Ec.

- composite sheet:

to be 0.85fck, where fck is the characteristic cylinder strength of the concrete.

with the rectangular-stress-block theory outlined above. The basic difference from the elastic behaviour of reinforced concrete beams is that the steel section in a composite beam is more than tension reinforcement, because it has a significant bending stiffness of its own.

M = 1.10, structural class 1 M = 1.20, structural class 2

3. In the plastic theory, the equivalent yield stress of concrete in compression is assumed

The concrete designer will be familiar with the method of transformed sections, and

M = 1.35, structural class 1 M = 1.50, structural class 2

M = 1.10

R = 1.1 = partial safety coefficient of fatigue resistance. Rk = characteristic value of fatigue resistance (experimental design). The ultimate limit state verifies the following load combination: G,jGk,j+Q,1Qk,1+Q,i0,iQk,i  1.35Gk,j

The formulae for the elastic properties of composite sections are more complex that those for

where

steel or reinforced concrete sections. The chief reason is that the neutral axis for bending may

Gk,j =

nominal values of dead loads

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Department of Architecture

131 Qk,1 =

nominal value of one live load

Qk,i =

nominal values of live loads

____________ kx =  [(c)2 + 2( c)] - (c) = relative height of the compressed part

G,j = 1.2 = partial safety coefficient of dead load

c = (As + Ar ) / (b d)

Q,1 = 1.5 = partial safety coefficient of live load Qk,1

= (1+Ø) Es / Ecd

Q,i = 1.5 = partial safety coefficient of live load Qk,i

x

= kx d = height of the compressed part in a cracked cross-section in the elastic state

0,i = 0.7 , for load classes A-D and F-G corresponding to SFS-ENV 1991-1-1, and for Plastic moment

snow loads =

1.0

, for load class E corresponding to SFS-ENV 1991-1-1

=

0.5

, for wind loads

Mu = plastic moment calculated on the basis of the ideal-plastic stress distribution (design strengths), bending strength =  u b d2 0.85 fcd

Other partial safety coefficients and combination coefficients can also be used if they do not compromise the total safety of the structure. In the ultimate limit state, the composite slab is designed for the following breaking types:

 u = relative bending strength = (1 -  /2)

1. Bending failure at the maximum moment 2. Shear failure at the end of the slab (Concrete shear failure may develop in heavily loaded, short slabs.)

7.3-3 Anchorage failure, anchorage fatigue resistance and anchorage strength In cases where the fatigue resistance to bending cannot be calculated on the basis of the plastic force balance, the anchorage fatigue resistance of the slab limits the level of the fatigue

7.3-2 Bending failure and bending strength

resistance to bending.

In tough slabs, the sheet cross-section reaching the full design strength fyd, the bending strength can be calculated on the basis of the plastic force balance.

In tough slabs, the bending strength MRd is always between the limits Mpa and Mp.Rd, where Mp.Rd

Variables deducted from the geometry and material properties d

plastic cross-section

= mechanical useful height of the slab

Mpa

ds (EA)s + dr (EA)r

Between the presented limits, the calculation is performed as a function of the anchorage durability

(EA)s + (EA)r

u.Rd of the connection of the sheet and the concrete.

= d - ecp

The breaking mechanism is assumed to change from anchorage failure to plastic bending failure

= distance of mass centres between the slab parts

when the anchoring length Ls exceeds the limit size Lsf

Mechanical reinforcement ratios Arb fryd + As fyd 

is the bending strength of the ribbed sheet only, based on the plastic theory

= ———————— ei

is the upper limit value, (  Mu ), of the bending strength, based on the

Lsf

= Ncf / (b u.Rd ),

Ncf

= min (As fyd,b hc 0.85 fcd) =greatest possible compression stress

= —————— with the lower surface under tension 0.85fcd b d

When Ls  Lsf , the shear connection is complete and a full plastic bending durability Mp.Rd can

Art fryd  t = 1,1————— with the upper surface under tension 0.85 fcd b drt

resultant of the concrete. be reached. When Ls < Lsf , the shear connection is incomplete, and the bending strength MRd is smaller than Mp.Rd, which can be calculated with the below formulas by setting Ls = Lsf.

Profiled Steel Sheeting


simultaneously; a tough slab can support an increasing load even after the start of the slip - A tough anchoring is one where the initial slip occurs before 90% of the breaking load has been reached. In SteelComp slabs, the anchorage has been proven to be tough by experiment. The anchorage strength u.Rd is an experimentally defined nominal shear strength on a design area formed as the product of the design width b and shear span Ls. The anchoring strength values are presented in Table 7.3-1.

Figure 7.3-1: Balance mechanism and effect of anchorage fatigue resistance on the highest moment reached with various values for the shear span MRd= z1 b Ls u.Rd + Mpa + z2 Arb fryd  Mp.Rd = Mu

Table 7.3-1: Nominal and design values of anchoring strengths for different sheet types (MPa) ————————————————————————————————— SteelComp

Galvanised sheet

sheet thickness

0.7

0.9

1.1

Plastic-coated sheet 0.7

0.9

1.1

—————————————————————————————————

z1

= ds - 0,5x

z2

= dr - 0,5x

u.Rk

0.601 0.844

0.806

0.5

0.626

0.636

x

= (b Ls u.Rd + Arb fryd ) / ( b 0.85 fcd )  hc

u.Rd

0.481 0.675

0.645

0.4

0.501

0.509

ds

= distance between the mass centre of the composite sheet and the

u.Rk = Nominal value of anchorage strengths

upper edge of the slab dr

—————————————————————————————————

= distance between the mass centre of the bar reinforcement and the

u.Rd = Design value of anchorage strengths

upper edge of the slab Arb

= cross-section area of the bar reinforcement for the design width b

fryd

= design strength of the bar reinforcement

fcd

= design strength of the concrete.

The safety coefficient of the design value of the anchorage strengths is 1.25.

7.3-4 Shear failure

In order to achieve sufficient safety against anchorage failure and bending failure, the design moment Md of the slab must not exceed MRd in any part of the slab. In edge spans where the slab ends to a free support, the slip between the sheet and concrete determines how the slab works. With a small stiffness of the anchorage connection, the slip can

The shear fatigue resistance Vcu of the slab for the design width b is calculated with the formula Vu

= 0,30 k (1 + 50 ) b d fctd ,  = (As + Arb ) / (b d),

k

= 1,6 - d  1, d [m]

have a significant effect on the bending in the service state. ENV 1994-1-1, part 1 defines brittle and tough behaviour on the basis of the slip properties: - Initial slip is defined to be 0.5 mm at the end of the edge span

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- In a brittle slab, the sheet slip can start and reach the breaking load almost

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132


7.3-5 Design of the slab for point and line load

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The distribution width of the point or line load supported by the slab is taken into account as follows: If the direction of the (point or) line load is parallel with the ribs of the composite sheets, the distribution width of the load can be calculated with the following formula: bm

= bp +2 (hc + hf),

Lp

is the distance of the centre of the force to the nearest support

L

is the span

When the main reinforcement of the slab has been designed bearing in mind the distribution widths of the point and line loads presented above, the distribution bar reinforcement of the slab must be at least 20% of the area corresponding to the necessary capacity of the main reinforcement (including the composite sheet). The bar distance of the distribution bar reinforcement must reach the lowest of the following values: s < 2.5d or 400 mm, where d is the mechanical useful height.

where bm

is the point or line load width across the span

hc

is the height of the upper part of the composite sheet in the slab

hf

is the thickness of the surface concrete (layer) of the slab

Figure 7.3-3: Shear-head design in slab Figure 7.3-2: Design of the slab for point and line loads The load width bp can be used as the distribution width bm when the point or line load is perpendicular to the ribs.

7.3-6 Reinforcement of the support The reinforcement of the permanent support of a composite slab is designed exactly the same

The functional width of the slab may not exceed the following values:

way as the reinforcement of a concrete slab. Distribution reinforcement can be used as necessary.

a) For bending

In fire design, the reinforcement of the upper surface must be placed through the entire slab, if the

- in single-span slabs and in outer spans of multi-span slabs

composite slab span does not contain reinforcement that could serve as support in a fire situation.

bem

= bm + 2 Lp[1-(Lp / L)]  slab width

- in multi-span slabs, inner spans bem

= bm + 1,33 Lp[1-( Lp / L)]  slab width

b) For transverse shear force bev where

= bm + Lp [1-( Lp / L)]  slab width,

Reinforcement of the span The span reinforcement of a composite slab must be designed in accordance with the calculation principles. The reinforcement of the slab can be positioned either between the ribs of the composite sheets or at a distance required by the protective concrete layer from the lower surface of the slab. In this case, only single bars can be used. Prefabricated reinforcement is

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installed with the distribution reinforcement placed directly on the composite sheets, and the main reinforcement is positioned between the ribs on level with the upper surface of the ribs. Thus the reinforcement type has an effect to the bending capacity of the slab span see Figure 7.3-4.

Figure 7.3-4: Positioning options for slab reinforcement. 7.3-7 Typical structural details of composite slabs The following standard structural details can be applied in the structural design of Steel composite sheet slabs. Connection between the composite sheet and a steel beam Composite sheets must always be positioned on supports with a minimum width of 50 mm in the longitudinal and lateral direction of the sheets. In the longitudinal direction of the profile, the sheets are fastened with a spacing of approximately 500 mm, and in the lateral direction of the profile, at every other rib bottom to the support structure either by bolting or by using self-tapping screws. The sheets can be positioned on the support either by leaving the sheets on the minimum support surfaces or by placing the sheet ends against each other. The sheets cannot be interlaced with each other. A protective lining must be placed at the sheet ends as a cast protection to prevent wet concrete from spreading inside the profile. Figure 7.3-5: Connecting composite slab to steel beam. Connecting composite slab to nonbearing partition wall structure

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12.7 15.8  d 0.4 0.5  D 25.0 31.3   hD 8.00.3 8.00.3   d 3.0 0.1 4.0 0.1  1  v 1.5 0.2 1.5 0.2 

 

b  bs   12 

  31.3  0.3  12.0   0.1  4.0   0.2  1.5  18.2

0.5

B3  2 2 b  12  h1  k  h2  bn  1 b   B1  B2   2

b  6  h1  k  h2 

bn

 

bs 

   h1  k  h2 h 

bs   b bs   12     h1  k  h2 h  

b  0.3  l b  B 

b 1

b b

bs 

1 2

2

 

  12 

0.3  l

 B1  B2 bs  h

B3

   h1  k  h2  2 

b

y

h

5

2  bm 

2 3

 bk  2  bm   6 

resp

b b

Figure 7.3-7

Figure 7.3-6: Connecting composite slab to steel beam. Connecting composite slab to nonbearing partition wall structure

Profiled Steel Sheeting

l

10 B 2

bk  2  bm 

 bm  bm  

h

    h1  k  h2  


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      1

zo

zc

r1  3  d

r2  1.5  d

r3  3  d

r1  40  mm

r3  6  h

r3  40  mm

Ai

Ai

Ao 

zco  zo zso  zo 

Ii

Wod

Woh

zs

 

Figure 7.3-8

n

 A ci  zco  A s  zso

A ci n

Io  A o   zo  2

Wc

 A s

Ic n

Ws

A ci n

2 2   zc  Is  A s   zs

Profiled Steel Sheeting

       

Figure 7.3-9

Ii yd

Ii yh

n Ii yc Ii zs

 

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7.3-8 Serviceability limit state for composite slabs

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137 Ecd

(EI)s = Es Is

The purpose of examining the serviceability limit state is to ensure that the design value of the load

(EA)s = Es As = axial stiffness of the composite sheet

effect (e.g., deflection or cracks) does not exceed the usability limits defined for the mentioned value. The following load combination is verified in the serviceability limit state:

G,j Gk,j+Q,1 Qk,1+Q,I 0,I Qk,i nominal values of dead loads

Qk,1 =

nominal value of one live load

Qk,i =

nominal values of live loads

r

= (1-kx) (1-3kx/8) Es/fyd = cracked stiffness

Kr

= ( y-) / ( y- r)  1 when  r <    s

Kr

= 0,8 [( y-) / ( y- s)]2  1 when  s <    y

r

= (1,2 + Ø) ctu i,  i= long-term

 ctu = 0,2 o/oo = break elongation of the concrete

where Gk,j =

y

G,j = 1.0 = partial safety coefficient of dead load

0,i =0.7 , for load classes A-D and F-G as in SFS-ENV 1991-1-1, and for snow loads , for load class E as in SFS-ENV 1991-1-1

=

0.5

, for wind loads

= (1-3kx/8) 2

= relative moment= Mk/(bd 0.85fcd)

s

= 0.2 y+0.8 r

e

= effective stiffness for calculating the elongation = Kr I (hi / d)3 +(1- Kr) r

Q,i = 1.0 = partial safety coefficient of live load Qk,i 1.0

= relative yield moment

Q,1 = 1.0 = partial safety coefficient of live load Qk,1

=

= 9250(0,8K + 8)1/3 = design coefficient of elasticity of the concrete

The

increase

(EI) =  e b d3 0.85 fcd , where  e = effective bending stiffness = 1100 c (1 + i ) / [(1+ Ø) K0,675 ], crack-free stiffness

K

= cubic strength of the concrete

Ø

= creep value

c

= partial safety coefficient of the concrete

i

= ei2 (EA)s (EA)c / [(EI)ni (EA)i] = composite stiffness coefficient

elongation,

caused by

=  f  cs/d (1-0.6 '/) L2,

where  f = 5/48 (1+( MA+MB )/ 10Mk)

stiffness in the following formulas.

i

the

calculated with the formula acs

Deflection in the serviceability limit state is calculated with numerical integration, using the

in

(EI)ni = (EI)c + (EI)s (EA)i = (EA)c + (EA)s (EI)c = Ecd b hi3 / [12(1 + Ø)] (EA)c = Ecd b hi/ (1+ Ø)

Profiled Steel Sheeting

the

shrinking

of

the

concrete,

is


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List of symbols in equations Cross-section geometry b

=

slab calculation width (= 1 m, unless there are concentrated loads

As

=

design area of the thin sheet

ds

=

useful height of the slab in relation to the sheet (distance between the

dr

=

useful height of the slab in relation to the reinforcement

hc

=

slab height on the thin sheet

hs

=

thin sheet height

hi

=

hs + hc

ecp

=

mass centre of the concrete member

Arb

=

bar reinforcement at the level of the upper sheet surface or near the thin

requiring the calculation of the distribution width)

upper slab surface and the mass centre of the sheet)

sheet Art

=

bar reinforcement near the upper slab surface

Material properties fyd

=

design strength of the thin sheet

fryd

=

design strength of the reinforcement

fcd

=

design strength of the concrete

fctd

=

design tensile strength of the concrete

Es

=

coefficient of elasticity of the sheet and reinforcement

Ecd

=

design coefficient of elasticity of the concrete = 9250(0.8K + 8)1/3

ď Ľcmax =

1.2

o/oo

= compressive strain needed to reach the compression strength of the concrete, when the stress-strain model is linearised to a fraction line and the stiffness of the concrete is represented by the design coefficient of elasticity Ecd

ď Ľcu

=

greatest compressive strain of the concrete that the bending stiffness can resist

Figure 7.3.8-1: Steel composite sheets as light weight beam

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138


Table 7.3-2

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139

Figure 7.3.8-2: Steel composite sheets as light weight beam

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Table 7.3-2 Table 7.3-2

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Table 7.3-3

Table 7.3-3

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Example 7.1 Composite steel and concrete joists Design and assessment imposed composite sheeting steel slab profile with span L, including assessment sheeting concrete joist span, as permanent formwork. Joist spacing is L1, the thickness of the concrete slab is h, variable load on the ceiling is v, joists are not supported during the installation. Material characteristics: PENV 1994-1-1 – Concrete C25/30 2

fctkom  1.40MPa  c  1.5

fcko  10MPa

fctk005  0.7fctm

fck  20MPa

fctk005  1.56MPa

   fcko 

fctm  fctkom 

Ecm  29GPa

fck

fctk095  1.3fctm

Limit bending stress for concrete strength: fcd  0.85

fck c

fcd  11.33MPa

Steel 235: fyp  235MPa  mo  1.1

fy  235MPa

fu  360MPa

 M1  1.1

 a  1.15

Span L  7m

L1  2.5m

hdef 

Ea  210GPa

L 24

hdef  0.29167 m

Figure 7.1-1: Charging scheme joists

Figure 7.1-2: Steel and concrete joist

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3

fctm  2.22MPa fctk095  2.89MPa

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Table 7.3-3

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142


The difference between the old and the new standard is 7%

a) Sheeting rib profile 11 002R, reverse position (concrete filled narrow ribs)

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143

b s1  84mm

b s2  116mm

h 2  80mm b s1  s

4

4

Ia  131  10 mm

h2  80mm

3

3

W ael  28.72  10 mm

h  h1  h2p

h 2p 

2

L1  2.0m

s  63.1mm

g 2  g dstale  v d b

h2 h 2p  0.02942m

b s1  b s2

h  0.07942 m

 b  25kN m

h 1  50mm

3

m1 

g odvsz  g osvsz 

2

g odvsz  0.15295m

kN

m2 

Self-weight of concrete (razed thick ribs): godc   gosc

  1.35

g sstale  g osvsz b  g osc b

g dstale  g odvsz b  g odc b

1

g 1  5.08338m

kN

L1

3

4

 g2

L2

3

4

 2M b L1  L2  M a L1

3

M b L2

g 2

Mb   2.03335 m kN

L2

4

3

 g3

L3

4

 2M c L2  L3  M d L3

Mc   2.03335 m kN

Moments at the mid span of the beam:

Permanent - profile 12 101:

gosc   b h

g 1

 Mb     Find  Mb Mc  Mc 

Load considering the width of the sheet: b = 1m

  1.35

g 1  g 3

Given

Figure 7.1-3: Scheme to calculate the average thickness of the concrete slab.

2

g 3  g dstale  v d b

L3  2.0m

Support moments:

M c L2

g osvsz  0.1133kNm

L2  2.0m

godc  2.68043 m 1

g sstale  2.0988m

kN

m3 

2

kN 1

g dstale  2.83338m

1 8 1 8 1 8

2

 Ma  Mb    2  

2

 Mb  Mc   2  

2

 Mc  Md   2  

g 1 L1 

g 2 L2  

g 3 L3  

M ael  5.86887mkN

M b  2.03335mkN

m1  1.52501mkN

m2  0.50834mkN

m3  1.52501mkN M ael  M b

M el  5.4568mkN

M el  M b

kN

Variable load during concreting: 2

v s  1.50kNm

 v  1.5

2

v d  v s  v

v d  2.25m

kN

Assess the ultimate limit state: M ael  W ael

fyp a

Figure 7.1-4 M ael  5.86887mkN

Assessment of serviceability limit state (deflection only from permanent load)

Elastic moment calculated according to the old standards: Rd  190MPa

M el  W ael Rd

M el  5.4568mkN

Moment of the permanent load over support: M ael M el

 1.07551

Mb  

1 10

gsstale L1

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2

Mb   0.83952 m kN

M1 

1 11

gsstale L1

2

M1  0.7632 m kN


1 16

g sstale L2

2

The largest bending moment:

M 2  0.5247mkN

M sd 

The maximum deflection in the first span: xm  0.422L1

xm  0.844 m

alebo

xm  L1

1

33

xm  0.84307 m

16

1 8

godstale  vd L2

M sd  63.41892mkN

The required section modulus of composite sectional:

4

f 

gsstale L1

f  0.00066 m

185Ea Ia

W 

Max. allowable deflection will fmax 

1

250

L1

fmax  0.008m

f  fmax

Msd

W  0.00031035 m

 fy     1.15 

Propose IPE 270 3

Beam:

3

W ply  484  10 mm

Installation condition - acting a steel girder

M plRd 

W ply fyp

g odnosnik  0.48735m

kN

godvsz  gosvsz 

M 

Self-weight of concrete:

 L1  L2  gosc  gosc   2 

1

1

kN

kN

  1.35

godc  gosc 

kN

M plRd  M sd

2

godstale L

M  M sd

M

hnmon  87.86462MPa

W ely

hnmon  Rd

dnmon  87.86462MPa

4

kN

1

g odstale  g odnosnik  g odvsz  g odc g odstale  6.15411m

kN

5 gosstale L

f  0.01172 m

Ea Iy

384

The maximum deflection:

Variable load during concreting: vd   v vs

vs  3 m

1

kN

vd  4.2 m

1

kN

f max 

L 250

fmax  0.028 m

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f  fmax

Rd  210MPa

2

A  4590mm

M sd  63.41892mkN

M  37.69392 m kN

Serviceability Limit state: 1

g osstale  4.5586m

 v  1.4

8

dnmon  hnmon

f 

 L1  L2    2 

1

hnmon 

1

g odc  5.36085m

g osstale  g osnosnik  g osvsz  g osc

4

Stress on the beam in the extreme fiber cross-section:

 L1  L2    2 

vs  vs 

M  90.09mkN

M  W ely Rd

gosvsz  gosvsz 

g osc  3.971m

6

Iy  57.9  10 mm

Or

1

g odnosnik   g osnosnik

Profile 2102:

godvsz  0.30591 m

3

M plRd  98.90435mkN

a

Permanent - self-weight (estimate IPE 270):  1

3

W ely  429  10 mm

Ultimate limit state:

load:

g osnosnik  0.361kNm

3

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M 2 

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144


3. The loads transferred from the service stage composite beam load:

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145 The greatest shearing force:

Permanent - self-weight (estimate IPE 270): 1

godnosnik   gosnosnik

  1.35

gosnosnik  0.361kN m

Qd 

gosvsz  0.2266 m

1

kN

kN

gd L

Qd  47.73887kN

Carrying capacity of steel beam in shear:

Sheeting profile 12 101: gosvsz  gosvsz

2

Sectional assessment:

1

godnosnik  0.48735 m

1

godvsz  gosvsz 

godvsz  0.30591 m

tw  6.6mm h  270mm

1

kN

VplRd  A v

fy

2

A a  4590mm

2

A v  1.04h t w

A v  0.00185m

VplRd  218.65048kN

3 a

Self-weight concrete:

 L1  L2    2 

gosc   b h

godc  5.14642 m

godc  gosc 

  1.35

gosc  3.81216 m

1

kN

1

kN

gosstale  gosnosnik  gosvsz  gosc gosstale  4.39976 m

godstale  godnosnik  godvsz  godc

1

kN

godstale  5.93968 m

Figure 7.1-5: composite scheme.

1

kN 0.5VplRd  109.32524kN

Interacting slab width:

Variable load – long term: floors, ceilings  2 L1  L2 

 

v sdl  0.75kNm

2

 

 s  1.4

b eff  2

1

v ddl  v sdl  s

v ddl  2.1m

kN

b eff  1.75m

8

joists is determined to be the lowest value of terms  2 L1  L2   2kN m    2 

 s  1.4

b B

vduzitne  vsuzitne  s

1

vduzitne  5.6 m

g s  g osstale  v sdl  v suzitne

g d  g odstale  v ddl  v duzitne

8

2

gd L

1

g s  9.89976m

B  2m

b  0.3L

L  7m

0.3L  2.1m

kN

1

g d  13.63968m

kN

b n  0m

b  12 h 1  h 2p  b n

12 h 1  h 2p  b n  0.95304m

b  953mm

With composite concrete slab with steel beams may be considered if the condition:

The largest bending moment: 1

B  L1

For one row, shear connectors on the steel beam:

kN

In sum

Msd 

L

Interacting slab width with sheeting profile perpendicular to the longitudinal axis of the ceiling

- Service: vsuzitne

beam satisfies.

Qd  0.5VplRd

Msd  83.54302 m kN

h1  kh.2 

1 200

H

H  h1  h2  h

Profiled Steel Sheeting

kh2

h2p

H  0.4 m

h2p  0.02942 m 1 200

H  0.002 m

h1  h2p  0.07942 m h1  h2p 

1 200

H


Aa x 

IPE beam 270 satisfies.

fy

a

b eff0.85

According to the old standards

x  0.04729m

fck

M plRd´  A a Rd h 

c

x´

M plRd´  217.24162mkN

2

M plRd

IPE beam 270 satisfies.

M plRd´

M sd  83.54302mkN

M plRd´  M sd

 1.06365

Serviceability limit state: Assessment of composite beams on deflection provided a flexible action. Effective modulus of elasticity of concrete is considered the influence of creep: E´c  14500MPa

E´c  0.5Ecm

Working factor: Figure 7.1-6: Stress distribution in the cross section.

n´ 

The compression depth of the cross-section according to the old standards:

Ea

n´  14.48276

E´c

- For short-term effects x´ 

A a Rd

x´  0.08924m

b fcd

x x´

 0.52991

Eb  27GPa

Inaccuracies influence the thickness of the concrete slab, deposit of reinforcement in the

n k 

Ea

n k  7.77778

Eb

- The long-term effects

calculation of structures expressed  cr  3

coefficient  :  

1 H   1   600   0.002 mm  

fy

Epretv  0.5Eb

Eb

Epretv  13500MPa

g osstale  v sdl  5.89976m

M sd  83.54302mkN

M plRd  M sd

E´b  6750MPa

n d 

Ea E´b

n d  31.11111

n pretv 

Ea

n pretv  15.55556

Epretv

To simplify the calculations can be considered average value of the ratio of elastic modulus: 1

h  x    2

a

M plRd  231.06938mkN

1 1   cr

- according to PENV 1994-1-1

  0.71429

Assessment of bending (at the moment of resistance) according to ENV 1994-1-1 MplRd  Aa

E´b 

n 

kN

gosstale  vsdlnd  vsuzitne nk gosstale  vsdl  vsuzitne

Profiled Steel Sheeting

1

v suzitne  4 m

n  21.68327

kN

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Position of the neutral axis (carrying capacity in shear):

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146


The neutral axis position (serviceability limit state):

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Department of Architecture

147

Aa 

h

2

e 

 h1  h2 

Aa 

h1 1 h1 beff n´ 2

fmax  0.02333m

300

PENV 1994-1-1

e  0.12861 m

1 h1 beff n´

L

fmax 

f´ 

5

 vsdl  vsuzitneL4

384

Ea Ii´

f´  0.00498m

f  fmax

Ideal concrete area: Short term load: A bik 

b h1 nk

2

A bik  6126.42857mm

Long term load: A bid 

n  21.68327

PENV 1994-1-1

1 n pretv

e

e´  0.16894m

b h1  A a

A bi´  h 1

 0.76131

The moment of inertia of ideal cross-section: 4

Ii  0.0002094m

A bi 

h1 b n

2

A bi  2197.54646mm

2

A bi´  3063.21429mm

2

A ik  10716.42857mm

Effective area of welded mesh is neglected: 2

 h 1  h 2  e´ 

1

1 h  b h 13  b h 1  e´  1   2 

2

n pretv  12

5

vsdl  vsuzitneL4

384

Ea Ii

4

Ii´  0.00016436m

Ii Ii´

 1.27405

A id  A bid  A a

A id  6121.60714 mm

2

PENV 1994-1-1

Deflection due to Accidental load: f 

b n pretv

A ik  A bik  A a

PENV 1994-1-1

2

b  0.953m

Position of the neutral axis:

2  h1   h 11  3 Ii  Iy  A   h 1  h 2  e  b eff h 1  b eff h 1  e   n´  12 2 2   2

h

2

A bid  1531.60714mm

PENV 1994-1-1

 h1  h b h 1   A a   h 1  h 2 n pretv  2  2  1

Ii´  Iy  A 

nd

The average value of the ratio of elastic modulus:

Figure 7.1-7

e´ 

b h1

f  0.00391 m

A i´  A bi´  A a

Profiled Steel Sheeting

A i´  7653.21429 mm

2

A i  A bi  A a

2

A i  6787.54646mm


b  0.953m

h 1  0.05m

2

A a  0.00459m

Department of Architecture

Static moment to the axis y1: h 2  0.08m

Distance center of gravity of the compression area of the concrete section from the center of gravity of the cross section of a steel composite beams

rbo 

h 2

 h2 

h1

rbo  240mm

2

Syk  A bik rbo  A a 

h

2

h

2

3

Figure 7.1-8: Schematic cross-section of an ideal neutral axis.

Syk  1470342.85714mm

The moment of inertia of ideal cross-section of the priority axis y-y: Syd  A bid rbo  A a 

h

2

Syi  A bi rbo  A a 

PENV 1994-1-1

h

2

Syi´  A bi´ rbo  A a 

h

2

h

2

h

2

h

 2

3

Syd  367585.71429mm

3

Syi  527411.14920767mm

Iyk  Iy  A a rok  A bik  rbo  rok 

2

2

2

2

Iyd  Iy  A a rod  A bid  rbo  rod  2

3

Syi´  735171.42857mm

Iyi  Iy  A a ro  A bi  rbo  ro 

2

2

Iyi´  Iy  A a ro´  A bi´  rbo  ro´ 

Short-term load: Syk Aik

Syd A id

Syi´ A i´

2

4

Iyi´  163720223.436mm

ideal composite beams - short-term y hk  rok 

rod  60.04726mm

ro 

Syi Ai

ro  77.70277mm

h

y hk  2.20456mm

2

Section modulus: for W yodk 

n k  7.77778

PENV 1994-1-1 ro´ 

4

Iyi  143497369.508mm

The distance of the lower fibers of steel parts from the neutral axis of the cross-section of an rok  137.20456mm

Long-term load: rod 

4

Iyd  124048057.758mm

PENV 1994-1-1

The position of center of gravity:

rok 

4

Iyk  209044542.292mm

Iyk y dk

3

W yodk  767968.55641mm

3

ro´  96.06048mm

W yohk  94823742.04082mm

Composite beam: Iyk

W ybhk  H

h 2

3

W ybhk  1635774.64736mm  rok

Profiled Steel Sheeting

W yohk 

Iyk y hk

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148


STRUCTURAL ENGINEERING ROOM

Department of Architecture

149

The distance of the lower fibers of steel parts from the neutral axis of the cross-section of an ideal composite beams

PENV 1994-1-1 W yod´ 

long term y hd 

Iyi´

3

W yoh´ 

W yod´  769713.64679246mm

yd

Iyi´

3

W yoh´  2857384.52707mm

yh

Composite beams:

h

 rod

2

y hd  0.07495m

Section modulus: for W yodd 

n d  31.11111

Iyd y dd

3

W yohd 

W yodd  635989.75953482mm

Iyd

Iyi´

W ybh´ 

3

W ybh´  969105.52523959mm

h

H

 ro´

2

y hd

3

W yohd  1655016.90748mm

Service stage (design value)

Composite beam:

Short-term:

Iyd

W ybhd  H

h 2

3

W ybhd  605251.99925704mm

1

v duzitne  5.6m

 rod

The distance of the lower fibers of steel parts from the neutral axis of the cross-section of an h 2

 ro

y d  0.2127m

The distance of the upper fibers of steel parts from the neutral axis of the cross-section of an ideal composite beams:

y h 

h 2

Long-term: 1

g odstale  v ddl  8.03968m

ideal composite beams y d 

kN

kN

1 2 M1  vduzitne L 8

M1  34.3 m kN

M pr  M 1  M 2

M pr  83.54302mkN

M2 

1  godstale  vddl L2 M2  49.24302 m kN 8

The normal stress for composite beams:  ro

y h  0.0573m

Short term:

Section modulus (average value) for n  21.68327

W yod 

Iyi yd

n k  7.77778 3

W yod  674637.99689334mm

3

W yoh 

Iyi yh

 d 

M 1  34.3mkN

M2

 h 

Wyodd

d  44.66329MPa

W yoh  2504438.08776mm

M2

 b 

Wyohd

h  0.36172MPa

M2 nd  Wybhd

b  2.69597MPa

Long-term: Composite beam: Iyi

W ybh  H

h 2

3

n d  31.11111

W ybh  766147.83983007mm  ro

M 2  49.24302mkN

d 

M2 W yodd

d  77.42737MPa

Profiled Steel Sheeting

h 

M2 W yohd

h  29.75378MPa

b 

M2 n d  W ybhd

b  2.61513MPa


n  21.68327  d 

M pr  83.54302mkN

Mpr

 h 

Wyod

d  123.83384MPa

Mpr

 b 

Wyoh

h  33.35799MPa

Mpr n   Wybh

b  5.0289MPa

Figure 7.1-9: The resulting stress in composite sectional.

PENV 1994-1-1 d´ 

M pr

h´ 

W yod´

d´  108.53779MPa

M pr

b´ 

W yoh´

h´  29.23758MPa

calculation of composite member:

M pr

For composite sheeting steel beams are used shear connectors which welded to the steel beam

n pretv  W ybh´

with simultaneous welding with sheeting steel profile.

b´  5.54183MPa

s  0.0631m

Control of the stress at the bottom of a steel beam: The total stress at the bottom of a steel beam calculated as the stress at the bottom of a steel beam in the installation stage, and serves as the steel beam and the tension in the bottom of the

if

2

h2 s

 1.0

2

h2 s

Rt  110MPa

then

 2.53566

r3  200mm

m=1 d 

4r3 X 2h 2  Rt m s

The largest bending moment: 1 2 gd L 8

Msd 

steel beam, the steel beam which is composite.

h 2  0.08m

M sd  83.54302mkN od  dnmon  d

od  211.69847MPa

Rd  210MPa

od  Rd

oh  hnmon  h

oh  121.22261MPa

Rd  210MPa

od  Rd

does not satisfies

PENV 1994-1-1

The greatest shearing force: Qd 

1 gd L 2

Qd  47.73887kN

o´d  dnmon  d´

o´d  196.40241MPa

Rd  210MPa

Rd  o´d

o´h  hnmon  h´

o´h  58.62704MPa

Rd  210MPa

Rd  o´h

Qmin  0kN

satisfies satisfies

The stress in the concrete slab: b 

M pr n  W ybh

b  5.0289MPa

b  5.0289MPa

ad   fcd

ad  8.09524MPa

Figure 7.1-10: The average value of the horizontal shear force

According to NAD b´ 

M pr n pretv  W ybh´

b´  5.54183MPa

b´  5.54183MPa

b  ad

satisfies

Vpriem 

rb 

Qd  Qmin 2

h1  h   h 2    ro 2 2

Vpriem  23.86943kN

rb  0.1623m

Profiled Steel Sheeting

A b  h 1 b

2

A b  0.04765m

STRUCTURAL ENGINEERING ROOM

Average value:

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150


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151

The horizontal shear force per unit length contact with steel and concrete of the cross-section

PRk2  0.29d

2

fck Ecm

PRk2  88.34297kN

PRk1  77.9115kN

PRd 

PRk1 v

PRd  59.93192kN

shall be determined according to installation procedure. X 

Vpriem A b rb

1

X  59.32631m

n Iyi

kN

The total horizontal shear force, which carried over one shear connector: X r3

1 m

 11.86526kN

The required diameter of the shank: 4r3 X 2h 2

 0.01866m

 Rt m s

surface of one shear connectors:

d  20mm

2

A t  

d

2

4

A reduced strength in the joist plate:

A t  0.00031m

kt 

Carrying capacity of one shear connector: U 

s

2h 2

U  13.62862kN

A t Rt

Nael  A a Rd

U r3

Nael  963.9kN

1

 68.14311m

kN

Nb  Nael

U r3

X

Fct  Nb

0.7 b o h t  h p Nr h p

hp

Ntrnov 

Na  A a

fy a

Nf  56.69139 Ntrnov  70.72615

U

L

Na  937.95652kN

ks

at  0.5

L Nf

PENV 1994-1-1 Diameter of shear connectors: d  0.02m

h t  h 1  h 2  10mm

s  0.0631m

h 2  0.08m

Nr  1

fu  310MPa b o  s

 v  1.3 h p  h 2

Carrying capacity of shear connectors in full slab: PRk1  0.8fu

d 4

PRdr  16.54496kN

Nc  Na

Fcf  Nc

Fcf  937.95652kN

in half beam

at  0.06174m

Or that the entire length of the beam, the number of shear connectors:

at  0.04949m

Ntrnov

PRdr  PRd kt

The distance of shear connectors will be in half span of beam:

The distance between the shear connectors in the half span of beam will be:

at  0.5

kt  0.27606

The number of shear connectors in half beam:

The number of shear connectors in half beam: Fct

Figure 7.1-11:

2

PRk1  77.9115kN

PRk1 X

 1.31327m

Profiled Steel Sheeting

Nf 

Fcf PRdr


failure is perpendicular to the bed joints, or Md

The design strength of masonry is given by:

2   Wk  f L when the plan of bending is

parallel to bed joints.

– in compression

where  is the bending moment coefficient

– in shear

 is the partial safety factor loads

– in flexure

characteristic strength divided by the appropriate partial safety factor γM.

is the orthogonal ratio of the flexural strength= modified flexural strength parallel to bed joints/flexural strength perpendicular to bed joints

L

8.1 Unreinforced masonry walls subjected to vertical loading

is the length of the panel between supports

Wk is the characteristic wind load per unit area

It may be assumed that: – plane sections remain plane;

When a vertical load acts so as to increase the flexural strength in the parallel direction, the

– the tensile strength of the masonry perpendicular to the bed joints is zero;

orthogonal strength ratio may be modified by using a flexural strength in the parallel direction,

Allowance in the design should be made for the following:

the orthogonal strength ratio may be modified by using a flexural strength in the parallel

– long-term effects of loading;

direction of f kx   m gd

– second order effects;

where f kx is the characteristic flexural strength appropriate to the plane of bending

– eccentricities

 m is the partial safety factor for materials.

calculated from a knowledge of the layout of the walls, the interaction of the floors and the

gd is the design vertical dead load per unit area.

stiffening walls; – eccentricities resulting from construction deviations and differences in the material

The design load may be taken as the sum of the products of the component characteristic

properties of individual components

loads multiplied by the appropriate partial safety factor, f

At the ultimate limit state, the design vertical load on a masonry wall, NSd, shall be less than

8.1-2 Characteristic compressive, flexural and shear strength of masonry wall

or equal to the design vertical load resistance of the wall, NRd:

The characteristic compressive, flexural and shear strength will normally have been

NSd ≤ NRd

determined by test on masonry specimen. The characteristic compressive strength of the masonry should be normally determined by tests, or from a suitable relationship between the

8.1-1 Verification of unreinforced masonry walls Masonry wall subjected to mainly lateral loads should be verified to have a design

characteristics strength of unreinforced masonry and mortar strength. Moreover, most design

strength greater than the design load. In assessing the lateral resistance of masonry walls, it is

codes will contain values for compressive strength fk , flexural strength fkx and shear strength fvk.

essential that support conditions and continuity over supports are taken in account.

The characteristics flexural strength fkx should be used only in the design of masonry in bending.

The design of load bearing masonry is undertaken primarily to ensure an adequate margin of

In general, no direct tension should be allowed.

safety against the ultimate limit state being reached. This is generally achieved by ensuring that the design strength of the member is greater than or equal to the design load. The calculation of the design moment per unit height of to take into account the masonry properties referred to above, and may be taken as either Md

2  Wk  f L when the mode of

Masonry

STRUCTURAL ENGINEERING ROOM

8. Design strength of masonry

Department of Architecture

152


What is the advantages and disadvantages of the brick when used as structural

STRUCTURAL ENGINEERING ROOM

Department of Architecture

153 Table 8.1.2-2 Partial safety factors for the material m

materials:

control (see 6.9)

low strength - yield stress 3-20MPa - very weak in tension - brittle

medium Modulus of Elasticity 10000-25000MPa

heavy: ~1900 kg/m3

not waterproof

Usage

walls, piers, footings, retaining walls, arches, vaults (not so much today) •

A

B

C

Masonry (see

Category of manufacturing

I

1,7

2,2

2,7

note)

control of masonry units (see 3.1)

II

2,0

2,5

3,0

2,5

2,5

2,5

Resistance of wall ties anchorage, tensile, and compressive

use in compression flammable (very large sections ok) - when fire resistance not an issue

Usage

Category of construction

m

Advantages

Resistance of anchorage bond of reinforcing steel

1,7

2,2

-

Steel (referred as s )

1,15

1,15

-

framing, post & beam, trusses, panels, floors

The value of m for concrete infill should be taken as that appropriate to the Category of

small structures usually - glulam can be used for larger spans

manufacturing control of the masonry units in the location where the infill is being used The design value Xd of a material property is generally defined as

Partial safety factors for the material When verifying the stability in the case of accidental actions, m for masonry shall be

Xd

taken as 1,2. 1,5. and 1,8 for category A, B and C of construction control respectively, m for

Xk m where m is the partial safety factor for the material property. Design values for the

wall ties and anchorage bond shall be taken as given in Table 8.1.2-2 and s for steel shall be taken as 1,0

material properties, geometrical data and effects of actions, R, when relevant, should be used

Combinations of actions

to determine the design resistance Rd from:

 GjGkj  1  4Qk1 Fd

 F Fk

Gd

 G Gk

 GjGkj  1  2

Qki

Rd

( i 1)

Qd

 Q Qk

Ad

 A Ak

Pd

 P Pk

internal forces or moments)

Table 8.1.2-1 Partial safety factors for actions on building structures Permanent Variable actions (G)

Prestressing (p)

actions One with its

Sd  Rd

Sd is the design value of an internal force or moment (or of a respective vector of several

Where F, G, Q, A, P are the partial safety factors for the action.

(G)

R Xd  ad  ....

Rd is the corresponding design resistance. The compressive strength of masonry units to be used in design shall be the normalised compressive strength fb, calculated from the mean compressive strength fu and multiplied by

Others

the factor δ, to allow for the height and width of the units, as given in table 8.1.2-3

characteristic value Favourable effect

1,0

0

0

0,9

Unfavourable effect

1,35

1,5

1,5

1,2

Masonry

fb

 f u


Height of unit (mm)

Table 8.1.2-4: characteristic compressive strength fck, of concrete infill

Last horizontal dimension of unit (mm) 50

100

150

200

Strength class

50

0,85

0,75

0,70

-

-

C12/15

C16/20

C20/25

C25/30 or

of concrete

250or greater

stronger

2

fck (N/mm )

12

16

20

25

65

0,95

0,85

0,75

0,70

0,65

100

1,15

1,00

0,90

0,80

0,75

The characteristic shear strength of concrete infill, fcvk, that may be assumed in design is given

150

1,30

1,20

1,10

1,00

0,95

in table 8.1.2-5 for the relevant concrete strength classes.

200

1,45

1,35

1,25

1,15

1,10

Table 8.1.2-5

250 or greater

1,55

1,45

1,35

1,25

1,15

Strength class

C16/20

C20/25

C25/30 or

of concrete

Note : Linear interpolation is permitted.

stronger

2

fcvk (N/mm ) Values of factor 

0,27

0,33

0,39

0,45

Characteristic compressive strength of unreinforced masonry made using lightweight mortar

1,8 1,6

Height of unit (mm)

1,4 1,2

C12/15

The characteristic compressive strength of unreinforced masonry, fk made with group 1 masonry units and lightweight mortar, may be calculated using equation.

50

1

100

0,8

200

fk

250 or greater

0,6

0  65

K f b

Provided that fb is not taken to be greater than 15 N/mm2 and the thickness of the masonry is

0,4

equal to the width or length of the masonry units so that there is no longitudinal mortar joint

0,2

through all or part of the length of the wall

0 0

50

100

150

200

250

300

Table 8.1.2-6

Last horizontal dimension of unit (mm)

lightweight mortar class

Constant K Longitudinal mortar joint

Figure 8.1.2-1

through all or part of the

Properties of concrete infill

length of the masonry

For the purposes of specification, the characteristic compressive strength of the concrete infill, fck, is classified by the concrete strength class which relates to the cylinder/ cube strength

1

0,55

0,40

at 28 days, the strength classes normally used for concrete infill in reinforced masonry are given

2a

0,50

0,35

in table 8.1.2-4, together with the value of fck to be used in design

2b

0,45

0,30

3

-

0,40

Masonry

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Table 8.1.2-3 Values of factor δ

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Characteristic compressive strength of unreinforced masonry made using general

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purpose mortar

The characteristic compressive strength of unreinforced masonry made with general purpose mortar, with all joints to be considered as filled, may be calculated using equation (8.1): fk = K fb 0,65 fm 0,25 N/mm2

(8.1)

provided that fm is not taken to be greater than 20 N/mm2 nor greater than 2 fb, whichever is the Figure 8.1.2-2

smaller;

Characteristic compressive strength of unreinforced masonry made using thin layer

where: 2 0,10

K is a constant in (N/mm )

mortar

that may be taken as:

Walls without longitudinal joints

(1) The characteristic compressive strength of unreinforced masonry, fk, made with thin layer

0,60 for Group 1 masonry units

mortar, with all joints to be considered as filled and using Group 1 calcium silicate units and

0,55 for Group 2a masonry units

autoclaved aerated concrete units may be calculated using equation (8.2): fk = 0,8 fb 0,85

0,50 for Group 2b masonry units

(8.2)

Walls without longitudinal joints when the thickness of masonry is equal to the width or length

provided that:

of the masonry units so that there is no longitudinal mortar joint through all or part of the length

– the masonry units have dimensional tolerances such that they are suitable for use with thin

of the wall

layer mortars; – the normalized compressive strength of masonry units, fb, is not taken to be greater than 50

Walls with longitudinal joint

N/mm2;

0,50 for Group 1 masonry units

– the thin layer mortar has a compressive strength of 5 N/mm2 or more;

0,45 for Group 2a masonry units

– there is no longitudinal mortar joint through all or part of the length of the wall. - 72 -

0,40 for Group 2b masonry units The characteristic compressive strength of unreinforced masonry, fk, made with thin layer

0,40 for Group 3 masonry units;

mortar and using masonry units other than Group 1 calcium silicate units and autoclaved aerated

fb is the normalized compressive strength of the masonry units in N/mm ,

concrete units may be calculated using equation (8.1):

fm is the specified compressive strength of the general purpose mortar in N/mm2.

where:

when there is a longitudinal mortar joint through all or part of the length of the masonry

K is a constant in (N/mm2)0,10 that may be taken as:

2

Example for the relation between the compressive strength fk of masonry and the compressive

0,70 for Group 1 masonry units;

strength fb of the units and fm of the mortar (according equation (8.1) and with K = 0,60):

0,60 for Group 2a masonry units; 0,50 for Group 2b masonry units;

Masonry


Limiting dimensions of masonry panel

mortar

In a lateral loaded masonry panel the dimensions should be limited as follows

(1) The characteristic compressive strength of unreinforced masonry, fk, made with Group 1, 2a and 2b masonry units and lightweight mortar, with all joints to be considered as filled, may be calculated using equation (8.3): fk = K fb0,65 N/mm2 (8.3)

1. Panel supported on three edges: a. two or more sides continuous t ef H L or less 1350

provided that fb is not taken to be greater than 15 N/mm2 and that there is no longitudinal

b. all other cases t ef

mortar joint through all or part of the length of the wall.

2. Panel supported on four edges:

The value of K depends on:

a. three or more sides continuous t ef

– the density of the used lightweight mortar, – the type of the masonry units. Further regulations are given for:

• characteristic compressive strength of unreinforced masonry with unfilled vertical joints • characteristic compressive strength of shell bedded unreinforced masonry Characteristic shear strength of unreinforced masonry

The characteristic shear strength fvk of unreinforced masonry can be determined – from the results of tests on masonry, – by calculation in the following way: For general purpose mortar and when all joints may be considered as filled, fvk will not fall below the least of the values described below: fvk = fvko+0,4 d or

= 0,065 fb, but not less than fvko

where: fvko is the shear strength, under zero compressive stress d

is the design compressive stress perpendicular to the shear

For: – masonry with unfilled perpend joints, – shell bedded masonry, – thin layer mortar, – lightweight mortar, – there are similar regulations.

b. all other cases t ef

H L or less 1500

H L or less 2250

H L or less 1025

3. Panel simply supported at top and bottom t ef

Height 40

Height 12

4. Free standing wall t ef

Where tef is effective thickness of the masonry wall. In case 1 and 2, no dimension should exceed 50 times the effective thickness, tef -Masonry column

d = H / 17.5

for Circle Square and Rectangular column

Where H is vertical distance between lateral supports and d is thickness of column. -Masonry wall

d = H / 20

Walls may also have vertical lateral supports. Formula valid when lateral movement is prevented at top and bottom of wall, at right angels to wall, such restraint usually provided by floor construction. Wall has greater bending strength in the horizontal direction so that vertical supports would be preferred to horizontal supports. -Reinforced and prestressed masonry columns and walls -Masonry arch and fill

d = span / 25

- Masonry arch

d = span / 45

-Vault and domes

d = span / 55

d = H / 27.5

Domes have been build spanning up to 40m and stone vaults up to 20m

Cavity wall

deff = H / 20 where deff is greater than t1, t2 or 2/3 (t1+t2)

t1 and t2 are thicknesses of leaves of cavity wall which are tied together.

Masonry

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Characteristic compressive strength of unreinforced masonry made using lightweight

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157 Example 8.1

Roof frame

Design of masonry structures

 1.15 kN m

qsn

Check that the design load is less than the resistance capacity of the masonry Masonry wall Light height of storey

hv

Span

l

Material data

w

Floor actions

3.10 m

Mortar MVC 5,2.5, 1

12.5 kN m

Depth of masonry wall

h

450 mm

b

Height of masonry wall

hw

3.45  m

Peripheral beam

hv

0 m

3

j

( 1  3)

w 2

Roof tile

0.40 kN m

Battening

0.10 kN m

( 1  5)

1 2 3

2

0.2 kN m

g1L

qsn  f w w

-2

Design value of actions qpd j

qsd  w 1.553 1.4 5.5

4

4

5

2.8

-2

4

j

-2

kN m

2.8

( 1.25  1.1  1.4)

-2

 0.2 kN m

1.155 kN m

w

j

Self weight

 3.2 kN m

2

2

Variable action 2

 5.0 kN m  2.0 kN m Floor layer, dividing wall

 0.4 kN m

2

f

j

qpd 

characteristic live floor load

Loads from floor

( 1.35  1.4  1.1  1.25  1.4)

qp

 qpd j

-2

qp  12.3 kN m

j

Wide load Load area

zs

1 m

A

l h zs2 2

2

A  4.05 m 3

Self weight of peripheral beam

Nv

hhvb 1.1 24 kN m

Nv  0 kN

Self weight of masonry

Nm

hhwb 1.1 

Nm  21.347 kN

Loads from roof

kN m

qpn  f

5.5

Characteristic value of actionsqpnj

2 2 0.023 6.5 kN m 0.15 kN m 2 2 0.14 1 kN m 0.14 kN m 2 2 0.10 kN m 0.15 0.015 kN m 2 2 0.023 6.5 kN m 0.15 kN m 2 -2

Roof load

qsd w

2

 2.0 kN m

qs  15.25 kN m

Partial safety factors for f actions in building structers j

Permanent load

w 

 qsdw

roof tile Self weight

1

Load arrangement and load cases

Self weight of roof frame

 3.2 kN m

j 

1.0 m

Roof

 5.0 kN m

2

Floor layer, dividing wall

Classification of loads

2

Width of masonry wall

Drywall

 1.0 kN m Snow

Variable load

2

Floor

3

Unit weight of masonry wall

Wooden grid

2

w

Geometrical data of masonry

glasewood

qs

4.5 m

Masonry member CP

wooden board

Self - weight

2

Loads from floor

Ns

Aqs

Ns  61.773 kN

Np

Aqp

Np  49.815 kN

Characteristic action from individual storey in kN i Ni

Masonry

(0  5)

 Nv  Ns   Nv  Ns  Nm   2Nv  Ns  Np  Nm   2Nv  Ns  Np  2Nm   3Nv  Ns  2Np  2Nm   3Nv  Ns  2Np  3Nm


i

( N ) i i 

Nd  i

0

-61.773

1

-83.12

2

-132.935

3

-154.281

4

-204.096

5

-225.443

Nlt

kN

Nser

N11

2.Storey

N21

3. Storey

N31

1

N 3 m 1 Nd  Nm 3 3 1 Nd  Nm 5 3 Nd 

j

N21  147.166 kN N31  218.328 kN

Behaviour of action

Centric compression

Eccentricity

e

Area of section of masonry

Ap

Strength of masonry member

10 Mpa

1 2.5 5

j

 qsnj   qpnj

-2

Nser  17.35 kN m

j

1   j

Nlt  1.2 e   1   h  Nser 

klt  j

0.907 0.929

Action coefficient of section u 75  0.1 tmin tmin 450 mm u 120

u  1

Design value of resistance capacity of masonry Rd

Inputs for calculation of strength of masonry

hw 1000  h j

-2

Nlt  12.35 kN m

j

0.907

Rd

1

 qpnj

N11  76.004 kN

1

Calculation of resistance capacity of masonry

Grade of Mortar

2

Coefficient express the effect of the term load Klt

klt

Charakteristik actions at lower storey

1. Storey

1

j

Design actions and eccentricity, calculated at 1/3 height of masonry wall

Nombers of storeys

qsn  qsn 

1  j

8.853 8.853 7.667

( 1000 kPa  1300 kPa  1500 kPa)

Slenderness ratio

0 m 2

Ap  0.45 m

b h

j

j

( 0.13  0.13  0.1)

j

( 750  750  1000)

Grade of Mortar 1 2.5 5

1  j

8.853 8.853

Rd j j

( 1000 kPa  1300 kPa  1500 kPa) ( 0.9  0.9  0.92)

7.667

Design resistance capacity of section in centric compression For mortar grade 1

Nud1

i

Ap Rd  1 klt  u 1 1

For mortar grade 2.5

Nud2

i

Ap Rd  2 klt  u 2 2

Nud3

Ap Rd  3 klt  u 3 3

For mortar grade 5

i

Nlt

qlt

Nser

qser

Because the wide load is identic

Masonry

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Nd

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Design actions and design resistance of storeys separately

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Design action i 1 2 3

Nud1 

Ndim  i

kN

76.004 147.166

Design resistance

218.328

i

367.523 367.523

kN

367.523

Example 8.2

Mortar grade 1

5

Design of masonry structures

4

Column

n

3

Width of section

b

600 mm

xd

2

Depth of section

h

300 mm

1

Parial safety factor for development of cracks

0

0

200

400

600

Eccentricity

800

 Ndim    Nud1       kN  n  kN  d

Design force Resistance in kN

Mortar grade 2.5 Design action

Design resistance

5 4

i 1 2 3

Ndim  i

76.004 147.166

kN

218.328

N ud2

477.78

kN

i

477.78

n

3

xd

2

477.78

0

200

400

600

Design action Design resistance i 1 2 3

Ndim  i

76.004 147.166 218.328

kN

N ud3 0 0 0

i

Design compressive strength of mortar

Rd

1.5 MPa P10 M5

Design axial tension strength of masonry

Rtd

0.16 MPa

Design shear strength of masonry

Rqd

0.16 MPa

Design flextural strength of masonry

Rtfd

0.12 MPa

Design tension strength of masonry

Rt1d

0.12 MPa

1

Mortar grade 5

75  0.1 300 120

u

0.875 2 hw

lef h

1000 750

1

20.01

0.54  0.0148 0.05 

0.48  0.06 0.0148

5

0.53926

0.480888

4

1  0.85

0.15

n

3

xd

2

0

a

Parial safety factor for permanent actions klt

1 0

200

400

600

800

 Ndim    Nud3       kN  n  kN  d

The masonry wall is, therefore, satisfactory for permanent and variable loads

Masonry

 

1   a  1 

1.2 e   h 

klt

0.706

Design resistance of rectangular section Nu1

Design force kN Resistance in kN

hw

300 kN

Slenderness ratio of the cross-section

Design force kN Resistance in kN

750

Nd

800

 Ndim    Nud2       kN  n  kN  d

Clear heigh of pillar

0.24 m

Effective length of pillarlef 0

1.5

Design vertical load at the top pillar

u

1

e

r

 u  r klt

b hRtfd

 6 e  1     b 

Nu1

14.2965kN

1

2.6 m


The orthogonal ratio  of the flexural strength

The panel masonry wall is simply supported on three sides.

L is the length of the panel between supports

L

7 m

H is the height of the panel

H

5.25 m

 orth

 fh

Dead load

 fl

Material factor

 fm

1.5

W

0.9

0.56 

2

m

7

 wall

2.5

Calculate section modulus

kN 3

fpara

0.18 MPa

plane of failure perpendicular to bed joint

fperp

0.31 MPa

0.56

Calculate the design dead load

We assume

tef

0.1565m

 bm  0.75

L

tef

B

1 m

Z

1 2 Btef 6

3

Z  0.015 m

Required wind load capacity

Wk

 W  fl 1m

Design moment

Md

 Wk  flL

-1

Wk  0.504 kN m

2

Md  1.467 kN m

Calculation of design moment of resistance of panel

The wall thickness should be designed to withstand a crow pressure of

tef

H

Calculate moment resistance

m

Plane of failure parallel to bed joint

HL 1500

 orth  0.714

kN

Determination characteristic flexural strength

Effective thicknesstef

fperp

Calculate bending moment coefficients  bm

Determination the partial safety factors Horizontal imposed load

f

0.30m

kN 2

m

fperp

The design moment of resistance of masonry is Mds

Wloadcapacity

Mds 2

 L

 fm

Z

Mds  1.86 kN m

-1

Wloadcapacity  0.575 kN m

Required wind load capacity Wk

W  fl1m

-1

Wk  0.504 kN m

0.30m Check that whether the design load is less than capacity

On the other hand, no dimension to exceed 50xtef= 12.5m, it is satisfactory Calculate the design dead load go

tef  wall fl

-2

go  1.89 kN m

Increase flexural strength in the plan of failure parallel to bed joint

In the parallel direction f

 g H  1   o fm  2  fpara  tef

Check that whether the design load is less than capacity Compared to required design wind load capacity of 0.504 kN/m the panel is, therefore, satisfactory for wind load.

Wloadcapacity  Wk f  0.2213 MPa

Masonry

Ok

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Example 8.3

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9. Terminology acceptance acceptance test administrative analysis annex

accidental action accidental line load action action effect (S) action in silos and tanks

applicability

actions and environmental influences actions in geotechnical design actions on parapets actions on structures actions on structures exposed to fire aerodynamic coefficient braking force centrifugal force characteristic value of a permanent action characteristic value of a prestressing action characteristic value of a single axle load for a road bridge characteristic value of a single variable action characteristic value of an accidental action characteristic value of an action

application rules approval arrangement assessment assumptions authority authorized background to the Euro code Programme basis of design building building regulations buildings calculated properties calculation change made to nominal geometrical data for particular design purposes, e.g. assessment of effects of imperfections characteristic dimension characteristic value characteristic value of a geometrical property characteristic value of a material property

characteristic value of seismic action characteristic value of the concentrated load (wheel load) on a footbridge characteristic value of the dominant variable action characteristic value of the non-dominant variable action classification of actions coefficient for combination value of a variable action

accidental design situation accidental situation acknowledge principle acknowledged rules active fire protection measures advanced calculation model allowance for imperfections ambient gas temperature angle application of tolerances area area ratio assessment methods average yield strength

characteristic value of resistance check tests checking civil engineering civil engineering works clause client codes of standards collapse common terms used in the Structural Eurocodes competence completion

axial force

condition

axis along member

conformity; compliance

axis distance basic yield strength basis for the fatigue assessment of railway structures basis of design for footbridges basis of design for road bridges

consequences of failure consignment note construction drawings construction material construction works control conversion

conversion factor coordination

batten beams beam-to-column connections bearing

corrective maintenance; repair corrective step

Terminology

coefficient for frequent value of a variable action coefficient for quasipermanent value of a variable action collisions combination factor combination of actions combination values consequential indirect effects dead, imposed and environmental loads from structures design action-effect (Sd) design effect of destabilizing actions design effect of stabilizing actions design value of a permanent action design value of a prestressing action design value of a variable action design value of an accidental action design value of an action design value of effects of actions design value of seismic action direct action discharge loads distance between rail supports, length of distributed loads dynamic action dynamic analysis when there is a risk of resonance or excessive vibrations of railway structures - basis of supplementary calculations dynamic effects dynamic factors 1+ď Ş for actual trains

bearing bending bending and axial compression bending and axial force bending and axial tension bending and shear bending moment bending moment resistance boundary conditions braced frames and nonbraced frames bracing breadth; width bridge buckling length buckling length of a compression member cable stay calculation model calculation of stresses cantilever carriageway carriageway width for a road bridge, including hard shoulders, hard strips and marker strips central reservation centre plane

characteristic checking deflections by calculation


damage danger dated reference (to standards) definitions delivery derivation of design values design design assisted by testing design resistance (Rd) design value of a geometrical property design value of a material property design value of geometrical data design value of the resistance design values design working life design; dimensioning designer detailing deterioration

dynamic impact components for actual trains dynamic models of pedestrian loads eccentricity of vertical loads, eccentricity of resulting action (on reference plane) effects of actions (E) equivalent distributed loads from slipstream effects equivalent uniformly distributed load for axle loads on embankments factors defining representative values of variable actions fatigue load models

checks during construction classification and geometry classification of the control measures

duty environment environmental influences error

coefficient

estimate

coefficient of linear thermal expansion columns

Eurocode

compliance controls at delivery to the site

Eurocode Programme European Prestandard (ENV) European Standard (EN) execution execution of tests fabrication restrictions failure

fatigue loading filling loads fixed action

components of the displacement of a point compression configuration factor conformity control

free action

connections

fitness for use form of structure

free water pressures

construction details

function

frequent value of a variable action fundamental combination

construction height

fundamental requirements

construction products directive construction rules constructional detailing continuous beam

general

ground water pressures group of loads groups of traffic loads on road bridges horizontal forces characteristic values ice loading

(E.E.C.) directive distinction between principles and application rules document

imposed loads imposed prestress in ground anchors or struts in situ stresses in the ground

documentation

indirect action

durability

indirect permanent action

continuous frames control and maintenance of the completed structure control of design control of production and construction control of the different stages of the building process convective heat transfer coefficient cool down

general test procedures

geometrical data hazard imperfection importance factor inaccuracy

Terminology

indirect variable action interaction force due to braking interaction force due to deflection interaction force due to temperature interaction force due to traction (acceleration) interaction force transferred to the bearings length of the longitudinal distribution of a load by sleeper and ballast liquid properties load arrangement load case load model on embankments loading rate loads on silos due to particulate materials loads on tanks from liquids lower characteristic value of a permanent action lower design value of a permanent action magnitude of characteristic axle load (Load Model 1 ) on notional lane number i (i=1 ,2...) of a road bridge magnitude of the characteristic longitudinal forces (braking and acceleration forces) on a road bridge magnitude of the characteristic transverse or centrifugal forces on road bridges main variable actions modelling for fire actions modelling for structural analysis and resistance modelling in the case of dynamic actions modelling in the case of static actions

coordinates corrosion critical critical temperature critical temperature of structural steel cross section cross-section shapes cross-sectional area cross-sectional dimensions deck deflections deformation properties deformations density depth design assisted by testing design criteria

design fire

design fire load density

design graph design methods design situations design value of the applied axial force (tension or compression) design value of the applied internal bending moment

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criterion

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163 inspection

models of special vehicles for road bridges

inspection

movements and accelerations caused by earthquakes, explosions, vibrations and dynamic loads movements caused my mining movements due to creeping or sliding soil masses movements due to degradation, decomposition, selfcompaction and solution natural frequency of the unloaded bridge nosing force

inspection and testing insurance irreversible serviceability limit states level; standard liability limit state limit state design limitations limitations and simplifications longevity maintainability maintenance manufacture material material property measure of reliability

measurement member state method of construction mistake

number of notional lanes for a road bridge other representative values of variable and accidental actions particulater material properties pedestrian, cycle actions and other actions specifically for footbridges permanent action prestressing action quasi-permanent value of a variable action quasi-static action rail traffic action rail traffic actions and other actions specifically for railway bridges reduction factor for slipstream effects on simple horizontal surfaces adjacent to the track remaining area removal of load or excavation of ground representation of actions representative value of an action

design value of the applied shear force at the ultimate limit state design value of the applied torsional moment

destabilizing diagonal diameter

dimension

monitor National Application Document (NAD) nominal value nominal value of geometrical data nominal value, or a function of certain design properties of materials normative references objectives of the Eurocodes operable state; workable state operation; operating state; working state overloading owner

dimensions and axes of sections direct

paragraph

displacement distance distribution of the temperature eccentricity eccentricity of the load effective effective section elastic elastic analysis elastic critical load

parameter part part of structure partial factor associated with the uncertainty of the action and/or action effect model partial factor design participant performance placing; installation planning of tests population preliminary design preparation of material prescribe

elastic critical moment elastic global analysis elastic range elongation

principal classifications probability of survival

Terminology

procedure process

resulting action road traffic actions and other actions specifically for road bridges roads vehicles seepage forces

environmental conditions equilibrium conditions

seismic action

equilibruim moisture content

self-weight short term loading single action

equivalent horizontal forces equivalent surface Euler buckling load

slipstream effects from passing trains (aerodynamic effects) snow load space between distributed loads (load models SW) specific factor for slipstream effects on vertical surfaces parallel to the tracks speed in km/h static action surcharges swelling and shrinkage caused by vegetation, climate or moisture changes tandem system tandem system for load model 1 temperature effects, including frost heave terms relating to actions thermal climatic action traction (acceleration) force traffic loads train shape coefficient

European technical approval

uniformly distributed load for load model 1 upper characteristic value of a permanent action upper design value of a permanent action variable action vertical axle load

emissivity coefficient end moment loading

evaluation of test results expansion length experimental evidence experimental model extension external control external member

fabrication and erection fabrication tolerances factor failure in shear failure mode fastener fatigue fatigue assessment procedures fatigue strength field of temperature finite difference method fire behaviour fire class


project project specification property quality assurance quality control radius recommendation reduction factor reference period rejection reliability reliability (narrow sense) reliability condition reliability differentiation renovation repairability required; specified requirement resistance (R) responsibility reversible serviceability limit states robustness rule safety scope section service conditions serviceability serviceability limit state significant

vertical distance from rail level to the underside of a structure vertical distributed load vertical loads - characteristic values weights of soil, rock and water wheel load width of a notional lane for a road bridge wind actions wind force wind force compatible with railway traffic 0.1 % proof-stress of prestressing steel additional rules for high bond bars exceeding 32 mm in diameter analysis of slabs anchorage anchorage by mechanical devices anchorage methods anchorage of links and shear reinforcement anchorage zones for posttensioning forces anchorages and couplers anchorages and couplers for prestressing tendons anchorages and joints area of a prestressing tendon or tendons area of reinforcement in the compression zone at the ultimate limit state area of reinforcement within the tension zone basic anchorage length bond bond and anchorage bond conditions calculation of crack widths cases where calculations may be omitted cast in situ solid slabs characteristic 0.1 % proof-

fire compartment fire conditions fire exposure fire load density fire protection fire resistance fire resistance test fire risks fire safety fire safety engineering fire situation fire spread fire wall fixed value flange flat bottom flexural stiffness flow chart flow pattern

simplified verification for building structures site inspection special terms relating to design in general specific test procedures specification

specimen state of failure steering step step; measure strength test

fluidised material footbridge

structural member structural model structural system structure (load-bearing) structure

footpaths

subscripts

force

suitability supply symbols

foundations frame frame stability frames free flowing material funnel flow (or coreflow) gap gauge

(final) take-over task terms relating to actions terms relating to geometric data terms relating to material

Terminology

stress of prestressing steel characteristic axial tensile strength of concrete characteristic compressive cylinder strength of concrete at 28 days characteristic tensile strength of prestressing steel characteristic tensile strength of reinforcement characteristic uniform elongation of reinforcement or prestressing steel at maximum load coefficient of friction between the tendons and their ducts compressive strain in the concrete compressive strain in the concrete at the peak stress fc compressive strength of concrete compressive stress in the concrete compressive stress in the concrete at the ultimate compressive strain ď Ľcu concentrated forces concrete concrete cover continuity and anchorage control of cracking without direct calculation controls prior to concreting and during prestressing corbels creep function at time t cross-sectional area of shear reinforcement curvature at a particular section deep beams deformation properties design value of concrete cylinder compressive strength design value of the damage

generation of the fire geometrical average geometrical imperfections geometry global analysis

global structural analysis global structural analysis (for fire) global structure grading system gross gross area guidelines for loading tests hard shoulder hard strip headroom heat flux heat transfer heating rates height high temterature creep higher hogging moment hogging moment resistance hole homogenizing silo

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production

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165 properties test test conditions test evaluation test location test to failure testing laboratory; test establishment; test centre theory type of building or civil engineering works type of construction types of tests ultimate limit state uncertainty undated reference (to standards) unsafe state use user verification verification by the partial factor method verification of static equilibrium and strength verification of structures susceptible to vibrations verifications of serviceability visual inspection warning

horizontal and vertical

indicator (fatigue) design value of the secant modulus of elasticity design yield strength of stirrups detailing provisions determination of prestressing force determination of the effects of prestressing determination of the effects of time dependent deformations of concrete diameter of a reinforcing bar or of a prestressing duct diaphragms

spacing horizontal reinforcement

length of diagonal

idealisation of the structure

limiting values

initial force at the active end of the tendon immediately after stressing

load bearing criterion

isolated precast members

load bearing failure

lap splices for bars or wires

load bearing function

indirect indirect fire actions inferior inferior; lower

laps for welded mesh fabrics made of high bond wires largest nominal maximum aggregate size

load bearing resistance

initial

lever arm of internal forces

load level

limit states of cracking

load-bearing member

limit states of deformation

long

limitation of damage due to accidental actions

longitudinal shear force

longitudinal reinforcement

loss of equilibrium

mean value of axial tensile strength of concrete mean value of concrete cylinder compressive strength mean value of the prestressing force at time t, at any point distance x along the member minimum dimensions modulus of elasticity of

low cohesion

hopper horizontal horizontal displacement horizontal distance to the track centre horizontal screen ignition temperature in mid-span indices (replace by numeral)

ductility characteristics durability requirements effective span of a beam effects of prestressing at the ultimate limit states effects of prestressing under service conditions elongation of reinforcement or prestressing steel at maximum load equivalent diameter of a bundle of reinforcing bars fabrication, assembly and placing of the reinforcement factors for frequent values factors for quasi-permanent values final value of creep coefficient flange

inspection gangway

flexural reinforcement

internal flow

forces associated with change in direction formwork and falsework general detailing arrangements grouting and other protective measures length

internal moments and forces

inner

insulation material integer integrity criterion

load introduction area

internal internal control

kick load lateral-torsional buckling of beams lattice girder joints partial factor (safety or

Terminology

lower

serviceability) partial factor associated with the uncertainty of the resistance model and the dimensional variations partial factor for a material property partial factor for a material property, also accounting for model uncertainties and dimensional variations partial factor for accidental actions partial factor for actions, also accounting for model uncertainties and dimensional variations partial factor for permanent action partial factor for permanent actions in accidental design situations partial factor for permanent actions in calculating lower design values partial factor for permanent actions in calculating upper design values partial factor for prestressing actions partial factor for prestressing actions in accidental design situations partial factor for the resistance, including uncertainties in material properties, model uncertainties and dimensional variations partial factor for variable actions partial factors for materials

lower flange

partial safety factors for materials

major axis marker strip

particular cases particular details for seismic


oven-dry density of concrete in kg/m3 overall width of a crosssection, or actual flange width in a T beam perimeter of concrete crosssection, having area Ac permissible curvatures placing of the tendons plain or lightly reinforced concrete post-tensioning precast element prestressing devices prestressing force prestressing steel pre-tensioning proportioning of ties reinforced concrete walls reinforcement ratio for longitudinal reinforcement reinforcement ratio for shear reinforcement reinforcing steel relaxation removal of formwork and false work required anchorage length rib and block floors sandwich panel secant modulus of elasticity of normal weight concrete second moment of area of a concrete section shear reinforcement spacing of bars spacing of stirrups strength classes of concrete stress-strain diagram structural analysis of walls and plates loaded in their own plane

design mass mass mass density; unit mass (kg/m3)

particular details for structural fire design passive fire protection measures patch load

material property

per unit length

maximum mean mechanical actions

perpendicular persistent design situation physical properties

mechanical properties member analysis member analysis (for fire) methods for checking stresses minimum minimum thickness

pier pitch plane flow plastic analysis

minor axis modulus of elasticity moisture content

plastic hinge mechanism plastic resistance to axial compression plastic theory plate point of zero moment

moment resistance

Poisson’s ratio

movable inspection platform multi-axial stresses mass flow

ponding position of neutral axis prefabricated member

natural fire net net area net heat flux

preloading force pressures prestress process standards

neutral axis depth

profile depth

neutral bending axis noise barrier nominal nominal temperature-time curve nominal yield strength non-dimensional slenderness ratio

proportional limit protected members protection systém radiative heat flux radius of gyration ratio

structural steel surface condition surface finish surface reinforcement tangent modulus of elasticity of normal weight concrete at a stress of c = 0 and at time t tangent modulus of elasticity of normal weight concrete at a stress of c = 0 and at 28 days technological properties temporary works inserts tensile strength

notional lanes

tensile strength of prestressing steel tensile strength of reinforcement tensioning of the tendons ties time at initial loading of the concrete time being considered time dependent effects tolerances for concrete cover tolerances for construction purposes thermal actions

number of ..

thermal conductivity thermal elongation; thermal expansion thermal gradient thermal insulation thermal insulation criterion thermal material properties thermal resistance thermal response thin walled circular silo tolerances torsion thermal actions standard temperature-time

Terminology

non-separating wall non-structural member non-sway frame normal normal erection tolerances

reaction rectangular axes rectangular sections reduction coefficient  regular spacing of axies

normal stress

regulation

normal temperature design notation

representative resistance of crosssections resistances of crosssections of beams response of structural member resulting emissivity

objectives outbreak of the fire outer overall depth of a crosssection overlap parallel parapets partial

road restraint systems room temperature root radius

structural analysis of beams and frames structural behaviour structural element [member]

sagging moment resistance

structural failure structural failure of wall in the fire situation structural fire design structural mechanics structural models for overall analysis structural response structural steel subassembly subassembly analysis (for fire) sub-frame superior

second order effects section and member design

rotation safety barrier safety requirements sagging moment

secant modulus second moment of area

section modulus section properties semi-continuous frames separating function separating member sequence of construction shear shear force shear load

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reinforcement or prestressing steel normal weight concrete

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167 curve stiff stiffener stiffness (shear, rotational stiffness) stiffness factor (I/L) storey strength strengthening and repair of buildings stress ratio stress-strain relationship strong axis structural analysis standard temperature-time curve stiff stiffener stiffness (shear, rotational stiffness) stiffness factor (I/L) storey strength strengthening and repair of buildings stress ratio tolerances with regard to structural safety torsional reinforcement total cross-sectional area of a concrete section transport and storage of the tendons transport, storage and fabrication of the reinforcement tolerances with regard to structural safety torsional reinforcement total cross-sectional area of a concrete section Â

support support and boundary conditions support arrangements support conditions sway frames and non-sway frames system length tabular data

shear resistance shear stress shift of centroidal axis silo silo forms simple frames slender silo

tandem systĂŠm tangent modulus tank technical delevery conditions temperature

slenderness slenderness ratio slip slip factor

temperature analysis temperature- dependent material property temperature differentials

spacing span specific heat

temperature distribution temperature field temperature rise temperature-time curve

spread of the fire sprinkler systems squat silo stabilizing

tensile strength tensile test

standard standard fire classe

tension force tension members

standard fire duration standard fire resistance

transverse reinforcement transverse reinforcement parallel to the concrete surface tying systĂŠm ultimate bond stress ultimate compressive strain in the concrete

slope

transport and storage of the tendons transport, storage and fabrication of the reinforcement tolerances with regard to structural safety torsional reinforcement transport and storage of the tendons

Terminology


Terminology STRUCTURAL ENGINEERING ROOM

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10. Realisation Projects 1. Club AFAD 2. Technical Device Design for Department of Conservation and Restoration 3. Design of Digital Art Library for AFAD Authors: Assoc. Prof. Dipl. Ing. Sabah Shawkat, MSc, PhD. Msa. Dipl. Ing. Richard Schlesinger, PhD.

STRUCTURAL STRUCTURALENGINEERING ENGINEERINGROOM ROOM

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PRESENT STATE CANTEEN

Realisation Projects


EXTERIOR PLACE FOR NEW GALLERY EXTENSION

Realisation Projects

STRUCTURAL ENGINEERING ROOM

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PLAN 1. FLOOR PRESENT STATE

Realisation Projects


Realisation Projects STRUCTURAL ENGINEERING ROOM

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174


STRUCTURAL ENGINEERING ROOM

Department of Architecture 175

PLAN 1. FLOOR NEW DESIGN

Realisation Projects


Realisation Projects STRUCTURAL ENGINEERING ROOM

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PLAN KITCHEN

SECTION 1

SECTION 2

Realisation Projects


MAIN ENTRANCE

Realisation Projects DRESSING ROOM DRESSING ROOM

KITCHEN

KITCH E N O P E R AT I O N D E S I GN

S U P P LY FOR KITCHEN

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WC, HYGIENE

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GALLERY KITCHEN BAR CANTEEN

Realisation Projects


SHAPE OF EXISTING MASONRY ROOF

GALLERY 3. FLOOR

Realisation Projects

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PLAN 1. FLOOR

PLAN 1. FLOOR- STRUCTURAL DESIGN

PLAN 2. FLOOR

PLAN 2. FLOOR- STRUCTURAL DESIGN

Realisation Projects


Realisation Projects STRUCTURAL ENGINEERING ROOM

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Realisation Projects


Realisation Projects STRUCTURAL ENGINEERING ROOM

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STRUCTURAL ENGINEERING ROOM

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Realisation Projects


Realisation Projects STRUCTURAL ENGINEERING ROOM

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STRUCTURAL ENGINEERING ROOM

Department of Architecture 187

Realisation Projects


SECTION AA

Realisation Projects

STRUCTURAL ENGINEERING ROOM

SECTION BB

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STRUCTURAL ENGINEERING ROOM

Department of Architecture 189

SECTION AA

SECTION BB

Realisation Projects


SECTION AA

Realisation Projects

STRUCTURAL ENGINEERING ROOM

SECTION BB

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STRUCTURAL ENGINEERING ROOM

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Realisation Projects


SECTION BB

TUBE LEGEND air inlet tube air intake tube

Realisation Projects

STRUCTURAL ENGINEERING ROOM

SECTION AA

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Realisation Projects


Realisation Projects STRUCTURAL ENGINEERING ROOM

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STRUCTURAL ENGINEERING ROOM

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DIGITAL LIBRARY BASEMENT- PRESENT CONDITION

NEW DESIGN- PLAN

Realisation Projects


DIGITAL LIBRARY OSB PANELS ROOF PLAN

SECTION AA

Realisation Projects

STRUCTURAL ENGINEERING ROOM

DIGITAL LIBRARY TIMBER PROFILED ROOF PLAN

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SECTION- VIEW E

SECTION- VIEW D

STRUCTURAL ENGINEERING ROOM

SECTION- VIEW B

SECTION- VIEW C

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SECTION- VIEW A

DIGITAL LIBRARY BASEMENT PLAN

Realisation Projects


Department of Architecture

SECTION- VIEW D

SECTION- VIEW A

Realisation Projects

SECTION- VIEW C

SECTION- VIEW B

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9 788026 313007

Profile for Archineer

Element Design to Shape a Structure II  

Author: Sabah Shawkat

Element Design to Shape a Structure II  

Author: Sabah Shawkat

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