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 4.214 Verify the stresses in concrete and in steel 4. Limitation of Stress
1
4. 1 Verification at the Serviceability Limit States
22
of a Tbeam
1
Example 4.215 Verify the position of the neutral axis of a Tbeam 24
4.1.1 Permissible Stresses
1
Example 4.216 Design the flexural reinforcement for the Ibeam 25
Example 4.1.11 Verification of the stresses in concrete and in steel
2
Example 4.217 Verify the stresses in concrete
3
Example 4.218 Verify the stresses in concrete
26
and in steel of a Tbeam
reinforcement 4.2 Service Load StressStraightLine Theory 4.2.1 Analysis for StressesSection cracked and elastic Example 4.2.1 Compute the stresses in concrete and steel
and in steel of a Tbeam
3 4
Example 4.219 Verify the stresses of concrete
29
and steel of a Tcolumn
reinforcement for a given bending moment Example 4.22 Compute the specified load stress in concrete
27
5
Example 4.220 Verify the stresses in concrete
30
and in steel of a Tbeam
and steel Example 4.23 Compute the stress in concrete and steel
7
Example 4.221 Design the flexural reinforcement for the T beam 32
Example 4.24 Check the stresses of concrete and steel
8
Example 4.222 Assessment of tension parts of PPRCB according 33 to limit state of crack width
for a given bending moment and axial load Example 4.25 Verify the stresses in concrete and in steel
9
Example 4.223 Assessment the risk of longitudinal cracks due to 39
11
Example 4.224 Limitation of concrete stress due to
increased pressure stress concrete
reinforcement Example 4.26 Verify the stresses in concrete and in steel
an increased creep
reinforcement Example 4.27 Check the stresses in concrete and in
11 5. Deformation Behaviour of Reinforced Concrete Beams
steel reinforcement Example 4.28 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 4.29 Verify the stresses in concrete and in steel
14
5.1.2 Loading and Instrumentation Details
Example 4.210 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 4.211 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 4.212 Determine the entire area of steel reinforcement
19
5.3.1 Formation, Development and Width of Cracks
Example 4.213 Compute the steel and concrete stresses
20
5.3.2 Evaluation of Cracks
54 55 55 58 59 59
Example 6.2.13 Static scheme of rafter with overhangs from left 117
shearing crack of reinforced concrete beam Example 52 To assess the crack width perpendicular
Example 6.2.14 Rafter with overhanging ends of the
65 66
perpendicular cracks Example 54 The calculation of the stress in the reinforcement
69
after full cracking Example 55 Calculation of shear crack widths on a reinforced
71
73 Example 5.41 Calculation of deflections for rectangular
Example 6.2.17 Calculation of the thickness t and stress
119
Example 6.2.18 Design dimensions of the timber elements
119
Example 6.2.19 Design dimensions of the timber elements
120 121
Example: 6.31 Calculation of carrying capacity of the column
124
Example 6.32 Calculate the required width of the support
125
of beam on column 125
dimensions of the elements
75
Example 6.34 Assessment of wooden column
127
75
bending moments Example 5.44 The calculation of the stress in the reinforcement
119
Example 6.33 Top chord of truss beam be designed
for rectangular crosssection Example 5.43 Evaluation of deflections due to shear forces and
118
Example 6.2.16 Buckling calculation
73
reinforced concrete beam Example 5.42 Detailed calculation of the coefficient Ď‡
Example 6.2.15 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 53 Assessment according to limit state the widths of
116
7. Profiled steel sheeting
128
7.1 Structural steel
128
76
7.11 Profiled steel sheeting
after cracking and crack width determination
128
84
7.2 The permanent load
129
Example 5.46 The calculation of loads using trapezoidal rules
90
7.3 Methods of analysis and design
130
Example 5.47 Calculation of ideal load
93
7.31 Ultimate limit state
Example 5.48 Calculation of the deflection by the ideal of loads
94
7.32 Bending failure and bending strength
131
7.33 Anchorage failure, anchorage fatigue resistance and
131
Example 5.45 Theorem of reciprocity of virtual work
6. Behaviour and Conception of Timber Structures Example 61 : Compute the stress in Rafter for a given bending
anchorage strength
95 96
moment and external force at the section Example 62 : 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
7.34 Shear failure
132
7.35 Design of the slab for point and line load
133
7.36 Reinforcement of the support
133
7.37 Typical structural details of composite slabs
134
7.38 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
8.11 Verification masonry Example 6.2.11 Dimensioning reinforcementofin unreinforced piles 115
walls
8.12 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 6.2.12 Static scheme rafters as a simple beam
63
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Example 51 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
8.11 Verification of unreinforced masonry walls
152
8.12 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
stepbystep 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 wellmanaged 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'
 nonlinear 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
bd c
´c 0.6 fck.cyl overall w idth of a crosssection [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: 4.1.11
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
crosssection figure 4.1.1.11. Verify the stresses in concrete and in steel reinforcement. The section properties: Characteristic value of concretecylinder compressive strength (MPa):
f ck
25 MPa
Characteristic yield stress of reinforcement (MPa):
f yk
410 MPa
Crosssection ( 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 crosssection (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: 4.1.1.11
2
Moment of inertia of the transformed cracked crosssection 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
nMs
s2
nMs
x d1 Icr
66.76MPa
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.
120.59MPa
s2 0.8f yk
Using the design aid in figure B3B3.3
out
or
0.228 m 2
Icr
From the graf figure B3B3.3 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 4.1.11: Pure bending with tension and compression reinforcement in a rectangular
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2
4.2 Service Load StressStraightLine Theory Stress in tension reinforcement (MPa): s2
´ s2
whichever is lesser
122.55 MPa
s2 0.8 f yk
4.2.1 Analysis for StressesSection 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 straightline 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: 4.1.1.12
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
bx
3
3
n As ( d x )
2
from which
< 0,8.fyk
1000
0
Fs
2
bx
The moment of inertia of the section, Icr is given by:
Fc z= .d
4000
bc
bx
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: 4.1.1.13
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. 4.2.11. 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 crosssection (m): d2
2
bd
250 kN m
0.4 m
and
Ms
410 MPa
Crosssection ( m): b
0.45
7.27681 MPa
Bending moment at the section ( MN m): Ms
2
10
bd
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 crosssection ( 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: 4.2.11
Limitation of Stress
Solving
x
0.268
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Example 4.21: A reinforced concrete beam of rectangular section has the crosssection
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4
Example 4.22: The crosssection 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 4.2.21a. 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
bx
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
Crosssection ( m):
Tension steel stress: s
fck
Characteristic yield stress of the steel (MPa):
b
dx
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 crosssection ( m): Fc z
bx 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
nc
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, microcracks or high levels of creep, where they could result in unacceptable
A1
nc
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
nt
4
0.0004 m
2
A2
t
nt
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 B3B3.3
transformed section, shown in figure 4.2.21b.
Ratio of tension reinforcement ():
The modular ratio
Es
n
n 15
Ec
about the neutral axis (see figure 4.2.21b): 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
bd
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 crosssection about neutral axis ( m ) Icr
bx 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 dx 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 4.2.21c, the concrete stress at compression
b2
1.19
form ():
0.23385 m
compressive forces Fc in concrete and Fs in steel, about tension reinforcement):
bx
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 bd
s1
115.25 MPa
or
s1
n b2
s2
198.19 MPa
or
s2
n bc
s1
dx 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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6
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7
Effective depth of a crosssection ( 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 4.23: The member shown in figure 4.2.31 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 d1 3
z
0.56155 m
As
0.00153 m
2
Assumptions: Characteristic value of concrete cylinder compressive strength (MPa): fck
90 M
Figure: 4.2.21
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
Crosssection ( m): b
0.35 m
h
Figure: 4.2.31
0.70 m
Distance of the tension reinforcement from the extreme fibre of the crosssection (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 bh
100
0.80122
Ms 2
bd
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 crosssection ( m):
1.69062
From the diagram we obtain ´ bc
246 MPa
sall
Distance of the tension reinforcement from the extreme fibre of the crosssection (m): d2
Using the design aid in diagram
0.6 fyk
Crosssection ( 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 4.24: Compute the steel area to the crosssection shown in figure 4.2.41 and check
2
The entire provided steel area ( m ) will then be:
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Example 4.25: Pure bending with tension and compression reinforcement in a rectangular
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crosssection figure 4.2.51. Verify the stresses in concrete and in steel reinforcement. The section properties: Characteristic value of concretecylinder compressive strength (MPa): fck
25 MPa
Characteristic yield stress of reinforcement (MPa): fyk
410 MPa
Crosssection ( m):
b
0.4 m
d
0.55 m
Bending moment at the section ( MN m): Ms
Figure: 4.2.41
Distance of the centroid tension and compression reinforcement from the extreme fibre of the
Lever arm of the sectional forces (m):
z
d1
crosssection (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 stressbc
2 Ns bx
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
dx Icr
s2
120.59 MPa
allowable
Using the design aid in diagram
A2
1.18
100
bd
From diagram, follows
Fig: 4.2.51 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
bd
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 crosssection about the neutral axis ( m ) Icr
bx
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 4.2.51b, 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 4.2.61b. 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
bx
3
n A2 ( d x )
3
2
Icr
0.00559 m
2
Figure 4.2.52: Isometric view of reinforced concrete beam
Example 4.26:
Use the section properties from example 4.2.5, the beam is without
compression reinforcement shown in figure 4.2.61. 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 crosssection 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
dx Icr
s2
123.11 MPa
s2 sall
Example 4.27: A rectangular member has the dimensions figure 4.2.71 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
Crosssection: h
Figure: 4.2.61 bx
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 crosssection (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 crosssection 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
bh 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: 4.2.71
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
bh 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 4.28: A rectangular column has the dimensions shown in figure 4.2.81. 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 crosssection (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
cy
x
0.173 m 3
Section modulus of the transformed cracked crosssection 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 py 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 crosssection is partially compressed
410 MPa
Crosssection (m): b
e
22.30 cm
2
S
K
bx
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: 4.2.81
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 4.29: A rectangular column has the dimensions shown in figure 4.2.91. Verify the stresses in concrete and in steel, the crosssection 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
Crosssection (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 crosssection (m): d
h d2
d
Figure: 4.2.91
1.03 m
We find point e on the outside of the crosssection, thus the crosssection 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 crosssection about the neutral axis: 2
bh 2
v1 v2
n A2 d A1 d1
S v1
Bo
0.57 m K
h v1
v2
bx
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 crosssection 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 py q
allowable
Stress in compression and tension reinforcement (MPa):
c d1 90
c d1
bx
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 crosssection 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
Crosssection (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: 4.2.101
effective depth of the crosssection (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
bh
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 crosssection is partially compressed.
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Example 4.210: A rectangular member has the dimensions as shown in figure 4.2.101. Verify
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Other way of solution is the compressive stress in concrete as follows (MPa):
4
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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.
yc
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 ) bx
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
Crosssection (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 crosssection (m): d2
0.05 m
d1
0.05 m
Effective depth of the crosssection (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
bx
Example 4.211: A rectangular member has the dimensions as shown in figure 4.2.111. The
3
y py 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 crosssection 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 crosssection, 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 crosssection is entire in compression.
Bo n A1 A2 v2
4
Moment of inertia of the transformed uncracked section ( m ): 3
bh
I
3
Figure: 4.2.111
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
bh 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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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 crosssection (m): d1
0.04 m
d2
0.06 m
Effective depth of the crosssection (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: 4.2.121
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
Crosssection (m): b
h
The entire area of tension reinforcing longitudinal bars:
Example 4.212: Determine the entire area of steel reinforcement as shown in figure 4.2.121
fck
e
h d2
sall
246 MPa
2
Example 4.213: A Tbeam having the crosssectional dimensions shown in figure 4.2.131
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):
CrossSection (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 Tbeam (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 crosssection (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 crosssection (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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20
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21 Es, Ec
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 4.2.131c
section. The neutral axis may lie in the flange as shown in Fig. 4.2.131b (that is x < hf ) or it may pass through the web as in Fig. 4.2.131c ( 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 4.2.131d, 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 bx
b bw x hf
and
The moment equation gives: Ms
bx
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
dx
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: 4.2.131 To locate the neutral axis, first assuming that it lies in the flange, equating the moments of areas about neutral axis figure 4.2.131b 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 crosssection 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 crosssection 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 4.214: Verify the stresses in concrete and in steel of a Tbeam figure 4.2.141 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 crosssection, 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 4.2.131b, 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 crosssection 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 Tbeam (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
STRUCTURAL ENGINEERING ROOM
The compressive stress in the concrete at the top fibres of the section (MPa):
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22
d2
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23
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 4.2.141: 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 4.215: Verify the position of the neutral axis of a Tbeam section figure 4.2.151 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 crosssection area will be in compression.
B o v2
fyk
Igg´ Bo v2
Ns
o
Bo
Total depth of the Tbeam (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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24
Example 4.216: Design the flexural reinforcement for the Ibeam section figure 4.2.161
STRUCTURAL ENGINEERING ROOM
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25
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
Crosssection (m):
Figure: 4.2.151 2
Area of compression reinforcing bars ( m ): A1
0m
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
dx 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.
bx
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 crosssection (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 4.217: Verify the stresses in concrete and in steel of a Tbeam section figure 4.2.171 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 Tbeam (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: 4.2.161
2
b hf d Mt
The neutral axis will then be in the web of the Tbeam.
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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26
Example 4.218: Verify the stresses in concrete and in steel of a Tbeam section figure 4.2.18
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27
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 Tbeam (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 ): 0m
90 cm
Flange depth (m):
Figure: 4.2.171
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 ): bx
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
dx 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 Tbeam 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 Tbeam 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
bx
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 Tsection. p
q
3bc
2
bw
bw 2bc
3
bw
3
y py 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
yc
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
dx x
Figure 4.2.181: 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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28
Example 4.219: Verify the stresses of concrete and steel of a Tcolumn figure 4.2.191 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 Tbeam (m) is: d
Figure: 4.2.191 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 Tbeam (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 Tbeam (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: 4.2.201
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
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Example 4.220: Verify the stresses in concrete and in steel of a Tbeam section figure 4.2.20
2
Area of tension reinforcing bars ( m ):
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30
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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 Tsection.
p
q
3bc
2
2bc
b
3
bw
bw
3
b
2
bw
bw
3
y py q
Figure 4.2.202: 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
dx 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
yc
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 Tbeam.
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 Tbeam is (m): d
Distance of the tension reinforcement from the extreme fibre of the crosssection (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 Tbeam. 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 4.221: Design the flextural reinforcement for the T beam figure 4.2.211 caused by bending moment Msand axial tension load Ns .
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32
STRUCTURAL ENGINEERING ROOM
Department of Architecture
33 Provided:
Example 4.222: 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 quasipermanent 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: 4.2.221 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: 4.2.211
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
Crosssectional 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 crosssection: 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 crosssection 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
Crosssectional 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
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35 Description of fullsectional acting
Ideal characteristics of a fullyacting sectional:
Area of the concrete section
Area ideal crosssection:
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 crosssection:
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 crosssection:
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 nonlinear stress distributions (self –equilibrating effects); the
following rules may be applied.
When the crosssection 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 nonuniform normal stress distribution by bond forces into the concrete crosssection 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= 0.41.0 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
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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 crosssection
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 crosssection 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 crosssection 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 crosssection:
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
Crosssection 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 longterm and shortterm 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 longacting, 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 onetime 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 crosssection 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 quasipermanent load combination will
pure bending Effective crosssectional 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 crosssection 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:
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38
STRUCTURAL ENGINEERING ROOM
Department of Architecture
39 Example 4.223
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 4.2.231: Values of ideal crosssection 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 35MPa
fctm 3.2MPa 2
Ecm 33.5GPa 2
As1 8.042cm
As2 12.56cm
Es 200GPa
Ep 200GPa
Md 2.35MN m
Nd 1.20MN
Np1 4.20MN
Limitation of Stress
,
eh 0.5m
2
Ap 33.98cm
e
Es Ecm
The moment of inertia of ideal crosssection
ap 0.85m
as1 0.05m
as2 0.05m
bs 0.15m
bh 0.4m
bd 0.3m
H 1m
hh 0.2m
hs 0.45m
hh1 0.1m
hd1 0.05m
hd 0.2m
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
22.8478319MPa
c2
1.7410728MPa
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 crosssection
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 crosssection
2
0.5 bs H
4
Distance resultant horizontal forces of gravity ideal crosssection
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 crosssection 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 crosssection 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 0.5H 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 crosssection. Because EC2 is considering a uniform value of modulus of elasticity for concrete and tendons Es singlevalue ratio e
Limitation of Stress
Es Ecm
.
Ep We can when calculating the ideal sections to a
STRUCTURAL ENGINEERING ROOM
crosssectional dimension
Department of Architecture
40
The crosssection 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 prestress 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 crosssection. Because EC2 is considering a uniform value of modulus of elasticity for concrete and tendons Es singlevalue ratio e c1
Ep We can when calculating the ideal sections to a
G k j
P 2 1 Qk 1
Es Ecm
22.8478319MPa
Example 4.224: Limitation of concrete stress due to an increased creep
fctm 3.2MPa
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
1.7410728MPa
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 quasipermanent load combinations
prestressing force it is Np. Because the quasipermanent load exceeds 50% of the total load
fctm 3.2MPa
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
0.6f 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 0.9Q k 1 0.6fck
21MPa
c1
22.8478319MPa
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
0.6f ck we can strengthen the transverse reinforcement element or increase the thickness
of the concrete cover.
Limitation of Stress
Figure: 4.2.241
15.
As2 0.001256m
bs 0.4m
Np 2.8MN
Md 0.14MN m
H ap 0.15m
2
2
Ap 25.49cm H 1 m
Variable crosssection with cracks
Calculation of the lower (compression) edge crosssection 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 crosssection
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 crosssection
3 1
x 0 m
Given 3
Ii bs H
e 0.2 m
Np
The height of compression zone of crosssection
2
bs H0.5H e As1 as1 As2 H as2 Ap H ap agi Ai
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As1 8.02cm
c1
16.5076358MPa
c1
fctm
bs x0.5x e As1 as1 As2 H as2 Ap H ap
The moment of inertia of ideal crosssection with cracks
3 1
I bs x
12
fctm
c2
6.283225MPa
fctm be assumed with weakened crosssection (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
14.2127021MPa
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Checking terms of reliability
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c1
14.2127021MPa
0.45fck
15.75MPa
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
0.8fyk
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
0.75fpk
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 stressstrain 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 shortterm loading. That is, the plane sections remain plane under shortterm 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 shortterm. It was found that the contribution of shear effect to shortterm 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 nonnegligible 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 51: 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 tensionstiffening 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 WilliotMohr translocation polygons. Strains of the upper and lower choards and the diagonals of the truss analogy system
Figure 52: 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 shortterm 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 51) where the relationships of the midspan
most cases.
deflection am, av, atot versus the corresponding loading force F (Figure 51) 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 momentcurvature 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 VV ds ds Ec Jc G Ac
where M , V are the values of the inner force from the unit force in the crosssection 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 crosssection
The shape coefficient has a value of 1.2 for the rectangular section.
Figure 53: Deflection of beam due to shearing deformation and rotation The coefficient depends on the shape of the crosssection. 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 crosssection shape hs h
Relative area of crosssection
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
bh j
zd
h zh
3
3
wd
h
i4
hh h s h dd d s d 3
J zd
zh
3
hh
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 crosssection regarded to ultimate crosssection 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
Mis
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: 54 Dimension of the crosssection for the evaluation of coefficient
n
avexp
iVis
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 crosssection with regard to the dimensions of h, b
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s  length of measurement base,
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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 55: 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 "momentcurvature" as a basis. A similar method was applied for the calculation of the reduction of the shear stiffness, the shear stiffness of the transformed crosssection and the
shear stiffness in the whole range of the loading force GAc were applied.
Figure 56: 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: 57
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 hMiL
i 1
si ci Ec Ic
n
ck
hMiL1
i 1
m
n
s
k ck 2
k
i 1
iG Ac ViL
s
2
s
ck
i 1
m
iG Ac ViL1
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: 58
own method of calculation. (Figure 57  Figure 58).
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 shortterm 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 tensionstiffening 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 kbid d
20 MPa
The behaviour of reinforced concrete structures under service conditions can be
25 MPa
a Deflection d Effective depth of a crosssection (m b Overall width of a crosssection (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 nonnegligible redistribution of the stress resultants.
Figure: 59
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 shortterm 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 kbid 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: 510
40 MPa
ks
The present experimental research quantifies the effect of shortterm, 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) Isection (Hanečka, Križma, Ravinger, Shawkat, 1994), (Križma, Hanečka, Shawkat, 1993), (Križma, Shawkat, Bolha, 1995) and Tsection [11]. An experimental study aimed at the deformation behaviour of reinforced concrete beams subjected to longterm 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
beamsPPCB, steel fibre reinforced concrete beamsSFRCB.
beams had a clear span of 1120 mm and were tested under two concentrated loads. The Isection
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 crosssections, the beam RCB(5) of an Isection and the beam RCB(6) of a Tsection. The partial prestressed beam PPCB as well as the beam SFRCB reinforced by steel fibres had an Isection: 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 crosssection of the beams are in (fig. 5.1.11). 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 5.1.11: The crosssection of the tested beams
Figure 5.1.21: Side view of test beams with the network of bases for mechanical gauges (A B,C,D,Elocation of the hydraulic arms)
Deformation Behaviour of Reinforced Concrete Beams
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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. 5.1.21). 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. 5.3.21. 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 level0.5 The beams (Isection) 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 5.1.22: 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. 5.1.31) 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 crosssection
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 (Isection 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 crosssection with regard to the ultimate crosssection 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 crosssection. Its value in the linear
Relative integrals for the calculation of coefficient
state is 1,2 for the rectangular section, 1,86 for the Isection and 1,46 for the Tsection. The detailed calculation of the coefficient p for all section shapes is given in the example of the
1
Tsection. Relative dimensions of the crosssection with regard to the dimensions of h, b (according to Fig. 1) bh d
0.6b s bd b
0.25h s s
0.25b d bs b
h
b s b bh b
0.6h h d
0.1h 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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53 1
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 crosssection 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 i1 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
i1
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
i1
( 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 ofcr 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 5.1.31: 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 crosssectional 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:
rj1 is the increase of curvature
divided into four series I to IV. The crosssection of series I and III was 150/200 mm the shear
Mj,k is the bending moment of crosssection j at loading level k.
reinforcement 8, the longitudinal reinforcement 28 in the compression zone, and 216 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 crosssections, 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 crosssection j at the loading level s is: 1
U
v j
2
s
Vj k 1 Vj k j
k 2
where: Figure 5.2.11: Crosssections 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 crosssection 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 crosssection. 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 Aj
2
dA
A bz
where: G is shear rigidity A is the surface of the crosssection 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 coefficientcr. 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 crosssection:
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 (WillotMohr 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 momentcurvature 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 5.2.21: The portion of strain energy due to shear forces on total internal energy
Figure 5.31: Crosssections 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 shortterm 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 5.3.21: Diagram bending moment versus curvature and diagram shear force versus shear strain
Figure 5.3.22: Stresses of concrete versus measurment basis due to loading level
Figure: 5.3.21
Deformation Behaviour of Reinforced Concrete Beams
STRUCTURAL ENGINEERING ROOM
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. 5.3.21. 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 5.3.22: The final process of cracks development along the beam
Deformation Behaviour of Reinforced Concrete Beams
Figure 5.3.23: Geometry of beams according to SBETA
Figure: 5.3.24
Deformation Behaviour of Reinforced Concrete Beams
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Example 51: We calculate the distance of the first an inclined shearing crack of reinforced
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63
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: 5.11
crosssectional 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 crosssection:
2
Ast
Figure: 5.12
The depth of ideal crosssection:
2
Ii
h2
bh
3
x i x i h n Ast d x i
2 Asc xi asc2
Ii
4
0.00356m
Sectional area and the crosssection 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 52: 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 agi2 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 5.21
unacceptable cracking or deformation.
Bending moments Mu dw st
6.0 kN m 1.2 for
M slt
2.5 kN m
M sst
Longterm 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 crosssection 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 crosssection: Ai
slt
Crack width perpendicular to the centreline of the elements:
Percentage reinforcement crosssection: A st
slt
xr
1.33
tb
h
2 n M slt
0 mm
0.09305m
he
agi
h ast
he
0.07m
Figure 5.22
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 agi2 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
Longterm and shortterm 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
crosssectional dimension
120 kN
M slt
225 kN m
Nsst
50 kN
M sst
80 kN m
The longterm 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 53: Reinforced concrete columns with concrete C30/25 and reinforcement with
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A bp
from shortterm components of the load from the difference between the overall and longterm
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 xr2 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 xr2 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
longterm 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 shortterm components of the load establishing from the difference between the overall and longterm effects:
Figure 5.31 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 54: 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
Crosssectional dimension: b
250 mm
h
Percentage reinforcement crosssection:
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 crosssection:
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 agi2 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 crosssection:
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 crosssection 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 5.41 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
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The coefficient the cover layer is the core relationship:
2
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Example 55: Calculation of shear crack widths on a reinforced concrete beam according to
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CEB  FIP.
hef
Crosssectional 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 crosssection: 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 crosssection with a crack:
ss
Vn Vcd
1 bmin d ss sin ( ) ( sin ( ) cos ( ) )
ss
200.06MPa
ss
40 MPa
Figure 5.51
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 crosssection (greater shear stress)
5.4 Methods
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Example 5.41: 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 WilliotMohr 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
Xi1 X1i Xi1 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 crosssection 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 WillotMohr diagram on the upper and lower chord X0i
X0( i 1) 0i
Y0i
X1i
X1( i 1) 1i
Y1i
Figure 5.4.11: Side view of reinforced concrete beam
First Truss Analogy Equations for the calculation of the WillotMohr 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
X0i1 X1( i) X0i1 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
X1i1 X0i cot 3i csc Y1i1 X1( i) X0i cot 2i csc
Y1i
XOi [m]
Y1n
YOi [m]
X1i [m]
Y1i [m]
0.000025
7.4905E05
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 5.42: Detailed calculation of the coefficient for rectangular crosssection
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75 Relative integrals for the calculation of the coefficient
Relative dimensions of the crosssection 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 5.43: 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 crosssection 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 crosssection regarded to ultimate crosssection fibers
h
60
0
Figure 5.4.21: Dimension of the crosssection
Relative area of crosssection
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
140mm
h
3
3
147mm
Deformation Behaviour of Reinforced Concrete Beams
m
m m m m m m m
1.2
L34h
u4h u3h
L34h
0.000089m
3
4
34
L34h
L34 L34h
L34h
0.13991m
4d4h 3h3d
L34d
u4d u3d
L34d
0.00033m
Example 5.44: 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.
4d3d 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 33: 3v
u 3d u 3h h 3dh3h
Shear strain of depth 44:
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 43: 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 5.4.41: 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 crosssection
l
1.12m
hs
0.147m
s
0.14m
Ec G b
30.94GPa 0.435E c 15cm
Deformation Behaviour of Reinforced Concrete Beams
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Determination of :
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76
Depth of the crosssection
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77
Area of the crosssection Moment of inertia of the crosssection
Ic
Point load
1
Coefficient of the shape of the crosssection
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
3.374mm
i
The vertical displacements at each bases due to shear forces of fictitious truss system at the upper chord
1.2 1.29m
Average strain on the bases of the fictitious truss system
M2
0.21 m
i
51.52 kN
Fs
0.0 m
M6
a mexp
80kN
2
F s1
M1
The evaluation of the deflection due to the bending moment at the middle of the beam span
160kN
F s2 Fs
3
12
F s1
Service load:
at the upper chord
The bending moments along the beam due to unit force
h 20cm 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
0.00107m1 2 0.01832m1 7
0.02459m1 0.03061m1
3 8
0.0191m1 0.01876m1
4 9
d h i i
0.02211m1 0.00296m1
h 5
10 3 0.02007m1
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
0.881mm
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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Area of the crosssection after the full
STRUCTURAL ENGINEERING ROOM
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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 crosssection 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
STRUCTURAL ENGINEERING ROOM
The determination of the moment of inertia of the crosssection in state II after
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81
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 cJ 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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82
6. By means of the coefficients , which express the reduction of the moment of inertia of the
STRUCTURAL ENGINEERING ROOM
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83
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 cJ 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 crosssection 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 cJ 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 crosssection 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 5.45: 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
Crosssectional area: A 736 cm
2
Sectional moment of inertia: J c 0.002311258 m
4
Figure 5.4.42: Diagram bending moment versus curvature Load action on RC beam Fs2
280 kN
1.16
The coefficient of crosssectional shape:
1.85
The load action Fs2 F
Figure 5.4.43: 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
Department of Architecture
84
d
STRUCTURAL ENGINEERING ROOM
Department of Architecture
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 5.4.51 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 1Q1 2Q2 3Q3 4Q4 5Q5 6Q6 7Q7 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
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Vd19 2.392 mm
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Calculation of the theoretical deflection along the beam from the effect of shear forces
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yqteor
Q1i
GA 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
GA
9
Q111
GA 5
GA
9
GA
GA
10
Q19
4
Q13
GA
Q14 GA
2
GA
Q12 GA
10
GA
7
Q12
4
2
6
Q15
3
3
Q12
5
Q18
2
Q110 GA
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 GA GA5 GA GA9 G A
cr2
cr6
cr10
GA2 GA GA6 GA
cr3
cr7
GA10 G A
cr11
GA3 GA GA7 GA
cr4
cr8
GA11 G A
cr1
GA4 GA GA8 GA
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 crosssection 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 10Q110 s
Eq11 11Q111 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 7Q17 4 8Q18 9Q19 10Q110 11Q111
Eq 0.4687 m kN
Deformation Behaviour of Reinforced Concrete Beams
Example 5.46: 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 twomass inertia, in the quarter span Ii 2Ia figure 5.4.61.
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 5.4.61b is drawn triangular moment diagram with the greatest ordinate pL
Mmax
21.9% Eq
We are looking for angular rotation in supports a a b and deflection at the center of the
, figure 5.4.61c is a diagram
4
on span of secondary beams S
0.21944
M
L
4
M EI
, figure 5.4.61d on the diagram
M EI
as the load
and figure 5.4.61. 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 5.4.61: Simple beam with discontinuous variable inertia.
Deformation Behaviour of Reinforced Concrete Beams
STRUCTURAL ENGINEERING ROOM
The total internal energy is the sum of energy from the effect of bending and shear energy of
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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 5.4.62
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
STRUCTURAL ENGINEERING ROOM
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Example 5.47: Calculation of ideal load W Calculation of loads in ideal load crosssection we done according to figure 5.4.71. 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 cde
f 4
where
is parabola a f is the peak of the whole parabola. For variable parabola is valid the
same relationship as the
ql 8
2
(mid span moment is reduced to a quarter).
Figure 5.4.72: 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 5.4.71
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 crosssection 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 5.4.81: 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 midspan 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
1.37E06
1.71E05
2.94E05
1.98E05
2.73E05
1.19E04
1.76E04
2.73E04
1.90E06
1.55E05
7.18E04
8.58E04
1.03E03
1.19E03
1.26E03
1.32E03
1/i = curvature
3.61E06
1.09E05
4.68E03
5.70E03
7.17E03
8.89E03
9.75E03
1.08E02
dsi [m]
0.0425
0.0425
0.03875
0.03875
0.03875
0.03875
0.03875
0.03875
Wi
1.54E07
4.63E07
1.81E04
2.21E04
2.78E04
3.44E04
3.78E04
4.19E04
ri [m]
0.62375
0.58125
0.541
0.501875
0.463125
0.424375
0.385625
0.34687
Wi . ri
9.58E08
2.69E07
9.82E05
1.11E04
1.29E04
1.46E04
1.46E04
1.45E04
Element h_d
21_38
24_39
41_54
44_55
45_58
48_59
49_62
52_63
eh ed
3.50E04
4.94E04
5.45E04
6.43E04
6.49E04
6.57E04
6.61E04
6.63E04
1.37E03
1.44E03
1.48E03
1.50E03
1.44E03
1.45E03
1.45E03
1.45E03
1/i = curvature
1.17E02
1.31E02
1.38E02
1.46E02
1.42E02
1.43E02
1.43E02
1.43E02
dsi [m]
0.03875
0.03875
0.041667
0.041667
0.041667
0.04167
0.041667
0.041667
Wi
4.52E04
5.09E04
5.74E04
6.07E04
5.92E04
5.97E04
5.97E04
5.97E04
ri [m]
0.308125
0.269375
0.2292
0.1875
0.145733
0.10412
0.0625
0.020833
Wi . ri
1.39E04
1.37E04
1.32E04
1.14E04
8.63E05
6.22E05
3.73E05
1.24E05
Ra [  ]
6.35E03
ra [ m ]
0.56
Total deflection f [m]
2.06E03
Deformation Behaviour of Reinforced Concrete Beams
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Example 5.48: 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 bendingtype 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 crosssections 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 61: 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 crosssection:
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 62: Truss configuration with bolted metal plate connections Example 61: 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 62:
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 crosssection: 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 crosssection: 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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Moment of inertia of the crosssection: 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 EJ
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 6.11: 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 crosssection Glued laminated timber column: a = H / 22.5 Glued laminated timber beam: h = span / 16
Figure 6.12:
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 6.13: 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 framework 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 frameworks 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 6.13. 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 6.14: Layout of roof constructions subjected to horizontal loadwind
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 6.16: Plan elevation of primary structural systems and secondary structural systems
Figure 6.15: 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 Isection 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 6.17: Plan elevation of primary structural systems and secondary structural systems
Behaviour and Conception of Timber Structures
Figure 6.18: a) The transformation of shear forces, b) The transformation of shear forces to
Figure 6.110: Trusses with steel rod
compression respectively. Traction (tension) and bending, c) buckling, d) Tension for bending, e) Compression for bending.
Figure 6.19: The three articulated lattice frame
Figure 6.111: 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 6.18.
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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: 6.113 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: 6.114 Figure: 6.112
Behaviour and Conception of Timber Structures
With hardwood block and steel plates on both sides
Cablestayed bridges
With members notched to accommodate Isection plus top plate to resist tension
Cablestayed bridges with single pylon tied back to end support
Cablestayed bridge with single raking pylon
Cablestayed 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
Aframe pylon as three legged trestle
Cablestayed 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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Collar conections:
Trussed beams
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103 Trussedbeam 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€“ 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€“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 twopin frames
h=1/101/15
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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Threepin 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
Threepin 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 ThreePin Frames Glued laminated timber
Transverse stability is ensured by the frame with its fixedbase 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 ThreePin 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 Tbox and solidweb sections of arches. glued laminated timber circular arc asymmetric arch
Arches for Bridges Driveâ€“through Driveover
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€“ in side
crosssections 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 crosssection 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
castinplace
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€“ 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 crosssections 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€“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
Xa
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
bX
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
116
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 12MPa J
d fcomposite
Edr Jdr
1 m
b1 Fwood a1
4
A 16 fyd
z
bb
3
Concrete
nsmaller
N16
HX
bdr hdr Fc
M Ec X1 fcd J Ewood
h
f
where a and b is crosssection dimension of purlins or rafters
b1
ho
Behaviour and Conception of Timber Structures
V b z
V
1 2
ql 2
fdeflection
1.2
q L
8 G A
STRUCTURAL ENGINEERING ROOM
Table 6.11: Load, Span, Dimensions, and Axis Spacments for Soft timber, Class I, and II
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108
STRUCTURAL ENGINEERING ROOM
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109
Fc X
Ec Es
2
Xa
F
bX
Table 6.1 2: Dimensions, Area, Section Modulus, Second Moment of Area for Timber
2
b Fo a
b1
X1
b (cm) 8
a1
HX 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 235MPa 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
nsmalle
crosssection need to be composite
Nmax
V = shearing force
0.8N max
4
Nf
2
31.6d
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 6.115: 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 6.116
Table: 6.14 Figure 6.117
8.00
Figure 6.118
9.00
10.00
Figure 6.119
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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111
Table: 6.15 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 6.16: 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‚Ľ
Figure 6.120: Membrane structures
Figure 6.121: 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 Solidweb members, support on different levels Solidweb 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 Isection joists on metal hangers and metalweb 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 onsite 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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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: 6.21
Figure: 6.22
Radialsymmetrical arrangements  Symmetrical arrangement Cumulative arrangement of radial frames Figure: 6.23 Proposal roof using steel elements
Behaviour and Conception of Timber Structures
Advantages •
Figure: 6.24
good in tension & compression  medium strength  yield stress 1050MPa (as used)
•
low weight 6001000kg/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 800015000MPa  large deformations  creeps
•
hard to create rigid joints  pinned and semirigid joints
•
strength time dependent  varies across grain (except for plywood, particleboard)
•
swells with moisture
•
small spans (except for glulam)
Figure: 6.25
Figure 6.26: Proposal roof using steel elements
Figure 6.27:
Figure 6.28: 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
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115 Example 6.2.11: 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
250kPa
soil G2 :  Coefficient of action conditions:
 root length:
lk
 Loadbearing 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 6.2.1.12: 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 6.32 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 3.2.12 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 6.32. 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: 6.2.1.11
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 6.2.12: 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 midspan
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
crosssectional 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: 6.2.1.21
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 6.2.13: Static scheme of rafter with overhangs from left
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117 Example 6.2.14: 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: 6.2.1.31
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 perpendL 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 perpendL 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 perpendx 1
R ab x 1
Mb
Figure: 6.2.1.41 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 6.2.1.52: The proposal dimensions of the roof truss, and the L (m) and a in (mm)
M d Lk
D perpend sin sin
D perpendsin90o 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: 6.2.1.51
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 6.2.1.53: Proposal dimensions purlin of the truss system with two stationary stands
90o 2
Behaviour and Conception of Timber Structures
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Example 6.2.15: Rafter as a continuous beam
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118
Example 6.2.16: Buckling calculation
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119
Example 6.2.17:
Figure: 6.2.1.61 Calculation and assessment of the strength of the strut N
84.15kN
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
159.28kN
Figure: 6.2.1.71
crosssectional dimension
2
bp 22cm hp 22cm Ap bp hp Ap 0.0484m Navrh Slenderness ratio and the value of which can be found in Table 6.32 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 6.32, calculate the stress as follows
h
Calculation of the thickness t, respectively. t1:
stress in crosssection in the pressure II
F 56b
F
cos 2
p
N b IId
F cos
x
b II
F cos
8 t
We see the result (stress in crosssection in the pressure) are almost the same. Example 6.2.18:
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 3.2.12, calculate the stress as follows 1
0.489
hamb
H
Ahamb
hamb 2.2123MPa
We see that the result is almost the same. Figure: 6.2.1.81
Behaviour and Conception of Timber Structures
t
t1
p b
t
cos F
70b
t
4 t2
F 70b
II
kx Fut
h
M Wut
Aeo
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: 6.2.1.83
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 gL
cm
pII
Dx A
M Wx
0.85
daN 2
cm
Where Dx is compressive normal force Example 6.2.19: 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 crosssection will have dimensions and sectional module: 150mmx220mm
Wx
2
b h 6
3
Wx 0.00129 m
Figure: 6.2.1.82
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 crosssection the dimensions of the square section the design strength in compression wood Six an angular crosssection i min
Table 6.31: calculating the necessary area and moment of inertia of the column square section rectangular crosssection circular crosssection
lf
The required crosssectional area
i l f 21.7a
l f 18.75D A
N f cod D
0.001l f
2
lf 18.75
a
Eight an angular crosssection
0.102a
i min
0.102R
75 lf 21.7b
0.001l f
f cod
A
1.13
D
A
N
2
A
N
A b
c
h
0.001l f h
c A
b
21.7
b
Figure 6.31: 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 6.31.
lf 21.7
Table 6.32: Coefficient of buckling
The moment inertia
STRUCTURAL ENGINEERING ROOM
Department of Architecture
121
l f 18.75D
lf
4
2.13
D i min
J
lf 18.75
D
75
l f 21.7a Jrequired
D
i
2
lf fcod 3100 N
4
a 1.86 J And must be lf a 21.7
i min
l f 21.7b
0.289a
4
i
the radius of gyration
lf
the buckling length of the column of figure 6.31
D
the diameter of the circular crosssection
h
the height of the rectangular crosssection
4
b
1.86
b i min
J c
lf 21.7 0.289b
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 6.32: Buckling coefficient of wood
Figure: 6.33
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
STRUCTURAL ENGINEERING ROOM
Table 6.33: Value for different values of slenderness
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122
STRUCTURAL ENGINEERING ROOM
Department of Architecture
123
b h b
5
1
7
2
6 Mx 2
dovoleneh
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 crosssection 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 140I 220
I 240I 600
U120 to U 160
U180 to U320
8
9 10
6
78
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 6.34: The values of coefficients for the different crosssectional forms are presented in Crosssectional
N1 a1 N2 a2
qy ls
2
ls
8
qx
l 2 c
Wpot
Mx My c
pII
qy l1x l2x m 2
Figure: 6.35
2
Jmin Jmin
Figure: 6.34
nPs 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
12Jmin
2
MB
0.16m
hs
0.20m
Ls
3.2m
The calculation of slenderness the column with a safety factor
Ls
64.879
0.289bs
75
nehrozi vzper
1.61
Stress of timber in compression parallel to the fibers in the wood is class II 8.5MPa
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: 6.36
161kN
The calculation of the stress on the lower surface of the column Fmax
stlpkolmo
Example: 6.31
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: 6.3.11 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 6.32: 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 pdovolenebs
X 0.503m
Calculation of the thickness of the plates have to check the stress s sdovolene
140MPa
hplatnicky 8mm
Figure: 6.3.21
assessment of stress s
Fmax bs hplatnicky
Example 6.33: 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: 6.3.31 Data: section height h, l is the length of the element which is shown in figure, F is the axial Figure: 6.3.12
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
h2 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 crosssection 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 xx
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
Vs1 2 a
T 13.908 kN
Minimum anchoring nails into the wooden cross section in the realization connection having 8 klinca
Slenderness in the xx x
w
A wooden base then transferred greatest shear force
radius of gyration Ix
y
The connection between pad wooden elements waist
The crosssectional moment of inertia Ix
20.353
Control the stress
The control stress F Askut
m 2 x c 1 2
klincaLklinca
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
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Lklinca
260 mm
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m is the number of crosssections within the rod of direction yy
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126
Example 6.34: 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: 6.3.41 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 crosssectional 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 (SFSENV 199411). The dimensioning of composite sheets adheres to the instructions laid out in Eurocode 3, part 1.3 (SFSENV 199313:1996). The load specifications and safety coefficients are based on Eurocode 1 (ENV 19911:1994).
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 12x106 per oC. but for simplicity the value 10x106 per oC (as for
0.003, and at higher strain it crushes, losing almost all its compressive strength. It is very brittle
reinforcement and normaldensity 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
7.11 Profiled steel sheeting o
Coefficient of linear thermal expansion, 10x10 per C.
Steel composite sheets are profiled steel sheets designed for use in concretesteel 3
Normaldensity – 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 lightweightaggregate 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
plasticcoated 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 reentrant 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 unpropped.
a reentrant 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 threequarters, 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
 Selfweight 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:  Selfweight 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
 Selfweight 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 selfweight 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 singlespan or a continuous composite slab. When the anchoring strength is sufficient, the crosssection 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.
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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 hotrolled and coldrolled 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 400500 N/mm2. The minimum breaking stretch for hotrolled 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 coldrolled steel is 3.5%.
agents containing chloride salts. The hardened concrete does not cause corrosion problems to the 7.31 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 crosssection 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 reinforcedconcrete 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 rectangularstressblock 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,i0,iQk,i 1.35Gk,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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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 crosssection in the elastic state
0,i = 0.7 , for load classes AD and FG corresponding to SFSENV 199111, and for Plastic moment
snow loads =
1.0
, for load class E corresponding to SFSENV 199111
=
0.5
, for wind loads
Mu = plastic moment calculated on the basis of the idealplastic 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.)
7.33 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
7.32 Bending failure and bending strength
resistance to bending.
In tough slabs, the sheet crosssection 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 crosssection
= 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.
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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 7.31.
Figure 7.31: 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 7.31: Nominal and design values of anchoring strengths for different sheet types (MPa) ————————————————————————————————— SteelComp
Galvanised sheet
sheet thickness
0.7
0.9
1.1
Plasticcoated 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
= crosssection 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.
7.34 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 199411, 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
7.35 Design of the slab for point and line load
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133
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 7.33: Shearhead design in slab Figure 7.32: 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.
7.36 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 singlespan slabs and in outer spans of multispan 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 multispan 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 7.34.
Figure 7.34: Positioning options for slab reinforcement. 7.37 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 selftapping 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 7.35: Connecting composite slab to steel beam. Connecting composite slab to nonbearing partition wall structure
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134
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135
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 7.37
Figure 7.36: 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
Department of Architecture
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 7.38
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 7.39
Ii yd
Ii yh
n Ii yc Ii zs
STRUCTURAL ENGINEERING ROOM
136
7.38 Serviceability limit state for composite slabs
STRUCTURAL ENGINEERING ROOM
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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
= (1kx) (13kx/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= longterm
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 AD and FG as in SFSENV 199111, and for snow loads , for load class E as in SFSENV 199111
=
0.5
, for wind loads
= (13kx/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 ], crackfree 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 (10.6 '/) 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
Department of Architecture
List of symbols in equations Crosssection 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 stressstrain 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 7.3.81: Steel composite sheets as light weight beam
Profiled Steel Sheeting
STRUCTURAL ENGINEERING ROOM
138
Table 7.32
STRUCTURAL ENGINEERING ROOM
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139
Figure 7.3.82: Steel composite sheets as light weight beam
Profiled Steel Sheeting
Table 7.32 Table 7.32
Profiled Steel Sheeting
STRUCTURAL ENGINEERING ROOM
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140
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Table 7.33
Table 7.33
Profiled Steel Sheeting
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 199411 – 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 7.11: Charging scheme joists
Figure 7.12: Steel and concrete joist
Profiled Steel Sheeting
3
fctm 2.22MPa fctk095 2.89MPa
STRUCTURAL ENGINEERING ROOM
Table 7.33
Department of Architecture
142
The difference between the old and the new standard is 7%
a) Sheeting rib profile 11 002R, reverse position (concrete filled narrow ribs)
STRUCTURAL ENGINEERING ROOM
Department of Architecture
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
Selfweight 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 7.13: 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 7.14 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
Profiled Steel Sheeting
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
Selfweight 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
Profiled Steel Sheeting
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 crosssection:
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  selfweight (estimate IPE 270): 1
3
W ely 429 10 mm
Ultimate limit state:
load:
g osnosnik 0.361kNm
3
STRUCTURAL ENGINEERING ROOM
M 2
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144
3. The loads transferred from the service stage composite beam load:
STRUCTURAL ENGINEERING ROOM
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145 The greatest shearing force:
Permanent  selfweight (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
Selfweight 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 7.15: 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 7.16: Stress distribution in the cross section.
n´
The compression depth of the crosssection according to the old standards:
Ea
n´ 14.48276
E´c
 For shortterm 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 longterm 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 199411
0.71429
Assessment of bending (at the moment of resistance) according to ENV 199411 MplRd Aa
E´b
n
kN
gosstale vsdlnd vsuzitne nk gosstale vsdl vsuzitne
Profiled Steel Sheeting
1
v suzitne 4 m
n 21.68327
kN
STRUCTURAL ENGINEERING ROOM
Position of the neutral axis (carrying capacity in shear):
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146
The neutral axis position (serviceability limit state):
STRUCTURAL ENGINEERING ROOM
Department of Architecture
147
Aa
h
2
e
h1 h2
Aa
h1 1 h1 beff n´ 2
fmax 0.02333m
300
PENV 199411
e 0.12861 m
1 h1 beff n´
L
fmax
f´
5
vsdl vsuzitneL4
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 199411
1 n pretv
e
e´ 0.16894m
e´
b h1 A a
A bi´ h 1
0.76131
The moment of inertia of ideal crosssection: 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 vsuzitneL4
384
Ea Ii
4
Ii´ 0.00016436m
Ii Ii´
1.27405
A id A bid A a
A id 6121.60714 mm
2
PENV 199411
Deflection due to Accidental load: f
b n pretv
A ik A bik A a
PENV 199411
2
b 0.953m
Position of the neutral axis:
2 h1 h 11 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 199411
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 7.17
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 7.18: Schematic crosssection of an ideal neutral axis.
Syk 1470342.85714mm
The moment of inertia of ideal crosssection of the priority axis yy: Syd A bid rbo A a
h
2
Syi A bi rbo A a
PENV 199411
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´
Shortterm load: Syk Aik
Syd A id
Syi´ A i´
2
4
Iyi´ 163720223.436mm
ideal composite beams  shortterm 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 199411 ro´
4
Iyi 143497369.508mm
The distance of the lower fibers of steel parts from the neutral axis of the crosssection of an rok 137.20456mm
Longterm load: rod
4
Iyd 124048057.758mm
PENV 199411
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
STRUCTURAL ENGINEERING ROOM
148
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Department of Architecture
149
The distance of the lower fibers of steel parts from the neutral axis of the crosssection of an ideal composite beams
PENV 199411 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:
Shortterm:
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 crosssection of an h 2
ro
y d 0.2127m
The distance of the upper fibers of steel parts from the neutral axis of the crosssection of an ideal composite beams:
y h
h 2
Longterm: 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
Longterm: 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 7.19: The resulting stress in composite sectional.
PENV 199411 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 199411
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 7.110: 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
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Average value:
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151
The horizontal shear force per unit length contact with steel and concrete of the crosssection
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 199411 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 7.111:
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
– longterm 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
8.12 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
8.11 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
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8. Design strength of masonry
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What is the advantages and disadvantages of the brick when used as structural
STRUCTURAL ENGINEERING ROOM
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153 Table 8.1.22 Partial safety factors for the material m
materials:
control (see 6.9)
•
low strength  yield stress 320MPa  very weak in tension  brittle
•
medium Modulus of Elasticity 1000025000MPa
•
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 8.1.22 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:
GjGkj 1 4Qk1 Fd
F Fk
Gd
G Gk
GjGkj 1 2
Qki
Rd
( i 1)
Qd
Q Qk
Ad
A Ak
Pd
P Pk
internal forces or moments)
Table 8.1.21 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 8.1.23
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 8.1.24: 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 8.1.25 for the relevant concrete strength classes.
200
1,45
1,35
1,25
1,15
1,10
Table 8.1.25
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 8.1.26
Last horizontal dimension of unit (mm)
lightweight mortar class
Constant K Longitudinal mortar joint
Figure 8.1.21
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 8.1.24, 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 8.1.23 Values of factor δ
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Characteristic compressive strength of unreinforced masonry made using general
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155
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 8.1.22
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
STRUCTURAL ENGINEERING ROOM
Characteristic compressive strength of unreinforced masonry made using lightweight
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156
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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 zs2 2
2
A 4.05 m 3
Self weight of peripheral beam
Nv
hhvb 1.1 24 kN m
Nv 0 kN
Self weight of masonry
Nm
hhwb 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
Aqs
Ns 61.773 kN
Np
Aqp
Np 49.815 kN
Characteristic action from individual storey in kN i Ni
Masonry
(0 5)
Nv Ns Nv Ns Nm 2Nv Ns Np Nm 2Nv Ns Np 2Nm 3Nv Ns 2Np 2Nm 3Nv Ns 2Np 3Nm
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
STRUCTURAL ENGINEERING ROOM
Nd
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158
Design actions and design resistance of storeys separately
STRUCTURAL ENGINEERING ROOM
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159
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 crosssection
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 hRtfd
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 Btef 6
3
Z 0.015 m
Required wind load capacity
Wk
W fl 1m
Design moment
Md
Wk flL
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
HL 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 fl1m
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
STRUCTURAL ENGINEERING ROOM
Example 8.3
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STRUCTURAL ENGINEERING ROOM
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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 nondominant 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 beamtocolumn 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 actioneffect (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 crosssection shapes crosssectional area crosssectional 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 quasipermanent value of a variable action quasistatic 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
selfweight 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 % proofstress 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 (loadbearing) structure
footpaths
subscripts
force
suitability supply symbols
foundations frame frame stability frames free flowing material funnel flow (or coreflow) gap gauge
(final) takeover 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 crosssectional 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
loadbearing 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 midspan 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 quasipermanent 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 lateraltorsional 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
ovendry 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 posttensioning precast element prestressing devices prestressing force prestressing steel pretensioning 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 stressstrain 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 multiaxial 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 temperaturetime curve nominal yield strength nondimensional 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 temperaturetime
Terminology
nonseparating wall nonstructural member nonsway 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) subframe 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 semicontinuous 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 stressstrain relationship strong axis structural analysis standard temperaturetime 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 crosssectional 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 crosssectional area of a concrete section Â
support support and boundary conditions support arrangements support conditions sway frames and nonsway 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 temperaturetime 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.
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PRESENT STATE CANTEEN
Realisation Projects
EXTERIOR PLACE FOR NEW GALLERY EXTENSION
Realisation Projects
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PLAN 1. FLOOR PRESENT STATE
Realisation Projects
Realisation Projects STRUCTURAL ENGINEERING ROOM
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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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Realisation Projects
Realisation Projects STRUCTURAL ENGINEERING ROOM
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Realisation Projects
SECTION AA
Realisation Projects
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SECTION BB
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SECTION AA
SECTION BB
Realisation Projects
SECTION AA
Realisation Projects
STRUCTURAL ENGINEERING ROOM
SECTION BB
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Realisation Projects
SECTION BB
TUBE LEGEND air inlet tube air intake tube
Realisation Projects
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SECTION AA
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Realisation Projects
Realisation Projects 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
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DIGITAL LIBRARY TIMBER PROFILED ROOF PLAN
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SECTION VIEW E
SECTION VIEW D
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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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