DESIGN OF REINFORCED CONCRETE MEMBERS

Sabah Shawkat

Verification of stress on reinforced concrete cross - section subjected to bending and bending moment with axial load in serviceability limit state can easily be performed using Table I. Three different diagrams, bi-linear ( β´  0), bi-linear ( β´ = 0) , and parabola-rectangle stress - strain for reinforcing steel and two diagrams, bi-linear ( β´  0), parabola-rectangle for concrete stress distribution are used for the determination of reinforcement to concrete cross-section for any concrete strength. Calculations are demonstrated by examples using Tables II, III, IV, V. 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 the Table VI can be used.

Preface This textbook is designed to teach students but it can serve as a reference for practising engineering or researchers as well. The text offers a simple, comprehensive, and methodical presentation of the basic concepts in the analysis of members subjected to axial loads, and bending. Many practical worked examples are included and design aids in the form of tables are presented.

The major quantities which are considered in the design calculations, namely the loads, dimensions, and material properties, are subject to varying degrees of uncertainty and randomness. Further, there are idealizations and simplifying assumptions used in the theories of structural analysis and design. There are also several other variable and often unforeseen factors that influence the prediction of ultimate strength and performance of a structure, such as construction methods, workmanship and quality control, probable service life of the structure, possible future change of use, frequency of loadings, etc. Reinforced concrete structures and structural components are designed to have sufficient strength and stability to withstand the effects of factored loads, thereby satisfying the safety requirements. The design for serviceability limits is made at specified (service) loads (generally, the design is first made for strength and the serviceability limits are checked). Primary considerations in structural design are safety and serviceability. Safety requires that the structure remains without damage under the normal expected loads. Serviceability requires that under the expected loads the structure performs satisfactorily with regard to its intended use, without discomfort to the user due to excessive deflection, permanent deformations, cracking etc. This textbook contains design aides such as tables and diagrams which enable very simple and rapid determination of reinforcement in a given rectangular concrete cross-section subjected to bending moment or bending moment with axial force. Design rules of this textbook are based on the concept of Concrete Structures Euro-Design.

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. My sincere thanks to my friends Maher Shamar ( Athens, Greece), Muhsen liony (Sacramento, U.S.A.) and Henrieta Fazekašová for their suggestions, discussions of general approach and other assistance. My special thanks belong to Ján Cesnak, the reviewer of the text for his suggestions and helpful recommendations as well as to Ľudovít Fillo and Imrich Tužinský for their professional advice. Finally, I am grateful to my wife Ivana for her help, constructive criticism, patience and encouragement that have made this project possible. I dedicate this book to her and to my daugther Tamara Shirin.

Bratislava, 2000

Author

About the author The author Sabah Shawkat is a university teacher at the Faculty of Architecture of the Slovak Technical University in Bratislava, Slovakia. He received his Ph.D. in 1993 from the Faculty of Civil Engineering, Slovak Technical University. He performs research in the field of deformation behaviour of reinforced concrete and prestressed beams. He has published numerous articles in professional journals in the field of calculation of the strain energy and the degradation of shear and bending rigidity after the full development of cracks in reinforced concrete and partially prestressed beams. Sabah Shawkat also practises as a licenced engineer.

The author whishes to express his gratitude to Mr. Maher Shamar from Athens in Greece for financial support that enabled publishing of this book.

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, now known or hereafter invented, including photocopying and recording, or in any information storage and retrieval system, without permission in writing from the author.

1. Introduction

The purpose of structural analysis is the establishment of the distribution of either internal forces and moments, or stresses, strains and displacements over the structure as a whole or a part of it. It is the aim of each design procedure - on condition that the structure or member is actually constructed according to the design - to minimize the probability of failure without unreasonable expenditure. Further more serviceability during the lifetime of the structure should be guaranteed with adequate probability. Therefore two limit states are defined: a) Serviceability Limit State (SLS) b) Ultimate Limit State (ULS) a) Serviceability Limit State (SLS) Analyses carried out in connection with serviceability limit states will normally be based on the linear elastic theory. In this case it will be satisfactory to assume the stiffness for members based on the stiffness of the uncracked cross-section, when cracking of the concrete has a significant unfavourable effect on the performance of the structure or member considered, it shall be taken into account in the analysis. The serviceability requirements concern: - the functioning of the construction works or parts of them, - the comfort of people, - the appearance. b) Ultimate Limit State (ULS) Depending on the specific nature of the structure, the limit state are considered and on the specific conditions of design or execution, analysis for the ultimate limit states may be linear elastic with or without redistribution, non-linear or plastic. Ultimate limit states concern:

- the safety of the structure and its contents - the safety of people. Ultimate limit states which may require consideration include: - loss of equilibrium of the structure or any part of it, considered as a rigid body, - failure by excessive deformation, transformation of the structure or any part of it into a mechanism, rupture, loss of stability of the structure or any part of it, including supports and foundations, - failure caused by fatigue or other time-dependent effects. When considering an ultimate limit state of rupture or excessive deformation of a section, member or connection (fatigue excluded), it should be verified that Sd  Rd Where Sd is the design value of internal forces or moments (e.g. M, N, V, T for bending moment, axial force, shear force, tortional moment) produced by the actions F (loads, prestressing forces and so called indirect actions such as imposed deformations from the settlment of supports or from

temperature and shrinkage effects) and Rd is the corresponding design resistance, associating all structural properties with the respective design values. The index d makes clear that we have to calculate the corresponding design values of S and R. In cases where the ultimate state is characterised by the loss of statical equilibrium, the condition is given in the form: Ed dst  Ed stb Where Ed dst stands for the destabilising influences and Ed stb for the stabilising influences. Examples for such cases are piers during the construction phase of symetrical free cantilever beam or the buckling of slender columns.

1. 1 Strain Diagrams in the Ultimate Limit State The allowable strain diagrams which are possible in accordance to the stress - strain diagrams in EC 2 are shown in Figure 1. 1. The strain diagrams are based on the assumption of Bernoulli s hypothesis. The strain limits of the stress-strain diagrams for concrete and steel result in different domains for the strain diagrams

in the design of cross sections. The stress state in a cross section is defined by the chosen strain diagram of the materials. With the assumption of an ideal bond the strain diagram in Figure 1.1 governs not only the concrete compression stresses but also the steel stress for reinforcement in the cross-section. The compressive strain of reinforcing steel caused by creep and shrinkage of concrete are normally negligble in the ultimate limite state. In the following a short explanation of the five domains shown in Figure 1.1 is given:

Figure 1.1 Strain distributions of reinforced concrete cross - section with respect to the limitation of the steel strains Assumption: Îľ s2( A) = 0.01

Zone 1: The entire section is in tension, the neutral axis lies outside the section, normally reinforcement yields. Zone 2: The neutral axis lies within the section, there exists thus a compression and a tension zone. The maximum strain in the concrete is less than the limiting value of 0.0035, thus the strength of the concrete is not exhausted. The boundary between zone 2 and 3 is defined by a strain diagram where both the maximum concrete strain ( ε c = 0.0035) and reinforcement strain ε s = 0.01 are present. Zone 3: The concrete compression strain at the upper fibre is 0.0035 (point B). The steel strain lies between 0.01 and ε yd, the strain corresponds to the design strength of the steel fyd. This means that the stress of reinforcing steel is fully exhausted in tension. The boundray between 3 and 4 is called the balanced condition. Zone 4: The strain in the steel at failure lies between ε yd and 0.00 that means the steel stress at the ultimate limit state is thus less than fyd. Zone 4a: All the steel (except prestressing steel) is in compression. Some small part of the section remains in tension. Zone 5: The entire section (with exception of possibly existing prestressing steel) is in compression. All strain profiles pass through point C. The maximum compressive strain of concrete is between 0.0035 - with ε c = 0 at the lower rim and 0.002 for centric axial compressive load. Point C is the point where the line BO (which defines the boundary between the sections partially in tension and sections in compression only) intersects the line characterized by ε c = 0.002 = const. The distance of this point from the utmost compressive fibre is equal to 3 / 7 of the total depth of the section.

Both assumption a and b for the steel strain limitation (point A) ( in Figure 1.1) result in slightly different strain diagrams.

Zone 4 and parts of zone 3 characterize the transition from dominant bending moments to dominant axial compression force. These areas cover the cases in which strong pure bending and moments with compression force require a deep situation of the neutral axis. In cases of pure bending or dominant bending with compression force this uneconomic area can be avoided by placing compression reinforcement which leads to a higher neutral axis and thus to full exhaustion of the tension reinforcement. Zone 3 with relatively low neutral axis cannot be used for dominant bending if the rules for the limitation of the relative depth of the compression zone (x / d) depending on the chosen redistribution of moments are obeyed. Thus the necessity of additional compression reinforcement depends also on the limits for (x / d).

1. 2 Stress - strain Diagram for Concrete The stress - strain diagram for concrete (the conventional stress distribution in the compression zone) is presented normally by the parabola - rectangle curve. According to the design concept with partial safety factors the maximum stress value is Îą ď&#x192;&#x2014; fcd (Figure 1.2). The factor Îą = 0.85 takes account of the concrete strength under sustained load which is smaller than the strength under short term loading. The non proportional interdependence between stress and strain of the concrete in the compression zone according to Figure 1.2 represents an approximation to the real stress distribution in the ultimate limit state. For simplification reasons this distribution can also be used in cases with the strain diagrams acc. to Figure 1.1 - for the neutral axis in high positions. In these cases only a part of the parabola - rectangular diagram gives the stress distribution in the compression zone which is limited by the relevant strain at the upper fibre.

Figure 1.2 Parabola - rectangular diagram for concrete stress distribution

Other idealized stress-diagrams may be used, provided they are effectively equivalent to the parabola-rectangular diagram, e.g. the bi-linear diagram (Figure 1.3) or the rectangular stress block (Figure 1.4). The following design aids are based exclusively on the prabola-rectangular diagram because of its universal applicability and its good approximation to test results. For quick calculation and for predesign of members the rectangular stress block is a useful tool. In this diagram Îľ cu max is taken as 0.0035

Figure 1.3 Bilinear stress-strain diagram for concrete

Figure 1.4 Rectangular diagram

The design concrete strength is defined by fcd =

fck γc

.

The design diagram is derived from the chosen idealized diagram by means of a reduction of the stress ordinate of the idealized diagram by a factor ( α / γc ), in which

γc is the partial safety factors for concrete γc = 1.5 α is a coefficient taking account of long term effects on the compressive strength and of unfavourable effects resulting from the way the load is applied. The additional reduction factor α for sustained compression may generally be assumed to be |0.85|. A rectangular stress distribution (as given in FIgure 1.4) may be assumed. The α - factors as given for idealized diagram are valid, except that it should be reduced to |0.80|, when the compression zone decreases in width in the direction of the extreme compression fibre.

1. 3 Stress-strain Diagram for Reinforcing Steel The characteristic value fyk for the yield stress is defined as 5% - fractile of the yield stress ( fy). The design values fyd and ftd for ultimate limit state result from the characteristic values, divided by the safety factors γs:

fyd =

fyk γs

ftd =

ftk γs

According to EC 2 the safety factor γs = 1.15 in case of basic combinations and γs = 1.0 in case of accidental combinations. Three different stress-strain diagrams for reinforcing steel are used (Figure 1.5). a) Assumption of a horizontal ideal-plastic branch ( β´ = 0) with a steel strain limitation to ε uk = 0.01 fyk ftk b) Assumption of a branch ascending from to but with a steel strain limitation to ε uk = 0.01 γs γs

c) Assumption of parabolic-rectangle with a steel strain limitation ε uk = 0.01

Figure 1.5 Three different stress-strain diagrams for reinforcing steel 1. 4 Basic Values for the Design Resistance to Moments and Axial Forces 1.4.1 Basic Assumptions (Figure 1.6) a) Plane sections remain plane (hypothesis of Bernoulli), that means strain distributions in the section are linear. b) Perfect bond between concrete and (reinforcing and bonded prestressing) steel. In each fibre is valid ε cs = ε s with ε cs as "smeared" concrete strain (including crack opening) and ε s as steel strain.

c) The tensile strength of concrete including the tension stiffening effect is neglected (so called "naked" state II). d) Simplified design stress - strain diagrams for concrete and steel which take account of the elastic plastic behaviour of the materials approximately. In EC 2 different diagrams for concrete and steel are given.

The ultimate compressive strain in concrete ( 3.5 promile) for bending and for normal force with great eccentricity must be reduced to ( 2.0 promile) for centric normal force. The strain diagram is to be turned around point C in Figure 1.1, which distance from the most compressed fibre is 3 / 7 of the total height of the concrete section.

Figure 1.6

1. 5 Verification at the Serviceability Limit States Limitation of Stress under Serviceability Conditions The provisions at the ultimate limit states in Eurocode 2 may lead to excessive stresses in concrete, reinforcing steel. These stresses may, as a consequence, adversely affecte the appearance and

performance in service conditions and the durability of concrete structures. Key-words in this context are: - non-linear creep of concrete due to excessive compressive stresses,

- increased permeability of the concrete surface due to micro - cracking of concrete around the reinforcing bars, - yielding of steel in service condition leading to cracks with unacceptable width, - cracks parallel to the reinforcing bars, - excessive deflection of concrete members caused by high stresses, cracking and creep of the members, 1. 5. 1 Permissible Stresses In reinforced concrete structures, longitudinal cracks parallel to the reinforcing bars may occur if the stress level in the concrete under the rare combination of actions ΣGk j  Pk  Qk 1  Σψ0 i  Qk i exceeds a critical value. Such cracking may lead to a reduction in durability.

In detail EC 2 requires the verification of stresses, associated with the following permissible values, for the following reasons: 1) For preventing the development of longitudinal cracks, the concrete compressive stresses under rare combinations of actions are limited to 0.6  fck in absence of other methods (for example, increasing the concrete cover of confinement of the compression zone). σc  0.6  fck 2) For preventing excessive creep, the concrete compressive stresses under quasi - permanent combinations of actions ΣGk j  Pk  Σψ2 i  Qk i are limited to 0.45  fck ,

σc  0.45  fck 3) for reinforcing steel subjected to loads and restraints, σs  0.8  fyk 4) for reinforcing steel subjected to restraints only, σs  fyk

Symbols h overall depth of a cross-section d effective depth of a cross-section b overall width of a cross-section d2 d1 distance of the tension and compression reinforcement from the extreme fibre of the cross-section μ relative moment ρ reinforcement ratio α reduction factor for concrete compressive strength, resp. limiting values for the related neutral axis depth x = α  d depth of compression zone z = β  d lever arm of the internal forces A2 As area of reinforcement in the tension zone A1 area of reinforcement in the compression zone at the ultimate limit state ρ2 = ρ1 =

A2 bd A1 bd

geometrical ratio of tension reinforcement geometrical ratio of compression reinforcement

ρtot = ρ1  ρ2 total geometrical ratio of reinforcement Nsd design value of the applied axial force (tension or compression) Msd design value of the applied internal bending moment fck characteristic compressive cylinder strength of concrete fcd or σbc design value of concrete cylinder compressive strength fy yield stress of reinforcement fyd design yield strength of reinforcement

fyk characteristic yield stress of reinforcement ε b compressive strain in concrete

ε cu ultimate compressive strain in concrete ε c1 compressive strain in concrete at the peak stress fc ε u elongation of reinforcement at maximum load ε uk characteristic uniform elongation of reinforcement at maximum load ε s2 strain in tension reinforcement for section analysis ε s1 strain in compression reinforcement for section analysis σs stress in tension steel γc partial safety factor for concrete material properties, γc = 1.5 γs partial safety factor for reinforcement material properties, γs = 1.15 Es modulus of elasticity of reinforcement, Es = 200  GPa Ec Ec28 tangent modulus of elasticity of normal weight concrete at a stress of σc = 0 and at 28 days λ slenderness ratio : λ =

lo i

i radius of gyration: lo effective length of column P prestressing action Pk characteristic value of a prestressing action G permanent action Gd design value of a permanent action

Gkj characteristic value of a permanent action Q variable action Qd design value of a variable action Qk1 characteristic value of the dominant variable action Qki characteristic value of the non - dominant variable action i

Analysis for Stresses-Section cracked and elastic

Sections designed for strength (Ultimate Limit States) under factored loads must be checked for serviceability (deflection, crack width, etc.) under the specified loads. Under service loads, flexural members are generally in the cracked phase with linear distribution of strains and stresses.The computation of stresses using the straight-line theory (elastic theory) for cracked section is discussed below. Figure I.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 Fig. I.1b. The centroid of the transformed section locates the neutral axis.

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 σbc = Ms 

x Icr

.

The same results could be obtained more simply and directly considering the static equilibrium of internal forces and external applied moment. The internal resultant forces in concrete and steel will then be: Fc = b  x  The inner lever arm will be: z = d 

σbc

Fs = A2  σs2

2

x 3

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

M = Fc  z

M = Fs  z

or

M = b  x

σbc

z

2

or

M = A2  σs  z

3

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

from which

σbc = M 

x Icr

and

Icr =

σs =

bx

M A2  z

3

 n  As  ( d  x)

2

Example I. 1: A reinforced concrete beam of rectangular section has the cross-section dimensions shown in Fig. I. 1. Compute the stresses in concrete and steel reinforcement for a given bending moment. Assumptions: Characteristic value of concrete cylinder compressive strength (MPa): fyk = 410  MPa

Characteristic yield stress of the steel (MPa): Bending moment at the section ( MN  m): Cross-section ( m):

b = 0.4  m

fck = 25  MPa

Ms = 250  kN  m

h = 0.90  m

Distance of the tension reinforcement centroid from the extreme fibre of the cross-section (m): d2 = 0.02  m Effective depth of a cross-section ( m): d = h  d2 Diameter: ϕ = 20  mm

Number of bars:

d = 0.88  m

nϕ = 5 2

The entire area of tension reinforcing bars ( cm ):

2

As = nϕ  π 

Figure I. 1

ϕ

4

2

As = 15.70  cm

Using the design aid in Table I.1

Ratio of tension reinforcement (-):

From Table I.1 we obtain

ρ=

As bd

 10

2

σ´bc = 7.27681  MPa

μ=

Ms

ρ = 0.45

σ´s = 247.47  MPa

and

μ = 0.80708

2

bd The allowable concrete stress (MPa):

σcall = 0.6  fck

σcall = 15  MPa

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

σbc = μ  σ´bc

σbc = 5.87  MPa

The steel stresses (MPa):

whichever is lesser

σsall = 0.6  fyk

Stress in tension reinforcement (MPa):

σbc  σcall = allowable

σsall = 246  MPa

σs = μ  σ´s

σs = 199.72  MPa

σs  σsall = allowable

whichever is lesser

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

n=

Es Ec

n  15

Es, Ec is modulus of elasticity of steel, or concrete respectively To locate the neutral axis (centroid) of the transformed section, we have to equate the moments of areas about the neutral axis.

2

b

x

2

 n  As  ( d  x) = 0

Solving

Using the design aid in Table I.1

x = 0.268

α = 0.309 x = αd

x = 0.272  m

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

3

Icr =

bx

 n  As  ( d  x)

3

2

Icr = 0.01139  m

4

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 

 2 

σbc

d 

x

=0

3

Tension steel stress:

Solving

σs = n  σbc 

dx x

σbc = 5.87  MPa

σs = 199.34  MPa

as before

as before

With the distribution of concrete and steel stresses shown in Fig. I.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:

Fc = b  x 

z=d

σbc

x 3

2

Fc = 316.71  kN

z = 0.78936  m

Fs = As  σs

Fs = 306.69  kN

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

M=

1 2

 

 σcall  MPa  b  x   d 

x

3

M = 643.92  kN  m

Example 1.2: The cross-section of a beam with compression reinforcement are shown in Fig. 1.2a. The beam is subjected to a specified load moment. Compute the specified load stress in concrete and steel. The material properties are Assumpted: fck = 30  MPa

Characteristic value of concrete cylinder compressive strength (MPa): fyk = 410  MPa

Characteristic yield stress of the steel (MPa): Bending moment at the section ( MN  m): Cross-section ( m):

b = 0.3  m

Ms = 0.185  MN  m h = 0.60  m

Distance of the centroid of the tension reinforcement from the extreme fibre of the cross-section (m): d1 = 0.05  m Effective depth of a cross-section ( m):

d = h  d2

d2 = 0.05  m d = 0.55  m

Compression reinforcement: Diameter: ϕc = 16  mm

Number of bars:

nϕc = 2 2

2

The entire area of compression reinforcing bars ( m ):

A1 = nϕc  π 

ϕc

4

 10

4

A1 = 0.0004  m

2

Tension reinforcement: Diameter: ϕt = 25  mm

Number of bars:

nϕt = 4 2

The entire area of tension reinforcing bars ( m ):

A2 = nϕt  π 

ϕt

2

4

 10

4

A2 = 0.00196  m

2

The specified load stresses are computed by the straight line theory applied to the cracked transformed section, shown in Fig. I.1b.

The modular ratio n =

Es Ec

n  15

To locate the neutral axis (centroid) of the transformed section, equate the moments of areas about the neutral axis (see Fig. I.1b)

2

b

x

 n  A1  x  d1  n  A2  ( d  x) = 0

2

Solving,

x = 0.23385  m

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

Ms = b  x 

 2 

σbc

d 

x

  n  A1  3

σbc  x  d1

x

 d  d1

Solving,

σbc = 9.77  MPa

Referring to the stress distribution in Fig. I.1c, the concrete stress at compression reinforcement of compressive level, σc2 (MPa) is:

σb2 =

σbc x

 x  d1

σb2 = 7.68  MPa

4

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

Icr =

bx 3

2

2

 n  A2  ( d  x)  n  A1  d1  x

Compression and tension steel stress (MPa):

Icr = 0.00443  m

4

σs1 = n  Ms 

σs2 = n  Ms 

x  d1 Icr

d x Icr

Using the flexure formula,

σs1 = 115.25  MPa

or

σs1 = n  σb2

σs2 = 198.19  MPa

or

σs2 = n  σbc 

σc = Ms 

x

σc = 9.77

Icr

σs1 = 115.25  MPa

dx x

σs2 = 198.19  MPa

as above see σbc

MPa

Using the design aid in Table I.1

Ratio of tension reinforcement (-):

From Table I.1 we obtain

ρ=

A2 bd

 10

2

σ´bc = 5.279  MPa

ρ = 1.19 σ´s = 98.86  MPa

and

To apply design table (I.1) the transformed action Ms have to be brought into a dimensionless form (-):

μ=

Ms

μ = 2.03857

2

bd The allowable concrete stress (MPa):

σcall = 0.6  fck

σcall = 18  MPa

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

σbc = μ  σ´bc

σbc = 10.76  MPa

The allowable steel stresses (MPa):

whichever is lesser

σsall = 0.6  fyk

Stress in tension reinforcement (MPa): σs = μ  σ´s whichever is lesser

σs  σsall = allowable

σbc  σcall = allowable

σsall = 246  MPa σs = 201.53  MPa

Fig. I. 2 Example I. 3: The member shown in Fig. I. 3 is subjected to bending moment and external normal load. Compute the stress in concrete and steel. Assumptions: Characteristic value of concrete cylinder compressive strength (MPa): fck = 25  MPa Characteristic yield stress of reinforcement (MPa):

fyk = 410  MPa

Allowable stress of reinforcement (MPa): σsall = 0.6  fyk

σsall = 246  MPa

Bending moment at the section ( MN  m): Ms = 0.25  MN  m Axial load at the section ( MN): Cross-section ( m):

Ns = 0.15  MN

b = 0.35  m

h = 0.70  m

Distance of the tension reinforcement from the extreme fibre of the cross-section (m): d2 = 0.05  m Effective depth of a cross-section ( m): d = h  d2

d = 0.65  m

Combined actions have to be transformed in relation

 Ms

to the centre of gravity of the tension reinforcement ( MN  m):

M = Ns  

 Ns

 

 d 

h 



2 

M = 0.295  MN  m

Figure I. 3

Limiting values for the related neutral axis depth:

α=

90  M 2

b  d  σsall

1α 3α

α = 0.40822

Lever arm of the internal forces (m):

2

 

z = d1 

α

As =

Number of bars:

ϕt = 25  mm

Diameter: 2

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

The allowable concrete stress (MPa):

z = 0.56155  m

 M  N  1  s z  σsall

The entire required steel area ( m ) is:

nt = 4

3

A2 = nt  π 

σcall = 0.6  fck

ϕt

As = 0.00153  m

2

2

4

A2 = 0.001963  m

σcall = 15  MPa

2

Modular ratio

n=

Es

n = 15

Ec

Es, Ec is the modulus of elasticity of steel, or concrete respectively The compressive stress in the concrete at the top fibres of the section (MPa): A2 = 0.001963  m

σbc =

σsall n

α 1α

σbc = 11.31  MPa

Using the design aid in Table I.1

From the Table I.1 we obtain

ρ=

whichever is lesser

A2 bh

 100

ρ = 0.80122

σ´bc = 5.96  MPa

2

σbc  σcall = allowable

μ=

Ms 2

μ = 1.69062

bd σ´s = 143.36  MPa

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

σbc = σ´bc  μ

σbc = 10.08  MPa

Stress in tension reinforcement (MPa):

whichever is lesser

whichever is lesser σbc  0.6  fck = allowable σs = μ  σ´s

σs = 242.38  MPa

σs  σsall = allowable

Example I. 4: Compute the steel area to the cross-section shown in Fig. I. 4 and check the stresses of concrete and steel for a given bending moment and axial load. Assumptions: Characteristic value of concrete cylinder compressive strength (MPa): fck = 30  MPa Characteristic yield stress of reinforcement (MPa): fyk = 410  MPa Allowable stress of reinforcement (MPa): σsall = 0.6  fyk Cross-section ( m):

b = 0.35  m

h = 0.75  m

σsall = 246  MPa

Distance of the tension reinforcement from the extreme fibre of the cross-section (m): d2 = 0.05  m Effective depth of a cross-section ( m): d = h  d2

d = 0.7  m

Bending moment at the section( MN  m): Ms = 0.15  MN  m

Axial external load at the section ( MN):

Ns = 0.60  MN

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

 Ms

M = Ns  

 Ns

 

 d 

h 



2 

M = 0.345  MN  m

Limiting values for the related neutral axis depth:

α=

90  M 2

b  d  σsall

1α 3α

α = 0.40957

Figure I. 4

 

z = d1 

Lever arm of the internal forces (m):

2

The entire steel area ( cm ) is:

As =

α

z = 0.60443  m

3

 M  N   1  10 4  s z  σsall

2

As = 1.187  cm

Since the sign of the amount of reinforcement is negative, the reinforcement is not necessary and the concrete alone resists the bending moment and external axial load.

Depth of the compression zone (m): x =

 h Ms     3  2 Ns 

x = 0.375  m

xh

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

σbc =

2  Ns

σbc = 9.14  MPa

bx

whichever is lesser

σbc  0.6  fck

Example I. 5: Pure bending with tension and compression reinforcement in a rectangular cross-section Fig. I. 5. Verify the stresses in concrete and in steel reinforcement. The section properties: Characteristic value of concrete-cylinder compressive strength (MPa): Characteristic yield stress of reinforcement (MPa): fyk = 410  MPa Cross-section ( m):

b = 0.4  m

d = 0.55  m

Bending moment at the section ( MN  m): Ms = 0.15  MN  m

fck = 25  MPa

Distance of the centroid tension and compression reinforcement d2 = 0.05  m

from the extreme fibre of the cross-section (m): 2

Area of compression reinforcing longitudinal bars ( m ): 2

Area of tension reinforcing longitudinal bars ( m ):

A1 = 0.0008  m

A2 = 0.0026  m

d1 = 0.05  m

2

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 concret area displaced by the embedded bars A1  A2 .

n = 15 Where n =

Es Ec

is known as the modular ratio

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

The term Youngs modulus of elasticity has relevance only in the linear elastic part of a stress-strain curve.

Fig. I. 5

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:

2

b

x

2

 n  A2  ( d  x)  n  A1  x  d1 = 0

Depth of the compression zone (m):

2

b

x

2

 n  A2  n  A1  x  n  A2  d  n  A1  d1 = 0

x = 0.228  m

or

x = 0.228  m

As above.

4

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

Icr =

bx 3

2

2

 n  A2  ( d  x)  n  A1  d1  x

Icr = 0.006  m

4

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

σbc = Ms 

x

σbc = 5.70  MPa

Icr

Allowable concrete stress:

σbc  0.6  fck

The assumption was correct.

0.6  fck = 15  MPa

from which the streel stress is (MPa)

σs1 = n  Ms 

σs2 = n  Ms 

x  d1 Icr

d x Icr

σs1 = 66.76  MPa

σs2 = 120.59  MPa

Using the design aid in Table. I.1

ρ=

A2 bd

σs2  0.6  fyk = allowable

 100

ρ = 1.18

From Table I.1, follows

σ´bc = 5.27  MPa

μ=

Ms 2

σ´s2 = 98.86  MPa

μ = 1.23967

bd

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

σbc = σ´bc  μ

σbc = 6.53  MPa

whichever is lesser

σbc  0.6  fck = allowable

Stress in tension reinforcement (MPa):

σs2 = σ´s2  μ

σs2 = 122.55  MPa

whichever is lesser

σs2  0.6  fyk = allowable

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 Fig. I. 5b, 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.)

Example I. 6: Use the section properties from Example I. 5, the beam is without compression reinforcement shown in Fig. I. 6.Verify the stresses in concrete and in steel 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 of the transformed cracked of 3

cross-section about the neutral axis ( m ).

2

bx 2

 n  A2  x  n  A2  d = 0

x = 0.2442  m or 2

b

x

2

 n  A2  ( d  x) = 0

x = 0.2442  m

As above. Fig. I. 6

Having obtained x by solving this quadratic equation, one can determine the moment of inertia and other properties of the transformed section as in the preceding case. Alternatively, one can proceed from basic principles by accouting directly for the forces which act in the cross section these are shown in Fig. I. 6b. The concrete stress, with maximum value σbc at the outer edge, is distributed linearly as shown. The entire steel area A2 is subjected to the stress σs2 . 3

Icr =

bx 3

 n  A2  ( d  x)

2

Icr = 0.00559  m

2

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

σbc = Ms 

x Icr

σbc = 6.55  MPa MPa

σbc  0.6  fck = allowable

Stress in tension reinforcement (MPa): σs2 = n  Ms 

d x

The assumption was correct.

σs2 = 123.11  MPa

Icr

σs2  σsall

Example I. 7: A rectangular member has the dimensions (Fig. I. 7) subjected to a bending moment and external normal force. The entire section is in compression, and symmetrically reinforced, check the stresses in concrete and in steel reinforcement. The section properties Characteristic value of concrete cylinder compressive strength (MPa): fck = 30  MPa Characteristic yield stress of the steel (MPa): fyk = 410  MPa Cross-section:

Depth:

h = 0.60  m

Width:

b = 0.30  m

Distance of the tension and compression reinforcement from the extreme fibre of the cross-section (m): d2 = 0.05  m

d1 = 0.05  m

Effective depth of cross-section (m):

d = h  d2

d = 0.55  m

Area of reinforcing longitudinal bars:

A1 = 10.50  cm

Bending moment at the section (MN m):

Ms = 0.115  MN  m

Axial compression load at the section (MN):

Ns = 1.20  MN

2

A2 = A1

Fig. I. 7

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 concret area displaced by the embedded bars A1  A2 .

Where n =

n = 15

Es Ec

is known as the modular ratio.

2

Area of the transformed uncracked section ( m ):

Bo = b  h  n  A2  A1

Bo = 0.2115  m

2

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

bh v1 =

 n  A2  d  A1  d1

2

 v1 = 0.3  m

Bo

v2 = h  v1 Ms

if

Ns

v2 = 0.3  m

Igg´

Bo  n   A1  A2  v2

Then cross-section is entirement comprimee.

4

Moment of inertia of the transformed uncracked section ( m )

3

I=

bh 3

 n   A2  d  A1  d1 2

2

2   Bo  v1

I = 0.00737  m

4

or Igg´ =

b 3

  v1  v2 3

3

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

Igg´ = 0.00737  m

4

Ms Ns

Ms Ns

Igg´

= 0.09583  m

Bo  n   A1  A2  v2

Igg´

Bo  n   A1  A2  v2

= 0.10108  m

The assumption was correct.

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

σb1 =

Ns Bo

Ms  v1

σb1 = 10.35  MPa

I

σb1  0.6  fck = allowable

0.6  fck = 18  MPa

Allowable concrete stress:

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

σb2 =

Ns Bo

Ms  h  v2

σb2 = 0.99  MPa

I

σb2  0

Stress in compression reinforcement (MPa):

 Ns

σs2 = n  

 Bo

Ms  d  v1 I



 

σs2 = 26.58  MPa

σs2  0.6  fyk = allowable

Stress in compression reinforcement (MPa):

 Ns

σs1 = n  

 Bo

Ms  v2  d1 I



 

σs1 = 143.63  MPa

σs1  0.6  fyk = allowable

Example I. 8: A rectangular column has the dimensions shown in (Fig. I. 8). Check the stresses in concrete and reinforcement caused by a bending moment Ms and external normal tension load Ns. Section properties: fck = 30  MPa

Characteristic value of concrete cylinder compressive strength (MPa): fyk = 410  MPa

Characteristic yield stress of the steel (MPa): Cross-section (m):

b = 0.30  m

h = 0.60  m

Distance of the tension and compression reinforcement d2 = 0.05  m

from the extreme fibre of the cross-section (m): Effective depth of cross-section (m): d = h  d2

d1 = 0.0  m

d = 0.55  m

Bending moment at the section (kN m):

Ms = 110  kN  m

Axial tension load at the section (kN):

Ns = 210  kN 2

Area of compression reinforcing longitudinal bars ( cm ): 2

Area of tension reinforcing longitudinal bars ( cm ):

2

A1 = 0  cm

2

A2 = 22.30  cm

Fig. I. 8 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 concret area displaced by the embedded bars A1  A2 .

Where n =

n = 15

Es Ec

is known as the modular ratio

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

e=

Ms

e = 0.52381  m

Ns

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

c=e

h

c = 0.82381  m

2

2

p = 3  c  90 

A1 b

 c  d1  90 

A2 b

 ( d  c)

p = 2.21916  m

2

3

q = 2  c  90 

A1 b

3

y  py  q = 0

 c  d1

2  90 

A2 b

 ( d  c)

2

q = 1.16833  m

3

y = 0.65053  m

The depth of the compression zone (m): x = c  y

x = 0.173  m 3

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

2

S=

K=

bx 2

Ns S

 n  A1  x  d1  A2  ( d  x)

S = 0.0081  m

3

K = 25935.122

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):

σbc = K  x

σbc = 4.49  MPa

Allowable concrete stress:

σbc  0.6  fck = allowable

0.6  fck = 18  MPa

Stress in compression reinforcement (MPa): σs1 = n  K  x  d1 Stress in tension reinforcement (MPa):

σs2 = n  K  ( d  x)

σs1 = 67.411  MPa σs2 = 146.553  MPa

Example I. 9: A rectangular column has the dimensions shown in (Fig. I. 9). Verify the stresses in concrete and in steel, the cross-section caused by a bending moment Ms and normal

compression force Ns.

The section properties:

Characteristic value of concrete cylinder compressive strength (MPa): fck = 30  MPa Characteristic yield stress of the steel (MPa): fyk = 410  MPa Cross-section (m): b = 0.60  m

h = 1.1  m

Distance of the tension and compression reinforcement d2 = 0.07  m

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

Effective depth of the cross-section (m):

d = h  d2

Bending moment at the section (kN m):

Ms = 430  kN  m

Axial compression load at the section (kN):

Ns = 390  kN 2

Area of compression reinforcing longitudinal bars ( cm ): 2

Area of tension reinforcing longitudinal bars ( cm ):

d1 = 0.055  m

d = 1.03  m

2

A1 = 15.71  cm

2

A2 = 40.21  cm

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 concret area displaced by the embedded bars A1  A2 .

n  15

Where n =

Es Ec

is known as the modular ratio.

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

Ms Ns

e = 1.102  m

Fig. I. 9 We find point e on the outside of the cross-section, thus the cross-section is partially compressed c=e

h

c = 0.55256  m

2

We substitute for c a negative sign in case that the position of c is situated outside of the cross-section, in order to determine the values of p, q and y. Then we can compute the depth of the compression zone of concrete. 2

Area of the transformed uncracked section ( m ):

Bo = b  h  n  A2  A1

Bo = 0.74388  m

2

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):

2

bh v1 =

 n  A2  d  A1  d1

2

 v1 = 0.57  m

Bo

v2 = h  v1

Ms

if

Ns

v2 = 0.53  m

Igg´

Then the cross-section is partially compressed.

Bo  n   A1  A2  v2

4

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

Igg´ =

Ms Ns

Ms Ns

b 3

  v1  v2 3

3

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

Igg´

= 1.10256  m

Bo  n   A1  A2  v2

Igg´

2

3

q = 2  c  90 

A1 b

A1 b

2  90 

 c  d1

Compute the value of y (m): 3

y  py  q = 0

A2

 c  d1  90 

y = 1.0241

= 0.19682  m

The assumption was correct.

Bo  n   A1  A2  v2

p = 3  c  90 

Igg´ = 0.08582  m

b

 ( d  c)

A2 b

 ( d  c)

p = 0.18172  m

2

2

q = 1.26015  m

3

4

Depth of the compression zone (m):

x=y c

x = 0.47  m

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

2

S=

K=

bx

 n  A1  x  d1  A2  ( d  x)

2 Ns

S = 0.04284  m

3

K = 9104.38

S

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):

σbc = K  x

σbc = 4.29  MPa

σbc  0.6  fck = allowable

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

σbc =

Ns  x 2

bx

σbc = 4.29  MPa

 n  A2  ( d  x)  n  A1  d1  x

2

Allowable concrete stress:

0.6  fck = 18  MPa

Stress in compression and tension reinforcement (MPa):

σs1 = n  K  x  d1

σs2 = n  K  ( d  x)

σs1 = 56.88  MPa

σs2 = 76.26  MPa

Eample I. 10: A rectangular member has the dimensions as shown in ( Fig. I. 10). Verify the stresses in

concrete and in steel, the cross-section caused by a bending moment Ms and normal compression force Ns.

The section properties: Characteristic value of concrete cylinder compressive strength (MPa): Characteristic yield stress of the steel (MPa): Cross-section (m):

b = 0.50  m

fck = 25  MPa

fyk = 410  MPa

h = 1.0  m

Distance of the tension and compression reinforcement d2 = 0.05  m

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

effective depth of the cross-section (m): d = h  d2 Bending moment at the section (kN m): Axial compression load at the section (kN):

d1 = 0.05  m

d = 0.95  m

Ms = 90  kN  m Ns = 410  kN 2

Area of compression reinforcing longitudinal bars: A1 = 15.71  cm 2

Area of tension reinforcing longitudinal bars: A2 = 38.21  cm

Where n =

n = 15

Es Ec

is known as the modular ratio (-).

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

e=

Ms Ns

e = 0.219  m

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

h 2

e

c = 0.28  m

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

Fig. I. 10

Reinforced concrete sections are usually transformed into equivalent concrete sections.

2

Area of the transformed uncracked section ( m ):

Bo = b  h  n  A2  A1

Bo = 0.58088  m

2

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): 2

bh v1 =

2

v2 = h  v1

 n  A2  d  A1  d1 Bo

 v1 = 0.52615  m

v2 = 0.47385  m

Ms

if

Ns

Igg´

Then the cross-section is partially compressed.

Bo  n   A1  A2  v2

4

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

Igg´ =

Ms Ns

Ms Ns

  v1  v2

b

3

3

3

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

Igg´

= 0.21  m

Bo  n   A1  A2  v2

Igg´

2

3

q = 2  c  90 

A1 b A1 b

3

y  py  q = 0

A2

2  90 

 c  d1  90 

 c  d1

4

= 0.18  m

The assumption was correct.

Bo  n   A1  A2  v2

p = 3  c  90 

Igg´ = 0.05765  m

b

 ( d  c)

A2 b

 ( d  c)

p = 0.15928  m

2

2

q = 0.36745  m

3

y = 0.642  m

Depth of the compression zone ( m):

x=y c

x = 0.92  m

3

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

2

S=

K=

bx 2 Ns S

 n  A1  x  d1  A2  ( d  x)

S = 0.23  m

3

K = 1767.57

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):

σbc = K  x

σbc = 1.63  MPa

σbc  0.6  fck = allowable

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

σbc =

Ns  x

σbc = 1.63  MPa

2

bx

 n  A2  ( d  x)  n  A1  d1  x

2

0.6  fck = 15  MPa

Allowable concrete stress:

Stresses in compression and tension reinforcements (MPa):

σs1 = n  K  x  d1

σs2 = n  K  ( d  x)

σs1 = 23.143  MPa

σs1  0.6  fyk = allowable

σs2 = 0.718  MPa

σs1  0.6  fyk = allowable

Example I. 11: A rectangular member has the dimensions as shown in ( Fig.I. 11).The entire section is in compression, and unsymmetrically reinforced, check the stresses in concrete and in steel reinforcement. The section properties: Characteristic value of concrete cylinder compressive strength (MPa): Characteristic yield stress of the steel (MPa):

Cross-section (m): b = 0.30  m

fck = 25  MPa

fyk = 410  MPa

h = 0.60  m

Distance of the compression reinforcement from the extreme fibre of the cross-section (m): d2 = 0.05  m

d1 = 0.05  m

Effective depth of the cross-section (m): d = h  d2

d = 0.55  m

Ms = 70  kN  m

Bending moment at the section (kN m):

Axial compression load at the section (kN):

Ns = 1700  kN 2

Area of compression reinforcing longitudinal bars ( cm ): 2

Area of tension reinforcing longitudinal bars ( cm ):

2

A1 = 10.40  cm 2

A2 = 4.50  cm

Reinforced concrete sections are usually transformed into equivalent concrete sections.

n = 15

Where n =

Es Ec

is known as the modular ratio.

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

e=

Ms Ns

e = 0.041  m

h 6

= 0.1  m

e

Fig. I. 11 2

Area of the transformed uncracked section ( m ):

h 6

The entire section is in compression.

Bo = b  h  n  A2  A1

Bo = 0.20235  m

2

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): 2

bh v1 =

 n  A2  d  A1  d1

2

 v1 = 0.29  m

Bo

v2 = h  v1

v2 = 0.31  m

4

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

Igg´ =

3

  v1  v2 3

3

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

Igg´ = 0.00677  m

4

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

c=

h 2

 v1

c = 0.01093  m

When the position of c is inside of the cross-section, then the sign remains positive for the determination of p,q and y values in order to compute the depth of the compression zone x.

MG = Ms  Ns  c MG

if

Ns

MG = 51.41216  kN  m Igg´

Bo  n   A1  A2  v2

Then the cross-section is entire in compression.

4

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

I=

bh 3

 n   A2  d  A1  d1 2

2

2   Bo  v1

I = 0.00677

or

Igg´ =

MG Ns

MG Ns

b 3

  v1  v2 3

3

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

Igg´

= 0.03024  m

Bo  n   A1  A2  v2

Igg´

Bo  n   A1  A2  v2

Igg´ = 0.00677  m

4

= 0.09694  m

The assumption was correct.

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

σb1 =

Ns Bo

Ms  v1

σb1 = 11.38  MPa

I

σb1  0.6  fck = allowable

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

σb2 =

Ns Bo

Ms  h  v2

σb2 = 5.41  MPa

I

σb2  0.6  fck = allowable

Stresses in reinforcement (MPa):

 Ns

σs1 = n  

 Bo

 Ns

σs2 = n  

 Bo

Ms  v2  d1 I

Ms  d  v1 I



 



 

σs1 = 166.47  MPa

σs1  0.6  fyk = allowable

σs2 = 85.56  MPa

σs2  0.6  fyk = allowable

Example I. 12: Determine the entire area of steel reinforcement as shown in (Fig. I. 12) and check the stresses of concrete and steel for a known bending moment and external tension axial load The section properties: Characteristic value of concrete cylinder compressive strength (MPa): Characteristic yield stress of the steel (MPa):

fyk = 410  MPa

fck = 30  MPa

Allowable stress of steel (MPa): σsall = 0.6  fyk Cross-section (m): b = 0.30  m

σsall = 246  MPa

h = 0.60  m

Bending moment at the section (MN m):

Ms = 0.05  MN  m

External axial tension load at the section (MN):

Ns = 0.40  MN

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

d1 = 0.04  m

Effective depth of the cross-section (m): d = h  d2

e=

Ms

h 2

d = 0.54

e = 0.125  m

Ns

a1 =

d2 = 0.06  m

 e  d1

a2 = z s  a1

a1 = 0.385  m

z s = h  d1  d2

z s = 0.5  m

a2 = 0.115  m

The entire area of tension reinforcing longitudinal bars:

A1 =

Ns  a2

Provided: 2  ϕ16 Diameter:

σsall  z s

ϕ1 = 16  mm

2

A1 = 3.74  cm

A2 =

As1 = n1  π 

Ns  a1

Provided: 4  ϕ20 Diameter:

σsall  z s 2

A2 = 12.52  cm

Stresses in tension reinforcement (MPa):

As2 = n2  π 

ϕ1

ϕ2 = 20  mm

4

n1 = 2

2

4

ϕ2

Numbers of bars:

2

As1 = 4.02  cm

Numbers of bars:

2 2

As2 = 12.56  cm

n2 = 4

σs1 =

Ns  a2

d  d1  As1

σs2 =

Ns  a1

d  d1  As2

σs1 = 228.78  MPa

σs2 = 245.09  MPa

σsall = 0.6  fyk

σs1  0.6  fyk = allowable

σs2  0.6  fyk = allowable

σsall = 246  MPa

Fig. I. 12

Section properties:

Characteristic value of concrete cylinder compressive strength (MPa): Characteristic yield stress of the steel (MPa): Cross-Section (m):

b  0.30  m

fck  25  MPa

fyk  410  MPa

h  0.50  m

Bending moment at the section (MN m):

M  0.03  MN  m

External axial compression load at the section (MN):

N  0.30  MN

Example I. 13: A T-beam having the cross-sectional dimensions shown in (Fig. I. 13) is subjected to a specified load bending moment Ms. Compute the steel and concrete stresses. The section properties: Characteristic value of concrete cylinder compressive strength (MPa): Characteristic yield strength of reinforcing steel (MPa):

fck = 25  MPa

fyk = 400  MPa

The concrete dimensions:

Total depth of the T-beam (m):

h = 0.60  m

Width of the web (m):

bw = 0.25  m

Flange width (m):

b = 1.00  m

Flange depth (m):

hf = 0.10  m

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

d2 = 0.10  m

d = h  d2

Effective depth to the steel centroid (m): 2

d = 0.5  m 2

Area of reinforcing longitudinal bars ( cm ):

A2 = 42.00  cm

The bending moment at the section (kN m):

Ms = 250  kN  m

The modular ratio is:

n=

Es Ec

n = 15

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

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

The neutral axis may lie in the flange as shown in Fig. I. 13b (that is x < hf ) or it may pass through the web as in Fig. I. 13c ( x > hf ).

Fig. I. 13 To locate the neutral axis, first assuming that it lies in the flange, equating the moments of areas about neutral axis (Fig. I. 13b ) gives the equation:

2

b

x

 n  A2  ( d  x) = 0

2

Solving

x = 0.196  m

Since this is greater than hf = 100 mm, the assumption is not correct and the neutral axis passes through the web. Taking moments of areas for the case in Fig. I. 13c

2

bw 

x

2

 

 b  bw  hf   x 

hf 

  n  A2  ( d  x) = 0

Solving

2

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 Fig. I. 13d, the moment is given by the moment of the compressive forces abou the centroid of steel. Taking the compressive area as the difference between two rectangles bx

b  bw  x  hf

and

Ms = b  x 

Solving

 2 

σbc

d 

The moment equation gives:

x

   b  bw   x  hf 

3

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

σbc  

σbc = 6.64  MPa

Allowable concrete stress:

0.6  fck = 15  MPa

σbc  0.6  fck

The assumption was correct.

From the stress distribution diagram, σs2 = n  σbc 

dx x

Allowable steel stress (MPa):

σs2 = 133.25  MPa 0.6  fyk = 240  MPa

Alternatively, to use the flexure formula,

σbc =

M x Icr

σs2  0.6  fyk

The assumption was correct.

4

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

3

Icr = b 

x

3

b  bw  x  hf3 3

 n  A2  ( d  x)

2

Icr = 0.00805  m

4

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

σbc =

Ms  x

σbc = 6.64  MPa

Icr

as before

Stress in tension reinforcement (MPa):

and

σs2 = n  Ms 

( d  x) Icr

σs2 = 133.25  MPa

as before

Check the location of the neutral axis

σsall = 0.6  fyk

σsall = 240  MPa

If Mt  Ms 2

Then the neutral axis will then be in the web, otherwise the neutral axis will lie in the flange

 

b  hf   d  Mt =

hf 

  σsall

3

30  d  hf

Mt = 93.33  kN  m

Mt  M s

The assumption was correct.

The neutral axis of a T beam may be either in the flange or in the web, depending upon the proportions of the cross-section, the amount of tensile steel, and the strengths of materials. If the calculated depth to the neutral axis is less than or equal to the slab thickness hf , the beam 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 Fig. I. 13b, which shows a T beam with neutral axis in the flange. The compressive area is indicated by the shaded portion of the figure.

if

Ms  M t

neutral axis is inside of the flange

if

Ms  M t

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

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

Ms Ns

Igg´ Bo  v2

Where Ms is the bending moment at the section Ns is the external axial load at the section Igg´ is the moment of inertia of the transformed uncracked section Bo is the aera of the transformed uncracked section

For very small eccentricities e, the distance to the neutral axis x may exceed the depth h of the section. Then the entire cross-section area will be in compression, so that the stress in the steel A2 will also be compressive, though mostly of magnitude smaller than σsall.

It should be noted that a concentrically loaded member is one whose eccentricity e about the axis of the cross-sectoin is zero. For symmetrically reinforced sections the member is concentrically loaded when the load passes through the geometric center of the section. For an unsymmetrically reinforced section, with A1  A2, the load must pass through a point known as the plastic centroid. This point is the location of the resultant of the three internal forces, concrete force, steel force on the upper side, and steel force in the botom side of section.

Evidently, for reasons of equilibrium, the external load must act along the same line as the resultant of all internal forces, i.e., must pass through the plastic centroid. In a symmetrical section the center and the plastic centroid coincide.

Example I. 14: Verify the stresses in concrete and in steel of a T-beam (Fig. I. 14) caused by bending moment Ms and external compression force Ns. The section properties:

fck = 25  MPa

Characteristic value of concrete cylinder compressive strength (MPa):

Characteristic yield strength of reinforcing steel (MPa):

fyk = 400  MPa

The concrete dimensions Total depth of the T-beam (m):

h = 0.65  m

Width of the web (m):

bw = 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 cross-section (m):

d2 = 0.03  m

d = h  d1

Effective depth to the steel centroid (m): 2

d = 0.61 2

Area of reinforcing longitudinal bars ( cm ):

A1 = 9.42  cm

The bending moment at the section (kN m):

Ms = 62  kN  m

Axial external load at the section (kN):

Ns = 1300  kN

n = 15

Where n =

Es Ec

d1 = 0.04  m

is known as the modular ratio.

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

2

A2 = 3.39  cm

Fig. I. 14 Entire T beam section is in compression 2

Area of the transformed uncracked section ( m ):

Bo = bw  h  b  bw  hf  n  A1  A2

Bo = 0.22672  m

2

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):

 bw  h2  b  bw  hf2  v1 =    n   A1  d1  A2  d Bo  2 2  1

v2 = h  v1

v1 = 0.25905  m

v2 = 0.39095  m 4

Moment of inertia of the transformed uncracked section ( m ): bw   v1  v2 3

Igg´ =

3

3

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

Igg´ = 0.00973  m

4

Check the location of the neutral axis: Ms

if

Ns

Ms Ns

Igg´

The entire cross-section area will be in compression.

Bo  v2

Igg´

= 0.047  m

σo =

Bo  v2

Ns

= 0.109  m

σo = 5.73  MPa

Bo

The assumption was correct.

K=

Ms Igg´

K = 6367.781

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

σb1 = σo  K  v1

σb1 = 7.38  MPa

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

σb2 = σo  K  v2

σb2 = 3.24  MPa

Stresses in compression reinforcement (MPa): 0.6  fyk = 240  MPa

σs1 = n  σo  K  v1  d1 

σs2 = n  σo  K  d  v1 

σs1 = 106.93  MPa

σs2 = 52.48  MPa

σs1  0.6  fyk = allowable σs2  0.6  fyk = allowable

Example I. 15: Verify the position of the neutral axis of a T-beam section (Fig. I. 15) caused by the bending moment Ms.The section properties and material characteristic of concrete and steel are as follows: Assumptions: fck = 20  MPa

Characteristic value of concrete cylinder compressive strength (MPa): Characteristic yield strength of reinforcing steel (MPa): Total depth of the T-beam (m): h = 0.65  m bw = 0.25  m

Width of the web (m):

fyk = 410  MPa

Flange depth (m):

hf = 0.15  m

Flange width (m):

b = 1.1  m

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

d2 = 0.06  m

The centroid of the steel bars from the top of the T-beam is (m): Bending moment at the section (kN m): σcall = 0.6  fck 2

 

σcall = 12  MPa

b  hf   d  Mt =

d1 = 0.00  m d = h  d2

d = 0.59  m

Ms = 230  kN  m σsall = 0.6  fyk

σsall = 246  MPa

hf 

  σsall

3

30  d  hf

Mt = 249.07  kN  m

Mt  M s

neutral axis is in the flange

Fig. I. 15 2

A1 = 0  m

Area of compression reinforcing bars ( m ): 2

A2 = 0.0016  m

Area of tension reinforcing bars ( m ):

Where n =

n = 15

Es

2

2

is known as the modular ratio.

Ec

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

Depth of the compression zone (m):

2

bw 

x

2

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

hf

2

2

=0

x = 0.140  m 4

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

3

Icr =

bx 3

 b  bw 

x  hf3 3

2

2

 n  A2  ( d  x)  n  A1  d1  x

Icr = 0.00587

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

σbc = Ms 

x

σbc = 5.50  MPa

Icr

0.6  fck = 12  MPa

σbc  0.6  fck = allowable

0.6  fyk = 246  MPa

σs2  0.6  fyk = allowable

Stress in tension reinforcement (MPa):

σs2 = n  Ms 

d x Icr

σs2 = 264.44  MPa

Example I. 16: Design the flexural reinforcement for the I-beam section (Fig. I. 16) 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

b1 = 1.10  m

b2 = 0.60  m

h = 0.9  m

h1 = 0.10  m

h2 = 0.25  m

Cross-section (m):

bw = 0.20  m

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

d2 = 0.10  m

The centroid of the steel bars from the top of the beam is Bending moment at the section (kN m)

n  15

Where n =

Es Ec

d = h  d2

Ms = 1300  kN  m

is known as the modular ratio.

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

Allowable stress in concrete (MPa): σcall = 0.6  fck

σcall = 12  MPa

Allowable stress in steel (MPa): σsall = 0.6  fyk

k1 =

α1 =

σsall = 270  MPa

σsall n n n  k1

k 1 = 18

α1 = 0.45455

d1 = 0.05  m d = 0.8  m

Fig. I. 16 Depth of the compression zone (m): x = α1  d

x  h1

x = 0.36

Compression force acting in the concrete (kN):

 

 

Fc = bw  x  b1  bw  h1   2 

h1  σcall   x  2

Fc = 1367.86  kN

Inner lever arm of internal forces (m):

z=d

h1 2

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

b1  h1  bw  x  h1

z = 0.73  m

Stresses in compression reinforcement (MPa):

σs1 =

n  x  d1

x

 σcall

σs1 = 155.25  MPa 2

Area of reinforcement in the tension and compression zone ( m ): A1 =

M s  Fc  z

d  d1  σsall

A1 = 0.00593  m

A2 =

Fc  A1  σs1 σsall

2

A2 = 0.00593  m

2

Example I. 17: Verify the stresses in concrete and in steel of a T-beam section (Fig. I. 17) caused by the 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 = 450  MPa

Allowable stress of concrete (MPa): σcall = 0.6  fck

σcall = 12

Allowable stress of steel (MPa):

σsall = 0.6  fyk

σsall = 270  MPa

Bending moment at the section

Ms = 0.550  MN  m

Total depth of the T-beam (m):

h = 1.10  m

Flange depth (m):

hf = 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 cross-section (m):

d2 = 0.08  m

The centroid of the steel bars from the top of the beam is (m): Position of the neutral axis: if

2

 

b  hf   d  Mt =

Mt  M s

d1 = 0.0  m

d = h  d2

d = 1.02  m

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

hf 

  σsall

3

30  d  hf

Mt = 158.04  kN  m

Mt  M s

The assumption was correct.

Fig. I. 17 2

A1 = 0  m

Area of compression reinforcing bars ( m ): 2

A2 = 0.002198  m

Area of tension reinforcing bars ( m ):

n  15

Where n =

Es

2

2

is known as the modular ratio.

Ec

Es, Ec is the modulus of elasticity of steel, or concrete respectively. Depth of the compression zone (m): 2

bw 

x

2

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

hf

2

2

=0

x = 0.28  m

4

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

3

Icr =

bx 3

 b  bw 

x  hf3 3

2

2

 n  A2  ( d  x)  n  A1  d1  x

Icr = 0.02288  m

4

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

σbc = Ms 

x

σbc = 6.77  MPa

Icr

0.6  fck = 12  MPa

σbc  0.6  fck = allowable

0.6  fyk = 270  MPa

σs2  0.6  fyk = allowable

Stress in tension reinforcement (MPa):

σs2 = n  Ms 

d x Icr

σs2 = 266.10  MPa

Example I. 18: Verify the stresses in concrete and in steel of a T-beam section (Fig. I. 18) caused by bending moment Ms and external compression force Ns. The section properties fck = 20  MPa

Characteristic value of concrete cylinder compressive strength (MPa): Characteristic yield strength of reinforcing steel (MPa): fyk = 410  MPa Total depth of the T-beam (cm): Width of the web (cm):

h = 90  cm

Flange depth (cm):

hf = 15  cm

bw = 25  cm

Flange width (cm):

b = 110  cm

2

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

2

A2 = 15.70  cm

2

A1 = 15.70  cm

Distance of the tension and compression reinforcement from the extreme fibre of the cross-section ( cm):

d2 = 5.0  cm

d1 = 5.0  cm

The centroid of the steel bars from the top of the T-beam section ( cm) is: d = 75  cm Bending moment at the section (kN m):

Ms = 450  MN  m

Axial compression external load at the section (kN):

n = 15

Where n =

Es Ec

Ns = 850  kN

is known as the modular ratio.

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

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

Eccentricity (cm): e =

Ms Ns

 100

yG =

e = 52.94  m

bw  h  b  bw  hf

2

2  bw  h  b  bw  hf c = e  yG

yG = 31.43617  m

c = 21.50501  m

Since the external compression force applied a significant distance outside the cross-section, this causes that the section tension governs. Position of the neutral axis:

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.



2



H2 = b  bw  3  c  2  hf  hf  90  A1  c  d1  d1  A2  ( d  c )  d

2

H2 = 11742646.95

H3 = hf  3  c  b  hf  90  A1  c  d1  d1  hf  A2  ( d  c )  d  hf 

H3 = 6194870.01

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

p=

q=

3

3bc

2

bw

2bc bw

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

p = 12515.52657

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

q = 724139.69811

 3

3

 2

y  py  q = 0

y = 48.65  m

Depth of the compression zone (m):

x=y c

x = 28.94  m

Fig. I. 18 The location and distribution of the internal and external forces. Section modulus of a transformed T-beam section 2

S=

K=

bx

2

b  bw  x  hf2 2

Ns  1000

K = 0.315

100  S

 n  A1  x  d1  A2  ( d  x)

S = 26955.19

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

The compressive stress in the concrete at the top fibres of the section (MPa): σbc = K  x

σbc = 9.12  MPa

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

σbc =

Ns  10  x 2

bw  x 2

b  bw  2  x  hf  hf 2

σbc = 9.12  MPa as above

 n  A1  d1  x  n  A2  ( d  x)

σbc  0.6  fck = allowable

Stress in tension reinforcement (MPa):

or

σs2 = n  K  ( d  x) σs2 = n  σbc 

dx x

σs2 = 217.87  MPa MPa σs2 = 217.87  MPa as above

Example I. 19: Verify the stresses of concrete and steel of a T-column (Fig. I. 19) caused by the bending moment Ms and tension force Ns.

Fig. I. 19 Since the point c is inside of the section or between the reinforcement As1 and As2 and the axial load is a tension load entire section will then be in tension. The section properties: Characteristic value of concrete cylinder compressive strength (MPa): fck = 20  MPa Characteristic yield strength of reinforcing steel (MPa): fyk = 450  MPa

Total depth of the T-beam (m):

h = 0.85  m

Flange depth (m):

hf = 0.15  m

Width of the web (m):

bw = 0.25  m

Flange width (m):

b = 1.10  m

Allowable stress of concrete (MPa): σcall = 0.6  fck

σcall = 12

σsall = 0.6  fyk

Allowable stress of steel (MPa):

σsall = 270  MPa

Distance of the tension and compression reinforcement d2 = 0.05  m

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

The centroid of the steel bars from the top of the T-beam (m) is: d = h  d2 2

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

Axial external tension load at the section (kN):

e=

Ms Ns

2

As1 = 0.0006  m

2

3  ϕ20

Modular ratio

n = 15

2  bw  h  b  bw  hf

yG = 0.29  m

Ns = 370  kN

A1 =

Ns  d  yG  e

d  d1  σsall

A1 = 0.00052  m

2

2

Area of tension reinforcing bars ( m ):

Provided:

2

e = 0.224  m

Area of compression reinforcing bars ( m ):

Provided: 3  ϕ16

d = 0.8  m

Ms = 83  kN  m

Bending moment at the section (kN m):

Eccentricity (m):

bw  h  b  bw  hf

d1 = 0.05  m

As2 = 0.00094  m

A2 =

Ns σsall

 A1

2

Where n =

Es Ec

is known as the modular ratio.

A2 = 0.0008552  m

2

Area of the transformed uncracked section

Bo = bw  h  b  bw  hf  n  As1  As2

Bo = 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

v1 =

 bw  h2  b  bw  hf2    n   As1  d1  As2  d Bo  2 2  1



v2 = h  v1

v1 = 0.30738  m

v2 = 0.542  m

Stress in tension reinforcement (MPa):

σs1 =

Stress in tension reinforcement (MPa):

σs2 =

Ns  d  v1  e

d  d1  As1 

σs1 = 219.5045  MPa

Ns  v1  d1  e

d  d1  As2

σs2 = 252.27  MPa

Example I. 20: Verify the stresses in concrete and in steel of a T-beam section (Fig. I.20) caused by bending moment Ms and external compression force Ns. The section properties Characteristic value of concrete cylinder compressive strength (MPa): Characteristic yield strength of reinforcing steel (MPa): Allowable steel stress (MPa):

σsall = 0.6  fyk

fck = 30  MPa

fyk = 450  MPa σsall = 270  MPa

Total depth of the T-beam (m):

h = 1.05

Flange depth (m):

hf = 0.15  m

Width of the web (m):

bw = 0.30  m

Flange width (m):

b = 0.85  m

Fig. I. 20 The location and distribution of the internal and external forces. σsall  246

Distance of the tension reinforcement d2 = 0.05  m

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

The centroid of the steel bars from the top of the T-beam is (m): d = h  d2 Bending moment at the section ( MN  m):

Ms = 0.900  MN  m Ns = 0.250  MN

Axial compression external load at the section ( MN):

2

Area of reinforcement in the tension and compression zone ( m ): 2

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

yG =

A2 = 0.0035

bw  h  b  bw  hf

2

2  bw  h  b  bw  hf

yG = 0.4316

Eccentricity (m):

e=

Ms

e = 3.6  m

Ns

c = yG  e

c = 3.1684  m

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.





2

H2 = b  bw  3  c  2  hf  hf  90  A1  c  d1  d1  A2  ( d  c )  d

2

H2 = 1.43438

H3 = hf  3  c  b  hf  90  A1  c  d1  d1  hf  A2  ( d  c )  d  hf 

H3 = 0.93143

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

p=

q=

3bc

2

bw

2bc bw

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

p = 20.38778

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

q = 28.00861

 3

3

 2

3

y  py  q = 0

y = 3.52843  mm x=y c

Depth of the compression zone (m):

x = 0.36  m

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

if

Mt  M s

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

2

 

b  hf   d  Compute Mt (MN m):

Mt  M s

Mt =

hf 

  σsall

3

30  d  hf

Mt = 0.175  MN  m

The assumption was correct.

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

σbc =

Ns  x 2

bw  x 2

b  bw  2  x  hf  hf 2

σbc = 9.61  MPa

 n  A1  d1  x  n  A2  ( d  x)

σbc  0.6  fck = allowable

σbc which is lesser than allowable stress

Stress in tension reinforcement (MPa):

σs2 = n  σbc 

dx

σs2 = 256.40  MPa

x

σsall = 0.6  fyk

σs2 which is lesser than allowable stress σsall of reinforcement

σsall = 270  MPa

Example II. 9 : Design the flextural reinforcement for the T beam (Fig II. 9) caused by bending moment Ms and axial tension load Ns . Section properties: Characteristic value of concrete cylinder compressive strength (MPa): fck = 20  MPa fcd = 0.85 

Design value of concrete cylinder compressive strength (MPa):

Characteristic yield strength of reinforcing steel (MPa):

Design yield stress of reinforcement (MPa):

fyd =

fck

fcd = 11.33  MPa

1.5

fyk = 325  MPa

fyk

fyd = 282.60  MPa

1.15

Total depth of T beam (m):

h = 0.65  m

Flange depth (m):

hf = 0.15  m

Web width (m):

bw = 0.35  m

Flange width (m):

b = 1.20  m

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

d2 = 0.06  m

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

Bending moment at the section ( MN  m): Msd = 0.55  MN  m

d = h  d2

d1 = 0.05  m d = 0.59  m

Axial tension load at the section (MN):

The eccentricity (m):

e=

Position of the neutral axis:

2

 

b  hf   d  Mt =

Ms Ns

if

Nsd = 0.720  MN

Nsd  0

e = 0.764

Mt  M s

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

hf 

  fyd

3

30  d  hf

Mt = 0.312  MN  m

Mt  Msd 2

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

bw  h  b  bw  hf

The assumption was correct

2

yG = 0.235  m

2  bw  h  b  bw  hf

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

hf 2

from the upper edge (MN).

Ffl = b  bw  hf  fcd

Ffl = 1.445  MN

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

z=d

hf 2

z = 0.515  m

The ultimate moment of flange reffers to the tension reinforcement (MN m):Mfl = Ffl  z

Mfl = 0.74  MN  m

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

Mext = Msd  Nsd  d  yG

Mext = 0.80  MN  m

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

Msds = Mext  Mfl

μ=

Msds 2

bw  d  fcd

Msds = 0.061  MN  m

μ = 0.045

From the Design Table (II. 6) we obtain :

ρ = 0.01475 2

The required tension reinforcement A2 ( cm ) can be calculated from: A2 = ρ  bw  d  fcd  100 

Ns fyd

6  ϕ25

Provided:

As2 = nt  π 

ϕt

 10

4

2

A2 = 28.92  cm

Diameter:

ϕt = 25  mm

Number of bars:

nt = 6

2

As2 = 0.00295  m

4

2

Which is greater than A2

To locate the neutral axis, first assuming that it lies in the flange, equating the moments of areas about neutral axis gives the equation: 2

b

x

 n  As2  ( d  x) = 0

2

Solving

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:

2

bw 

x

2

 

 b  bw  hf   x 

hf 

  n  As2  ( d  x) = 0

2

Solving

x = 0.176  m

Example II. 1 The cross - section dimensions of a beam shown on Figure II. 1 are subjected to bending moment Msd. Determine the required tension reinforcement to the section.

Material data:

Characteristic value of concrete cylinder compressive strength (MPa):

fck = 25  MPa

Design value of concrete cylinder compressive strength (MPa): fcd = 0.85  Characteristic yield stress of reinforcement (MPa):

Cross-section ( m):

b = 0.3  m

fck 1.5

fcd = 14.16  MPa

fyk = 410  MPa

h = 0.5  m

Cover of reinforcement (m): d1 = 0.02  m

d2 = 0.02  m

Effective depth of a cross-section (m): d = h  d1 Design maximum bending moment ( MN  m):

d = 0.48  m

Msd = 0.15  MN  m

1) To apply the Design Table (II. 9, III. 9, IV. 9) the bending moment Msd has to be brought into a dimensionless form: μ=

Msd

μ = 0.15

2

b  d  fcd From the Design Table (II. 9) we obtain the reinforcement ratio:

ρ = 0.04147

2

The required tension reinforcement A2 ( cm ) is as follows: A2 = ρ  b  d  fcd  100

2) From the Design Table (III. 9):

ρ = 0.04581

2

The required tension reinforcement A2 ( cm ) is as follows: A2 = ρ  b  d  fcd  100 3) From the Design Table (IV. 9):

2

A2 = 8.456  cm

ρ = 0.04179

2

A2 = 9.340  cm

2

The required tension reinforcement A2 ( cm ) is as follows: A2 = ρ  b  d  fcd  100

The difference between Table (II. 9) and Table (III. 9) 8.456

for design of tension reinforcement is:

9.340 Diameter:

Provided:

ϕt = 18  mm

4  ϕ18

Which is greater than the value:

Number of bars:

As = nϕ  π 

ϕt

 0.90535

nϕ = 4

2

4

As  A2

Figure II. 1

2

As = 10.17  cm

2

A2 = 8.521  cm

Example II. 2 Determine the required tension reinforcement of reinforced concrete rectangular beam shown on Figure II. 2, subjected to bending moment Msd. Use the bi - linear diagram for steel ( β´ = 0) and the bi - linear diagram for concrete. Assumptions: Characteristic value of concrete cylinder compressive strength (MPa):

fck = 25  MPa

Design value of concrete cylinder compressive strength (MPa): fcd = 0.85  Characteristic yield stress of reinforcement (MPa):

Design yield stress of reinforcement (MPa):

fyd =

fck 1.5

fcd = 14.16  MPa

fyk = 245  MPa

fyk 1.15

fyd = 213.04  MPa

Cover of reinforcement (m): d2 = 0.03  m Cross - section (m):

h = 0.70  m

b = 0.30  m

Effective depth of a cross - section (m): d = h  d2 Design maximum bending moment (MN.m):

Msd = 0.4 MN  m

Msd

μ=

d = 0.67  m

μ = 0.21

2

b  d  fcd From the Design Table (III. 4) we obtain: ρ = 0.11191

2

The required tension reinforcement A2 ( cm ) is as follows: A2 = ρ  b  d  fcd  100

Diameter: ϕt = 30  mm

Number of bars:

Provided tension reinforcement:

Which is greater than the value:

As = nϕ  π 

As  A2

nϕ = 5 ϕt

2

4

2

As = 35.34  cm

2

A2 = 31.85  cm

Figure II. 2 Example II. 3 Determine the required tension reinforcement of a reinforced concrete rectangular column shown on Figure II. 3, subjected to bending moment Msd and tension axial force Nsd. Material data: Concrete (MPa):

Reinforcing steel (MPa):

Cross-section (m):

fck = 25  MPa

fcd = 0.85 

fyk = 300  MPa

fyd =

b = 0.25  m

fyk

fcd = 14.16  MPa fyd = 260.86  MPa

1.15

d2 = 0.06  m

Effective depth of a cross - section (m): d = h  d1

Bending moment (MN.m) :

1.5

h = 0.5  m

Cover of reinforcement (m): d1 = 0.04  m

Tension axial force (MN):

fck

Nsd = 0.300  MN

d = 0.46  m Nsd  0

Msd = 0.030  MN  m

We obtain the value of eccentricity (m):

e=

Msd Nsd

e = 0.1  m

h 6

= 0.08  m

e

h 6

 h  d  e  2 2   10 4

Nsd  

The required tension reinforcement A1 =

2

d  d1  fyd

A1 ( cm ) is as follows: Provided:

3  ϕ12

Diameter:

ϕc = 12  mm

Number of bars:

As1 = nϕc  π 

The required tension reinforcement

A2 =

2

A2 ( cm ) is as follows: Provided:

3  ϕ20

Diameter:

ϕc

2

As1 = 3.39  cm

4

As2 = nϕt  π 

2

A2 = 9.0  cm

Number of bars:

ϕt

nϕc = 3

2

 Nsd 4   10  A1   fyd 

ϕt = 20  mm

2

A1 = 2.46  cm

nϕt = 3

2

4

Figure II. 3

2

As2 = 9.42  cm

Or use Table III. 5 Material data: fck = 25  MPa

Concrete (MPa):

Reinforcing steel (MPa):

fyk = 300  MPa

b = 0.25  m

Cross-section (m):

fcd = 0.85 

fyd =

1.5

fyk 1.15

fyd = 260.86  MPa

d2 = 0.06  m

Effective depth of a cross-section (m): d = h  d1

Bending moment (MN.m) :

fcd = 14.16  MPa

h = 0.5  m

Cover of reinforcement (m): d1 = 0.04  m

Tension axial force (MN):

fck

d = 0.46  m

Nsd = 0.300  MN Msd = 0.030  MN  m

Combined actions have to be transformed in relation to the centre of gravity of the tension reinforcement:

h  d  2 2 

Msds = Msd  Nsd  

Msds = 0.027 MN  m

To apply the Design Table (III. 5) the bending moment Msds has to be brought into a dimensionless form:

μ=

Msds

μ = 0.03603

2

b  d  fcd ρ = 0.01407

From the Design Table (III. 5) we obtain :

2

The required tension reinforcement A2 ( cm ) is as follows:

A2 = ρ  b  d  fcd  100 

Nsd fyd

 10

4

2

A2 = 9.20  cm

As above

2

The required tension reinforcement A1 ( cm ) is as follows:

A1 =

Nsd fyd

4

 10  A2

2

A1 = 2.29  cm

As above

Conclusion: With the acting tension force Nsd  0 within the lower and the upper reinforcement layers no concrete compression zone appears. The tension force is carried completely by reinforcement and nothing by the concrete because the tensile strength of concrete is neglected.

Example II. 4 Reinforced concrete rectangular column on Figure II. 4 subjected to bending moment Msd and axial compression force Nsd is given. Determine the required tension reinforcement for the cross-section. Use the bi - linear diagram for steel ( β´ = 0) and bi - linear diagram for concrete. Material data: Concrete (MPa):

Reinforcing steel (MPa):

fck = 25  MPa

fcd = 0.85 

fyk = 300  MPa

fyd =

fck 1.5

fyk 1.15

Modulus of elasticity of reinforcement (GPa):

Es = 200  GPa

Design value of bending moment (MN.m):

Msd = 0.09  MN  m

Design value of axial compression force (MN):

Nsd = 0.1  MN

It is assumed that (m):

b = 0.3  m

fcd = 14.16  MPa fyd = 260.86  MPa

h = 0.4  m

Cover of reinforcement (m): d2 = 0.04  m Effective depth (m): d = h  d2

d = 0.36  m

Eccentricity due to action effects (m): e=

Msd Nsd

e = 0.9  m

h 6

= 0.06667  m

e

h 6

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

 

 h  d  2 2 

Msds = Msd  Nsd  

Msds = 0.106  MN  m

To apply the design table (III. 5) the transformed actions Msds has to be brought into a dimensionless form (-):

Msds

μ=

μ = 0.19245

2

b  d  fcd

From the Design Table (III. 5) we get :

0.0035

αlim =

0.0035 

fyk

ρ = 0.08269

α = 0.27

αlim = 0.72851

α  αlim

1.15  Es

The compression reinforcement is not necessary.

2

The required tension reinforcement A2 ( cm ) is as follows:

  Nsd  4   10    fyd  

2

A2 = ρ  b  d  fcd  100  

Provided:

3  ϕ20

Diameter:

As2 = nϕt  π 

A2 = 8.81  cm

ϕt = 20  mm

ϕt

Number of bars:

2

4

2

As2 = 9.42  cm

Figure II. 4

nϕt = 3

Example II. 5 Determination of the tension reinforcement of reinforced concrete rectangular column shown on Figure II. 5, subjected to bending moment Msd and tension force Nsd, use the bi - linear diagram for steel ( β´ = 0) and bi - linear diagram for concrete. Material data: Concrete (MPa):

Reinforcing steel (MPa):

fck = 25  MPa

fcd = 0.85 

fyk = 300  MPa

fyd =

fck 1.5

fyk 1.15

fcd = 14.16  MPa

fyd = 260.86  MPa

Modulus of elasticity of reinforcement (GPa): Es = 200  GPa Design value of bending moment (MN.m):

Msd = 0.06  MN  m

Design value of axial tension force (MN):

Nsd = 0.1  MN

It is assumed that (m): b = 0.20  m

Cover or reinforcement (m):

Effective depth (m):

h = 0.50  m

d2 = 0.04  m

d = h  d2

Eccentricity due to action effects (m):

d = 0.46  m

e=

Msd Nsd

e = 0.6  m

h 6

= 0.08333  m

e

h 6

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

 

 h  d  2 2 

Msds = Msd  Nsd  

Msds = 0.039  MN  m

To apply the Design Table (III. 5) the transformed actions Msds has to be brought into a dimensionless

form (-): μ=

Msds

μ = 0.06505

2

b  d  fcd

From the Design Table (III. 5) we obtain: 0.0035

αlim =

0.0035 

ρ = 0.0258

αlim = 0.72851

fyk 1.15  Es

2

The required tension reinforcement A2 ( cm ) will be:

A2 = ρ  b  d  fcd  100 

Provided:

 Nsd  4    10   fyd  

3  ϕ18

Diameter:

As2 = nϕt  π 

2

A2 = 7.19  cm ϕt = 18  mm ϕt

Number of bars:

2 2

As2 = 7.63  cm

4 2

The required tension reinforcement A1 ( cm ) is as follows:

A1 =

 Nsd 4  10   A2   fyd 

2

A1 = 3.36  cm

nϕt = 3

Figure II. 5 Example II. 6 Determine the tension reinforcement of a reinforced concrete rectangular column shown on Figure II. 6, subjected to bending moment Msd and compression force Nsd. Use the bi - linear diagram for steel ( β´ = 0) and bi - linear diagram for concrete. Material data:

Concrete (MPa):

Reinforcing steel (MPa):

fck = 25  MPa

fcd = 0.85 

fyk = 300  MPa

fyd =

Design value of bending moment (MN.m):

fck 1.5

fyk 1.15

b = 0.2  m

h = 0.55  m

Cover of reinforcement (m): d2 = 0.05  m Effective depth (m): d = h  d2

d = 0.5  m

Eccentricity due to action effects (m):

fyd = 260.86  MPa

Msd = 0.1  MN  m

Design value of axial compression force (MN): Nsd = 0.4  MN It is assumed that (m):

fcd = 14.16  MPa

Nsd  0

e=

Msd Nsd

h

e = 0.25  m

6

= 0.09167  m

e

h 6

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

 

 h  d  2 2 

Msds = Msd  Nsd  

Msds = 0.19  MN  m

To apply the Design Table (III. 5) the transformed actions Msds has to be brought into a dimensionless form (-): μ=

Msds

μ = 0.26824

2

b  d  fcd

From the Design Table (III. 5) we get:

ρ = 0.12235

2

The required tension reinforcement A2 ( cm ) is as follows:

  Nsd  4   10    fyd  

2

A2 = ρ  b  d  fcd  100  

Provided:

2  ϕ12

Diameter:

As2 = nϕt  π 

A2 = 1.99  cm

ϕt = 12  mm

ϕt

Number of bars:

2

4

2

As2 = 2.26  cm

nϕt = 2

Figure II. 6

Example II. 7 Determination of the tension reinforcement of reinforced concrete rectangular column shown on Figure II. 7, subjected to bending moment Msd and compression force Nsd by the bi - linear diagram for steel ( β´ = 0) and bi - linear diagram for concrete.

Material data: Concrete (MPa):

fck = 25  MPa

fcd = 0.85 

fck 1.5

fyk

Reinforcing steel(MPa): fyk = 300  MPa

fyd =

Modulus of elasticity of reinforcement (GPa):

Es = 200  GPa

1.15

Design value of bending moment (MN.m):

Msd = 0.1  MN  m

Design value of axial compression force (MN):

Nsd = 0.08  MN

fcd = 14.16  MPa fyd = 260.86  MPa

It is assumed that (m):

b = 0.3  m

Cover of reinforcement (m):

h = 0.55  m

d2 = 0.05  m

d = h  d2

Effective depth (m):

d = 0.5  m

Eccentricity due to action effects (m):

e=

Msd Nsd

e = 1.25  m

h 6

= 0.091  m

e

h 6

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

 

 h  d  2 2 

Msds = Msd  Nsd  

Msds = 0.118  MN  m

To apply the Design Table (III. 5) the transformed actions Msds has to be brought into a dimensionless form (-): μ=

Msds

μ = 0.11106

2

b  d  fcd From the Design Table (III. 5) we obtain:

αlim =

0.0035 0.0035 

fyk

ρ = 0.04478

αlim = 0.72851

1.15  Es

The compression reinforcement is not necessary.

α  αlim

2

The required tension reinforcement A2 ( cm ) is:

A2 = ρ  b  d  fcd  100 

 Nsd  4     10   fyd  

2

A2 = 6.54  cm

α = 0.14603

Provided:

4  ϕ116

Diameter:

As2 = nϕt  π 

ϕt  16  mm

ϕt

Number of bars:

nϕt = 4

2 2

As2 = 8.04  cm

4

Figure II. 7 Example II. 8 Determination of the tension reinforcement of a reinforced concrete rectangular column shown on Figure II. 8, subjected to bending moment Msd and compression force Nsd. Use the bi - linear diagram for steel ( β´ = 0) and the bi - linear diagram for concrete.

Material data:

Concrete (MPa):

Reinforcing steel (MPa):

fck = 25  MPa

fyk = 300  MPa

fcd = 0.85 

fyd =

fck 1.5

fyk 1.15

Modulus of elasticity of reinforcement (GPa):

Es = 200  GPa

Design value of bending moment (MN.m):

Msd = 0.4  MN  m

fcd = 14.16  MPa fyd = 260.86  MPa

Nsd = 0.1  MN

Design value of axial compression force (MN):

It is assumed that (m):

b = 0.3  m

h = 0.77  m

Cover of reinforcement (m): d2 = 0.07  m

d = h  d2

Effective depth (m):

d = 0.7  m

To apply the Design Table (III. 5) the transformed action Msd, has to be brought into a dimensionless form (-): Msd

μ=

μ = 0.19208

2

α = 0.27

b  d  fcd From the design table (III. 5) we obtain : μ1 = 0.19

ρ=

μ2 = 0.195

ρ1 = 0.0815

μ  μ2 μ  μ1  ρ1  ρ μ1  μ2 μ2  μ1 2

αlim =

0.0035 0.0035 

fyk  MPa

ρ2 = 0.08394

ρ = 0.08251

α  αlim

αlim = 0.72851

1.15  Es

The compression reinforcement is not necessary.

α  αlim

2

The required tension reinforcement A2 ( cm ) is as follows:

  Nsd  4   10    fyd  

A2 = ρ  b  d  fcd  100  

2

A2 = 20.71  cm

Diameter (mm):

ϕt = 25

Number of bars:

mm

2

Provided longitudinal reinforcement ( cm ): 5  ϕ25 2

Which is greater than the A2 ( cm ) value:

nϕt = 5

As2 = nϕt  π 

ϕt

2

4

 10

2

As2  A2

Figure II. 8

VI. Bearing capacity of reinforced concrete rectangular column in centric compression force and compression force with small eccentricity. Example VI. 1 Determine the bearing capacity of a rectangular column shown on Figure VI. 1. Material data and dimensions of the cross-section are given bellow.

Material data: Concrete (MPa):

fck = 25  MPa

Reinforcing steel (MPa): fyk = 300  MPa

fcd = 0.85 

fyd =

fck 1.5

fyk 1.15

Height of the column measured between centres of restraint (m):

fcd = 14.16  MPa fyd = 260.86  MPa l = 5.56  MPa

2

As2 = 24.54  cm

lo = 0.7  l

Effective length of column (m):

i = 0.288  b

Slenderness ratio:

λ=

lo = 3.892  m

i = 0.0864  m

lo

λ = 45.0463

i

Design value of bending moment (MN.m): Msd = 0.06  MN  m Design value of axial compression force (MN): Nsd = 1.800  MN b = 0.30  m

It is assumed that (MN):

h = 0.6  m

Cover of reinforcement (m): c = 0.05  m

Effective depth (m):

b´ = b  2  c

h´ = h  2  c

Eccentricity due to action effects (m):

e=

h 6

Msd

h

e = 0.033  m

Nsd

6

= 0.1  m

Centric compression force (compression simple), or compression force with

e

small eccentricity

Stress of concrete (MPa):

σc =

Nsd b´  h´

σc = 18  MPa

From the Table (VI. 3) we obtain the following dimensionless parameters: 2

The minimum amount of longitudinal reinforcement Asmin ( cm ) is given by: Asmin = 0.15 

Nsd

 fyk     1.15 

 10

4

2

Asmin = 10.35

cm

2

The required total longitudinal reinforcement A ( cm ) is as follows: 2

A = ρ  10  b´  h´

2

A = 26.95  cm

ρ = 2.6956

Diameter (mm):

ϕ = 22 mm

Number of bars:

2

2

As = n  π 

Provided longitudinal reinforcement ( cm ): 2

n=8

Which is greater than the A value ( cm ):

As  A

The reinforcement ratio is given by (-):

ρ=

From the Table (VI. 3) (MPa):

As b´  h´

ϕ

4

 10

 10

2

2

2

As = 30.41  cm

ρ = 3.04106

σb = 18.815  MPa

Design ultimate capacity of a cross-section may be determined (MN): Nud = σb  b´  h´

Nud = 1.8815  MN

Which is greater than the Nsd (MN) value: Nud  Nsd Nsd = 1.8  MN Figure VI. 1 Example VI. 2 Determine the bearing capacity of a rectangular column shown on Figure VI. 2. Material data and dimensions of the cross-section are given bellow.

Assumptions: Height of the column measured between centres of restraint (m):

Effective length of column (m):

lo = 0.7  l

lo = 3.024  m

i = 0.288  b

i = 0.0864  m

Slenderness ratio:

λ=

lo i

l = 4.32  m

λ = 35

Characteristic yield strength of the reinforcement (MPa):

fck = 25  MPa

fyk = 410  MPa

Design value of an applied axial force (compression) (MPa):

Design value of bending moment (MN.m):

Msd = 0.200  MN  m

Design value of axial compression force (MN):

Nsd = 2.700  MN

It is assumed that the cross-section (m): Cover of reinforcement (m):

Eccentricity due to action effects (m):

6

h

= 0.08  m

6

h = 0.50  m

c = 0.03  m

b´ = b  2  c m

Effective depth (m):

h

b = 0.40  m

h´ = h  2  c

e=

Msd

e = 0.07  m

Nsd

 e Centric compression force (compression simple) or compression force with small eccentricity

Stress of concrete (MPa):

σc =

Nsd

σc = 18.04  MPa

b´  h´

From Table VI. 3 we obtain the following dimensionless parameters: 2

ρ = 1.94325

The minimum amount of longitudinal reinforcement Asmin ( cm ) is given by: Asmin = 0.15 

Nsd

     1.15  fyk

 10

4

2

Asmin = 11.35  cm

2

The required total longitudinal reinforcement A ( cm ) is: 2

A = ρ  10  b´  h´ Diameter:

2

A = 29.07  cm

ϕ = 22 mm

Number of bars:

n=8

2

Provided longitudinal reinforcement ( cm ): 2

2

As = n  π 

Which is greater than the A value ( cm ):

As  A

The reinforcement ratio is given by (-):

ρ=

From the Table VI. 3 (MPa):

As b´  h´

ϕ

4

 10

 10

2

2

2

As = 30.41  cm

ρ = 2.0328

σb = 18.19  MPa

Design ultimate capacity of a cross-section may be determined (MN):

Nud = σb  b´  h´

Nud = 2.72  MN

Which is greater than the Nsd value: Nud  Nsd

Nsd = 2.7  MN

Figure VI. 2 Example VI. 3 Determine the bearing capacity of a rectangular column shown on Figure VI. 3. Material data and dimensions of the cross-section are given bellow.

Assumptions: Cover of longitudinal reinforcement (m): c = 0.02  m Depth of cross-section (m):

h = 0.60  m

h´ = h  2  c

h´ = 0.56  m

Width of cross-section (m):

b = 0.30  m

b´ = b  2  c

b´ = 0.26  m

Characteristic value of concrete cylinder compressive strength (MPa):

fc28 = 25  MPa

Characteristic yield strength of the reinforcement (MPa):

fyk = 410  MPa

Design value of an applied axial force (compression) (MN):

Nsd = 3.25  MN

l = 3.70  m

Height of the column measured between centres of restraint (m): Effective length of column (m):

lo = 0.7  l

lo = 2.59  m

i = 0.288  b

i = 0.0864  m

Slenderness ratio:

λ=

Stress of concrete (MPa):

From the Table VI. 3:

σc =

lo

λ = 30

i

Nsd b´  h´

σc = 22.32  MPa

ρ = 3.23846

2

The minimum amount of longitudinal reinforcement Asmin ( cm ) is given by:

Asmin = 0.15 

Nsd

     1.15  fyk

 10

4

2

Asmin = 13.67  cm

2

Determination of the longitudinal reinforcement of the column ( cm ): A = ρ  b´  h´  10

Diameter: ϕ = 25  mm

Number of bars:

n = 10

2

As = n  π 

Provided longitudinal reinforcement ( cm ): 2

2

Which is greater than the A value ( cm ):

As  A

The reinforcement ratio is given by (-):

ρ=

From the Table (VI. 3) (MPa):

As b´  h´

ϕ

4

 10

 10

2

2

σb = 22.67  MPa

Design ultimate capacity of a cross-section may be determined (MN):

2

2

A = 47.15  cm

2

As = 49.08  cm

ρ = 3.3714

Nud = σb  b´  h´

Nud = 3.30  MN

Which is greater than the Nsd value (MN):

Nud  Nsd

Nsd = 3.25  MN

Figure VI. 3

II. Rectangular cross - sections without compression reinforcement based on the bi - linear diagram for steel and a bi - linear diagram for concrete stress distribution for pure bending or bending moment with axial forces:

Figure II

III. Rectangular cross - sections without compression reinforcement for pure bending or bending moment with axial forces, based on the bi - linear diagram for steel ( Î˛Â´ = 0) and a bi - linear diagram for concrete:

Figure III iV.Rectangular cross - sections without compression reinforcement based on the parabolic-rectangular diagram for steel and a bi - linear diagram for concrete stress distribution for pure bending or bending moment with axial forces:

Contents

Symbols

Preface

1. Introduction

1

1.1 Strain Diagrams in the Ultimate Limit State

2

1.2 Stress-strain Diagram for Concrete

5

1.3 Stress-strain Diagram for Reinforcing Steel

6

1.4 Basic Values for the Design Resistance to Moments and Axial Forces 7

1.5 Verification at the Serviceability Limit States

8

1.5.1 Permissible Stresses

9

1.5.2 Service Load Stress - Straight - Line Theory

10

Example I. 1: A reinforced concrete beam of rectangular section is known. Compute the stresses in concrete and steel reinforcement for a given bending moment.

11

Example 1.2: The cross-section of a beam with compression reinforcement. The beam is subjected to a specified load moment. Compute the specified load stress in concrete and steel.

14

Example I. 3: The cross-section of a beam is subjected to bending moment and external normal load. Compute the stress in concrete and steel.

17

Example I. 4: Compute the steel area to the cross-section and check the stresses of concrete and steel for a given bending moment and axial load.

19

Example I. 5: Pure bending with tension and compression reinforcement in a rectangular cross-section is known. Verify the stresses in concrete and in steel reinforcement.

21

Example I. 6: Pure bending with tension reinforcement is known. Verify the stresses in concrete and in steel reinforcement.

24

Example I. 7: A rectangular member subjected to a bending moment and external normal force is known. The entire section is in compression, and

25

2 symmetrically reinforced check the stresses in concrete and in steel reinforcement. Example I. 8: A rectangular column. is known. Check the stresses in concrete and reinforcement caused by a bending moment M s and external normal tension load N s .

27

Example I. 9: A rectangular column is known. Verify the stresses in concrete and in steel, 30 the cross-section caused by a bending moment M s and normal compression force N s .

Example I. 10: A rectangular member. is known. Verify the stresses in concrete and in steel, the cross-section caused by a bending moment M s and normal compression force N s .

33

Example I. 11: A rectangular member is known. The entire section is in compression, and unsymmetrically reinforced, check the stresses in concrete and in steel reinforcement.

37

Example I. 12: Determine the entire area of steel reinforcement and check the stresses of concrete and steel for a known. bending moment and external tension axial load.

40

Example I. 13: A T-beam is subjected to a specified load bending moment M s . Compute the steel and concrete stresses.

42

Example I. 14: Verify the stresses in concrete and in steel of a T-beam caused by bending moment M s and external compression force N s .

46

Example I. 15: Verify the position of the neutral axis of a T-beam section caused by the bending moment M s .

49

Example I. 16: Design the flexural reinforcement for the I-beam section caused by the bending moment M s .

51

Example I. 17: Verify the stresses in concrete and in steel of a T-beam section caused by the bending moment M s and external compression force N s .

53

Example I. 18: Verify the stresses in concrete and in steel of a T-beam section caused by bending moment M s and external compression force N s .

54

Example I. 19: Verify the stresses of concrete and steel of a T-column caused by the bending moment M s and tension force N s .

57

Example I. 20: Verify the stresses in concrete and in steel of a T-beam section caused by bending moment M s and external compression force N s .

59

Design Table I. for verification of stresses in pure bending or in bending with axial force

62

2

3

Example II. 1: The cross-section dimensions of a beam are subjected to bending moment M sd. Determine the required tension reinforcement to the section.

68

Example II. 2: Determine the required tension reinforcement of a reinforced concrete rectangular beam subjected to bending moment M sd.

69

Use the bi-linear diagram for steel (  ´ 0) and the bi-linear diagram for concrete.

Example II. 3: Determine the required tension reinforcement of a reinforced concrete rectangular column subjected to bending moment M sd and tension axial force N sd.

71

Example II. 4: Reinforced concrete rectangular column subjected to bending moment M sd and axial compression force N sd is given. Determine the required tension

74

reinforcement for the cross-section. Use the bi-linear diagram for steel (  ´ 0) and the bi-linear diagram for concrete.

Example II. 5: Determine of the tension reinforcement of a reinforced concrete rectangular column subjected to bending moment M sd and tension force N sd.

76

Use the bi-linear diagram for steel (  ´ 0) and the bi-linear diagram for concrete.

Example II. 6: Determine the tension reinforcement of a reinforced concrete rectangular column subjected to bending moment M sd and compression

78

force N sd. Use the bi-linear diagram for steel (  ´ 0) and the bi-linear diagram for concrete.

Example II. 7: Determine of the tension reinforcement of reinforced concrete rectangular column subjected to bending moment M sd and compression force

80

N sd by the bi-linear diagram for steel (  ´ 0) and the bi-linear diagram for concrete.

Example II. 8: Determine of the tension reinforcement of a reinforced concrete rectangular column subjected to bending moment M sd and compression force N sd. Use the bi-linear diagram for steel (  ´ 0) and the bi-linear diagram for concrete.

82

Example II. 9 : Design the flextural reinforcement for the T beam caused by bending moment M s and axial tension load N s .

84

Design Table II. For rectangular cross-sections without compression reinforcement based on the bi-linear diagram for steel and the bi-linear diagram for concrete stress distribution for pure bending or bending moment with axial forces.

87

Design Table III. For rectangular cross-sections without compression reinforcement for pure bending or bending moment with axial forces, based on the bi-linear diagram for steel (  ´ 0) and the bi-linear diagram for concrete.

130

3

4

Design Table IV. For rectangular cross-sections without compression reinforcement based on the parabolic-rectangular diagram for steel and the bi-linear diagram for concrete stress distribution for pure bending or bending moment with axial forces.

170

Design Table V. For rectangular cross-sections without compression reinforcement based on the parabolic-rectangular diagram for concrete and the bi-linear diagram for steel stress distribution for pure bending or bending moment with axial forces.

213

Example VI. 1 - VI. 3: Determine the bearing capacity of a rectangular column.

232

Design Table VI. 1 - VI. 6 For bearing capacity of a reinforced concrete rectangular column in centric compression force and compression force with small eccentricity.

238

References

262

4

References [1] ACI: Cracking of concrete members in direct tension. ACI Journal, Vol. 83, January - February, 1986 [2] Aide - mémoire: Composants en béton précontraint. Bordas, Paris, 1979 [3] Beeby, A. W.: The Prediction of Crack Widths in Hardened Concrete, Cement and Concrete Association, London, 1979 [4] Bjarne, Ch. J.: Lines of Discontinuity for Displacements in the Theory of Plasticity of Plain and Reinforced Concrete, Magazine of Concrete Research, Vol. 27, No. 92, September, 1975 [5] Boulet, B.: Aide - mémoire du second oeuvre du batiment. Bordas, Paris, 1977 [6] Brooks, J. J., Neville, A. M.: A comparison of creep, elasticity, and strength of concrete in tension and in compression. Magazine of Concrete Research, Vol. 29, 1977 [7] CEB - Bull. 124/125 - F: Code modéle CEB - FIP pour les structures en béton. CEB, Paris, 1980 [8] CEB - Bull. 156 - F: Fissuration et déformations. École Polytechnique Fédérale de Lausanne, 1983. [9] CEB - FIP Model Code 1990, Comité Euro - International du Béton, 1991 [10] CEB - Bull. 159: Simplified methods of calculating short term deflections of reinforced concrete slabs. Paris - Lausanne, 1983 [11] CC. BA 68: Régles Techniques de conception et de calcul des ouvrages et constructions en béton armé. D.T.U. Paris, 1975 [12] Consenza, E., Greco, C.: Comparison and Optimization of Different Methods of Evaluation of Displacements in Cracked Reinforced Concrete Beams. Materials and Structures, No. 23, 1990 [13] Coates, R. C., Coutie, M. G., Kong, F. K.: Structural analysis, Second Edition, Hong Kong, 1980 [14] Davidovici, V.: Béton armé, aide - mémoire. Bordas, Paris, 1974

[15] Eibl, J.: Concrete Structures. Euro - Design Handbook. Karlsruhe, 1994 - 96 [16] Edward, G., Nawy, P.E.: Prestressed Concrete A fundamental Approach. Part 1, New Jersey, 1989 [17] Elvery, R., Shafi, M.: Analysis of shrinkage effect on reinforced concrete structural members. ACI Journal, Vol. 67, 1970

[18] Ghali, A., Faver, R.: Concrete Structures: Stresses and Deformations. Great Britain, 1986

[19] Goto, Y.: Cracks Formed in Concrete Around Deformed Tension Bars, Journal of the ACI, No. 68, April, 1971

[20] Gregor, J. G.: Reinforced Concrete, New Jersey, 1988 [21] Gvozdev, A. A.: Novoje v projektirovanii betonnych i železobetonnych konstrukcij. Moskva, 1978 [22] Goulet, J.: Résistance des matériaux. aide - mémoire, Bordas, Paris, 1976 [23] Gupta, A. K.: Unified Approach to Modelling Postcracking Membrane Behaviour of Reinferced Concrete, Journal of Structural Engineering, Vol. 115, No. 4, April, 1989 [24] Gupta, A. K.: Postcracking Behaviour of Membrane Reinforced Concrete Elements Including Tension-Stiffening. Journal of Structural Engineering, Vol.115, No. 4, April, 1989 [25] Hsu, T. T. C.: Torsion of reinforced concrete. Van Nostrand Reinhold, New York, 1984 [26] Ismail, M. A., Jirsa, J. O.: Bond deterioration in reinforced concrete subject to low cycle loads. ACI Journal, Vol. 69, June, 1972 [27] Klink, S. A.: Actual Elastic Modulus of Concrete. ACI Journal, September - October, 1985 [28] Leonhardt, F.: Reducting Shear Reinforcement in Reinforced Concrete Beams and Slabs, Magazine of Concrete Research, Vol. 17, No. 53, December, 1965 [29] Leonhardt, F.: Recommendations for the Degree of Prestressing - Prestressed Concrete Structures. FIP Notes 69, July - August, 1977 [30] Leonhardt, F.: Crack Control in Concrete Structures. ACI Journal, July - August, 1988 [31] Leonhardt, F.: Vorlesungen uber Massivbau. Vol. 4, 1978 [32] Lenkei, P.: Deformation capacity in reinforced concrete slabs. In: IABSE Colloquium - Plasticity in reinforced concrete, Copenhagen, 1979 [33] Leong, T. W., Warner, R. F.: Creep and shrinkage in reinforced concrete beams. Journal of the Structural Division, Vol. 96, March, 1970

[34] Placas, A., Regan, P.E.: Limit - state design for shear in rectangular and "T" beams. Magazine of Concrete Research, December, 1970 [35] Placas, A., Regan, P.E.: Shear Failure of Reinforced Concrete Beams. Journal ACI, October, 1971 [36] Rehm, G., Eligehausen, R., Mallee, R.: Limitation of Shear Crack Width in Reinforced Concrete Construction. University of Stuttgart, Heft 6, 1983

[37] Rehm, G.: Berechnung der Breite von Schubrissen in Stahlbetonbauteilen. Ausfsatz fur CEB. Stuttgart, 1977 [38] Sargin, M: Stress - strain relationships for concrete and the analysis of structural concrete sections. University of Waterloo, Study No. 4, 1971

[39] Saliger, R.: Der Stahlbetonbau. 8. Auflage. Franz Deuticke, Wien, 1956 [40] Schlaich, J., Scheef, H.: Concrete box - girder bridges. Structural Engineering Documents 1e. Stuttgart, January, 1982 [41] Shawkat, S.: Deformation of reinforced concrete beams. Proceedings of the RILEM International Conference Concrete Bridges, Ĺ trbskĂŠ pleso, 22. - 24. september 1997 [42] Vagner, V. W., Erlhof, G.: Praktische Baustatik. Teil 1, 2, 3, Stuttgart, 1977 [43] Vecchio, F. J.: Reinforced Concrete Membrane Element Formulations. Journal of Structural Engineering, Vol. 116, No. 3, March, 1990

# Design of Reinforced Concrete Members

Author: Sabah Shawkat

# Design of Reinforced Concrete Members

Author: Sabah Shawkat