1 The University of Hong Kong Department of Civil Engineering Theory and Design of Structures I Reinforced Concrete Design Scope • Rectangular singly and doubly reinforced beams • Elastic design • Limit state design concepts; material strength and loading • Flexural strength and shear strength of beams; one-way slabs References 1. BS8110: 1985, Structural use of concrete – Part 3: Design charts for singly reinforced beams, doubly reinforced beams and rectangular columns, British Standard Institution, London, 1985. 2. BS8110: 1997, Structural use of concrete – Part 1: Code of practice for design and construction, British Standard Institution, London, 1997. 3. Code of practice for structural use of concrete 2004, second edition, Buildings Department, Hong Kong, 2008. 4. Design of structural elements: concrete, steelwork, masonry and timber design to British standards and Eurocodes, 2nd ed., C. Arya, Spon Press, London, 2003. 5. Reinforced concrete design, 5th ed., W.H. Mosley, J.H. Bungey and R. Hulse, Macmillan Press, Basingstoke, 1999. 6. Reinforced concrete designer’s handbook, 10th ed., C.E. Reynolds and J.C. Steedman, E. & F.N. Spon, London, 1988. 7. Reinforced concrete design to BS8110: simply explained, A.H. Allen, E. & F.N. Spon, London, 1988. 8. Structural design in concrete to BS8110, L.H. Martin, P.C.L. Croxton and J.A. Purkiss, Edward Arnold, London, 1989. Introduction A plain concrete beam cannot support much loading because of the low tensile strength. The introduction of steel tension reinforcement can effectively strengthen a concrete beam. In a reinforced concrete (RC) beam, the concrete carries compression while the steel reinforcement mainly carries tension. Figure 1(a) Plain concrete beam under loading. f c f c ′ b d Z = b d 2 / 6 Figure 1(b) Reinforced concrete beam under loading. Section (cracked) C T a M = C⋅a = T⋅a
Transcript
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The University of Hong Kong Department of Civil Engineering
Theory and Design of Structures I Reinforced Concrete Design
Scope • Rectangular singly and doubly reinforced beams • Elastic design • Limit state design concepts; material strength and loading • Flexural strength and shear strength of beams; one-way slabs References 1. BS8110: 1985, Structural use of concrete – Part 3: Design charts for singly reinforced
beams, doubly reinforced beams and rectangular columns, British Standard Institution, London, 1985.
2. BS8110: 1997, Structural use of concrete – Part 1: Code of practice for design and construction, British Standard Institution, London, 1997.
3. Code of practice for structural use of concrete 2004, second edition, Buildings Department, Hong Kong, 2008.
4. Design of structural elements: concrete, steelwork, masonry and timber design to British standards and Eurocodes, 2nd ed., C. Arya, Spon Press, London, 2003.
5. Reinforced concrete design, 5th ed., W.H. Mosley, J.H. Bungey and R. Hulse, Macmillan Press, Basingstoke, 1999.
6. Reinforced concrete designer’s handbook, 10th ed., C.E. Reynolds and J.C. Steedman, E. & F.N. Spon, London, 1988.
7. Reinforced concrete design to BS8110: simply explained, A.H. Allen, E. & F.N. Spon, London, 1988.
8. Structural design in concrete to BS8110, L.H. Martin, P.C.L. Croxton and J.A. Purkiss, Edward Arnold, London, 1989.
Introduction A plain concrete beam cannot support much loading because of the low tensile strength. The introduction of steel tension reinforcement can effectively strengthen a concrete beam. In a reinforced concrete (RC) beam, the concrete carries compression while the steel reinforcement mainly carries tension.
Figure 1(a) Plain concrete beam under loading.
fc
fc′
b
d
Z = b d 2 / 6
Figure 1(b) Reinforced concrete beam under loading.
Section (cracked)
C
T
a
M = C⋅a = T⋅a
2
Properties of steel The stress is proportional to the strain up to the yield point. At the yield point, steel becomes plastic and the stress remains practically constant while the strain increases. Finally owing to work hardening, there is another increase in stress but not in proportion to strain. Properties of concrete It has no clearly defined yield point. Nor is stress ever proportional to strain exactly. However, that portion of the stress-strain curve below 1/3 of the ultimate strength is very nearly a straight line and is assumed to be so. Beyond 1/3 of the ultimate strength, the concrete is in the elasto-plastic state, and the stress is no longer proportional to strain. Methods of design The following methods of design are available: 1. Based on elastic theory (elastic theory method in CP114 and previous Hong Kong codes) 2. Based on ultimate load (load factor method in CP114 and previous Hong Kong codes) 3. Based on limit states design philosophy (BS8110, CP110 and present Hong Kong
concrete code) Elastic method When the elastic method is adopted, the structural members are reinforced so that at working load, the maximum stress in the concrete is a certain fraction of the cube strength and the maximum stress in the steel is a certain fraction of the yield stress. The ratio of the yield stress (or cube strength) to the permissible stress in steel (or concrete respectively) is called the stress factor of safety. When a structure is designed on an elastic basis with a stress factor of 2, it does not mean that the structure will carry twice the working load before it fails. The reason is that at stresses slightly greater than the permissible values, the concrete structure no longer behaves in a linear manner. Therefore, the stress factor of safety is no real guide to the true safety margin of the structure. Ultimate load design method When the ultimate load design method is adopted, the structure is designed so that the working load is some fraction of the ultimate load. The load that the structure can carry is calculated. The ratio of the ultimate load to the working load is called the load factor of safety, and hence ultimate load design method is often called load factor design method. In the previous Hong Kong concrete code and CP114, the load factor is 1.8.
stre
ss
strainConcrete
stre
ss
strain Steel
Y.P.
Figure 2. Stress-strain curves for concrete and steel.
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Elastic Design of an RC Section Notations:
=b breadth =h total depth =d effective depth =x neutral axis depth =sA area of tension reinforcement =cbp permissible compressive stress in concrete due to bending =stp permissible tensile stress in steel
== cse EEα modular ratio Taking moment about N.A. (uniform stress creates a resultant through N.A.)
( )xdAbx se −= α2
21
021 2 =−+ dAxAbx sese αα
( ) ( )( )b
dAbAAx sesese ααα 2)4(2 ++−
=
bdA
bA
bA sesese ααα 22
+
+−=
( )23
31 xdAbxI se −+= α
At top fibre, At steel level,
xIZc = ( ) e
s xdIZ
α−=
IMy
=σ
cbc
cb pI
MxZMf ≤== ( )
ste
sst p
IxdM
ZMf ≤
−==
α
Design resistance moment is the smaller of
xpIZpM cb
ccbc ==
( )xdpIZpM
e
stssts −
==α
Case 1: stst pf = , cbcb pf < Under-reinforced Case 2: stst pf < , cbcb pf = Over-reinforced Case 3: stst pf = , cbcb pf = Balanced
d
b
h
As
b
x
αe As
N.A.
Figure 3. Assumptions in elastic design.
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Example: Elastic design of reinforced concrete slab A 225mm thick reinforced concrete slab spans an effective distance of 7.5m between two brick walls. The slab is designed to support an imposed load comprising a uniformly distributed load and a transverse line load at mid-span. The main reinforcement is T20/150 and the secondary reinforcement is T12/300. The cover provided to the main reinforcement is 15mm. The screeding and ceiling finish together weigh 0.5kN/m2 and the density of reinforced concrete is assumed to be 24kN/m3. (Question 5, May 2000 Examination) (a) Using the elastic method, calculate the maximum safe bending moment per unit width
that can be carried by the slab if permissible tensile stress in steel = 250N/mm2; permissible bending stress in concrete = 13.3N/mm2; and modular ratio = 15.
(b) If the uniformly distributed imposed load is 5kN/m2, determine the maximum safe transverse line load that can be carried at mid-span.
(c) Calculate the maximum stresses in concrete and steel if the slab is subjected to a uniformly distributed imposed load of 5kN/m2 and a transverse line load of 6kN/m at mid-span. Draw a sketch showing how the stresses are distributed along the depth of the mid-span section.
Solution: (a) Maximum safe bending moment per unit width
Consider 1m width of slab. The dimensions are span = 7.5m, b = 1000mm, h = 225mm and cover = 15mm. The reinforcement is
Main reinft.: T20/150 Secondary reinft.: T12/300
Effective depth mm2002/2015225 =−−=d
Sectional area of reinft. ( ) /mmm2094150100020
422 =
=
πsA
Allowable stresses 2N/mm250=stp 2N/mm3.13=cbp Modular ratio 15=eα Let the neutral axis depth be x. Taking moment about the neutral axis,
( )( ) ( )( )( )xx −= 2002094151000 221
0628200031410500 2 =−+ xx mm0.85=x
The second moment of area is ( )( )( ) ( )( )( )23
31 0.852002094150.851000 −+=I
46 mm101.620 ×=
5
The section moduli are 36
6
mm10295.70.85101.620
×=×
==xIZc
( ) ( )( )36
6
mm103595.0150.85200
101.620×=
−×
=−
=e
s xdIZ
α
The maximum safe moment is the smaller of ( )( ) kNm02.97Nmm1002.973.1310295.7 66 =×=×=cM ( )( ) kNm88.89Nmm1088.89250103595.0 66 =×=×=sM
Hence the maximum safe moment is 89.88kNm and the section is under-reinforced. (b) Maximum safe transverse line load at mid-span
Self-weight ( )( )( ) kN/m4.5225.00.124 == Screeding and ceiling finish kN/m5.0= Total DL kN/m9.55.04.5 =+=
2kN/m5UDLL = Let kN/mKELL P= Moment caused by DL and LL
( )( )( ) ( ) ( )5.75.759.5 412
81 P++=
( )kNm/m875.164.76 P+= 88.89875.164.76 =+ P
Maximum kN/m06.7=P
(c) If 2kN/m5UDLL = and kN/m6KELL == P
Bending moment ( )( ) kNm/m89.876875.164.76 =+=M The stresses are
26
6
N/mm05.1210295.71089.87
=××
==c
cb ZMf
26
6
N/mm5.244103595.01089.87
=××
==s
st ZMf
85
200
12.05N/mm2
244.5N/mm2
Figure 4. Stress distribution (not to scale).
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Limit State Design of RC Structures The general philosophy of limit state design applies. The particular requirements are highlighted below. Unless otherwise stated, the design code is Code of Practice for Structural Use of Concrete 2004 Second Edition (the Hong Kong Concrete Code). Partial safety factors for load γf The partial safety factors for load γf for ultimate limit state (ULS) are shown in Table 1. A partial safety factor of γf = 1.0 is usually applied to all load combinations at the serviceability limit state (SLS). Define the following symbols for the common types of loading.
The design loads should be combined such that they give the most severe condition to the structure or cross section being considered. Whether the larger or smaller value should be used depends on which gives the more critical condition. For Load Combination 1 (i.e. dead + imposed), the design load effects at ULS should normally be calculated based on (1.4 Gk + 1.6 Qk). For Load Combination 2 (i.e. dead + wind), the corresponding design load effects should be calculated based on 1.4 (Gk + Wk). For Load Combination 3 (i.e. dead + imposed + wind), the corresponding design load effects should be calculated based on 1.2 (Gk + Qk + Wk). However in the rough check of stability of a building of breadth B and height H against overturning, the criterion should be 1.0 Gk × ½ B > 1.4 Wk × ½ H At SLS, the design load effects for Load Combinations 1, 2 and 3 are (Gk + Qk), (Gk + Wk) and (Gk + Qk + Wk) respectively. Partial safety factors for strength of material γm The partial safety factors for strength of materials γm for ULS are shown in Table 2.
Table 1. Load combinations and values of γf for ultimate limit state (Table 2.1 in Hong Kong Concrete Code)
Table 2. Values of γm for ultimate limit state (Table 2.2 in Hong Kong Concrete Code)
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d h
b At working load (stress condition for elastic design)
Beyond working load
At failure (stress condition for ultimate load design)
Figure 5. Stress development in a reinforced concrete section.
Stress development in RC section Because of the non-linear stress-strain relationship of concrete, the stress distribution in an RC section varies with the applied bending moment.
Flexural Strength of RC Sections The following notations are used:
=b breadth =h total depth =d effective depth of tension reinforcement =′d depth to compression reinforcement
=x neutral axis depth =sA area of tension reinforcement =′sA area of compression reinforcement =cuf characteristic strength of concrete =stf stress in tension reinforcement =scf stress in compression reinforcement =yf characteristic strength of reinforcement =ccε maximum compressive strain in concrete =stε tensile strain in tension reinforcement=scε compressive strain in compression reinforcement
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The following assumptions are made for the ultimate strength design of reinforced concrete sections: 1. The strain distribution across the section is linear. 2. The tensile strength of concrete is ignored. 3. The compressive strain of concrete is the criterion for failure of the RC beam section.
The ULS of the section is reached when the concrete strain at the extreme compression fibre εcc reaches a specified ultimate value of εcu where 0035.0=cuε for 60≤cuf MPa
and 6000006.00035.0 −×−= cucu fε for 60>cuf MPa. 4. From the short-term design stress-strain curve for normal-weight concrete as shown in
Figure 6, the maximum concrete compressive stress at failure is taken to be (0.67fcu)/γm, which is equal to 0.45fcu noting that γm = 1.5 for concrete in flexure.
5. At failure of the RC beam section, the distribution of concrete compressive stress may be defined by the idealized stress-strain curve in Figure 6 (i.e. parabolic plus rectangular) or the simplified rectangular stress block as shown in Figure 7.
6. The stresses in the reinforcement are derived from the stress-strain curve as shown in
Figure 8. The maximum steel stress is taken to be fy/γm, which is equal to 0.87fy noting that γm = 1.15 for reinforcement. Assuming a Young’s modulus of Es = 200000 N/mm2, the yield strain of the design stress-strain curve is therefore 20000087.0 yf .
7. Where a section is designed to resist flexure only, the lever arm should not be greater
than 0.95d, where d = effective depth (Clause 6.1.2.4(a) of Hong Kong Concrete Code).
fy / γm
fy / γm
Tension
Compression
Strain
Stre
ss
200 kN/mm2
Figure 8. Short-term design stress-strain curve for reinforcement (Fig. 3.9 of Hong Kong Concrete Code)
Strain StressNeutral Axis
Figure 7. Simplified stress block for concrete at ultimate limit state (Fig. 6.1 of Code of Practice for Structural Use of Concrete 2004 Second Edition)
εcu0.67 fcu / γm
x
0035.0=cuε for 60≤cuf MPa
6000006.00035.0 −−= cucu fε for 60>cuf MPa
s
s = 0.9x for ≤cuf 45 MPa s = 0.8x for 45 < ≤cuf 70 MPa s = 0.72x for 70 < ≤cuf 100 MPa
cmc u Ef γ34.1
Parabolic curve
Strain εcu
Stre
ss
mcuf γ67.0
Figure 6. Short-term design stress-strain curve for normal-weight concrete (Fig. 3.8 of Code of Practice for Structural Use of Concrete 2004 Second Edition)
+= 3.2146.3
m
cuc
fE
γ kN/mm2
for 20MPa ≤≤ cuf 100MPa
0035.0=cuε
for 60≤cuf MPa
6000006.00035.0 −−= c ucu fε
for 60>c uf MPa
Εc
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Figure 9 shows the cross-section of an RC member subjected to bending and the resultant strain diagram together with three different types of stress distribution in concrete. Classification of RC sections An RC section may be: (1) Under-reinforced; (2) Critical / balanced; or (3) Over-reinforced. (1) Under-reinforced section If the tension reinforcement is smaller than a certain amount known as the balanced steel content, the section is regarded as an under-reinforced section. Upon loading, the tension reinforcement will reach the yield strength before the concrete reaches its compression capacity. The steel strain must not be less than that corresponding to the yield stress when the section fails. After yielding has commenced, the stress in the steel remains constant while the strain increases. It is assumed that at failure, the steel stress is the design yield stress. The large plastic elongation of the tension reinforcement causes the concrete in the tension zone to crack. The serious cracking and excessive deflection serve as warning signals before the imminent failure of the member. This kind of tension failure is ductile and it allows redistribution of forces to other parts of the structure if it is statically indeterminate. (2) Balanced section A balanced or critical section has exactly the balanced steel content. At the failure of a balanced section, the concrete reaches the maximum strain εcu at the same time when the steel reaches the yield strain.
Figure 9. Section with strain diagrams and stress blocks.
(b) Rectangular
parabolic
Stress Blocks
(a) Triangular
Neutral axis
x
(c) Equivalent rectangular
s
Strains
ccε
scε
stε
d
d ' A'
s
As
Section
b
εcc < εcu
εst > yield strain
Fst
Fcc
Figure 10. Under-reinforced section.
εcc = εcu
εst = yield strain
Fst
Fcc
Figure 11. Balanced section.
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(3) Over-reinforced section If the tension reinforcement is more than a certain amount known as the balanced steel content, the section is regarded as an over-reinforced section. Upon loading, the concrete may reach its compression capacity before the tension reinforcement reaches its tensile strength. In an over-reinforced section, the concrete reaches the maximum strain εcu while the steel strain is still below the yield strain. Members with over-reinforced sections will fail suddenly in a brittle manner if the concrete is not properly confined. There may be little visible warning prior to failure. The characteristics of failure are the crushing of concrete in the compression zone, small deflection, and absence of cracking in the tension zone. The failure is called compression failure. To prevent a brittle failure without warning, some design codes (such as BS8110) specify that the design neutral axis depth x cannot exceed half of the effective depth d, no matter how much reinforcement is provided. The Hong Kong Concrete Code introduces the following restrictions: dx 5.0≤ for 45≤cuf N/mm2;
dx 4.0≤ for 7045 ≤< cuf N/mm2; or
dx 33.0≤ for 10070 ≤< cuf N/mm2 and no moment redistribution. Ultimate moment of resistance of a singly reinforced rectangular section For the design of most RC structures, it is usual to design for the ULS first, followed by checks to ensure that the structure is adequate for the SLS without excessive deflection or cracking. In most cases, the simplified rectangular stress block is used for design at the ULS. The following demonstrates how simple reinforced concrete sections can be analyzed. For simplicity, it is assumed that the concrete grade does not exceed 45 and therefore one has
0035.0=cuε and dx 5.0≤ .
Figure 13. Singly reinforced section with rectangular stress distribution (fcu ≤ 45).
Fcc
0.67fcu /γm ≅ 0.45 fcu
Stress distribution
Fst
s/2
z
s = 0
.9x
d
b
Section
As
x
εst
Strain distribution
Neutral axis
εcu
εcc = εcu
εst < yield strain
Fst
Fcc
Figure 12. Over-reinforced section.
11
(1) Under-reinforced section The bending of a singly reinforced section will induce a resultant tensile force Fst in the steel reinforcement and a resultant compressive force in the concrete Fcc that acts through the centroid of the effective area of concrete in compression. Assume a neutral axis depth x measured from the top extreme compression fibre, the depth of equivalent stress block is 0.9x. The resultant compressive force in concrete Fcc is given by
( ) ( ) xbfbxfxbfF cucu
m
cucc 402.09.0
5.167.09.067.0
===γ
Assuming that the steel reinforcement has yielded, the resultant tensile force Fst in the steel reinforcement is
sysy
sm
ysstst AfA
fA
fAfF 87.0
15.1=
=
==
γ
Equating the compression Fcc to tension Fst, we have sycu Afbxf 87.0402.0 =
bf
Afx
cu
sy164.2= (1)
It is necessary to check if the steel reinforcement has actually yielded. To ensure that the steel reinforcement has actually yielded, check if syst Ef87.0>ε which may be worked out as
sycu
cu
Efdx
87.0+<
εε
It shows that the maximum value of the neutral axis depth ratio x/d is dependent on fy, subject to additional requirement of the Hong Kong Concrete Code (Clause 6.1.2.4(b)) limiting the ratio x/d to a maximum of 0.33 to 0.5 depending upon the concrete cube strength fcu. The lever arm z can be worked out as xdxdz 45.02)9.0( −=−= (2) Applying Eq. (2) and assuming a concrete cube strength fcu not exceeding 40N/mm2 (thus
5.0/ ≤dx ) gives dddxdz 775.02/45.045.0 =−≥−= , i.e. dz 775.0≥ The Hong Kong Concrete Code (Clause 6.1.2.4(a)) also specifies that, where a section is designed to resist only flexure, the lever arm z should not be assumed to be greater than 0.95 times the effective depth d. Therefore the ratio z/d lies within the following limits. 95.0775.0 ≤≤ dz
x
εcc = εcu
εst > 0.87 fy / Es
d
Figure 14. Strain distribution of an under-reinforced section at failure.
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The ultimate moment M of the section can be worked out in terms of the concrete stress as zFM cc= zxbfcu402.0=
zzdbfcu
−
=45.0
402.0
( )zzdbfM cu −= 9.0 (3) The ultimate moment M of the section can also be worked out in terms of the steel stress as zFM st= zAf sy87.0= On substituting the lever arm z from Eq. (2) and the neutral axis depth x from Eq. (1), the above equation appears as ( )xdAfM sy 45.087.0 −=
×−=
bfAf
dAfcu
sysy 164.245.087.0
−=
bfAf
dAfMcu
sysy 974.087.0 (4)
Dividing Eq. (4) throughout by 2bd gives
−
=
dbA
ff
dbAf
dbM s
cu
ysy 974.0187.02
−=
cu
yy f
ff
dbM ρρ 974.0187.02 (5)
where the tension reinforcement ratio is defined as bdAs=ρ . Eq. (5) gives the relationship between M and ρ for the construction of part of the design charts. The limiting moment of resistance for a singly reinforced concrete section can be worked out by setting x/d = 0.5 or z/d = 0.775 and invoking Eq. (3) 2156.0 bdfM cu= (6) Considering both steel yielding and concrete crushing, the ultimate moment M based on steel yielding is given by
−=
bfAf
dAfMcu
sysy 974.087.0
while the moment M based on concrete crushing is given by 2156.0 bdfM cu= The actual ultimate moment is the smaller of the above two values. If the applied moment exceeds 2156.0 bdfcu , compression reinforcement is required. In the design of an RC section, it is necessary to establish whether it can be designed as a singly-reinforced section by checking the K value that is defined as 2dbfMK cu= . If K does not exceed 0.156, it can be designed as a singly reinforced section. It is normally
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designed as an under-reinforced section. If K exceeds 0.156, compression reinforcement is necessary. Alternatively, the section may be enlarged or the material strengths may be increased. (2) Balanced section At the failure of a balanced section, the concrete reaches the maximum strain ( 0035.0=cuε for 45≤cuf N/mm2) at the same time when the steel reaches the design yield strain of sy Ef87.0 . From the strain distribution shown in Figure 15, one can write
sy Efxd
x87.0
0035.0=
−
where Es is the Young’s modulus of steel, which is around 200,000 MPa. The neutral axis depth ratio x/d can therefore be worked out as
0035.087.0
0035.0+
=sy Efd
x
Equating the compression Fcc to tension Fst, we have sycu Afbxf 87.0402.0 = The tension reinforcement ratio for a balanced section defined as bdAsb =ρ can be worked out as
+
=
===
0035.087.00035.0462.0462.0
87.0402.0
syy
cu
y
cu
y
cusb Eff
fdx
ff
dfxf
dbAρ (7a)
The ultimate moment M of the section can be calculated from Eq. (5) by setting ρ to be ρb. Alternatively, in connection with the use of the simplified stress block, a simpler definition of a balanced section is one having a neutral axis depth ratio x/d of 0.5 while the steel reaches the design yield strain of sy Ef87.0 . The balanced tension reinforcement ratio can be calculated as
=
===
y
cu
y
cu
y
cusb f
fdx
ff
dfxf
dbA 231.0462.0
87.0402.0ρ (7b)
An RC section with tension reinforcement ratio ρ below ρb is under-reinforced, while that with tension reinforcement ratio ρ above ρb is over-reinforced. x
εcc = 0.0035
εst = design yield strain in steel = 0.87 fy / Es
d
Figure 15. Strain distribution of a balanced section at failure.
x
εcc = 0.0035
εst < 0.87 fy / Es
d
Figure 16. Strain distribution of an over-reinforced section at failure.
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(3) Over-reinforced section An over-reinforced section fails by crushing of concrete while the tension reinforcement remains elastic, i.e. ys ff 87.0< . The steel stress may be determined by considering the strain distribution diagram in Figure 16.
sxd
xε0035.0
=−
−
=x
xds 0035.0ε
−
==x
xdEEf sssst 0035.0ε
Equating the compression Fcc to tension Fst, we have
−
=x
xdAEbxf sscu 0035.0402.0
0)(0035.0402.0 2 =−+ dxAEbxf sscu (8) After solving for the neutral axis depth x from Eq. (8), the lever arm z can be worked out from Eq. (2). The ultimate moment M of the section can then be calculated from Eq. (3). Ultimate moment of resistance of a doubly reinforced rectangular section Compression reinforcement is required when 156.02 >= dbfMK cu . Depending on the reinforcement areas, their positions and strength, the tension and compression steel of a doubly reinforced section may or may not reach the design yield stress when the ultimate moment of the section is reached. However for convenience in design, it is normally assumed at first that all the reinforcement has yielded. This is subsequently modified in case some or all of the reinforcement does not reach the design yield stress. For simplicity, the area of concrete in compression has not been deducted to allow for the concrete displaced by the compression reinforcement. It is also assumed that the concrete grade does not exceed 45 and therefore one has 0035.0=cuε and dx 5.0≤ .
Figure 17. Doubly reinforced section with rectangular stress distribution (fcu ≤ 45).
d
b
Section
As
A's d'
x =
d/2
εst
Strain distribution
Neutral axis
εcu = 0.0035
εsc Fcc
0.67fcu /γm=0.45 fcu
Stress distribution
Fst
s = 0
.9x
z
Fsc
15
Assuming first that all reinforcement has yielded, the compressive force carried by concrete Fcc, the resultant compressive force Fsc in the compression reinforcement and the resultant tensile force Fst in the tension reinforcement are given as
( ) ( ) xbfbxfxbfF cucu
m
cucc 402.09.0
5.167.09.067.0
===γ
sysy
sm
ysscsc AfA
fA
fAfF ′=′
=′
=′= 87.0
15.1γ
sysy
sm
ysstst AfA
fA
fAfF 87.0
15.1=
=
==
γ
For equilibrium of the section in Figure 17, scccst FFF += so that with the reinforcement at yield sycusy AfxbfAf ′+= 87.0402.087.0
( )
bfAAf
xcu
ssy
402.087.0 ′−
= (9)
Once the neutral axis depth x is obtained, the strain distribution of the section can be defined. The stresses stf and scf of the tension and compression reinforcement respectively should be the design yield stress provided that the respective strains stε and scε satisfy the following conditions:
yst ff 87.0= if s
yst E
fx
xd 87.00035.0 ≥
−
=ε (10)
ysc ff 87.0= if s
ysc E
fx
dx 87.00035.0 ≥
′−
=ε (11)
If the above conditions are satisfied, the assumption that all the reinforcement has reached the design yield stress is correct. The ultimate moment M of the section can be obtained by taking moment about the centroid of the tension reinforcement as )()45.0( ddFxdFM sccc ′−+−= )(87.0)45.0(402.0 ddAfxdbxfM sycu ′−′+−= (12) When the above checks reveal that not all the reinforcement has yielded, the value of neutral axis depth x calculated from Eq. (9) is incorrect. The actual reinforcement stresses and neutral axis depth x have to be recalculated from the equilibrium equation and the strain diagram as follows:
bfAfAfx
cu
sscsst
402.0′−
= (13)
−
==x
xdEEf sstsst 0035.0ε (14)
′−
==x
dxEEf sscssc 0035.0ε (15)
16
Once the neutral axis depth x is obtained, the ultimate moment M of the section can be obtained by taking moment about the centroid of the tension reinforcement as )()45.0( ddFxdFM sccc ′−+−= )()45.0(402.0 ddAfxdbxfM sscu ′−′′+−= (16) In a doubly reinforced section, tension or compression failure may occur. In tension failures, the tension reinforcement yields, while in compression failures, the tension reinforcement remains elastic. In both types of failure, the compression reinforcement may or may not yield when the section fails. In the normal design process, the coefficient K is checked to see whether compression reinforcement is required. If K exceeds 0.156, it means that the concrete section alone is insufficient to resist the compression. Compression reinforcement is provided such that the neutral axis depth x remains at the maximum permitted value of 0.5d. Summary of procedures for flexural design of reinforced concrete beams ( 45≤cuf N/mm2) 1. Classification of section
Let the required ultimate moment be M. Work out K where 2dbfMK cu= • If 156.0≤K , no compression reinforcement is needed and the section may be
designed as a singly-reinforced section. • If 156.0>K , compression reinforcement is needed and the section must be designed
as a doubly-reinforced section. 2. Singly-reinforced section zFM cc= zxbfcu402.0=
( )[ ]9.0/25.05.0 Kdz −+= Having solved for z, the steel area As may be determined from zAfzFM syst 87.0==
zf
MAy
s 87.0=
17
3. Doubly-reinforced section )()45.0( ddFxdFM sccc ′−+−= )()45.0(402.0 ddAfxdbxf ssccu ′−′+−= Assume that 2/dx = 2156.0)45.0(402.0 bdfxdbxf cucu =− Assume that the compression steel has yielded, i.e. ysc ff 87.0= Therefore )(87.0156.0 2 ddAfbdfM sycu ′−′+=
)(87.0
156.0 2
ddfbdfMA
y
cus ′−
−=′
Check whether the compression steel has yielded or not. If
s
ysc E
fd
ddx
dx 87.02/
2/0035.00035.0 ≥
′−
=
′−
=ε ,
then the compression steel has yielded, i.e. ysc ff 87.0= . To compute the required tension steel, note the equilibrium condition scccst FFF += sycusy AfxbfAf ′+= 87.0402.087.0
y
sycus f
AfdbfA
87.087.0201.0 ′+
=
Shear Strength The shear force is numerically equal to the rate of change of bending moment. Hence the vast majority of structural members have to resist shear forces. Shear transfer in RC beams relies heavily on the tensile and compressive strength of concrete. A shear failure is in general non-ductile. To suppress this type of failure, the shear strength of a member must be somewhat in excess of that associated with the maximum flexural strength it can possibly develop. Types of shear failure The mechanism of shear failure in RC structures is complicated. Generally, there are five types of failure related to shear and flexure for RC beams. The behaviour of shear failure depends on the ratio of shear span av to the effective depth d (i.e. the shear-span / depth ratio av/d) where av is the distance between the point load and the nearest support.
18
Type I: av/d > 6 The bending moment is large compared to the shear force. The failure is similar to that expected in pure bending. Normally the tension reinforcement is close to yield or has already yielded. Only minimum shear reinforcement is required. Type II: 2 < av/d < 6 The initial flexural cracks become inclined early in the loading process. At failure, horizontal cracks will appear running along the tension reinforcement. These horizontal cracks reduce the shear resistance of the section by destroying the dowel force and also reduce the bond stress between the steel and concrete. Normally the tension reinforcement does not reach yield. Type III: av/d < 2 This is a typical form of shear failure. Flexural cracks do not develop but shear cracks at roughly 45° suddenly appear leading to collapse. Normally the tension reinforcement does not reach yield. Type IV: av/d ≅ 0 This arrangement normally leads to punching shear failure. The shear resistance of the section is at a maximum. The addition of shear reinforcement in the form of vertical links does not increase the shear resistance above the punching shear value. Type V: Shear reinforcement provided The shear reinforcement increases the shear resistance against failure types I, II and III. Diagonal cracks normally develop. At failure, the shear reinforcement and the longitudinal reinforcement yield, provided that the reinforcement is anchored well and the member is not over-reinforced.
Figure 18. Types of shear failure in reinforced concrete beams.
(b) Type II 6 > av/d > 2
V
d
Dowel force crack leading to bond failure
Compression zone fails
av
(a) Type I av/d > 6 V
d
Shear crack
Bending crack
Beam fails when compression zone crushes
av
(c) Type III av/d < 2
V
d Shear crack suddenly appears followed by failure in compression zone
av
(d) Type IV av/d = 0 Punching shear
V
(e) Type V With shear reinforcement
V Vertical links
Bent-up bars
Shear cracks
19
Components of shear resistance The shear resistance V of an RC section may be broken down into several components as follows (Figure 19(c)): • Concrete compression zone: shear carried by the uncracked concrete compression zone
Vcz (20-40%) • Aggregate interlock: vertical component of aggregate interlock across the diagonal crack
Va (35-50%) • Dowel action: shear force carried by the dowel action of longitudinal tension
reinforcement Vd (35-50%) • Shear reinforcement: shear force carried by shear reinforcement or web reinforcement
(stirrups / links, or bent-up bars) Vs (maximum 50%) The above can be expressed as ( ) scsdacz VVVVVVV +=+++= (19) where Vc represents the shear carried by the concrete ( )dacz VVV ++ . The interaction among these shear components is complicated. The design recommendations given in the common design codes (e.g. Hong Kong Concrete Code) are therefore a lower-bound fit to experimental results. Shear reinforcement The provision of shear reinforcement (Figures 19(a) and 19(b)) increases the ductility of the beam and considerably reduces the likelihood of a sudden and catastrophic failure, which often occurs in beams without shear reinforcement. Links or stirrups are the most common type of shear reinforcement. They are sometimes used in combination with bent-up bars. Only the use of links is discussed in this course.
Figure 19(a). Links or stirrups.
CL
Figure 19(b). Combined system with links and bent-up bars.
CL
Figure 19(c). Shear transfer in beam with web reinforcement.
Vcz
Vd
Vs
Concrete compression
Steel tension
Va
20
The stresses in the shear reinforcement are often analysed by the truss analogy. Links contribute to the shear strength by the following means: • Improving the contribution of the dowel action: A link can effectively support a
longitudinal bar that is being crossed by a flexural shear crack close to a link. • Suppressing flexural tensile stresses in the concrete by means of the diagonal
compressive force resulting from truss action. • Limiting the opening of diagonal cracks within the elastic range, thus enhancing and
preserving the shear transfer by aggregate interlock. • Providing confinement, when the links are sufficiently closely spaced, and thus
increasing the strength of the concrete core. • Preventing the breakdown of bond when splitting cracks develop in anchorage zones
because of dowel and anchorage forces. Derivation of shear stress in links The following additional notations are used:
=svA total cross sectional area of links at the neutral axis at a section =vb breadth of section =yvf characteristic strength of links (not to be taken as more than 460 N/mm2)
=vs spacing of links along the member =v design (average) shear stress at a cross section ( dbVv v/= ) =cv design concrete shear stress
To derive the design equations using the truss analysis, consider the equilibrium of the free body to the left of a section line parallel to the concrete ‘struts’. The shear resistance Vs is contributed by the tension svyvmsvyv AfAf 87.0=γ in the links that are crossed by the section line, where Asv is the area of both legs of each link and fyv is the characteristic strength of the links. ⋅= svyvs AfV 87.0 (Number of links crossed by section line)
( )
′−=
vsvyv s
ddAf βcot87.0
βcot87.0
′−=
vsvyv s
ddAf
( ) βcot87.0 vsvyvs sdAfV ≅
Figure 20. Truss analogy.
CLA
A
β
sv
bv d'
d
21
where β is the angle between the ‘compression strut’ and the axis of the beam. Experiments have led to the recommendation that β could be taken as 45°, giving
=
vsvyvs s
dAfV 87.0 (20)
From Eq. (19), cs VVV −= , and introducing dvbV v= and dbvV vcc = , the above equation becomes
( )v
svyvvc s
Afbvv
87.0=−
( )yv
cv
v
sv
fvvb
sA
87.0−
= (21)
Note that dbVv v/= is the average shear stress at a cross section and vc is the ultimate shear stress that can be resisted by the concrete. The design concrete shear stress vc can be obtained from Table 6.3 of Hong Kong Concrete Code and it is derived from
=
mv
sc ddb
Akkvγ140010079.0
4/13/1
21 (22)
where k1 = an enhancement factor equal to 2d/av for shear-span / depth ratio (av/d) less than 2
(Clause 6.1.2.5(g)); otherwise k1 = 1 k2 = ( ) 3/125cuf for concrete with cube strength fcu > 25MPa, but fcu should not be taken
as greater than 80 N/mm2
dbA
v
s100 = tensile steel percentage which should not be taken as greater than 3
d400 = a ratio that should not be taken as less than 1. The code should be consulted for
details. mγ = partial safety factor for strength of materials taken as 1.25
To avoid web crushing or compression zone failure, the code specifies that the average shear stress v should not exceed cuf8.0 or 7.0 N/mm2, whichever is the smaller. The provision of shear reinforcement should be in accordance with the following (Table 6.2 of Hong Kong Concrete Code). Note that the design shear resistance vr provided by minimum links is taken as 4.0=rv N/mm2 for 40≤cuf N/mm2 and ( ) 32404.0 cur fv = N/mm2 for
40>cuf N/mm2. (a) cvv 5.0< throughout the beam Provide minimum links in beams of structural importance. Minimum links may be omitted in members of minor structural importance. (b) ( )rcc vvvv +<<5.0 Provide minimum links equivalent to additional shear strength of vr, i.e.
yv
vvrsv f
sbvA87.0
≥
22
(c) ( ) curc fvvv 8.0<<+ or 7.0 N/mm2 Provide links according to
( )yv
cvvsv f
vvsbA87.0
−≥
Procedure for designing of beams for shear 1. Calculate the design shear stress at the section from dbVv v= . 2. Check whether the design shear stress v is within the upper limit (Clause 6.1.2.5(a)), i.e.
cufv 8.0< or 7.0 N/mm2. 3. Calculate the percentage of longitudinal tension reinforcement and determine the design
concrete shear stress vc from Table 6.3 of Hong Kong Concrete Code or the associated equation.
4. Determine the form and area of shear reinforcement, if any, from Table 6.2 of Hong Kong Concrete Code.
5. Choose suitable link size and spacing sv, ensuring that dsv 75.0≤ (Clause 6.1.2.5(d)). 6. At least 50% of the shear resistance provided by the steel should be in the form of links. Span-effective depth ratios Hong Kong Concrete Code specifies a set of basic span / effective depth ratios to control deflections. For spans less than 10m, the basic span / effective depth ratios are given in Table 3. These ratios are normally more critical for slabs.
Table 3. Basic span / effective depth ratios for rectangular sections with spans less than 10m Cantilever 7 Simply supported 20 Continuous 26
The above basic ratios are modified to account for the amounts of tension and compression reinforcement and the stresses there. The modification factor for tension reinforcement can be calculated by
Modification factor for tension reinforcement = ( )( ) 0.2
9.012047755.0 2 ≤
+−
+bdM
fs (23)
where M is the design ultimate moment at the centre of the span or at the support of a cantilever, and fs is the service stress of tension reinforcement there. They are also tabulated in Table 7.4 of Hong Kong Concrete Code. The design service stress in the tension reinforcement in a member may be estimated from
bprovs
reqsys A
Aff
β1
32
×=
where reqsA is the area of compression reinforcement required, provsA is the area of compression reinforcement provided and βb is the ratio
tionredistribubeforesectionatmoment
tionredistribuaftersectionatmoment=bβ
that is used with moment redistribution.
23
Similarly the modification factor for compression reinforcement can be calculated by Modification factor for compression reinforcement
= 5.1100
3100
1 ≤
′+
′+
bdA
bdA provsprovs (24)
where provsA′ is the area of compression reinforcement provided. Design of slabs Concrete slabs behave primarily as flexural members, and therefore the design is similar to that for beams. In addition, the shear stresses are usually low in a slab except when there are heavy concentrated loads. Compression reinforcement is seldom required except at supports. The common methods of analysis of slabs include (a) Elastic analysis (b) Yield line theory by Johansen (c) Strip method by Hillerborg In this course, only approximate elastic analysis will be covered. Depending on the aspect ratio (longer side / shorter side), a slab is classified as one-way or two-way slab. If the aspect ratio is greater than 2, the slab can be treated as a one-way slab that is assumed to be spanning in the direction of the shorter side only. If the aspect ratio is between 1 to 2, the slab should be treated as a two-way slab spanning in both directions. Only one-way slabs will be discussed here. One-way slabs The slab is often assumed to be built up of parallel strips (beams) of 1m width (i.e. b=1000mm). The calculation for the required main reinforcement follows that used in beam design. The area of tension reinforcement is zfMA ys 87.0= . A suitable bar diameter and spacing can be worked out by sAbar) one of (area 1000 = spacing maximum × . The main reinforcement is in the direction of the span. The main reinforcement should form the outer layer of reinforcement to give the maximum lever arm. The secondary reinforcement or distribution steel is required normal to the main direction. The minimum areas of reinforcement are as follows: 10013.0 bh for high yield steel (fy = 460 MPa) 10024.0 bh for mild steel (fy = 250 MPa) The calculation of shear stress v is similar to that of beams, i.e. cuv fdbVv 8.0≤= or 7.0 MPa It is difficult to provide shear reinforcement in a slab, and hence it is not advisable to use shear reinforcement in slabs (less than 200 mm thick). The shear stresses in slabs are seldom critical. Therefore for v < vc, no shear reinforcement is required.
24
Example 1: Design of a singly reinforced rectangular section A rectangular beam of breadth 260 mm and effective depth to tension reinforcement 440 mm is required to resist an ultimate bending moment 185 kNm. Determine the area of tension reinforcement As required given the characteristic material strengths are fy = 460 MPa and fcu = 30 MPa. Hence determine the shear reinforcement using mild steel with characteristic strength fyv = 250 MPa if the ultimate shear force is 150 kN. Design data: M = 185 kNm V = 150 kN b = 260 mm d = 440 mm
fcu = 30 MPa fy = 460 MPa fyv = 250 MPa
The K factor is
156.0123.030440260
101852
6
2 <=××
×==
cufbdMK
and therefore compression steel is not required. The lever arm z can be calculated as ( ) mm3689.025.05.0 =−+= Kdz The area of tension reinforcement required As is
26
mm125636846087.0
1018587.0
=××
×==
zfMA
ys
Provide 4T20, giving As prov = 1257mm2 O.K. The shear design is carried out as follows
MPa31.144026010150 3
=××
==db
Vvv
< MPa38.48.0 =cuf O.K.
10.1440260
1257100100=
××
=dbA
v
s
1909.0440400400
<==d
; take 1
( )8020.12530
25≤== cu
cu ff
25.1=mγ
( ) ( ) ( )
MPa31.1MPa69.025.1120.1110.179.01
2540010079.0 3/14/13/1
3/14/13/1
=<=
=
=
v
fddb
Avm
cu
v
sc γ
MPa09.14.069.0MPa31.1 =+=+>= rc vvv Let sv = 200 mm
( ) 2mm14825087.0
)69.031.1(20026087.0
=×
−××=
−≥
yv
cvvsv f
vvsbA
Provide R10 links at 200 mm spacing (Asv = 157 mm2) O.K.
25
Example 2: A singly reinforced rectangular section with different amounts of tension reinforcement A rectangular beam has a breadth 250 mm and an effective depth to tension reinforcement 460 mm. The concrete has a characteristic strength of 25 MPa. The steel reinforcement has a modulus of elasticity of 200,000 MPa and a yield strength of 320 MPa. Calculate the section moment capacity of the following tension steel areas: (a) balanced steel area; (b) 2500mm2; and (c) 5160mm2. Design data: b = 250 mm d = 460 mm fcu = 25 MPa fy = 320 MPa Es = 200000 MPa (a) Balanced section The neutral axis depth x can be worked out by considering the strain distribution as
( )sy Efxdx
87.00035.0
=−
mm3290035.087.0
0035.0=
+=
sy
s
EfdEx
The lever arm z is therefore mm31245.0 =−= xdz The balanced steel ratio ρb can be obtained from equilibrium consideration as sycu Afbxf 87.0402.0 = ( )( ) 0258.0462.0 == dxff ycubρ
2mm2969=sA The moment capacity M is kNm9.25787.0 == zAfM sy (b) Section with As = 2500 mm2 The tension reinforcement ratio ρ is 0258.00217.0 =<== bs bdA ρρ Therefore this section is under-reinforced and tension failure occurs. The neutral axis depth x can be worked out from equilibrium as sycu Afbxf 87.0402.0 = ( )( ) mm277164.2 == bAffx scuy The lever arm z is therefore mm33545.0 =−= xdz
26
The moment capacity M is kNm2.23387.0 == zAfM sy (c) Section with As = 5160 mm2 The tension reinforcement ratio ρ is 0258.00449.0 =>== bs bdA ρρ Therefore this section is over-reinforced and compression failure occurs. The neutral axis depth x can be worked out from equilibrium as
sscu AEx
xdbxf
−
= 0035.0402.0
( ) 00035.0402.0 2 =−+ dxAEbxf sscu 01066.136120005.2512 92 =×−+ xx mm367=x The lever arm z is mm29545.0 =−= xdz The moment capacity M is based on compression in concrete, namely kNm8.273)(9.0 2 =−= zdzbfM cu The steel strain is checked to make sure that it has indeed not yielded sscu Afbxf =402.0
( ) 001392.087.0000892.0402.0=<=== sy
ss
cu
s
ss Ef
AEbxf
Efε O.K.
The following graph shows the relationship between moment capacity and the amount of tension reinforcement.
Variation of moment capacity with amount of tension reinforcement
0
50
100
150
200
250
300
0.00 0.01 0.02 0.03 0.04 0.05Tension reinforcement ratio ρ
Mom
ent c
apac
ity (k
Nm
27
Example 3: A doubly reinforced rectangular section with different concrete grades A rectangular beam has a breadth of 280 mm and an effective depth to tension reinforcement of 510 mm. The depth to compression reinforcement is 50 mm. The areas of tension and compression reinforcement are 2580 mm2 and 650 mm2 respectively. The steel reinforcement has a modulus of elasticity of 200,000 MPa and a yield strength of 330 MPa. Calculate the moment capacity if the concrete has a characteristic strength of (a) 35MPa ( 0035.0=cuε ,
dx 5.0≤ and xs 9.0= ); and (b) 70MPa ( 0033.0=cuε , dx 4.0≤ and xs 8.0= ). Design data: b = 280 mm d = 510 mm (depth to tension reinft.) d' = 50 mm (depth to compression reinft.) As = 2580 mm2 A's = 650 mm2 fy = 330 MPa Es = 200000 MPa (a) Concrete cube strength fcu = 35 MPa Assuming that all reinforcement reaches design yield stresses, the forces acting on the section are N6.3939402.0 xbxfC cuc == N18661587.0 =′= sys AfC N74071887.0 == sy AfT Consider the equilibrium of forces TCC sc =+ 7407181866156.3939 =+x mm141=x Check steel strains:
001436.087.0002259.00035.0 =
=>=
′−
=′s
yys E
fx
dx εε
Therefore the compression steel has yielded.
001436.087.0009160.00035.0 =
=>=
−
=s
yys E
fx
xd εε
Therefore the tension steel has also yielded. The previous assumptions are valid. The moment capacity M is ( ) ( ) kNm9.33387.045.0402.0 =′−′+−= ddAfxdbxfM sycu
28
(b) Concrete cube strength fcu = 70 MPa Assuming that all reinforcement reaches design yield stresses, the forces acting on the section are N7.70033573.0)8.0)(5.1/67.0( xbxfbxfC cucuc === N18661587.0 =′= sys AfC N74071887.0 == sy AfT Consider the equilibrium of forces: TCC sc =+ 7407181866157.7003 =+x mm79=x Check steel strains:
001436.087.0001211.00033.0 =
=<=
′−
=′s
yys E
fx
dx εε
Therefore the compression steel has NOT yielded.
001436.087.0018004.00033.0 =
=>=
−
=s
yys E
fx
xd εε
Therefore the tension steel has yielded. The previous assumptions have to be amended. The steel compressive force Cs and neutral axis depth x are updated assuming that the compression steel remains elastic.
Therefore the tension steel has yielded. The revised assumptions are valid. The moment capacity M is ( ) ( ) kNm1.3514.03573.0 =′−′′+−= ddAExdbxfM ssscu ε
29
Example 4: Design of a reinforced rectangular slab Design an R.C. slab to span between two 230 mm brick walls 4.5 m apart c/c to carry a live load of 3.0 kN/m2. Assume the floor finishes and ceiling loads to weigh 1.0 kN/m2. The characteristic material strengths are fcu = 30 MPa, fy = 460 MPa (high-yield steel) and fy = 250 MPa (mild steel). Take the basic span-effective depth ratio as 20. Try the following two cases: (a) using high-yield steel; and (b) using mild steel. Design data: Span L = 4500 mm Wall thickness t = 230 mm Breadth b = 1000 mm Basic span / effective depth ratio = 20 fcu = 30 MPa fy = 460 MPa (high-yield steel) fy = 250 MPa (mild steel) Cover = 25 mm (for mild exposure) Density of R.C. = 24 kN/m3 Floor finishes & ceiling loads = 1.0 kN/m2 Live load = 3.0 kN/m2 (a) Using high yield steel Assume modification factor (m.f.) = 1.3 Minimum effective depth d = span / (20 × m.f.) = 173 mm Try effective depth d = 180 mm Assume bar diameter = 10 mm Overall depth of slab h = effective depth + bar diameter / 2 + cover = 210 mm Self-weight of slab = overall depth × density = 5.04 kN/m2 Total dead load Gk = 6.04 kN/m2 Live load Qk = 3.0 kN/m2 Ultimate load w = 1.4 Gk + 1.6 Qk = 13.26 kN/m2 For unit width of slab, M = wL2 / 8 = 33.55 kNm
Limiting span / effective depth ratio = 20 × 1.28 = 25.6 Actual span / effective depth ratio = 4500 / 180 = 25.0 < 25.6 O.K. The K factor is 156.0035.02 <== bdfMK cu Therefore compression steel is not required. The lever arm z is then determined: 960.09.025.05.0 =−+= Kdz > 0.95 Take 95.0=dz mm171=z The area of tension reinforcement As is
m/mm49087.0
2==zf
MAy
s
Provide T10 at 150mm centres As prov = 524 mm2/m O.K. The next step is to check for shear. At the face of support, the shear force V and shear stress v are worked out as follows. ( ) kN3.282 =−= tLwV MPa38.48.0157.0 =<== cuv fdbVv
291.01801000524100100
=×
×=
dbA
v
s
122.2180400400
≥==d
( )8020.12530
25≤== cu
cu ff
25.1=mγ
( ) ( ) ( )
MPa157.0MPa543.025.1120.122.2291.079.01
2540010079.0 3/14/13/1
3/14/13/1
=>=
=
=
v
fddb
Avm
cu
v
sc γ
Therefore no shear reinforcement is required.
31
The required amount of distribution steel is then determined. Area of transverse high yield steel reinforcement = 0.13 bh / 100 = 273 mm2/m Provide T10 at 250 mm centres, As prov = 314 mm2/m O.K. (b) Using mild steel Assume modification factor (m.f.) = 1.7 Minimum effective depth d = span / (20 × m.f.) = 132 mm Try overall depth of slab h = 180 mm Assume bar diameter = 12 mm Effective depth d = overall depth of slab – cover – bar diameter / 2 = 180 – 25 – 12 / 2 = 149 mm O.K. Self-weight of slab = overall depth × density = 4.32 kN/m2 Total dead load Gk = 5.32 kN/m2 Live load Qk = 3.0 kN/m2 Ultimate load w = 1.4 Gk + 1.6 Qk = 12.25 kN/m2 For unit width of slab, M = wL2 / 8 = 31.01 kNm Span / effective depth ratio: MPa40.12 =bdM Assume design service stress fs = 250 × 2/3 = 167 MPa Modification factor
( ) 0.267.19.0120
47755.0 2 <=+−
+=bdM
fs
Limiting span-effective depth ratio = 20 × 1.67 = 33.4 Actual span-effective depth ratio = 4500 / 149 = 30.2 < 33.4 O.K. The K factor is 156.0047.02 <== bdfMK cu Therefore compression steel is not required. The lever arm z is then determined: 945.09.025.05.0 =−+= Kdz mm141=z The area of tension reinforcement As is
m/mm101187.0
2==zf
MAy
s
Provide R12 at 100mm centres As prov = 1131 mm2/m O.K.
32
Re-check modification factor using the updated service stress in the tension reinforcement fs:
( )( )bprovsreqsys AAff β132
= where 1=bβ
MPa0.149=sf
( ) 0.274.19.0120
47755.0.. 2 <=+−
+=bdM
ffm s
Revised limiting span-effective depth ratio = 20 × 1.74 = 34.8 > 30.2 O.K. At the face of support, the shear force V and shear stress v are worked out as follows. ( ) kN2.262 =−= tLwV MPa38.48.0176.0 =<== cuv fdbVv
759.01491000
1131100100=
××
=dbA
v
s
168.2149400400
≥==d
( )8020.12530
25≤== cu
cu ff
25.1=mγ
( ) ( ) ( )
MPa176.0MPa784.025.1120.168.2759.079.01
2540010079.0 3/14/13/1
3/14/13/1
=>=
=
=
v
fddb
Avm
cu
v
sc γ
Therefore no shear reinforcement is required. The required amount of distribution steel is then determined. Area of transverse mild steel reinforcement = 0.24 bh / 100 = 432 mm2/m Provide R10 at 150 mm centres, As prov = 524 mm2/m O.K.
