RC2009 University of HongKong Reinforced concrete design

Theory and Design of Structures IReinforced 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

References1. BS8110: 1985, Structural use of concrete – Part 3: Design charts for

singly reinforced beams, doubly reinforced beams and rectangularcolumns, BSI, London, 1985.

2. BS8110: 1997, Structural use of concrete – Part 1: Code of practice for design and construction, BSI, 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• Steel reinforcement is introduced into a concrete

beam mainly to carry tension, thereby resulting in a reinforced concrete (RC) beam.

• Components: concrete and reinforcing bars (rebars)

Rectangular beam T-beam

Tensionreinft.

Singly reinforced beam

Tension reinft.

Doubly reinforced beam

Comp. reinft.

Stirrup carrier or link hanger

Stirrup or link

Introduction

Figure 1(a) Plain concrete beam under loading.

fc

fc′

b

d

Z = b d 2 / 6

*

*

Compressive strength

Compression

Tension

Modulus of rupture

Cracks & collapses!

Plain concrete beam (Unreinforced concrete beam)

Introduction

Figure 1(b) Reinforced concrete beam under loading.

T Section

(cracked)

C

a

M = C⋅a = T⋅a

Compression

Tension

Neutral axis

Load increases

N.A.

Concrete cracks

• Distinct yield point• Linear relationship within elastic range• Beyond the yield point, plastic deformation

followed by work hardening

Properties of steel

stre

ss

strain

Y.P.

Stress-strain curves for steel.

• No clearly defined yield point• Portion of the curve below 1/3 of the ultimate

strength is nearly linear• Beyond that, it becomes elasto-plastic

Properties of concrete

stre

ss

strainStress-strain curves for concrete

Methods of design1. Elastic theory (elastic method in CP114 and

previous HK codes)

2. Ultimate load / load factor method (load factor method in CP114 and previous HK codes)

3. Limit states design philosophy (BS8110, CP110 and present HK concrete code)

Elastic design of an RC section

Elastic method• 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 At working load

≤ fcu / (FOS)conc

≤ fy / (FOS)steel

Elastic methodAssumptions:• Plane sections remain plane after bending• The materials are linearly elastic • The tensile strength of concrete is ignored• For analysis, the “composite section”

(comprising concrete and steel) is considered as an equivalent section of concrete

Symbols:b = breadthh = total depthd = effective depthx = neutral axis depthAs = area of tension reinft.

pcb = permissible compressive stress in concrete due to bending

pst = permissible tensile stress in steelαe = Es / Ec = modular ratio

d

b

h

As

b

x

αe As

N.A.

Figure 3. Assumptions in elastic design.

Concrete in tension ignored

Equivalent area of steel reinft.

• The position of N.A. varies:

N.A.

Plain concrete section before cracking

Reinforced concrete section after cracking

N.A. N.A.

y

x y

x

Let the section be subjected to uniform stress σ = 1Total force

Taking moment about y-axis

Therefore

Similarly

∫∫=

dA

dAxx

AdAdAF === ∫∫σ

dAxdAxxF ∫∫ =⋅=⋅ σ

∫∫=

dA

dAyy

Determination of centroidRevision

d

b

h

As

b

x

αe As

N.A.


Taking moment about N.A. (uniform stress creates a resultant through N.A.)

021 2 =−+ dAxAbx sese αα

( ) ( )( )b

dAbAAx sesese ααα 2)4(2 ++−

=

bdA

bA

bA sesese ααα 22

+

+−=

( )xdAbx se −= α2

21

( )23

31 xdAbxI se −+= α

d

b

h

As

b

x

αe As

N.A.


At top fibre,At steel level,

xIZc =

cbc

cb pI

MxZMf ≤==

( ) es xd

IZα−

=

( )st

e

sst p

IxdM

ZMf ≤

−==

αI

My=σ

To convert stress from equivalent section

Design resistance moment is the smaller of

• Under-reinforced:

• Over-reinforced:

• Balanced:

xpIZpM cb

ccbc ==

( )xdpIZpM

e

stssts −

==α

pst pcb

Under-reinforced

Over-reinforced

Balanced

stst pf =

stst pf <

stst pf =

cbcb pf <

cbcb pf =

cbcb pf =

Simplified interpretation

ExampleElastic design of RC slab

ExampleA 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.

Unit width

Example(a) Using the elastic method, calculate the maximum safe

bending moment per unit width that can be carried by the slab ifpermissible tensile stress in steel = 250N/mm2;permissible bending stress in concrete = 13.3N/mm2; andmodular 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 widthConsider 1m width of slab. The dimensions arespan = 7.5m, b = 1000mm, h = 225mm, and cover = 15mm

The reinforcement isMain reinft.: T20/150Secondary reinft.: T12/300

Effective depth d = 225 – 15 – 20/2 = 200mm

d

Sectional area of reinft.

Allowable stresses

Modular ratio

( ) /mmm2094150100020

422 =

=

πsA

2N/mm250=stp 2N/mm3.13=cbp

15=eα

Solution

Average no. of bars in a 1m strip

Let the neutral axis depth be x.Taking moment about the neutral axis,

The second moment of area is

( )( ) ( )( )( )xx −= 2002094151000 221

0628200031410500 2 =−+ xx

mm0.85=x

( )( )( ) ( )( )( )2331 0.852002094150.851000 −+=I

46 mm101.620 ×=

Solution

The section moduli are

The maximum safe moment is the smaller of

Hence the maximum safe moment is 89.88kNm and the section is under-reinforced.

366

mm10295.70.85101.620

×=×

==xIZc

( ) ( )( )36

6

mm103595.0150.85200

101.620×=

−×

=−

=e

s xdIZ

α

( )( ) kNm02.97Nmm1002.973.1310295.7 66 =×=×=cM

( )( ) kNm88.89Nmm1088.89250103595.0 66 =×=×=sM

Solution(b) Maximum safe transverse line load at mid-spanSelf-weight Screeding and ceiling finishTotal DL of 1m stripUniformly distributed LLLet knife edged live load be

( )( )( ) kN/m4.5225.00.124 =kN/m5.0=kN/m9.55.04.5 =+=

Solution

2kN/m5UDLL =

kN/mKELL P=

7.5 m

P kN/m LL 5.9 kN/m2 DL 5 kN/m2 LL

Moment caused by DL and LL

Thereforegiving maximum

Solution

( )( )( ) ( ) ( )5.75.759.5 412

81 P++=

( )kNm/m875.164.76 P+=

88.89875.164.76 =+ PkN/m06.7=P

7.5 m

P kN/m LL 5.9 kN/m2 DL 5 kN/m2 LL

(c) If and Bending moment

The stresses are( )( ) kNm/m89.876875.164.76 =+=M

2kN/m5UDLL = kN/m6KELL == P

( )kNm/m875.164.76 PM +=

26

6

N/mm05.1210295.71089.87

=××

==c

cb ZMf

26

6

N/mm5.244103595.01089.87

=××

==s

st ZMf

Solution

85

200

12.05N/mm2

244.5N/mm2

Figure 4. Stress distribution (not to scale).

Elastic method• 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. Supposing

How about this? Will it fail? Usually not! Why not?

P

2P

fy / 2

fcu / 2

Elastic methodDrawbacks of elastic method

P 2P

Not twice corresponding stresses at P

2P

< fcu

< fy

P

fy / 2

fcu / 2Note: Not to scale

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.

Ultimate load design method• 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.

Load factor of safety = Pult / P

P

Pult Failure!

Limit State Design of RC Structures

Ultimate limit state (ULS)• collapseServiceability limit states (SLS)• Excessive deflection• Cracking• Vibration• Etc.

Limit state design of RC structures

• γf = 1.0 is usually applied to all load combinations at the serviceability limit state (SLS)

• For ultimate limit state (ULS), γf are as shown (HK and UK codes)

Partial safety factors for load γf

Limit state design of RC structuresChoice of partial safety factors for load γf

1.0 Gk × ½ B > 1.4 Wk × ½ H


1.0 Gk

1.4 Wk

To be checked at ULS

GkWk

Nominal Loads

H/2

B/2 B/2

H/2

• γm for ULS are as shown (HK and UK codes)Partial safety factors for strength of material γm


• Because of the non-linear stress-strain relationship of concrete, the stress distribution in an RC section varies with the applied bending moment.

Stress development in RC section

d h

b At working load (stress condition for elastic design)

Beyond working load

At failure (stress condition for ultimate load design)

Stress development in a reinforced concrete section.




σ

ε

ε σ

fcu

Small curvature and strains

45°



σ

ε

ε σ

fcu

Large curvature and strains

45°

Assumptions: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 is reached when the concrete strain at the extreme compression fibre εcc reaches a specified ultimate value of εcu where

for

for

0035.0=cuε MPa60≤cuf6000006.00035.0 −×−= cucu fε

MPa60>cuf


0

0.001

0.002

0.003

0.004

0 20 40 60 80 100Concrete grade

Ulti

mat

e co

ncre

te s

train

4. The maximum concrete compressive stress at failure is taken to be (0.67fcu)/γm, which is equal to 0.45fcunoting that γm = 1.5 for concrete in flexure.


The factor of 0.67 is to allow for difference between bending strength and cube strength.

Ideal case Reality

Strength over-estimated

cmc u Ef γ34.1

Parabol ic 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

To be enlarged …


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 2 0MPa ≤≤ cuf 100MPa

0035.0=cuε

for 60≤cuf MPa

6000006.00035.0 −−= c ucu fε

for 60>c uf MPa

Εc

5. At failure of the RC beam section, the distribution of concrete compressive stress may be defined by the idealized stress-strain curve or the simplified rectangular stress block.


s = 0.9x for ≤cuf 45 MPa s = 0.8x for 45 < ≤cuf 70 MPa s = 0.72x for 70 < ≤cuf 100 MPa

Strain Stress Neutral Axis

Figure 7. Simplified stress block for concrete at ULS

εcu0.67 fcu / γm

x

s 0035.0=cuε for 60≤cuf MPa

6000006.00035.0 −−= cucu fε for 60>cuf MPa

Simplified Idealized

To be enlarged …


s = 0.9x for ≤cuf 45 MPa s = 0.8x for 45 < ≤cuf 70 MPas = 0.72x for 70 < ≤cuf 100 MPa

Strain Stress Neutral Axis

Figure 7. Simplified stress block for concrete at ULS

εcu 0.67 fcu / γm

x

s 0035.0=cuε for 60≤cuf MPa

6000006.00035.0 −−= cucu fε for 60>cuf MPa

Simplified Idealized

6. 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 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)

• BS8110: 1985 & HK Code: γm = 1.15; fy/γm = 0.87fy

• BS8110: 1997γm = 1.05; fy/γm = 0.95fy

7. Where a section is designed to resist flexure only, the lever arm should not be greater than 0.95d, where d = effective depth.


Section with strain diagrams and stress blocks

Strains

ccε

scε

stε

d

d ' A'

s

As

Section

b

(b) Rectangular

parabolic

Stress Blocks

(a) Triangular

Neutral axis

x s

(c) Equivalent rectangular

Classification of RC sections

Classification of RC sectionsAn RC section may be• Under-reinforced• Critical / balanced• Over-reinforced

Ben

ding

mom

ent M

Curvature φ

Over-reinforced

Under-reinforced

Under-reinforced section• The tension reinforcement is smaller than a certain

amount known as the balanced steel content.• Upon loading, tension reinforcement will reach the

yield strength first.• It is assumed that at failure, the steel stress is the design

yield stress.• This tension failure is ductile.• Sequence:

(a) steel yielding and concrete cracking; and(b) then concrete crushing

εcc < εcu

εst > yield strain

Fst

Fcc

Under-reinforced section

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.

εcc = εcu

εst = yield strain

Fst

Fcc

Balanced section.

εcc = εcu

εst < yield strain

Fst

Fcc

Over-reinforced section.

Over-reinforced section• The tension reinforcement is more than a certain

amount known as the balanced steel content.• Upon loading, concrete reaches its compression

capacity first.• Concrete reaches the maximum strain εcu while the steel

strain is still below the yield strain.

Over-reinforced section• Over-reinforced sections will fail suddenly in a brittle

manner, i.e. compression failure.• To prevent a brittle failure without warning, most

design codes impose certain restrictions.• For example, in the Hong Kong Concrete Code

for ;for ; orfor and no moment

redistribution

dx 5.0≤dx 4.0≤dx 33.0≤

2N/mm45≤cuf2N/mm7045 ≤< cuf

2N/mm10070 ≤< cuf

Ultimate moment of resistance of a singly reinforced rectangular section

Design of RC sections• Consider ULS first, followed by SLS.• Simplified rectangular stress block is used.• For simplicity in demonstration, it is assumed that the

concrete grade does not exceed 45, and therefore

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

0035.0=cuε dx 5.0≤

To be enlarged …


An under-reinforced section at failure (fcu ≤ 45).

Fcc

0.67fcu /γm ≅ 0.45 fcu

Stress distribution

Fst

s/2

z

s = 0

.9x

Strain distribution

xεcc = εcu

εst > 0.87 fy / Es

d

Under-reinforced section• The resultant compressive force in concrete Fcc is

given by

( ) ( ) xbfbxfxbfF cucu

m

cucc 402.09.0

5.167.09.067.0

===γ

An under-reinforced section at failure (fcu ≤ 45).

Fcc

0.67fcu /γm ≅ 0.45 fcu

Stress distribution

Fst

s/2

z

s = 0

.9x

Strain distribution

xεcc = εcu

εst > 0.87 fy / Es

d

Under-reinforced section• Assuming that the steel reinforcement has yielded,

the resultant tensile force Fst

• Equating the compression Fcc to tension Fst

sysy

sm

ysstst AfA

fA

fAfF 87.0

15.1=

=

==

γ

sycu Afbxf 87.0402.0 =

bfAf

xcu

sy164.2= (1)

• It should have yielded. Check ifor

• 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 limiting the ratio x/d to a maximum of 0.33 to 0.5 depending upon the concrete cube strength fcu.


x

εcc = εcu

εst > 0.87 fy / Es

d

Strain distribution of an under-reinforced section at failure

syst Ef87.0>εsycu

cu

Efdx

87.0+<

εε

• The lever arm z can be worked out as

• Applying Eq. (2) and assuming a concrete cube strength fcu not exceeding 40N/mm2 (x/d ≤ 0.5)

• When 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, thus


xdxdz 45.02)9.0( −=−= (2)

dddxdz 775.02/45.045.0 =−≥−=

dz 775.0≥

Fcc

0.67fcu /γm ≅ 0.45 fcu

Stress distribution (fcu ≤ 45)

Fst

s/2

z

s = 0

.9x

95.0775.0 ≤≤ dz

• The ultimate moment M of the section can be worked out in terms of the concrete stress as

• The ultimate moment M of the section can also be worked out in terms of the steel stress as


zFM cc= zxbfcu402.0=

zzdbfcu

−

=45.0

402.0

( )zzdbfM cu −= 9.0 (3)

zFM st= zAf sy87.0=

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x

• 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)

bfAf

xcu

sy164.2= (1)

xdxdz 45.02)9.0( −=−= (2)

• Dividing Eq. (4) throughout by bd2 gives

where the tension reinforcement ratio is ρ = As/bd


−

=

dbA

ff

dbAf

dbM s

cu

ysy 974.0187.02

−=

cu

yy f

ff

dbM ρρ 974.0187.02 (5)

−=

bfAf

dAfMcu

sysy 974.087.0 (4)

• 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)


( )zzdbfM cu −= 9.0 (3)

2156.0 bdfM cu= (6)

• Considering both steel yielding and concrete crushing, the ultimate moment M based on steel yielding is

while the moment M based on concrete crushing is

• The actual ultimate moment is the smaller of the above two values. If the applied moment exceeds 0.156fcubd2, compression reinforcement is required.

−=

bfAf

dAfMcu

sysy 974.087.0

2156.0 bdfM cu=

Balanced section

A balanced section at failure (fcu ≤ 45).

Fcc

0.67fcu /γm ≅ 0.45 fcu

Stress distribution

Fst

s/2

z

s = 0

.9x

Strain distribution

x

εcc = εcu = 0.0035

εst = 0.87 fy / Es

d

• At the failure of a balanced section, the concrete reaches the maximum strain at the same time when the steel reaches the design yield strain of 0.87fy/Es, and thus

Balanced section

sy Efxdx

87.00035.0

=− 0035.087.0

0035.0+

=sy Efd

x

A balanced section at failure (fcu ≤ 45).

Fcc

0.67fcu /γm ≅ 0.45 fcu

Stress distribution

Fst

s/2

z

s = 0

.9x

Strain distribution

x

εcc = εcu = 0.0035

εst = 0.87 fy / Es

d • Equating the compression Fcc to tension Fst

• The tension reinforcement ratio for a balanced section

• The ultimate moment M of the section can be calculated from Eq. (5) by setting ρ to be ρb

Balanced section


+

=

===

0035.087.00035.0462.0

462.087.0

402.0

syy

cu

y

cu

y

cusb

Efff

dx

ff

dfxf

dbAρ

(7a)

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x

−=

cu

yy f

ff

dbM ρρ 974.0187.02 (5)

• Alternatively, a simpler (but approximate) 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 0.87fy/Es, and thus

Balanced section

x εcc

εst = design yield strain in steel = 0.87 fy / Es

d

Strain distribution of a balanced section at failure (fcu ≤ 45)

x/d = 0.5

=

===

y

cu

y

cu

y

cusb f

fdx

ff

dfxf

dbA 231.0462.0

87.0402.0ρ

(7b)

Over-reinforced section

An over-reinforced section at failure (fcu ≤ 45).

Fcc

0.67fcu /γm ≅ 0.45 fcu

Stress distribution

Fst

s/2

z

s = 0

.9x

Strain distribution

xεcc = εcu

εst < 0.87 fy / Es

d

• An over-reinforced section fails by crushing of concrete while the tension reinforcement remains elastic, i.e. fs < 0.87fy

• The steel stress can be determined as follows:


x εcc = 0.0035

εst < 0.87 fy / Es

d

Strain distribution of an over-reinforced section at failure

sxdx

ε0035.0

=−

−

=x

xds 0035.0ε

−

==x

xdEEf sssst 0035.0ε

• Equating the compression Fcc to tension Fst,

• 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).


−

=x

xdAEbxf sscu 0035.0402.0

0)(0035.0402.0 2 =−+ dxAEbxf sscu (8)

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x xdxdz 45.02)9.0( −=−= (2)

( )zzdbfM cu −= 9.0 (3)

Ultimate moment of resistance of a doubly reinforced rectangular section

Doubly reinforced rectangular section• Compression reinforcement is required when

K = M / fcubd2 > 0.156• Other alternatives: increase in size, increase in concrete

grade, or their combinations• Assume first that all reinforcement has yielded. Later

modify 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 for demonstration purpose that the concrete grade does not exceed 45.

Doubly reinforced rectangular section

Doubly reinforced section with rectangular stress distribution (fcu ≤ 45)

d

b

Section

As

A's d'

Des

irabl

e 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

• Assume first that all reinforcement has yielded


d

b

Section

As

A'sd'

Des

irabl

e x

= d/

2

εst

Strain distribution

Neutral axis

εcu = 0.0035

εsc Fcc


Stress distribution

Fst

s = 0

.9x

z

Fsc

( ) ( ) 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

so that with the reinforcement at yieldscccst FFF +=

sycusy AfxbfAf ′+= 87.0402.087.0

( )bfAAf

xcu

ssy

402.087.0 ′−

= (9)

Fcc



Fst

s = 0

.9x

z

Fsc • The stresses fst and fsc of the tension and compression reinforcement respectively should be the design yield stress provided that the respective strains εst and εscsatisfy the following conditions:


d

b

Section

As

A'sd'

x

εst

Strain distribution

Neutral axis

εcu = 0.0035

εsc Fcc


Stress distribution

Fst

s = 0

.9x

z

Fsc

yst ff 87.0=

ysc ff 87.0=

s

yst E

fx

xd 87.00035.0 ≥

−

=ε

s

ysc E

fx

dx 87.00035.0 ≥

′−

=ε

if

if

(10)

(11)

• If the previous conditions are satisfied, the assumption that all the reinforcement has reached the design yield stress is correct. Thus the ultimate moment M is


d

b

Section

As

A's d'

x

εst

Strain distribution

Neutral axis

εcu = 0.0035

εsc Fcc


Stress distribution

Fst

s = 0

.9x

z

Fsc

)()45.0( ddFxdFM sccc ′−+−=

)(87.0)45.0(402.0 ddAfxdbxfM sycu ′−′+−= (12)

• When the previous checks reveal that not all the reinft. has yielded, the value of NA depth x calculated from Eq. (9) is incorrect and has to be recalculated


d

b

Section

As

A'sd'

x

εst

Strain distribution

Neutral axis

εcu = 0.0035

εsc Fcc


Stress distribution

Fst

s = 0

.9x

z

Fsc

bfAfAfx

cu

sscsst

402.0′−

=

−

==x

xdEEf sstsst 0035.0ε

′−

==x

dxEEf sscssc 0035.0ε

(13) (14)

(15)Apply as appropriate

• The ultimate moment M of the section is


d

b

Section

As

A's d'

x

εst

Strain distribution

Neutral axis

εcu = 0.0035

εsc Fcc


Stress distribution

Fst

s = 0

.9x

z

Fsc


)()45.0(402.0 ddAfxdbxfM sscu ′−′′+−= (16)

Summary of procedures for flexural design of RC beams (fcu ≤ 45MPa)

Flexural design of RC beams1. Classification of section

Let the required ultimate moment be M.Work out K where K = M / fcubd2

• If K ≤ 0.156, no compression reinforcement is needed and the section may be designed as a singly-reinforced section.

• If K > 0.156, compression reinforcement is needed and the section must be designed as a doubly-reinforced section. Alternatively the section size may be increased, or the concrete grade may be increased.

Flexural design of RC beams2. Singly-reinforced section

zFM cc=

zxbfcu402.0=

zzdbfcu

−

=45.0

402.0

( )zzdbfcu −= 9.0

( )( )dzdzdbfM cu //19.02 −=

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x

Flexural design of RC beams

Substituting K = M / fcubd2

Having solved for z, the steel area As is determined from

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x

( )( )dzdzdbfM cu //19.02 −=

09.0/)/()/( 2 =+− Kdzdz

( )9.0/25.05.0/ Kdz −+=

( )[ ]9.0/25.05.0 Kdz −+=

zAfzFM syst 87.0==zf

MAy

s 87.0=

Flexural design of RC beams3. Doubly-reinforced section

Assume that x = d / 2

Assume that the compression steel has yielded

Therefore


)()45.0(402.0 ddAfxdbxf ssccu ′−′+−=

2156.0)45.0(402.0 bdfxdbxf cucu =−

ysc ff 87.0=

Fcc



Fst

s = 0

.9x

z

Fsc

)(87.0156.0 2 ddAfbdfM sycu ′−′+=

)(87.0156.0 2

ddfbdfMA

y

cus ′−

−=′

Flexural design of RC beamsCheck whether the compression steel has yielded or not. If

then the compression steel has yielded, i.e.To compute the required tension steel, note the equilibrium condition

Fcc



Fst

s = 0

.9x

z

Fsc

s

ysc E

fd

ddx

dx 87.02/

2/0035.00035.0 ≥

′−

=

′−

=ε

ysc ff 87.0=

scccst FFF +=

sycusy AfxbfAf ′+= 87.0402.087.0

y

sycus f

AfdbfA

87.087.0201.0 ′+

=

Shear Strength

Shear Strength• The shear force is numerically equal to the rate of

change of bending moment, i.e. dM/dx = V.• 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• Generally, there are five types

of failure related to shear and flexure for RC beams.

• The behaviour of shear failure depends on the shear-span / depth ratio (av/d) where av is the distance between the point load and the nearest support.

P P

av avL

Bending moment

P.av

Shear force

P

-P

Diagonal compression

Diagonal tension

v

vv

v

v

vv

v

An element at neutral axis

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

Types of shear failure

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 reinft. is close to yield or has

already yielded.• Only minimum shear reinforcement is required.

(a) Type I av/d > 6 V

d

Shear crack

Bending crack

Beam fails when compression zone crushes

av

• 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.

(b) Type II 6 > av/d > 2

V

d

Dowel force crack leading to bond failure

Compression zone fails

av

Type II: 2 < av/d < 6• The initial flexural cracks

become inclined early in the loading process.

Type III: av/d < 2• 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.

(c) Type III av/d < 2

V

d

Shear crack suddenly appears followed by failure in compression zone

av

Type IV: av/d ≅ 0• A 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.

(d) Type IV av/d = 0 Punching shear

V

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.

(e) Type V With shear reinforcement

V Vertical links

Bent-up bars

Shear cracks

Components of shear resistance

• Concrete compression zoneshear carried by the uncracked concrete compression zone Vcz (20-40%)

• Aggregate interlockvertical component of aggregate interlock across the diagonal crack Va (35-50%)

( ) scsdacz VVVVVVV +=+++=

Shear transfer in beam with web reinforcement.

Vcz

Vd

Vs

Concrete compression

Steel tension

Va Aggregate interlock

(19)

Components of shear resistance

• Dowel actionshear force carried by the dowel action of longitudinal tension reinft. Vd (35-50%)

• Shear reinforcementshear force carried by shear reinft. or web reinft. (stirrups / links, or bent-up bars) Vs (max 50%)

( ) scsdacz VVVVVVV +=+++=

Shear transfer in beam with web reinforcement.

Vcz

Vd

Vs

Concrete compression

Steel tension

Va

Dowel bar

Shear reinforcement

Links or stirrups.

CL

Combined system with links and bent-up bars.

CL

Bent-up bar

• The provision of shear reinforcement increases the ductility of the beam and considerably reduces the likelihood of a sudden and catastrophic failure.

Shear reinforcementContribution of links to shear strength:• 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

A flexural shear crack

CL

Truss analogy

CL

Shear reinforcementContribution of links to shear strength:• 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

C L

Shear reinforcementContribution of links to shear strength:• Preventing the breakdown of bond when splitting

cracks develop in anchorage zones because of dowel and anchorage forces

Splitting

Derivation of shear stress in links

Truss analogy

CLA

A

β

sv

bv d'

d

total cross sectional area of links at the N.A. breadth of sectioncharacteristic strength of links (not to be taken as more than 460 N/mm2)spacing of links along the memberdesign (average) shear stress at a cross section design concrete shear stress

svA

vb

yvf

vsv

cv

Derivation of shear stress in links

Truss analogy

CLA

A

β

sv

bv d'

d

• The shear resistance Vs is contributed by the tension in the links that are crossed by the section line

⋅= svyvs AfV 87.0

( )

′−=

vsvyv s

ddAf βcot87.0 βcot87.0

′−=

vsvyv s

ddAf

( ) βcot87.0 vsvyvs sdAfV ≅

( no. of links crossed by section line)

⇒ 2AA

A

Derivation of shear stress in links CL

A

A

β

sv

( ) βcot87.0 vsvyvs sdAfV ≅

( )vsvyvs sdAfV 87.0=

cs VVV −=dvbV v=dbvV vcc =

( ) vsvyvvc sAfbvv 87.0=−

( )yv

cv

v

sv

fvvb

sA

87.0−

=

• Experiments have led to the recommendation that β could be taken as 45°, giving

• From (19), and introducing

(20)

sc VVV += (19)AppliedResisted by concrete

(21)

The design concrete shear stress vc can be obtained from Table 6.3 of Hong Kong Concrete Code

where

=

mv

sc ddb

Akkvγ140010079.0

4/13/1

21

an enhancement factor equal to 2d/av for shear-span / depth ratio (av/d) less than 2; otherwise k1 = 1

for concrete with cube strength fcu > 25MPa, but fcu should not be taken as greater than 80 N/mm2

=1k

( ) 3/12 25cufk =

Shear design

(22)

The design concrete shear stress vc can be obtained from Table 6.3 of Hong Kong Concrete Code

where

=

mv

sc ddb

Akkvγ140010079.0

4/13/1

21

tensile steel percentage which should not be taken as greater than 3a ratio that should not be taken as less than 1.partial safety factor for strength of materials taken as 1.25

Shear design

(22)

=dbA

v

s100

=d/400=mγ

• To avoid web crushing or compression zone failure, the average shear stress v should not exceed or 7.0 N/mm2, whichever is the smaller.

• The provision of shear reinft. should be in accordance with the following (Table 6.2 of HK Concrete Code)

• Note that the design shear resistance vr provided by minimum links is taken asvr = 0.4 N/mm2

for fcu ≤ 40 N/mm2

vr = 0.4 (fcu/40)2/3 N/mm2

for fcu > 40 N/mm2

cuf8.0

0.0

0.2

0.4

0.6

0.8

1.0

0 20 40 60 80 100Concrete grade

Des

ign

shea

r res

ista

nc

Shear design

There are three cases to be considered in shear design.(a) v < 0.5 vc throughout the beam

Provide minimum links in beams of structural importance. Minimum links may be omitted in members of minor structural importance.

(b) 0.5 vc < v < (vc + vr)Provide minimum links equivalent to additional shear strength of vr, i.e.Setting

Shear design

yv

vvrsv f

sbvA87.0

≥

( )vsvyvs sdAfV 87.0= (20)dbvV vrs =

There are three cases to be considered in shear design.(c) (vc + vr) < v < 0.8√fcu or 7.0 N/mm2

Provide links according to (21), namely

Shear design

( )yv

cvvsv f

vvsbA87.0

−≥

Min. links preferred

Min. links

Links depending on shear stress

Prohibited!

0 0.5 vc

vc vc + vr

0.8√fcu or 7.0 N/mm2

Procedure for design for shear1. Calculate the design shear stress at the section from

.2. Check whether the design shear stress v is within the

upper limit, i.e. 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.

dbVv v=

cufv 8.0<

Procedure for design for shear5. Choose suitable link size and spacing sv, ensuring that

.6. At least 50% of the shear resistance provided by the

steel should be in the form of links.

dsv 75.0≤

Deflection control DL, LL, etc

M

Fst

Mild steel High yield steel

Low fs & high As

High fs & low As

High εs, high curvature & high deflection

Low εs, low curvature & low deflection

Parameters:• Breadth b• Effective depth d• Steel area As

• A set of basic span / effective depth ratios are used to control deflections. These ratios are normally more critical for slabs.

• The basic ratios are modified to account for the amounts of tension and compression reinforcement and the stresses there.

Basic span / effective depth ratios for rectangular sections with spans less than 10m

Cantilever 7Simply supported 20Continuous 26

Deflection control

• The modification factor for tension reinforcement can be calculated by

• The design service stress in the tension reinforcement in a member may be estimated from

• Similarly, the modification factor for compression reinforcement

Deflection control

( )( ) 0.2

9.012047755.0 2 ≤

+−

+bdM

fs

bprovs

reqsys A

Aff

β1

32

×=

tionredistribubeforesectionatmomenttionredistribuaftersectionatmoment

=bβ

5.1100

3100

1 ≤

′+

′+

bdA

bdA provsprovs

(23)

(24)

• Concrete slabs behave primarily as flexural members, and therefore the design is similar to that for beams.

• The shear stresses are usually low in a slab except when there are heavy concentrated loads.

• Compression reinforcement is seldom required except at supports.

• Common methods of analysis:- Elastic analysis- Yield line theory by Johansen- Strip method by Hillerborg

Design of slabs

Design of slabs

1000

Unit width

No. of bars required = Area of one bar

Max. spacing = No. of bars required

1000

= 1000 × area of one bar

Ast

Ast

Two-way slab

Main reinft. Secondary

reinft.

One-way slab

• The slab is often assumed to be built up of parallel strips (beams) of 1m width (i.e. b=1000mm).

• The area of tension reinforcement is • A suitable bar diameter and spacing can be worked out

by • Min. reinforcement (esp. secondary reinft.):

for high yield steel (fy = 460 MPa) for mild steel (fy = 250 MPa)

• Shear stress• Avoid using shear reinforcement in slabs

One-way slabs

zfMA ys 87.0=

sAbar) one of (area 1000 = spacing maximum ×

10013.0 bh10024.0 bh

cuv fdbVv 8.0≤=

Examples

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 MPaif the ultimate shear force is 150 kN.

260mm

440m

m

156.0123.030440260

101852

6

2 <=××

×==

cufbdMK

( ) mm3689.025.05.0 =−+= Kdz

Example 1Design data:

M = 185 kNm V = 150 kN b = 260 mmd = 440 mm fcu = 30 MPa fy = 460 MPafyv = 250 MPa

Determine if compression reinforcement is requiredK factor

and therefore compression steel is not requiredLever arm zArea of tension reinft. required As

26

mm125636846087.0

1018587.0

=××

×==zf

MAy

s

Example 1Area of tension reinft. required As = 1256mm2

Select suitable arrangement of reinforcementProvide 4T20, giving As prov = 1257mm2 O.K.

MPa31.144026010150 3

=××

==db

Vvv

Example 1Shear design:Determine the average shear stress

Check the average shear stress to avoid web crushing or compression zone failure

O.K.7.00MPaMPa38.48.0MPa31.1 <=<= cufv

10.1440260

1257100100=

××

=dbA

v

s

1909.0440400400

<==d

(Take 1)

( )8020.12530

25≤== cu

cu ff

25.1=mγ

Determine the design concrete shear stress vc

=

m

cu

v

sc

fddb

Avγ1

2540010079.0

3/14/13/1

Example 1

(Use As provided)

( ) ( ) ( )

MPa31.1MPa69.025.1120.1110.179.0

125

40010079.0

3/14/13/1

3/14/13/1

=<=

=

=

v

fddb

Avm

cu

v

sc γ

Example 1

MPa09.14.069.0MPa31.1 =+=+>= rc vvvCheck if minimum links are sufficient

Links more than the minimum amount are necessary

Example 1

( ) 2mm14825087.0

)69.031.1(20026087.0

=×

−××=

−≥

yv

cvvsv f

vvsbA

Determine the amount of links requiredLet sv = 200mm

Provide R10 links at 200 mm spacing

(Asv prov = 157 mm2) O.K.

Check if the spacing is acceptable:

200mm ≤ 0.75×440mm = 330mm O.K.

Example 1Asv req’d = 148 mm2

Provide R10 links at 200 mm spacing

(Asv prov = 157 mm2) O.K.

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.

Example 2: A singly reinforced rectangular section with diff. amounts of tension reinft.

250mm

460m

m

Example 2

Design data:Breadth b = 250 mmEffective depth d = 460 mmConcrete cube strength fcu = 25 MPaSteel yield strength fy = 320 MPaYoung’s modulus of steel Es = 200,000 MPa

250mm

460m

m

(a) Balanced section1. Determine the neutral axis depth x and lever arm z

The neutral axis depth x can be worked out by considering the strain distribution:

The lever arm z is therefore:

( )sy Efxdx

87.00035.0

=−

mm3290035.087.0

0035.0=

+=

sy

s

EfdEx

mm31245.0 =−= xdz

Strain distribution

x

εcc = εcu = 0.0035

εst = 0.87 fy / Es

d

2. Determine the balanced steel areaFrom equilibrium considerations:

The balanced steel ratio ρbcan be determined as

3. Determine the moment capacity


( )( ) 0258.0462.0 == dxff ycubρ

2mm2969=sA

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x

kNm9.25787.0 == zAfM sy

(a) Balanced section (b) Section with As = 2500 mm2

1. Determine the type of reinforced concrete sectionThe tension reinforcement ratio ρ is

Therefore this section is under-reinforced and tension failure occurs.

0258.00217.0 =<== bs bdA ρρ

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x

(b) Section with As = 2500 mm2

2. Determine the neutral axis depth x and lever arm zThe neutral axis depth x can be worked out by considering equilibrium as

The lever arm z is therefore

3. Determine the moment capacityThe moment capacity M is


( )( ) mm277164.2 == bAffx scuy

mm33545.0 =−= xdz

kNm2.23387.0 == zAfM sy

(c) Section with As = 5160 mm2

1. Determine the type of reinforced concrete sectionThe tension reinforcement ratio ρ is

Therefore this section is over-reinforced and compression failure occurs.

0258.00449.0 =>== bs bdA ρρ

Strain distribution

x εcc = εcu

εst < 0.87 fy / Es

d

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x


2. Determine the neutral axis depth x and lever arm zThe neutral axis depth x can be worked out by considering equilibrium as

The lever arm z is therefore

sscu AEx

xdbxf

−

= 0035.0402.0

( ) 00035.0402.0 2 =−+ dxAEbxf sscu

01066.136120005.2512 92 =×−+ xx

mm367=x

mm29545.0 =−= xdz

Fcc

0.67fcu /γm ≅ 0.45 fcu


Fst

s/2

z

s = 0

.9x


3. Determine the moment capacityThe moment capacity M is based on compression in concrete, namely

The steel strain is checked to make sure that it has indeed not yielded

kNm8.273)(9.0 2 =−= zdzbfM cu

sscu Afbxf =402.0

( ) 001392.087.0

000892.0402.0

=<

===

sy

ss

cu

s

ss

EfAE

bxfEfε

O.K.

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

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 reinft. of 510 mm. The depth to compression reinft. is 50mm. The areas of tension and compression reinft. are 2580 mm2 and 650 mm2

respectively. The steel reinft. 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 (εcu = 0.0035, x ≤ 0.5d and s = 0.9x); and

(b) 70MPa (εcu = 0.0033, x ≤ 0.4d and s = 0.8x).

Example 3: Doubly reinforced sectionDesign data:

b = 280 mmd = 510 mm (depth to tension reinft.)d' = 50 mm (depth to compression reinft.)As = 2580 mm2

A's = 650 mm2

fy = 330 MPaEs = 200000 MPa

280mm

510m

m

50m

m

(a) Concrete strength fcu = 35 MPa1. Initial estimation of internal forces

Assuming that all reinforcement reaches design yield stresses, the forces acting on the section are

2. Consider the equilibrium of forces

N6.3939402.0 xbxfC cuc ==

N18661587.0 =′= sys AfC

N74071887.0 == sy AfT

TCC sc =+

7407181866156.3939 =+xmm141=x

Cc



T s =

0.9

x

z

Cs

3. Check validity of assumptionsCheck steel strains:

Therefore the comp. steel has yielded.

Therefore the tension steel has also yielded. The previous assumptions are valid.

(a) Concrete strength fcu = 35 MPa

( ) 001436.087.0

002259.00035.0

==>

=

′−

=′

syy

s

Efxdx

ε

ε

( ) 001436.087.0

009160.00035.0

==>

=

−

=

syy

s

Efx

xd

ε

ε

x

εst

Strain distribution

Neutral axis

εcu = 0.0035

εsc

Cc



T

s = 0

.9x

z

Cs

4. Moment capacityThe moment capacity M is

( ) ( )0.402 0.45 0.87

333.9 kNmcu y sM f bx d x f A d d′ ′= − + −

=

(a) Concrete strength fcu = 35 MPa

(b) Concrete strength fcu = 70 MPa1. Initial estimation of internal forces

Assuming that all reinforcement reaches design yield stresses, the forces acting on the section are

2. Consider the equilibrium of forces Cc


Stress distribution (45 < fcu ≤ 70)

T

s = 0

.8x

z

Cs

N18661587.0 =′= sys AfC

N74071887.0 == sy AfT

TCC sc =+

7407181866157.7003 =+xmm79=x

N7.70033573.0)8.0)(5.1/67.0( xbxfbxfC cucuc ===

3. Check validity of assumptionsCheck steel strains:

Therefore the comp. steel has NOT yielded.

Therefore the tension steel has yielded. The previous assumptions have to be amended.

(b) Concrete strength fcu = 70 MPa

x

εst

Strain distribution

Neutral axis

εcu = 0.0035

εsc

( ) 001436.087.0

001211.00033.0

==<

=

′−

=′

syy

s

Efxdx

ε

ε

( ) 001436.087.0

018004.00033.0

==>

=

−

=

syy

s

Efx

xd

ε

ε

4. Amendment of assumptionsThe steel compressive force Cs and neutral axis depth x are updated assuming that the compression steel remains elastic.


x

xdxEAEAC ssssss

710145.2429000

)0033.0(

×−=

′−′=′′= ε

N7.7003 xCc =

N740718=T

x

εst

Strain distribution

Neutral axis

εcu = 0.0035

εsc Fcc


Stress distribution(45 < fcu ≤ 70)

Fst

s = 0

.8x

z

Fsc

4. Amendment of assumptions


TCC sc =+

010145.23117187.7003 72 =×−− xx

mm82=x

xCs710145.2429000 ×−=

N7.7003 xCc =

N740718=T

Cc



T

s = 0

.8x

z

Cs

5. Re-check validity of assumptions Check steel strains:

Therefore the compression steel has NOT yielded.

Therefore the tension steel has yielded. The revised assumptions are valid.


( ) 001436.087.0

001288.00033.0

==<

=

′−

=′

syy

s

Efxdx

ε

ε

( ) 001436.087.0

01722.00033.0

==>

=

−

=

syy

s

Efx

xd

ε

ε

x

εst

Strain distribution

Neutral axis

εcu = 0.0035

εsc

6. The moment capacity M is


( ) ( )kNm1.351

4.03573.0=

′−′′+−= ddAExdbxfM ssscu ε

Cc



T

s = 0

.8x

z

Cs

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.

4500 mm Main reinforcement

Secondary reinforcement

Cross section of a slab

Example 4: Design of slabDesign data:Span L = 4500 mmWall thickness t = 230 mmBreadth b = 1000 mmBasic span / eff. depth ratio = 20fcu = 30 MPafy = 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 steel1. Estimate the effective depth of slab

Assume modification factor (m.f.) = 1.3Minimum effective depth d = span / (20 × m.f.)

= 173 mmTherefore, try effective depth d = 180 mm

2. Estimate the self-weight of slabAssume bar diameter = 10 mmOverall depth of slab h= effective depth + bar diameter / 2 + cover = 210 mmSelf-weight of slab = overall depth × density

= 5.04 kN/m2

d

3. Estimate the design loadsTotal dead load Gk = 1.0 + 5.04 = 6.04 kN/m2

Live load Qk = 3.0 kN/m2

Ultimate load w = 1.4 Gk + 1.6 Qk= 13.26 kN/m2

4. Estimate the design moment per unit width at mid-spanM = wL2 / 8

= 33.55 kNm

(a) Using high yield steelFloor finishes & ceiling loads

Self-weight of slab

5. Check the span / effective ratio

Assume design service stress fs = 460 × 2/3 = 307 MPaModification factor

Limiting span / effective depth ratio= 20 × 1.28= 25.6

Actual span / effective depth ratio = 4500 / 180 = 25.0 < 25.6 O.K.

Hence the span / effective depth ratio is satisfactory.

(a) Using high yield steel

MPa04.12 =bdM

( ) 0.228.19.0120

47755.0 2 <=+−

+=bdM

fs

6. Determination of the lever arm zThe K factor is

Therefore compression steel is not required.The lever arm z is then determined:

> 0.95Take giving

7. The area of tension reinforcement As is

Provide T10 at 150mm centresAs prov = 524 mm2/m O.K.


156.0035.02 <== bdfMK cu

960.09.025.05.0 =−+= Kdz95.0=dz mm171=z

m/mm49087.0 2== zfMA ys

Provide T10 at 150mm centresAs prov = 524 mm2/m O.K.


Results obtained by truncation

8. Check for shearAt 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

8. Check for shear

Therefore no shear reinforcement is required.


25.1=mγ

( ) ( ) ( )

MPa157.0MPa543.025.1120.122.2291.079.0

125

40010079.0

3/14/13/1

3/14/13/1

=>=

=

=

v

fddb

Avm

cu

v

sc γ

9. Distribution steelThe required amount of distribution steel is then determined.Area of transverse high yield steel reinforcement

= 0.13 bh / 100 = 273 mm2/mProvide T10 at 250 mm centres, As prov = 314 mm2/m

O.K.


(b) Using mild steel1. Estimate the effective depth of slab

Assume modification factor (m.f.) = 1.7Minimum effective depth d = span / (20 × m.f.)

= 132 mmTry overall depth of slab h = 180 mmAssume bar diameter = 12 mmEffective depth d= overall depth of slab - cover - bar diameter / 2= 180 – 25 – 12 / 2 = 149 mm O.K.

2. Estimate the self-weight of slabSelf-weight of slab = overall depth × density

= 4.32 kN/m2

Different approach

3. Estimate the design loadsTotal dead load Gk = 1.0 + 4.32 = 5.32 kN/m2

Live load Qk = 3.0 kN/m2

Ultimate load w = 1.4 Gk + 1.6 Qk= 12.25 kN/m2

4. Estimate the design moment per unit width at mid-spanM = wL2 / 8

= 31.01 kNm

Floor finishes & ceiling loadsSelf-weight of slab

(b) Using mild steel

5. Check the span / effective ratio

Assume design service stress fs = 250 × 2/3 = 167 MPaModification factor

Limiting span / effective depth ratio= 20 × 1.67= 33.4

Actual span / effective depth ratio = 4500 / 149 = 30.2 < 33.4 O.K.

Hence the span / effective depth ratio is satisfactory.


MPa40.12 =bdM

( ) 0.267.19.0120

47755.0 2 <=+−

+=bdM

fs

6. Determination of the lever arm zThe K factor is

Therefore compression steel is not required.The lever arm z is then determined:

7. The area of tension reinforcement As is

Provide R12 at 100mm centresAs prov = 1131 mm2/m O.K.


156.0047.02 <== bdfMK cu

945.09.025.05.0 =−+= Kdzmm141=z

m/mm101187.0 2== zfMA ys

Substantially more reinforcement provided!

Provide R12 at 100mm centresAs prov = 1131 mm2/m O.K.



8. Re-check modification factorRe-check modification factor using the updated service stress in the tension reinforcement fs:

where

Revised limiting span / effective depth ratio= 20 × 1.74 = 34.8 > 30.2 O.K.


( )( )bprovsreqsys AAff β132

= 1=bβ

MPa0.149=sf

( ) 0.274.19.0120

47755.0.. 2 <=+−

+=bdM

ffm s

9. Check for shearAt 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

9. Check for shear

Therefore no shear reinforcement is required.


25.1=mγ

( ) ( ) ( )

MPa176.0MPa784.025.1120.168.2759.079.0

125

40010079.0

3/14/13/1

3/14/13/1

=>=

=

=

v

fddb

Avm

cu

v

sc γ

10.Distribution steelThe required amount of distribution steel is then determined.Area of transverse mild steel reinforcement

= 0.24 bh / 100 = 432 mm2/mProvide R10 at 150 mm centres, As prov = 524 mm2/m

O.K.



The End

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RC2009 University of HongKong Reinforced concrete design