AZ024 Lecture 3 (rev[1] a)

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-Page 1-November 2006 (rev.a)

Topic: Reinforced Concrete (II)- Design in RC Beam and slab-1

1.0 Introduction to Structural Element beams

Beams Horizontal members carrying roof and floor loads. They resist loads in

bending, shear and bond and may be simply supported or continuous. For in-situ

construction beam are often flanged, of T or L shape where part of the floor slab acts

with the beam.

2.0 General Terms in Limit state design (BS 8110(1995))

Characteristics loads:- the working or service loads, classified into dead, imposed and

wind loads. They have a low probability of being exceed during the life of the

structure.

Characteristics strengths:- the strength of materials below which not more than 5% of

test results fall. For concrete this is the cube strength at 28days and for reinforcement,

the yield stress.

Design load:- the characteristics loads multiplied by the partial factors of safety for

the load

Design strength:- the characteristics strengths divided by partial factors of safety for

materials

Limit state:- States where the structure has become unfit for use. The main limit state

are:

(1)ultimate limit state: to satisfy this the strength must be adequate to carry the

loads. Account must also be taken of stability

(2)serviceability limit state: to satisfy these both deflection and cracking must

not be excessive.

Limit State Design

The condition of a structure when it becomes unserviceable is called a Limit State

Loads and partial factor of safety

Characteristics dead load, KG

This is the weight of the structure complete with finishes, fixtures and partitions.

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Characteristics imposed load, KQ

This load depends on the use of the Buildings,

Characteristics wind load, KW

This is defined and calculated in accordance with code of practice, Ch.V, Part 2. (it is

out of scope in this course)

The design load

= characteristics Load x Partial factor of safety for loads

= KKF

Where K takes account of (1) possible overloads (2) inaccurate assessment of the

effects of loading and unknown stress distribution within the structure.

Materials and partial factors of safetyThe characteristics strength is defined as the cube strength of concrete CUf at 28 days,

and the yield of reinforcement yf , below which not more than 5% of the test results

fall.

The resistance of sections is based on the design strength

Design strength = characteristics strength / partial factors of safety for materials

=M

Kf

Table 1 Design Load

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Table 2 Design Strengths

The maximum design stress in the concrete is given by CUCU ff 67.05.1/ = where the

factor 0.67 takes account of the ratio between the characteristics cube strength and the

bending strength in a flexural member.

Assumption takes in single reinforced rectangular beams

The ultimate moment of resistance of a section is based on the assumption given in

the BS8110.

(1)The strains in the materials derived assuming that plane sections remain plane.(2)The stresses in the concrete are derived using either (a) the design stress-strain

curve given in BS8110 with 5.1=m , or (b) a uniform compressive stress of

CUf45.0 over the whole compressive zone.

(3)Depth of the stress block = 0.9 x depth of the neutral axis = x9.0

(4)The tensile strength of the concrete is ignored

(5)The stresses in the reinforcement are derived from the stress-strain curve given in

BS8110 where .15.1=m

On the basis of the above assumptions, the strain and stress diagrams for a beam

section are shown below:-

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h =overall depth of the section

d =effective depth = depth of centre line of the steel

b =breath of the beam

x =depth to the neutral axis

SA =area of steel in tension

C =strain in concrete

T =strain in steel

stf =stress in the steel in tension

The term balance design refers to a beam with the maximum ultimate moment of

resistance, for example, with sufficient steel to causes the neutral axis to be at its

maximum depth of 1/2d.

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Considering the rectangular stress diagram,

The concrete stress =CUf45.0

The steel stress = yf87.0

C(force in the concrete in compression) = CUCU bdfdbf 20.09.05.045.0 =

T(force in the steel in tension) = syAf87.0

Z = level arm = ( ) ddd 775.09.04

=

RCM (moment of resistance with respect of the concrete) = ZC

= 22156.0775.09.05.045.0 kbdfbdddbf CUCU ==

where CUfk 156.0=

RSM (moment of resistance with respect of the steel) = ZAfTZ sy87.0=

Let RSRC MM=

ZAfbdf syCU 87.0156.02=

Percentage of steel in the tension

( )%

23

775.087.0

156.0100100 2

y

CU

y

CUs

f

f

bddf

bdf

bd

Ap =

=

=

Point to notes

(1)the nominal cover should always be at least equal to the size of the bar and in the

case of bundles of 3 or more bars should be equal to the size of a single bar of

equivalent areas

(2)further recommendations regarding cover are given in BS8110. These depend on

conditions of exposure and concrete grade. For example, for grade 25 concrete for

mild exposure, for example, completely protected against the weather except

during construction, the cover given is 20mm. The cover is lower with higher

grades of concrete and greater when condition of exposure become more severe.

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Doubly reinforced beams

If the concrete alone cannot resist the applied moment, reinforcement can be added to

strengthen the beam section in compression.

Design formulae for doubly reinforced beams are given in BS8110. These are based

on the followings:-

(1)rectangular stress block with the depth to the neutral axis, dx2

1=

(2)stress in concrete in compression = CUf45.0

(3)stress in reinforcement in compression = yf87.0

(4)stress in reinforcement in tension = yf87.0

Note: SC = force in steel in compression

T = force in steel in tension

BS 8110 states that the formulae should not be used whend

d' is >0.2. If this

requirement is not met, the stress in the compression steel will not reachyf87.0 . The

moment of resistant of the concrete 2156.0 bdfM CURC =

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However, if this is less than M (ultimate moment), the compression steel resists

( )RCMM

The force in steel in compression

( )( )'dd

MMC RCs

=

Area of compression steel =y

s

sf

CA

87.0

'=

For equilibrium,

SC CCT +=

( ) ( )'

'

87.0225.0

87.05.045.0

SyCU

SyCU

Afbdf

Afbdf

+=

+=

Area of Tension Steel =As = Force / Stress =y

f

T

87.0

Example 1:-

A simply support rectangular beam of 7m span carried a uniformly distributed loadwhich includes a self-weight of 4kN/m and imposed load of 3kN/m. the breath of the

beam is 300mm. Find the depth of the beam and the steel area required for balanced

using Grade 30 concrete and mild steel reinforcement (yield stress,

2/250 mmNfy = )

Solution:

Design Load = mkNQG KK /4.10)3(6.1)4(4.16.14.1 =+=+

Ultimate Moment, mkNwLMU /7.638)7(4.10

8

22

===

Considering the rectangular stress diagram,

The stress of concrete =CUf45.0

The stress of steel = yf87.0

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C(force in the concrete in compression)

= CUCU bdfdbf 20.0)9.0)(5.0)((45.0 =

T= Force in the steel in tension = syAf87.0

Z= ( ) ddd 775.022

9.0 =

MRC(Moment of resistance with respect to concrete)

where CUfk 156.0= .

MRS (moment of resistance with respect to steel)

zAfTM syZRS 87.0==

But RSRC MM =

2156.087.0 bdfzAf CUsy =

Percentage of steel in the tension

( )%

23

775.087.0

156.0100100 2

y

CU

y

CUs

f

f

bddf

bdf

bd

Ap =

=

=

For balance design, (when the depth x to the neutral axis is dx 5.0= ,

Moment of resistance of concrete,

2

156.0)775.0)(9.0)(5.0(45.0 bdfddbfM CUCURC ==

Moment of resistance of concrete RCM Ultimate moment (Mu)

mmd

d

bdfCU

212

.107.63)300)(30(156.0

107.63156.0

62

62

Moment of resistance of steel RSM Ultimate moment (Mu)

22156.0)775.0)()(9.0)(5.0(45.0 kbdbdfddbf

zc

cuCU ===

=

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2

6

6

1783

107.63)212)(775.0)(250)(87.0(

107.63)87.0(

mmA

A

fA

s

s

yS

=

From steel area table,

Combination Steel Area (mm2) Comment

4 nos. bar 25 dia. 1960 ok

6 nos. bar 20 dia. 1890 ok

9 nos. bar 16 dia. 1801 economical, but need to put in several

positions.

1 nos. bar 50 dia 1960 Min 2 nos. of steel bar for shear stress

Then from steel area table,

6 nos. 20mm dia. Steel bar given 22 17831890 mmmmAs >= (O.K.)

Alternatives

n x area of one bar > 1783mm2

( ) 178342>Dn whereD = Diameter of bar (in mm)

If the diameter of the bar is choose, say 20mm diameter, then the nos. of the bar

required can be easily calculated as follows:-

( )

6

67.5

1783204

2

=

>

>

n

n

n

Example 2:-

A rectangular beam is simply supported over a span of 6m and carries a dead load

including self-weight of 9kN/m and an imposed load of 6kn/m. the beam is 210mm

wide and 310mm effective depth and the inset of the compression steel is 40mm.

Design the steel for mid span of the beam for grade 25 concrete and high yield stress

reinforcement ( )/2502

mmNfy = and mild steel links or stirrups ( )/2502

mmNfy = .

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Solution:

Design Load = mkNQG KK /2.22)6(6.1)9(4.16.14.1 =+=+

kNmwL

MU 9.998)6(2.22

8

22

===

Design of reinforcement for bending moment

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For balance design, (when the depth x to the neutral axis is 0.5d),

kNmbdfddbfM CUCURC 7.78)10(7.78156.0)775.0)(9.0)(5.0(45.0

62====

Since RCU MM > , compression reinforcement is required.

Thus, doubly reinforced design is required.

Cs= Force in steel in compression

( )( )

( )( )

kNdd

MM RCU 6.891040310

7.789.99' 3

=

=

=

Area of steel in compression,

23 412)250(87.0/)10(6.8987.0

mmf

CA

y

SS ===

For equilibrium, Tensile force = Compressive Force

kNfbdfCCT yCUSC 450)412(87.09.0)5.0(45.0 =+=+=

Area of tension steel,

yS fTA 87.0/= =23 1863)250(87.0/)10(450 mm=

Proposed number of Tension Steel

6 No. 20mm dia. Steel bars given 21890mmAS =

Proposed Number of Compression Steel

2 No. 20mm dia. Steel bars given 2628mmAS =

-END-

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Appendix: Steel Area Table

AZ024 Lecture 3 (rev[1] a)

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