rc deflection & cracking

7/27/2019 RC Deflection & Cracking

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A simply supported rectangular reinforced concrete beam is as shown inFIGURE 1.1. Given the following design data, determine what depth, h, ofthe beam will ensure that deflection criterion is not violated? Take fireresistance to be 1.5 hours.

i) Nominal cover to main reinforcement = 20 mm

ii) Dead load on beam due to finishes = 0.75 kN/m2iii) Imposed load on beam = 3 kN/miv) Unit weight of concrete = 24 kN/m3

v) Concrete strength, fcu = 25 N/mm2

vi) Reinforcing bar strength fy, = 460 N/mm2

(a) Longitudinal section of the beam

(b) Transverse section XX of the beam

2

h

250 mm

5000

X

X

b

h


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FIGURE 1.1

Solution

From the basicdepth

spanratio given in Table 3.10 of BS 8110: Part 1: 1985,

and assuming a modification factor of 1.2

mm

xd

xd

Span

3.208

2.120

5000

2.120

min

min

=

=

=

Thus initial depth of the beam cdh ++=2

min

Assume that 16 mm bar is used, cover of 20 mm for 1.5 hour fire

resistance, then we have

mm

h

250

3.236

202

163.208

=

++=

For fire resistance of 1.5 hour the minimum breadth of the beam fromFigure 3.2 of BS 8110: Part 1:1985 is 200 mmThus use a beam of mmbmmh 200250 ==

Then effective depth is mmd 222202

16

250 ==Loading computation

Dead load Self weight =24 x 0.25 x 0.2 kN/m=1.2 kN/m

Finishes =0.75x[0.25+0.2]x2=0.675 kN/m

Total D.L., gk =1.2+0.675 = 1.875 kN/mImposed Load qk = 3 kN/mTotal ultimate Load n = 1.4gk + 1.6qk

= 1.4x1.875 + 1.6x3=7.425 kN/mApplied ultimate bending moment for simply supported condition, M

kNm

x

nlM

2.23

8

5425.7

82

2

=

=

=

Design ultimate bending moment of resistance, Mu

kNmxxxx

dbfM cuu

44.381022220025156.0

156.0

62

2

=

=

=

3


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Since Mu > M, provide only tension reinforcement only.

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Design of tensile reinforcement

K=cufbd

M2

=

25222200

102.232

6

xx

x

= 0.094

Lever arm,Z= d[0.5+ )9.0

25.0(K ]

= 222[0.5+ )9.0

094.025.0(

= 0.881 x 222 < 0.95d= 195.65 mm

Area of tensile reinforcement

As = Zf

M

y87.0

=65.19546087.0

102.23 6

xx

x

= 296 mm2

Check adequacy of reinforcement

Ast(min) = bh100

13.0=

100

13.0x 200 x 250

= 65 mm2

Ast(max) = bh100

4

= 100

4

x 200 x 250= 2000 mm2

Thus provide, 3Y12 339.3 mm2

Check deflection

2bd

M=

2

6

222200

102.23

x

x= 2.35 MPa

Service stress, fs =bprovs

reqs

yA

Af

1

8

5

)(

)(

Equation 8 (BS8110 Part 1)

=1

1

3.339

296460

8

5xxx

= 250.8 MPa

Modification factor, M.F.

M.F. = 0.55 +)9.0(120

)477(

2bd

M

fs

+

2.0 Equation 7 (BS8110 Part 1)

= 0.55 +)35.29.0(120

)8.250477(

+

= 0.55 + 0.58= 1.13 < 2.0

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Allowable span/d ratio = 20 x 1.13 = 22.6Actual span/d ratio = 5000/222 = 22.5

The beam depth is satisfactory with respect to deflection. Thus provide a

minimum depth of mmh 250= for the beam.

11.2.5 Deflection Calculation

The calculation of deflection in reinforced concrete is spelt out in BS 8110:Part 2 Sections 3.6 and 3.7.

Procedure of calculating deflections in reinforced concrete elements is asfollows:

Step 1: Determine the moment acting on the element

Use elastic analysis using gross concrete section and taking serviceability

limit loads, viz. kk QG 0.10.1 + with appropriate loading patterns. Noredistribution of moment is allowed.

Step 2: Determine the curvatures of the element

BS 8110: Part 2 Section 3.6 outlines the procedure of determining thecurvature due to the loading. It differentiates the curvature for a crackedconcrete section from that of an uncracked concrete section, then adopts

the one that gives the larger value.

(i) To determine the curvatures of an r.c. element that hascracked Method one

Assumptions Refer to section 3.6 page 3/2 of BS 8110: Part 2.

N A

SECTION STRAINS STRESSES FORCES

Figure 1.2: Section properties of a cracked r.c. element (Figure 3.1(a) ofBS 8110: Part 2)

6

fst

fct

fcc

d/

d/

d

b

h

Ast

Asc

d-xst

sc

cc

x

fsc

Ts

Tc

Cs

Cc


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NOTE: h is the overall depth of the sectionx is the depth from the compression face to the neutral

axis

d Effective depth of the sectionfcc is the maximum compressive stress in the concretefst is the tensile stress in the reinforcementEs The modulus of elasticity of the reinforcementEc The short term modulus of elasticity of the concrete

Curvature:xr

cc ==

1; from strain diagram, Eqn. 7 of BS 8110:

Part 2

Compression strain in concrete:c

cccc

E

f= ; from Hookes Law

Compression stress in concrete: ccccc Ef =

Tensile strain in steel:s

st

ccstE

f

x

xd=

= ; from stress diagram.

Tensile stress in concrete:

=

x

xhff ccct

'; from stress diagram

Tensile stress in concrete at level of steel:

=

x

xdff ccct ; Where ctf is

fixed at 1.00 N/mm2 in short term loading and 0.55 N/mm2 in long-termloading.From the BS 8110 assumptions, curvature at mid span of a beam or forcantilever at the support section is given by:

( ) s

st

c

cccc

b Exd

f

xE

f

xr ====

1Which is given as equation 7 in the BS 8110.

The Procedure

(a) Construct the strain, stress and force diagram for the section asshown in Figure 1.2 above. Note that for long-term loads theeffective value of Ec should be used.

Let e = Modular ratio =c

s

E

E

(b) Express the reinforcement stresses in terms of concrete stresses

cc

sce

ccc

scs

cc

sc

E

E

f

f

== but from the strain diagram

x

dx

cc

sc/

=

thus;

=

x

dxff ccesc

/

Similarly;

=

x

xdff ccest

(c) Compute the internal forces

bxfC ccc 5.0=

==

x

dxAfAfC scccescscs

/

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==

x

xdAfAfT stcceststs

[ ][ ]

=xd

xhbfT ctc

2

5.0

(d) From the equilibrium of internal forces

0=+ cssc TTCCThe respective values in the above equation can be substituted fromequations in step (c) above

(e) Taking moment about the N.A. for the internal forces

( ) ( ) ( )xhTxdTdxCxCM csscI +++=3

2

3

2 /

The respective values in the above equation can be substituted fromequations in step (c) above.

(f) Equating the external moment computed in Step 1 and the internal

moment computed in Step 2 (e). That is moment due to internalforces must balance the moment due to external loads.

Thus there are two polynomial functions in terms of two unknowns,fcc and x, which can be solved by successive trials to obtain thedesired values.

Note that the area of concrete occupied by the reinforcement hasnot been deducted in the expression above.

(ii) To calculate the curvatures of an r.c. element that hascracked Method two

An alternative equation defining is given by:

; where M is the moment at the sectionconsidered I is the second moment of

area.The Procedure

(a) Transform the reinforced concrete section into a pure concrete

section as shown in Figure 1.3 below

8

b

IE

M

r cb== 1

d/eAsc


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R.C. Section Transformed section

Figure 1.3: Transformed r.c. section

(b) Compute the second moment of area of the transformed sectionabout x-x

( ) ( ) ( )

( ) ( ) 22/3

22/

2

3

3

1

212

1

xdAdxAbx

xdAdxAx

bxbxI

stesce

stescexx

++=

++

+=

(c) Establish the reduced moment of resistanceConsider the concrete in the tension zone. The moment of

resistance of the concrete in tension is ( )xhTM cc =3

2. Where Tc

was computed in step 2.cR MMM = . Where M was computed in step 1

(d) Compute the curvature

Otherwise calculate;xc

Rc

IE

M

x=

x

Rc

I

xMf =

c

ccE

f

= ; where Ec depends on whether the loads applied are short-

term or long-term.

Thenxr

c=1

(iii) To calculate the curvatures of an r.c. element that is un-cracked

Assumptions See BS 8110: Part 2 section 3.6(b) page 3/3

9

Dcc

eAst

2(h-x)/3

Tc

x-d/

d-x

x

fct

d/

d

b

h

Ast

Asc

xc

R

b IE

M

r==

1

fccd/

Asc

d-xst

sc

cc

x

fsc

Ts

Tc

Cs

Cc


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SECTION STRAIN STRESSFORCE

DIAGRAM DIAGRAM DIAGRAMDIAGRAM

Figure 1.4: Section properties of a uncracked r.c. element (Figure3.1(b) of BS 8110:

Part 2), with xDcc3

1= and )(3

2xdDct =

(a) Establish the equivalent area of the transformed section[ ]stscee AAbhA ++=

(b) Find the depth of the neutral axis, taking moment of area about thecompression fibres

( )

e

stsce

A

dAdAh

bx

++=

/2

2

(c) Find the second moment of area about the neutral axis

( ) ( )2/223

212dxAxdAx

hbh

bhI scestexx ++

+=

(d) Find the curvature

xxcIE

M

r=1

(iv) Curvature Due to Long Term Loads

(a) Creep: See BS 8110: Part 2 Section 7.3(b) Shrinkage: See BS 8110: Part 2 Section 7.4

Step 3: Calculation of Deflection from Curvatures

The deflected shape of a member is related to the curvatures by Eqn 10 ofBS 8110: Part 2, thus:

2

21

dx

ad

rx= ; Equation 10 (BS 8110 Part 2)

where,x

r

1is the curvature at x and as computed above

a is the deflection at x.

10

Dct

fct

fst

d/

d

b

h

Ast


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Deflections may be computed directly from this equation by calculatingthe curvatures at successive sections along the member, and using anumerical integration technique.

Alternatively, the simplified scheme given by Eqn 11 of BS 8110: Part 2can be used

brkla

12= ; Equation 11 (BS 8110 Part 2)

Wherebr

1is the curvature at mid-span or, for cantilever,

at the support sectionl is the effective span of the memberkis a constant that depends on shape of the bending

moment diagram.

The various coefficients of k for common shapes of bending momentdiagram are given in Table 3.1 of BS 8110: Part 2 page 3/5.

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11.3 CRACK CONTROL AND CALCULATION IN REINFORCEDCONCRETE STRUCTURES

11.3.1 Introduction

The presence of the crack does not necessarily mean that the structurewill collapse, but it is a signal that the behaviour of the structureshould be considered more carefully.

Cracking may detract from the appearance of the structure, permitingress of moisture and lead to corrosion of steel. Excessive cracking andwide cracks will thus affect the durability of the structure.

An accurate prediction of the initiation of cracks and the subsequentmonitoring of their behaviour during the structures response to loading

play a significant role in the prediction of the general safety of thestructure.

11.3.2 Sequence of Crack Formation

RC elements experience cracks in the tension face when the tensilestrength of the concrete is exceeded. The tensile strength of concrete isgenerally less than 20% of the compressive strength.

Primary cracks form first and as applied moment increases secondarycracks follow. The width of secondary crack at a point on the surface of a

beam is affected by:1 Surface strains which can be found by analyzing the section

assuming the section remains plane in accordance with Bernoullisbeam theory

2 The distance of the point from a point of zero crack width. Thepoints of zero crack width are the neutral axis and the surface oflongitudinal reinforcing bars. The larger the distance of point fromthe points of zero crack width, the larger the cracks width will be asshown in Figure 1.1 below.

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Figure 2.1: Critical locations for cracking in a beam are:

At A equidistant between the neutral axis and the barsurface.

At B equidistant between the bar. At C on the corner of the beam.

11.3.3 Code Provision for Cracks

The following sections in BS 8110: Part 1, Sections 2.2.3; 2.2.3.4.1;3.12.11.2 and BS 8110: Part 2: Sections 3.8.1 and Section 8 deal withcrack phenomena in reinforced concrete structures.

11.3.4 Sources of Cracks in Reinforced Concrete

(i) Incorrect detailing is the most common source of cracking in RCstructures. For example, wrong or no placement of movement jointduring detailing and construction. Movement joints are provided toreduce or eliminate cracking by permitting relative movement tooccur. Movement joints are covered in BS 8110: Part 2, Section 8.The various types of movement joints as shown in Figure 1.2 are:

Contraction/expansion joints

Sliding joints

Hinged joints

Settlement joints;

Figure 2.2 Joints in R.C. Structures

(ii) Errors in design calculation and detailing;(iii) Reinforcing bar spacing limits in tension zones;

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(iv) Amount of reinforcing bars;(v) Poor construction methods such as:

Incorrect placement of steel leading to insufficient cover

In adequate cover to reinforcement

Poor curing - loss of water can cause shrinkage cracking.During curing the concrete should be kept damp and covered;

(vi) External physical and mechanical factors such as:(a) Restraint of elements against movement causes cracking. The

movement might be due to elastic deformation, creep underconstant load, shrinkage on drying and settlement offoundations; changes in temperature and moisture content.

(b) Overloading the structure - overloading may be brought aboutby change of function of the building or room i.e. classroomchanged to library increases Imposed Loading which was notcatered for in the earlier design.

(c) Structure settlement namely: Differential settlement of foundation can cause

cracking,

Shrinkage of clays from ground dewatering,

Drying out in droughts,

Tree roots causing disruptions,

Ground movement from nearby excavations.

11.3.5 Crack Control

Cracking should be kept within reasonable limits. A maximum surfacecrack width of 0.3 mm is appropriate, as stipulated in BS 8110: Part 1Section 3.12.11.2, and can be achieved if correct detailing is done. Thiscan be controlled by adhering to detailing rules with regard to bar spacingin zones where the concrete is in tension. However, if this limit will impairthe efficiency of the structure a smaller value may be more appropriate.For instance for water tightness values of 0.2 mm or even 0.1 mm may beadopted.

BS 8110: Part 1 Section 2.2.3.4.1 specifies two methods of crack control.(i) Limiting maximum bar spacing in the tension zone of

members under normal situations - maximum spacing is specified inBS 8110: Part 1 Section 3.12.11.2 Table 3.30.

(ii) Use of the formula provided in BS 8110: Part 2 Section3.8 for assessing the design crack width, in special cases.

In addition, however, the following should be complied with:(a) Allowance should be made for stiffening effects of concrete in the

tension zone, and for creep and shrinkage;(b) Minimum areas of steel to control cracking in beams as per BS

8110: Part 1 Section 3.12.5.3, Table 3.27;

(c) Bar spacing control rules to limit crack width to 0.3mm, thus:

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Bars of diameter less than 0.45 of the largest diameter shouldbe ignored except when considering beam side face bars.

Refer to Figure 2.3 below for bar spacing control to limit crackwidth.

Figure 2.3: Illustration of bar spacing rules

Bar spacing rules;

ba should not be greater than values specified in Table 3.30

of BS 8110 Part 1

sb should not be greater than 250 mm

Clear spacing, ba y

b

f

75000 300; where

tionredistribubeforeMoment

tionredistribuafterMomentb =

Forb = 0; maximum ba = 300 mm, for fy = 250 N/mm2 and

maximum ba = 160 mm for fy = 460 N/mm2

An alternative method of computing ba is:

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Clear spacing, ba sf

47000 300; where fs is the service stress

determined using Equation 8Table 3.10 (BS 8110: Part 1)

bprovs

reqsy

s XA

Af

f

1

3

2

.= Equation 8 of BS 8110: Part 1; 1997

ca should not be greater than 0.5 of ba

Ifh exceeds 750 add bars as indicated in Figure 2.3 above, onthe right hand side (as per Section 3.12.11.2.6 of BS 8110: Part 1:1997). Otherwise the side reinforcement is not necessary, as on theleft hand side.

The size of the side bar should not be less than that specifiedin BS 8110: Part 1 Section 3.12.5.4

y

b

f

ba

The maximum clear spacing for slabs is given in BS 8110: Part1 Section 3.12.11.7. Other control measures are outlined in3.12.11.8 for slabs.

11.3.6 Calculation of Crack Width

11.3.6.1 General

The equations for assessing crack width for flexure and direct tension are

stipulated in BS 8110: Part 2 Section 3.8. The computation of the crackwidth for flexure and direct tension shall be treated separately in thiscourse.

11.3.6.2 Crack width calculation for reinforced concreteelements in flexure

The design crack width, crw , at any point on the surface of the tension

zone of the element under flexure is calculated using Equation 12 of BS8110: Part 2, thus:

+=

xh

ca

aw

cr

mcrcr

min21

3

Eqn. 12 (BS 8110: Part 2)

Where; cra is the distance of the point considered to the

surface of the nearest longitudinal bar,m is the average strain at the level where thecracking is being considered,cmin is the minimum cover to the tension steel,h is the overall depth of the element,x is the depth of the neutral axis.

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This is provided that the strains, st, in the tension reinforcement is not

greater than

s

y

E

f8.0, viz. the steel has not yielded.

Determination of m

An approximate method for the determination of m is given in BS 8110Part 2 Section 3.8.3.For a rectangular tension zone, use Equation 13 given in BS 8110 Part 2Section 3.8.3, thus;

( ) ( )

( )

=

xdAE

xaxhb

ss

tm

3

'

1 Eqn. 13 (BS 8110: Part 2)

Where; 1 Strain at level being considered,

calculated while ignoring the stiffening effectof the concrete in the tension zone.

'a Distance from the compression face to thepoint at which the crack width is beingconsidered/calculated.

bt Width of the section at the centroid of thetension steel.

Es Modulus of elasticity of the reinforcementAs Area of tension reinforcementThe term within the square brackets defines thetension stiffening effect.

If the whole section is in tension an effective value of (hx) can beestimated by interpolation between the following conditions:

(a) Where the n.a. is at the compression face (h-x) = h, i.e.x=0(b) For axial tension with (h-x) = 2h

NOTE1. If m < 0 the section is not cracked.

2. The modulus of elasticity of the concrete is taken as 0.5 of theinstantaneous value (see Table 7.2 Part 2 BS 8110).

3. For high drying shrinkage, viz. > 0.0006, m should beincreased by adding 50% of the shrinkage strains. Otherwise,in normal cases shrinkage may be neglected.

Implications of Equation 13 of BS 8110: Part 2

The following can be illustrated diagrammatically as shown in Figure 2.4

overleaf:

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1. Along the bottom edge in the region of maximum tension forwhich 'a = h

m becomes( )

( )

=xdAE

xhb

ss

t

m3

2

1 , a constant value. So in the crack

width formula cra is the only variable and as the maximum value ofcra is midway between bars, the maximum crack width occurs at this

point.

2. Immediately below the reinforcing bar, cra is a minimum and

equals to cmin, so crack width = 3 cmin m and is the minimum value.

3. At the corner cra = ca which is greater than cmin so the crack

in wider.

4. Moving up the side of the beam; m decreases linearly from a maximum at the corner of the beam

to zero at the n.a.

The value of cra decreases to a minimum at the level of the

reinforcement and then increases up to the level of n.a. If nofurther longitudinal bars are encountered.

It will be found that the maximum value of the crack width occurs

at about 1/3 of the distance between the longitudinal steel andthe n.a.

Note that these comments assume that the maximum tensile stress is inthe span of the element. Where the maximum stress occurs in the top ofthe element, e.g. over support, the diagram will be inverted.

19


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Figure 2.4: Significant factors in calculating crack width.

11.3.6.2Calculation of crack width for reinforced concreteelements in direct tension

Direct tension as a predominant force is unlikely to occur in normalbuilding structures, but direct tension forces in combination with bendingmoments may occur. In this case the neutral axis will be calculated takinginto account the tension, following which the calculations for flexural crackwidth will be as before.

The limit of flexure being predominant is where the neutral axis is at thetop face of the member shown in Figure 2.4, viz. when x=0. The equationfor the crack width now becomes

+=

h

ca

aw

cr

mcr

cr

min21

3

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And, the tension stiffening effectdAE

hb

ss

t

3

2

=

When the whole section is in tension it is suggested that the modificationgiven in 3.8.3(b), viz. the parameter (h x) = 2h for axial tension is used.

The equation for the crack width then becomes;mcrcr aw 3= (for cmin = cra ) and

ss

tm

AE

hb

3

21 = ; for (d-x) (h-x) and (

'a -x) = (h-x).

where; As is the total area of steel inthe section, equally dividedbetween the two faces.

The most obvious case of axial tension is the hoop tension in a circulartank containing water. In the Code for liquid-retaining structure themaximum crack width is 0.2mm. Designers of these structures areinterested primarily in serviceability limit state of cracking.A calculation isrequired at ultimate limit state, but if the walls are cracked and water isleaking out, the tank is unserviceable and a factor of safety at ultimate isirrelevant.

A designer can select an arrangement of bars to suit a particular wallthickness, tensile force, cover and crack width requirement.

11.3.7 Examples of Crack Width Calculation

11.3.7.1 Crack width calculation for a rectangular beam

Calculate design crack widths at critical positions for the internal supportsection of the following two-span continuous beam for which 15%redistribution has been allowed at the ultimate limit state.

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Figure 2.5:

Assuming a concrete cover of 45 mm and =40 mm, from the diagram itcan be seen that the distance between the bars is 130mm, which is lessthan the value given in Table 3.30 (BS8110 Part 1). The formula below the

table gives the upper limit to be 138.6mm. The distance to the corner ofthe beam is 68.5 mm, which is also within the limit (0.5 ba ).

(1) Calculate internal support moment due to service load.

kNmX

Ms 4638

10)1522( 2=

+=

(2) Calculate properties of equivalent transformed section with2

ce

EE =

228mm

kNEc= from Table 7.2 in Part 2 so 214

mm

kNEe=

1414

200===

c

se

E

E

17.0690300

251014==

X

Xso e

( )[ ] 427.02 =++= eeed

x

mmzandmmxso 8.5916.294 ==

(3) Calculate average surface strain at top of beam

( )226

3688.07.3118.5912510

10463

mmNf

mmN

X

X

zA

Mf y

s

ss =


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Figure 2.6:

001558.010200

7.3113===

XE

f

s

ss

1 = 00179.0001558.04.395

4.455==

= Xxd

xhsh

00169.0

000104.000179.0

4.3952510102003

)4.455(30000179.0

)(3

)(

3

2

2

==

=

=

XXXX

xdAE

xhb

ss

thmh

(4) Calculate crack widths at critical position

(a) Top of beamMaximum crack width occurs midway between bars, where

( ) mmam 8420856022 =+= )(40min facetopmmc =

)3.0(36.0

4.455

408421

107.1843

21

3

3

min

mmmm

XXXxh

ca

aw

cr

mcrcr

>=

+

=

+

=

(b) At corner of beam mmac 5.68=

)3.0(31.0

4.455

405.6821

107.15.683

21

3

3

min

mmmm

XXX

xh

ca

aw

cr

mcrcr

>=

+

=

+

=

(c) On side of beam

Critical position approximately (d-x)/3 from reinforcement, which is 263

mm from neutral axis

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mma 2.5586.2636.294' =+=

00098.00017.04.455

6.263'==

= Xxh

xamhm

( ) mmadacr 5.12420)12965(20)(65222'2 =+=+=

)(45min facesidemmc =

)3.0(27.0

4.455

455.12421

108.95.1243

21

3

4

min

mmmm

XXX

xh

ca

aw

cr

mcrcr


25/27

Figure 2.7: (a) Section: (b) transformed section (c) crack locations anddimensions (d) stress diagram

The moment of inertia about the neutral axis is

48

223

1022.13

1.219226699.3561913

9.801450

mmX

XXXIxx

=

++=

The stress in the tension steel is:

2

8

6

8.284

1022.13

4.151.219106.111

mmN

X

XXXft

=

=

The strain in the tension steel is:

3

3

10424.1

10200

8.284

=

=

X

Xs

Neglect the stiffening effect of the concrete in tension in the flange of theT-beam.

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27/27

5

31078.3

)9.80300(1472102003

)9.806.189()9.80350(250

=

XXXX

The average strain at the crack location is:33 10668.010)038.0706.0( == XXm

The design surface crack width at C where acr = 108.7mm is:

mm

XXX

xh

ca

aw

cr

mcr

cr

14.0

9.80350

5.377.10821

10668.07.1083

21

3

3

min

=

+

=

+=

All crack width are less than 0.3mm and are thus satisfactory

11.3.7.3 Crack width calculation for an element in direct tension

The hoop tension force in the wall of a tank is 570kN/m. The wall is 200thick, cover to reinforcement is 40mm, and the limiting crack width is0.2mm. Check that T16 at 150 centres each face is satisfactory.

mmmATotal

s

2

2680=

00106.0

102002680

105703

3

1 ==XX

X

00025.02680102003

200100023

==XXX

XXgstiffenninTensioning

00081.000025.000106.0 ==m

8187548 22 =+=cra

mmXXwcr 197.000081.0813 ==

rc deflection & cracking

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