session 3 - sls cg apr 2012a

Serviceability Limit States

www.eurocode2.info

1 © MPA The Concrete Centre

Outline

• Crack control and limitations

• Crack width calculations

• Crack width calculation example

• Crack width calculation problem

• Restraint cracking

• Deflection calculations

• Deflection calculation example


Crack control and limitations

© MPA The Concrete Centre 4

EC 2: Concise:

(Flexural) Crack Width Limits (Table 7.1N) Table 7.1(N) 10.2

Exposure class RC or unbonded PSC

members

Prestressed members

with bonded tendons

Quasi-permanent load Frequent load

X0,XC1 0.3 0.2

XC2,XC3,XC4 0.3

XD1,XD2,XS1,XS2,

XS3

Decompression


EC 2: Concise:

(Flexural) Crack Width Control Cl. 7.3.3 10.2

Crack control may be achieved:

• Limiting the maximum bar diameter using Table 7.2

• Limiting the maximum bar spacing using Table 7.3

(this table is not applicable for restraint loading)

• Calculating cracks to ensure they are within limits


EC 2: Concise:








Crack width calculations


Basis

Figure 7.2

Slab soffit

(h - x)

Neutral axis

Crack width w

Actual crack width

Section

Crack width vs spacing

Crack width predicted by Expressions (7.8) & (7.11)

Crack width predicted by Expressions (7.8) & (7.14)

5(c + /2)


Crack width calculation

The crack width may be calculated from:

wk = sr,max (εsm - εcm)

where

sr,max = maximum crack spacing

εsm = mean strain in reinforcement

εcm = mean strain in concrete between cracks


EC2 Exp (7.8)

Maximum crack spacing

sr,max = 3.4c + 0.425k1k2Φ /ρp,eff

c = (nominal) cover to the longitudinal reinforcement

k1 = factor to take account on bond properties

= 0.8 for high bond bars

= 1.6 for plain bars

k2 = factor to take account of strain distribution

= 0.5 for flexure

= 1.0 for pure tension

Φ = bar diameter

When spacing > 5(c + Φ/2);

sr,max = 1.3(h – x)


(εsm - εcm) Difference in concrete and reinforcement strain

Strain diagram

εcs ≤ εcult

S

wk = s .(εsm - εcm)

(sr,max includes fos of 1.7)


Cra

ck


For flexure,(εsm - εcm) may be calculated from:

where:

σs = stress in the tension steel calculated using the cracked concrete section

kt = factor that accounts for the duration of loading

= 0.6 for short-term load

= 0.4 for long-term load

αe = Es/Ec = modular ratio

s

s

s

effpeeffp

effctts

cmsmEE

fk

6.0

)1( ,

,

,



where cont.

ρp,eff, is the effective reinforcement ratio.

ρp,eff = As /Ac,eff

where

As = area of tension reinforcement

Ac,eff = effective area of concrete in tension around the reinforcement

hc,ef = Min{2.5(h - d); (h - x)/3; h/2}

Figure 6.12: Typical examples of effective concrete tension area

hd Effective

tension area

Beam

hc ,eff

Effectivetension area

hc ,eff

d

d

h

hc ,eff

hc ,eff

Effectivetension areafor this face

Effectivetension areafor this face

Slab

Member in tension

h = lesser of 2.5(h-d), (h-x)/3 or h/2c ,eff


(Flexural) Crack width calculation

example


(Flexural) Crack width calculation

Calculate the design flexural crack

for the beam shown.

MQP = 650 kNm

Concrete class C25/30

As = 3770 mm2

(∞,t0) = 2.63

h =

1000

d =

930

3 No H40 bars

(d – x/3)

Fc

Fs

x

d = 400


Crack width example

Step 1 – Calculate effective modulus

From Table 3.1, Ecm = 31 kN/mm2

Ec,eff = Ecm/(1 + (∞,t0)) = 31 / (1 + 2.63) = 8.54 kN/mm2

Step 2 – Calculate the stress in the tension steel : find x

Taking moments about neutral axis:

b x2/2 = αe As (d – x)

400 x2/2 = 200/8.54 x 3770 (930 – x)

This has the solution, x = 457 mm


Crack width example

Step 3– Calculate stress in the tension steel

Taking moments about the level of force in the concrete:

σs = MQP/(d – x/3)As

= 650 x 106 /((930 – 457/3) x 3770)

= 222 MPa


Step 4– Calculate difference in concrete and reinforcement strains

Crack width example

kt = 0.4 (long-term loading)

fct,eff = fctm = 2.6 MPa (Table 3.1)

αe = Es/Ecm = 200 / 31 = 6.45

hc,ef = Min{2.5(h - d); (h - x)/3; h/2}

= Min{2.5(1000 - 930); (1000 - 457)/3; 1000/2}

= Min{175; 181; 500} = 175 mm

s

s

s

effpeeffp

effctts

cmsmEE

fk

6.0

)1( ,

,

,


Crack width example

Ac,eff = 175 x 400 = 70 000 mm2

ρp,eff = As /Ac,eff = 3700 / 70 000 = 0.0539

001.0

00067.010200

97.19222

10200

2226.0

10200

)0539.045.61(0539.0

6.24.0222

3

33

cmsm


Crack width example

Step 5 – Calculate the maximum crack spacing


c = 1000 – 930 - 40/2 = 50 mm

k1 = 0.8 (ribbed bars)

k2 = 0.5 (flexure)

Φ = 40 mm

sr,max = 3.4 x 50 + 0.425 x 0.8 x 0.5 x 40 /0.0539

= 296 mm < 5(c + Φ/2) = 350 mm

Step 6 – Calculate crack width

wk = 0.0010 x 296 = 0.30 mm


EC 2: Concise:








Say 110 mm

cf (400 – 2 x 40 -40) /3

= 93 mm . . . .OK

Say

435/ 1.4 =

310 MPa

d =

192

Workshop problem

Calculate the design flexural crack for the slab shown.

MQP = 85 kNm


As = 2010 mm2/m

Assume the depth to neutral axis, x = 63.5 mm

HINT: You can start the calculation from step 3

h =

250

H16 bars @ 100 mm CTRS

(d – x/3)

Fc

Fs

x


Workshop problem

Step 3– Calculate stress in the tension steel

Taking moments about the level of force in the concrete:

σs = MQP/(d – x/3)As

= 85 x 106 /((192 – 63.5/3) x 2010)

= 247.5 MPa


Step 4– Calculate difference in concrete and reinforcement strains

Workshop problem

kt = 0.4 (long-term loading)

fct,eff = fctm = 3.2 MPa (Table 3.1)

αe = Es/Ecm = 200 / 34 = 5.88

hc,ef = Min{2.5(h - d); (h - x)/3; h/2}

= Min{2.5(250 - 192); (250 – 63.5)/3; 250/2}

= Min{145; 62.2; 125} = 62.2mm

Ac,eff = 62.2 x 1000 = 62 200 mm2

s

s

s

effpe

effp

effctts

cmsmEE

fk

6.0

)1( ,

,

,


Workshop problem

Ac,eff = 62.2 x 1000 = 62 200 mm2

ρp,eff = As /Ac,eff = 2010 / 62 200 = 0.0323

0010.0

00074.010200

2.476.247

10200

6.2476.0

10200

)0323.088.51(0323.0

2.34.06.247

3

33

cmsm


Workshop problem

Step 5 – Calculate the maximum crack spacing


c = 50 mm

k1 = 0.8 (ribbed bars)

k2 = 0.5 (flexure)

Φ = 16 mm

sr,max = 3.4 x 50 + 0.425 x 0.8 x 0.5 x 16 /0.0323

= 254.2 mm < 5(c + Φ/2) = 290 mm

Step 6 – Calculate crack width

wk = 0.0010 x 254 = 0.25 mm


Restraint cracking


Restraint cracking

Movement occurs not only

due to loading but also

due to:

• Early thermal effects

• Shrinkage

– drying

– autogenous,

– seasonal/long term

temperature drop

That becomes a problem when

there is Restraint:

• Edge:

– adjacent slab pours,

– wall on base,

– adjacent wall pours

• End:

– infill bays,

– large area ground slabs

(friction, foundations),

– piled slabs

• Internal (not covered . .

members > say 750 mm th.)


Restraint cracking

End restraint: restrained strain & stresses

Original poured length

Free contraction: R = 0

Partial restraint R = 0.0 to 1.0

Restrained strain: Stresses induced

Fully restrained: R =1.0

Restrained strain: Large stresses

induced


r = Raxfree

= K1 { cT1 +ca R1 + cT2 R2 + cd R3}

where:

K1 = allowance for creep = 1.0 to BS EN 1992-3 or = 0.65 to CIRIA C660

c = coefficient of thermal expansion (typical design value 12 m)

T1 = Peak to ambient temperature oC (See CIRIA C660). (e.g. 500 mm thick wall formed using 18 mm ply, using C30/37 concrete with 40%ggbs = 29oC)

ca = Autogenous shrinkage strain – (typical design values using C30/37 concrete 15 m @ 3days, 50 m long term)

R1, R2, R3 = appropriate restraint factor for the short-term, medium term and long term see figure L1 of BS EN 1992-3 (includes for creep) or calculated for base wall restraint in accordance with CIRIA C660 (excludes for creep)

1) Cracking will occur if r ctu

Restraint cracking

CIRIA C660: Cl 3.2


r = Raxfree

= K1 { cT1 +ca R1 + cT2 R2 + cd R3}

where:

cd = drying shrinkage strain, a function of time, thickness, RH, cement Class (BS EN 1992-1-1 or CIRIA C660) (e.g. 500 mm thick wall, using C30/37 concrete with 40% ggbs ≡ Class N = 340 m)

T2 = long-term drop in temperature after concreting. Recommended values: of 20oC for concrete cast in the summer and 10oC for concrete cast in winter. (See CIRIA C660),

ctu = tensile strain capacity of the concrete. A function of concrete strength and type of aggregate used. (Typical design values of 76 m @ 3 days and 108 m for 28 days may be used for initial calculations. See CIRIA C660.


Restraint cracking (cont’d)

CIRIA C660: Cl 3.2


Restraint factors

Table 1 – Values of restraint factor R for a particular pour

configuration

0,8 to 1,0 Infill bays, i.e. rigid restraint

0,2 to 0,4 Suspended slabs

0,3 to 0,4 at base

0,1 to 0,2 at top

Massive pour cast onto existing concrete

0,1 to 0,2 Massive pour cast onto blinding

0,6 to 0,8 at base

0,1 to 0,2 at top

Thin wall cast on to massive concrete base

R Pour configuration

BS EN 1992-3

Annex L

Beware: effects

of creep

included

usually 0.5


Restraint cracking

CS TR 67

Short term load strength

Long term load strength

Stress due to early thermal –

allowing for creep

Stress due to early

thermal & drying

shrinkage

Stress due to

early thermal &

shrinkage &

seasonal

34

r = Raxfree

= K1 { cT1 +ca R1 + cT2 R2 + cd R3}


Restraint cracking

CIRIA C660: Cl 3.2

Short term

(≡ 3 days)

Medium

term

(≡ 28 days)

long term

(≡ > 10000

days)


BS EN 1992-1-1 Exp (7.8)


2) Minimum reinforcement,

(with respect to restraint to movement):

As,min = kc kAct (fct,eff / fyk)

3) Controlled cracking:

Crack width wk = sr,max cr

where

Maximum crack spacing sr,max = 3.4c + 0.425 (k1/p,eff)

Crack inducing strain cr ……

Restraint cracking

CIRIA C660: Cl 3.2

BS EN 1992-1-1 Exp (7.1)

Where:

kc = coeff. for stress distribution = 1.0 for full tension

k = coeff. for thickness 1.0 for h < 300 mm and 0.75 for h > 800 mm

(interpolation allowed)

Act = area of concrete in the tension zone just prior to onset of cracking.

Most often based on full thickness of the section. (

fct,eff ≡ fctm = mean tensile strength when cracking may be first expected to

occur (Typical design values for a C30/37 concrete, 1.73 MPa @ 3 days

and 2.9 MPa @ 28 days See BS EN 1992-1-1

fyk = 500 MPa


Restraint cracking

Watchpoints:

• Ensures rebar does not yield

• Typically 0.58% for C30/37 in a 300 mm wall

• 0.8 factor on fct,eff for sustained loading? or 0.67 to TR 59

• Early age only? B Hughes

• Revised CIRIA C660?


BS EN 1992-1-1 Exp (7.8)


2) Minimum reinforcement,

(with respect to restraint to movement):

As,min = kc kAct (fct,eff /w fyk)

3) Controlled cracking:

Crack width wk = sr,max cr

where

Maximum crack spacing sr,max = 3.4c + 0.425 (k1/p,eff)

Crack inducing strain cr …… ≡ (sm - cm) . . . . . .

Restraint cracking

CIRIA C660: Cl 3.2

BS EN 1992-1-1 Exp (7.1)

As before where:

c = nominal cover, cnom in mm

k1 = 0.8 for high bond bars (CIRIA C660 suggests a value 1.14 to account

for poor bond conditions)

k2 = 1.0 for tension (e.g. from restraint), 0.5 for bending, (1 + 2)/ 21 for

combinations

= bar diameter, mm

p,eff= As/Ac,eff ,

where Ac,eff is calculated for each face = min0.5h or 2.5(c + 0.5) x b

BS EN 1992-1-1 Exp (7.11)



Strain diagram

εcs ≤ εcult

S

wk = s .(εsm - εcm)



Cra

ck


Crack inducing strain, cr

a) Edge restraint:- early thermal effects

cr = k cT1 +ca R – 0.5 ctu

b) Edge restraint:- long term restraint effects

cr = k { cT1 +ca R1 + cT2 R2 + cd R3} – 0.5 ctu

c) End restraint

cr = 0.5 e kckfct,eff 1 + (1/e ) /Es

where:

kc = coeff. for stress distribution = 1.0 for full tension

k = coeff. for thickness 1.0 for h < 300 mm and 0.75 for h > 800 mm

fct,eff = fctm for long-term effects, 28 day value considered to be reasonable e.g. 2.9 Mpa

for C30/37. NB Possible 0.8 factor for sustained load in CIRIA C660

e = modular ratio, Es/Ec. Typical values are 6 @ 3 days, 7 @ 28 days and 12 long-

term. When cracking occurs, no creep has taken place so a modular ratio of

7 should be used.

= ratio of total area of reinforcement to the gross section in tension.

Note that this different from p,eff.

BS EN 1992-3 Exp (M.1)

CIRIA C660: Cl 3.2

CIRIA C660: Cl 3.2

Can be critical!

Deflection calculations


Deflection limits

Deflections are limited for the following reasons:

1. Excessive deflections are unsightly and alarming. EC2 restricts

total deflections to span/250.

2. To avoid damage to cladding, partitions and finishes due to

increments in deflection following their construction. EC2 limits

deflections after construction of finishes to span/500.

3. Both construction tolerances and deflections need to be

considered in the design of fixings for cladding systems and

partitions. In practice it can be difficult to separate

construction tolerances from deflections.


Deflection limits

The EC2 deflection limits are guidelines. It is the designers

responsibility to agree suitable deflection limits with his client

taking into account the nature of the structure and finishes.


Introduction

Four factors need to be considered in the calculation of

deflections

1. criteria defining the limiting deflections

2. appropriate design loads

3. appropriate design material properties

4. means of predicting behaviour

Deflections cannot be predicted exactly before construction since

neither the loading or material properties are accurately known at

the design stage.


Calculated assuming concrete has no tensile strength

Actual behaviour

Deflection calculation

Deflection

Lo

ad

Calculated assuming no cracking


Basic behaviour

= (II) + (1 - )(I)

where:

= deformation parameter considered

(e.g. strain, curvature)

I is the calculated uncracked parameter

II is the calculated cracked parameter

= ‘distribution coefficient’ allowing for tension stiffening at a

section.

e.g. total curvature = S(cracked curvature + uncracked curvature)

for each effect considered


(1 - )S S S

Basic behaviour

Steel stress

s2

s1

Concrete stress

S

Crack Crack Crack

0

Idealised steel stress

Where

= ‘distribution coefficient’ allowing for

tension stiffening at a section.


Basic behaviour

where:

= ‘distribution coefficient’

= 1 - (sr/s)2

where:

= coefficient taking account of the influence of the

duration of the loading or of repeated loading on

the average strain

= 1.0 for first loading

= 0.5 for long-term loading

But always use 0.5

s = stress in tension steel based on cracked section

sr = stress in tension steel based on cracked section at first cracking

NB sr/s ≡ Mcr/M for flexure


Basic behaviour

= 1 - (first crack result/cracked analysis result)2

Deflection

Load

no cracking

Actual

cracked Mcr

MEd

= 0.0 for un-cracked sections

= 1.0 for fully cracked sections

(in theory)


Concrete material properties

for deflection calculation

It is only possible to estimate concrete material properties at the

design stage. Actual material properties may differ significantly

from those assumed in design. Therefore, it is prudent to assume a

range of material properties in deflection calculations.

EC2 relates all the concrete properties required for deflection

prediction to the concrete grade and cement type. In practice,

properties are influenced by the aggregate type, curing etc.

Mean values should be used for the tensile strength and elastic

modulus of concrete to obtain a best estimate of the actual

deflection.


Concrete tensile strength

Deflections in slabs depend significantly on the effective concrete

flexural strength which governs the cracking moment. The flexural

strength of concrete is calculated from the peak failure load of

unreinforced concrete beams with engineers bending theory.

EC2 defines the flexural strength of concrete as follows:

ffl=(1.6-h/1000)fctm>fctm

where fctm = 0.3fck2/3 is the mean tensile strength of concrete which

can be estimated indirectly in the splitting test. fck is the

characteristic cylinder compressive strength of concrete.

The flexural strength is greater than the tensile strength since the

tensile stress distribution is not linear at failure as assumed in its

derivation.


Flexural strength of concrete

The flexural strength is greater due the assumptions implicit in its

derivation as illustrated below.

Stress at

peak load

fct

Strain Stress assumed in

calculation ffl

ffl>fct


Concrete tensile strength

In reinforced concrete structures the effective flexural strength of

concrete is reduced by tensile stresses induced by restraint of

shrinkage by reinforcement and restraining elements such as stiff

columns and shear walls.

It is conservative to use the tensile strength fctm in deflection

calculations.

The How to Leaflet suggests that the design value concrete tensile

strength for a low restraint layout is taken as the mean of the

tensile and flexural strengths.


Long-term deflections

Three additional factors must be considered in the long term

calculation of deflections.

1.Loading

2.Creep

3.Shrinkage


Design loads: Permanent loads

In concrete structures, deflections increase with time under sustained load. The greater part of the deflection normally occurs under sustained loads. Therefore, long-term deflections are calculated under a best estimate of the sustained load during the lifetime of the structure.

The design load for calculating long-term deflections is the permanent load:

Permanent load = Gk + 2Qk

where

Gk = dead load

Qk = imposed load

Recommended values for 2 are:

0.3 for residential and offices,

0.6 for parking

0.8 for storage


Design loads: Frequent load

Cracking is irreversible. Therefore, it is prudent to calculate long-

term deflections using a modified flexural strength which

corresponds to the worst cracking during the lifetime of the

structure. The How to Leaflet suggests that the frequent load

combination is used to calculate the deflection affecting cladding.

The frequent load is given by:

Frequent load = Gk + 1Qk

Recommended values for 1 are:

0.5 for residential and offices,

0.7 for parking

0.9 for storage


Time dependent deformation

Creep is the continuous deformation of a member under

sustained load.

Shrinkage consists of autogenous (due to hydration) and

drying shrinkage. 57 © MPA The Concrete Centre

Creep

EC2 uses the effective modulus method to model creep in which

creep is modelled as a delayed elastic strain.

The creep strain at time t is given by:

cc(t) = (t0) *

where

(t0) = the strain at the time of first loading t0 and

= /Ec(t0)

where

Ec(t0) is the elastic modulus of the concrete at time t0.

* = the true creep coefficient.

= EC2[Ec(t0)/Ec28]


Ec

EC2 defines the creep coefficient in

terms of the 28 day tangent modulus of

concrete, Ec

Ec = 1.05 Ecm

where

Ecm = secant modulus

= 22[fcm/10]0.3 See table 3.1

59 © MPA

EC2

Annex B or . . .

Figure 3.1


Creep

So total strain:

(t) = (t0) (1+ *)

= [/Ec(t0)] (1+ *)

= /Eceff

where

Eceff = effective elastic modulus

= Ec(t0)/(1+ *)

For practical purposes

Eceff = Ec28/(1+ EC2)


In practice, there are usually several loads placed at different times.

In that case long term modulus, ELT for n serviceability loads, Wi :-

ELT = SW/{ (W1/Eceff,1) + (W2/Eceff,2) + (W3/Eceff,3) + . . . . +(Wn/Eceff,n)}

Shrinkage induced curvature

Shrinkage induces curvatures in asymmetrically reinforced

sections that can increase deflections by as much as 25%.

The reinforcement restrains the shortening of the member due to

shrinkage which induces tension in the concrete. Consequently,

the cracking moment is reduced.


Tensile stress

Shrinkage induces a curvature

that is given by:

1/rcs = cseS/I

where

c = free shrinkange strain

e= Es/Eceff

S = Ase = the first moment of area of the reinforcement

about the centroid of the transformed section

I = second moment of area of the section

EC2 extends the distribution coefficient approach to cover cracked

sections by applying to Scr/Icr and (1- ) to Suncr/Iuncr.


Accuracy of deflection

calculations Many factors influence the accuracy of deflection calculations

including:

1. actual loading relative to design loading

2. early age striking and loading from constructing slabs above

3. differences between actual and assumed material

properties

4. Composite action between floor slabs and floor screeds and

partitions

5. Temperature effects


Use of finite element analysis

to calculate deflections

Two approaches are commonly used:

1.Cracked section analysis in which the plate stiffness is reduced

to account for cracking

2.Elastic analysis with reduced stiffness to allow for cracking

creep and shrinkage. In this case, the effective E value can be

taken as:

E*ceff=0.5Ec/(1+)

where the factor of 0.5 accounts for the effects of cracking and

shrinkage


Deflection

The deflection may be calculated:

Either by calculating the curvatures (due to load, shrinkage,

creep) at a number of sections and then double

integrating numerically

Or by the simplified formula:

δ = kL2(1/r)

k depends on the shape of the bending moment

diagram.

Both methods are described in detail in How to design concrete

structures using Eurocode 2: Deflection


Rigorous method


Rigorous method

This is the approach used in the “Rigourous” RC Spreadsheets


Rigorous method TCC41R


Simpler method (outline)


Essentially add curvature due to SLS moments:

shrinkageuncrackedshrinkagecrackedshrinkage rrr ,,

)(

11

11

To curvature due to shrinkage:

momentsslsuncrackedslsmomentscrackedmomentssls rrr ,,,.

)(

11

11

δ = kL2(1/r)

Calculate deflection:

shrinkageslsmoments rrr

111

So total curvature:

k from chart – depends on shape of BMD

Simpler method (in detail)


Simpler method


Deflection calculation

example


Estimate the long-term deflection for the beam

shown.

Span = 9.5 m

MQP = 200 kNm


As = 2450 mm2

xc = 329 mm

Icr = 7976 x 106 mm4

xu = 350 mm (ignoring reinforcement)

Iu = 8575 x 106 mm4 (ignoring reinforcement)

(∞,t0) = 2.8

εcs = 470 x 10-6

Worked example

h =

700

d =

600

5 No H25 bars

d = 300


Deflection calculation example

Step 1 – Calculate cracking moment

fctm = 2.6 MPa (Table 3.1)

If uncracked section properties are used, Mcr = 57.3 kNm

Section is cracked, therefore:

ζ = 1 – 0.5(57.3/200)2 = 0.95

u

uctmcr

xh

IfM

9.0

kNm3.57

350700

1085756.29.0 6

crM



Step 2 – Calculate flexural curvature

Ec,eff = Ecm/(1 + (∞,t0)) = 31 / (1 + 2.8) = 8.15 kN/mm2

ueffc

QP

u IE

M

r ,

1

mm/1086.21085751015.8

102001 6

63

6

ur

mm/1008.31079761015.8

102001 6

63

6

,

ceffc

QP

c IE

M

r

mm/1007.31086.2)95.01(1008.395.0

1)1(

11

666

ucn rrr



Step 3 – Calculate shrinkage curvature

where:

Sc = As (d – x) = 2450 (600 – 329) = 664 x 103 mm3

Su = As (d – x) = 2450 (600 – 350) = 612.5 x 103 mm3

I

S

recs

1

mm/1096.0107976

10664)15.8/200(104701 6

6

36

sr

mm/1082.0108575

105.612)15.8/200(104701 6

6

36

sur

mm/1095.01082.0)95.01(1096.095.0

1)1(

11

666

ucn rrr



Step 4– Calculate deflection

Total curvature = 3.07 x 10-6 + 0.95 x 10-6 = 4.02 x 10-6 /mm

For a simply supported slab, k = 0.104

δ = kL2(1/r)

= 0.104 x 9500 2(4.02 x 10-6)

= 37.8 mm


session 3 - sls cg apr 2012a

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