507 39 solutions-instructor-manual ch12 drcs

8/12/2019 507 39 Solutions-Instructor-manual Ch12 DRCS

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Design of RC Structures Dr. N. Subramanian

1

Chapter 12

Exercise 12.1 Calculation of transformed section properties

A reinforced concrete beam of span 6 m, has a rectangular cross-section as shown in Fig. 12.

23. Assuming M 25 concrete and Fe 415 steel, compute the moment of inertia of bothuncracked and cracked transformed section.

Fig. 12.23 Beam of Exercise 12. 1

Solution

Step 1: M ater ial proper ties for M 25 concrete

Modulus of elasticity (cl.6.2.3.1) 2/000,252550005000 mm N f E ck c

Modular ratio, m = 825000/102/ 5c s E E

Note: Modular ratio as per cl.B-2.1.2 and Table 21 of code is cbcm 3/280

98.10)5.83/(280 (greater than 8 as this takes into account the effects of creep)

Modulus of rupture MPa f cl f ck cr 5.3257.07.0)2.2.6.(

Step 2: Approximate cracki ng moment (assuming gross section)

Section modulus 3622

1067.66

4002506

mmbD

Z

Cracking moment, kNm Z f M cr cr 66 1033.231067.65.3

Step 3: Uncr acked transformed section

Area of tension steel22 18474/283 mm A

st


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2

Area of compression steel 22 9424/203 mm A sc

Transformed area of tensile steel = (m-1)A st = (8-1)1847 = 12,929 mm2

Transformed area of compression steel = (m-1)A sc = (8-1)942 = 6,594 mm 2

Note: The compression area of steel is taken as (m-1)A sc and not as (1.5m-1)A sc asconsidered in stress calculation, because the factor 1.5m, which takes into account creepeffects is not applicable in short term deflection calculations.

The centroid of transformed section can be located as shown in Table 12.20

Table 12.20 Determination of centriod

Part Area (mm 2) y top (mm) Ay top (mm2)

Concrete 250 x 400 = 100 x 10 3 200 20 x 10 6

Compressionsteel

6,594 48 0.3165 x 10 6

Tensionsteel

12,929 348 4.5 x 10 6

A= 119,523 Ay top=24.817 x 106

mm ytop 63.207119,52310817.24 6

The moment of inertia of the section can be calculated as shown in Table 12.21 (Note that themoment of inertia of the steel layers about their centroids are negligible).

Table 12.21 Moment of Inertia of uncracked transformed section

Part Area(mm 2)

)(mm y I own axis (mm ) 42 mm y A

Concrete 100 x 10 200-207.63 = -7.63

93 1033.112

bD

5.82 x 10

Compression steel 6,594 207.63-48 =159.63

- 168.03 x 10

Tension steel 12,929 207.63-348 = -140.37

- 254.75 x 10

2

y A I I own gr =1.7586 x 109

= 428.6 x 10


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3

Note that the M.I. of uncracked transformed section is 32% larger than the gross moment ofinertia of concrete alone.

Step 4: Cracked transformed section

Let us assume that the neutral axis in this case, is below compression steel. The transformedareas of steel are as below:

Compression steel (m-1)A sc= (8-1) 942= 6594 mm2

Tension steel mA st = 8 x 1847= 14776 mm 2

The centroid is calculated as below, assuming depth of neutral axis as x

Table 12.22 Depth of neutral axis of cracked transformed sectionPart Area (mm 2) )(mm y )( 3mm y A

Compression zoneof concrete

250 x x/2 200 x2/2 = 125x 2

Compression steel 6594 x-48 6594 x - 316.51 10 3

Tension steel 14776 x-348 14776 x - 5.142 106

By definition, at x, 0 y A . Hence we have,

04366896.1702 x x

Thus, mm x 3.1402

55.45196.1702

43668496.17096.170 2 (from top fibre)

Moment of inertia of the cracked section may be computed as shown in Table 12.23.

Table 12.23 Moment of Inertia of uncracked transformed section

Part Area(mm 2)

y(mm) I own axis (mm4) Ay 2(mm 4)

Compression zone ofconcrete

250 x140.3 =35075

140.3 /2 = 70.15 63 10535.5712bx 172.605 x10

Compression steel 6594 140.3-48 = 92.3 - 56.176 x10


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4

Tension steel 14776 140.3-348 = -207.7

- 637.426 x10 6

462 10742.923 mm Ay I I ownaxiscr

In this beam, I cr is approximately 52.5% of the moment of inertia of uncracked transformedsection and 69.5% of the concrete section alone. This indicates that the stiffness reducesconsiderably due to cracking.

Exercise 12.2

A one-way slab of effective span 4.2 m is subjected to a total load, inclusive of self weight, of10 kN/m 2 and is reinforced with 10 mm bars at 125 mm c/c in the short span and withdistributors of 8 mm bars at 200 mm c/c. if the total depth of slab is 200 mm with effectivedepth 165 mm, calculate the maximum short-term deflection as per IS 456, assuming M20concrete and Fe 415 steel.

Solution

Step 1: Calcul ate bending moments

Span, L = 4.2 m, f y = 415 MPa, f ck = 20 MPa, working load = 10 kN/m2, D =200 mm, d = 165

mm

Maximum bending moment at mid-span

= wL 2/8 = kNm05.228/2.410 2 per m width

Step 2: Calcul ation of m, E c and f cr

Modulus of Elasticity of concrete, E c (clause 6.2.3.1)

= MPa f ck 223602050005000

Es =2 x 105 MPa, Hence, Modular ratio, 9.8

22360102 5

c

s

E E

m

Modulus of rupture f cr (clause 6.2.2) = 2/13.3207.07.0 mm N f ck

Step 3: Check f or Cr acking moment

Moment of inertia, 4633 10667.66612/200100012/ mmbD I gr


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6

= < L/250 = 4200/250 =16.8 mmHence the slab is safe.

Just for comparison, let us calculate the deflection based on gross sectionMaximum short-term deflection, assuming uncracked section

= Thus we see that, in this case, the deflection calculated based on cracked section is 4.5 timeslarger than the value calculated based on uncracked section!

Exercise 12.3

If the beam shown in Exercise 1 is subjected to a superimposed service load (including deadload) of 17.5 kN/m and a central concentrated load of 30 kN, calculate the instantaneous

deflection as per IS 456 and ACI 318 methods [Hint: ,481

3845 34

eff eff EI PL

EI wL

where w is

the udl and W is the concentrated load]

Solution:

From Exercise 12.1, L = 6m, gr I =1.7586 x 109 mm 4 and I cr = 923.742 x 10 6 mm 4, M cr =23.3

kNM, x = 140.3 mm, Service load, w = 17.5 kN/m and P = 30 kN,2/000,252550005000 mm N f E ck c

Step 1: Determine effective moment of inertia, I eff

Maximum moment, M = wL 2/8 +PL/4 =

f

wcr

cr eff

b

b

d

x

d

z

M

M I

I

12.1

with gr eff cr I I I , z = d-x/3 = 348 140.3/3 = 301.23 mm

K = f

w

bb

d x

d z

1 =

mm 4 < I cr

Hence I eff = mm 4

Step 3: Determine the short-term deflection as per IS 456The sort-term deflection as per IS 456


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7

eff eff EI PL

EI wL 34

481

3845

= ( ) = 12.78 + 5.85 = 18.63 mm < 6000/250 = 24 mm

Step 4: Short term deflection as per ACI 318

gr cr a

cr gr

a

cr eff I I M

M I

M M

I

331

= (11.78 + 917.55) x 10 6 = 929.33 x 10 6 mm 4> Icr

The sort-term deflection as per ACI

eff eff EI PL

EI wL 34

481

3845

= ( ) = 12.71 + 5.81 = 18.52 mm < 6000/250 = 24 mm

Exercise 12.4

For the beam of exercise 12.3, determine the maximum long-term deflection due to creep andshrinkage and check whether the total deflection is within code limits. Assume ultimate

shrinkage strain of 0.0004 and ultimate creep coefficient of 1.6.

Solution:

Given : From Exercise 12.1 and 12.3, A st = 1847 mm2 and A sc = 942 mm

2 , b = 250 mm, d=348 mm, I eff = mm 4, C t = 1.6, and 0004.0cs

Step 1: Deflection due to shrinkage (clause C-3.1) for simply supported beams

2125.0 L sh sh

Where D

k cs sh

4 where 0004.0cs


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9

icpicp

,

eff cceff cccpi I E

PL I E

wL 34, 48

1384

5 = ( )

= 17.46 + 7.98 = 25.44 mm

18.63 (from Exercise 12.3)

Hence, Total long-term deflection = 18.52 + 2.09 + 6.92 = 27.53mm

Permissible deflection [Cl. 22.2(a)] = L/ 250 = 6000/250 = 24 mm < 27.53 mm

Note: In this case, permissible deflection is slightly less than the calculated long-termdeflection (14.7%). There is no need to revise the section, as we have assumed in thecalculation that all the applied imposed loads are permanent loads. If we assume that only60% of imposed loads only are sustained, the deflection will be within limits. w

Exercise 12.5

For the one-way slab of Exercise 2, determine the long-term deflection due to creep andshrinkage and check whether the total deflection is within code limits.

Solution:From Exercise 12.2, L = 4.2 m, gr I =666.667 x 10

6 mm 4, M cr =20.87 kNM, M= 22.05 kNm,

Service load, w = 10.0 kN/m 2/360,222050005000 mm N f E ck c

Step 1: Deflection due to shrinkage (clause C-3.1) for simply supported beams

2125.0 L sh sh

Where D

k cs sh

4 . Assuming 0003.0cs

; p c =0

p t - p c =0.38

Hence 0.172.04 t

ct

p

p pk for 0.25 p t pc < 1.0


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10

438.038.0

38.072.04 k

71057.6200

0003.0438.0 sh

mm sh 45.142001057.6125.0 27

Step 2: L ong term deflection due to creep

As per clause C-4.1 of IS 456

Let us assume that the age at loading as 28 days. Hence from clause 6.2.5.1, C t = 1.6

2/8600

6.11

22360

1

mm N

C

E E

t

ccc

3108.31651000

628 bd A st ,

From Table 12.3

[( ) ] = ( )-1] = 56.33 mm2

3

)(3

xd mAbx

I st cr

= 23 )33.56165(62826.23)33.56(3

1000

= 173.56 x 106 mm

4 < Igr = 666.667 x 10

6 mm

4

Maximum moment, M = wL 2/8 +PL/4 =

f

wcr

cr eff

bb

d x

d z

M M

I I

12.1


K =

f

w

bb

d x

d z

1 =


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11

mm 4 < Igr

icpicp

,

eff cccpi I E

wL4, 384

5 = = 17.58 mm 12.16 (from Exercise 12.2)

Hence, Total long-term deflection = 12.16 + 1.45 + 5.42 = 19.03 mm

Permissible deflection [Cl. 22.2(a)] = L/ 250 = 4200/250 = 16.8 mm < 19.03 mm

Hence it may be necessary to increase the depth or provide compression reinforcement, inorder to keep the long term deflection within limits. Also note that we have assumed in thecalculation that all the applied imposed loads are permanent loads. If we consider only 60%of imposed loads as permanent, the calculated deflection will be smaller.

Exercise 12.6

The interior span of a continuous beam is shown in Fig. 12.24. The beam has the following basic dimensions: b = 380 mm, D = 700 mm, d = 640 mm, and d = 60 mm. Negativereinforcement at left support is 5 32 dia. bars (area = 4021 mm 2) and compression steel is 2-

32 dia. bars (1608 mm2

). Negative reinforcement at right support is 4 32 dia. bars (area =3216 mm 2) and compression steel is 2- 32 dia. bars (1608 mm 2). The span is 6 m and thesuperimposed load, inclusive of self weight is 175 kN/m. The end moment at left support is500 kNm and at right support is 400 kNm. Calculate the deflection at service load using M 25concrete and Fe 415 grade steel.

Fig. 12.24 Continuous beam of Exercise 12.6

Solution

Step 1: Calculation of I gr and M cr

Youngs modulus of concrete, MPa f E ck c 250002550005000

Modular ratio, 825000/102/ 5 c s E E m

The gross moment of inertia of the section at mid-span and end-span is same and is


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12

4633

1067.861,1012

70038012

mmbD

I gr

Effective depth, d = 700 -60 = 640 mm

d= 60 mm

Step 2 Calcul ate M cr and I cr at mid-span

2/5.3257.07.0 mm N f f ck cr

kNm Nmm y

f I M

b

cr gr cr 62.1081062.108)2/700(

5.31067.861,10 66

Moment of inertia of cracked section at mid-span

Ast = 4 x 804 = 3216 mm2

Asc = 2 x 804 = 1608 mm2

0132.0640380

3216bd A st , 0066.0

6403801608

' bd A sc

094.0640/60/ d d

From Table 12.3 of the book

)1(1(21 2 mmd d

mmmmd x

0066.070132.08094.00066.070132.0820066.070132.08 2 = 0.4928 - 0.1518 = 0.341

Neutral axis depth, x = 0.341 x 640 = 218.24 mm

223 )()1()(3

d x Am xd mAbx I sc st cr

= 223 )6024.218(16087)24.218640(32168)24.218(3

380

= 6175 x 10 6 mm 4 < Igr = 10861.67 mm4.

Step 3 Calcul ate I cr at Suppor t

Ast = 5 x 804 = 4020 mm2


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13

Asc = 2 x 804 = 1608 mm 2

0165.0640380

4020bd A st , 0066.0

6403801608

' bd A sc

094.0640/60/ d d


)1(1(21 2 mmd d

mmmmd x

0066.070165.08094.00066.070165.0820066.070165.08 2 = 0.5518 - 0.1782 = 0.3736

Neutral axis depth, x = 0.3736 x 640 = 239.08 mm

223

)()1()(3

d x Am xd mAbx

I sc st cr

= 223 )6008.239(16087)08.239640(40208)08.239(3

380

= 7261.25 x 10 6 mm 4 < I gr = 10861.67 mm4.

Step 4 Calcul ate Service Level B.M .

Total load on beam = 175 kN/m

Using Table 12 of IS 456 to determine the bending moments

At middle of interior span

B.M. = kNm75.39316

6175 2

B.M. at the left end = 500 kNm (given)

B.M. at the right end = 400 kNm (given)

Step 5: Cal cul ate ef fective moment of inertia

(a) At mi d-spanLever arm distance, z = d-x/3 = 640-218.24/3 = 567.25 mm



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14

bb

d x

d z

M M

I I

wcr

cr meff

12.1, but I cr Ieff < I gr

= cr I

6

6

109.59431

64024.218

1640

25.56775.39362.108

2.1106175

(b) At suppor t in th e left endLever arm distance, z = d-x/3 = 640-239.08/3 = 560.3 mm

cr seff I I

66

, 108.67171

64008.239

1640

3.560500

62.1082.1

107261

(c) At suppor t in the right end

bb

d x

d z

M M

I I

wcr

cr meff

12.1,

but I cr I eff < I gr

= cr I

66

106.59291

64024.218

1640

25.5670.400

62.1082.1

106175

Step 5: Calcul ate average effective moment of i nerti a for end span

kNmwL M M f f 52512/617512/ 22

21

M1 = 500 kNm and M 2 = 400 kNm

The factor k 2 in Table 25 of IS 456 is found as

857.05252

400500

21

212

f f M M

M M k

Value of k 1 from Table 25 of IS 456 is 0.24

0121

1, )1(2 X k

X X k I aveff

= 66

109.5943)24.01(2

10)6.59298.6717(24.0

= 6106155 mm 4

Note: As per the weighted average method of ACI 318,


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15

4662,1,,, 108.605710)]6.59298.6717(15.09.59437.0[15.07.0 mm I I I I eff eff meff aveff Although both IS code and ACI code methods yield similar results, ACI method is simpler touse.

Step7: Calculate Short-term deflection

)312/(2.1910615525000

6000175384

5384

56

44

Lmm EI wL

eff

Hence the deflection is within allowable limits.

Exercise 12.7

Determine the maximum probable crack width for the one-way slab of Exercise 12.2.

Solution:From Exercise 12.2, f y = 415 MPa, f ck = 20 MPa, M = 22.05 kNm, D = 200 mm, d = 165mm, A st = 628 mm 2.

Step 1: Calculate cracked moment of inertia

As per Clause B-2.1.2(d) and Table 21 of IS 456,


st mAb

B = = 0.1195

[( ) ] = ( )-1] = 44.84 mm2

3

)(3

xd mAbx

I st cr

= 23 )84.44165(62833.13)84.44(3

1000

= 150.92 x 10 6 mm 4

Step 2: Calcul ate Str ain in concrete at the location where crack wi dth i s calcul ated

MPa Note: the stress just exceeds to the allowable stress of 230 MPa, as per working stressmethod.


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16

As per Annex F of IS 456,

45

2

1032.5)84.44165(6281023

)84.44200(1000)(3

))(('

xd A E xa x Db

s s

4431 1078.91032.51051.1'

m

Step 5- Cr ack width as per I S code

Now, s= 125 mm, clear cover = c min = 200 - (165 + 5) = 30 mm, c = 30 +5 = 35 mm

Distance from the surface of the bar to Point 2

a cr =[(0.5 s)2 + c 2]0.5 d b/2 = [(125/2)

2+ 35 2]0.5 -5 = 66.63 mm

Design surface crack width as per Annex F of IS 456 is

mm

kd Dca

aW

cr

mcr cr 133.0

)84.44200()3063.66(

21

1078.963.663

)()(

21

)3( 4

min

< 0.3 mm (Normal exposure)

Hence the crack width is within allowable limits.

Exercise 12.8For the T-beam designed in Example 5.27, calculate the following:a. Short-term deflection due to service loads

b. Long-term deflection due to creepc. Long-term deflection due to shrinkaged. Maximum possible crack width

Verify whether the calculated deflection and crack widths are within the code stipulatedlimits.

Solution:

The cross-section of T-beam as per Example 5.27 is given as below:


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17

Fig. 12.25 Beam of Example 5.27

Working load, w = 40 kN/m, L = 8 m, M = 40 x 8 2/8 = 320 kNm , f y = 415 MPa, and f ck = 20MPa

Step 1: Calculate modular ratio, modulus of rupture2/360,222050005000 mm N f E ck c

Es =2 x 105 MPa, Hence, Modular ratio, 9.8

22360102 5

c

s

E E

m



C .G. distance from bottom tension fibre = = 409mm

Moment of inertia,

= 9.209 x 10 9 mm 4

40910209.913.3 9

t

gr cr cr y

I f M = 70.47 x 10 6 Nmm /m = 70.47 kNm /m < M= 320 kNm

Hence the beam will crack and we need to calculate cracked moment of inertia.

Step 3: Calculate cracked moment of Inertia, I cr

( )

=100.59 mm


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18

( ) =

= 4980 x 106

mm4

< Igr = 9209 x 106

mm4


f

wcr

cr eff

bb

d x

d z

M M

I I

12.1


K = f

w

bb

d x

d z

1 =

mm 4 < I cr

Step 3: Determine the short-term deflection as per IS 456

The sort-term deflection as per IS 456

eff EI wL4

3845

= = 22.62 mm < 8000/250 = 32 mm(b) Deflection due to shrinkage (clause C-3.1) for simply supported beams

2125.0 L sh sh

Where D

k cs sh

4 . Assuming 0003.0cs

; p c =0

p t - p c =1.893

Hence 0.165.04 t

ct

p p p

k for p t pc 1.0

894.0893.1

893.165.04 k

74 10876.4550

0003.0894.0 D

k cs sh

mm sh 9.3800010876.4125.0 27


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19

(c ) L ong term deflection due to cr eep

Step 1: Determine Cracked moment of inertia


Let us assume that the age at loading as 28 days. Hence from clause 6.2.5.1, C t = 1.6

2/86006.11

223601

mm N C

E E

t

ccc

( )

=137.94 mm

( ) =

= 5529.6 x 10 6 mm 4 < Igr = 9209 x 106 mm 4


f

wcr

cr eff

bbd xd z M M

I I

12.1


K = f

w

bb

d x

d z

1 =

mm 4 < I cr

Step 3: Compute the deflection

icpicp ,


20/26


21/26


21


55

2

1069.8)113500(28391023

)113550(300)(3

))(('

xd A E xa x Db

s s

3531 10259.11069.810346.1'

m


Considering a point mid-way between the two 28 mm bars and assuming a side clear cover of32 mm at the ends, clear distance between bars = [300- (4 x 32 + 2 x 28)]/3 = 38.67 mm.Hence s = 38.67 + 28 = 66. 67 mm, clear cover = c min = 550 - (500 +28/2) = 36 mm, c = 36 +14 = 40 mm


a cr =[(0.5 s) 2 + c 2]0.5 d b/2 = [(66.67/2) 2+ 40 2]0.5 - 14 = 38.07 mm


mm

kd Dca

aW

cr

mcr cr 142.0

)113500()3607.38(

21

10259.107.383

)()(

21

)3( 3

min



Exercise 12.9

The Interior span of a typical level of multi-storey office building has L x and L y as 6.2 m and5.5 m. Evaluate the in-service vibration response of the floor and its acceptability for thefollowing data: thickness of slab, D = 180 mm; concrete strength, f ck = 25 MPa; Poissonsratio = 0.2; Live load = 4 kN/m 2

Solution

Let us assume simply supported boundary condition as it is more conservative

As per clause 6.2.3.1,


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22

ck c f E 5000 = 2/000,25255000 mm N

Ecd = 1.25 x 25,000 = 31,250 N/mm 2

g = 9810 mm/sec 2

r = L y/Lx = 6.2/5.5 = 1.127

Hence from Table 12.15, )127.11(57.1)1(57.1 221 r k = 3.564

Approximately 10% of the full live load may be considered for vibration calculation.

2/4.00.41.0 mkN w LL

Dead load, 0.12518.0 DLw (Partition load) = 5.5 kN/m2

mmmbd

I X /10486.0121801

1246

33

(i) Using approximate formula

mm Lmm slab 7.15350/34.92500010486.0620010)4.05.5(

3845

6

43

[Clause 23.2 (b)]

89.534.918

n f Hz > 5 Hz (Clause C-3 IS 800:2007)

(ii) Using Eqn. 12.41

sec/1019.162109.5)2.01(12

9810180250,31)1(12

2632

3

2

3

mmw

g D E c cd

Hz Hz k Lc

f y

n 504.156200564.31019.162

2

6

12

Hence the vibration characteristics of the slab are with in limits.

Exercise 12.10 Crack width in beams

A beam shown in Fig. 12.25 is subjected to a service bending moment of 106 kNm. Using M20 concrete and Fe 415 steel calculate the crack width at the following locations

a. At a point 100 mm below the neutral axis at the side of the beam b. At a point mid-way between bars at the tension facec. At the bottom corner


23/26


23

Fig. 12.26 Beam of Exercise 12.10

Solution

For the given beam:

Ast = 4 314 = 1256 mm 2, D = 530 mm, d = 490 mm , clear cover = c min = 530-490-10 = 30mm, s = 65 mm, M = 106 kNm,

Step 1: D etermi ne E c , m, and f cr

Modulus of Elasticity of concrete, E c (clause 6.2.3.1)

= MPa f ck 223602050005000

Es =2 x 10 5 MPa,

As per Clause B-2.1.2(d) and Table 21 of IS 456,



Moment of inertia, 4933 10102.312/53025012/ mmbD I gr

5305.010310213.3 6

t

gr cr cr y

I f M = 36.64 x 10 6 Nmm /m

= 36.64 kNm /m < M DL+LL = 106 kNm

Hence the secion is cracked. It has to be noted that the value of M cr calculated above is notconservative and some authors suggest that the value of M cr should be taken as 0.7 times thevalue calculated as per IS 456.

Thus M cr = 0.7 x 36.64 = 25.65 kNm

Step 3 : Calculation of Cracked moment of inertia


24/26


24

As per Table 12.3 of the book

Ast = 1256 mm2

Neutral axis depth

B Bd x /121 , where st mA

b B = = 0.0149

/0.0149 = 198 mmHence as per Table 12.3 of the book, the cracked moment of inertia,

23

)(3

xd mAbx

I st cr = = 2074.4 x 106mm 4

Step 4 Strain in concrete at the location where crack width is calculated (Point 2 inFig.12.25)

MPa


45 10252.1)198490(12561023

)198530)(198530(250)(3

))(('

xd A E xa x Db

s s

3431 10006.110252.110131.1'

m


Now, s= 65 mm, clear cover = c min = 530 - (490 + 10) = 30 mm, c = 30+10 = 40 mm


a cr =[(0.5 s)2 + c 2min]

0.5 d b/2 = [(65/2)2+ 40 2]0.5 -10 = 41.54 mm



25/26


25

mm

kd Dca

aW

cr

mcr cr 12.0

)198530()3054.41(

21

10006.154.413

)()(

21

)3( 3

min



(b) Crack width at Point 3 (corner)

Side cover = (250- 3 x 65)/2 =27.5 mm


a cr = =[(27.5)2+ 40 2]0.5 -10 = 38.54 mm


mm

kd Dca

aW cr

mcr cr 11.0

)198530()3054.38(

21

10006.154.383

)()(

21

)3( 3

min

< 0.3 mm


(c) Crack width at Point 1 (100 mm below NA)

Distance of the point from the extreme compression fibre = 169.24 + 100 = 269.24 mm

Strain in this location,

The strain reduction due to the stiffening effect if concrete in the tension zone

35 10027.0)198490(12561023

)19824.269)(198530(250)(3

))(('

xd A E xa x Db

s s

3331 10314.010027.010341.0'

m

Vertical distance between this point to the upper surface of the bar is = 490-10 -198 -100 =

182 mmHorizontal distance from the vertical surface of beam to center of bar

= Cover +d b/2 = 30 +10 = 40 mm

Vertical distance from Point 1 to centre of bar = 490-(198+100) = 192 mm

Hence a cr = - 20/2 =186.12 mmDesign surface crack width as per Annex F of IS 456 is


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507 39 solutions-instructor-manual ch12 drcs

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