507 39 solutions-instructor-manual ch11 drcs

8/12/2019 507 39 Solutions-Instructor-manual Ch11 DRCS

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Design of RC Structures Dr. N. Subramanian

1

SOLUTION MANUAL -CHAPTER 11

Exercise 11.1 Design of Flat plate

Design the interior panel of a flat plate supported on columns spaced at 6 m in both

directions. The size of column is 450 mm square and the imposed load on the panel is 3

kN/m2. The floor slab is exposed to moderate environment. Assume floor finishing load as 1

kN/m2and height of each floor as 3.2 m. Use M30 concrete and Fe 415 grade steel.

Solution

Step 1: Select thi ckness of slab

As the slab experiences moderate environment, choose cover = 30 mm (Table 16 of IS 456)

As per Clause 31.2.1 and 24.1 of the code,

L/D ratio with Fe 415 steel = 0.9 x (0.8 x 40) = 28.8

Minimum depth = mmmmspan

1252088.28

6000

8.28 (min. as per Code)

Let us assume a total depth of 210 mm

Assuming 12 mm dia. bars

Effective depth, d= 210 - 6 - 30 = 174 mm

Step 2: Calculate Loads

Self weight of slab = 0.21 x 25 = 5.25 kN/m2

Weight of finishing = 1.00 kN/m2

Imposed load = 3.00 kN/m2

-------------------

Total working load, w = 9.25 kN/m2

-------------------

Design factored load, wu= 1.5 x 9.25 = 13.88 kN/m2

Clear spacing between the columns, Ln= L1c1= 6.00.45 = 5.55 m

Total design load on the panel, kNLLwW nu 2.462655.588.132

Step 3: Calculate Bending Moments

Sum of the positive and average negative B.M. (Clause 31.4.2.2)

panelkNmWL

M no /65.3208

55.52.462

8

Interior Panel


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2

Panel negative design moment (Clause 31.4.3.2)

= 0.65 Mo=0.65 x 320.65 = 208.42 kNm/6 m

Panel positive design moment

= 0.35 Mo=0.35 x 320.65 = 112.23 kNm/6 m

As per clause 31.4.3.3, the relative stiffness of the columns and the slab determine the

distribution of the negative and positive moments in the exterior panel.

The building is not restrained against lateral sway, hence the effective height of column can

be taken as 1.2 times the clear height (Clause E-1 of IS 456). Assuming height of each floor

as 3.2 m, effective length of column is

L = HD = 3.20.21= 2.99 m

Le= 1.2 x 2.99 = 3.59 m

Relative stiffness of column

444

10519.959.312

45.0

12

ee

cc

L

h

L

IK

Relative stiffness of slab panel

433

10718.7612

21.06

12

sese

ss

L

bt

L

IK

47.210718.7

10519.924

4

s

c

cK

K

Imposed load / dead load = 3/(5.25 +1) = 0.48

L2/L1= 1; Hence from Table 17 of IS 456, c,min= 0

c> c,min; Hence stiffness is sufficient.

factor that decides the relative distribution of bending moment between negative andpositive bending moment as per Clause 31.4.3.3 of IS 456 is

405.147.2

11

11

c

End span

In an end span,

Exterior negative bending moment coefficient = 0.65 / = 0463

Interior negative bending moment coefficient = 0.750.1/ = 0.679

Positive bending moment coefficient = 0.630.28/ = 0.431


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3

The corresponding end panel bending moments

Negative bending moment at outer support = 0.463 x 320.65 = 148.46 kNm/panel

Negative bending moment at inner support = 0.679 x 320.65 = 217.72 kNm/panel

Positive bending moment in the panel = 0.431 x 320.65 = 138.20 kNm/panel

The bending moments are to be distributed between the column and middle strips as shown

below in Table 11.15 (clause 31.5.5)

Width of column strip = 0.5 x 6 = 3 m

Table 11.15 Bending Moments and Ast

Column Strip in kNm/3

m

Ast, mm Middle strip in

kNm/3 m

Astmm

Interior panel

Negative

moment

0.75 x 208.42 = 156.32 2706+

#12 @ 120c/c

52.1 902*

#10 @ 240 c/c

Positive

moment

0.60 x 112.23 = 67.34 1166

#12 @ 230c/c

44.89 777*

#10 @ 240 c/c

Outer Panel

Negative at

exterior

support

1.0 x 148.46 = 148.46 2570+

#12 @ 120

c/c

0 -

Negative at

inner support

0.75 x 217.72 =163.29 2808+

#12 @ 120c/c

54.43 942*

#10 @ 240 c/c

Positive at

panel

0.60 x 138.20 = 82.92 1435

#12 @ 230

c/c

55.28 957*

#10 @ 240 c/c

Step 4: Check slab depth for bending

The thickness of the slab is controlled by the absolute maximum bending moment. From the

above table, the critical bending moment occurs at the interior support of the outer panel

column strip and is

Mcip= 163.29 kNm / 3 m

The depth required to resist this bending moment


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4

dmmKbf

Md

ck

c

114

303000138.0

1029.1635.0

65.0

chosen = 174 mm

Hence the adopted depth in satisfactory and the slab is under reinforced.

Effective depth for upper layer of reinforcement = 17412 = 162 mm

Step 5: Design of reinforcement

Let us use this effective depth to calculate reinforcement

ck

yststyu

f

f

bd

AdAfM 187.0

Hence for maximum negative moment at the interior support of the outer panel column strip

30415

1743000117441587.01029.163 6 stst

AA

Simplifying, we get

01029.1637.62822665.1 62 stst AA

Solving we get Ast= 2808 mm2

The same result may be got by using design aids in Appendix C-Table C.3

798.117430001029.163

2

6

2

bdMu

From Table C.3 and for fy= 415 N/mm2

pt = 0.539; Ast= (0.539/100) x 174 x 3000 = 2814 mm2

We may also use the approximate formula

26

28271744158.0

1029.163

8.0mm

df

MA

y

ust

(0.7% increase over exact results)

From Table 96 of SP 16, provide 12 mm bars at 120 mm c/c at topface of slab over the

columns in the column strip (Area provided = 2826 mm2)

The reinforcement at different locations may be calculated by using proportionate spacing

compared with the other bending moments as shown below; for Example

Spacing of 12 mm bars at outer support at top = 120 x 163.29 / 148.46 = 132 mm (provide

120 mm)

Spacing at middle span at bottom = 120 x 163.29/82.92 = 236 mm, say 230 mm

Spacing at inner support at top = 120 x 163.29/156.32 = 125 mm (provide 120 mm)

Minimum reinforcement


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Fig. 11.50 Reinforcement details of Flat plate of Exercise 11.1

Exercise 11.2 Flat slab with drop panels

Design the interior panel of a building with flat slab roof having a panel size of 7 m x 7 m

supported by columns of size 600 mm x 600 mm. Take live load as 4.0 kN/m2and weight of

finishes as 1.0 kN/m2. Use M25 concrete and Fe 415 steel. Assume mild environment.

Solution

Step 1: Select thi ckness of slab


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7

For mild environment, minimum cover = 20 mm which can be reduced by 5 mm if we use

bars less than 12 mm (as per Table 16 of IS 456)

As per Clause 31.2.1 and 24.1 of the Code, for flat slab with Fe 415 steel

L/D = 0.8 x 40 = 32

Thus, total depth

= span / 32 = 7000 /32 = 218.75 mm, say 220 mm > 125 mm (minimum as per code)

Adopt a cover of 20 mm and assume 16 mm dia bars.

Hence effective depth = 220 - 25 - 8 = 187 mm

Adopt a depth of 220 mm and d =187mm

Step 2: Size of dr op panel

As per Clause 31.2.2 of code drop should not be less than 7000/3 = 2333 mm

Minimum depth of drop panel = mmDs 552204

1

4

1

Provide a drop panel of depth of 60 mm and size 3500 3500 mm.Take total depth at drop panel = 220 + 60 = 280 mm > 1.25 x 220 = 275 mm

Width of column strip = width of middle strip = 7000 / 2 = 3500 mm

Step 3: Calculation of loads

Self weight of slab = 0.22 x 25 = 5.5 kN/m2in middle strip

Dead load due to extra thickness of slab at drops = 0.06 x 25 = 1.5 kN/m2

Live load = 4.0 kN/m2

Finishes = 1.0 kN/m2

----------------

Total working load w = 12 kN/m2

Design factored load, wu= 1.5 x 12 = 18 kN/m

2

Clear span, Ln= 70.6 = 6.4 m

Design Load = W = wuL2Ln= 18 x 7 x 6.4 = 806.4 kN

Step 4: Calculation of Bending Moment

Design static moment, Mo(Clause 31.4.2.2) = kNmWLn 12.645

8

4.64.806

8

As per Clause 31.4.3.2

Total negative design moment = 0.65 Mo = 0.65 x 645.12 = 419.33 kNm


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Total positive design moment = 0.35 Mo= 0.35 x 645.12 = 225.80 kNm

The above moments are distributed into column and middle strips as shown in Table 11.16,

as per Clause 31.5.5.1 & 31.5.5.3

Table 11.16 Moment in column and middle strips Exercise 11.2

Type Column strip, kNm Middle strip, kNm

Negative moment 0.75 x 419.33 = 314.5 104.83

Positive moment 0.60 x 225.8 = 135.48 90.32

Step 5: Check slab depth for bending

Thickness of slab required at drops

mmbf

Md

ck

u 161350025138.0

105.314

138.0

5.06

5.0

Total depth provided at drop = 220 + 60 = 280 mm with effective depth = 247 mm and

effective depth provided at middle strip = 187 mm. Hence depth provided is adequate and the

slab is under reinforced.

Step 6: Check for punching shear

(a)The critical section is at a distance d/2 = 247 /2 = 123.5 mm from the face of thecolumnThe perimeter of critical section, b0 = 4(a + d) = 4 (600 + 247) = 3388 mm

The shear force on this plane is

kNdadaLLwV cu 1.869)847.07(18))(( 22

Nominal shear stress = 2

0

/04.12473388

10001.869mmN

db

Vu

Shear strength of concrete (Clause 31.6.3) = ksc

ks=0.5 +c< 1; c= 0.5/0.5 = 1; Hence ks=1

c= 1 x 0.25fck= 0.2525 = 1.25 N/mm2> 1.04 N/mm2

Hence the slab is safe in punching shear and no need to provide punching shear

reinforcement

(b)Shear strength at a distance d/2 from drop also has to be checked. It will be safe as thedrop size is large and hence the shear force at that section will be considerably

reduced.



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9

(a)Maximum negative reinforcement in column stripMu= 205.3 kNm, effective depth at drop = 247 mm

Hence 473.12473500

105.3142

6

2

bd

Mu

From Table 3 of SP 16 we get pt= 0.4404 2

38072473500100

4404.0mmAst to be

provided in 3500 mm width

Required spacing of 16 mm bars = (201 / 3807 ) x 3500 = 185 mm

Henceprovide 16 mm dia. bars at 180 mm c/c at the top face of the slabover the

columns in the column strip (Astprovided = 3909.5 mm2)

(b)For positive moment in column stripMu= 135.48 kNm, Effective depth of slab = 189 mm

26

21821874158.0

1048.135mmAst

Minimum 22 21829242203500100

12.0mmmmAst

Required spacing mmDmm 440218035002182

113

Provide 12 mm dia. bars at 180 mm c/c at bottom in the column strip

(c)For negative moment in middle stripMu= 104.83 kNm, Effective depth of slab = 187 mm

26

16891874158.0

1083.104mmAst

Required spacing for 12 mm bars = (113/1689) x 3500 = 234 mm < 440 mm

Provide 12 mm dia. bars @ 230 mm c/cat top in the middle strip

(d)For positive moment in middle stripMu= 90.32 kNm; Effective depth of slab = 187 mm

2145583.104/32.901689 mmAst ; Spacing of 12 mm bar = 271 mm

Provide 12mm dia. bars at 270 mmc/c at the bottom of middle strip

Since the span is same in both directions, same reinforcement may be provided in

both directions. The reinforcement detailing as per Fig.16 of IS 456 code is shown in

Fig.11.51.

Integrity reinforcement

Integrity reinforcement (Eqn.11.23)


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10

23

21 122141587.0

1077185.0

87.0

5.0mm

f

LLwA

y

us

Provide 2 numbers 28 dia. bars, with a length of 2Ld(2 x 1128 =2256 mm) passing through

the column cage each way.

Fig. 11.51 Reinforcement details of Flat Slab of Exercise 11.2

Exercise 11.3 Flat slab with drop and column head

Redesign Exercise 11.2 assuming the flat slab is supported by circular column of 500 mm

diameter and with suitable column head.

Solution

Step 1: Equivalent Column Head and Design Load

Let the diameter of column head = 0.25L = 0.25 x 7 = 1.75 m


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11

The circular section may be considered as an equivalent square of size a,

i.e., 2275.14

a

Solving, we get a = 1.55m

Hence Ln= 71.55 = 5.45 m

kNLLwW nu 7.68645.57182

Step 2: Calculation of bending moment

kNmWL

M no 8.4678

45.57.686

8

Total negative moment = 0.65 x 467.8 = 304 kNm

Total positive moment = 0.35 x 467.8 = 163.8 kNm

The distribution of the above positive and negative moments in the column and middle strip

is done as shown in Table 11.17.

Table 11.17 Distribution of positive and negative moments

Type Column strip, kNm Middle strip, kNm

Negative moment 0.75 x 304 = 228 kNm 76 kNm

Positive moment 0.60 x 163.8 = 98.3 kNm 65.5 kNm

Width of middle strip = width of column strip = 3500 mm

Step 3: Check depth f or bending

Required depth for resisting bending moment

mmbf

Md

ck

u 4.137350025138.0

10228

138.0

5.06

5.0

Effective depth provided at drop = 247 mm and at middle strip = 187 mm. Hence depth is

sufficient for bending and slab is under reinforced.


The critical section is at a distance of d/2 = 247/2 mm from the face of the column

Diameter of the critical section = 1750 + 247 = 1997 mm

Perimeter of the critical section = d = x 1997 = 6274 mm

Shear on this section kNVu 62.825

4

997.1718

22


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12

Nominal shear stress, 23

/533.02476274

1062.825mmNv

As per clause 31.6.3 of IS 456

cksc fk 25.0

175.1/75.1;15.0 ccsk ; Hence ks=1

22 /533.0/25.12525.0 mmNmmNc

Hence slab is safe in punching shear and no shear reinforcement is necessary.


The reinforcement at different locations has been calculated as per Table 11.18

Table 11.18 Reinforcement at various locations for Exercise 11.3

Bending moment Mu, kNm d, mm At, mm2 Diameter and

spacing of bars

Negative moment in

column strip

228 247 2780 12 mm dia at 140

mm c/c

Positive moment in

column strip

98.3 187 1583 10 mm dia at 170

mm c/c

Negative moment inmiddle strip

76 187 1224 10 mm dia at 220mm c/c

Positive moment in

middle strip

65.5 187 1055 10 mm dia at 260

mm c/c

Note:Minimum steel = (0.12/100) x 3500 x 220 = 924 mm

As the span is same in both directions, provide same reinforcement in both directions (of

course the effective depth will be smaller, for example, in the middle strip 187-12 = 175 mm

in the other direction). Reinforcement detail will be similar to that shown in Fig.11.51,including integrity reinforcement.

Exercise 11.4

Design slab reinforcement at exterior column of Exercise 11.1 for moment transfer between

slab and column and check for combined stresses. Also calculate the moments to be carried

by columns.

Solution

Step 1: Determine porti on of Muto be transferred by flexur e


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13

Moment at exterior column = 148.46 kNm.

It should be noted that as per ACI 318, this value is only 0.26 320.65 = 83.37 kNm. Hencewe will consider the moment as 83.37 kNm only in this Exercise, or otherwise we may get

uneconomical results.

From Exercise 11.1, we have, effective depth = 174 mm, total depth = 210 mm

Portion of unbalanced moment transferred by flexure ufM

From Fig.11.21 (Case c)

mmd

ca 5372

174450

21

mmdcb 6241744502

From Eqn.11.6, we get

618.0

624

537

3

21

1

3

21

15.05.0

b

af

kNmMuf 52.5137.83618.0

Effective transfer width = mmDc s 1080210345032

Step 2: Determine area of reinforcement

576.11741080

1052.512

6

2

bd

Mu

From Table 3 of SP16, for Fe 415 steel

pt = 0.475; At = (0.475/100) x 1080 x 174 = 893 mm2

We need 9 Nos 12 # bars @ 135 mm c/c

We have already provided 12 # bars at 120 mm c/c in the column strip. Hence no additional

reinforcement is necessary.

Step 3: F raction of unbalanced moment carr ied by eccentr ic shear

382.0618.011 fv

kNmMuv 85.3137.83382.0

Step 4: Properties of cr itical section for shear

bo=(2a + b) = (2 x 537 + 624) =1698 mm

2452,295174)6245372()2( mmdbaAc


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14

From Table 11.7 of the book,

a

badbada

c

J

6

)2()2(2 32

=5376

)452,295(174)6242537(1745372 22

=58. 372x106 mm4,

Similar result can be obtained by using Table 11.9.

With c1/d = 450/174 = 2.59 and c2/c1= 1, we get f2 = 5.5678

J/c = 2f2d3 = 2 x 5.5678 x 1743 = 58.663 x 106 mm4

Step 5: Check for combined stresses

Gravity load shear to be carried by exterior column

kNLLw

V uu 84.2492

6688.13

2

21

Combined stresses (Eqn.11.13)

6

6

max,10663.58

1085.31

452,295

100084.249

J

cM

A

Vuv

c

uv = 0.846 + 0.543 = 1.389 N/mm

2

2

min, /303.0543.0846.0 mmNv

Design punching shear stress (Eqn.11.12) ckc f25.0 =1.25 N/mm2< 1.389 N/mm2

Hence the slab is not safe to transfer the combined stresses.

With shear studs as shear reinforcement,

= 332.38 kN

Thus,

249.84/0.75-332.38 = 0.74 kN

Hence provide nominal shear studs. Use eight stud rails [as shown in Fig. 11.27(a) of the

book] each with three 10 mm diameter studs with 32 mm diameter heads. The spacing of

shear studs should be less than d/2= 174/2 =87 mm. Use 85 mm spacing.

Area provided by the inner row of shear studs Av= 8 x Ab= 8 x 78.5 =628 mm2

Assuming only one line of shear studs are crossed by potential critical shear crack nearest the

column

Note that the maximum value of Vnallowed with headed shear studs is


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664.7 kN > 249.84 kN

Step 6: F actored moments in columns

(a)For interior columns, using Eqn.11.8

)/11'''5.008.0 2222

c

ndnldu

LLwLLwwM

Since the spans are equal, Ln= Ln. Thus, we get

)/11(

)5.0(08.0 2

2

c

nlu

LLwM

We already have 1+1/c= 1.405 and wl= 3 kN/m2, L2 = 6 m, and Ln= 5.5 m from

Exercise 11.1.

Hence kNmMu 5.15405.1

)5.5635.0(08.0 2

With same column size and length above and below the slab

Mc=15.5/2 = 7.75 kNm

This moment should be combined with factored axial load (for each storey) for the design

of interior columns.

(b)Exterior columnsTotal exterior negative moment from slab must be transferred directly to the column.

Mu=83.37 kNm

Mc=83.37/2 = 41.69 kNm

This moment is to be combined with the factored axial load (for each storey) for the

design of exterior columns.

Exercise 11.5

Assume that a corner column of size 400 mm x 800 mm supporting a 175 mm thick flat slab

with effective depth = 150 mm. Factored moment due to gravity loads at the face of thecolumn is 80 kNm and factored shear force at the face of the column is 200 kN and shear

force at the edge is 25 kN. Check the stresses due to combined forces assuming that the slab

is made using M25 concrete and Fe 415 steel.

Solution

We have c1= 400 mm, c2= 800 mm and d =150 mm

The use of Eqn.11.13 requires the calculation of shear and moment relative to the centroid of

the critical section (see Fig. 11.52)


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16

Fig. 11.52 The edge column and critical section

mmdcb 4752/1504002/11

mmdcb 95015080022

Perimeter of critical section

mmbbb 190095047522 210

Distance from centroid to extreme fibre of critical section (Alexander and Simmons 2005)

mmb

bC

o

AB 75.1181900

47522

1

mmCbC ABCD 25.35675.1184751

mmd

Cd ABe 75.437575.1182

Polar moment of the shear perimeter

2

0

3

1

3

1

63

2ABdCb

dbdbJ

= 233

75.11815019006

150475

3

1504752

= 6.965 x109 mm4

Step 2: M oment at centr oid

Moment at centroid of critical section is given by

eedgeuespanufaceuu d

cVdVMM

2

1,,,

= 80 + 200 x 0.0437525 (0.4/20.04375)


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17

= 80 + 8.753.91 = 84.84 kNm

Shear at the centroid, Vu= 200 + 25 = 225 kN

Shear stress

J

CM

db

V ABuv

o

uv

68.0

950

475

3

21

1

3

21

15.05.0

2

1

b

bf ; 32.068.01 v

Hence 29

63

/252.1463.0789.010965.6

75.1181084.8432.0

1501900

10225mmNv

Design punching shear stress 22 /252.1/25.12525.0 mmNmmNc

Hence slab is safe to carry the shear and bending moment. Let us check the stress using the

approximate formula (Eqn.11.17)

dcc

dMV faceuuv

)(

)4/(65.0

21

,

=

150)800400(

)1504/(10801022565.0 63

= 1.294 N/mm2

Note:The approximate method is simple to apply and predicts the stresses with reasonable

accuracy (in this case, the percentage of error is only 3.35%)

Exercise 11.6

A flat plate panel of dimension 5 m x 6 m, supported by columns of size 450 mm x 450 mm

has a slab thickness of 150 mm and designed for a working (total) load of 9 kN/m2. Check the

safety of the slab in punching shear and provide shear reinforcement, if required. Assume

M20 concrete and Fe 415 steel.

Solution

Assuming that 10 mm dia. bars are used and the cover is 25 mm

Effective depth, d = 150255 = 120 mm

Factored design load = 1.5 x 9 = 13.5 kN/m2


Critical perimeter bo = 4 (c + d) = 4(450 + 120) = 2280 mm

kNVu 6.400)12.045.0(565.13 2

Stress due to punching shear 23

/464.1

1202280

106.400mmN

db

V

o

u


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18

Allowable punching shear stress

cksc fk 25.0

1)5.0( cs

k ; 1450

450

c ; Hence k

s=1

22 /464.1/118.12025.01 mmNmmNc

Hence shear reinforcement has to be provided.

Note:Increasing the strength of concrete may also solve this problem. Thus 464.125.0 ckf

or 22

/3.3425.0

464.1mmNfck

Hence we may have to use M35 concrete to keep the punching shear stress within limits.

Step 2: Check suitabil ity of shear rein forcement

As per clause 31.6.3.2 of the code, check whether stress is below 1.5c

22 /464.1/677.1118.15.15.1 mmNmmNc

Hence shear reinforcement can be used.

Note: If stress exceeds 1.5c, we need to increase the depth of the slab.

Step 3: Shear to be carr ied by rein forcement (Clause 31.6.3.2)

Shear stress assumed to be carried by concrete = 2/559.0118.15.05.0 mmNc

Shear force carried by concrete = 0.559 x bod = 0.559 x 2280 x 120/103= 152.94 kN

Shear to be carried by reinforcement = 400.60152.94 = 247.66 kN

Note: As per ACI, shear force carried by concrete =

kNdbf ock 6.13710/12022802075.015.015.0 3

Step 4: Total area of shear sti r rups (Eqn.11.18)

Maximum spacing of stirrups = 0.75d (SP 24-1983) = 0.75 x 120 = 90 mm

Adopt sv= 60 mm d/2 as suggested by ACI 318

Assuming fy= 415 N/mm2

233

343120

60

41587.0

1066.247

87.0

1066.247mm

d

s

fA v

y

sv

Required stirrup area on each side of column = 343/4 = 86 mm2

With 2 legged Ustirrups as shown in Fig.11.25 (c) and (d) of the book


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19

Area of each leg = 86/2 = 43 mm2

Provide 2 legged 8 mm dia. stirrupswith area 50.2 mm2

Alternatively, using Table 62 of SP-16, we get (per each side of column)

cmkNd

Vu /16.512

4/66.2474/

Fe 415 grade 8 mm dia. stirrups at 70 mm c/c give 5.185 kN/cm

Note that spacing between adjacent legs should not excess 2d = 2 x 120 = 240 mm

Step 5: Length up to which sti r rups are to be provided

As per SP 24:1983, shear reinforcement should be provided up to a section where shear stress

does not excess 2/559.0118.15.05.0 mmNc or where )5.0( cou dbV

Let this distance be a from the face of the column [see Fig. 11.25(d) of the book]

For a square column

)2450(4 abo

Thus, 559.0120)]2450(4[106.400 3 a

Solving 10432a mm or a= 738 mm

Provide the first stirrups at 60 mm from each face of the column and provide 11 sets of 8 mm

dia. stirrups of width 300 mm, depth 100 mm, spaced at 70 mm c/c at each face of column

[similar to the arrangement given in Fig.11.25(d) of the book].

Exercise 11.7

For the slab in Exercise 11.6, design headed shear studs as shear reinforcement, instead of

shear stirrups.

Solution

Step 1: I ni tial area of shear studs

ACI code provisions are used as IS code does not have provisions for headed shear studs.

From Exercise 11.6,

Size of column = 450 mm x 450 mm, d = 120 mm, bo= 2280 mm, fck= 20 N/mm2, Factored

design load = 1.5 x 9 = 13.5 kN/m2, Vu= 400.6 kN

Shear force carried by concrete =

kNdbf ock 48.20610/12022802075.01225.0225.0 3

Shear force to be carried by headed shear studs = 400.60206.48 = 194.12 kN

The maximum value of shear allowed with headed-shear studs is =


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20

60.40015.73410/1202280206.06.0 3 kNdbf ock kN

Hence chosen column size and slab depth can be used.

From Exercise 11.6, 22 /509.12075.045.045.0/464.1 mmNfmmNdb

V

cko

u

Adopt spacing of studs = 0.75d = 0.75 x 120 = 90 mm

Assuming fyt= 415 N/mm2

233

2.403120

90

41587.0

1012.194

87.0

1012.194mm

d

s

fA v

y

sv

Headed stud area for each side of column = 403.2/4 = 100.8 mm2

With 2 studs at a section as shown in the plan of Fig. 11.27(a) of the book,

Area of each stud = 100.8/2 = 50.4 mm2

Provide 8 mm diameter stud of length 100 mm(Area = 50.26 mm2)

Step 2: Length up to which studs are requi red

At the outer critical section, the shear stress should be less than 2515.0

)2450(4 abo

Thus, 2075.015.0120])2450(4[106.400 3 a

5.241)2450(106.400 3 a

Solving a= 854 mm

Provide 10 sets of 8 mm diameter headed studs on each face, each with a length 100 mm,

head diameter of 25 mm, at a spacing of 90 mm, with g = 200 mm ( 2d); place the first setof headed stud at a distance of d/2 = 60 mm from the face of the column . The arrangement

should be as shown in Fig.11.27 of the book.

Step 3: Check shear strength at the inner cr itical section

The outermost studs are at a distance of 870 mm [60 + 9 x 90] from the face of the column

and outer critical section is located at d/2 outside the outer shear stud.

The perimeter of this peripheral line

( ] mm

The area inside this line =[

mm2= 3.2823 m2

The factored shear force at the critical section =


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21

Shear stress =

0.447 N/mm

2

This shear stress on the outer critical section should be limited to

N/mm2> 0.447 N/mm2

Hence the slab is safe.

Exercise 11.8

Design a waffle slab for an internal panel if a floor system that is constructed in a 6.3 m squaremodule and subjected to a live load of 3 kN/m

2. Use M25 concrete and Fe 415 steel. The slab is to be

supported on square columns of size 750 mm x 750 mm and constructed using removable forms ofsize 900 mm x 900 mm x 300 mm. The slab is subjected to mild exposure.

Solution

Step 1: Proporti oning of the slab elements

As per clause 24.1 and 31.2.1 of IS 456, L/d = 0.9(0.8 x 40) = 28.8

Required minimum effective depth = 6300 /28.8 = 219 mm

Assume that the waffle slab is as shown in Fig.11.53.

Fig. 11.53 Waffle slab of Exercise 11.8


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22

Let us choose a GFRP dome pan of size 900 mm x 900 mm, with void plan of 750 mm x 750

mm, joist (rib) width 150 mm and depth 300 mm.

Let us provide 50 mm thick top slab.

As per Clause 30.5 of IS 456

Rib width should be greater than 65 mm; provided 150 mm c/c of ribs should be less than 1.5 m ; provided 0.9 m Depth of rib: Less than 4 times width, i.e. 4 x 150 = 600 mm; provided 300 mm. Depth of topping: Slab greater than 50 mm or less than (1/10) x 900 = 90 mm (Table

3.17 of BS 8110)- provided 50 mm.

Total depth = 350 mm > 219 mm

Hence proportions chosen are as per code requirements.

From Table 16 of IS 456, cover for mild exposure = 20 mm

Let the solid portion of slab over column be three modules wide, i.e., 3 x 900 + 150 = 2850

mm wide in both directions.

Step 2: Calculation of loads

Self weight

f

wf

f

wfcs

b

bDD

b

bDw 2

=

900

150

1000

75350

900

1502

1000

7525 = 3.98 kN/m

2

Weight of solid head = 0.35 x 25 = 8.75 kN/m2

Hence self weight of slab = 2/14.63.6

)85.23.6(98.385.275.8mkN

Finishes = 0.86 kN/m2(assumed)

Imposed load = 3.00 kN/m2

Total load w = 10.00 kN/m2

Factored load, wu= 10.00 x 1.5 = 15.0 kN/m2

Step 3: Calculate bending moments

The direct design method is applicable for this system. Imposed load/ dead load = 3.0/7.0 =

0.43

L2/L1= 6.3/6.3 = 1.0

Hence from Table 17 of IS 456, c,min= 0

Hence effect of pattern load need not be considered.


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23

mLLwL

M nn 55.575.030.6;

8

2

20

Thus, kNmM 3578

5.53.60.15 2

0

Column strip

Negative bending moment = - 0.65 x 0.75 x 357 = 174 kNm/3.6m

Positive bending moment = 0.35 x 0.60 x 357 = 75 kNm for 4 ribs

Middle strip

Negative bending moment = -0.65 x 0.25 x 357 = 58 kNm/3.6m

Positive bending moment = 0.35 x 0.40 x 357 = 50 kNm/3.6m

Step 4: Check for punching stress

(a)In the solid head portionAssuming 12 m bars, the effective depth = 350206 = 324 mm

bo=4(c1+d) = 4 (750 + 324) = 4296 mm

kNVu 6931000

3247503.60.18

2

2

2

0

/498.03244296

1000693 mmNdbVu

v

cksc fk 25.0 with ks= 1 for square column

= 22 /498.0/25.12525.01 mmNmmN

Hence safe in punching shear

(b)Shear stress at the ribsNominal shear stress is checked at a distance d from the edge of solid head

Width of solid head = 3 x 0.9 + 0.15 = 2.85 mon both directions

bo = 4(2.85 + 0.324) = 12.696 m

Total shear at this section = [6.32- (2.85 + 0.324)2] x 18.0 = 539 kN

This shear is resisted by 12 ribs of 150 mm width and 324 mm effective depth

Hence 22 /25.1/924.032415012

1000539mmNmmNv

Hence the adopted waffle slab is safe in punching shear


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24

Note: Punching shear reinforcement in waffle slabs should be avoided.

Step 5: Check depth to resist bending moment

mmmmkbf

Md

ck

324133252850138.0

101745.0

65.0

Hence the depth adopted is adequate and slab is under reinforced


The bending moment of 174 kNm is to be shared by 4 ribs. Hence bending moment per rib =

174/4 = 43.5 kNm

For M25 concrete and Fe 415 steel, 45.32

lim, bd

Mu(Table D of SP-16)

HenceMu,lim= 3.45 x 150 x 3242/106 = 54.3 kNm > 43.5 kNm

Hence the ribs can be designed as singly reinforced (underreinforced rectangular section)

MPabd

Mu 763.2324150

105.432

6

2

From Table 3 of SP 16, we get for Fe 415 and M25 concrete, pt= 0.901,

2438324150

100

901.0mmAst

Provide 2 Nos 20 dia. bars with area = 628 mm2

For positive bending moment,

Moment in each rib = 75 / 4 = 18.75 kNm

MPabd

Mu 19.1324150

1075.182

6

2

From Table 3 of SP 16, we get pt= 0.35

217032415010035.0 mmAst

Provide 2 numbers of 12 dia. bars with area = 226 mm2

Note:for positive moment, the ribs may be considered as T-beams to reduce the

reinforcement.

Design of ribs in the middle strip

For negative moment

Only 3 ribs carry this moment. Hence moment in each rib = 58/3 = 19.33 kNm

Henceprovide 2 Nos 12 dia. bars.


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25

For positive moment, each rib carries 50 / 3 = 16.67 kNm

Henceprovide 2 Nos 12 dia. bars.

Check for shear in r ibs

Maximum shear force at ddistance for support,

mkNdLwV nuu /34.43)342.05.55.0(0.18)5.0(

Shear force carried by rib = 43.34 x 0.9 = 39 kN

Nominal shear stress, 2/80.0324150

100039mmN

bd

Vuv

%292.1324150

628100100

bd

As

From Table 19 of IS 456 for M25 concrete 2/707.0 mmNc

22

max /80.0/1.3 mmNmmNc

Hence spacing of 8 mm stirrups

324150707.0100034.43

3242.5041587.087.0

bdV

dAfs

cu

svy

v

= 654 mm

Maximum spacing of stirrups (Clause 26.5.1.5) = 0.75 x 324 = 243 mm or 300 mm

Hence provide 8 mm dia. stirrups at 240 mm c/c

The reinforcement details for the rib in the column strip are shown in Fig.11.54 (for the sake

of clarity slab reinforcements are not shown in the figure).

Reinforcement in topping slab

Minimum reinforcement has to be provided in the 50 mm thick topping slab.

mmmmAst /601000100

5012.0 2min,

The spacing of rods should not be greater than one half of the c/c distance of rib, or 300 mm.

Provide 5 mm Torkari bars at 200 mm c/c, in the center of the topping slab both ways (Area

provided = 98 mm2/m)


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Fig. 11.54 Reinforcement detail in the rib including shear reinforcement

Reinforcement in the solid portion of width 2850 mm

Minimum Ast= 0.12 x (350/100) x 1000 = 420 mm

2

/mm

Provide 10 mm dia. bars at 180 mm c/c both ways (Ast= 436 mm2) at top and bottom of the

solid portion.

507 39 solutions-instructor-manual ch11 drcs

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