507 39 solutions-instructor-manual ch10 drcs

Design of RC Structures © Oxford University Press India DR. N. Subramanian

SOLUTION MANUAL

CHAPTER 10

Exercise 10.1 (Simply supported two-way slab)

The slab of a residential building of size 5 m x 6.5 m is simply supported on all the four sides on

230 mm walls. Assuming an imposed load of 3 kN/m2, and load due to finishes of 1.0 kN/m

2,

design the floor slab. Use M 25 concrete and Fe 550 steel. Assume mild exposure.

Solution

Step 1: Thickness of slab and effective spans

Cover for mild exposure (Table 16, Note 1 of IS 456) = 15 mm

Lx = 5 m , Ly = 6.5 m

Since the aspect ratio, i.e., the ratio Ly/Lx = 6.5/5= 1.3 < 2, we should design the slab as a two-

way slab.

L/D ratio of simply supported slab (as per Clause 24.1 of IS 456) for Fe 415 steel

= 0.8 x 30 =24

(Note that this is valid up to Lx = 3.5 m only as per code)

Hence D = 5000/24 = 208 mm

Provide D = 200 mm.

Assuming 10 mm dia bars are used, from Table 16 of IS 456, cover for mild exposure and M 25

concrete = 15 mm. Hence, dx = 200- 15 -5 = 180 mm, dy = 180-10 = 170 mm

Effective Span

The effective span of the slab in each direction = clean span + d (or width of support whichever

is smaller)

Thus effective Span

Lx = 5000 + 180 = 5180 mm; Ly = 6500 + 170 = 6670 mm

Hence r = Ly/Lx = 6670 / 4180 = 1.29

Step 2: Loads on slab

Self-weight of slab = 0.2 x 25 = 5.0 kN/m2

Weight of finishes (Given) = 1.0 kN/m2

Imposed load = 3.0 kN/m2

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

Total load, w = 9.0 kN/m2

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


Step 3: Design moments (for strips at mid-span, with 1m width in each direction)

For Ly/Lx = 1.29, From Table 10.2 (Table 27 of the code)

αx = 0.0921

αy = 0.0554

Hence 36.3318.55.130921.0 22 xuxx LwM kNm/m

07.2018.55.130554.0 22 xuyy LwM kNm/m

Check depth for maximum B.M. 2

max 138.0 bdfM ck

mmmmd 18098100025138.0

1036.335.0

6

Hence the depth adopted is adequate and the slab is under-reinforced.

Step 4: Design of Reinforcement

121.0180100025

1036.3368.644.12.1

68.644.12.1

2

6

2

dbf

M

d

x

ck

uu

mmd

xdz 94.170121.0416.01180416.01

26

40894.17055087.0

1036.33

87.0mm

zf

MA

y

st

We may also use the approximate formula,

4215501808.0

1036.33

8.0

6

y

ust

df

MA mm

2 ≈ 408 mm

2

From Table 96 of SP 16, provide 10 mm dia. bars at 190 mm c/c (Ast = 413 mm2); spacing < 3d,

Hence crack width will be controlled.

Similarly for long direction

080.0170100025

1007.2068.644.12.1

68.644.12.1

2

6

2

dbf

M

d

x

ck

uu

mmd

xdz 34.16408.0416.01170416.01

26

26034.16155087.0

1007.20

87.0mm

zf

MA

y

st

From Table 96 of SP 16

Provide 8 mm dia. bars at 190 mm c/c (Ast = 265 mm2)


The detailing is shown in Fig.10.54 with alternate bars bent up at 0.1Lx and 0.1Ly in the short

and long direction respectively. (Note at support in the long direction the 8 mm bars are provided

at 380 mm c/c; spacing ≈ 3 x 170 = 510 mm; Hence OK)

Fig. 10.54 Reinforcement detailing for slab of Example 10.1

Step 5: Check for deflection

Let us check the deflection in short direction, since it is critical.

%234.01801000

421100100

bd

Ap st

t

MPaf s 8.237413

40841558.0

0.2)/1log(625.000322.0225.0

1

ts

tpf

k

Modification factor kt from Fig. 4 of the code = 1.68

Basic span to depth ratio for simply supported slab = 20 (Clause 23.2.1)

Allowable L/d = 20 x 1.68 = 33.6

Provided span / depth ratio = 5180/180 = 28.78 < 33.6

Hence assumed depth is enough to control deflection.

Step 6: Check for shear

Average effective depth d = (180 + 170)/2 = 175 mm

The maximum shear force occurs at a distance of effective depth from the face of support.

mkNdLwV xnuu /54.3218.018.55.05.135.0


MPav 181.0)1801000/(1054.32 3

For pt = 0.234, τc for M 25 concrete (Table 19 of IS 456) = 0.349 MPa

kτc > 0.181 MPa

Hence slab is safe in shear

Note: It is clearly seen that shear will not be critical in two-way slabs subjected to uniformly

distributed loads.

Step 7: Check for cracking

Steel more than 0.12% in both directions.

Spacing of steel < 3d = 3 x 180 = 540 mm or 300 mm in both directions.

Diameter of steel reinforcement < 200/8 = 25 mm

Hence no calculation required for cracking.

Step 8: Check for development length

As shown in chapter 9, Example 9.1, as per clause 26.2.3.3(d)

Check should be made to see

013.1 L

V

ML

u

nd

It is found that 10 mm dia. bar is satisfactory

Length of embedment available at support

= 230 – clear side cover = 230-25 =205 mm > Ld/3

mmLd 403)6.14.1(4

1041587.0

Ld/3 = 135 mm < 205 mm

Hence development length is sufficient to develop bond.

Exercise 10.2 (Use of Marcus correction)

Compute the design moments for the slab analyzed in Exercise 1 using the Marcus correction.

Solution

Step 1: Moments with out Marcus correction:

From Exercise 10.1

For Ly/Lx = 1.29, From Table 10.2 (Table 27 of the code)

αx = 0.0921

αy = 0.0554

Hence 36.3318.55.130921.0 22 xuxx LwM kNm/m

07.2018.55.130554.0 22 xuyy LwM kNm/m

Step 2: Moments with Marcus correction:


Marcus correction factor,

4

2

16

51

r

rCC yx =

4

2

29.11

29.1

6

51 1- 0.368 = 0.632

Hence Mx = 33.36 x 0.632 = 21.08 kNm/m

My = 20.07 x 0.632 = 12.68 kNm/m

Thus, 36% reduction in moment is possible by taking into account torsional effects and corner

restraint.

Exercise 10.3

Redesign the slab given in Exercise 1, assuming that the corners of the slab are prevented from

lifting up, by wall loads due to the floor above.

Solution

From Exercise 10.1,

Lx = 5180 mm; Ly = 6670 mm; dx = 180 mm and dy =170 mm, fy =550 N/mm2 and fck = 25

N/mm2

Factored load wu = 13.5 kN/m2

Ly/Lx = 1.29

Step 1: Design moment (considering 1m width in each direction at mid-span)

As the ends of the slabs are restrained 2

xuxux LwM where αx may be taken as per Table 26 of the code (case 9)

Hence 0783.02.13.1

2.129.1)072.0079.0(072.0

x

Hence 36.2818.55.130783.0 2 uxM kNm/m

Note that, due to the restraints, the B.M. has reduced from 33.36 kNm/m to 28.36 kNm/m (as

obtained from Example 10.1), i.e. a reduction of 16%.

From Table 26, Case 9, we get, αy = 0.056

Hence 29.2018.55.13056.0 22 xuyvy LwM kNm/m

(This value is slightly higher than the non-constraint value of 20.07 kNm/m obtained in Example

10.1.)

Step 2: Design of Reinforcements

Short span

102.0180100025

1036.2868.644.12.1

68.644.12.1

2

6

2

dbf

M

d

x

ck

uu


mmd

xdz 36.172102.0416.01180416.01

26

33836.17255087.0

1036.28

87.0mm

zf

MA

y

st

From Table 96 of SP 16,

Provide 10 mm dia @ 230 mm c/c (Ast = 341 mm2

, pt = 0.189 %)

Max permitted spacing = 3 x 180 = 540 mm or 300 mm > 230 mm

For Long span

081.0170100025

1029.2068.644.12.1

68.644.12.1

2

6

2

dbf

M

d

x

ck

uu

mmd

xdz 27.164081.0416.01170416.01

26

25827.16455087.0

1029.20

87.0mm

zf

MA

y

st

From Table 96 of SP 16

Provide 8 mm dia bars @190 mm c/c (Ast = 265 mm2)

Maximum permitted spacing = 3 x 170 = 510 mm or 300 mm > 190 mm

Step 3: Corner Reinforcement

As the slab is torsionally restrained, corner reinforcement as per clause D-1.8 should be provided

for a distance of Lx/5 = 5180/5 = 1036 mm in both directions in meshes at top and bottom (four

layers).

Area of torsion reinforcement = 0.75 of area required for the maximum mid span moment

= 0.75 x 338 = 254 mm2

Provide 8 mm dia. bars @ 200 mm c/c (Ast = 251 mm2) both ways at top and bottom at each

corner over an area of 1040 mm x 1040 mm, i.e., 6 U – shaped bars as shown in Fig. 10.55


Fig. 10.55 Reinforcement detailing of slab for Exercise 10.3

Step 4: Check for deflection control

pt = 0.189 %

fs ≈ 0.58 x 415 = 240.7 MPa

Modification factor

0.2)/1log(625.000322.0225.0

1

ts

tpf

k

=

(

)

(L/d)max = 1.84 x 20 = 36.8

(L/d)provided = 5180/180 = 28.8 < 36.8

Hence the assumed depth is enough to control deflection.

Check for shear will not be critical as shown in Exercise 10.1


Steel more than 0.12% in both directions

Spacing of steel < 3d or 300 mm in both direction

Diameter of steel bar < 200 /8 = 25 mm

Hence cracking will be within acceptable limits.


Exercise 10.4

Find the bending moment coefficients αx and αx for a slab having two long edges discontinuous,

i.e. Case 6 in Table 10.4, with r =1.5, using Equations 10.17-10.19.

Find the bending moment coefficients αx and αx for a slab having all edges continuous, i.e. Case

1 in Table 10.4, with r =1.75, using Equations 10.17-10.19.

Solution

For this case, the number of discontinuous edges, Nd =2

Hence from Eqn. 10.17 we get, 1000

)5.1224( 2

ddy

NN

= 034.01000

6424

7/3 C C s2s1 (continuous edges)

1 C C l2l1 (discontinuous edges)

√ √

From Eqn. 10.18 we get,

078.0

)11(

)055.3)(5.1/034.018(3

9

2))(/18(3

9

222

21

21

ll

ssy

xCC

CCr

Thus the values of moment coefficients for this slab are as follows:

In the short direction

Positive at mid-span = 0.078 (0.068)

Negative at edges = 0 (0)

In the long direction

Positive at mid-span = 0.034 (0.035)

Negative at edges = (4/3) 0.034 = 0.0453 (0.045)

Note: The values given in brackets are those obtained from Table 26 of the code.

Exercise 10.5: Design of two-way slabs with two adjustment edges continuous

Design a rectangular slab panel of size 4.5 m by 6 m in which one long edge is discontinuous.

Assume that the slab supports imposed load of 4 kN/m2 and a floor finish of 1 kN/m

2. The slab is

subjected to mild exposure and is made of M 25 concrete and Fe 415 steel.


Solution:

Step 1: Thickness of slab and effective span

As the slab is subjected to mild exposure, from Table 16 of the code, for M 25, nominal cover =

20 mm. We shall use Canadian Code formula to estimate the minimum thickness of slab.

Assuming αm = 2.0 and β = Ly/Lx = 6/4.5 = 1.33

mmfL

Dm

yn150

233.1430

)1000/4156.0(6000

430

)1000/6.0(

also mmPerimeter

D 150140

)45006000(2

140

Let us adopt D = 150 mm,

Using 10 mm bars, dx = 150-20-5 = 125 mm, dy = 125-10 = 115 mm

As the size of supporting beams are not given, assume effective spans as 4.5 m and 6.0 m

Step 2: Load on slabs

Self weight of slab = 0.15 x 25 = 3.75 kN/m2

Weight of finishes = 1.00 kN/m2


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

Total load = 8.75 kN/m2

Factored load wu = 8.75 x 1.5 = 13.13 kN/m2

Step 3: Design moments (for strips of 1m width at each direction)

For Ly/Lx = 1.33, From Table 26 of Code, row 3 (one long edge discontinuous),

αx for negative moment at continuous edge at short and long span is 0.059 and 0.037

respectively. Similar positive moment coefficients at mid span are 0.045 and 0.028 respectively

for short and long span.

Factored negative bending moments in the short and long span are

69.155.413.13059.0059.0 22 xunx LwM kNm/m

84.95.413.13037.0037.0 22 xuny LwM kNm/m

Required effective depth for resisting the bending moment

mmmmbf

Md

ck

u 12567100025138.0

1069.15

138.0

5.06

5.0

Hence the adopted depth is sufficient and the slab is under-reinforced.


Step 4: Design of negative reinforcement

Approximate mmmdf

MA

y

st /3484151258.0

1069.15

8.0

26

Alternatively, let us use Table 2 of SP 16,

with 00.11251000

1069.152

6

2

bd

M u

From Table 3 of SP 16, for fck = 25 N/mm2 and fy = 415 N/mm

2, we get

pt = 0.291% and Asx2 = mmm /364100

1251000291.0 2

From Table 96 of SP 16, provide 10 mm dia. bars at 210 mm c/c (Ast = 374 mm2), in the short

span direction at top face at support. Also spacing < 3d = 375 mm. Hence spacing is adequate.

Reinforcement in long direction

App Ast = mmmdf

M

y

/2584151158.0

1084.9

8.0

26

Alternatively,

744.01151000

1084.92

6

2

bd

Mu

From Table 3 of SP 16, for fy = 415 N/mm2, we get

pt = 0.2142% and mmmmmmAsy /258/246100

11510002142.0 22

2

From Table 96 of SP 16, provide 8 mm dia. bars at 200 mm c/c (Ast = 251 mm2)

Spacing < 3d = 345 mm

Step 5: Design for positive reinforcement

The area of reinforcement required for positive moment in short direction can be computed

proportionately, based on the bending moment coefficient, as the effective depth is same

mmmAsx /278059.0/364)045.0( 2

1

Provide 8 mm dia. bars at 180 mm c/c (Ast = 279 mm2)

Similarly required area of steel for positive moment in the long direction

mmmAsy /186037.0/246028.0 2

1

From Table 96 of SP16, Provide 8 mm @ 270 mm c/c (Ast = 209 mm2/m)

180100/100015012.0min, stA mm2/m (# 8 @ 270 mm c/c = 186 mm

2/m)

Max. Spacing = 3 x 115 = 345 mm; greater than the adopted spacing.


Step 6: Detailing of reinforcement

The detailing of reinforcement using straight bars is shown in Fig.10.56.

Fig.10.56 Reinforcement detailing for Exercise 10.5


From Table 10.5, shear force coefficient for Ly/Lx = 1.33 for one long edge discontinuous case is

βvx = 0.477

Shear force = 0.477 wuLx = 0.477 x 13.13 x 4.5 = 28.18 kN

The maximum shear force occurs at a distance ‘d’ from the face of support ,

54.2613.13125.018.2818.28 xdwV kN

Nominal shear stress, 2/212.0

1251000

100054.26mmN

bd

Vc

This stress is less than the min. value in Table 19 of IS 456 for M25 concrete(0.29 MPa)

Hence the slab is safe in shear.

Step 8: Check for deflection control

pt = 0.299%

fs = 0.58 x 415 = 240.7 MPa

Modification factor

0.2)/1log(625.000322.0225.0

1

ts

tpf

k

=

(

)

Basic L/D = 32 (clause 24.1), Max L/D = 32 x 1.49 = 47.68

(L/D) provided = 4500 / 150 = 30 < 47.68

Hence no additional calculation for deflection is necessary.



As the spacing of reinforcement are < 3d or 300 mm, steel more than 0.12% in both direction

have been provided, and diameter of bar < 150/8 = 18.75 mm. Hence additional calculation for

cracking is not required.

Example 10.6 Continuous two-way slab

Design a continuous two-way slab system shown in Fig.10.57. It is subjected to a imposed load

of 3 kN/m2, and surface finish of 1 kN/m

2. Consider M 25 concrete, grade Fe 415 steel and

moderate environment. Also assume that the supporting beams are 230 x 500 mm.

Fig. 10.57

Solution

Due to symmetry only one quadrant of the floor system has to be designed i.e. slabs S1 to S4

Step 1: Thickness of slabs

Let us assume uniform thickness for all slabs – it will be governed by the corner slab S1 with two

adjacent edges discontinuous

m

yn fLD

430

)1000/6.0(

or

140

perimeter

ss

bb

fIE

IE

For this calculation, let us take Ds= 4200/30 ≈ 140 mm

Considering an L beam, with a slab projection of 360 mm (500-140) beyond the beam web,

The distance of centroid of beam from top of the slab


232

3

2

140195140360

12

140360)195250(500230

12

500230

bI

= 3.6135 x 109 mm

4

46

3

1083.62812

140

2

5500mmIs

;75.51083.628/106135.3 69 f As αf > 2, we may assume αm = 2.0;

30.12.4/5.5/ xy LL

138230.1430

)1000/4156.0(5500

D mm or 139

140

)55004200(2

mm

Assume D = 150 mm

From Table 16 of IS 456, for M 20 and moderate environment with db = 10 mm

Clear cover = 30 mm

Hence dx = 150 - 30 – 5 = 115 mm and dy = 115- 10 = 105 mm

Step 2: Load on Slab


Weight of finishes = 1 kN/m2

Imposed load = 3 kN/m2

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

Total load, w = 7.75 kN/m2


Step 3: Design Moments

Moments = Moment coefficients (α,β) x wu x Lx2

Where α and β are moment coefficients from Table 26 of IS 456 for slabs S1 to S4. They are

shown in Fig.10.48 (S1 – case 4, S2 – case 3, S3 – case 2, S4 – case 1)


Fig. 10.58 Moment coefficients and moments

Note that the negative moments at the common edges of slabs are not equal. For more accurate

design, we should distribute the unbalanced moments in proportion to the relative stiffness of

slabs meeting at the common edge and also modify the span moments by adding half of the

distributed moments, as shown in Example 10.6 of the book. In design offices, the design

negative moment in the common edges is usually taken as the larger of the two values obtained

from either side of the support. Even though this may increase the reinforcement slightly, it

simplifies the calculations considerably.

Step 4: Design of Reinforcement

The bending moment and corresponding area of required reinforcements are tabulated below

Slab S1

dx =115 mm

dy =105 mm

B.M./kNm Mu/bd2 pt %

Table 3 of

SP 16 with

min 0.12%

As mm2 Reinforcement

(a) Short direction

+10.05 0.760 0.219 252 #8 @ 190 (265

mm2)

-13.33 1.01 0.2942 338 #8 @ 150 (335

mm2)

(b) Long direction

+7.18 0.651 0.1863 196 #8@250 (201

mm2)

-9.64 0.874 0.2532 266 #8 @ 190 (265


mm2)

Slab S2 (a) Short direction

+9.03 0.683 0.196 225 #8 @ 220(228

mm2)

-11.69 0.884 0.2412 277 #8 @ 180 (279

mm2)

(b) Long direction

-9.64 0.874 0.2532 266 #8 @ 180 (279

mm2)

+5.74 0.521 0.1488 156 #8 @ 300 (168

mm2)


-10.46 0.791 0.2283 263 #8 @ 180 (279

mm2)

+8.00 0.605 0.1725 198 #8 @ 240 (209

mm2)

(b) Long direction

+5.74 0.521 0.1488 156 #8 @ 300 (168

mm2)

-7.59 0.688 0.1974 207 #8 @ 240 (209

mm2)


-9.64 0.874 0.2532 291 #8 @ 170 (296

mm2)

+7.39 0.559 0.1587 182 #8 @ 270 (186

mm2)

(b) Long direction

-6.56 0.595 0.1695 178 #8 @ 270 (186

mm2)

+4.92 0.446 0.1259 132 #8 @ 300 (168

mm2)

Area of steel in edge strips = 0.12% of gross sectional area

= mmm /1801000150100

12.0 2 (say #8 @250 mm c/c, Ast = 201 mm2/m)

Torsion Reinforcement

Torsion reinforcement when both edges are discontinuous,

= ¾ x Mid-span reinforcement = ¾ 252 = 189 mm2 , provide #8 @ 250 c/c, (Ast = 201 mm

2)

Torsion reinforcement when only one edge is discontinuous

= 3/8 x Mid-span reinforcement = 3/8 225 = 84 mm2, provide #8 @ 300 c/c (Ast = 168 mm

2)

This reinforcement should be provided for a length of Lx/5 = 4200/5 = 840 mm


Step 5: Check for Shear

From Table 10.4, shear force coefficient for Ly/Lx = 5.5/4.2 = 1.3, for two adjacent sides

continuous is βvx = 0.50

kNLwV xuvxsx 42.242.463.1150.0

Maximum shear force at a distance 135 mm (d) from the face of support

V =24.42 – 0.115 x 11.63 = 23.08 kN

Nominal shear stress, 2/20.0

1151000

100008.23mmN

bd

Vv

This stress is less than the minimum value in Table 19 of IS 456 for M 25 concrete

Hence slab is safe in shear.

Step 6: Check for deflection

pt at mid span = %23.01151000

100265

fs = 240.7 MPa

Modification factor

0.2)/1log(625.000322.0225.0

1

ts

tpf

k

=

(

)

Max L/D = 28 x 1.67 = 46.7

L/D provided = 4000 / 150 = 28 < 46.7

Hence deflection will be within limits.


Spacing of reinforcement are less than 3d (3 x 115 = 345 mm) or 300 mm and steel provided is

more than 0.12% of gross cross-section in both directions. Dia. of bars < 150/8 = 18.75 mm.

Hence no separate check for cracking is required.

Step 8: Detailing of reinforcement

Detailing of reinforcement in various middle stirrups and edge strips should be done as per

Fig.10.21 and 10.23 of the book. Note that for practical convenience only two bar spacing may

be adopted.


Exercise 10.7

Design a circular slab of diameter 5.5 m subjected to an imposed load of 5 kN/m2. Assume that

the slab is simply supported and is in severe environment. Use fck = 25 MPa and fy = 415 N/mm2.

Solution

Step 1: Compute the loads

The usual span/depth ratio will be in the range 25 to 40.

Let us adopt L/D =32.5

Thickness of slab, D = 5500/32.5 = 169.2 mm

Adopt 200 mm thickness

Self weight = 0.20 x 25 = 5.0 kN/m2

Floor finish (assumed) = 1.0 kN/m2


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


Factored load = 1.5 x 11.0 = 16.5 kN/m2

Step 2: Calculation of Bending Moment

Assuming ν = 0, as per Eqn.10.27,

22

16

3ar

wM r

22316

arw

M t

B.M. at edge (a = r)

Mr = 0 and Mt = 2/16 wr2

B.M. at centre (a = 0)

2

16

3wrM r and

2

16

3wrM t

Hence max. moment at centre = 4.2375.25.1616

3 2 kNm/m

Note: According to yield line theory,

Ultimate moment = 8.206

75.25.16

6

22

rwu kNm/m

Hence by using yield line theory we will achieve economy.


Step 3: Check depth of slab

For fy = 415 N/mm2

Required depth mmkbf

Md

ck

u 78251000138.0

108.205.0

6

Selected overall depth is 200 mm. The slab is under-reinforced.

The slab is in severe environment, Hence cover = 45 mm (Table 16 of IS 456)

Let us assume diameter of bar = 12 mm

Provided effective depth in one direction = 200 - 45 – 6 = 149 mm

Effective depth in the other direction = 149 – 12 = 137 mm

Step 4: Calculate area of reinforcement

The bending moment in the circumferential direction is same as that of the radial direction.

The area of reinforcement is

mmmdf

MA

y

ust /420

4151498.0

108.20

8.0

26

The same result may be obtained by calculating Mu/(bd2) and using Table 3 of SP16.

From Table 96 of SP 16, provide 12 mm dia. bars @ 260 mm c/c (Area = 435 mm2) in both the

directions.

Maximum allowed spacing: least of 411 mm (3 x d ) or 300 mm. Hence spacing is within limits

to control cracking.

Overall diameter of slab is chosen as 5.7 m and reinforcement details are as shown in

Fig.10.24(c) of the book.

Step 5: Check for Shear

The critical section for shear is at a distance d from support. The shear force at critical section is

kNdR

wVu 43.20137.02

75.25.16

2

The actual width of slab at a distance d from support is slightly less than a unit width at the

support because of the radial coordinates. However, let us check the shear stress using unit width

as an approximation.

Thus nominal shear stress of 2

3

/15.01371000

1043.20mmN

bd

Vuv

This is less than (min)c as per Table 19 of IS 456, for M 25 concrete. Hence slab is safe in

shear.


Exercise 10.8 Design of Circular Well cap

Design of well cap to support a circular pier of diameter 2 m. Assume that the internal diameter

of well is 7 m and that the load on the pier is 600 kN. Use Fe 415 steel and M 35 concrete

Fig. 10.59 Well cap

Solution

Step 1: Calculations of loads

As the well cap will be subjected to alternate wetting and drying, we assume that it is subjected

to severe exposure and hence minimum cover as per Table 2 and 16 of IS 456 is 45 mm. As we

are using M 35 concrete, it may be reduced by 5 mm. Hence use cover = 40 mm. Assume L/D =

20; hence D = 7000/20 = 350 mm. With 20 mm bars, effective depth d = 350 – 40 - 10 = 300

mm, effective radius of slab, R = 7/2 + 0.30 = 3.80 m

Let the overall outside dia = 2 3.80 = 7.6 m

Factored self weight wc = 1.5 x 0.35 x 25 = 13.13 kN/m2

Factored imposed load, wI = 1.5 x 600 = 900 kN

Step 2: Calculate Bending Moment

The maximum radial bending moment for this case at the face of concentrated central load as per

Timoshenkno and Krieger, 1959 (pp.67) is

The diameter of pier = 2 m; Hence, a = 1 m

2

22

14

1

416

3

R

a

a

RLog

WRwM n

Icr


=

2

2

8.3

11

4

1

1

8.3

4

900

16

8.313.133nLog

= 35.55 + 71.62(1.335 - 0.233) = 35.55 + 78.93 = 114.48 kNm/m

The maximum circumferential bending moment at the edge of pier support, that is 1000 mm

from the centre is given by

2

22

34

1

416

3

R

a

a

RLog

WRwM n

Ict

=

28.3

1325.0

1

8.3

4

90055.35 nLog

= 35.55 + 71.62 (1.335 + 0.733) = 183.66 kNm/m > Mr

Step 3: Check depth for bending

Required depth for a balanced section

mmbf

Md

ck

195351000138.0

1066.183

138.0

5.06

5.0

However provide an overall depth of 400 mm in order to resist the shear due to the pier. Using

20 mm bars, d = 400 – 40 – 10 = 350 mm

Step 4: Calculate reinforcement

Area of reinforcement

15814153508.0

1066.183 6

stA mm

2/m

Provide 20 mm bars at 190 mm c/c spacing (Ast, provided = 1653 mm2) in the circumferential

direction

The effective depth of reinforcement in the radial direction = 350 - 20 = 330 mm

Required Ast = 10454153308.0

1048.114 6

mm

2/m

Provide 20 dia. bars at 300 mm c/c (Ast, provided = 1047 mm2) at 1000 mm from centre

Minimum reinforcement = 22 10454801000400

100

12.0mmmm

Maximum spacing = 3d > 300 mm

The reinforcement should be provided as shown in Fig. 10.24(b) of the book.

Note: As the depth of slab is greater than 200 mm, provide minimum steel, i.e., 12 mm @ 230

mm at the top of the slab also as temperature and shrinkage steel.



Shear force = 24.1430.12

900

2

a

p kN/m

Nominal shear stress = 273.0)3501000(

105.95 3

N/mm

2

Design shear strength τc for M 35 with minimum steel, from Table 19 of IS 456,

= 0.29 N/mm2 > 0.273 N/mm

2

Hence the slab is safe for shear.

Step 6: Check for punching shear (clause 31.6.2 and 31.6.3):

Critical section for punching is located at d/2 from the column, Hence the critical diameter

= ( )

Punching shear stress =

N/mm

2

Limiting punching shear stress, √ =1.47 N/mm2

Hence the slab is safe for punching shear

Example 10.9

A simply supported semi circular slab of 3.5 m radius is subjected to a uniformly distributed

imposed load of 3 kN/m2 and floor finish of 1 kN/m

2. Assuming moderate environment, design

the slab with Fe 500 steel and M 20 concrete.

Solution

Step 1: Conversion to equivalent rectangle

The semi circular slab can be idealized by rectangle as shown in Fig.10. 60. The short side of the

rectangle may be selected as 0.867 times the radius of the circle; i.e. = 0.867 x 3500 = 3035 mm

Fig. 10.60 Semi-circular slab

Hence we need to design a rectangular slab of size 7000 x 3035 mm as Ly > 2Lx, we need to

design it as a one-way slab.


Step 2: Calculate Loads and B.M.

L/d = 20 (Simply supported slab as per Clause 23.2.1)

D = 3035/20 = 151.75 mm, we may reduce this by taking into account the reinforcement ratio.

Hence, assume D = 140 mm, effective depth with 10 mm bars, with moderate environment, d =

140-30-5 = 105 mm

Self weight = 0.14 x 25 = 3.5 kN/m2

Floor finish = 1.0 kN/m2


_____________

Total = 7.5 kN/m2


Let effective span = 3.035+ 0.105 =3.14 m

86.138

14.325.11

8..

22

Lw

MB u kNm/m


Required depth for resisting bending moment

mmbf

Md

ck

u 71201000138.0

1086.13

138.0

5.06

5.0

Hence slab is under-reinforced.


Required area of steel

mmmdf

MA

y

ust /330

5001058.0

1086.13

8.0

26

Provide 10 mm bars at 230 mm c/c in short direction. (Table 96 of SP 16) Ast, provided = 341

mm2

Minimum reinforcement for shrinkage = 0.12 x 105 x 1000/100 = 126 mm2/m

Provide 6 mm @ 220 mm c/c in long direction (Ast provided = 128 mm2)

Check for cracking: provided spacing ≤ 3 x 105 = 315 mm or 300 mm

Hence cracking will be controlled.


Maximum shear force at a distance d from the face of support

48.16105.02

14.325.11

2

d

LwV x

uu kN/m


Nominal shear stress = 23

/157.01051000

1048.16mmN

bd

Vu

Min. shear strength as per Table 19 for M 20 concrete = 0.28 N/mm2 > 0.157 N/mm

2

Hence slab safe in shear.

Example 10.10 Slab with opening

A simply supported slab with effective short and long spans of 4 m and 6 m respectively is

subjected to an imposed load of 3.0 kN/m2. An opening of 500 mm x 500 mm is to be provided

in the slab as shown in Fig.10.55. Design the slab using M 25 concrete and Fe 415 steel.

Fig. 10.61 Slab with opening

Solution

Step 1: Assume depth of slab

Assuming mild environment, cover = 20 mm (Table 16 of Code)

L/d ratio for simply supported slab (Clause 24.1) = 28

Hence, d = L/28 =4000 /28 = 143 mm

Assume overall depth of 150 mm

With 10 mm bars, effective depth = 150 – 20 – 5 = 125 mm

Step 2: Calculate loads & B.M.


Floor finish (assumed) = 1.00 kN/m2


_____________


Factored Load = 1.5 x 7.75 = 11.63 kN/m2

Lx/Ly = 6 / 4 = 1.5

From Table 27 of the code, αx = 0.104, αy = 0.046

Mx =0.104 x 11.63 x 42 = 19.35 kNm/m

My =0.046 x 11.63 x 42 = 8.56 kN/m



Required depth mmmmbf

Md

ck

u 12575251000138.0

1035.19

138.0

5.06

5.0

Hence depth assumed is sufficient.


Area of reinforcement is short span mmmdf

M

y

u /4664151258.0

1035.19

8.0

26

Provide 10 mm at 160 mm c/c (Ast, provided = 491 mm2)

Area of reinforcement in long direction =

mmm /224415101258.0

1056.8 26

Provide 8 mm @ 220 mm c/c (Ast, provided = 228 mm2)

Spacing of bars < 3 x d = 3 x 125 = 375 mm or 300 mm

Hence cracking will be within permissible limits.

Minimum renforcement = (0.12 / 100) x 125 x 1000 = 150 mm2 < 228 mm

2

Step 5 : Check for shear

Load transferred to short span, u

yx

y

x wLL

Lw

22

2

mkN /05.8

46

)63.11(622

2

Maximum shear force at a distance d from the face of support

mkNdL

wV xxu /1.15125.0

2

405.8

2

Nominal shear stress = 22

3

/29.0/12.01251000

101.15mmNmmN

bd

Vu

(Table 19 of IS 456)

Hence slab safe in shear.

Alternate bars should be bent up at 0.2 x span to resist any secondary bending moment occurring

at support.

Step 6: Detailing at opening

The size of opening is 500 mm and the thickness of slab is 150 mm. The opening is not at the

critical zone of the slab. In short span, 500 / 160 = 3 numbers 10 dia. bars will be intercepted by

the opening and in the long span, 500/220 = 2 numbers 8 dia. Bars. Hence provide 2 Nos. 10 mm

bar at each edge along the long and short span direction. Similarly provide two 8 mm bar at the

four diagonal corners of the opening as shown in Fig.10.62.


Fig. 10.62 Reinforcement around opening

Example 10.11

A two-way continuous slab of size 4 m by 8 m, and 300 mm thickness has a central point load of

85 kN due to some electrical equipments of size 0.9 m by 1.8 m. Determine the BM and shear

due to the concentrated load using Pigeaud’s method. Assume a wearing coat of thickness 40

mm.

Solution

Step 1: Calculate u and v

)04.03.0(29.0 u = 1.58 m

)04.03.0(28.1 v = 2.48 m

Step 2: Determine the coefficients using Pigeaud’s curves

4.04

58.1

xL

u

;

31.08

48.2

yL

v

;

0.24

8

x

y

L

L

Using Fig.10.29 of the book, mx = 0.142; my = 0.06

Step 3: Calculate the moments

)06.02.0142.0(85 yxx mmPM = 13.09 kNm

)142.02.006.0(85 xyy mmPM = 7.51 kNm

Allowing a reduction of 20% for continuity, we get Mx = 10.47 kNm and My = 6.01 kNm per

meter width


Step 4: Calculate maximum shear force:

Short span: V = P (Ly-v/2)/(uLy) = 85(8-1.24)/(1.588) = 45.46 kN/m

Long span: V = P (Lx-u/2)/(vLx) = 85(4-0.79)/(2.484) = 27.51 kN/m

Example 10.12

Consider a rectangular slab of size 8 m by 6 m with one of its longest sides free and other three sides

simply supported. The reinforcement in two perpendicular directions are such that mx = 12 kNm/m

and my = 15 kNm/m. Find its collapse load.

Solution

mx = 12 kNm/m and my = 15 kNm/m. Hence μ = 15/12 = 1.25.

Ratio of short side to long side, α = 6/8 =0.75, L = 8 m

For Mode 1 failure (see Table 10.6)

2

94

24 2

22

Lwm u

uu w

w479.32

75.0

25.194

25.124

8)75.0(12

2

22

Hence wu = 3.45 kN/m2

For Mode 2 failure, 2

2

2

243

24

Lwm u

uu w

w041.3

75.02

25.1

75.04

25.13

24

812

2

2

2

Hence wu = 3.95 kN/m2

We must take the lower value of the ultimate load, since we are using an upper bound approach.

Hence the first mode of failure with two positive yield lines govern the failure, with wu = 3.45

kN/m2. Note that it has been shown already by Jones that when ,2/ 2 the first mode will

govern; otherwise the second mode of failure will govern. For our case,

222.275.0/25.1/ 22 ; hence, the first mode governs, which is evident by the calculations

also.


Example 10.13

Design the rectangular slab given in Exercise 10.1 using the yield line theory and compare the

results.

Solution

Given: L = 6.5 m; B = 5 m, I.L. = 3 kN/m2; Load due to finishes = 1.0 kN/m

2; M25 concrete and

Fe 550 steel. Mild exposure.

Step 1 Assume depth of slab

As per clause 24.1 of IS 456, L/d ratio for simply supported slab = 35 x 0.8 =28

Hence depth of slab = L/28 = 5000/28 =178.5 mm

Let us adopt D = 180 mm, Assuming cover = 25 mm, and dia. of bars used as 10 mm,

d =180 -25-5 = 150 mm

Step 2 Calculate Loads

Self-weight of slab = 0.1825 = 4.5 kN/m2


Floor finish (say) = 1.00 kN/m2

Total service load = 8.5 kN/m2

Design factored load, wu = 1.5 x 8.5 = 12.75 kN/m

2

Step 2 Calculate B.M. and shear

Ultimate bending moment as per yield line theory

24

2LwMm u

u = 28.1324

575.12 2

kNm/m

Ultimate shear, 875.31575.125.05.0 LwV uu kN

Step 3 Check limiting moment capacity of slab

22

lim, 150100025138.0138.0 bdfM cku 77.62 kNm

Mu < Mu,lim. Hence the section is under reinforced.

Step 4 Calculate Reinforcement

59.01501000

1028.132

6

2

bd

Mu

From Table 3 of SP 16, we get for M 25 concrete, pt = 0.168%

2521501000100

168.0stA mm

2

Provide 10 mm diameter bars at 300 mm c/c bothways (Ast = 262 mm2)

Note that we obtained 10 mm dia. bars at 190 mm c/c in Exercise 10.1


Step 5 Check for shear stresses

Nominal shear stress =

1501000

31875

bd

Vu 0.213 N/mm2

1501000

262100100

bd

Ap st

t = 0.175

From Table 19 of IS 456,

csk ks0.30 N/mm2

Hence the slab is safe in shear.

507 39 solutions-instructor-manual ch10 drcs

Documents

twoway slab

slab selfweight of slab

thickness of slab

slab of example

floor slab

effective span lx

design of reinforcement

long direction