Solved example 17.1 The items 1 to 28 below constitutes the Navigation chart here

The details about how to use the Navigation chart is given 1 Total depth 2 Effective depth in short direction dx

#3

3 Effective depth in long direction dy 4 Effective span in short direction lx

#5

5 Effective span in long direction ly 6 Ratio r = ly/lx

#7

7 Check for dia. of main bars 8 Total factored load per sq.m

#9

#4 #6 #8 #13

9 Factored Bending moments Mux and Muy 10 Effective depth required in x direction

#16

11 Ast required in x direction 12 Ast required in y direction

#19

13 Spacing required in x direction 14 Spacing required in y direction

#21

15 Spacing provided in x direction 16 Spacing provided in y direction

#25

17 Ast provided in x direction 18 pt provided in x direction

#28

19 Ast provided in y direction 20 pt provided in y direction

#31

21 Check for distributor bars Checks

#34

22 Check for concrete cover 23 Check for maximum bar diameter

#35

24 Check for spacing of bars 25 Check for minimum area of steel

#37

26 Check whether pt less than ptlim 27 Check for deflection control

#39

28 Check for shear

#43

#17 #20 #22 #26 #30 #33

#36 #38 #40

The roof of a room with internal dimensions 3.7m x 4.8m consist of a simply supported slab. The walls of the room is of brick masonry 23 cm thick all around. The LL is 3 kN/sq.m. and the loads of finishes is 1 kN/sq.m. Design the roof slab. Use M20 concrete and Fe 415 steel. Assume mild exposure conditions. Solution: Assuming 'mild' exposure conditions, Grade of concrete Grade of steel

f ck fy w DL , finishes

=

20

=

415

=

1

N / mm

2

N / mm 2 kN / m2

kN / m

3

kN / m2

w DL , finishes w LL

=

3

kN / m3

kN / m2 #1

1

#2

2

Effective span lx,ly, Effective depth dx,dy and Total depth D : Clear shorter span

=

3700

Add 150 mm

=

3850

c/c distance between supports

=

3930

Lesser of the above two values

=

3850 Details here Eq.17.9

d provided should be greater than or equal to 3850/30 =

128.33 Details here

For 'mild' exposure conditions, we can provide a clear cover of 20 mm Assume 8 mm diameter bars at a clear cover of 20 mm. Then total depth D= 128.33 +4 +20 =

152.33

=

155

#3

3

Then dx = 155-20-4

=

131

#4

4

dy = 155-20-8-4

=

123

#5

5

=

3831

c/c distance between the supports =

3930 #6

6

Provide D

Short direction: clear span + effective depth So lesser of the above = lx

=

3831

=

4923

c/c distance between the supports =

5030

Long direction: clear span + effective depth So lesser of the above = ly

=

4923

#7

7

=

1.285

#8

8

#9

9

ratio r: r = ly/lx

This is less than 2. So we can design the slab as a two way slab Check for diameter of main bars: Dia. should not be greater than D/8 D/8

cl.26.5.2.2 =

19.38

So 8mm dia. is OK Calculation of loads: For this we use

Eq.17.11

(1) Self wt. of tread slab

kN / m

2

=

3.88

kN / m2

(2) Finishes

=

1

kN / m2

(4) Live load

=

3

kN / m

=

7.88

kN / m

2

=

11.813

kN / m

2

25D

#10

10

(given)

#11

11

(given)

#12

12

#13

13

('D' is in meter)

2

We take the sum of the above three items from #10

to

#12 w

Factored load wu(load factor 1.5) Calculation of

α x ,α y

Eq.17.7

We will use the equations directly:

[ ]

=

0.0915

#14

14

[ ]

=

0.0554

#15

15

#16

16

1 r4 αx= 8 1+r 4 1 r2 α y= 8 1+r 4

We will cross check using the method of interpolation: Table 27, Cl.D-2.1 ratio r

=

r

1.2850

αx

1.2000

0.0840

1.2850

0.0917

1.3000

0.0930

close to the value in

#14

close to the value in

#15

αy

r 1.2000

0.0590

1.2850

0.0556

1.3000

0.0550

Calculation of factored Bending moments

M ux =α x w u l 2x

=

M uy =α y w u l 2x

=

M ux , M uy 15.86

kNm

9.6 kNm

Design of steel for Mux For the design, the following data are available to us:

b

=

1000

mm

D

=

155

mm

#3

Effective depth dx

=

131

mm

#4

M ux

=

15.86

kNm

#16

Calculation of effective depth required: Eq.3.27

Eq.G 1.1c of the code, same as

(

)

x umax x umax 2 M ulim=0.36 1−0.42 b d f ck d d For Fe415 steel,

x umax d

=

0.4791

Table 3.1

Eq.4.20 and 4.21

All known values are substituted in eq.3.27 to get

M ulim = R d 2 where

R=0.36

(

)

x umax x 1−0.42 umax b f ck d d

Putting all the known values, we get R

M ulim =

So Equating

2755.4

M ux 15.86

to

=

2755.4

x d2

M ulim

we get

6

x 10 =

2755.4 d

=

75.86

This is the effective depth required, and it is less than the dx provided So we can proceed to find the area of steel Calculation of

A st:

#4

xd

2

mm

#17

17

Consider again G.1.1.c of the code

M u =0.36

(

)

xu x 1−0.42 u b d 2 f ck d d

Substituting Mu for Mulim and xu for xumax we get,

(

)

xu xu 2 M u =0.36 1−0.42 b d f ck d d Substituting all the known values we get,

15.86

x 106 =

0.36

(

)

xu x 1−0.42 u x d d

343220000

This is same as

0.36

(

xu x 1−0.42 u d d

)

=

0.0462

This can be rearranged as 0.1512

( )

2

xu − 0.36 d

( )

xu + d

0.0462 = 0

This is a quadratic equation of the form:

a x 2 +b x 2 +c=0 Where

a=

0.1512

0.1

b=

-0.36

0.32

c=

0.0462

So solutions of this equation are: Solution 1:

1.64

Solution 2:

2.2448 0.1361

Out of the two solutions, solution 2 is the appropriate one because xu can only be a fraction of d So

xu d

=

0.1361

Now we can find Ast from the eq.G.1.1.a of the code:

#18

18

x u 0.87 f y A st = d 0.362 f ck b d

A st

So

=

357.54

mm 2

#19

19

#20

20

We can also use table 2 of SP16 to determine Ast:

M u / bd 2

=

M u / bd

A st =

2

0.9240

pt

0.9000

0.2640

0.9240

0.2717

0.9500

0.2800

( )

pt bd 100

=

This is close to the value in

355.89

mm

2

#19

The same procedure is used to find the steel for Muy, and it is calculated as 225.59

Details here

Required spacing of bars in x direction:

Eq.5.1

Eq.5.2

mm 2

ϕ

=

8

2 Ab = π ϕ 4

=

50.27

=

140.59

#21

21

222.82

#22

22

( )

mm mm 2

( )

A st =

1000 Ab s

Here Ast is obtained from #19

Sx

Required spacing of bars in y direction: Here Ast is obtained from #20

Sy

=

Now we look at the provisions given by the code regarding the spacing of bars in a slab. Minimum spacing to be provided between bars of a slab

cl.26.3.2 of the code. According to this clause, the spacing should not be less than the largest of the following: 1. Diameter of the bars

=

8

mm

=

25

mm

2. 5mm more than the nominal maximum size of aggregates

So the spacing should not be less than 25mm

#23

23

Maximum spacing allowable between bars of the slab : Details here cl26.3.3.b(1) of the code According to this clause, the spacing should not be more than the smallest of the following: 1. Three times the effective depth of the slab = 3 x d 3 x dx

=

3 x 3y

=

mm 369 mm 393

2 So the spacing should not be more than

300

mm

300

mm

#24

24

mm 220 mm

#25

25

#26

26

#27

27

Based on the results from #21

to

#24

We can provide: Sx

=

Sy

=

140

Check whether the minimum area of steel as specified by the code is provided: Details here

cl. 26.5.2 of the code

According to this clause, when Fe 415 is used, Ast should not be less than 0.0012 Ag 0.0012Ag = 0.0012bD

=

186

mm 2

8

mm

Ast actually provided in x direction: Using Eqs. 5.1 and 5.2 again,

ϕ

=

Sx

Ab = π ϕ2 4

Eq.5.1

Eq.5.2

( )

( )

A st =

1000 Ab s

=

140

=

50.27

=

359.04

Ast provided is not less than that in

mm mm 2 mm 2

#27

Hence OK

#28

28

#29

29

#30

30

#31

31

#32

32

#33

33

pt actually provided in the x direction:

pt =

100 A st bd

=

0.27

Ast actually provided in y direction: Using Eqs. 5.1 and 5.2 again, =

8

Sy

=

220

=

50.27

=

228.48

Ab = π ϕ2 4

Eq.5.1

Eq.5.2

ϕ

( )

( )

A st =

1000 Ab s

Ast provided is not less than that in

mm mm mm 2 mm 2

#27

Hence OK

pt actually provided in the x direction:

100 A st pt = bd

=

0.19

Distribution bars: cl.26.5.2

Details here

Two way slabs do not require distributor bars because bars are present in the two perpendicular directions. But we will check whether the bars provided will satisfy the requirements of distributor bars.

Area of distribution bars to be provided = 0.0012Ag = 0.0012bD From

#28

=

186

and

mm 2

#34

34

#35

35

#36

36

Hence OK

#37

37

and

#32

#38

38

and

#33 #39

39

#31

Ast provided in both directions is greater than this. Hence OK Maximum spacing allowable between distribution bars of the slab : cl 26.3.3.b(2) of the code According to this clause, the spacing should not be more than the smallest of the following: 1. Five times the smaller effective depth = 5 x dx

=

2 So the spacing should not be more than

655

mm

450

mm

450

mm

The spacing provided in both directions is less than this. Hence OK

mm At this stage we can perform the various checks 1. Concrete cover: This has been determined based on the provisions given by the code.

#2

2. Maximum dia. of the bars of the slab: This has been checked in

#9

3. Spacing of bars: From

#23

the spacing should not be less than 25 mm

The spacing in either direction is greater than this. From

#24

Hence OK

the spacing should not be greater than 300mm

The spacing in either direction is less than this. 4. Minimum area of flexural reinforcement: This was checked for both x and y in

#29

5. Check whether pt is less than pt lim Details here For Fe 415 steel and M20 concrete, pt lim = 0.961 pt for the present case is obtained using the equation: Pt in each direction is obtained in Both the values are less than 0.961

#30 Hence OK

6. Check for deflection control Details here

For this we refer cl. 23.2.1 of the code

For a two way slab, the deflection check is to be done for the short span lx We can first write the following data:

mm 2 357.54 mm 2 359.04 mm

Effective span

3831

Ast required Ast provided

#6 #19 #28

For members with span upto 10m, having tension reinforcement only,

l / d =( l / d ) basic k t

#40

40

#41

41

For simply supported members, (l/d) basic

=

20

To find kt, we can use the expression given in SP24

1

k t=

0.225+0.00322 f s −0.625 log 10

(

bd 100 As

)

This expression can be rearranged into the form:

k t=

1 0.225+0.00322 f s +0.625 log 10 ( p t )

f s =0.58 f

y

[

Area of steel required Area of steel provided

]

To find fs: Ast required

=

357.54

Ast provided

=

359.04

fs

=

239.7

N / mm

pt

=

0.27

#30

Substituting pt and fs in

#41 we get kt

Substituting kt in

=

1.55

=

30.98

#40 l/d

2

l / d provided

=

131

=

29.24

<

30.98

Divide 3831

by

29.24

Hence OK

#42

42

#43

43

#44

44

#45

45

Details here

7. Check for shear We can first write the following data: wu

=

11.81

Clear span in short direction = lxn

=

3700

Effective depth d= (dx + dy)/2

=

127

pt in y direction

=

0.19

Vu = wu(0.5lxn – d)

=

20.35

kN / m

2

#13

mm #33

kN

We have to determine value of design shear strength of concrete from table 19 of the code.

pt

τc

0.1500

0.2800

0.19

0.3086

0.2500

0.3600

This can be increased by multiplying with 'k' D= 155mm. This falls between 150 and 300. So k = [1.6-(0.002 x155)]

=

1.29

=

50558.68

So the contribution from concrete

k τc b d

N 50.56 kN

= This is greater than the value in

#44

Hence safe.

This completes the design of the two way slab The drawings showing the arrangements of bars is given in main section

here

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