cd2_lect3_2-10-2011

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e n orce oncre e es gne n orce oncre e es gn

Dr. Nader Okasha

Lecture 3

Calculation of the depth of two wayslabs

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Methods of limiting deflections in two way slabs:

Two methods are given by the ACI for controlling deflections:ACI 9.5.3.4

1) by calculating the deflection and comparing it with code

specified values given in ACI Table 9.5(b).y prov ng m n mum va ues or e mem er c ness as

give in the next slides.

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Minimum thickness of two wa slabs to control deflection

Limiting thickness

(mm)

Deflection control

method

Case

125*Table 9.5(c) No INTERIOR

Beams 0.0mα =a e . c

125ACI Eq 9-12

0.2mα ≤

0.2 2α < ≤

90ACI Eq 9-132 mα <

* 100 mm for slabs with drop panelsACI 9.5.3

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ACI 9.5.3

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Definition of β and l n

longer clear span

shorter clear span β =

l n = Clear span of the panel considered in the long direction measured face to face of support.

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Definition of α fm

b b

s s E I α =

α 4

α1

α 2

o u us o e as c y o eam

bE Modulus of elasticity of slabs

=

=α3

s

I Moment of inertia for beam b

I Moment of inertia for slab

=

=

1 2 3 4α α α α α + + +

=

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Definition of I b

and I s

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Definition of the equivalent beam:

be = b w +X

ACI 13.2.4

be = b w +2X

X= min(hb, 4 hf )

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Definition of the standard drop panel:

ACI 13.2.5

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ACI 9.5.3ge eam requ remen :

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Example 1

Solution:

n

n

l 760 45.7 714.3

l 714.3

cm= − =

min . .

30 30Use h = 24 cm

= = =

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Example 2

Determine the minimum thickness for the slab shown to satisfy ACI

deflection requirements for panels A and B. All columns are

. .

6.8 m

6.5 m A B

6.8 m

6.2 m 6.0 m 6.2 m

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Solution:

1- Determine initial depth of slab:

An initial value can be assumed. Assume slab is not su orted

with interior beams and use Table 9.5(c):

nl 680 30 650cm= − =n 650 19.7

33 33

Try h = 18 cm

inh cm= = =

2- Determine initial depth of beam:

Use Table 9.5(a):

68036.8

18.5 18.5in

lh m m= = =

3- Determine dimensions of equivalent beam

=

94 cm

b, f - ,

be = b w +2X = 30 + 2(32) =94 cm18 cm

cm

30 cm

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Solution:

4- Determine the centroid of the T-section beam:

30 32 32 / 2 94 18 32 18 / 2 A y + +.

30(32) 94(18)cm

A= = =

+∑

94 cm

18 cm

50 cm Y

30 cm

5- Determine the moment of inertia of the T-section beam:

3 21 (30)(32) (30)(32)(32 / 2 31.95)12

I = + −

3 2

4

1(94)(18) (94)(18)(32 18 / 2 31.95)

12

510409.4cm

+ + + −

=

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Solution:

6- Determine the moment of inertia of the slab:

In direction 1:

Direction 1

6.8 m2

(600 620)610

2l cm

+= =

3 41 (610)(18) 29646012

I cm= =

. m

In direction 2:

Direction 2

6.8 m2

3 4

6652

1

(665)(18) 323190

l cm

I cm

= =

= =

6.2 m 6.0 m 6.2 m

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Solution:

7- Determine αfm:

Due to s mmetr α = α α = α

6.8 m1 510409.4 1.72α = =

2

510409.41.57

323190α = = α

2

. m

1 2 1.652 fm

α α α

+= =

α3

α 4

α1

6.8 m

longer clear span=

8- Determine β :

6.2 m 6.0 m 6.2 m

shorter clear span

650 301.0877

600 30 β

−= =

−

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Solution:

9- Determine hmim:

ACI Eq 9-120.2 2mα < ≤

y

n

420

0.8 (650 30) 0.81400 1400

f

l

⎛ ⎞ ⎛ ⎞

+ − +⎜ ⎟ ⎜ ⎟

( ) ( )m. .

36 5 0.2 36 5 1.0877 1.65 0.2cm cm

β α = = = >

+ − + × −

Try 16 and repeat solution:

1 2

517955 5179552.49 2.28

208213.3 226986.7α α = = = =

1 2 2.42

fm

α α α

+= =

y

n

4200.8 650 30 0.8

f l

⎛ ⎞ ⎛ ⎞+ − +

q -mα >

14.9 936 9 36 9(1.0877)

h cm cm β

= = = >+ +

hmin

= 15 cm

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Solution:

1- Determine initial depth of slab:

Tr h = 15 cm as found from Panel A

2- Determine initial depth of beam:

Try h = 50 cm as before

-

X=min(hb, 4 hf ) = min(50-15=35 , 4(15)=60)=3565 cm

Y

e = w +X = 30 + 3 =6 cm

15 cm

50 cm

30 cm- -

30(35)(35) / 2 65(15)(35 15 / 2)29.54

30 35 65 15

A yY cm

A

+ += = =

+

∑

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Solution:

5- Determine the moment of inertia of the L-section beam:

3 2

3 2

1(30)(35) (30)(35)(35 / 2 29.54)

12

1

I = + −

−

65 cm

4

.12

441441cm=15 cm

50 cm

-

Since h = 15 cm, the moment of inertia for the interior beams is found in a manner

similar to slide 16:

4521342 I cm=

15 cm30 cm

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Solution:

6- Determine the moment of inertia of the slab:

In direction 1:

Direction 1

6.8 m(600 620)

Interior side:

+= =2

3 4

2

1(610)(15) 171563

12 I cm= =

. m

2

620 30325

2 2

Exterior side:

l cm= + =Direction 2

6.8 m3 41 (325)(15) 91406

12 I cm= =

6.2 m 6.0 m 6.2 m

2

(650 680)665l cm

+= =

In direction 2:

3 41(665)(15) 18703112 I cm= =

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Solution:

7- Determine αfm:

Due to s mmetr α = α .

1

5213423.04

171563

α = =

6.8 m2 2.79

187031

4414414.83

α

α

= =

= =

6.5 m1 2 3

91406

23.36

4 fm

α α α α

+ += =

α1

α 4

α3

6.8 m8- Determine β :

6.2 m 6.0 m 6.2 m

longer clear span

shorter clear span β =

−1.051

620 30

β = =−

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Solution:

9- Determine hmim:

y 4200.8 f l ⎛ ⎞ ⎛ ⎞+ −

ACI Eq 9-133.36 2mα = >

n .1400 1400

15 936 9 36 9(1.051)

h cm cm β

⎝ ⎠ ⎝ ⎠= = = >+ +

hmin = 15 cm

cd2_lect3_2-10-2011

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