Slab_Design_Example.pdf

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CE 433, Fall 2006 Slab Design Example 1 / 6

Design a one-way slab for an exterior bay of a multi-story office building using the

information specified below.

Plan View of Floor System

Assume partitions will not be damaged by deflections. Round slab thickness up to the nearest.

wSD = 20 psf

wLL = 80 psf

f'c = 4000 psi

fy = 60,000 psi

span = 80 ft

beam spacing = 16 ft

t_stab = ?

b_w of beam = 16 in

Lbeam Lbeam Lbeam

Beam

S

pacing

Beam

Spacing

Beam

Spacing

Beam

Spacing

Exterior Span Interior Span Exterior Span

slab

span


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1) Slab Thickness, t.

Use ACI Table 9.5(a): hmin = L / 28,

L = distance from center of support to center of support = beam spacing = 16 ft

hmin = (16 x 12/) / 28 = 6.857 , use h = 7.0

2) Design Flexure Reinforcement at each Section

- - - - - - - - - - - - - Section @ Midspan - - - - - - - - - -

2.1 Flexural Strength

2.1.1 Mu

ACI moment coefficient = 16 (ACI 8.3.3)

ftk

u

klf

u

beamofwn

u

ksfksfklf

u

kcfself

LSDself

u

nuu

MM

ftbspacingbeamspanclearl

klfww

klfw

wwww

lwM

==

====

=++=

==

++=

=

46.3,

16

)'67.14()257.0(

67.141/"12

"16'16

257.0),'1)(08.0(6.1)]'1)(02.0(0875.0[2.1

0875.0)/'"12

"7)(

/'"12

"12)(150.0(

6.1)(2.1

16

2

2

2.1.2 As for Mn > MuUsing spreadsheet, can use:

#4 @ 18, As = 0.133 in2

/ ft

#3 @ 9, As = 0.147 in2

/ ft

Use #3 @ 9 (better distribution), As = 0.147 in2

/ ft

Calc. Mn

dmax = 7 (0.75 + (3/8)) = 6.06, use d = 6.0

FH = 0, C = T

0.85 fc a b = As fy , Assume steel yields

0.85 (4 ksi) a (12) = (0.147 in2) (60 ksi), a = 0.216


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"0.6

003.

85.0

"216.0

003.0

003.0003.0

+=

+=

s

s

t dy

s = 0.0678

Therefore:

Steel has yielded (s > y = 0.002)

= 0.90 (et > 0.005)

Mn = As fy (d a/2) = (0.90)(0.147 in2)(60 ksi) [ (6.0 0.216/2)] 1/12

Mn = 3.89 k-ft

Mn = 3.89 k-ft > 3.46 k-ft = Mu, OK

2.2 Check for Ductile Failure

Under-reinforced if As < As_min

As_min = 0.0018 b h (ACI 7.12.2.1)

As_min = 0.0018 (12)(7), As_min = 0.151 in2

As = 0.147 in2

< 0.151 in2

= As_min , NG Use #3 @ 8, As = 0.165 in2

Over-reinforced ift < 0.004 (ACI 10.3.5)

Re-compute t since As has changed

0600.0,

6

003.0

85.0

243.0

003.0

243.0),60)(165.0()12()4(85.0

85.0:

003.0003.0

2

'

=+

=

==

=

+=

tt

ksiinksi

ysc

t

y

in

inaina

fAabfa

dy

t = 0.0600 > 0.004, OK

2.3 Check Distribution of Reinforcement

max spacing = min[ 3h, 18 ] ACI 7.12.2.2

max spacing = min[ 3 x 7, 18] = 18

spacing = 8 < 18, OK


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- - - - - - - - - - - - - Section @ Ext. Face of 1st Interior Support - - - - - - - - - -

2.1 Flexural Strength

2.1.1 Mu

10

)'67.14(

)257.0(10

22klfn

uu

l

wM == Mu = 5.53 k-ft

2.1.2 As for Mn > MuCarry of mid-span reinforcement full length of beam and through supportsAs: of #3 @ 8 = #3 @ 16, As = 0.083 in

2/ ft

Using spreadsheet, can use:

#5 @ 18, As = 0.207 in2

/ ft

#3 @ 6, As = 0.220 in2

/ ft

Use #3 @ 6 (better distribution), As = 0.220 in2 / ft

Calc. Mn

dmax = 7 (0.75 + (3/8)) = 6.06, use d = 6.0

d = 0.75 + (3/8 in) = 0.9378 use d = 1.0

from spreadsheet: yt = 0.381

a = b1 yt = 0.85 (0.381) a = 0.329

"0.6

003.

"381.0

003.0

003.0003.0

+=

+=

s

s

t dy

s = 0.0442

Therefore:

Steel has yielded (s > y = 0.002)

= 0.90 (es

> 0.005)

00487.0,"0.1"381.0"381.0

003.0

'

003.0

''

'

=

=

=

ss

t

s

t dyy

s = 0.00415Note: since d > yt , top steel is in tension

d y


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fs = es Es = (0.00487)(29,000 ksi) = 141 ksi > 60 ksi, use fs = 60 ksi

Cs = As fs = (0.083 in2)(60 ksi) Cs = 4.98 k

Mn = [As fy (d a/2) + Cs(d a/2)]= 0.90[(0.220 in2)(60 ksi)(6.0 0.446/2)

+ (4.98k)(1.0 0.446/2)] 1/12 Mn = 6.01 k-ft

Mn = 6.01 k-ft > 5.53 k-ft = Mu, OK

2.2 Check for Ductile Failure

Under-reinforced if As < As_min

As_min = 0.0018 b h (ACI 7.12.2.1)

As_min = 0.0018 (12)(7) = 0.151 in

2

As = 0.220 in

2< 0.151 in

2= As_min , OK

Over-reinforced ift < 0.004 (ACI 10.3.5) t = 0.0372 > 0.004, OK

2.3 Check Distribution of Reinforcement

max spacing = min[ 3h, 18 ] ACI 7.12.2.2

max spacing = min[ 3 x 7, 18] = 18

spacing = 8 < 18, OK

3. Temperature & Shrinkage Reinforcement

As_temp = 0.151 in2

(= As_min from pg 3/6)

max spacing = 18 Use #3 @ 8 (As = 0.165 in2) for Temp. & Shrink. Steel

(transverse to the flexural steel)

Sketch of Elevation View on following page.


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Slab_Design_Example.pdf