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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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CE 433, Fall 2006 Slab Design Example 2 / 6
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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CE 433, Fall 2006 Slab Design Example 3 / 6
"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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CE 433, Fall 2006 Slab Design Example 4 / 6
- - - - - - - - - - - - - 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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CE 433, Fall 2006 Slab Design Example 5 / 6
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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