Civil Engineering Reference
In-Depth Information
where d = 7.25 + 2.25 = 9.50 in.
6. Slab Reinforcement
Required slab reinforcement is easily determined using a tabular form as follows:
A
s
=
M
u
/4d
(in.
2
)
b
1
(in.)
b
2
(in.)
A
s
3
(min)
(in.
2
)
Span Location
M
u
(ft-kips)
No. of
#4 Bars
No. of
#5 Bars
END SPAN
Column
Strip
Ext. Negative
146.75
-51.4
120
120
7.25
7.25
5.06
1.77
1.84
0.00
26
9
17
9
Positive
75.9
120
7.25
2.62
1.84
14
9
Int. Negative
194.3
-20.1
120
120
9.5
7.25
5.11
0.69
2.32
0.00
26
9
17
9
Middle
Ext. Negative
2.5
168
7.25
0.09
2.57
13
9
Strip
Positive
50.6
168
7.25
1.74
2.57
13
9
Int. Negative
43.1
168
7.25
1.49
2.57
13
9
INTERIOR SPAN
Column
Positive
53.2
120
7.25
1.83
1.84
10
9
Strip
Middle
Positive
40
168
7.25
1.38
2.57
13
9
Strip
1
Column strip = 0.5(20 x 12) = 120 in. (see Fig. 4-9b)
Middle Strip = (24 x 12) - 120 = 168 in.
2
Use average d = 8.5 - 1.25 = 7.25 in.
At drop panel, d = 7.25 + 2.25 = 9.50 in. (negative moment only)
3
A
s(min)
= 0.0018 bh
s
max
= 2h < 18 in = 2(8.5) = 17 in.
7. Check slab reinforcement at interior columns for moment transfer between slab and column. Shear strength
of slab already checked for direct shear and moment transfer in Step (2)(b). Transfer moment (unfactored)
at 1st-story due to wind, M
w
= 111.56 ft-kips.
Fraction transferred by flexure using ACI Eqs. (13-1) and (9-6):
M
u
= 0.60(1.6
111.56) = 107.1 ft-kips
A
s
= M
u
/4d = 107.1/(4
9.50) = 2.82 in.
2
For No. 5 bars, 2.82/0.31 = 9.1 bars, say 10-No. 5 bars
Must provide 10-No. 5 bars within an effective slab width = 3h + c
2
= 3(10.75) + 16 = 48.3 in.
Provide the required 10-No. 5 bars by concentrating 10 of the column strip bars (18-No. 5) within the 4-ft
slab width over the column. Distribute the other 8 column strip bars (4 on each side) in the remaining
column strip width. Check bar spacing:
48/9 spaces = ± 5.3 in.
(120 - 48)/7 spaces = ± 10.3 in. < 18 in. O.K.
Search WWH ::
Custom Search