Civil Engineering Reference
In-Depth Information
Depth Required for Two-Way Shear (ACI Equations 11.33 and 11.34 not shown as they do
not control)
2
42.5
12
V
u
2
at right column
480
(5.85)
406.6 k
4
06,60
0
(0.75)(4
3000)(4
42.5)
d
14.56
22.5
OK
29.25
40.4
144
V
u
2
at left column
304
(5.85)
255.9 k
25
5,900
d
(0.05)(4
3000)(2
29.25
40.5)
15.73
25.5
OK
Design of Longitudinal Steel (Space is not taken to revise depth)
M
u
729.5 ft- k
M
u
(12)(729,500)
(0.90)(99)(22.5)
2
194.1 psi
bd
2
from Table A.12 by interpolation
0.00337
A
s
(0.00337)(99)(22.5)
7.51 in.
2
M
u
171.3 ft-k (computed from right end of shear diagram, Figure 12.21)
say 10 #8 (7.85 in.
2
)
M
u
(12)(171,300)
(0.90)(99)(22.5)
2
bd
2
45.6 psi
p
min
3
3000
60,000
Use larger of 200/
f
y
0.00333 or
0.00274
A
s
(0.00333)(99)(22.5)
7.42 in.
2
Use 8 #9 (8.00 in.
2
)
Design of Short-Span Steel under Interior Column
d
2
Assuming steel spread over width
column width
(2)
22.5
2
20
(2)
42.5
Referring to Figure 12.22 and calculating
M
u
:
q
u
480
8.25
58.18 k/ft
3.29
2
M
u
(3.29)(58.18)
314.9 ft- k
M
u
(12)(314,900)
(0.90)(42.5)(22.5)
2
195.1 psi
bd
2
p
0.00339 from Appendix Table A.12
A
s
(0.00339)(42.5)(22.5)
3.24 in.
2
Use 6 #7 (3.61 in.
2
)
Development lengths check out OK.
