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8.1. A rectangular beam is 16 in. wide and 29 in. deep. For
c
f′=
4000 psi and
y
f=
60,000 psi,
determine the required spacing of No. 4 (No. 13) closed stirrups for a factored shear of 80 kips and
factored torsional moment of 50 ft-kips. The stirrup centroid is located 1.75 in. from each concrete
face. The effective depth is 26.5 in.
( )
( ) ( )
2
22
16 29 464 in ; 2 16 29 90 in.
λ 0.75 1.0 4000 464 90 12,000 9.45 ft-kips < Must design for torsion
cp cp
th c cp cp u
Ap
T fAp T
ff
=×= = + =
′
==×=
( ) ( )
22
16 2 1.75 12.5 in.
29 2 1.75 25.5 in.
12.5 25.5 319 in ; 0.85 0.85 319 271 in
2 =2 12.5 25.5 =76 in.
o
o
oh o o o oh
h oo
x
y
A xy A A
p xy
= −× =
= −× =
= =× = = = ×=
=++
Check stresses:
()
( )
2
2
2
22
2
8 10
1.7
80 50 12 76 0.75 10 4000
16 26.5 1.7 319
0.324 ksi 0.474 ksi Okay
u uh c cc
ww
oh
V Tp V ff
bd bd
A
ff
′′
+ ≤ +=
××
+≤
××
≤
Calculate stirrup spacing: Use
θ
= 45 deg.
50 12 0.0246
2 cot 2 0.75 271 60 1
Area of 2 legs 2 0.0246 0.0492
u
t
oy
Ts s
As
Af
ss
fθ
×
= = =
× × ××
=×=
( ) ( )
0.75 2 1.0 4000 16 26.1
2λ 40.2 kips
1000
80 40.2 0.0334
0.75 60 26.5
2 0.0492 0.0334 0.0826
c cw
uc
v
yt
tv
V fbd
V Vs s
As
fd
AA s s s
ff
f
f
×× × ×
′
= = =
−−
= = =
××
+= + =
For No. 4 (No. 13) bars,
2 2 0.20 0.40 in.
tv
AA+=× =
, giving
0.40 0.0826 4.82 in.s= =
Check minimum steel requirement:
min
min
76 26.5
9.5 in.; 12 in.; 13.25 in.
88 22
0.75 0.75 4000 16 4.82 0.06 in.
60,000
50 50 16 4.82 0.064 in.
60,000
h
cw
v
yt
w
v
yt
pd
s ss
fbs
Af
bs
Af
≤ = = ≤ ≤= =
′××
≥= =
××
≥= =
Use s = 4.5 in.
