Civil Engineering Reference
In-Depth Information
Solution
Check
l
n
/
d
, evaluate factored shear force,
V
u
,
d=h-
cover-1/2 dia. of bar =44-
0.5-0.625=42.875 in (1089 mm),
l
n
/
d
=69/42.875=1.609<5; hence treat as a
deep beam.
Factored shear force
V
u
and resisting capacity
V
c
From Eqn 11.30 (ACI Code)
Shear force due to shear reinforcement
V
s
: from Eq 11.31 (ACI Code)
=15690 l
b
(69.8 kN)
s
=
d
/5 =42.875/5=8.6 in (218 mm)<18 in (457 mm)
s
2
=
d
/3 =42.875/3=14.3 in (363 mm)<18 in (457 mm)
Minimum
A
v
=0.00 15
bs
=0.0015 (1.5) (8.6)=0.019 in
2
(12.3 mm
2
)
Provided
A
v
=0.04 in
2
(19.3 mm
2
)
Minimum
A
vh
=0.0025
bs
2
=0.0025 (1.5) (14.3)=0.054 in
2
(34.6 mm
2
)
Provided
A
vh
=0.067 in
2
(43.0 mm
2
)
Total shear force:
V
u
=
V
c
+
V
s
=61 880+15 690=77 570 lb (345.0 kN)
Test,
V
u
=900001b (400.3 kN).
