Page 214 - Civil Engineering Formulas

CONCRETE FORMULAS 149

where V total design shear force
u
capacity reduction factor 0.85
d 0.8l w
h overall thickness of wall
l horizontal length of wall
w
The shear V carried by the concrete depends on whether N , the design axi-
u
c
al load, lb (N), normal to the wall horizontal cross section and occurring simul-
taneously with V at the section, is a compression or tension force. When N is
u
u
a compression force, V may be taken as 2 f c hd , where f c is the 28-day
c
2
strength of concrete, lb/in (MPa). When N is a tension force, V should be tak-
c
u
en as the smaller of the values calculated from
N u d
f
V c 3.3 2 c hd (5.113)
4l w
l w (1.25 2 c 0.2N u /l w h)
f
V c hd 0.6 2 c M u /V u l w /2 (5.114)
f
This equation does not apply, however, when M /V l /2 is negative.
u
w
u
When the factored shear V is less than 0.5 V , reinforcement should be
c
u
provided as required by the empirical method for bearing walls.
When V exceeds 0.5 V , horizontal reinforcement should be provided with V
s
u
c
A f d/s , where s spacing of horizontal reinforcement and A reinforcement
2
v y
2
v
area. Also, the ratio of horizontal shear reinforcement to the gross concrete area of
h
the vertical section of the wall should be at least 0.0025. Spacing of horizontal shear
bars should not exceed l /5, 3h, or 18 in (457.2 mm). In addition, the ratio of vertical
w
shear reinforcement area to gross concrete area of the horizontal section of wall does
not need to be greater than that required for horizontal reinforcement but should not
be less than
h w
n 0.0025 0.5 2.5 (5.115)
l w
( h 0.0025) 0.0025
where h total height of wall. Spacing of vertical shear reinforcement should not
w
exceed l /3, 3h, or 18 in (457.2 mm).
w
In no case should the shear strength V be taken greater than 10 f c hd at
n
any section.
Bearing stress on the concrete from anchorages of posttensioned members with
adequate reinforcement in the end region should not exceed f calculated from
b
A b
f b 0.8 f c 0.2 1.25 f ci (5.116)
BA b
f
f b 0.6 2 c A b f c (5.117)
BA b