Answer:
the nominal shear capacity of the section.=61505. 94ibor
61.51k
Step-by-step explanation:
given that
beam width = 14 in
effective depth = 20.5 in
fc = 3000psi
fyt = 60,000 psi
norminal share capacity vn = vc + vs
vc = 2d√ft bd
= 2 x1 x √3000 x14 x 20.5
=31439.27
vs = Avt + fytd/s
= 2 x 0.11 x 60,000 x 20.5/9
= 30066.67ib
therefore
vn = vc + vs
= 31439.27 + 30066.67
= 61505. 94ib
or
61.51k