A rectangular beam is 14 in. wide and has an effective depth of 20.5 in. The beam contains No. 3 (No. 10) stirrups spaced at 9 in. For f c = 3000 psi and f yt = 60,000 psi, calculate the nominal shear capacity of the section.

Question

Beatrice Shah

Engineering

19 December, 15:13

A rectangular beam is 14 in. wide and has an effective depth of 20.5 in. The beam contains No. 3 (No. 10) stirrups spaced at 9 in. For f c = 3000 psi and f yt = 60,000 psi, calculate the nominal shear capacity of the section.

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Answers (1)

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Gavin Combs · Answer 1

the nominal shear capacity of the section.=61505. 94ibor

61.51k

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