Question

If the stirrup spacing is equal to 0.75 times the effective depth of an RC beam. then the shear capacity of stirrup steel is equal to

where $f_y$ is strength and $A_{sv}$ is cross-sectional area of the stirrup steel.

• A. $1.25\left(f_y{A}_{sv}\right)$
• B. $1.16\left(f_y{A}_{sv}\right)$
• C. $1.08\left(f_y{A}_{sv}\right)$
• D. $1.00\left(f_y{A}_{sv}\right)$

Answer 1

The correct option is B $1.16\left(f_y{A}_{sv}\right)$

$\because$ shear capacity $\left(V_{us}\right)$

$=\frac{0.87f_y{A}_{sv}d}{S_v}$

$\therefore$ $V_{us}=0.87f_y{A}_{sv}\times \frac{d}{0.75d}$

$=1.16\left(f_y{A}_{sv}\right)$