Question

If the effective depth of an $RCC$ beam is $4/3$ times the stirrups spacing. Then the shear capacity of stirrup steel is equal to:

• A. $\frac{29}{25}{f}_y{A}_{sv}$
  
• B. $\frac{29}{23}{f}_y{A}_{sv}$
  
• C. $\frac{4}{3}{f}_y{A}_{sv}$
  
• D. $\frac{13}{11}{f}_y{A}_{sv}$
  

Answer 1

The correct option is A $\frac{29}{25}{f}_y{A}_{sv}$


Given, $d=\frac{4}{3}{S}_v$
Shear capacity $=0.87f_y{A}_{sv}\cdot \frac{d}{S_v}$
$=0.87f_y{A}_{sv}\times \frac{4}{3}$
$=\frac{29}{25}{f}_y{A}_{sv}$
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