Question

A reinforced concrete beam of rectangular cross-section of breadth 230 mm and effective depth 400 mm is subjected to a maximum factored shear force of 120 kN. The grades of concrete, main steel and stirrup steel are M-20, Fe-415 and Fe-250 respectively. For the area of main steel provided, the design shear strength ${\tau }_c$ as per IS : 456-2000 is 0.48 N/mm${}^2$. The beam is designed for collapse limit state.

In addition, the beam is subjected to a torque whose factored value is 10.90 kN-m. The stirrups have to be provided to carry a shear (kN) equal to

• A. 50.42
• B. 130.56
• C. 151.67
• D. 200.23

Answer 1

The correct option is C 151.67

Torque = 10.9 kN-m
Shear force
V = 120 kN
Equivalent shear force,

$V_e=V+\frac{1.6T}{B}$

$=120+\frac{1.6\times 10.9}{0.230}$

= 195.826 kN
= 195826 N

Stirrups have to be provided to carry a shear force
$=V_e-V_c$
$=V_e-{\tau }_c\times bd$
$=195826-0.48\times 230\times 400$
$=151666N$
$=151.67kN$