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.

The spacing (mm) of 2-legged 8 mm stirrups to be provided is

• A. 40
• B. 115
• C. 250
• D. 400

Answer 1

The correct option is B 115

Breadth, b = 230 mm
Effective depth, d = 400 mm
Factored shear force
$V_u=120kN$
Design shear strength,
${\tau }_c=0.48N/m{m}^2$
Area of stirrup,

$A_{sv}=\frac{\pi d^2}{4}\times 2=\frac{\pi \times 8^2}{4}$

$=100.53m{m}^2$

Nominal shear stress,

${\tau }_v=\frac{120\times 1000}{230\times 400}$

$=1.304N/m{m}^2$

Design shear force,

$V_s=({\tau }_V-{\tau }_C)bd$

$=(1.304-0.48)\times 230\times 400$

= 75840 N
$V_s=0.87f_y.A_{sv}.\frac{d}{S}$
Spacing,

$S=\frac{0.87\times 250\times 100.53\times 400}{75840}$
= 115.32 mm

Checking from minimum reinforcement criteria

$\frac{A_{sv}}{b.S_v}>\frac{0.4}{0.87f_y}$

$\frac{100.53}{230.S_v}>\frac{0.4}{0.87\times 250}$

$S_v<237.66$ mm

Hence O.K.