Question

A singly reinforced concrete beam has rectangular section 230 mm wide x 460 mm effective depth. At a section under consideration there are two 16 mm diameter bent-up bars at ${45}^o$and 6 mm diameter two legged stirrups at a spacing of 200 mm centre to centre. The grade of concrete is M20. The reinforcement is HYSD 415. The design shear strength of concrete is 0.6 N/mm${}^2$and maximum shear stress is 2.8 N/mm${}^2$. Using limit state method, determine the value of factored shear force to be permitted at the section.

• A. 157.4

Answer 1

The correct option is A 157.4
Given b = 230 mm, d = 460 mm
Design shear strength of concrete (${\tau }_{nc}$)
= 0.6 N/mm${}^2$
$\therefore$ shear resistance of concrete ($V_{nc}$)
$={\tau }_{nc}\times b\times d$
$=\frac{0.6\times 230\times 460}{1000}=63.48kN$
Shear resistance of 2 number of 16 mm dia. bars bent up at ${45}^o$is given by
$V_{nsb}=0.87\times f_y\times A_{sb}sin{45}^o$
$=0.87\times 415\times \frac{2\times \pi \times {16}^2}{4}\frac{1}{\sqrt{2}}\times \frac{1}{1000}$
= 102.66 kN
Shear resistance of 6 mm dia. 2 legged spaced at 200 mm is,

$V_{nsv}=\frac{0.87f_y{A}_{sv}d}{S}$
$A_{sv}=2\times \frac{\pi }{4}\times 6^2=56.55m{m}^2$
d = 460 mm
S = 200 mm
$\therefore V_{nsv}=\frac{0.87\times 415\times 56.55\times 460}{200}$
= 46.96 kN

According to I.S code, the shear resistance of the bent up should not exceed 50% of the total shear required to be carried by both vertical stirrups and bent up bars.

$\therefore V_{snb}=V_{nsv}=46.96kN$ $\therefore (V_{nsb}\ngtr \frac{V_{ns}}{2}$)
Hence factored shear force permitted at the given section

$=V_{nc}+V_{nsb}+V_{nsr}$
$=63.48+46.96+46.96=157.4$ kN