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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 45oand 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/mm2and maximum shear stress is 2.8 N/mm2. Using limit state method, determine the value of factored shear force to be permitted at the section.
  1. 157.4

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Solution

The correct option is A 157.4
Given b = 230 mm, d = 460 mm
Design shear strength of concrete (τnc)
= 0.6 N/mm2
shear resistance of concrete (Vnc)
=τnc×b×d
=0.6×230×4601000=63.48kN
Shear resistance of 2 number of 16 mm dia. bars bent up at 45ois given by
Vnsb=0.87×fy×Asbsin45o
=0.87×415×2×π×162412×11000
= 102.66 kN
Shear resistance of 6 mm dia. 2 legged spaced at 200 mm is,

Vnsv=0.87fyAsvdS
Asv=2×π4×62=56.55 mm2
d = 460 mm
S = 200 mm
Vnsv=0.87×415×56.55×460200
= 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.

Vsnb=Vnsv=46.96 kN (VnsbVns2)
Hence factored shear force permitted at the given section

=Vnc+Vnsb+Vnsr
=63.48+46.96+46.96=157.4 kN

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