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Question

A RC beam of rectangular cross - section of breadth 250 mm and effective depth 450 mm is subjected to a maximum factored shear force of 150 kN. The grade of concrete is M25 and grade of main steel and stirrups are Fe415 and Fe250 respectively. For the area of main steel provided, the design shear strength τcas per IS 456: 2000 is 0.56N/mm2. If the beam is designed for collapse limit state then the spacing of 2 - legged 10 mm vertical stirrups to be provided is mm

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Solution

Given, factored (ultimate) design shear force, Vu=150kN
Also, ultimate shear capacity of concrete,
Vuc=τcbd
=0.56×250×450N=63kN
Ultimate shear force to be resisted by shear reinforcement,
Vus=VuVuc
=15063=87kN
For vertical stirrups, spacing is given as
Sv=0.87fyASDdVus
=0.87×250×2×π4×102×45087×103=176.71mm

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