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An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strength τc of concrete as 0.62 N/mm2 and maximum allowable sher stress τc,max in concrete as 2.8 N/mm2 . If two legged 10 mm diameter vertical stirrups of Fe 250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per limit state method will be_________.
  1. 8.2

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Solution

The correct option is A 8.2
Given, Vu=200kN,τc=0.62 MPa,
τc max=2.8MPa
τv=Vubd=200×103250×350=2.286 MPa

As τc max=2.8MPa
and design shear force =(τvτc)bd
=(2.2860.62)×250×350
= 145.775 kN
Spacing of shear reinforcement calculation:
Vus=145775
=0.87×250×2×π4×102×350Sv
Sv=82.03 mm
Spacing for minimum shear reinforcement
AsvbSv0.40.87fy
Sv0.87fyAsv0.4b
Sv341.65 mm

Spacing should be minimum of
(i) 0.75d = 26.25 cm (ii) 8.2 cm
(iii) 34.16 cm (iv) 30 cm

So spacing will be 8.2 cm

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