Question

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 ${\tau }_c$ of concrete as 0.62 N/mm${}^2$ and maximum allowable sher stress ${\tau }_{c,max}$ in concrete as 2.8 N/mm${}^2$ . 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_________.

• A. 8.2

Answer 1

The correct option is A 8.2
Given, $V_u=200kN,{\tau }_c=0.62$ MPa,
${\tau }_{cmax}=2.8$MPa
$\Rightarrow {\tau }_v=\frac{V_u}{bd}=\frac{200\times {10}^3}{250\times 350}=2.286$ MPa

As ${\tau }_{cmax}=2.8$MPa
and design shear force =(${\tau }_v-{\tau }_c$)bd
$=(2.286-0.62)\times 250\times 350$
= 145.775 kN
Spacing of shear reinforcement calculation:
$V_{us}$=145775
$=0.87\times 250\times 2\times \frac{\pi }{4}\times {10}^2\times \frac{350}{S_v}$
$\Rightarrow S_v=82.03$ mm
Spacing for minimum shear reinforcement
$\frac{A_{sv}}{b{S}_v}\ge \frac{0.4}{0.87f_y}$
$\Rightarrow S_v\le \frac{0.87f_y{A}_{sv}}{0.4b}$
$\Rightarrow S_v\le 341.65$ 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