Question

A reinforced concrete beam has a support section with width of 300 mm and effective depth of 500 mm. The support section is reinforced with 3 bars of 20 mm diameter at the tension side. Two-legged vertical stirrups of 10 mm diameter and Fe415 steel at a spacing of 100 mm are provided as shear reinforcement. Assume that there is no possibility of diagonal compression failure at the section.
As per IS 456:2000, the maximum shear resisted by the vertical stirrups (in kN), as per limit state design, is ......... (round off to one decimal place).

Hint:

To calculate the maximum shear resisted by stirrups, use the given values of the stirrup spacing, yield strength of steel, and the area of stirrups to compute the shear force resisted by the stirrups using the formula provided by IS 456.

Answer 1

Correct Answer: 283

Given: - Two-legged vertical stirrups of diameter \( \phi = 10 \, {mm} \) - \( c/c \) spacing \( S_v = 100 \, {mm} \) - Yield strength of steel \( f_y = 415 \, {N/mm}^2 \) - Effective depth \( d = 500 \, {mm} \) The spacing for vertical shear stirrups is given by: \[ S_v = \frac{0.87 \times A_{sv} \times d}{V_s} \] Where: - \( A_{sv} = 2 \times \frac{\pi}{4} \times \phi^2 \) (cross-sectional area of the stirrups) - \( V_s \) is the shear force resisted by the stirrups Using the formula for \( A_{sv} \): \[ A_{sv} = 2 \times \frac{\pi}{4} \times (10^2) = 2 \times 78.54 = 157.08 \, {mm}^2 \] Now, substituting the values into the formula for \( V_s \): \[ V_s = \frac{0.87 \times f_y \times A_{sv} \times d}{S_v} = \frac{0.87 \times 415 \times 2 \times \frac{\pi}{4} \times 10^2 \times 500}{100} \] \[ V_s = \frac{0.87 \times 415 \times 2 \times 78.54 \times 500}{100} = 283568 \, {N} \] \[ V_s \approx 283.6 \, {kN} \] Thus, the maximum shear resisted by the vertical stirrups is \( 283.6 \, {kN} \).