Question

A reinforced concrete beam with rectangular cross section (width = 300 mm, effective depth = 580 mm) is made of M30 grade concrete. It has 1% longitudinal tension reinforcement of Fe 415 grade steel. The design shear strength for this beam is 0.66 N/mm\(^2\). The beam has to resist a factored shear force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of Fe 415 grade steel is \(\underline{\hspace{2cm}}\) mm. (round off to the nearest integer)

Hint:

To determine the spacing of stirrups, calculate the required shear strength using the beam dimensions and then use it to find the appropriate spacing based on the reinforcement properties.

Answer 1

Correct Answer: 100

To calculate the spacing of the stirrups, we can use the formula for the design shear strength of a reinforced concrete beam:
\[ V_{\text{c}} = 0.66 \cdot b \cdot d \] where \( b \) is the width of the beam, and \( d \) is the effective depth. The total shear force that needs to be resisted is 440 kN, and the design shear strength is given as 0.66 N/mm\(^2\). Using this information, we first calculate the required shear strength, and then find the appropriate spacing for the stirrups based on the dimensions and the reinforcement properties. After performing the calculation, the required spacing of the stirrups is: \[ \text{Spacing of stirrups} = 102 \, \text{mm}. \] Thus, the spacing of the stirrups is \( \boxed{102} \, \text{mm} \).