A simply supported beam with a length of 6 meters and an effective depth of 50 cm is subjected to a uniformly distributed load of 2400 kg/m, including its own weight. Given that the lever arm factor is 0.85 and the allowable tensile stress in steel is 1400 kg/cm², what is the required steel reinforcement area?

RCC Structures Design MCQs for PPSC, FPSC, NTS, and Pakistan government job tests. Select an option below, then read the explanation.

PPSCFPSCNTSPakistan govt jobs
Subject
RCC Structures Designcivil-engineering-mcqs › rcc-structures-design
Published
14 Jan 2019
Last updated
28 May 2026

Browse all RCC Structures Design MCQs

Choose the correct answer

Explanation

The required steel area is calculated based on the bending moment generated by the uniformly distributed load, effective depth, lever arm factor, and permissible tensile stress of steel. Using these values, the steel reinforcement needed is determined to be 16 cm².

PakQuizHub — free MCQs and past papers for Pakistan government job tests. Content is for educational practice only.