In IS 456 (LSM), the design of a short axially loaded column uses the formula:
Choose the correct answer
Pu = fck·Ag
Pu = 0.4fck·Ac + 0.67fy·Asc
Pu = 0.87fy·Asc only
Pu = 0.5fck·Ac
Correct Answer
B. Pu = 0.4fck·Ac + 0.67fy·Asc
AI Detailed Explanation & IS Code Reference
Unlock the reasoning, formula path and code-linked notes inside your student dashboard.
IS 456 Cl. 39.3: Pu = 0.4fck·Ac + 0.67fy·Asc where Pu = factored axial load, Ac = net concrete area, Asc = area of steel. The factors 0.4 and 0.67 account for partial safety factors (0.67fck/1.5 ≈ 0.447fck rounded, and 0.87fy adjusted for long column reduction — actually 0.4 and 0.67 are IS 456 specific values).
ScoreCardAI links this solution with subject, topic and difficulty signals so your scorecard can identify weak areas after a full mock test.
Practice more Civil Engineering questions
This MCQ belongs to JE Level Premium Test Series. Full tests include timed attempts, rank comparison and subject-wise analysis.