As per IS 456: 2000, the design strength of a short axially loaded compression member is expressed as _____, Where, fy = characteristic strength of the compression reinforcement, Asc = area of longitudinal reinforcement for columns, fck = characteristic compressive strength of the concrete, Ac = Area of concrete.
- A[0.87 fck Ac + fsc Asc]
- B[0.45 fsc Asc + fck Ac]
- C[0.87 fsc Asc + fck Ac]
- D[0.4 fck Ac + 0.67 fy Asc]Correct
Solution
Concept
For a short axially loaded column, IS 456 gives Pu = 0.4 fck Ac + 0.67 fy Asc. This includes partial safety factors and allows for minimum eccentricity.
Formula
Pu = 0.4 fck Ac + 0.67 fy Asc
Step-by-step solution
1. Design stress in concrete under compression = 0.67 fck / 1.5 = 0.446 fck, taken as about 0.45 fck. 2. Design stress in steel = 0.87 fy; in compression it is about 0.75 fy. 3. For minimum eccentricity, IS 456 reduces both values by about 10%: 0.45 fck becomes 0.4 fck, and 0.75 fy becomes 0.67 fy. 4. So Pu = 0.4 fck Ac + 0.67 fy Asc, which is option D.
Correct answer: D) [0.4 f_ck A_c + 0.67 f_y A_sc]