Design of Columns MCQ with answer and solution

Concept

For an axially loaded short column, IS 456 gives Pu = 0.4 f_ck A_c + 0.67 f_y A_sc.

Formula

P_u = 0.4 f_ck (A_g − A_sc) + 0.67 f_y A_sc; p = (A_sc / A_g) × 100

Step-by-step solution

1. A_g = 400 × 400 = 160000 mm² 2. P_u = 1600 kN = 1600 × 10³ N 3. 1600 × 10³ = 0.4 × 20 × (160000 − A_sc) + 0.67 × 415 × A_sc 4. 1600000 = 1280000 − 8 A_sc + 278.05 A_sc 5. 320000 = 270.05 A_sc, so A_sc = 1185 mm² 6. p = 1185 / 160000 × 100 = 0.74 %

Explanation

Concrete takes 0.4 f_ck on its net area and steel takes 0.67 f_y. Put the load equal to the sum of both and find the steel area. The steel is about 0.74 % of the gross area, which is within the 0.8 % to 6 % range only after checking the minimum, but the calculated value is 0.74 %.

Why the other options are wrong

A (0.64) comes from an arithmetic error. B (7.40) and D (6.40) are about 10 times too large and are above the 6 % practical limit.

Shortcut

A_sc ≈ (P_u − 0.4 f_ck A_g) / (0.67 f_y − 0.4 f_ck) = 320000 / 270 ≈ 1185 mm².

Common mistake

Using gross area for concrete without deducting steel, or using 0.45 f_ck and 0.87 f_y (beam values) instead of 0.4 f_ck and 0.67 f_y.