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Columns and Struts MCQs – Part 1

15 questions. Tap an option to check it instantly, then read the solution.

Practice SSC JE and other civil engineering exam questions on columns and struts: Euler's buckling theory, crippling (critical) load, effective length for different end conditions (theory and IS 800 / IS 456 values), slenderness ratio and radius of gyration, short columns under eccentric load, the middle-third rule and the core (kern) of rectangular and circular sections, plus a few related questions the book places in this chapter. Tap an option to check it, then open the full explanation for each question.

Question 1Easy
As per Euler's theory, buckling of a long column occurs when ______ (P = applied axial load, Pcr = crippling load by Euler's theory).
Full explanation

Answer: B. P > Pcr

Below Pcr the straight shape is stable; once the axial load exceeds Pcr the column cannot stay straight and buckles.

Question 2Medium
Consider a rectangular column cross-section of size 4 m × 2 m. Identify the correct statement.
Full explanation

Answer: A. The point load shall be applied within the rhombus located at the centre with diagonals 1.33 m × 0.66 m to ensure no tensile stress develops in the column cross-section.

Core of a b × d rectangle = rhombus with diagonals b/3 and d/3 = 4/3 = 1.33 m and 2/3 = 0.67 m.

Question 3Easy
As per Euler's theory, for which of the following end conditions is the longest effective length of a column obtained?
Full explanation

Answer: C. When one end of the column is fixed and the other end is free

Fixed–free: le = 2L, the largest of the four standard cases.

Question 4Medium
The actual length of a prismatic compression member is 3 m. Calculate its effective length if one of its ends is restrained against both translation and rotation, and the other end is restrained against rotation but free to translate.
Full explanation

Answer: B. 3.6 m

IS 800 Table 11: fixed + (rotation restrained, translation free) → le = 1.2L = 1.2 × 3 = 3.6 m.

Question 5Easy
The maximum value of the radius of gyration is obtained if the material ______.
Full explanation

Answer: D. is placed farthest from the centroid

r = √(I/A); for the same area, I grows with the square of the distance from the centroid, so r is largest when the material is farthest away.

Question 6Medium
Four columns are identical in all parameters except end conditions: (i) both ends hinged, (ii) both ends fixed, (iii) one end fixed and the other free, (iv) one end fixed and the other hinged. Arrange them in ascending order of load-carrying capacity.
Full explanation

Answer: D. iii, i, iv, ii

P ∝ 1/le²: iii (0.25) < i (1) < iv (2) < ii (4) in units of π²EI/L².

Question 7Easy
What is the expression for the crippling load of a column with both ends hinged?
Full explanation

Answer: C. P = π²EI/l²

Both ends hinged: le = l, so P = π²EI/l² (the basic Euler load).

Question 8Easy
What is the relation between equivalent length (Le) and actual length (l) for a column with one end fixed and the other free?
Full explanation

Answer: D. Le = 2l

A fixed–free column buckles as a quarter sine wave, i.e. half of a pin-ended column of length 2l: Le = 2l.

Question 9Easy
A column of length L has flexural rigidity EI; one end is fixed and the loaded end is free. The critical load will be:
Full explanation

Answer: B. Pc = π²EI/(4L²)

le = 2L → Pc = π²EI/(2L)² = π²EI/(4L²).

Question 10Easy
The maximum axial load which is just sufficient to keep a column in a small deflected shape is called:
Full explanation

Answer: D. All options are correct

Crippling load, buckling load and critical load are three names for the same Euler load.

Question 11Easy
The ratio of the length of a column to the minimum radius of gyration of its cross-sectional area is known as:
Full explanation

Answer: A. slenderness ratio

λ = le / rmin is the slenderness ratio.

Question 12Easy
The equivalent length of a column of length L having both ends hinged is:
Full explanation

Answer: B. L

The pin-ended column is the reference case: le = L.

Question 13Easy
The critical load for a column of length L hinged at both ends and having flexural rigidity EI is given by:
Full explanation

Answer: B. Pc = π²EI/L²

Euler: Pc = π²EI/L² for a pin-ended column.

Question 14Easy
The equivalent length of a column of length L having one end fixed and the other end hinged is:
Full explanation

Answer: D. L/√2

Fixed–hinged: le = L/√2 ≈ 0.707L.

Question 15Easy
The equivalent length of a column of length L having both ends fixed is:
Full explanation

Answer: C. L/2

Fixed–fixed: inflection points at L/4 from each end, so le = L/2.