Structures · MCQ practice Columns and Struts MCQs – Part 2
15 questions. Tap an option to check it instantly, then read the solution.
Part 2 of 3 Question 16 Easy
Slenderness ratio of a long column is:
A area of cross-section divided by radius of gyration B area of cross-section divided by least radius of gyration C radius of gyration divided by area of cross-section D length of column divided by least radius of gyration
Answer: D. length of column divided by least radius of gyration
λ = (effective) length / least radius of gyration.
Question 17 Easy
The effective length of a column effectively held in position and restrained in direction at one end, but neither held in position nor restrained in direction at the other end, is:
A L B 0.67L C 0.85L D 2L
Answer: D. 2L
Held and restrained = fixed; neither held nor restrained = free. Fixed–free → le = 2L.
Question 18 Medium
A column has effective length L when both its ends are fixed. What will be the new effective length if one end becomes hinged?
A L B 0.5L C 1.41L D 2L
Answer: C. 1.41L
Actual length l = 2L. Fixed–hinged: le = l/√2 = 2L/√2 = √2 L ≈ 1.41L.
Question 19 Easy
Euler's crippling load for a column of length L with one end fixed and the other hinged is:
A π²EI/L² B 4π²EI/L² C π²EI/(4L²) D 2π²EI/L²
Answer: D. 2π²EI/L²
le = L/√2 → P = π²EI/(L²/2) = 2π²EI/L².
Question 20 Medium
The slenderness ratio of a 5 m long column hinged at both ends and having a circular cross-section of diameter 16 cm is:
A 31.25 B 62.5 C 100 D 125
Answer: D. 125
r = d/4 = 4 cm; le = 500 cm; λ = 500/4 = 125.
Question 21 Medium
When both ends of a column are fixed, the crippling load is F. If one end of the column is made free, the crippling load will change to:
A F/4 B F/2 C F/16 D 4F
Answer: C. F/16
Fixed–fixed le = L/2; fixed–free le = 2L. Ratio of le = 4, so load ratio = 1/16.
Question 22 Easy
A column of length L with both ends fixed may be considered equivalent to a column of length ______ with both ends hinged.
A L/8 B L/2 C L/4 D L
Answer: B. L/2
Fixed–fixed le = L/2, i.e. a pin-ended column of length L/2.
Question 23 Easy
According to Euler's column theory, the crippling load for a column of length l fixed at both ends is ______ the crippling load for a similar column hinged at both ends.
A equal to B 4 times C 2 times D 8 times
Answer: B. 4 times
4π²EI/l² ÷ π²EI/l² = 4.
Question 24 Easy
A column with maximum equivalent length has:
A both ends hinged B both ends fixed C one end fixed and the other end hinged D one end fixed and the other end free
Answer: D. one end fixed and the other end free
Fixed–free: le = 2l, the largest.
Question 25 Easy
The equivalent length of a column of length l fixed at both ends is:
A 0.5l B 0.7l C l D 2l
Answer: A. 0.5l
Fixed–fixed: le = l/2 = 0.5l.
Question 26 Easy
Euler's formula is valid for:
A short columns only B long columns only C both short and long columns D none of the above
Answer: B. long columns only
Euler's formula ignores direct crushing, so it applies only to long (slender) columns that fail by buckling.
Question 27 Easy
The buckling load for an axially loaded column with both ends fixed is given by:
A π²EI/l² B 2π²EI/l² C 4π²EI/l² D π²EI/(4l²)
Answer: C. 4π²EI/l²
le = l/2 → P = 4π²EI/l².
Question 28 Easy
The allowable stress in a long column can be increased by increasing the:
A radius of gyration B eccentricity C slenderness ratio D length of the column
Answer: A. radius of gyration
Larger r → smaller λ = le/r → higher allowable (buckling) stress.
Question 29 Easy
A compression member is termed a column or strut if the ratio of its effective length to its least lateral dimension is more than:
A 1 B 2 C 3 D 5
Answer: C. 3
IS 456: a compression member with le/least lateral dimension > 3 is a column; ≤ 3 is a pedestal.
Question 30 Easy
If a column's ends are effectively held in position and restrained against rotation at both ends, the effective length is:
A 2L B L/2 C 0.707L D L
Answer: B. L/2
Held in position + restrained against rotation at both ends = fixed–fixed → le = L/2.
Part 2 of 3