Fixed Beam MCQs – Part 1

Practice SSC JE and other civil engineering exam questions on fixed beams: fixed-end moments for a central, eccentric and uniformly distributed load, mid-span moment, points of contraflexure, moments due to sinking of a support, and the IS 800 design bending strength of a laterally unsupported beam that the book places in this chapter. Tap an option to check it, then open the full explanation for each question.

Question 1Medium

Which is the correct expression for the design bending strength of a laterally unsupported beam (IS 800: 2007)? fbd = design bending compressive stress, βb = Ze/Zp for semi-compact sections (1.0 for plastic and compact), Ze = elastic section modulus, Zp = plastic section modulus.

Full explanation

Answer: D. Md = βb Zp fbd

IS 800 clause 8.2.2: Md = βb·Zp·fbd.

Question 2Medium

A beam of span 5 m, fixed at A and B, carries a point load of 50 kN at 2 m from A. The fixed-end moments at supports A and B respectively are:

Full explanation

Answer: B. 36 kN·m anticlockwise and 24 kN·m clockwise

MA = Wab²/L² = 50 × 2 × 9/25 = 36 kN·m; MB = Wa²b/L² = 50 × 4 × 3/25 = 24 kN·m. At A the wall's moment on the beam is anticlockwise, at B clockwise.

Question 3Easy

For a fixed beam of span L carrying a uniformly distributed load w over the entire span, the bending moment at mid-span is:

Full explanation

Answer: B. wL²/24 sagging

Free (simply supported) moment wL²/8 minus fixing moment wL²/12 = wL²/24 sagging.

Question 4Easy

What will be the end moment of a fixed beam of length l carrying a uniformly distributed load of w kN/m?

Full explanation

Answer: D. ±wl²/12

Fixed-end moment for a full UDL = wl²/12 (equal at both ends, opposite senses).

Question 5Medium

A point load at mid-span of a fixed beam of uniform section produces fixed-end moments of 60 kN·m. The same load spread uniformly over the entire span will produce fixed-end moments equal to:

Full explanation

Answer: C. 40 kN·m

WL/8 = 60 → WL = 480. Spread as UDL: wL²/12 = WL/12 = 40 kN·m.

Question 6Medium

For a fixed beam AB of length L, what moments MA and MB are induced when the right support sinks by Δ (EI = flexural rigidity)?

Full explanation

Answer: A. MA = 6EIΔ/L², MB = 6EIΔ/L²

Relative settlement Δ with both ends kept from rotating gives equal end moments 6EIΔ/L² (hogging at one end, sagging at the other).

Question 7Medium

A fixed beam carries a uniformly distributed load over its entire span L. The points of contraflexure occur on either side of the centre at a distance of:

Full explanation

Answer: C. L/(2√3)

Mx = wLx/2 − wx²/2 − wL²/12 = 0 → x = (L/2)(1 ± 1/√3); distance from centre = L/(2√3) ≈ 0.289L.