Structural Analysis and Strength of Materials MCQ with Solutions (NEC Chapter 4)

Chapter 4: Structural Mechanics MCQ

235 questions with answers. Practise by sub-section or take a timed test.

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Structural Mechanics MCQ: what this set covers

Structural mechanics rewards speed with standard results: maximum bending moments, deflection coefficients, Euler loads, carry-over factors and influence line ordinates. The questions here mix those results with short numerical problems, and each explanation shows the working.

There are 235 questions in this chapter, each with the answer and a short explanation. Choose a sub-section, pick how many questions you want, and either see the answer straight away or take it as a timed test. Your mistakes are saved in your browser, so you can come back and practise only those.

Code	Sub-section	Questions here
4.1	Shear force and bending moment	32
4.2	Stress and strain analysis	32
4.3	Theory of flexure and columns	31
4.4	Determinate structures I	32
4.5	Determinate structures II	32
4.6	Indeterminate structures	32

Sample questions with answers

A few questions from each sub-section are shown below. Tap a question to see the answer. Use the practice tool above for the full set.

4.1 Shear force and bending moment

A simply supported beam of span L carries a point load W at midspan. The maximum bending moment is:

WL/2
WL/4
WL/8
WL/16

Answer: B. Each reaction is W/2, so the moment under the load is (W/2)(L/2) = WL/4.

The maximum bending moment in a simply supported beam of span L carrying a UDL of w per metre over the whole span is:

wL^2/12
wL^2/24
wL^2/8
wL^2/2

Answer: C. Reactions are wL/2; at midspan M = (wL/2)(L/2) - w(L/2)(L/4) = wL^2/8.

4.2 Stress and strain analysis

Hooke's law, stating that stress is proportional to strain, holds good up to the:

elastic limit
proportional limit
upper yield point
ultimate stress

Answer: B. Linear proportionality between stress and strain ends at the limit of proportionality; the elastic limit lies slightly beyond it.

The relation between modulus of elasticity E, modulus of rigidity G and Poisson's ratio mu is:

E = 2G(1 + mu)
E = 3G(1 + mu)
E = 2K(1 + mu)
E = 3K(1 + 2mu)

Answer: A. E = 2G(1 + mu). The corresponding bulk modulus relation is E = 3K(1 - 2mu).

4.3 Theory of flexure and columns

The equation of simple bending is:

M/I = sigma/y = E/R
M/J = tau/r = G/L
M/I = y/sigma = R/E
M/Z = E/y = sigma/R

Answer: A. M/I = σ/y = E/R, where y is the distance from the neutral axis and R the radius of curvature.

Which of the following is an assumption of the theory of simple bending?

The beam is initially curved
Plane sections remain plane after bending
E in tension differs from E in compression
Shear stress is uniform over the section

Answer: B. Simple bending assumes an initially straight, homogeneous beam with equal E in tension and compression, and that transverse plane sections remain plane, so strain varies linearly with y.

4.4 Determinate structures I

For a plane pin-jointed truss with m members, j joints and r support reactions, the degree of static indeterminacy is:

m + r - 3j
m + r - 2j
3m + r - 3j
2j - m - r

Answer: B. Each joint gives 2 equilibrium equations, so Ds = (m + r) - 2j. Ds = 0 is determinate, Ds > 0 indeterminate and Ds < 0 a mechanism. (m - 2j + 3 gives only the internal indeterminacy, which equals Ds when r = 3.)

A plane truss has 11 members, 7 joints and 3 support reactions. It is:

statically determinate
indeterminate to degree 1
indeterminate to degree 2
unstable

Answer: A. Ds = m + r - 2j = 11 + 3 - 14 = 0, so it is statically determinate (assuming a stable arrangement).

4.5 Determinate structures II

An influence line diagram shows the variation of:

a response function at a fixed section as a unit load moves across the structure
bending moment along the span for a fixed load
deflection along the span for a fixed load
shear force along the span for a UDL

Answer: A. An ILD is drawn for one fixed section or reaction; each ordinate gives the value of that response when a unit load stands at that point. A BMD instead shows moment at all sections for one fixed loading.

The influence line for the reaction at support A of a simply supported beam AB is:

a rectangle of height 1
a triangle with ordinate 1 at A and 0 at B
a parabola with maximum at midspan
a triangle with ordinate 1 at midspan

Answer: B. With the unit load at x from A, RA = (L - x)/L: a straight line from 1 at A to 0 at B.

4.6 Indeterminate structures

The fixed-end moment at each support of a fixed beam of span L carrying a UDL of w per metre over the whole span is:

wL^2/8
wL^2/12
wL^2/24
wL^2/16

Answer: B. FEM = wL²/12 (hogging) at each end; the midspan moment is wL²/8 - wL²/12 = wL²/24 (sagging).

The fixed-end moment at each support of a fixed beam of span L carrying a central point load W is:

WL/4
WL/12
WL/8
WL/16

Answer: C. FEM = WL/8 at each end and the midspan moment is WL/4 - WL/8 = WL/8.

Full mock tests

When you have covered every chapter, take a full NEC licence mock test or Lok Sewa model test. The NEC mock follows the current NEC pattern of 100 one-mark questions, the format used in the model question sets NEC published in November 2025.