Engineering Mechanics MCQ Practice Set — 12 Questions with Answers
12 exam-oriented Engineering Mechanics multiple-choice questions with the correct answer and a clear explanation for each. Frequently asked in AE Level Civil Engineering, JE Level Civil Engineering. Solve the full set below for free — no login required.
- 1Centroid and Moment of InertiaEASY
The centroid of a semicircular area of radius R from its diameter is:
A4R/(3π)BR/2C2R/3DR/πAnswer: A. 4R/(3π)
Explanation: Centroid of semicircular area from diameter: ȳ = 4R/(3π). For a semi-circular arc (wire): ȳ = 2R/π. For hemisphere (volume): ȳ = 3R/8.
- 2Centroid and Moment of InertiaEASY
Second moment of area (moment of inertia) of a rectangle (b wide, d deep) about its centroidal axis parallel to width:
Abd³/12Bbd³/3Cb³d/12Dbd²/6Answer: A. bd³/12
Explanation: Ixx (centroidal, parallel to b) = bd³/12. About base: bd³/3. Ixx = bd³/12 and Iyy = b³d/12. Parallel axis theorem: I = Ic + A×d² where d=distance between axes.
- 3Euler Critical Buckling LoadMEDIUM
For a column of length L with both ends pinned, the Euler critical buckling load Pcr is:
APcr = pi x E x I / L^2BPcr = pi^2 x E x I / L^2CPcr = pi^2 x E x A / L^2DPcr = E x I / LAnswer: B. Pcr = pi^2 x E x I / L^2
Explanation: Euler (1744): Pcr = pi^2 EI / (KL)^2, where KL = effective length (K = 1.0 for both ends pinned, 0.5 for both ends fixed, 0.7 for fixed-pinned, 2.0 for fixed-free/cantilever). Slenderness ratio lambda = KL/r (r = sqrt(I/A)). Euler formula valid only for slender columns (lambda > 100 for steel). Critical stress sigma_cr = pi^2 E / lambda^2.
- 4Castigliano Strain EnergyMEDIUM
Castigliano's second theorem is used to find the deflection (or rotation) at a point in a structure. It states that the deflection delta_i at the point of application of force P_i is:
Adelta_i = partial U / partial P_i (partial derivative of total strain energy U with respect to that force)Bdelta_i = U / P_iCdelta_i = P_i x L / AEDdelta_i = M_i / EIAnswer: A. delta_i = partial U / partial P_i (partial derivative of total strain energy U with respect to that force)
Explanation: Castigliano's second theorem: delta_i = dU/dP_i, where U = total strain energy stored in the structure. For beams: U = integral(M^2/(2EI) dx). Method: compute M in terms of P_i (actual or dummy), differentiate w.r.t. P_i, integrate dU/dP_i. If no force at desired point, add dummy force Q, find dU/dQ, then set Q = 0.
- 5EquilibriumEASY
Which of the following about rigid body under coplanar forces is correct?
Aonly vertical force equilibrium is enough for every bodyBsum of horizontal forces, vertical forces and moments must be zeroCequilibrium requires accelerationDmoments are ignoredAnswer: B. sum of horizontal forces, vertical forces and moments must be zero
Explanation: Static equilibrium requires zero resultant force and moment. Correct option: B.
- 6D Alembert PrincipleMEDIUM
D'Alembert's principle converts a dynamics problem into an equivalent statics form by adding:
AAn inertia force opposite to accelerationBA pore pressure force in soilCA bending moment at every pin jointDA chlorine residualAnswer: A. An inertia force opposite to acceleration
Explanation: Adding inertia force lets dynamic equilibrium be written like static equilibrium.
- 7Bow Notation Graphical TrussMEDIUM
In the Bow notation system for graphical truss analysis (force polygons):
AForces are labeled only at jointsBSpaces between forces in the space diagram are labeled with capital letters; member forces are identified by the two letters on either side of the member; the force polygon (Maxwell diagram) is drawn using these letter pairsCOnly reaction forces are labeledDMembers are numbered not letteredAnswer: B. Spaces between forces in the space diagram are labeled with capital letters; member forces are identified by the two letters on either side of the member; the force polygon (Maxwell diagram) is drawn using these letter pairs
Explanation: Bow notation: capital letters assigned to each space between external forces and inside each panel of the truss (in the space diagram). Each member is identified by two letters (the spaces on each side). In the force polygon (Cremona/Maxwell diagram), vectors are drawn using the same letters. This enables systematic graphical solution of all member forces.
- 8D'Alembert PrincipleMEDIUM
D'Alembert's principle converts a dynamic problem into an equivalent static problem by:
AIgnoring acceleration entirelyBAdding a fictitious inertia force (-ma) to the real applied forces; the system is then in dynamic equilibrium (sigma F + (-ma) = 0)CUsing energy methods onlyDApplying the principle of virtual workAnswer: B. Adding a fictitious inertia force (-ma) to the real applied forces; the system is then in dynamic equilibrium (sigma F + (-ma) = 0)
Explanation: D'Alembert: F_net - ma = 0, written as sigma F = 0 if the inertia force (-ma) is treated as a real force. This allows free body diagram + static equilibrium methods to solve dynamics problems. Particularly useful for systems with constraints (connected bodies, pulleys, rotating systems). Equivalent to Newton's second law F_net = ma.
- 9Coefficient of RestitutionMEDIUM
The coefficient of restitution (e) for a direct central impact between two bodies is defined as:
ARatio of impulse forces during impactBRatio of relative velocity of separation to relative velocity of approach along line of impact (e = v2 - v1) / (u1 - u2)CRatio of masses of the two bodiesDEnergy retained after impact / energy before impactAnswer: B. Ratio of relative velocity of separation to relative velocity of approach along line of impact (e = v2 - v1) / (u1 - u2)
Explanation: e = (v2 - v1)/(u1 - u2), where u = pre-impact velocity, v = post-impact velocity (along line of impact). e = 1: perfectly elastic (no energy loss); e = 0: perfectly plastic (they stick together); 0 < e < 1: real impacts. Energy lost = m1 m2 (1-e^2)(u1-u2)^2 / (2(m1+m2)).
- 10Belt Friction Capstan EquationMEDIUM
For a flat belt over a fixed pulley, the ratio of tensions T1/T2 (tight side/slack side) as a function of angle of wrap theta and coefficient of friction mu is:
AT1/T2 = 1 + mu x thetaBT1/T2 = e^(mu x theta) (capstan equation)CT1/T2 = mu x thetaDT1/T2 = 1/e^(mu theta)Answer: B. T1/T2 = e^(mu x theta) (capstan equation)
Explanation: Capstan equation: T1/T2 = e^(mu x theta), where theta = angle of wrap in radians, mu = coefficient of friction between belt and pulley. This is the Euler-Eytelwein formula. For rope over a fixed peg: ratio of tensions = e^(mu x theta). Used for power transmission belts, capstans, band brakes.
- 11Principle of Virtual WorkMEDIUM
The Principle of Virtual Work states that for a system in static equilibrium:
AThe actual work done by all forces equals zeroBThe virtual work done by all external forces and reactions on a virtual displacement consistent with constraints equals zero (sum of delta_W = 0)COnly internal forces do virtual workDVirtual work equals kinetic energyAnswer: B. The virtual work done by all external forces and reactions on a virtual displacement consistent with constraints equals zero (sum of delta_W = 0)
Explanation: Principle of Virtual Work: for equilibrium, sum(F_i . delta_r_i) = 0 for any virtual displacement delta_r consistent with constraints. Useful for finding unknown reactions or forces without solving full equilibrium equations. Especially powerful for mechanisms and structural analysis (complementary virtual work for deflections).
- 12Engineering MechanicsMEDIUM
A set of forces whose resultant is zero are known as
AEquilibrium forcesBCollinear forcesCCoplanar forcesDConcurrent forcesAnswer: A. Equilibrium forces
Explanation: Forces in equilibrium have a zero resultant.