Nepal Engineering Council · Mechanical Engineering · Chapter 4
Engineering Mechanics and Strength of Material
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192 questions in 6 syllabus topics.
4.1 Applied mechanics
32 questions · AMeE0401
1. In mechanics, a body is idealised as a particle when
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Think about size relative to the problem, not about mass.
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Answer: C. its dimensions are negligible compared with the other dimensions in the problem
A particle has mass but its size and shape do not affect the analysis, so all forces can be taken as concurrent at one point.
2. A rigid body is one in which
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Statics ignores deformation of the body.
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Answer: A. the distance between any two points remains constant under load
A rigid body is an idealisation in which deformation is ignored, so distances between its points never change.
3. Strength of materials differs from rigid-body statics mainly because it treats bodies as
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Which idealisation keeps internal effects?
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Answer: D. deformable, so internal stresses and strains are of interest
In deformable-body mechanics the change of shape and the internal force distribution (stress and strain) are the objects of study.
4. The number of independent equilibrium equations for a general coplanar force system acting on a rigid body is
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Two force components plus one moment.
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Answer: B. 3
In 2D: ΣFx = 0, ΣFy = 0 and ΣM = 0, giving three independent equations.
5. For a rigid body under a general three-dimensional force system, the number of independent scalar equilibrium equations is
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Forces and moments, three directions each.
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Answer: B. 6
Three force equations (ΣFx, ΣFy, ΣFz) and three moment equations (ΣMx, ΣMy, ΣMz) are available.
6. Lami's theorem applies to a body in equilibrium under
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Count the forces and check whether lines meet at a point.
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Answer: D. three concurrent coplanar forces
Lami's theorem states that for three concurrent coplanar forces in equilibrium each force is proportional to the sine of the angle between the other two.
7. Varignon's theorem states that the moment of a force about a point is equal to
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It is about moments of components.
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Answer: A. the algebraic sum of the moments of its components about the same point
Varignon's theorem lets us find a moment by summing the moments of the force's components about the point.
8. Which statement about a couple is TRUE?
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A couple is a free vector.
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Answer: C. Its moment is the same about every point in the plane
A couple has zero resultant force and a free-vector moment equal to F × d, independent of the moment centre.
9. The principle of transmissibility states that the external effect of a force on a rigid body
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Only sliding along one line is allowed.
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Answer: A. remains unchanged if the force is moved along its line of action
Sliding a force along its line of action does not change the equilibrium or motion of a rigid body (external effect).
10. A two-force member is in equilibrium only if the two forces are
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A non-collinear equal and opposite pair gives a moment.
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Answer: D. equal in magnitude, opposite in direction and collinear
Equal and opposite forces that are not collinear would form a couple, so the member would not be in equilibrium.
11. The angle of friction is the angle between the normal reaction and the
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It involves the total reaction.
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Answer: B. resultant of the normal reaction and limiting friction force
tan φ = F/N = μ, where φ is the angle made by the total reaction with the normal at the point of impending motion.
12. The angle of repose of a rough inclined plane is numerically equal to the
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Both satisfy tan = μ.
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Answer: B. angle of limiting friction
A body just begins to slide down when the plane angle θ satisfies tan θ = μ = tan φ, so repose angle = friction angle.
13. Compared with the limiting static friction, the kinetic friction between the same two surfaces is generally
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It is easier to keep a box sliding than to start it.
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Answer: D. slightly smaller
Experiment shows μk ≤ μs, so the force needed to keep sliding is a little smaller than that required to start it.
14. Newton's second law of motion states that the net force on a body is equal to
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Momentum, and its time rate.
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Answer: A. the rate of change of its linear momentum
F = d(mv)/dt, which reduces to F = ma for constant mass.
15. Newton's law of gravitation states that the attractive force between two masses is
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Inverse-square law.
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Answer: C. proportional to the product of the masses and inversely proportional to the square of the distance
F = G m1 m2 / r², with G = 6.674 × 10⁻¹¹ N·m²/kg².
16. A force is called conservative if the work done by it
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Think of gravity versus friction.
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Answer: C. depends only on the end positions and not on the path
Gravity and ideal spring forces are conservative; friction is not because its work depends on the path length.
17. The principle of work and energy states that the work done by all forces acting on a particle equals the change in its
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Net work changes motion energy.
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Answer: D. kinetic energy
W(net) = ½mv₂² − ½mv₁², which is the work–energy theorem.
18. The SI unit of linear impulse is equivalent to
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Impulse and momentum share units.
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Answer: B. kg·m/s
Impulse = F·t = N·s = kg·m/s, the same dimensions as momentum.
19. In a perfectly plastic (perfectly inelastic) collision the coefficient of restitution is
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Bodies stick together.
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Answer: B. 0
The bodies move together after impact, so relative separation velocity is zero, giving e = 0.
20. Two forces of 30 N and 40 N act at a point at right angles to each other. The magnitude of their resultant is
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Use Pythagoras.
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Answer: D. 50 N
R = √(30² + 40²) = 50 N.
21. A force of 50 N acts at a perpendicular distance of 0.6 m from a point O. The moment of the force about O is
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Force times perpendicular distance.
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Answer: A. 30 N·m
M = F × d = 50 × 0.6 = 30 N·m.
22. A 20 kg block rests on a horizontal floor with μs = 0.3. Taking g = 9.81 m/s², the minimum horizontal force to start it moving is
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N = mg on a horizontal floor.
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Answer: C. 58.9 N
F = μN = 0.3 × 20 × 9.81 = 58.9 N.
23. A block just begins to slide on a rough plane when the plane is tilted to some angle. If μs = 0.5 the angle is nearly
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Use tan θ = μ.
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Answer: C. 26.6°
tan θ = μ → θ = tan⁻¹ 0.5 = 26.6°.
24. A weight of 100 N hangs from two light strings, each making 30° with the horizontal ceiling. The tension in each string is
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Vertical equilibrium with sin 30° = 0.5.
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Answer: A. 100 N
2T sin30° = W → T = 100/(2×0.5) = 100 N.
25. A 2 kg body starts from rest and is acted on by a constant net force of 10 N over a distance of 5 m. Its final speed is
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Work equals gain in kinetic energy.
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Answer: B. 7.07 m/s
Work = 10×5 = 50 J = ½·2·v² → v = √50 = 7.07 m/s.
26. A 0.2 kg ball strikes a wall at 15 m/s and rebounds along the same line at 10 m/s. The magnitude of the impulse on the ball is
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The velocity changes sign.
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Answer: B. 5.0 N·s
Impulse = m(v₂ − v₁) = 0.2(10 − (−15)) = 5 N·s.
27. At a height equal to the earth's radius above the surface, the acceleration due to gravity is (surface value 9.81 m/s²)
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Distance from the centre doubles.
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Answer: D. 2.45 m/s²
Distance from centre is 2R, so g' = g/2² = 2.45 m/s².
28. A 60 kg person stands on a weighing scale in a lift accelerating upward at 2 m/s². Taking g = 9.81 m/s², the scale reads a force of
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Apparent weight rises in an upward-accelerating lift.
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Answer: A. 708.6 N
N − mg = ma → N = 60(9.81 + 2) = 708.6 N.
29. A 2 kg block moving at 6 m/s collides with a stationary 4 kg block and they stick together. The common velocity is
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Conserve linear momentum.
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Answer: C. 2 m/s
Momentum conservation: 2×6 = (2+4)v → v = 2 m/s.
30. A force of 100 N acts at 30° above the horizontal. Its horizontal component is
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Use the cosine for the component adjacent to the angle.
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Answer: C. 86.6 N
Fx = 100 cos30° = 86.6 N.
31. A 10 kg block is pushed up a 30° incline (μ = 0.2) by a force parallel to the plane. The force needed for uniform motion is (g = 9.81 m/s²)
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Weight component plus friction both oppose motion.
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Answer: A. 66.0 N
F = mg(sin30° + μcos30°) = 98.1(0.5 + 0.1732) = 66.0 N.
32. A spring of stiffness 2000 N/m is compressed by 0.1 m from its free length. The energy stored in it is
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Half k x squared.
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Answer: D. 10 J
U = ½kx² = ½ × 2000 × 0.01 = 10 J.
4.2 Theory of elasticity
32 questions · AMeE0402
33. Engineering (normal) stress is defined as
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Force per unit area.
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Answer: C. internal axial force divided by the original cross-sectional area
σ = P/A₀ with units of N/m² (Pa); the ratio ΔL/L is strain.
34. Linear strain is
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Ratio of lengths.
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Answer: A. a dimensionless quantity
Strain = change in length / original length, a ratio of two lengths, hence dimensionless.
35. Hooke's law states that, within the proportional limit,
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Linear region of the stress-strain curve.
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Answer: D. stress is directly proportional to strain
σ = Eε, the slope of the linear part of the stress–strain curve being Young's modulus E.
36. The modulus of elasticity of mild steel is approximately
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Greater than aluminium by a factor of about three.
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Answer: B. 200 GPa
Steel has E ≈ 200–210 GPa, aluminium about 70 GPa and cast iron about 100 GPa.
37. Poisson's ratio is defined as the ratio of
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Transverse over axial.
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Answer: B. lateral strain to longitudinal strain (in magnitude)
ν = −ε_lateral/ε_longitudinal; for most metals ν ≈ 0.25–0.35.
38. For an isotropic elastic material, the theoretical upper limit of Poisson's ratio is
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Think of an incompressible material such as rubber.
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Answer: D. 0.5
From K = E/[3(1−2ν)], a positive bulk modulus requires ν < 0.5; ν = 0.5 corresponds to an incompressible material.
39. The relation among Young's modulus E, shear modulus G and Poisson's ratio ν is
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Two-G times one plus Poisson.
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Answer: A. E = 2G(1 + ν)
The standard relation for an isotropic material is E = 2G(1+ν).
40. The relation among Young's modulus E, bulk modulus K and Poisson's ratio ν is
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Compare with the shear relation, which has (1+ν).
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Answer: C. E = 3K(1 − 2ν)
K = E/[3(1−2ν)] is equivalent to E = 3K(1−2ν).
41. Bulk modulus is defined as the ratio of
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It relates pressure and volume change.
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Answer: A. hydrostatic (volumetric) stress to volumetric strain
K = p/(ΔV/V), measuring resistance to uniform compression.
42. For a rectangular bar under axial load, the volumetric strain is equal to
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Small strains add.
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Answer: D. the sum of the three mutually perpendicular linear strains
For small strains εv = εx + εy + εz = ε(1−2ν) for uniaxial loading.
43. If a bar is free to expand when heated uniformly, the thermal stress developed is
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What restrains the bar?
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Answer: B. zero
Thermal stress develops only when expansion is prevented; a free bar just changes length by αLΔT.
44. Resilience is the
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Recoverable energy.
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Answer: B. strain energy stored in a body when it is stressed within the elastic limit
Resilience is recoverable strain energy; proof resilience is its value at the elastic limit.
45. A load applied suddenly (without impact) on an elastic bar produces a stress that is
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Equate work done to strain energy.
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Answer: D. twice that produced by the same load applied gradually
Work done by the load Pδ equals the strain energy ½Pmaxδ, so the sudden load gives σ = 2P/A.
46. Strain energy per unit volume of a bar stressed uniformly to σ within the elastic range is
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Half stress times strain.
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Answer: A. σ²/(2E)
U/V = ½σε = σ²/(2E), the area under the linear stress–strain curve.
47. The modulus of resilience of a material is represented by the area under the stress–strain diagram up to the
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Recoverable energy only.
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Answer: C. elastic (proportional) limit
Modulus of resilience is the strain energy per unit volume at the elastic limit.
48. Compared with a ductile material, a brittle material generally shows
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Think of glass or cast iron.
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Answer: C. little or no yielding and small strain at fracture
Brittle materials such as cast iron fracture with little plastic deformation and no appreciable necking.
49. Factor of safety is defined as the ratio of
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It is greater than 1.
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Answer: D. ultimate (or yield) stress to the allowable working stress
FOS = failure stress/working stress, which is always greater than 1.
50. A rod of cross-sectional area 100 mm² carries an axial tensile load of 10 kN. The normal stress is
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1 N/mm² = 1 MPa.
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Answer: B. 100 MPa
σ = P/A = 10 000 N / 100 mm² = 100 N/mm² = 100 MPa.
51. A steel bar 2 m long, area 200 mm², E = 200 GPa carries a tensile force of 20 kN. Its elongation is
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Use δ = PL/AE with consistent N and mm.
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Answer: B. 1.0 mm
δ = PL/(AE) = 20 000 × 2000 /(200 × 200 000) = 1.0 mm.
52. A steel bar (E = 200 GPa, α = 12 × 10⁻⁶ /°C) is fixed rigidly between two supports and heated by 50 °C. The compressive stress developed is
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σ = EαΔT.
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Answer: D. 120 MPa
σ = EαΔT = 200 000 × 12×10⁻⁶ × 50 = 120 MPa.
53. A free bar 2 m long with α = 12 × 10⁻⁶ /°C is heated through 60 °C. Its increase in length is
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Free expansion: αLΔT.
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Answer: A. 1.44 mm
δ = αLΔT = 12×10⁻⁶ × 2000 × 60 = 1.44 mm.
54. A bar is stretched so that its longitudinal strain is 0.001. If ν = 0.3, the magnitude of lateral strain is
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Multiply by Poisson's ratio.
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Answer: C. 0.0003
ε_lat = ν ε = 0.3 × 0.001 = 0.0003.
55. A bar of material with ν = 0.3 has an axial strain of 0.001 under uniaxial stress. The volumetric strain is
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Volumetric strain = ε(1 − 2ν).
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Answer: C. 0.0004
εv = ε(1−2ν) = 0.001 × 0.4 = 0.0004.
56. For a material with E = 200 GPa and ν = 0.25 the bulk modulus is
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K = E/3(1−2ν).
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Answer: A. 133.3 GPa
K = E/[3(1−2ν)] = 200/(3×0.5) = 133.3 GPa.
57. For the same material (E = 200 GPa, ν = 0.25) the shear modulus is
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G = E/2(1+ν).
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Answer: B. 80 GPa
G = E/[2(1+ν)] = 200/2.5 = 80 GPa.
58. A bar is stressed to 100 MPa within the elastic limit. For E = 200 GPa the strain energy stored per unit volume is
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σ²/2E in SI units.
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Answer: B. 25 kJ/m³
u = σ²/2E = (100×10⁶)²/(2×200×10⁹) = 25 000 J/m³ = 25 kJ/m³.
59. A load of 5 kN is suddenly applied on a bar of cross-section 100 mm². The maximum stress developed is
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Twice the gradual-load stress.
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Answer: D. 100 MPa
Sudden loading gives σ = 2P/A = 2 × 5000/100 = 100 MPa.
60. A weight of 2 kN falls through 6 mm onto the collar of a 10 m long vertical rod (A = 200 mm², E = 200 GPa). Treating the static elongation correctly, the maximum instantaneous stress is nearly (σmax = σst[1+√(1+2h/δst)])
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First find the static stress and the static elongation.
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Answer: A. 60 MPa
σst = 2000/200 = 10 MPa, δst = σstL/E = 0.5 mm; factor = 1+√(1+2×6/0.5) = 6, so σmax = 60 MPa.
61. A vertical steel rod 100 m long hangs under its own weight (ρ = 7850 kg/m³, E = 200 GPa). Its elongation is nearly
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Self-weight elongation is half the value for a tip load of equal weight.
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Answer: C. 1.93 mm
δ = ρgL²/(2E) = 7850×9.81×100²/(2×200×10⁹) = 1.92×10⁻³ m = 1.92 mm.
62. A steel with yield strength 250 MPa and E = 200 GPa has a modulus of resilience of
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σy²/2E.
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Answer: C. 156.25 kJ/m³
σy²/2E = (250×10⁶)²/(2×200×10⁹) = 1.5625×10⁵ J/m³ = 156.25 kJ/m³.
63. A steel has an ultimate strength of 400 MPa. If the allowable working stress is 160 MPa, the factor of safety is
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Ratio of failure stress to working stress.
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Answer: A. 2.5
FOS = 400/160 = 2.5.
64. A 250 mm gauge length extends by 0.5 mm under load. The strain is
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Change in length over original length.
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Answer: D. 0.002
ε = ΔL/L = 0.5/250 = 0.002.
4.3 Strength of materials
32 questions · AMeE0403
65. The centroid of a triangle lies at a distance from its base equal to
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Medians divide in 2:1.
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Answer: C. one-third of its height
The centroid is at the intersection of the medians, h/3 above the base.
66. The centroid of a semicircular area of radius R from its diametral base is at
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About 0.42R.
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Answer: A. 4R/(3π)
ȳ = 4R/3π ≈ 0.424R from the diameter for a semicircular area.
67. The parallel axis theorem for area moment of inertia states
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Distance enters squared.
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Answer: D. I = I_G + A d²
The moment of inertia about any axis parallel to the centroidal axis equals I_G plus area times square of the separation.
68. The perpendicular axis theorem (for plane laminae) states that
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Polar = sum of two rectangular.
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Answer: B. the moment of inertia about an axis normal to the plane equals the sum of the moments of inertia about two perpendicular axes in the plane
J = Iz = Ix + Iy for a plane area about axes meeting at the point where the normal axis passes.
69. The moment of inertia of a rectangle of width b and depth h about its centroidal axis parallel to b is
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Depth is cubed.
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Answer: B. bh³/12
I_xx = bh³/12 about the centroidal axis; bh³/3 is about the base.
70. The polar moment of inertia of a solid circular section of diameter d is
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Twice the diametral moment of inertia.
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Answer: D. πd⁴/32
J = Ix + Iy = 2 × πd⁴/64 = πd⁴/32.
71. Radius of gyration of an area A about an axis with moment of inertia I is
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Square root of I over A.
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Answer: A. √(I/A)
k = √(I/A), the distance at which the whole area could be concentrated to give the same I.
72. The mass moment of inertia of a solid uniform disc of mass M and radius R about its central axis perpendicular to the disc is
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Ring has MR².
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Answer: C. ½MR²
I = ½MR² for a solid disc/cylinder; MR² is for a thin ring.
73. Section modulus of a beam section is defined as
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Used with M to get maximum bending stress.
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Answer: A. moment of inertia divided by the distance from neutral axis to the extreme fibre
Z = I/y_max, so σ_max = M/Z.
74. In a simply supported beam, the bending moment is maximum at the section where
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dM/dx equals the shear force.
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Answer: D. the shear force is zero or changes sign
Since dM/dx = V, M has an extremum where V = 0.
75. The point of contraflexure in a beam is the point where
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Hogging changes to sagging.
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Answer: B. the bending moment changes sign
At the point of contraflexure the curvature reverses, so M passes through zero and changes sign.
76. The relation between load intensity w, shear force V and bending moment M is
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Integrate load to get shear, shear to get moment.
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Answer: B. dV/dx = −w and dM/dx = V
The slope of the shear diagram is the negative of the load intensity and the slope of the moment diagram is the shear force.
77. The Euler–Bernoulli bending equation is
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Do not mix it up with torsion.
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Answer: D. M/I = σ/y = E/R
For pure bending, M/I = σ/y = E/R; T/J = τ/r = Gθ/L is the torsion equation.
78. A plane truss with j joints and m members is a perfect (just-stiff) truss when
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Basic triangle: j = 3, m = 3.
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Answer: A. m = 2j − 3
Three members form the basic triangle with 3 joints, and each additional joint adds two members: m = 2j − 3.
79. In a loaded plane truss a joint of three members, two of which are collinear and no external load acts at the joint, shows that the third member is
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Resolve normal to the two collinear members.
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Answer: C. a zero-force member
ΣF perpendicular to the collinear members shows the third member's force must be zero.
80. The torsion equation for a circular shaft is
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Uses shear modulus G.
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Answer: C. T/J = τ/r = Gθ/L
Shear stress varies linearly with radius, τ = Tr/J, and the twist is θ = TL/GJ.
81. For the same weight and material, a hollow circular shaft compared with a solid shaft is
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Where is material most effective?
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Answer: D. stronger in torsion
Material near the axis is lightly stressed; moving it outward increases J and hence torque capacity for the same mass.
82. A rectangular section 100 mm wide and 200 mm deep has a moment of inertia about its centroidal horizontal axis of
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bh³/12.
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Answer: B. 66.67 × 10⁶ mm⁴
I = bh³/12 = 100 × 200³/12 = 66.67 × 10⁶ mm⁴.
83. For the same 100 mm × 200 mm section, the moment of inertia about its base is
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Parallel axis theorem with d = 100 mm.
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Answer: B. 266.67 × 10⁶ mm⁴
I = I_G + Ad² = 66.67×10⁶ + 20 000 × 100² = 266.67×10⁶ mm⁴ (= bh³/3).
84. The polar moment of inertia of a solid circular shaft of diameter 100 mm is
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πd⁴/32.
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Answer: D. 9.82 × 10⁶ mm⁴
J = πd⁴/32 = π × 100⁴/32 = 9.82 × 10⁶ mm⁴.
85. An inverted-T section consists of a flange 100 mm × 20 mm at the bottom and a web 20 mm × 80 mm on top, centred. The centroid height above the base is
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Area-weighted mean of the part centroids.
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Answer: A. 32.2 mm
ȳ = (2000×10 + 1600×60)/3600 = 32.2 mm.
86. A simply supported beam of span 6 m carries a UDL of 10 kN/m over the whole span. The maximum bending moment is
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wL²/8.
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Answer: C. 45 kN·m
Mmax = wL²/8 = 10 × 36/8 = 45 kN·m at mid-span.
87. A cantilever 2 m long carries a UDL of 5 kN/m over its entire length. The bending moment at the fixed end is
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wL²/2.
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Answer: C. 10 kN·m
M = wL²/2 = 5 × 4/2 = 10 kN·m (hogging).
88. A cantilever of length 2 m carries a point load of 1 kN at its free end. E = 200 GPa and I = 8 × 10⁶ mm⁴. The tip deflection is
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PL³/3EI with N and mm.
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Answer: A. 1.67 mm
δ = PL³/(3EI) = 1000×2000³/(3×200 000×8×10⁶) = 1.67 mm.
89. A simply supported beam 4 m long (E = 200 GPa, I = 8 × 10⁶ mm⁴) carries a UDL of 1 kN/m. The maximum deflection at mid-span is
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5wL⁴/384EI.
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Answer: B. 2083.33 mm
δ = 5wL⁴/(384EI) = 5×1×4000⁴/(384×200 000×8×10⁶) = 2.08 mm (w = 1 N/mm).
90. A solid circular shaft of diameter 50 mm transmits a torque of 500 N·m. The maximum shear stress is
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τ = 16T/πd³.
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Answer: B. 20.4 MPa
τ = 16T/(πd³) = 16 × 500 000/(π × 50³) = 20.4 MPa.
91. A shaft transmits 20 kW at 300 rpm. The torque is
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P = 2πNT/60.
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Answer: D. 636.6 N·m
T = 60P/(2πN) = 60×20 000/(2π×300) = 636.6 N·m.
92. A 1 m long solid shaft of 50 mm diameter (G = 80 GPa) carries 500 N·m torque. The angle of twist is nearly
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Convert radians to degrees.
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Answer: A. 0.58°
θ = TL/(GJ) = 500×10³×1000/(80 000 × 0.6136×10⁶) = 1.02×10⁻² rad = 0.58°.
93. A beam of 100 mm × 200 mm section is subjected to a bending moment of 10 kN·m about its horizontal centroidal axis. The maximum bending stress is
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σ = My/I.
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Answer: C. 15 MPa
σ = My/I = 10×10⁶ × 100/(66.67×10⁶) = 15 MPa.
94. A solid uniform disc of mass 10 kg and radius 0.2 m rotates about its central axis. Its mass moment of inertia is
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½MR².
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Answer: C. 0.20 kg·m²
I = ½MR² = 0.5 × 10 × 0.04 = 0.20 kg·m².
95. A hollow shaft has outer diameter 100 mm and inner diameter 80 mm. Its polar moment of inertia is
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Subtract the inner circle's J.
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Answer: A. 5.80 × 10⁶ mm⁴
J = π(D⁴ − d⁴)/32 = π(10⁸ − 4.096×10⁷)/32 = 5.80×10⁶ mm⁴.
96. A symmetrical two-member roof truss has members inclined at 30° to the horizontal and carries a 10 kN vertical load at the apex. The force in each inclined member is
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Vertical equilibrium at the apex.
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Answer: D. 10 kN
2F sin30° = 10 → F = 10 kN (compressive).
4.4 Theory of machines
32 questions · AMeE0404
97. In a kinematic pair, if the two elements have surface (area) contact and the relative motion is purely turning, sliding or screw, the pair is called a
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Surface versus line or point contact.
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Answer: C. lower pair
Lower pairs have surface contact (revolute, prismatic, screw, cylindrical, spherical); higher pairs have line or point contact.
98. A cam and follower or a pair of meshing gear teeth is an example of a
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Not surface contact.
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Answer: A. higher pair
The elements touch along a line or at a point, so the pair is a higher pair.
99. Which of the following is a lower pair?
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A sliding pair has surface contact.
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Answer: D. Piston and cylinder
A piston in a cylinder is a sliding (prismatic) pair with surface contact; the others have line/point contact.
100. The Kutzbach (Grübler) equation for the degrees of freedom of a planar mechanism with n links, j lower pairs and h higher pairs is
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A fixed frame removes one link.
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Answer: B. F = 3(n − 1) − 2j − h
Each free link has 3 DOF (one fixed), each lower pair removes 2, each higher pair removes 1.
101. A kinematic chain consisting of four links connected by four revolute pairs is called a
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All joints are revolute.
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Answer: B. four-bar chain (4R)
Four turning pairs on four links form the four-bar linkage, designated 4R.
102. A single slider-crank chain has pairs of the type
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Slider is the only sliding pair.
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Answer: D. three revolute and one prismatic (3R-1P)
Crank–frame, crank–connecting rod and connecting rod–slider pairs are revolute; slider–frame is prismatic.
103. The double slider-crank chain (2R-2P) includes which of the following mechanisms?
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Two sliders, two turning pairs.
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Answer: A. Oldham's coupling and elliptical trammel
Scotch yoke, Oldham's coupling and the elliptical trammel are inversions of the double slider-crank chain; the others come from the single slider-crank.
104. Whitworth quick-return mechanism is an inversion of the
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The fixed link is the shortest one.
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Answer: C. single slider-crank chain with the crank fixed
Fixing the crank (shortest link) of the slider-crank chain gives the Whitworth mechanism.
105. According to Grashof's law, a four-bar linkage has at least one link capable of making a complete rotation if
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Shortest plus longest.
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Answer: A. sum of the shortest and longest links ≤ sum of the other two
Grashof condition: s + l ≤ p + q.
106. The number of instantaneous centres in a mechanism of n links is
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Choose 2 links out of n.
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Answer: D. n(n − 1)/2
Each pair of links has one instantaneous centre, giving nC2 = n(n−1)/2.
107. According to Kennedy's theorem, three bodies in relative plane motion have
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A collinearity statement.
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Answer: B. three instantaneous centres lying on a straight line
Kennedy's (Aronhold–Kennedy) theorem: the three centres of any three rigid bodies lie on one line.
108. The Coriolis component of acceleration arises when
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Sliding on a rotating link.
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Answer: B. a point moves along a link that is itself rotating
It equals 2ωv, the product of twice the link's angular velocity and the sliding velocity along the link.
109. The velocity of point B relative to point A on the same rigid link is always
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Distance between points is fixed.
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Answer: D. perpendicular to the line AB
Since the distance AB is constant, relative velocity has only a rotational component ω × AB, perpendicular to AB.
110. For a point moving on a circular path with angular velocity ω and angular acceleration α, the centripetal acceleration is directed
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Radial vs tangential.
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Answer: A. towards the centre of the circle and has magnitude ω²r
Radial (centripetal) component = ω²r towards the centre; tangential component = αr.
111. Klein's construction is used to determine
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Slider-crank graphical method.
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Answer: C. the velocity and acceleration of points in a slider-crank mechanism
Klein's construction is a graphical method for velocity and acceleration of the slider-crank mechanism.
112. The inertia force on a body of mass m having acceleration a (D'Alembert's principle) is
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Opposite to acceleration.
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Answer: C. −ma, acting opposite to the acceleration
Adding a fictitious force −ma converts the dynamic problem into one of static equilibrium.
113. A Peaucellier mechanism is used to
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Straight-line motion.
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Answer: D. generate an exact straight-line motion
It is a lower-pair linkage producing exact straight-line motion by inversion of a circle.
114. In a pin-jointed four-bar chain, if the shortest link is fixed, the mechanism obtained is a
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Both adjacent links rotate fully.
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Answer: B. double-crank (drag-link) mechanism
With the shortest link fixed (Grashof chain) both links adjacent to it are cranks.
115. In a slider-crank mechanism when the crank is at 90° to the line of stroke, the angular velocity of the connecting rod is
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Locate the instantaneous centre.
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Answer: B. zero
The instantaneous centre of the rod is at infinity (normal to slider path and crank line are parallel), so ω_rod = 0 at that instant.
116. Ackermann steering gear is an application of
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Think of the steering trapezium.
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Answer: D. a four-bar linkage (4R) which approximates correct steering geometry
Its trapezium linkage makes the inner wheel turn more than the outer to nearly avoid scrubbing.
117. A planar mechanism has 6 links and 7 lower pairs (revolute). Its degrees of freedom are
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Kutzbach's equation.
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Answer: A. 1
F = 3(n−1) − 2j = 3(5) − 14 = 1.
118. A planar mechanism has 8 links, 9 lower pairs and 1 higher pair. Its degrees of freedom are
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Each higher pair removes one DOF.
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Answer: C. 2
F = 3(7) − 18 − 1 = 2.
119. The number of instantaneous centres in a six-link mechanism is
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n(n−1)/2.
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Answer: C. 15
n(n−1)/2 = 6×5/2 = 15.
120. A crank of length 100 mm rotates uniformly at 300 rpm. The linear velocity of the crank-pin is
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Convert rpm to rad/s first.
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Answer: A. 3.14 m/s
ω = 2π×300/60 = 31.42 rad/s; v = ωr = 31.42 × 0.1 = 3.14 m/s.
121. For the same crank (r = 100 mm, 300 rpm), the centripetal acceleration of the crank-pin is
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ω²r.
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Answer: B. 98.7 m/s²
a = ω²r = (31.42)² × 0.1 = 98.7 m/s².
122. A link AB of length 0.5 m rotates at 4 rad/s. The velocity of B relative to A is
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v = ωr.
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Answer: B. 2.0 m/s
v(BA) = ω × AB = 4 × 0.5 = 2.0 m/s, perpendicular to AB.
123. A slider moves outward at 0.5 m/s along a link rotating at 10 rad/s. The Coriolis acceleration is
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2ωv.
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Answer: D. 10 m/s²
a_c = 2ωv = 2 × 10 × 0.5 = 10 m/s².
124. A reciprocating part of mass 5 kg has an acceleration of 20 m/s². The magnitude of its inertia force is
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F = ma.
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Answer: A. 100 N
F = ma = 5 × 20 = 100 N.
125. A piston of diameter 100 mm is subjected to a gas pressure of 2 MPa. The gas force on the piston is
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Pressure times piston area.
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Answer: C. 15.71 kN
F = pA = 2×10⁶ × π/4 × 0.1² = 15 708 N = 15.71 kN.
126. In a quick-return mechanism the cutting stroke takes place while the crank turns through 200° and the return stroke through 160°. The quick-return ratio (time of cutting to time of return) is
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Same angular speed, so time ratio equals angle ratio.
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Answer: C. 1.25
Ratio = 200°/160° = 1.25.
127. A slider-crank crank of length 150 mm rotates at 120 rpm. At the instant the crank is perpendicular to the line of stroke, the crank-pin speed is
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Convert rpm to rad/s first.
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Answer: A. 1.88 m/s
ω = 2π×120/60 = 12.57 rad/s; v = ωr = 12.57 × 0.15 = 1.88 m/s.
128. In a Whitworth quick-return mechanism the cutting stroke corresponds to 220° of crank rotation. The ratio of cutting time to return time is
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Return stroke angle is the remainder of 360°.
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Answer: D. 1.57
Return angle = 360 − 220 = 140°; ratio = 220/140 = 1.57.
4.5 Mechanism
32 questions · AMeE0405
129. The gyroscopic couple acting on a rotor of moment of inertia I spinning at ω and precessing at ωp (axes perpendicular) is
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Product of the three quantities.
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Answer: C. I ω ωp
C = Iωωp, maximum when the spin axis is perpendicular to the precession axis.
130. The main function of a governor is to
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Governor versus flywheel.
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Answer: A. maintain the mean speed of an engine within limits under varying load
A governor controls fuel/steam supply to keep mean speed nearly constant; a flywheel smooths cyclic fluctuation.
131. A governor is said to be isochronous when its equilibrium speed is
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Range of speed is zero.
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Answer: D. the same for all radii of rotation of the balls
For an isochronous governor the speed range is zero; any displacement of the sleeve results in the same speed (but it is unstable).
132. Compared with a Watt governor, a Porter governor is
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The extra central mass is the difference.
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Answer: B. able to operate at higher speeds because of the central load on the sleeve
The central dead load increases the equilibrium speed and the sleeve movement for a given radius change.
133. The function of a flywheel is to
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Cyclic fluctuations, not load changes.
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Answer: B. reduce speed fluctuations within each cycle by storing and releasing kinetic energy
A flywheel absorbs excess energy when torque exceeds the load and returns it when the torque is deficient.
134. The coefficient of fluctuation of speed is defined as
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Maximum speed variation over mean.
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Answer: D. (ωmax − ωmin) / ωmean
Cs = (ω1 − ω2)/ω, where ω is the mean speed.
135. For complete static balancing of several masses rotating in one plane about an axis, the condition is that
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A force polygon closes.
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Answer: A. the vector sum of the centrifugal forces is zero
Σ m r ω² = 0 (equivalently Σ m r = 0) is the force condition.
136. For dynamic balance of masses rotating in different transverse planes, in addition to the force polygon closing, it is required that
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Couples as well as forces.
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Answer: C. the couple polygon (Σ m r l) closes
Dynamic balance requires both Σ m r = 0 and Σ m r l = 0 (couples about a reference plane).
137. In a reciprocating engine the primary unbalanced force along the line of stroke is
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Component along the stroke line.
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Answer: A. m ω² r cos θ
The primary component of the inertia force is mω²r cosθ; the secondary is mω²r cos2θ/n.
138. A cam follower with uniform velocity motion has
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The velocity jumps from 0 to a constant value.
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Answer: D. infinite acceleration at the beginning and end of the stroke
The velocity changes abruptly at the ends of the stroke, giving theoretically infinite acceleration and shock.
139. Among the standard follower motions, the one which gives zero acceleration at the start and end of rise (smooth, no abrupt change) is
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Zero acceleration at the dwell transitions.
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Answer: B. cycloidal motion
Cycloidal motion has zero acceleration at both ends, so there is no sudden force change.
140. The pressure angle of a cam–follower pair is the angle between
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Angle of the force transmitted relative to motion.
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Answer: B. the common normal at the point of contact and the direction of follower motion
A large pressure angle increases side thrust on the follower; the usual limit is about 30° for translating followers.
141. In simple harmonic motion the acceleration of the moving point is
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Restoring nature.
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Answer: D. directly proportional to its displacement from the mean position and directed towards it
a = −ω²x, so the acceleration is maximum at the extremes and zero at the mean position.
142. In an open belt drive the ratio of tensions on tight and slack sides (flat belt, no centrifugal effect) is T1/T2 =
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Capstan equation.
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Answer: A. e^(μθ)
The limiting ratio of tensions for belt slipping on a pulley of angle of contact θ is e^{μθ}.
143. A chain drive differs from a belt drive in that it
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Positive engagement.
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Answer: C. has no slip and gives a positive (constant average) velocity ratio
Chain links engage with sprocket teeth; there is no slip, though speed varies slightly because of the polygonal effect.
144. The law of gearing states that for constant velocity ratio the common normal at the point of contact of the teeth must
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The pitch point is fixed.
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Answer: C. always pass through the pitch point on the line of centres
The common normal must cut the line of centres at a fixed pitch point for a constant angular velocity ratio.
145. In a simple gear train the idler gear
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Cancels in the ratio.
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Answer: D. changes the direction of rotation of the driven gear without changing the speed ratio
The idler's tooth number cancels in the train value but reverses direction.
146. The module of a gear is defined as
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Diameter per tooth.
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Answer: B. pitch circle diameter divided by number of teeth
m = d/z in mm, and circular pitch p = πm.
147. A rotor of moment of inertia 0.5 kg·m² spins at 3000 rpm and precesses at 0.2 rad/s about an axis perpendicular to the spin axis. The gyroscopic couple is
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Convert rpm to rad/s first.
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Answer: B. 31.4 N·m
ω = 2π×3000/60 = 314.16 rad/s; C = Iωωp = 0.5 × 314.16 × 0.2 = 31.4 N·m.
148. A flywheel of moment of inertia 10 kg·m² runs at a mean speed of 300 rpm with Cs = 0.04. The maximum fluctuation of energy is
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ΔE = I ω² Cs.
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Answer: D. 395 J
ΔE = Iω²Cs = 10 × (31.42)² × 0.04 = 395 J.
149. A flat belt passes over a pulley with angle of contact 180° and μ = 0.3. The ratio T1/T2 at the point of slipping is
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θ must be in radians.
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Answer: A. 2.57
T1/T2 = e^{μθ} = e^{0.3π} = e^{0.942} = 2.57.
150. The tensions on tight and slack sides of a belt are 1000 N and 400 N and the belt speed is 10 m/s. The power transmitted is
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Use the difference of tensions.
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Answer: C. 6.0 kW
P = (T1 − T2)v = 600 × 10 = 6000 W = 6 kW.
151. A belt weighs 0.5 kg per metre length and runs at 20 m/s. The centrifugal tension is
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Tc = mv².
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Answer: C. 200 N
Tc = m v² = 0.5 × 400 = 200 N.
152. A belt of mass 0.4 kg/m has a maximum allowable tension of 1200 N. The belt speed at which maximum power is transmitted is
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Optimal condition Tc = T/3.
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Answer: A. 31.6 m/s
v = √(T/3m) = √(1200/1.2) = 31.6 m/s (centrifugal tension = T/3).
153. A pinion of 20 teeth rotating at 900 rpm drives a gear of 60 teeth. The speed of the gear is
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Speeds are inversely proportional to teeth.
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Answer: B. 300 rpm
N2 = N1 × z1/z2 = 900 × 20/60 = 300 rpm.
154. In a compound gear train, gear 1 (20 T) meshes with gear 2 (60 T); gear 3 (15 T) fixed on the shaft of gear 2 meshes with gear 4 (45 T). If gear 1 runs at 900 rpm, gear 4 runs at
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Multiply driver teeth over driven teeth for each stage.
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Answer: B. 100 rpm
N4 = 900 × (20/60) × (15/45) = 100 rpm.
155. A cam rotating at 10 rad/s gives a follower SHM rise of 40 mm during 60° of cam rotation. The maximum follower velocity is
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SHM follower: π s ω / 2θ.
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Answer: D. 0.6 m/s
vmax = π s ω/(2θ) = π × 0.04 × 10/(2 × π/3) = 0.6 m/s.
156. For the same cam and follower (s = 40 mm, ω = 10 rad/s, θ = 60° of SHM rise) the maximum follower acceleration is
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Acceleration formula with θ in radians.
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Answer: A. 18 m/s²
amax = π² s ω²/(2θ²) = π² × 0.04 × 100/(2 × (π/3)²) = 18 m/s².
157. A cam rotating at 20 rad/s lifts a follower 30 mm in 90° of rotation using uniform acceleration and retardation. The maximum follower velocity is
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Maximum velocity is twice the mean velocity.
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Answer: C. 0.764 m/s
t = (π/2)/20 = 0.0785 s; vmax = 2s/t = 0.06/0.0785 = 0.764 m/s.
158. A point executes SHM given by x = 0.05 sin(10t) metres. The maximum velocity is
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Amplitude times ω.
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Answer: C. 0.5 m/s
v = dx/dt = 0.5 cos(10t), so vmax = Aω = 0.05 × 10 = 0.5 m/s.
159. A Watt governor runs at 60 rpm. The height of the governor is
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h = g/ω².
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Answer: A. 0.248 m
h = g/ω² = 9.81/(2π)² = 0.2485 m.
160. A mass of 2 kg is attached at 0.3 m radius on a shaft. To balance it statically by a single mass placed at 0.2 m radius diametrically opposite, the balancing mass required is
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Balance m r products.
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Answer: D. 3 kg
m1 r1 = m2 r2 → m2 = 2 × 0.3/0.2 = 3 kg.
4.6 Mechanics of solid
32 questions · AMeE0406
161. A structure is statically determinate if
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Statics only.
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Answer: C. all support reactions and internal forces can be found from the equilibrium equations alone
Determinate structures need only statics; indeterminate ones also need compatibility (deformation) conditions.
162. A propped cantilever beam (fixed at one end and simply supported at the other) is statically indeterminate to the degree of
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Count reactions minus 3.
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Answer: A. one
It has 4 reaction components (3 at the fixed end, 1 at the prop) against 3 equations of equilibrium, so degree = 1.
163. A beam fixed at both ends under vertical loading has, in general planar analysis, a degree of static indeterminacy of
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Six reactions.
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Answer: D. three
Six reaction components minus three equilibrium equations gives 3 (reduces to 2 if axial force is absent).
164. To solve a statically indeterminate beam, in addition to the equilibrium equations we use
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Deformation conditions.
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Answer: B. compatibility conditions based on deformation
The extra equations come from geometric compatibility (e.g. zero deflection at a redundant support) with the force–deformation relations.
165. Principal planes are the planes on which
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Zero shear.
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Answer: B. the shear stress is zero
Principal planes carry only normal (principal) stresses; shear vanishes on them.
166. In a plane-stress state, the maximum in-plane shear stress is equal to
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Mohr circle radius.
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Answer: D. half the difference of the principal stresses
τmax = (σ1 − σ2)/2, which equals the radius of Mohr's circle.
167. In Mohr's circle for plane stress, the coordinates of the centre are
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Average normal stress.
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Answer: A. ((σx + σy)/2, 0)
The centre lies on the σ-axis at the average normal stress; the radius is √(((σx−σy)/2)² + τxy²).
168. The principal planes and the planes of maximum shear stress are inclined to each other at
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Half of the Mohr angle.
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Answer: C. 45°
On Mohr's circle the angle between the principal stress points and maximum shear points is 90°, i.e. 45° in the physical body.
169. A cylindrical shell is classified as thin-walled when the ratio of wall thickness to internal diameter is
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Small ratio.
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Answer: A. less than about 1/20
For t/d < 1/20 the hoop stress can be taken uniform across the wall.
170. In a thin cylindrical shell subjected to internal pressure, the hoop stress is
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Compare pd/2t with pd/4t.
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Answer: D. twice the longitudinal stress
σh = pd/2t and σl = pd/4t, so σh = 2σl.
171. A thin spherical shell of diameter d and thickness t under internal pressure p has a circumferential stress of
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Same as the longitudinal stress in a cylinder.
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Answer: B. pd/(4t)
Cutting through a diameter gives p(πd²/4) = σ(πdt), so σ = pd/4t.
172. The longitudinal joint of a thin cylindrical boiler shell is made stronger than the circumferential joint mainly because
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Which stress acts across the longitudinal seam?
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Answer: B. the hoop stress is twice the longitudinal stress
A longitudinal joint resists the hoop stress, which is the larger of the two.
173. In the Lamé analysis of a thick-walled cylinder under internal pressure, the maximum hoop stress occurs
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Largest where the radius is least.
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Answer: D. at the inner surface
Hoop stress decreases from the inner to the outer radius; its maximum value at r = ri exceeds the internal pressure.
174. For a thick cylinder with internal pressure p and no external pressure, the radial stress at the inner surface is
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Boundary condition on the loaded surface.
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Answer: A. −p (compressive, magnitude p)
The radial stress must equal the applied pressure on the surface, acting as compression, and is zero at the free outer surface.
175. A non-circular prismatic bar under torsion shows warping of cross-sections, so the simple formula τ = Tr/J
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Plane sections no longer remain plane.
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Answer: C. is not applicable
Plane sections do not remain plane; Saint-Venant's theory or the membrane analogy is used instead.
176. In the membrane (soap-film) analogy for torsion of a non-circular bar, the shear stress is proportional to
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Slope, not height.
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Answer: C. the slope of the membrane
Prandtl's analogy: stress ∝ slope, and twisting torque ∝ twice the volume under the membrane.
177. For a rectangular bar in torsion, the maximum shear stress occurs
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Corners have zero stress.
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Answer: D. at the middle of the longer side
The corners are stress-free; the maximum occurs at the midpoints of the longer sides.
178. At a point in a plane-stressed body σx = 80 MPa, σy = 20 MPa and τxy = 40 MPa. The maximum principal stress is
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Use σ = avg ± R.
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Answer: B. 100 MPa
σ1 = 50 + √(30² + 40²) = 50 + 50 = 100 MPa.
179. For the same stress state (σx = 80, σy = 20, τxy = 40 MPa) the minimum principal stress is
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Average minus radius.
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Answer: B. 0 MPa
σ2 = 50 − 50 = 0.
180. For the same stress state (σx = 80, σy = 20, τxy = 40 MPa) the maximum in-plane shear stress is
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Radius of Mohr's circle.
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Answer: D. 50 MPa
τmax = R = √(30² + 40²) = 50 MPa.
181. A bar is under a uniaxial tensile stress of 100 MPa. The normal stress on a plane whose normal makes 30° with the axis is
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σn = σcos²θ.
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Answer: A. 75 MPa
σn = σ cos²θ = 100 × 0.75 = 75 MPa.
182. For the same uniaxial stress of 100 MPa the shear stress on the plane whose normal is at 30° to the axis is
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τ = (σ/2) sin2θ.
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Answer: C. 43.3 MPa
τ = σ sinθ cosθ = 100 × 0.5 × 0.866 = 43.3 MPa.
183. A thin cylinder with internal diameter 500 mm and wall thickness 10 mm carries an internal pressure of 2 MPa. The hoop stress is
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pd/2t.
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Answer: C. 50 MPa
σh = pd/2t = 2 × 500/(2×10) = 50 MPa.
184. For the same thin cylinder (d = 500 mm, t = 10 mm, p = 2 MPa) the longitudinal stress is
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Half of the hoop stress.
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Answer: A. 25 MPa
σl = pd/4t = 25 MPa.
185. A thin spherical vessel of internal diameter 1000 mm and wall thickness 10 mm carries an internal pressure of 2 MPa. The stress in the wall is
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pd/4t.
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Answer: B. 50 MPa
σ = pd/4t = 2 × 1000/(4×10) = 50 MPa.
186. What minimum wall thickness is required for a thin cylindrical shell of diameter 800 mm to withstand 1.5 MPa internal pressure if the permissible hoop stress is 80 MPa?
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t = pd/2σ.
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Answer: B. 7.5 mm
t = pd/(2σ) = 1.5 × 800/(2 × 80) = 7.5 mm.
187. A thick cylinder of internal radius 100 mm and external radius 200 mm is subjected to an internal pressure of 40 MPa only. The hoop stress at the inner surface is
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Lamé's formula at r = ri.
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Answer: D. 66.7 MPa
σh(inner) = p(ro² + ri²)/(ro² − ri²) = 40 × 50 000/30 000 = 66.7 MPa.
188. For the same thick cylinder (ri = 100 mm, ro = 200 mm, p = 40 MPa internal) the hoop stress at the outer surface is
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Lamé's formula at r = ro.
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Answer: A. 26.7 MPa
σh(outer) = 2p ri²/(ro² − ri²) = 2 × 40 × 10 000/30 000 = 26.7 MPa.
189. A thin-walled closed tube of uniform thickness 2 mm encloses a mean area of 2500 mm² and carries a torque of 400 N·m. The shear stress (Bredt–Batho) is
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τ = T/2At.
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Answer: C. 40 MPa
τ = T/(2At) = 400 000/(2 × 2500 × 2) = 40 MPa.
190. A continuous beam rests on four supports (one hinge and three rollers) under vertical loads, with possible horizontal reaction at the hinge included. The degree of static indeterminacy is
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Reactions minus 3.
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Answer: C. 2
Reactions = 4 vertical + 1 horizontal = 5; equations = 3; degree = 2.
191. For a thin cylinder (d = 500 mm, t = 10 mm, p = 2 MPa, E = 200 GPa, ν = 0.3), the increase in diameter is nearly
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Hoop strain times diameter.
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Answer: A. 0.106 mm
ε_h = (σh/E)(1 − ν/2) = (50/200 000)(0.85) = 2.125×10⁻⁴; δd = ε d = 0.106 mm.
192. For the same cylinder (d = 500 mm, t = 10 mm, p = 2 MPa, E = 200 GPa, ν = 0.3) the volumetric strain is
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εv = 2ε_h + ε_l.
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Answer: D. 4.75 × 10⁻⁴
εv = 2εh + εl = (pd/4tE)(5 − 4ν) = (2×500/(4×10×200 000))×3.8 = 4.75×10⁻⁴.