Nepal Engineering Council · Civil Engineering · Chapter 4
Structural Mechanics
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184 questions in 6 syllabus topics.
4.1 Shear forces and bending moments
32 questions · ACiE0401
1. A bending moment that causes compression in the top fibres of a simply supported beam is called
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Picture the beam bending like a smile.
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Answer: B. sagging, taken as positive in the usual convention
Sagging curvature shortens the top fibres (compression) and lengthens the bottom fibres; sagging moment is conventionally positive.
2. At any section of a loaded beam, the slope of the bending moment diagram is equal to
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Differentiate the moment with respect to x.
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Answer: B. the shear force at that section
From equilibrium of a small element, dM/dx = V.
3. At any section of a beam, the slope of the shear force diagram equals
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Go one step up the chain: load, shear, moment.
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Answer: A. the negative of the downward load intensity at that section
Vertical equilibrium of a small element gives dV/dx = -w, where w is the downward load intensity.
4. The bending moment in a beam is maximum (or minimum) at a section where
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Think of dM/dx = 0.
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Answer: D. the shear force is zero or changes sign
Since dM/dx = V, M has a stationary value where V passes through zero.
5. For a simply supported beam carrying a udl over its entire span, the shear force and bending moment diagrams are respectively
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Each integration raises the degree by one.
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Answer: B. a straight inclined line and a parabola
Load is constant, so V is linear in x and M is quadratic (parabolic) in x.
6. In a portion of a beam carrying no load, the shear force diagram and bending moment diagram are respectively
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Zero load means the shear cannot change.
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Answer: C. constant (horizontal) and linearly varying
With w = 0, V is constant, and M, the integral of V, varies linearly.
7. When a concentrated couple is applied at a section of a beam, the shear force diagram and the bending moment diagram show respectively
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A couple has no net force.
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Answer: C. no change and a sudden jump equal to the couple
A pure couple has no vertical resultant, so V is continuous; the bending moment jumps by the magnitude of the couple.
8. A point of contraflexure in a beam is the section at which
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Think of sagging turning into hogging.
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Answer: B. the bending moment changes its sign
At a point of contraflexure the curvature reverses, so the bending moment passes through zero with a change of sign.
9. The bending moment at the free end of a cantilever carrying only a udl over the entire span is
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No load lies beyond the free end.
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Answer: D. zero
Taking moments of the forces on the free side of a section at the free end gives no lever arm, so M = 0; the maximum wL²/2 is at the fixed end.
10. For a beam carrying a uniformly varying load (zero at one end and maximum at the other), the shear force and bending moment diagrams are respectively
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Integrate a linear load twice.
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Answer: D. a parabola and a cubic curve
Load is linear in x, so V is quadratic and M is cubic.
11. The area of the shear force diagram between two sections of a beam is equal to
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Integrate dM/dx = V.
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Answer: D. the change in bending moment between those sections
Integrating dM/dx = V between two sections gives M2 - M1 = area under the shear diagram.
12. The area of the loading diagram between two sections of a beam is equal to
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Integrate dV/dx = -w.
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Answer: A. the change in shear force between those sections
Integrating dV/dx = -w gives V2 - V1 = -(area of load diagram).
13. The axial force at a section of a member is the
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Normal to the cross-section means along the axis.
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Answer: A. algebraic sum of components, along the member axis, of all forces on one side of the section
Axial force is the resultant force normal to the cross-section, i.e. the sum of force components along the member axis on one side of the section.
14. The principle of superposition for beam actions (SF, BM) is valid only when
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Effects must add linearly with loads.
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Answer: D. the material is linearly elastic and deformations are small
Superposition requires a linear relationship between loads and effects, which holds for linear elastic behaviour and small displacements.
15. At an internal hinge in a beam, the bending moment is
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A hinge is free to rotate.
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Answer: C. zero
A hinge cannot transmit moment, so the bending moment at an internal hinge is zero.
16. Which of the following is classified as a live (imposed) load on a floor as per IS 875?
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Which load can change with time?
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Answer: A. weight of occupants and movable furniture
Live loads are movable or variable loads such as occupants and furniture; the others are permanent dead loads.
17. For a simply supported beam with no external couple at its ends, the bending moment at each support is
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Pins and rollers do not resist rotation.
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Answer: C. zero
A pin or roller support cannot resist moment, so M = 0 at the ends when no couple is applied there.
18. At the section where a downward point load acts on a beam, the shear force diagram shows
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Check vertical equilibrium of a small element.
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Answer: D. a sudden drop equal to the magnitude of the load
Vertical equilibrium across a downward point load gives a drop in V (left to right) equal to the load; M changes slope but remains continuous.
19. 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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Use wL²/8.
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Answer: B. 45 kN·m
Mmax = wL²/8 = 10×6²/8 = 45 kN·m.
20. A simply supported beam of span 8 m carries a point load of 40 kN at midspan. The bending moment at midspan is
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Use PL/4.
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Answer: A. 80 kN·m
M = PL/4 = 40×8/4 = 80 kN·m.
21. A simply supported beam of span 5 m carries a 30 kN point load at 2 m from the left support. The maximum bending moment is
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Find the left reaction first.
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Answer: C. 36 kN·m
R_A = 30×3/5 = 18 kN; M under load = 18×2 = 36 kN·m.
22. A cantilever of length 4 m carries a udl of 12 kN/m over its full length. The magnitude of the bending moment at the fixed end is
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Resultant load acts at mid-length.
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Answer: D. 96 kN·m
M = wL²/2 = 12×4²/2 = 96 kN·m (hogging).
23. A cantilever of length 3 m carries a 20 kN point load at the free end together with a udl of 5 kN/m over the full length. The fixed-end moment is
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Add the two moments about the fixed end.
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Answer: A. 82.5 kN·m
M = 20×3 + 5×3²/2 = 60 + 22.5 = 82.5 kN·m.
24. A simply supported beam of span 6 m carries a triangular load varying from zero at the left support to 12 kN/m at the right support. The right reaction is
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The resultant acts at two-thirds of the span from the zero end.
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Answer: B. 24 kN
Total load W = ½×12×6 = 36 kN acting at 4 m from the left support. R_B = W×(2/3) = 24 kN.
25. A beam ABC of total length 8 m is simply supported at A and B (6 m apart) with an overhang BC of 2 m, and carries a udl of 10 kN/m over the entire length. The bending moment at support B has magnitude
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Consider only the overhanging part.
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Answer: B. 20 kN·m
Taking the free overhang side, M_B = wℓ²/2 = 10×2²/2 = 20 kN·m (hogging).
26. For the same beam ABC (6 m span AB, 2 m overhang BC, udl of 10 kN/m on the whole 8 m), the point of contraflexure lies within AB at a distance from A of
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Set M(x) = 0 for x > 0.
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Answer: C. 5.33 m
R_A = 80−53.33 = 26.67 kN. M(x) = 26.67x − 10x²/2 = 0 gives x = 5.33 m.
27. A simply supported beam of span 10 m carries a udl of 5 kN/m over the left 6 m only. The shear force is zero at a distance from the left support of
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Find R_A, then set V(x) = 0 within the loaded length.
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Answer: C. 4.2 m
R_B = 30×3/10 = 9 kN, R_A = 21 kN. V = 21 − 5x = 0 gives x = 4.2 m.
28. For the same beam (10 m span, 5 kN/m on the left 6 m), the maximum bending moment is
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Evaluate M at the section of zero shear.
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Answer: A. 44.1 kN·m
Mmax = 21×4.2 − 5×4.2²/2 = 44.1 kN·m at x = 4.2 m.
29. A simply supported beam (hinge at left, roller at right) carries a 50 kN load at midspan inclined at 60° to the beam axis. The axial force in the beam between the hinge and the load is
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Resolve the load along the beam axis.
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Answer: B. 25 kN
Only the hinge resists axial force: N = 50cos60° = 25 kN.
30. A simply supported beam of span 4 m carries a clockwise couple of 20 kN·m at midspan. The magnitude of each vertical reaction is
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Couple = reaction × span.
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Answer: D. 5 kN
Moment equilibrium: R×4 = 20, so R = 5 kN (equal and opposite).
31. A simply supported beam of span 10 m carries a udl of 6 kN/m over the whole span and a 20 kN point load at 4 m from the left support. The shear force just to the right of the point load is
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Include the load once you cross it.
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Answer: C. -2 kN
R_A = 30 + 20×6/10 = 42 kN. V = 42 − 6×4 − 20 = -2 kN.
32. The bending moment at a distance x from the left end of a simply supported beam is M = 20x − 5x² (kN·m, x in m, 0 ≤ x ≤ 4 m). The maximum bending moment is
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Set dM/dx = 0.
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Answer: A. 20 kN·m
dM/dx = 20 − 10x = 0 gives x = 2 m; M = 40 − 20 = 20 kN·m.
4.2 Stress and strain analysis
30 questions · ACiE0402
33. Young's modulus of elasticity is defined as the ratio of
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Think of the slope of the straight part of the σ–ε curve.
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Answer: D. stress to strain within the proportional limit
E = σ/ε, constant for a given material in the linear (Hookean) range.
34. The theoretical upper limit of Poisson's ratio for an isotropic elastic material is
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Zero volume change gives the limit.
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Answer: C. 0.5
From K = E/[3(1−2ν)] > 0 and G > 0, ν lies between −1 and 0.5; ν = 0.5 corresponds to an incompressible material.
35. On the stress–strain curve of mild steel, the point at which strain increases appreciably with no increase in stress is called the
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Look for the plateau.
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Answer: C. yield point
At the yield point the material flows plastically at nearly constant stress.
36. Hooke's law is strictly valid up to the
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The stress–strain curve must be a straight line.
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Answer: C. proportional limit
Stress is proportional to strain only up to the proportional limit.
37. For materials such as aluminium or high-tensile steel that show no definite yield point, the yield stress is taken as the
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Draw a line parallel to the elastic slope.
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Answer: C. 0.2% proof stress
The 0.2% offset method defines an equivalent yield stress for such materials.
38. The area under the complete stress–strain curve of a material up to fracture represents its
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Energy per unit volume until failure.
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Answer: C. toughness
Area under σ–ε is energy absorbed per unit volume to failure, i.e. toughness; resilience is the area up to the elastic limit.
39. Principal planes are the planes on which
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Defined by what is absent.
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Answer: B. the shear stress is zero
By definition, principal planes carry only normal (principal) stresses and no shear stress.
40. The planes of maximum shear stress are inclined to the principal planes at
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Half of the 90° between principal planes.
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Answer: A. 45°
Maximum shear occurs at 45° to the principal planes, where τ = (σ1 − σ2)/2.
41. If σ1 and σ2 are the principal stresses (σ1 > σ2) in a plane stress state, the maximum in-plane shear stress is
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It is the radius of the circle.
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Answer: D. (σ1 − σ2)/2
τmax = (σ1 − σ2)/2, equal to the radius of Mohr's circle.
42. In Mohr's circle for plane stress, the distance of the centre of the circle from the origin along the σ-axis is
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Average normal stress.
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Answer: D. (σx + σy)/2
The centre lies at the average normal stress (σx + σy)/2; the radius is √[((σx − σy)/2)² + τxy²].
43. For a plane stress element, which quantity remains the same on any two mutually perpendicular planes?
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It is the first invariant.
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Answer: B. the sum of the normal stresses
σx + σy = σ1 + σ2 is an invariant of the stress transformation.
44. A material element subjected to pure shear τ has principal stresses of
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Centre of the Mohr circle is at the origin.
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Answer: B. +τ and −τ, on planes at 45° to the shear planes
For σx = σy = 0, R = τ and the centre is at the origin, so σ1 = τ and σ2 = −τ at 45°.
45. The complementary shear stresses on two mutually perpendicular planes of an element are
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Moment equilibrium of the element.
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Answer: B. equal in magnitude and both point towards or both away from the common edge
Moment equilibrium of the element requires τxy = τyx with the senses as stated.
46. In the simple torsion theory of circular shafts, one of the assumptions is that
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Circular sections do not warp.
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Answer: B. plane transverse sections remain plane and radii remain straight after twisting
The theory assumes plane sections remain plane, radial lines remain straight, material is linear elastic and homogeneous.
47. In a circular shaft under pure torsion, the shear stress is
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It varies linearly with r.
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Answer: A. zero at the axis and maximum at the outer surface
τ = Tr/J varies linearly with the radius r.
48. For the same weight, length and material, a hollow circular shaft compared with a solid shaft is
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Where is the stress high?
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Answer: D. stronger in torsion because its material is placed farther from the axis
A larger polar moment J is obtained for the same area when material lies at a larger radius.
49. A ductile circular bar twisted to failure normally fails along a section
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Ductile fails in shear.
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Answer: B. perpendicular to its axis, by shear
Ductile material is weaker in shear, and maximum torsional shear acts on the transverse plane; brittle material fails along a 45° helix by tension.
50. A steel rod of diameter 20 mm carries an axial tensile load of 100 kN. The normal stress in the rod is
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Area of a circle is πd²/4.
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Answer: B. 318.3 MPa
σ = P/A = 100000/(π×20²/4) = 318.3 N/mm² = 318.3 MPa.
51. A bar of length 2 m and cross-section 500 mm² carries an axial pull of 50 kN. For E = 200 GPa, the elongation is
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Mind the units (N, mm).
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Answer: C. 1 mm
δ = PL/AE = 50000×2000/(500×200000) = 1 mm.
52. For a material with E = 200 GPa and Poisson's ratio 0.3, the modulus of rigidity is
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E = 2G(1+ν).
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Answer: C. 76.9 GPa
G = E/[2(1+ν)] = 200/(2×1.3) = 76.9 GPa.
53. At a point, σx = 60 MPa, σy = -20 MPa (compressive) and τxy = 30 MPa. The major principal stress is
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Centre plus radius of Mohr's circle.
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Answer: B. 70 MPa
σ1 = (σx+σy)/2 + √[((σx−σy)/2)²+τ²] = 20 + √(40²+30²) = 20 + 50 = 70 MPa.
54. For the same stress state (σx = 60 MPa, σy = -20 MPa, τxy = 30 MPa), the maximum shear stress is
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It equals the radius of Mohr's circle.
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Answer: A. 50 MPa
τmax = radius = √(40²+30²) = 50 MPa.
55. A bar is under uniaxial tension σ = 100 MPa. The normal stress on a plane whose normal makes 30° with the bar axis is
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Use σ cos²θ.
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Answer: A. 75 MPa
σn = σ cos²θ = 100×cos²30° = 75 MPa.
56. At a point, σx = 80 MPa, σy = 20 MPa and τxy = 40 MPa. The angle of the major principal plane with the x-plane is
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tan 2θ = 2τ/(σx − σy).
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Answer: A. 26.6°
tan2θ = 2τ/(σx−σy) = 80/60 = 1.333, 2θ = 53.1°, θ = 26.6°.
57. A solid circular shaft of diameter 60 mm transmits a torque of 2 kN·m. The maximum shear stress is
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τ = 16T/πd³.
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Answer: D. 47.2 MPa
τ = 16T/(πd³) = 16×2×10⁶/(π×60³) = 47.2 MPa.
58. A shaft transmits 100 kW at 300 rpm. The torque transmitted is
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P = Tω, ω = 2πN/60.
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Answer: A. 3183 N·m
T = 60P/(2πN) = 60×100000/(2π×300) = 3183 N·m.
59. A solid shaft of diameter 50 mm and length 2 m is subjected to a torque of 1 kN·m. For G = 80 GPa, the angle of twist is
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θ = TL/GJ in radians, then convert.
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Answer: D. 2.33°
J = πd⁴/32 = 613592 mm⁴; θ = TL/GJ = 0.0407 rad = 2.33°.
60. A hollow circular shaft has outer diameter 100 mm and inner diameter 80 mm. The torque it can transmit for an allowable shear stress of 50 MPa is
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T = τJ/R with J of the annulus.
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Answer: A. 5.8 kN·m
J = π(D⁴−d⁴)/32 = 5796238 mm⁴; T = τJ/R = 50×5796238/50 = 5.8 kN·m.
61. A steel bar is rigidly fixed at both ends and heated through 40°C (α = 12×10⁻⁶ per °C, E = 200 GPa). The thermal stress developed is
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σ = EαΔT.
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Answer: D. 96 MPa
σ = EαΔT = 200000×12×10⁻⁶×40 = 96 MPa (compressive).
62. A bar is subjected to a uniaxial tensile stress of 120 MPa. The maximum shear stress in the bar is
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Use σ2 = 0.
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Answer: A. 60 MPa
τmax = (σ1 − σ2)/2 = 120/2 = 60 MPa, acting on planes at 45°.
4.3 Theory of flexure and columns
30 questions · ACiE0403
63. A beam segment is in a state of pure bending when it is subjected to
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Uniform moment means dM/dx = 0.
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Answer: B. a constant bending moment with zero shear force
Pure bending means uniform moment along the length (V = dM/dx = 0), as between the two loads of a four-point bending test.
64. In the theory of simple bending, which of the following is an assumption?
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Strain varies linearly with distance.
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Answer: D. plane sections before bending remain plane after bending
Navier's hypothesis: plane sections remain plane, leading to linear strain variation across the depth; the material is linearly elastic and homogeneous.
65. The flexure formula is written as
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Remember: moment, stress, curvature.
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Answer: B. M/I = σ/y = E/R
Bending stress σ = My/I and curvature 1/R = M/EI give M/I = σ/y = E/R.
66. For a homogeneous beam of linearly elastic material in pure bending, the neutral axis passes through the
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Net axial force is zero.
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Answer: D. centroid of the cross-section
Equilibrium of axial force (∫σ dA = 0) requires the neutral axis to pass through the centroid.
67. The section modulus of a beam section is defined as
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Think of σmax = M/Z.
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Answer: C. the moment of inertia divided by the distance of the extreme fibre from the neutral axis
Z = I/ymax, so that σmax = M/Z.
68. The flexural rigidity of a beam is given by
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The denominator in M/EI.
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Answer: D. EI
Flexural rigidity (flexural stiffness) is the product of modulus of elasticity and the second moment of area; the curvature is M/EI.
69. The differential equation of the elastic curve of a beam under small deflection is
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Curvature equals second derivative.
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Answer: A. EI d²y/dx² = M
Curvature ≈ d²y/dx² = M/EI; successive differentiation gives V = EI y‴ and w = EI y⁗ (with signs).
70. A circular log of diameter D is to be cut into a rectangular beam of the maximum bending strength. The ratio of depth to width of the section is
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Maximise bd² with b² + d² = D².
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Answer: B. √2
Maximising Z = bd²/6 subject to b² + d² = D² gives b = D/√3 and d = D√(2/3), so d/b = √2.
71. For a rectangular beam section subjected to a vertical shear force, the shear stress distribution over the depth is
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Zero where the first moment Aȳ vanishes.
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Answer: C. parabolic, maximum at the neutral axis and zero at the extreme fibres
τ = VAȳ/(Ib) gives a parabola with τmax = 1.5 V/(bd) at the neutral axis.
72. The Euler formula for the buckling load of a long column is valid only if
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Ideal column assumptions.
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Answer: C. the column is initially straight, axially loaded and stresses stay within the elastic limit
Euler's theory idealises a perfectly straight, centrally loaded, homogeneous elastic column.
73. The slenderness ratio of a column is the ratio of
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Buckling is about the weakest axis.
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Answer: C. effective length to least radius of gyration
λ = Le/rmin, since buckling occurs about the axis of least stiffness.
74. The effective length of a column of length L fixed at one end and free at the other is
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Think of a flagpole.
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Answer: A. 2L
A fixed–free column behaves like half of a pinned–pinned column of length 2L, so Le = 2L.
75. According to Euler's theory, the critical buckling load of a column is
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Le appears squared in the denominator.
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Answer: A. inversely proportional to the square of the effective length
Pcr = π²EI/Le².
76. The Euler buckling load of a slender column does not depend on
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It is a stability, not strength, criterion.
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Answer: B. the yield strength of the material
Pcr = π²EI/Le² depends on stiffness (E, I) and geometry, not on strength.
77. Euler's formula is not applicable to short columns because
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Compare σcr with the proportional limit.
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Answer: A. the critical stress it predicts exceeds the proportional limit, so the column fails by crushing or inelastic buckling
For small slenderness ratio σcr = π²E/λ² would exceed the elastic limit, so the elastic buckling assumption fails.
78. A column with a rectangular cross-section b × d (b < d) and pinned at both ends will buckle
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Pcr depends on least I.
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Answer: B. about the axis of least moment of inertia, i.e. in the direction of the smaller dimension
Buckling takes place about the axis having minimum I, which governs Pcr.
79. A rectangular beam 200 mm wide and 400 mm deep is subjected to a bending moment of 80 kN·m. The maximum bending stress is
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Z = bd²/6.
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Answer: A. 15 MPa
σ = M/Z = 6M/(bd²) = 6×80000000/(200×400²) = 15 MPa.
80. A timber beam of section 150 mm × 300 mm (depth) has an allowable bending stress of 10 MPa. Its moment of resistance is
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M = σ × Z.
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Answer: A. 22.5 kN·m
Z = bd²/6 = 150×300²/6 = 2250000 mm³; M = σZ = 22.5 kN·m.
81. A beam has flexural rigidity EI = 20000 kN·m². The radius of curvature produced by a uniform bending moment of 50 kN·m is
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R = EI/M.
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Answer: D. 400 m
1/R = M/EI so R = EI/M = 20000/50 = 400 m.
82. A thin steel strip (E = 200 GPa) is bent to a radius of curvature of 1 m. The bending stress at a fibre 1 mm from the neutral axis is
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Use σ/y = E/R.
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Answer: D. 200 MPa
σ = Ey/R = 200000×1/1000 = 200 MPa.
83. A simply supported beam of span 4 m carries a central point load of 60 kN. For E = 200 GPa and I = 8×10⁷ mm⁴, the central deflection is
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Use PL³/48EI in N and mm.
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Answer: C. 5 mm
δ = PL³/(48EI) = 60000×4000³/(48×200000×80000000) = 5 mm.
84. A cantilever of length 3 m carries a udl of 10 kN/m (= 10 N/mm). For E = 200 GPa and I = 10⁸ mm⁴, the free-end deflection is
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wL⁴/8EI.
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Answer: D. 5.06 mm
δ = wL⁴/(8EI) = 10×3000⁴/(8×200000×100000000) = 5.06 mm.
85. A cantilever of length 2 m with EI = 5000 kN·m² carries a point load of 20 kN at the free end. The slope at the free end is
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θ = PL²/2EI.
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Answer: B. 0.008 rad
θ = PL²/(2EI) = 20×2²/(2×5000) = 0.008 rad.
86. A simply supported concrete beam, span 6 m, carries a udl of 12 kN/m. Take E = 25 GPa and I = 3.125×10⁹ mm⁴. The maximum deflection is
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5wL⁴/384EI.
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Answer: D. 2.59 mm
δ = 5wL⁴/(384EI) = 5×12×6000⁴/(384×25000×3125000000) = 2.59 mm.
87. A pin-ended steel column of length 3 m has E = 200 GPa and least I = 2×10⁶ mm⁴. Its Euler buckling load is
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Pcr = π²EI/L².
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Answer: C. 438.6 kN
Pcr = π²EI/L² = π²×200000×2×10⁶/3000² = 438.6 kN.
88. For the same column (L = 3 m, E = 200 GPa, I = 2×10⁶ mm⁴) fixed at the base and free at the top, the Euler load is
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Use effective length 2L.
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Answer: A. 109.7 kN
Le = 2L = 6 m; Pcr = π²EI/Le² = 109.7 kN, a quarter of the pinned value.
89. A column of effective length 3 m has a rectangular section 50 mm × 100 mm. Its slenderness ratio is
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Use the smaller dimension: r = b/√12.
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Answer: A. 207.8
rmin = 50/√12 = 14.43 mm; λ = Le/rmin = 3000/14.43 = 207.8.
90. The Euler critical stress for a column of slenderness ratio 100 with E = 200 GPa is
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σcr = π²E/λ².
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Answer: C. 197.4 MPa
σcr = π²E/λ² = π²×200000/100² = 197.4 MPa.
91. The ratio of Euler's buckling load of a column fixed at both ends to that of the same column pinned at both ends is
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Compare effective lengths L/2 and L.
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Answer: B. 4
Fixed–fixed has Le = L/2, so Pcr ∝ 1/Le² is four times that of Le = L.
92. A rectangular beam section 200 mm × 300 mm carries a shear force of 60 kN. The maximum shear stress is
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Maximum is 1.5 times the average.
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Answer: B. 1.5 MPa
τmax = 1.5×V/(bd) = 1.5×60000/(200×300) = 1.5 MPa.
4.4 Determinate structures-1
31 questions · ACiE0404
93. A plane pin-jointed truss with m members, r support reactions and j joints is statically determinate (and stable) when
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Two equations per joint.
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Answer: B. m + r = 2j
Each joint gives two equilibrium equations, so 2j equations solve m + r unknowns when m + r = 2j.
94. The strain energy stored in a bar of length L, area A and modulus E under an axial load P is
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Work done by a gradually applied load is ½Pδ.
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Answer: A. P²L/(2AE)
U = ½Pδ with δ = PL/AE gives U = P²L/(2AE).
95. The strain energy stored in a beam due to bending only is given by
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Energy is quadratic in the load.
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Answer: B. ∫ M² dx / (2EI)
U = ∫ (M²/2EI) dx over the length of the beam.
96. Castigliano's first theorem states that the partial derivative of the total strain energy with respect to an applied load gives
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Energy derivative means displacement.
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Answer: C. the displacement at the point of application of that load, in its direction
δi = ∂U/∂Pi for linear elastic structures.
97. Maxwell's reciprocal theorem states that for a linearly elastic structure
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Cause and effect are interchangeable.
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Answer: C. the deflection at A due to a unit load at B equals the deflection at B due to a unit load at A
δAB = δBA, a special case of Betti's law.
98. In the unit load (virtual work) method, the deflection of a beam at a point is obtained from
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Real moment times virtual moment over EI.
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Answer: A. δ = ∫ (M m / EI) dx, where m is the moment due to a unit load at that point
External virtual work (1×δ) equals internal virtual work ∫ m (M/EI) dx.
99. To find the rotation (slope) at a point of a beam by the unit load method, the virtual load applied at that point is
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Work-conjugate quantity.
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Answer: C. a unit couple
The virtual load must be work-conjugate to the required displacement: a couple for rotation, a force for translation.
100. In the conjugate beam method, a real fixed end corresponds in the conjugate beam to
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Conjugate shear = slope, conjugate moment = deflection.
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Answer: D. a free end
At a real fixed end slope and deflection are zero, so the conjugate shear and moment are zero: a free end. A real free end becomes a fixed end.
101. In the moment-area method, the first theorem states that the change in slope between two points on the elastic curve is equal to
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Area under the M/EI diagram.
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Answer: D. the area of the M/EI diagram between those points
θAB = ∫ (M/EI) dx between A and B.
102. In the double integration method, the boundary conditions at the fixed end of a cantilever are
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What does a fixed support restrain?
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Answer: C. deflection = 0 and slope = 0
A fixed end prevents both translation and rotation.
103. Macaulay's method is particularly convenient for finding the deflection of beams
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One equation for the full span.
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Answer: C. carrying point loads and partial udls, using a single equation for the whole span
Macaulay's bracket notation writes one expression for M over the full span, so only two constants of integration appear.
104. For a truss, the deflection of a joint by the unit load method is given by
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Real force × virtual force × length / AE.
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Answer: A. Σ (F u L) / (AE)
δ = Σ F u L/AE, where F is the real member force and u is the member force due to a unit load at the joint.
105. Which of the following statements about strain energy is correct for a linearly elastic structure?
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U is a quadratic function.
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Answer: D. Strain energy is proportional to the square of the load, so it cannot be superposed
U ∝ P²; strain energies due to separate loads cannot simply be added because of the cross-term.
106. The introduction of one internal hinge in a rigid structure
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A hinge cannot carry moment.
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Answer: A. gives one additional equation of condition (M = 0 at the hinge)
A hinge releases moment transfer, giving one extra equation ΣM = 0 for the part on one side.
107. A structure that has fewer restraints than are necessary for equilibrium under general loading is
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Not enough supports.
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Answer: A. unstable (a mechanism)
Insufficient restraints allow rigid-body motion or collapse mechanism, so the structure is unstable.
108. In the virtual work method, the virtual force system must
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Which conditions does a force system satisfy?
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Answer: B. satisfy the equilibrium conditions of the structure
The principle of virtual forces uses an equilibrated virtual force system acting through real displacements.
109. In a portal frame with hinged bases carrying a horizontal load at beam level, the horizontal displacement of the beam level is called
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A frame moves sideways.
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Answer: D. sway
Lateral translation of the beam level of a frame under lateral load is sway.
110. A plane pin-jointed truss has 8 joints, 14 members and 3 support reaction components. Its degree of static indeterminacy is
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D = m + r − 2j.
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Answer: B. 1
D = m + r − 2j = 14 + 3 − 16 = 1.
111. A propped cantilever has a fixed support at A and a roller at B. If one more roller support is added at C, the degree of static indeterminacy of the continuous beam becomes
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Count reactions minus 3.
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Answer: C. 2
Reactions = 3 (fixed) + 1 + 1 = 5; equilibrium equations = 3; D = 5 − 3 = 2.
112. A single-bay portal frame with hinged supports at both bases (4 reaction components, no internal hinge) is
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Four reactions against three equations.
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Answer: C. statically indeterminate to the first degree
r − 3 = 4 − 3 = 1.
113. A closed rectangular rigid-jointed frame (a closed ring) is internally statically indeterminate to the degree of
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One cut releases three forces.
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Answer: D. 3
A closed ring must be cut once to become a determinate (open) structure; each cut releases 3 internal forces (N, V, M).
114. A bar of area 500 mm² and length 2 m carries a gradually applied axial load of 100 kN. For E = 200 GPa, the strain energy stored is
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U = P²L/2AE in N and mm.
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Answer: A. 100 N·m
U = P²L/(2AE) = (100000)²×2000/(2×500×200000) = 100000 N·mm = 100 N·m.
115. A cantilever of length 2 m with EI = 4000 kN·m² carries a 10 kN load at the free end. The free-end deflection is
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Use PL³/3EI.
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Answer: B. 6.67 mm
δ = PL³/(3EI) = 10×2³/(3×4000) = 0.00667 m = 6.67 mm.
116. A simply supported beam of span 6 m with EI = 20000 kN·m² carries a central load of 40 kN. The central deflection is
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PL³/48EI.
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Answer: D. 9 mm
δ = PL³/(48EI) = 40×6³/(48×20000) = 0.009 m = 9 mm.
117. A simply supported beam of span 6 m with EI = 30000 kN·m² carries a udl of 20 kN/m. The slope at the supports is
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wL³/24EI.
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Answer: A. 0.006 rad
θ = wL³/(24EI) = 20×6³/(24×30000) = 0.006 rad.
118. For three members of a truss, real forces F (kN), unit-load forces u and lengths L (m) are (60, 1.0, 4), (−45, −0.75, 3) and (75, 1.25, 5). The remaining members carry u = 0. For AE = 2×10⁵ kN (all members), the joint deflection is
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Only members with u ≠ 0 contribute.
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Answer: B. 4.05 mm
ΣFuL = 240 + 101.25 + 468.75 = 810 kN·m; δ = ΣFuL/AE = 0.00405 m = 4.05 mm.
119. A simply supported beam of span 5 m with EI = 15000 kN·m² carries a central point load of 48 kN. The slope at a support (moment-area result) is
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PL²/16EI.
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Answer: D. 0.005 rad
θ = PL²/(16EI) = 48×5²/(16×15000) = 0.005 rad.
120. A cantilever of length 4 m and EI = 10000 kN·m² carries a couple of 30 kN·m at the free end. The free-end deflection is
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Moment diagram is constant.
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Answer: C. 24 mm
δ = ML²/(2EI) = 30×4²/(2×10000) = 0.024 m = 24 mm.
121. Using the unit load method, the free-end deflection of a cantilever of length 3 m, EI = 5000 kN·m², carrying a udl of 8 kN/m is
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The answer is the standard wL⁴/8EI.
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Answer: B. 16.2 mm
M = −wx²/2, m = −x: δ = ∫ wx³/(2EI) dx = wL⁴/(8EI) = 16.2 mm.
122. The strain energy of a cantilever under tip load P is U = P²L³/(6EI). For P = 12 kN, L = 2 m and EI = 3000 kN·m², Castigliano's theorem gives the tip deflection as
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Differentiate with respect to P.
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Answer: A. 10.67 mm
δ = ∂U/∂P = PL³/(3EI) = 12×2³/(3×3000) = 10.67 mm.
123. The strain energy stored in a linearly elastic bar under a load P is U. If the load is doubled, the strain energy stored becomes
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U is proportional to P².
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Answer: B. 4U
U ∝ P², so doubling P multiplies U by 4.
4.5 Determinate structures-2
30 questions · ACiE0405
124. An influence line for a function (reaction, shear or moment) is a graph that shows
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One section, one moving load.
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Answer: C. the variation of that function at a fixed section as a unit load moves across the structure
An IL is plotted with the position of the unit load as abscissa and the value of the function at one fixed location as ordinate.
125. The main difference between a bending moment diagram and an influence line for bending moment is that
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Think about what the horizontal axis represents.
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Answer: A. the BMD gives moments at all sections for fixed loads, whereas the IL gives the moment at one section for a moving load
BMD: abscissa = section, loads fixed. IL: abscissa = load position, section fixed.
126. Müller-Breslau principle states that the influence line for a force (or moment) in a structure is to the same scale as
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Remove the restraint, give a unit displacement.
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Answer: C. the deflected shape obtained by removing the restraint corresponding to that force and giving a unit displacement in its direction
Removing the restraint and introducing a unit displacement corresponding to the force gives the IL ordinates as deflections.
127. The influence line for the left reaction of a simply supported beam of span L is
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Unit load at A is carried fully by A.
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Answer: D. a straight line with ordinate 1 under the left support and 0 under the right support
R_A = (L − x)/L for a unit load at x from A, linear from 1 to 0.
128. The influence line for shear force at a section of a simply supported beam consists of
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V jumps by the unit load when it crosses the section.
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Answer: A. two parallel inclined lines with a vertical jump of unity at the section
The IL for V shows two lines parallel to the IL for the reaction, separated by a jump of 1 at the section.
129. The influence line for bending moment at a section of a simply supported beam is
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The IL is linear between the supports and the section.
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Answer: B. a triangle with its apex under the section
M = x(L − a)/L for x ≤ a and a(L − x)/L for x ≥ a: a triangle with peak ab/L at the section.
130. The units of ordinates of an influence line for a bending moment and for a shear force are respectively
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Divide the function by the unit load.
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Answer: C. length and dimensionless
IL ordinates are the function per unit load: moment/force = length, shear/force = dimensionless.
131. For a simply supported beam carrying a single moving concentrated load, the absolute maximum bending moment occurs
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Maximise ab/L.
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Answer: A. at midspan, when the load is at midspan
M at a section is maximum when the load is on it; the largest value of ab/L is L/4 at midspan, giving PL/4.
132. To obtain the maximum positive shear at a section of a simply supported beam due to a udl longer than the span that can be placed anywhere, the load should cover
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Load where the IL is positive.
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Answer: B. only the portion of the span where the IL ordinates are positive
Maximum effect arises when loading covers the entire positive area of the IL and none of the negative area.
133. The maximum shear force at the support of a simply supported beam due to a moving point load occurs when the load is
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Maximum IL ordinate.
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Answer: B. at the support (just on the span)
The IL of the support shear has the maximum ordinate (1) at the support.
134. A two-hinged arch is
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Count reactions minus three.
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Answer: A. statically indeterminate to the first degree
Four reaction components against three equilibrium equations gives one redundant, taken as the horizontal thrust.
135. A three-hinged arch is
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The hinge gives M = 0.
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Answer: D. statically determinate, the third hinge providing the additional equation
Four reactions with 3 equilibrium equations + 1 condition (M = 0 at the crown hinge).
136. A parabolic arch carrying a udl over its entire horizontal span has
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Funicular shape.
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Answer: D. only axial compression at every section, with no bending moment
A parabola is the funicular (equilibrium) shape for a uniformly distributed load on the horizontal projection, so M = 0 throughout.
137. In the analysis of a two-hinged arch, the horizontal thrust is obtained by applying the condition that
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Compatibility at a hinge.
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Answer: D. the horizontal displacement of the support is zero, i.e. ∂U/∂H = 0
H is the redundant; since the hinged supports do not move apart, the horizontal displacement is zero, i.e. ∂U/∂H = 0 (least work).
138. The bending moment at any section of a two-hinged arch is expressed as
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Thrust reduces beam moment.
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Answer: A. M = M0 − H y, where M0 is the moment in the equivalent simple beam and y is the rise at that section
The thrust H, acting at height y above the line of the supports, reduces the beam moment M0 by Hy.
139. In a two-hinged arch, an increase in temperature
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Expansion is resisted.
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Answer: B. increases the horizontal thrust
Expansion is restrained by the hinged supports, producing additional horizontal thrust H = EIαTL/∫y²ds.
140. The effect of rib shortening in a two-hinged arch is to
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Shortening lets the supports move closer.
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Answer: C. reduce the horizontal thrust
Axial compression shortens the rib, which reduces the span-wise restraint and hence the horizontal thrust.
141. For a simply supported beam of span 10 m, the ordinate of the influence line for the left reaction under a unit load at 3 m from the left support is
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R_A = (L − x)/L.
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Answer: A. 0.7
R_A = (L − x)/L = (10 − 3)/10 = 0.7.
142. For a simply supported beam of span 12 m, the maximum ordinate of the influence line for bending moment at a section 4 m from the left support is
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ab/L.
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Answer: D. 2.67 m
ab/L = 4×8/12 = 2.67 m, occurring under the section.
143. For the beam of span 12 m, a single 100 kN load moves across the span. The maximum bending moment at the section 4 m from the left support is
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Load times the peak ordinate.
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Answer: C. 266.67 kN·m
M = P×ab/L = 100×2.67 = 266.67 kN·m, with the load on the section.
144. For the same section, a udl of 20 kN/m longer than the span covers the entire span. The bending moment at the section is
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Load intensity times area of the IL.
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Answer: D. 320 kN·m
M = w × area of IL = 20×½×12×2.67 = 320 kN·m.
145. A single 100 kN load moves over a simply supported beam of span 12 m. The maximum positive shear at the section 4 m from the left support is
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Positive part of the V-IL lies to the right of the section.
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Answer: B. 66.7 kN
The positive IL ordinate just right of the section is (L − a)/L = 0.67; V = 100×0.67 = 66.7 kN.
146. A udl of 20 kN/m, 5 m long, can occupy any position on the beam of span 12 m. The maximum positive shear at the section 4 m from the left support is
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Place the load just right of the section.
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Answer: A. 45.83 kN
Load is placed from the section for 5 m: IL ordinates 0.667 to 0.25; area = 5×(0.667+0.25)/2 = 2.292 m; V = 20×2.292 = 45.83 kN.
147. Two moving loads of 60 kN and 40 kN, 3 m apart (60 kN leading at the left), cross a simply supported beam of span 10 m. The maximum shear force at the left support is
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Put the heavier load at the support.
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Answer: C. 88 kN
Place 60 kN at the support and 40 kN at 3 m: V = 60 + 40×(10−3)/10 = 88 kN.
148. A parabolic arch of span 24 m and rise 4 m is three-hinged and carries a udl of 20 kN/m over the whole span. The horizontal thrust is
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H = M0(crown)/h.
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Answer: B. 360 kN
H = wL²/(8h) = 20×24²/(8×4) = 360 kN.
149. A three-hinged parabolic arch of span 20 m and rise 4 m carries a 120 kN point load at the crown. The horizontal thrust is
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Take moments about the crown hinge.
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Answer: C. 150 kN
M0 at crown = PL/4 = 600 kN·m; H = M0/h = 600/4 = 150 kN.
150. A two-hinged parabolic arch (constant EI) has span 20 m and rise 4 m, and carries a 80 kN central point load. Using H = 25PL/(128h), the horizontal thrust is
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Substitute directly.
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Answer: B. 78.1 kN
H = 25×80×20/(128×4) = 78.1 kN, which is less than the three-hinged value 100 kN.
151. A single 25 kN load moves along a cantilever of length 4 m. The maximum bending moment (magnitude) at the fixed support is
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Maximum IL ordinate is at the free end.
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Answer: D. 100 kN·m
The IL for fixed-end moment has ordinate −x (x from the fixed end), maximum 4 m at the free end: M = 25×4 = 100 kN·m.
152. At a section of a two-hinged arch the equivalent simple-beam bending moment is 300 kN·m, the horizontal thrust is 100 kN and the rise of the arch axis at the section is 2.5 m. The bending moment in the arch at that section is
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M = M0 − Hy.
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Answer: B. 50 kN·m
M = M0 − Hy = 300 − 100×2.5 = 50 kN·m.
153. At a section of an arch where the tangent makes 30° with the horizontal, the beam shear is 40 kN and the thrust is 100 kN. The normal thrust at the section is
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Resolve V and H along the tangent.
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Answer: A. 106.6 kN
N = V sinθ + H cosθ = 40sin30° + 100cos30° = 106.6 kN.
4.6 Indeterminate structures
31 questions · ACiE0406
154. In the flexibility (force) method of analysis, the primary unknowns are
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Unknowns are forces.
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Answer: C. the redundant forces, found from compatibility of displacements
The flexibility method releases redundants to get a determinate primary structure and restores compatibility to find them.
155. In the stiffness (displacement) method of analysis, the primary unknowns are
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Unknowns are displacements.
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Answer: C. the joint displacements (rotations and translations), found from equilibrium of joints
The stiffness method writes joint equilibrium equations in terms of unknown joint displacements.
156. The slope-deflection equation for a member AB with end A rotation θA, end B rotation θB and chord rotation ψ is
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Coefficients 2EI/L, 2θA + θB, −3ψ.
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Answer: D. M_AB = (2EI/L)(2θA + θB − 3ψ) + FEM_AB
This is the standard slope-deflection equation for a prismatic member with clockwise-positive rotations and chord rotation ψ.
157. Which of the following effects is neglected in the basic slope-deflection method?
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Only bending is considered.
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Answer: B. axial and shear deformations of members
The method considers only flexural deformations; axial and shear deformations are neglected.
158. The carry-over factor for a prismatic member with the far end fixed is
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Half the applied moment is carried over.
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Answer: D. 1/2
Moment M applied at the near end induces M/2 at the fixed far end.
159. The flexural stiffness of a prismatic member, rotated at one end while the far end is fixed, is
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Unit rotation with the far end fixed.
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Answer: D. 4EI/L
The moment needed to produce unit rotation at the near end with far end fixed is 4EI/L; with far end hinged it is 3EI/L.
160. The distribution factor of a member at a joint in the moment distribution method is
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Share of the joint's total stiffness.
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Answer: A. its stiffness divided by the sum of the stiffnesses of all members meeting at the joint
DF = k/Σk, so the distribution factors at a joint add up to 1.
161. In the moment distribution method, the distribution factor for a member at a fixed support is
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Support does not rotate.
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Answer: B. zero
A fixed support does not rotate, so it behaves as infinitely stiff and absorbs the unbalanced moment without distributing it back.
162. When the far end of a prismatic member is hinged, the modified stiffness used in moment distribution is
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Hinged ends carry no moment.
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Answer: B. 3EI/L, with no carry-over to the hinged end
A hinged far end cannot sustain moment, so k = 3EI/L and no moment is carried over.
163. In a continuous beam, to obtain the maximum positive bending moment at midspan of a particular span due to live load, the udl should be placed on
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Alternate spans loaded.
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Answer: A. that span and on every alternate span
Live load on that span produces sagging; loading adjacent spans reduces it, but loading the next-but-one spans increases it.
164. To obtain the maximum hogging moment at an interior support of a continuous beam due to a live udl, the load should be placed on
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Look at the sign of the IL in adjacent spans.
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Answer: C. the two spans adjacent to the support and then alternate spans beyond them
The IL for support moment is negative in the two adjacent spans and alternates in sign thereafter.
165. The number of plastic hinges required to convert a structure with r degrees of static indeterminacy into a collapse mechanism is generally
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One for each redundant plus one.
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Answer: C. r + 1
A determinate structure needs one hinge to become a mechanism; each redundant needs one more hinge, so r + 1.
166. As per the uniqueness theorem of plastic analysis, the true collapse load is the load that
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Three conditions are needed.
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Answer: D. satisfies the equilibrium, mechanism and yield conditions simultaneously
The true collapse load is the one for which equilibrium, mechanism and yield (M ≤ Mp) conditions are all satisfied; the kinematic method gives an upper bound and the static method a lower bound.
167. The shape factor of a section is defined as
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Ratio of the two section moduli.
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Answer: B. the ratio of plastic section modulus to elastic section modulus
S = Zp/Z = Mp/My; it is 1.5 for a rectangular section.
168. A flexibility coefficient fij is defined as
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Displacement per unit force.
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Answer: D. the displacement at i due to a unit force at j
Flexibility coefficients are displacements per unit force; stiffness coefficients are forces per unit displacement.
169. A uniform settlement of supports, or a temperature change, will cause internal stresses in
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Redundant restraints resist deformation.
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Answer: C. statically indeterminate structures only
Determinate structures can adjust freely to such movements; indeterminate ones are restrained, so stresses develop.
170. The fixed-end moment of a beam of span 5 m carrying a udl of 24 kN/m and fixed at both ends is
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wL²/12.
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Answer: B. 50 kN·m
FEM = wL²/12 = 24×5²/12 = 50 kN·m.
171. A fixed beam of span 6 m carries a 80 kN point load at 2 m from the left end A. The fixed-end moment at A is
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Larger moment at the end nearer to the load.
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Answer: A. 71.11 kN·m
M_A = Pab²/L² = 80×2×4²/6² = 71.11 kN·m.
172. At a joint B of a continuous beam, member BA (length 4 m) and member BC (length 6 m) have the same EI and both far ends A and C are fixed. The distribution factor for BA is
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DF = k/Σk with k = 4EI/L.
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Answer: D. 0.6
k_BA = 4EI/4 = EI, k_BC = 4EI/6 = 0.667EI; DF_BA = 1/(1+0.667) = 0.6.
173. An unbalanced moment of 100 kN·m is distributed at a joint, and the distribution factor of a member is 0.6 with its far end fixed. The moment carried over to the far end is
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Carry-over is half of the distributed moment.
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Answer: B. 30 kN·m
Distributed moment = 0.6×100 = 60 kN·m (opposite sign); carry-over = ½×60 = 30 kN·m.
174. A propped cantilever of span 6 m (fixed at A, roller at B) carries a udl of 10 kN/m. The fixed-end moment at A is
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wL²/8.
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Answer: B. 45 kN·m
R_B = 3wL/8 = 22.5 kN; M_A = wL²/8 = 10×6²/8 = 45 kN·m (hogging).
175. A propped cantilever of span 6 m, fixed at A and with a roller at B, carries a central point load of 40 kN. The reaction at B is
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Compatibility: deflection at B is zero.
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Answer: A. 12.5 kN
By the flexibility method, R_B = 5P/16 = 5×40/16 = 12.5 kN (and M_A = 3PL/16 = 45 kN·m).
176. A two-span continuous beam with equal spans of 6 m is simply supported at the ends and carries a udl of 10 kN/m on both spans. The middle support reaction is
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Each span carries 5wL/8 on the middle support.
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Answer: A. 75 kN
M_B = wL²/8 by the three-moment equation, R_B = 5wL/4 = 5×10×6/4 = 75 kN.
177. A prismatic member AB of length 4 m, EI = 20000 kN·m², has end A fixed and end B rotating by θB = 0.002 rad without chord rotation. Using slope-deflection, the moment at B is
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M = 4EIθ/L.
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Answer: A. 40 kN·m
M_BA = (2EI/L)(2θB) = (2×20000/4)(2×0.002) = 40 kN·m (= 4EIθB/L; the carry-over to A is 20 kN·m).
178. One support of a beam of span 5 m fixed at both ends settles by 10 mm. For EI = 30000 kN·m², the fixed-end moment induced is
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M = 6EIΔ/L².
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Answer: D. 72 kN·m
M = 6EIΔ/L² = 6×30000×0.01/5² = 72 kN·m.
179. A rectangular steel section 100 mm wide and 200 mm deep has a yield stress of 250 MPa. Its plastic moment capacity is
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Zp = bd²/4 for a rectangle.
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Answer: C. 250 kN·m
Zp = bd²/4 = 1000000 mm³; Mp = σy Zp = 250×1000000 = 250 kN·m.
180. A beam of span 5 m fixed at both ends has a plastic moment capacity Mp = 100 kN·m and carries a central point load. The collapse load is
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Three hinges: two ends and centre.
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Answer: C. 160 kN
Hinges form at both ends and at the centre: W(L/2)θ = Mp(θ+2θ+θ), so Wc = 8Mp/L = 160 kN.
181. A fixed beam of span 6 m with Mp = 90 kN·m carries a udl. The collapse load intensity is
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Hinges at both ends and mid span; wcL²/16 = Mp.
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Answer: B. 40 kN/m
Hinges at both ends and midspan: external work = wL²θ/4, internal work = 4Mpθ, so w_c = 16Mp/L² = 16×90/6² = 40 kN/m.
182. A portal frame with fixed bases, neglecting axial deformations, has a degree of kinematic indeterminacy of
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Rotations plus sway.
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Answer: A. 3
Two joint rotations (at the beam–column joints) plus one independent sway translation: 3.
183. A two-hinged parabolic arch of span 30 m and rise 5 m carries a udl of 15 kN/m over the whole span. The horizontal thrust is
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H = wL²/8h.
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Answer: C. 337.5 kN
For a parabolic arch under full udl, the arch is funicular and H = wL²/(8h) = 15×30²/(8×5) = 337.5 kN.
184. A two-hinged parabolic arch (constant EI) of span 24 m and rise 4 m carries a udl of 20 kN/m on the left half only. The horizontal thrust is
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Half the loading gives half the thrust.
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Answer: A. 180 kN
H = wL²/(16h) = 20×24²/(16×4) = 180 kN, half of the full-span value.