Structures · MCQ practice Bending and Shear Stress MCQs – Part 2
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Part 2 of 5 Question 16 Medium
The ratio of the section modulus of a circular section of diameter D to that of a square section of side B (with B = D) is:
A 2π/15 B 3π/16 C 3π/8 D π/16
Answer: B. 3π/16
Z_circle = πD³/32, Z_square = D³/6. Ratio = (π/32)(6) = 3π/16.
Question 17 Easy
In a structure, cables and wires are generally used:
A to resist shear stress B as compression members C as flexural members D none of these
Answer: D. none of these
Cables and wires are flexible: they can only carry tension, so none of the listed uses is correct.
Question 18 Easy
The maximum shear stress (q_max) in a rectangular beam is:
A 1.25 times the average B 1.50 times the average C 1.75 times the average D 2.0 times the average
Answer: B. 1.50 times the average
τmax = (6V/bd³)(d²/4) = 1.5V/(bd) = 1.5 τavg.
Question 19 Easy
The radius of gyration of a rectangular section (depth D, width B) about a centroidal axis parallel to the width is:
A D/2 B D/√3 C D/(2√3) D D/(4√3)
Answer: C. D/(2√3)
r = √(I/A) = √((BD³/12)/(BD)) = D/√12 = D/(2√3).
Question 20 Medium
For a simply supported circular beam with a central load W, the maximum bending stress is equal to the direct tensile stress produced by the same load. The ratio of length to diameter is:
A 1/8 B 1/4 C 1/2 D 1/3
Answer: C. 1/2
σb = (WL/4)/(πd³/32) = 8WL/(πd³); σt = 4W/(πd²). Equal ⇒ 2L/d = 1 ⇒ L/d = 1/2.
Question 21 Easy
About the principal axes of a section:
A the sum of the moments of inertia is zero B the difference of the moments of inertia is zero C the product of inertia is zero D none of these
Answer: C. the product of inertia is zero
Principal axes are the pair of perpendicular axes about which the product of inertia I_xy = ∫xy dA is zero.
Question 22 Easy
For a beam, the term M/EI is:
A stress B rigidity C curvature D shear force
Answer: C. curvature
From M/I = E/R: 1/R = M/(EI), the curvature.
Question 23 Easy
Of several prismatic beams of equal length and the same material, the one that can carry the maximum load in flexure is the one with the maximum:
A depth of section B area of cross-section C section modulus D moment of inertia
Answer: C. section modulus
Moment of resistance M = σ·Z. Same material (same σ) ⇒ capacity ∝ Z.
Question 24 Easy
The section modulus of a beam section may be obtained by dividing its moment of inertia by:
A the depth of the section B the distance of the extreme fibre from the neutral axis C the maximum tensile stress at the section D the maximum compressive stress at the section
Answer: B. the distance of the extreme fibre from the neutral axis
Z = I / y_max, where y_max is the distance of the extreme fibre from the neutral axis.
Question 25 Easy
In a beam, at a section carrying a shear force F, the shear stress is maximum at the:
A neutral surface B topmost fibre C bottommost fibre D mid-depth
Answer: A. neutral surface
τ = FAȳ/(Ib). The first moment Aȳ is largest at the neutral axis, so τ is maximum there (for usual sections).
Question 26 Easy
The centre of gravity of a thin hollow cone lies on the axis of symmetry at a height of:
A one-half of the total height above the base B one-third of the total height above the base C one-fourth of the total height above the base D none of these
Answer: B. one-third of the total height above the base
Thin hollow cone (curved surface only): h/3 above the base. Solid cone: h/4.
Question 27 Easy
The shape of a cable under a load uniformly distributed along the horizontal span is:
A parabolic B catenary C circular D triangular
Answer: A. parabolic
Load uniform per horizontal length ⇒ moment like a beam with UDL (parabola) ⇒ cable shape y = wx(L − x)/(2H), a parabola.
Question 28 Easy
The moment of inertia of a triangular section about an axis through its centre of gravity (parallel to the base) is:
A bh³/12 B bh³/36 C bh³/32 D none of these
Answer: B. bh³/36
I_G = bh³/36. About the base: bh³/12; about the apex: bh³/4.
Question 29 Medium
For the strongest rectangular beam that can be cut from a circular log, the ratio of width to depth is:
A 0.303 B 0.404 C 0.505 D 0.707
Answer: D. 0.707
Maximise Z = b(d² − b²)/6 ⇒ d² = 3b² ⇒ h = √2·b ⇒ b/h = 1/√2 = 0.707.
Question 30 Easy
The maximum shear stress at the neutral axis due to a shear force F on a rectangular section of area A is:
A F/A B F/2A C 3F/2A D 2F/3A
Answer: C. 3F/2A
τmax = 1.5 τavg = 1.5F/A = 3F/(2A).
Part 2 of 5