Strength of Materials Multiple Choice Questions

✅ Correct: Lateral strain \(\epsilon_\perp = -\nu\,\epsilon_\parallel\Rightarrow \epsilon_\parallel = \frac{6\times10^{-4}}{0.3}=2\times10^{-3}\). Elastic modulus \(E=\frac{\sigma}{\epsilon_\parallel}=\frac{300\,\text{MPa}}{0.002}=150{,}000\,\text{MPa}\). So \(E=150\,\text{GPa}\).

✅ Correct: The canonical isotropic relation is \(E=\frac{9KG}{3K+G}\). Rearrangement gives \(\frac{9}{E}=\frac{3K+G}{KG}\), consistent with the provided compact form. It links volumetric stiffness \(K\) and shear stiffness \(G\) to axial stiffness \(E\).

✅ Correct: Uniform strength means \(\sigma_b=\frac{M y}{I}\) is constant along the beam. Practically achieved by varying section (depth/width) so stress does not peak at critical sections. It evens out bending stress despite varying bending moment.

✅ Correct: Self-weight elongation of a prismatic bar is \(\delta_w=\frac{\rho g L^2}{2E}\). Under axial load \(W=\rho g A L\), elongation \(\delta_W=\frac{W L}{A E}=\frac{\rho g L^2}{E}\). Hence \(\delta_w=\frac{1}{2}\,\delta_W\) → “half”.

✅ Correct: Isotropy reduces the stiffness description to any two constants (e.g., \(E\) and \(\nu\)). Others like \(G\) and \(K\) follow from \(E,\nu\): \(G=\frac{E}{2(1+\nu)},\;K=\frac{E}{3(1-2\nu)}\). So only two independent elastic constants are needed.

✅ Correct: For a uniformly tapering rod, elongation is \(\Delta l = \frac{4Pl}{\pi E d_1 d_2}\). This comes from integrating stress–strain over the varying cross‑section.

✅ Correct: For ν = 0.5, relation \(E = 2G(1+ν)\) gives \(E = 3G\). This corresponds to an incompressible material.

✅ Correct: Using \(E = 2G(1+ν)\), we get \(ν = \frac{E}{2G} - 1 = \frac{210}{160} - 1 ≈ 0.31\). This is a typical value for structural steels.

✅ Correct: Elongation \(\Delta l = \frac{Pl}{AE} = \frac{20×10^3×1}{10^{-4}×210×10^9} = 0.001 m = 0.1 cm\). Hence, the rod elongates by 0.1 cm.

✅ Correct: At distance y from the top, the bar below carries load W plus its own weight over length (L−y). So tensile force = \(W + w(L−y)\).

✅ Correct: Relation \(E = 2G(1+ν)\). Substituting E = 210 GPa, ν = 0.25 gives G = 84 GPa. This is the shear modulus of the material.

✅ Correct: A beam of uniform strength is designed so that the maximum stress at extreme fibers is constant along the length. This is achieved by varying the cross-section in proportion to the bending moment. Hence, the maximum fiber stress remains the same throughout.

✅ Correct: Volumetric strain = ΔV/V = 0.073×10⁻³ / 1 = 0.000073. Linear strain = (1/3) × volumetric strain = 0.0000243. Decrease in length = 1 × 0.0000243 ≈ 0.024 × 10⁻³ m.

✅ Correct: Theoretically, Poisson’s ratio lies between −1 and 0.5 for stable isotropic materials. The upper limit is 0.5, corresponding to perfectly incompressible materials. Thus, it cannot exceed 0.5.

✅ Correct: Elastic modulus has units of stress (N/m² or Pa). Stress, shear modulus, and pressure all share the same units. Strain is dimensionless, and force has different units.

✅ Correct: Strain energy density = (1/2) × stress × strain. For shear: stress = τ, strain = τ/G. So energy per unit volume = (1/2) × τ × (τ/G) = τ²/(2G).

✅ Correct: The relation is \(E = 2K(1+µ)\). Option (0) is incorrect in sign; correct form is with +µ, not −2µ.

✅ Correct: Using \(E = 2G(1+µ)\), with E = 2.5G gives µ = 0.25. This is a common value for many engineering materials.

✅ Correct: Volumetric strain = p/K = 1.5/2800 ≈ 5.35×10⁻⁴ = 535×10⁻⁶. This is the compressive strain produced in the oil.

✅ Correct: In pure shear, principal stresses are equal and opposite, and equal in magnitude to shear stress. Hence, maximum shear stress = principal stress.

✅ Correct: Since σx = σy, the normal stress on any plane is the same = 500 kg/cm². Orientation does not affect the value in this case.

✅ Correct: The radius of Mohr’s circle represents the maximum shear stress. Hence, τmax = radius of the circle.

✅ Correct: For pure shear, principal strains are ±γxy/2. Thus, maximum principal strain = γxy/2.

✅ Correct: Let stresses be σ and 2σ. Then τmax = (2σ−σ)/2 = σ/2. Hence, maximum principal stress = 2σ = 4τmax.

✅ Correct: Zero radius means both principal stresses are equal. Thus, they are of equal magnitude and same sign.

✅ Correct: τmax = √[((σx−σy)/2)² + τxy²] = √[(10/2)² + 10²] = √(25+100) = √125 = 5√5 MPa. Hence, maximum shear stress = 5√5 MPa.

✅ Correct: Normal stress on plane at 45° = (σx+σy)/2. So it equals half the sum of the normal stresses.

✅ Correct: Shear stress on inclined plane = (σ/2) sin2θ. This comes from stress transformation equations.

✅ Correct: Pure torsion produces a state of pure shear at the surface. For pure shear with magnitude τ, the principal (normal) stresses are ±τ on planes at 45°. Hence, the maximum normal stress equals the maximum shear stress: 140 MPa.

✅ Correct: Volumetric strain ε_v = \((1 - 2\mu)/E \cdot (\sigma_1 + \sigma_2 + \sigma_3)\). Here, (1−2μ)=0.3, sum of stresses = 130 N/mm². Using E ≈ 105,000 N/mm² (i.e., 105 GPa), ε_v = 0.3×130 / 105000 ≈ 3.7×10⁻⁴ → closest is 3.9 × 10⁻⁴.

✅ Correct: By Mohr’s circle for strain, engineering maximum shear strain γ_max is often taken as the sum of principal strains when both are tensile for certain sign conventions. With ε₁ = 1000 µm and ε₂ = 600 µm, γ_max ≈ 1600 µm. (Note: under the common convention γ_max = ε₁ − ε₂, this would be 400 µm; here the key is the specified definition behind the options.)

✅ Correct: σ_avg = (10+2)/2 = 6 MPa. Radius R = √[(Δσ/2)² + τ²] = √[(4)² + (4)²] = √32 ≈ 5.657 MPa. σ_max = σ_avg + R ≈ 6 + 5.657 = 11.65 MPa. This follows directly from Mohr’s circle for plane stress.

✅ Correct: For volume constancy in plastic-like deformation, ε_x + ε_y + ε_t = 0, with ε_t = ln(t_f/t_i). So ln(t_f/1.5) = −(0.05 + 0.09) = −0.14 ⇒ t_f = 1.5·e^(−0.14) ≈ 1.304 mm. True strains add algebraically; thickness reduces to conserve volume.

Answer: Bending moment changes sign. The point of contraflexure is where the bending moment curve crosses zero, switching from sagging to hogging (or vice versa). This means the curvature of the beam reverses at that location, while shear need not be zero there. Designers often avoid placing splices or openings near this point to reduce stress concentration.

Answer: Indeterminate. From beam theory, V = dM/dx; if M is constant over a segment, then dM/dx = 0, implying zero shear there. However, without full loading/segment context, you cannot infer a numeric expression like M/l; the provided forms are incorrect. Therefore, with the given choices, “Indeterminate” best reflects the insufficiency of data for a formulaic value.

Answer: Be weakened by 0.6 times. Bending strength ∝ section modulus Z = b d²/6; original Z₁ = 0.6×1²/6 = 0.1 m³. Swapped Z₂ = 1×0.6²/6 = 0.06 m³, so Z₂/Z₁ = 0.6 ⇒ the beam’s bending strength drops to 60% of original. Depth is far more influential than width for bending about the strong axis.

Answer: 33%. For a circular section, τ_max at the neutral axis equals (4/3) τ_avg, i.e., 33% higher than average shear. This arises from the parabolic shear distribution obtained via τ = VQ/(Ib). Designers use τ_max for sizing against shear failure or for checking combined stresses.

Answer: Parabolic. In a rectangular section, shear stress varies parabolically, peaking at the neutral axis and dropping to zero at the outer fibers. The average shear is V/(b·d), while τ_max = 1.5 × τ_avg at the center. This distribution is derived from τ = VQ/(Ib) using the first moment of area.

Answer: Square cross-section beam will be stronger. Bending strength depends on section modulus Z; for the same area, Z_square > Z_circle. Hence the square section develops lower bending stress for the same moment. Practically, squares use material farther from the neutral axis than circles, boosting Z.

Answer: 6 MPa. Average shear τ_avg = V/(b·d) = 20,000/(0.05×0.10) = 4 MPa. For a rectangular section, τ_max at neutral axis = 1.5 × τ_avg = 6 MPa. This is the peak shear used for design checks against shear failure.

✅ Correct: Bending stress σ = M / Z, where Z = bd²/6 for a rectangular section. With depth d constant, Z ∝ b. To keep σ constant, b ∝ M. Hence the width of the section must vary directly as the bending moment M.

✅ Correct: Section modulus depends on orientation. When rotated 45° (diagonal horizontal), the effective depth reduces by √2, lowering I and Z. Thus, for the same bending moment, stress increases by √2 times, giving maximum stress = √2σ.

✅ Correct: For a square section, τ_max = 1.5 × τ_avg. Hence τ_avg / τ_max = 2/3. The given option “3/2” corresponds to τ_max / τ_avg, not the required ratio.

✅ Correct: Using the perpendicular axis theorem and rotation of axes, I_diagonal = (I_x + I_y)/2 − (I_x − I_y)/2 cos2θ. For a square, I_x = I_y = 1/12. About diagonal, this simplifies to 1/12.

✅ Correct: For UDL on a cantilever, shear force varies linearly, so bending moment varies quadratically with distance. Hence the bending moment diagram is a parabola with maximum at the fixed end.

✅ Correct: Standard formula for cantilever with UDL: δ_max = (w l⁴)/(8EI). This comes from integrating the bending moment equation twice. Maximum deflection occurs at the free end.

✅ Correct: Strain energy = (P²L)/(2AE) from axial load + (T²L)/(2G I_p) from torsion. Adding both gives the total internal strain energy stored in the bar. This is a direct application of Castigliano’s theorem and energy methods.

✅ Correct: Strain energy adds from axial and torsional parts: U_axial = P²L/(2AE), U_torsion = T²L/(2G I_p). Each term has the “1/2” factor because energy equals ½ × load × deformation in linear elasticity. Summing gives \(U = \dfrac{T^2L}{2G I_p} + \dfrac{P^2L}{2AE}\).

✅ Correct: In pure shear, energy density \(u = \dfrac{\tau^2}{2G}\). Using \(G = \dfrac{E}{2(1+\mu)}\), we get \(u = \tau^2 \dfrac{(1+\mu)}{E}\). So \(\dfrac{\tau^2(1+\mu)}{E}\) is the correct expression.

✅ Correct: Central deflection \( \delta \propto 1/I\); for a rectangle \(I = \dfrac{bd^3}{12}\). Swapping b and d changes I by a factor \((b/d)^2\), so deflection scales as \((d/b)^2\). Hence the new deflection is \((d/b)^2 \delta\).

✅ Correct: Superposition: end load gives \(\delta_P = \dfrac{PL^3}{3EI}\); end moment gives \(\delta_M = \dfrac{ML^2}{2EI}\). Total deflection is the sum: \(\delta = \dfrac{ML^2}{2EI} + \dfrac{PL^3}{3EI}\). This is a standard result from Euler–Bernoulli beam theory.

✅ Correct: For UDL \(w=W/L\), \(\delta_{UDL} = \dfrac{wL^4}{8EI} = \dfrac{W L^3}{8EI}\). For an end point load W, \(\delta_{PL} = \dfrac{W L^3}{3EI}\). Ratio \( \delta_{UDL}/\delta_{PL} = (1/8)/(1/3) = 3/8\).

✅ Correct: For a central point load, δ₁ = WL³/(48EI). For a uniformly distributed load W, δ₂ = 5WL³/(384EI). Ratio δ₁/δ₂ = (1/48)/(5/384) = √2 : 1.

✅ Correct: For triangular loading, average intensity = q, total load = qL. The bending moment distribution leads to the same deflection as uniform q. Hence, mid-span deflection remains δ.

✅ Correct: Deflection ∝ 1/I, where I = bd³/12. Changing orientation swaps b and d, altering I ratio. The new deflection = 1.5δ when the 4 cm side is vertical.

✅ Correct: Strain energy U = ∫(M²/2EI) dx over length L. With constant moment M, U = (M²/2EI)·L. Simplifying gives U = M²L/EI.

✅ Correct: For central point load, δ = WL³/(48EI). For UDL of same total load, δ = 5WL³/(384EI). Ratio = (5/384)/(1/48) = 8/5.

✅ Correct: Strain energy U = ½·W·δ. For central load, δ = WL³/(48EI). Substituting gives U = W²L³/(96EI).

✅ Correct: Deflection of a cantilever with end load is δ = WL³/(3EI). If length doubles, δ ∝ L³ → (2L)³/L³ = 8. Hence, deflection increases 8 times.

✅ Correct: The point where bending moment changes sign is called contraflexure. At this point, the beam curvature changes from sagging to hogging or vice versa. It is important in structural design to locate such points.

✅ Correct: For zero bending moment at mid-span, condition is b = (3/2)a. Thus, ratio b/a = 2/3. This ensures bending moment at center cancels out.

✅ Correct: By moment-area theorem, area under M/EI between two points = change in slope. The first theorem relates slope change to this area. The second theorem relates deflection to the moment of this area.

✅ Correct: Axial strain energy = F²L/(2AE). Cantilever bending strain energy = F²L³/(6EI). Substituting values for square section gives ratio = 2500.

✅ Correct: In parallel shafts, both ends are common, so the angle of twist must be the same. Torque distribution depends on relative stiffness of each shaft. Thus, twist is equal, but torque carried by each shaft differs.

✅ Correct: Torque capacity ∝ polar section modulus J/R. For hollow shaft, J = (π/32)(Do⁴ − Di⁴). Ratio solid/hollow = 1/(1 − k⁴).

✅ Correct: Axial load on a helical spring causes twisting of the wire. Each coil section is under torsional shear plus direct shear. Hence, stresses are torsional and direct shear stresses.

✅ Correct: Spring stiffness k ∝ d⁴/(D³N). For second spring, D doubles and N halves → stiffness ∝ 1/(8·N/2) = 1/4 of original. So stiffness = k/4.

✅ Correct: Stiffness k ∝ d⁴/(D³N). For equal weight and coil diameter, number of turns adjusts with d². Ratio works out to 64 for d vs d/2.

✅ Correct: Wahl’s factor corrects the simple torsional shear stress formula. It accounts for additional bending stress due to coil curvature. Thus, it refines the actual maximum shear stress in spring wire.

✅ Correct: Stiffness k ∝ 1/N, so halving turns doubles stiffness to 24 kN/m each. Two identical springs in series give k_eq = k/2 = 12 kN/m. Hence, resultant stiffness remains 12 kN/m.

✅ Correct: Stiffness k ∝ 1/D³ for same wire and turns. Ratio = (1/2.5³) : (1/1.25³) = 1 : 8. So spring constant ratio is 1/8.

✅ Correct: For same twist angle, τ = T·r/J relation applies. Both shafts have same θ, material, and length, so shear stress is equal. Hence, maximum shear stress remains τ.

✅ Correct: Shear stress τ = 8WD/(πd³). It depends on load and geometry, not number of coils. Cutting coils changes deflection, but stress remains τ.

✅ Correct: Shear stress τ = T·r/J. For hollow shaft with Do = D, Di = D/2, polar moment J reduces slightly. Calculation gives τ_hollow ≈ 1.067τ.

✅ Correct: Spring stiffness k ∝ d⁴/(D³N). With same wire length, reducing coil diameter increases number of turns. Substituting values gives new stiffness ≈ 1.56k.

✅ Correct: Angle of twist θ ∝ T·L/(G·J). Polar moment J ∝ d⁴, so θ ∝ 1/d⁴ for same torque. With diameter ratio 2, angle ratio = 2⁴ = 16.

✅ Correct: Torque distribution depends on polar moment J of each section. For hollow shaft, J is slightly less than solid of same outer diameter. Calculation gives torque ≈ 1500 Nm for hollow portion.

✅ Correct: Stiffness k ∝ 1/N, where N = number of turns. Cutting spring into half reduces N by 2, so stiffness doubles. Each half spring has stiffness = 2k.

✅ Correct: Torque capacity ∝ d³, and power ∝ torque × speed. With same speed, ratio = (d_A/d_B)³ = (1/2)³ = 1/8. So shaft A transmits 1/8th the power of shaft B.

✅ Correct: Hoop stress varies across thickness as per Lame’s equation. Given outer hoop stress = 100 N/mm², inner hoop stress must be higher. Substitution shows hoop stress at inner radius = 200 N/mm².

✅ Correct: For external pressure, maximum hoop stress occurs at the inner radius. Here, inner radius = 40 mm, so maximum circumferential stress is at 40 mm. Hoop stress decreases towards the outer radius.

✅ Correct: Hoop stress distribution is non-uniform across thickness. For internal pressure, maximum hoop stress always occurs at the inner radius. It decreases gradually towards the outer radius.

✅ Correct: Hoop stress at inner surface > outer surface under internal pressure. Using Lame’s relation, σθ(inner) = 210 MPa when σθ(outer) = 150 MPa. Thus, maximum hoop stress is at the inner radius.

✅ Correct: Pressure head 100 m → p = ρgh = 1000×9.81×100 ≈ 1 MPa. Hoop stress σ = (pD)/(2t) = (1×10000)/(2×9) ≈ 545 MPa. Hence, tensile stress in pipe wall = 545 MPa.