Question 1

Bearing stress control in dams: limiting base stress against permissible soil stress  For a masonry/gravity dam, to prevent crushing of masonry at the base, compare the maximum base stress with the soil’s permissible compressive stress. Choose the c…

Accepted Answer

Concept Foundation safety requires that the maximum compressive stress at the dam base must not exceed the allowable bearing capacity (permissible compressive stress) of the supporting soil/rock and masonry. Design check σmax, base ≤ σperm. Final Answer Less than or equal to the permissible stress.

Question 2

Strain measurement devices: purpose of a strain rosette (commonly misspelled as resetter)  A strain rosette comprises multiple gauges at known angles bonded at a point on a surface. Identify what it directly measures. Choose the correct option.

Accepted Answer

Concept A strain rosette holds multiple uniaxial strain gauges (e.g., 0°/45°/90°). Each gauge measures normal strain along its own axis. Use From these measured linear strains, one computes principal strains and in-plane shear strain via transformation equations. The device does not physically relieve strain. Final Answer Linear (normal) strains along the individual gauge directions.

Question 3

Thermal stress in an unconstrained bar: heating a free steel bar from 15°C to 40°C  A steel bar is heated from 15°C to 40°C and is perfectly free to expand (no constraint). State the stress induced in the bar as a result of heating. Choose the corre…

Accepted Answer

Concept/Approach Thermal strain = α ΔT. If the bar is completely unrestrained, it undergoes free expansion, so no internal force develops. Result Thermal stress arises only when expansion/contraction is restrained. With no restraint: σ = 0. Final Answer No stress induced.

Question 4

Strain energy in a solid circular shaft under torsion: expression in terms of τ and C  A solid circular shaft is subjected to torsion; let τ denote maximum shear stress and C the shear modulus (modulus of rigidity). Express the total strain energy s…

Accepted Answer

Approach (energy method) Total strain energy U = ∫ (τ² / (2 C)) dV. For a solid circular shaft in pure torsion, τ varies linearly with radius: τ(r) = τmax(r/R). Step-by-step integration Let R = outer radius, L = length. dV = 2π r L dr.U = ∫0R [ (τmax² r² / R²) / (2 C) ] · (2π r L dr )= (τmax² /(2 C R²)) · 2π L ∫0R r³ dr= (τmax² /(2 C R²)) · 2π L · (R⁴/4)= (τmax² /(4 C)) · (π R² L)= (τ² /(4 C)) × Volume. Final Answer τ² / (4 C) × Volume.

Question 5

Basic definition in mechanics of materials: deformation per unit length  In axial loading, the ratio of change in length to the original length is termed as what? Choose the correct term.

Accepted Answer

Definitions Strain ε = ΔL / L (dimensionless), the deformation per unit length.Stress σ or τ = Force / Area (units of pressure). Final Answer Strain (engineering strain).

Question 6

Column buckling: direction of failure based on principal second moments of area  A vertical column has two principal moments of inertia Ixx and Iyy about orthogonal centroidal axes. Identify the direction in which the column will tend to buckle firs…

Accepted Answer

Concept Euler buckling shows that the critical load for a prismatic column is Pcr = π² E I / (K L)². For a given column (same E, L, end condition K), the smaller principal second moment of area I gives the smaller critical load and hence buckling occurs first about that weaker axis. Why not the others? (a) & (b) mention directions relative to the load line, not stiffness; (c) maximum inertia would raise Pcr, resisting buckling; (d) correctly identifies the weak-axis buckling. Final Answer about the axis with minimum moment of inertia

Question 7

Beam support classification: is “supported at both ends” a simply supported beam?  A beam carried by a pin (hinge) at one end and a roller at the other is termed a simply supported beam. Evaluate the statement: “A beam supported at its both ends is …

Accepted Answer

Definition A simply supported beam has no end fixity: typically a pin at one end and a roller at the other (or two supports that do not restrain rotation). That is indeed “supported at both ends”. Reasoning The statement claims the opposite and is therefore incorrect. Final Answer False

Question 8

Thin cylindrical shell under internal pressure: ratio of longitudinal strain to volumetric strain  For a thin cylinder (diameter d, thickness t) under internal pressure p, with hoop and longitudinal stresses as in thin-shell theory, express (longitu…

Accepted Answer

Given / Thin-shell stresses Hoop stress: σh = p d / (2 t)Longitudinal stress: σl = p d / (4 t) = σh/2 Strains (Hooke’s law with Poisson effect) Longitudinal strain: εl = (1/E)(σl − ν σh)Hoop strain: εh = (1/E)(σh − ν σl) Volumetric strain for thin cylinder ΔV/V ≈ εl + 2 εh Step-by-step Let σl = S ⇒ σh = 2Sεl = (S/E)(1 − 2ν)εh = (S/E)(2 − ν)Volumetric strain: εv = εl + 2εh = (S/E)[(1 − 2ν) + 2(2 − ν)] = (S/E)(5 − 4ν)Ratio εl/εv = (1 − 2ν)/(5 − 4ν) Final Answer (1 − 2ν) / (5 − 4ν)

Question 9

Impact versus gradual loading: compare induced stresses in a linear elastic member  A body (within elastic range) is loaded suddenly versus the same load applied gradually (static). How do the resulting stresses compare? Choose the correct relation.

Accepted Answer

Concept For a suddenly applied load (no initial velocity, no damping), the maximum strain/stress in a linear elastic member is twice the value produced by the same load applied gradually. Derivation sketch Gradual loading: σ = P/A.Sudden loading: strain energy ½ k δ² equals work P δ; solving gives δ = 2(P/k) and σ = 2(P/A). Final Answer twice

Question 10

Basic definition check: compressive stress and strain under axial shortening  When a body is subjected to two equal and opposite pushes and tends to shorten in length, the nature of the stress and strain induced is compressive. State whether this st…

Accepted Answer

Concept Axial loads that push ends together reduce length and create compressive normal stress (σ < 0 by sign convention) and compressive strain. Final Answer True

Question 11

Definition of bending moment at a section of a beam  Complete the definition: The bending moment at a point on a beam is the algebraic ________ of all the moments of the forces about that point (on either side). Choose the correct word.

Accepted Answer

Definition Bending moment at a section equals the algebraic sum of the moments of external forces (including reactions) about that section, considering either side of the cut. Final Answer sum

Question 12

Deflection comparison: point load at midspan vs UDL (same total load) Beam A: simply supported, span L, central point load W. Beam B: simply supported, uniformly distributed load with total load W. Find δ_A/δ_B, the ratio of the…

Accepted Answer

Formulas Point load at midspan: δmax = W L³ / (48 E I)UDL of intensity w = W/L: δmax = 5 w L⁴ / (384 E I) = 5 (W/L) L⁴ /(384 E I) = 5 W L³ /(384 E I) Compute ratio δA/δB = [W L³ /(48 E I)] / [5 W L³ /(384 E I)] = (1/48) ÷ (5/384) = 384/240 = 8/5. Final Answer 8/5

Question 13

Stress distribution in simple bending of beams  Under the assumptions of simple (Euler–Bernoulli) bending, how does normal bending stress vary from the neutral axis to the outer fibers? Choose the correct variation.

Accepted Answer

Concept From σ = M y / I, the bending stress is proportional to distance y from the neutral axis. Hence it varies linearly, zero at the NA and maximum at the extreme fibers. Final Answer linearly

Question 14

Punching a plate: maximum permissible hole diameter from shear versus punch crushing  A plate of thickness t is punched. The plate’s maximum shear stress is one-fourth of the punch’s maximum crushing stress. Find the maximum hole diameter that can b…

Accepted Answer

Given limits Let allowable shear of plate = τ and allowable crushing of punch = σc, with τ = σc/4. For a punched hole of diameter d Shear force capacity (plate): P = τ (π d t)Crushing stress on punch: σcr = P / (π d² / 4) Set crushing at limit σc = [τ (π d t)] / (π d² / 4) = τ (4 t / d)Using τ = σc/4 ⇒ σc = (σc/4)(4 t / d) = σc (t / d)⇒ d = t Final Answer t

Question 15

Deflection in closely coiled helical springs: effect of number of turns Two closely coiled helical springs A and B are identical in all respects except that spring A has half as many turns as spring B. Find the ratio of their deflections δ_A

Accepted Answer

Formula Deflection of a close-coiled helical spring: δ = (8 P D³ n)/(G d⁴). For identical D, d, G and same load P, δ ∝ n. Apply If nA = ½ nB, then δA/δB = nA/nB = 1/2. Final Answer 1/2

Question 16

Power transmission and shaft sizing: compare diameters for equal power at different speeds  Two circular shafts 'A' and 'B' transmit the same power. Shaft 'A' runs at 250 rpm and shaft 'B' runs at 300 rpm. Decide which shaft must have the larger dia…

Accepted Answer

Given data Equal power P is transmitted by shafts 'A' and 'B'.Speeds: NA = 250 rpm, NB = 300 rpm.Same material and allowable shear stress (design criterion). Concept/Approach Power P = T·ω, so torque T = P/ω. Lower speed ⇒ lower ω ⇒ higher T for the same power.For a solid circular shaft at allowable shear stress τ: T = (π/16)·τ·D³ ⇒ D ∝ T1/3. Step-by-step ω ∝ N, thus TA/TB = ωB/ωA = NB/NA = 300/250 = 1.2.Therefore DA/DB = (TA/TB)1/3 ≈ (1.2)1/3 > 1.Hence shaft 'A' (slower) needs the larger diameter. Common pitfall Confusing high speed with larger diameter; actually, for equal power, lower speed requires higher torque and thus a larger diameter. Final Answer False — shaft 'A' must have the greater diameter.

Question 17

Thermal stress due to cooling: nature of stress developed in a bar cooled to −5°C  A straight uniform bar is cooled down to −5°C. Assume the bar's ends are fully restrained against axial movement so that free contraction cannot occur. Identify the n…

Accepted Answer

Assumption The bar is restrained longitudinally (standard design assumption unless “free to expand/contract” is stated). If unrestrained, thermal stress would be zero. Concept Cooling tends to contract the bar by α·ΔT. Restraint prevents contraction, inducing a tensile stress to balance the prevented strain. Step If restrained, σ = E·α·|ΔT| (tension under cooling). Final Answer Tensile stress.

Question 18

Hooke's law within elastic limit: relation between stress and strain  For a material loaded within its elastic (proportional) limit, state how stress varies with strain. Choose the most appropriate completion.

Accepted Answer

Concept Hooke's law: within the proportional limit, σ = E·ε ⇒ stress is directly proportional to strain. Final Answer Proportional to.

Question 19

Bending theory: stress at the neutral axis of a beam cross-section  Define the neutral axis of a beam cross-section under pure bending. Identify the bending stress value at the neutral axis.

Accepted Answer

Concept In pure bending: σ = (M/I)·y, where y is the distance from the neutral axis (NA).At the NA, y = 0 ⇒ σ = 0. Final Answer Zero.

Question 20

Tension test terminology: stress level where extension accelerates with small load increase  From a standard stress–strain curve in tension, identify the stress at which the extension increases more rapidly than the increase in load.

Accepted Answer

Concept At the yield point, a small increase in load produces a disproportionate increase in strain (onset of plastic flow in ductile materials). Notes Elastic limit: end of linear behavior; Ultimate: maximum nominal stress; Breaking: fracture. Final Answer Yield point.

Strength of Materials Questions