Question 1

Pure torsion of circular shaft: Under a twisting moment, every cross-section of the shaft is primarily subjected to which type of stress?

Accepted Answer

Introduction / Context:Shafts transmit torque in machines (motors, gearboxes, rotors). The dominant stress under pure torsion for circular shafts is shear, which defines the torsional strength and angle of twist.Given Data / Assumptions:Solid or hollow circular shaft.Pure torque (no bending, no axial load).Linear elastic, isotropic material; Saint-Venant torsion holds.Concept / Approach:In circular shafts, warping is negligible and cross-sections remain plane. The shear stress varies linearly with radius: tau(r) = T * r / J where T is torque and J is polar second moment of area. Normal stresses (tension/compression) arise only if bending or axial loads are present.Step-by-Step Solution:Identify loading: twisting moment T only.Use torsion formula: tau_max at outer radius r_o is tau_max = T * r_o / J.Angle of twist: theta = T * L / (G * J), confirming shear-dominated behaviour (G is shear modulus).Verification / Alternative check:Mohr’s circle for pure shear shows principal normal stresses ±tau at 45°, but the applied stress state on the cross-section is shear.Why Other Options Are Wrong:Tension/Compression/Bending are not produced by pure torque in a circular shaft.Hydrostatic stress requires equal normal stresses in all directions, unrelated to torsion.Common Pitfalls:Confusing principal normal stresses at 45° with applied shear on the cross-section; applying noncircular-shaft concepts where warping matters.Final Answer:Shear stress

Question 2

Combined stress state: A body has direct tensile stress σ = 300 MPa (one plane) and simple shear τ = 200 MPa. What is the maximum in-plane shear stress?

Accepted Answer

Introduction / Context:When both normal and shear stresses act, the maximum shear stress must be computed using stress transformation or Mohr’s circle. This governs yielding under Tresca criterion and influences fatigue in multiaxial states.Given Data / Assumptions:Plane stress with σx = 300 MPa, σy = 0 MPa, τxy = 200 MPa.Linear elasticity; sign conventions standard.We seek the maximum in-plane shear stress.Concept / Approach:Maximum in-plane shear is the radius of Mohr’s circle: tau_max = √[ ((σx − σy)/2)^2 + τxy^2 ]. This reduces to the pure-shear value σ/2 only when τxy = 0.Step-by-Step Solution:Compute half-difference: (σx − σy)/2 = (300 − 0)/2 = 150 MPa.Apply formula: tau_max = √(150^2 + 200^2) MPa.Evaluate: 150^2 = 22500; 200^2 = 40000; sum = 62500.tau_max = √62500 = 250 MPa.Verification / Alternative check:Construct Mohr’s circle with points (300, 200) and (0, −200); the circle radius equals 250 MPa, confirming the calculation.Why Other Options Are Wrong:−100 MPa: shear magnitude cannot be negative; sign denotes sense only.300 or 400 MPa: overestimates; not aligned with the geometric radius.200 MPa: equals applied shear but ignores accompanying normal stress.Common Pitfalls:Using σ/2 blindly; neglecting the influence of τxy; arithmetic slips when squaring and adding.Final Answer:250 MPa

Question 3

Simply supported beam of span l under uniformly distributed load w (per unit length): The location of maximum deflection and its value—does the maximum occur at midspan?

Accepted Answer

Introduction / Context:Serviceability design limits deflection to protect finishes and function. For common loading cases, knowing where the maximum deflection occurs avoids unnecessary full integrations.Given Data / Assumptions:Straight, prismatic beam; simply supported ends.Uniformly distributed load w along the whole span.Elastic analysis; constant E and I.Concept / Approach:The classical solution for a simply supported beam with full-span UDL gives maximum deflection at the midspan (x = l/2) with value:delta_max = 5 w l^4 / (384 E I). Symmetry of loading and supports ensures the extremum is at the center.Step-by-Step Solution:Write bending moment: M(x) = (w/2)(l x − x^2).Curvature: d^2y/dx^2 = M(x)/(E I).Integrate twice, apply boundary conditions y(0) = 0, y(l) = 0.Evaluate deflection at x = l/2 → delta_max = 5 w l^4 / (384 E I).Verification / Alternative check:Use standard beam tables or Castigliano’s theorem; both return the same midspan maximum and value.Why Other Options Are Wrong:“False” contradicts the symmetric solution.Linear varying w changes the position; but here w is uniform.End fixity is a different support condition; not applicable.Cantilevers have different deflection profiles; the maximum is at the free end, not midspan.Common Pitfalls:Mixing formulas for point loads and UDLs; forgetting the 5/384 factor; confusing simply supported with fixed-ended results.Final Answer:True

Question 4

Cantilever beam with linearly varying load  A cantilever beam of length l carries a triangular distributed load that varies from zero at the free end to w per unit length at the fixed end. Identify the correct nature of the bending moment diagram (B…

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Introduction / Context:This question checks understanding of how distributed load shapes translate into shear force and bending moment diagrams in strength of materials. Recognizing whether the bending moment diagram (BMD) is linear, parabolic, or cubic is critical for sizing beams and predicting maximum stresses and deflections.Given Data / Assumptions:Cantilever beam of length l, fixed at one end and free at the other.Transverse load varies linearly along the span: zero at the free end, w per unit length at the fixed end.Small-deflection beam theory; sign conventions standard.Concept / Approach:For any beam: load intensity w(x) integrates to shear V(x), and shear integrates to bending moment M(x). If w(x) is uniform (constant), V is linear and M is parabolic. If w(x) is linear (triangular), then V is quadratic in x and M is cubic in x.Step-by-Step Solution:Let w(x) = k x (with x measured from the free end toward the fixed end).Shear force: V(x) = ∫ w(x) dx = ∫ k x dx = (k/2) x^2 + C → quadratic.Bending moment: M(x) = ∫ V(x) dx = ∫ (k/2) x^2 dx = (k/6) x^3 + C′ → cubic.Therefore the BMD is a cubic curve (not a parabola).Verification / Alternative check:Dimensional progression rule: increase the polynomial order by one upon integration. Triangular load (order 1) → shear (order 2) → moment (order 3). This quick rule confirms the cubic nature.Why Other Options Are Wrong:Agree: incorrect because it calls the BMD parabolic. The BMD is cubic. The options about dependence on elastic modulus or shear shape are distractors; the curve family depends directly on the load distribution and basic equilibrium relations.Common Pitfalls:Confusing the UDL case (parabolic M) with triangular loading (cubic M); mixing the coordinate origin; forgetting that shear is the integral of load and moment is the integral of shear.Final Answer:Disagree

Question 5

Hollow circular shaft – torque capacity expression  For a hollow circular shaft of outer diameter D and inner diameter d subjected to torsion, which expression gives the transmitted torque T in terms of maximum shear stress τ_max?

Accepted Answer

Introduction / Context:Design of shafts under torsion requires the correct relationship among torque, section geometry, and shear stress. For hollow shafts, capacity depends on the polar section modulus rather than simple area.Given Data / Assumptions:Hollow circular shaft with outer diameter D and inner diameter d.Maximum shear stress in the material is τ_max.Elastic torsion theory applies (linear elastic, Saint-Venant assumptions).Concept / Approach:The torsion formula is T / J = τ / R = G θ / L. The torque–stress relation uses the polar second moment J and radius to the outer fiber R = D/2. The polar section modulus Z_p = J / R directly gives T = τ_max * Z_p.Step-by-Step Solution:Polar second moment: J = (π/32) * (D^4 - d^4).Outer radius: R = D / 2.Polar section modulus: Z_p = J / R = [(π/32) * (D^4 - d^4)] / (D/2) = (π/16) * (D^4 - d^4) / D.Torque capacity at τ_max: T = τ_max * Z_p = (π/16) * τ_max * (D^4 - d^4) / D.Verification / Alternative check:For a solid shaft (d = 0), the formula reduces to T = (π/16) * τ_max * D^3, which is the standard solid-shaft result (since J = π D^4 / 32 and Z_p = J/R = π D^3 / 16).Why Other Options Are Wrong:(a) omits division by R and thus gives J scaled by τ rather than T; (c) uses area terms D^2 - d^2 which are not appropriate for torsion; (d) has an incorrect constant; (e) is dimensionally inconsistent.Common Pitfalls:Confusing polar second moment with area; forgetting to divide by the outer radius; mixing diameter and radius.Final Answer:T = (π/16) * τ_max * (D^4 - d^4) / D

Question 6

Effect of section modulus on bending stress  If the section modulus Z of a beam is increased while bending moment M remains the same, how will the maximum bending stress change?

Accepted Answer

Introduction / Context:Beam design frequently leverages the relationship σ = M / Z. Increasing section modulus by changing cross-section shape or size is a primary strategy to cut bending stress for a given moment.Given Data / Assumptions:Bending moment M is fixed by loading and span.Section modulus Z is altered via geometry (e.g., larger depth, flanges).Linear elastic bending; plane sections remain plane.Concept / Approach:Maximum bending stress in elastic range is σ_max = M / Z. This is an inverse proportionality: higher Z means lower σ for the same M.Step-by-Step Solution:Start with σ = M / Z.Increase Z → denominator increases.Therefore σ decreases proportionally.No sign change occurs; stress nature (tension/compression) is set by loading, not Z.Verification / Alternative check:Example: if Z doubles, σ halves. This is why I-sections (large Z for given area) are efficient for bending resistance.Why Other Options Are Wrong:Not change: contradicts σ = M / Z. Increase: opposite of the formula. Become zero or reverse sign: impossible without changing M or load sense.Common Pitfalls:Confusing section modulus Z with second moment I; Z = I / c, so increasing depth c while adjusting I is key to stress reduction.Final Answer:decrease

Question 7

Helical spring deflection ratio – effect of mean coil diameter  Two closely coiled helical springs A and B are made of the same material, have the same number of active turns, and the same wire diameter. Under the same axial load W, the mean coil di…

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Introduction / Context:Spring sizing depends strongly on geometry. For closely coiled helical springs, deflection varies with the cube of the mean coil diameter, making diameter a dominant factor in compliance.Given Data / Assumptions:Springs A and B: same material (same shear modulus G), same number of active turns n, and same wire diameter (common symbol often d_w).Both carry the same axial load W.Mean coil diameters: D_A = 2 D_B.Concept / Approach:The deflection of a closely coiled helical spring under axial load is δ = 8 W D^3 n / (G d_w^4). With all factors except D identical, δ ∝ D^3.Step-by-Step Solution:δ_A / δ_B = (D_A^3) / (D_B^3).Given D_A = 2 D_B → δ_A / δ_B = (2^3) = 8.Therefore δ_B / δ_A = 1 / 8.Verification / Alternative check:Dimensional check: the formula shows cubic dependence on D, consistent with torsional deformation of coils.Why Other Options Are Wrong:1/4, 2, 4, 8 result from linear or squared assumptions about diameter influence; the correct relationship is cubic.Common Pitfalls:Confusing wire diameter with mean coil diameter; using outer instead of mean diameter; ignoring inactive end coils (not affecting the ratio here since n is the same).Final Answer:1/8

Question 8

Axial deformation and linear strain of a rectangular bar  A rectangular bar of length l, breadth b, and thickness t is subjected to an axial tensile pull P. What is the linear (engineering) strain ε along the length in terms of P, b, t, and modulus …

Accepted Answer

Introduction / Context:Linear strain connects applied load to material stiffness and cross-section. This relationship underpins elongation predictions for bars, tie rods, and machine members subjected to direct tension.Given Data / Assumptions:Prismatic bar with constant rectangular cross-section A = b * t.Axial tensile load P; homogeneous, isotropic material.Linear elastic behavior (Hooke's law) and small strains.Concept / Approach:Normal stress σ = P / A. Engineering strain ε = σ / E in the linear elastic range. Substituting A gives ε = (P / (b * t)) / E = P / (b * t * E).Step-by-Step Solution:Area: A = b * t.Stress: σ = P / A = P / (b * t).Hooke’s law: ε = σ / E = [P / (b * t)] / E.Therefore: ε = P / (b * t * E).Verification / Alternative check:Elongation Δl = (P * l) / (A * E). Dividing by original length l yields ε = Δl / l = P / (A * E) = P / (b * t * E), matching the result.Why Other Options Are Wrong:(b) inverts the relationship; (c) mixes incompatible terms; (d) ignores area and is dimensionally incorrect; (e) gives displacement-like units rather than dimensionless strain.Common Pitfalls:Forgetting to use cross-sectional area; confusing strain with elongation; mixing true and engineering strain definitions.Final Answer:ε = P / (b * t * E)

Question 9

Thermal stress in a restrained bar – dependence on area  In a prismatic bar fully restrained against expansion and subjected to a uniform temperature rise ΔT, the induced thermal stress will

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Introduction / Context:Thermal stresses arise when free thermal expansion or contraction is prevented. Designers must know whether bar size alters the induced stress under full restraint to select materials and allowances properly.Given Data / Assumptions:Uniform temperature change ΔT.Bar is perfectly restrained axially (no net change in length).Linear elastic material with Young’s modulus E and thermal expansion coefficient α.Concept / Approach:Free thermal strain would be ε_free = α ΔT. Full restraint enforces ε_total = 0, so mechanical strain must cancel thermal strain. Hence ε_mech = −α ΔT, and the induced stress is σ = E * ε_mech = −E α ΔT, independent of area.Step-by-Step Solution:Free strain: ε_free = α ΔT.Restraint condition: ε_total = ε_mech + ε_free = 0.Therefore ε_mech = −α ΔT.Stress from Hooke’s law: σ = E * ε_mech = −E α ΔT (sign indicates compression for heating).Verification / Alternative check:Although the reaction force depends on area (F = σ * A), the stress level σ does not. Two bars of different areas but same material and ΔT develop the same thermal stress under identical perfect restraint.Why Other Options Are Wrong:(a) confuses force with stress; (c) moment of inertia matters for bending, not pure axial thermal stress; (d) zero stress occurs only when expansion is unrestrained; (e) length influences free expansion, but not the stress under perfect restraint.Common Pitfalls:Mixing up stress and force; assuming partial restraint behaves the same as full restraint; neglecting temperature gradients which can induce bending.Final Answer:does not depend on the cross-sectional area of the bar

Question 10

Definition of neutral axis in beam bending  The line formed by the intersection of the neutral layer with any normal cross-section of a beam is called the neutral axis. State whether this statement is correct.

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Introduction / Context:Understanding the neutral axis helps visualize stress distributions in bending. At the neutral axis, fiber strain and bending stress are zero, transitioning from compression to tension across the section.Given Data / Assumptions:Beam under pure or combined bending (elastic range).Plane sections remain plane; material is homogeneous and isotropic.Concept / Approach:The neutral layer (or surface) is the locus of points within the beam that experience zero longitudinal strain during bending. Intersecting this surface with a cross-section produces a line: the neutral axis.Step-by-Step Solution:Strain distribution in bending is linear: ε = (y / R).At y = 0 (neutral layer), ε = 0 and hence σ = E ε = 0.Intersection with a cross-section gives the neutral axis, lying typically through the centroid for homogeneous beams under pure bending.Verification / Alternative check:For symmetric homogeneous sections, the neutral axis passes through the centroidal axis. For non-symmetric sections, it still exists but its position relative to geometric axes depends on stiffness distribution.Why Other Options Are Wrong:Limiting the definition to cantilevers, prismatic, or symmetrical sections is unnecessary; the concept applies broadly in elastic bending.Common Pitfalls:Confusing neutral axis with zero shear line; assuming neutral axis always aligns with a geometric symmetry axis in composite or nonhomogeneous beams.Final Answer:True

Question 11

Principal normal stress under uniaxial tension  A body is subjected only to a direct tensile stress σ (no other stresses). The maximum normal stress in the body equals:

Accepted Answer

Introduction / Context:Principal stresses represent extreme normal stresses at a point. For basic stress states, being able to identify principal values without constructing Mohr’s circle is a key skill in mechanics of materials.Given Data / Assumptions:Only one nonzero stress component: σ_x = σ (tension); σ_y = 0; τ_xy = 0.Continuum is homogeneous; small deformations.Concept / Approach:Principal stresses for a 2D state satisfy τ_n = 0 on principal planes. With σ_x = σ, σ_y = 0, τ_xy = 0, the principal stresses are simply σ_1 = σ and σ_2 = 0. Therefore, the maximum normal stress equals σ.Step-by-Step Solution:General principal stress formula: σ_{1,2} = (σ_x + σ_y)/2 ± sqrt[((σ_x − σ_y)/2)^2 + τ_xy^2].Substitute σ_x = σ, σ_y = 0, τ_xy = 0 → σ_{1,2} = σ/2 ± sqrt[(σ/2)^2] = σ and 0.Hence maximum normal stress = σ.Verification / Alternative check:Mohr’s circle reduces to a circle with center at σ/2 on the σ-axis and radius σ/2. The rightmost point corresponds to σ.Why Other Options Are Wrong:σ/2 confuses center value with principal maximum; 2σ is impossible without other stresses; zero is only the minor principal stress; dependence on Poisson’s ratio is irrelevant to principal stress magnitude here.Common Pitfalls:Mixing true and engineering stress; assuming lateral contraction affects principal stress magnitude; forgetting that shear is zero on principal planes.Final Answer:the direct tensile stress σ

Question 12

Unriveted (solid) plate strength per pitch length in riveted joint design  For a plate of thickness t, pitch p, and allowable tensile stress σ_t, what is the strength of the unriveted (solid) plate per pitch length?

Accepted Answer

Introduction / Context:In riveted joint design, capacities per pitch length are compared for several failure modes. The strength of the unriveted (solid) plate is the baseline tensile capacity if no rivet hole weakened the section.Given Data / Assumptions:Plate thickness t and pitch p between rivets.Allowable tensile stress in the plate σ_t.Linear elastic, uniform stress distribution assumed for nominal capacity comparison.Concept / Approach:The unriveted (solid) plate per pitch carries tension across the full pitch area A_solid = p * t. Its tensile capacity equals allowable stress times this area.Step-by-Step Solution:Solid area per pitch: A_solid = p * t.Allowable tensile capacity: F = σ_t * A_solid.Therefore, solid plate strength per pitch = p * t * σ_t.Verification / Alternative check:Compare with net section capacity (p − d) * t * σ_t used for plate tearing around the rivet; the solid capacity is larger because no hole reduces the area.Why Other Options Are Wrong:(a) and (d) relate to bearing or shear modes; (c) gives net section (tearing) strength, not the unriveted plate; (e) uses circular area, irrelevant for plate tension over pitch width.Common Pitfalls:Confusing solid, net, and bearing capacities; mixing symbols σ_t (tension), σ_c (bearing/crushing), and τ (shear).Final Answer:p * t * σ_t

Question 13

Columns under compressive loading: What do we call a column that primarily fails by direct compressive stress (crushing) rather than by elastic buckling?

Accepted Answer

Introduction / Context:Columns may fail either by direct compressive crushing or by elastic instability (buckling). Correctly identifying the governing failure mode is essential for sizing members and choosing the right formula (crushing stress versus Euler or Rankine theory).Given Data / Assumptions:Prismatic column under axial compression.Low slenderness ratio (small effective length to radius of gyration).Material behaves elastically up to crushing strength.Concept / Approach:Failure mode depends primarily on slenderness ratio. Short, stocky members (small slenderness) develop high compressive stresses before they can deflect laterally; thus they fail by direct compression (crushing). Long, slender members fail by buckling at stresses well below crushing strength. Intermediate columns require combined formulas (e.g., Rankine–Gordon).Step-by-Step Solution:Classify by slenderness ratio: lambda = Le / r.If lambda is small (stocky), buckling load is very high and crushing governs.Therefore, a column failing by direct stress is termed a short column.Verification / Alternative check:Compare Euler load P_cr ∝ 1 / (Le^2) with material crushing load P_c = A * sigma_c. For small Le, P_cr ≫ P_c, so crushing occurs first, confirming the classification as a short column.Why Other Options Are Wrong:Long column: dominated by buckling, not direct crushing.Medium column: mixed behavior; not necessarily direct-stress failure.Weak column: vague term; weakness is not the defining criterion.Common Pitfalls:Using Euler's formula for stocky members; ignoring end conditions that change effective length; forgetting to check material crushing stress first for low slenderness ratios.Final Answer:short column

Question 14

Thick cylinder under internal pressure p: What is the maximum radial stress at the inner surface of a thick cylindrical shell (sign convention: tensile positive)?

Accepted Answer

Introduction / Context:Lame's theory for thick-walled cylinders provides radial and hoop stresses that vary across the thickness. Knowing boundary values, especially at the bore where stresses are critical, is fundamental for pressure vessel design and safety assessment.Given Data / Assumptions:Thick-walled, cylindrical geometry; internal pressure p only.Elastic, isotropic material; plane stress in the cylinder wall.Sign convention: tensile positive, compressive negative.Concept / Approach:Lame's solution gives radial stress sigma_r and hoop stress sigma_theta as functions of radius. Boundary condition at the inner radius r_i is sigma_r(r_i) = -p (compressive), and at the outer radius r_o (if no external pressure) sigma_r(r_o) = 0. Thus the maximum magnitude of radial stress occurs at the inner surface and equals the internal pressure but compressive in sign.Step-by-Step Solution:Apply boundary: sigma_r(r_i) = -p.Across thickness, sigma_r increases from -p at r_i to 0 at r_o.Therefore, maximum radial stress (by magnitude) at the inner surface is -p (compressive).Verification / Alternative check:Physical reasoning: internal pressure pushes the wall inward; on the inner face the fluid contact produces compressive normal stress equal to p on the solid surface, matching sigma_r = -p.Why Other Options Are Wrong:Zero or +p contradict boundary conditions.2p or p/2 are not consistent with Lame's boundary values for radial stress.Common Pitfalls:Confusing radial stress with hoop stress (which is tensile and typically larger in magnitude); mixing sign conventions; assuming thin cylinder relations where radial stress is approximated as negligible across thickness.Final Answer:-p (compressive)

Question 15

Weld design (double fillet, lap/tee joint): Using the conventional leg size s and weld length l, the tensile strength (capacity) of a double-fillet welded joint is approximated as:

Accepted Answer

Introduction / Context:Design of fillet welds often uses simplified strength expressions in terms of leg size s and weld length l for quick sizing. For symmetric double-fillet joints (e.g., lap or tee with fillets on both sides), the effective area is roughly doubled compared to a single fillet, leading to a doubled tensile capacity under direct load sharing.Given Data / Assumptions:Double fillet of leg size s on both sides, length l.Allowable tensile stress for weld metal = σt.Conventional simplified design (leg-size basis), ignoring detailed throat coefficient in this question's formulation.Concept / Approach:In basic textbook treatments aligned with these options, the tensile capacity is taken proportional to weld effective area. For a single fillet, capacity is approximately s * l * σt; for a double fillet, area doubles, so capacity ≈ 2 * s * l * σt. (More rigorous practice uses the throat thickness t_throat = 0.707 s, giving 2 * 0.707 s * l * σ_allow, but such detail is not reflected in the given options.)Step-by-Step Solution:Single fillet capacity (simplified) ≈ s * l * σt.Double fillet → two such areas in parallel → 2 * s * l * σt.Therefore, the tensile strength of the double-fillet joint is 2 s * l * σt.Verification / Alternative check:Throat-based check: using t_throat = 0.707 s gives capacity ≈ 2 * 0.707 s * l * σ_allow = 1.414 s l σ_allow, close to but slightly less than 2 s l σt when σ_allow definitions differ. The problem's answer set aligns with the simplified leg-size expression.Why Other Options Are Wrong:0.5 s l σt and s l σt correspond to smaller capacities (single fillet or half value).0.707 s l σt is a single-fillet throat approximation, not double fillet.2.0 s l σt duplicates option C; both indicate the same correct value.Common Pitfalls:Confusing leg size with throat thickness; mixing allowable weld metal stress with base metal allowable; forgetting that two fillets share the load.Final Answer:2 s * l * σt

Question 16

Thin cylindrical pressure vessel: For a cylinder of diameter d and thickness t under internal pressure p, what is the longitudinal (axial) membrane stress in the shell?

Accepted Answer

Introduction / Context:Thin-walled pressure vessels develop two principal membrane stresses: circumferential (hoop) and longitudinal (axial). Correct formulas are vital for sizing thickness and checking allowable stress compliance in boilers, tanks, and pipelines.Given Data / Assumptions:Thin cylinder, t ≪ d.Internal pressure p; ends closed.Uniform membrane stresses; neglect radial stress.Concept / Approach:Equilibrium on a free-body of half the vessel gives hoop stress sigma_h = p d / (2 t). Equilibrium on the end cap gives longitudinal stress sigma_L = p d / (4 t). The longitudinal stress is half the hoop stress in a thin cylinder with closed ends.Step-by-Step Solution:Hoop: sigma_h = p d / (2 t).End equilibrium: p * (π d^2 / 4) balanced by 2 * sigma_L * t * (d / 2) * π → sigma_L = p d / (4 t).Hence the longitudinal stress is p d / (4 t).Verification / Alternative check:Check ratio: sigma_h / sigma_L = 2, consistent with thin cylinder theory.Why Other Options Are Wrong:p d / t and 2 p d / t are excessive (do not match equilibrium).p d / (2 t) is the hoop, not longitudinal, stress.p d / (6 t) has no basis in thin-cylinder derivation.Common Pitfalls:Using hoop formula for both directions; forgetting that end closures create axial stress; applying thick-cylinder relations to thin shells.Final Answer:p d / (4 t)

Question 17

Plane stress with in-plane shear: Given normal stresses σx and σy on perpendicular planes and shear τxy, what is the maximum normal (principal) stress?

Accepted Answer

Introduction / Context:Transformations of plane stress determine principal stresses and directions. The maximum normal stress controls yielding under maximum normal stress criteria and is widely used to assess safety margins in flat plates and machine elements.Given Data / Assumptions:Plane stress state: σx, σy, τxy known.Homogeneous, isotropic, linear elastic material.Standard sign conventions.Concept / Approach:From Mohr's circle or transformation equations, principal stresses areσ1,2 = (σx + σy)/2 ± sqrt( ((σx − σy)/2)^2 + τxy^2 )The maximum normal stress is σ1, obtained using the plus sign.Step-by-Step Solution:Compute average normal stress: (σx + σy)/2.Compute radius R = sqrt( ((σx − σy)/2)^2 + τxy^2 ).Maximum principal stress: σ1 = (σx + σy)/2 + R.Verification / Alternative check:Special cases: if τxy = 0, σ1 = max(σx, σy); if σx = σy, σ1 = σx + |τxy|, matching Mohr's circle for pure shear superposed on hydrostatic stress.Why Other Options Are Wrong:(σx + σy)/2 is the center value, not the maximum principal stress.|σx − σy|/2 is valid only when τxy = 0 and then equals the radius, not σ1 directly.sqrt( σx^2 + σy^2 + τxy^2 ) is not a principal-stress formula.(σx − σy)/2 + τxy is not invariant and lacks proper vector combination.Common Pitfalls:Dropping the 1/2 inside the square root; mixing up radius and diameter on Mohr's circle; sign mistakes for τxy.Final Answer:(σx + σy) / 2 + sqrt( ((σx − σy) / 2)^2 + τxy^2 )

Question 18

Earth pressure on a retaining wall: The total horizontal thrust per metre length (active condition) on a wall of height h retaining cohesionless soil (unit weight w, angle of repose φ) is

Accepted Answer

Introduction / Context:Retaining structures must resist lateral earth pressures. The Rankine active earth pressure theory provides a convenient closed-form for cohesionless backfill, frequently used in preliminary design and checks.Given Data / Assumptions:Backfill is dry, cohesionless, with unit weight w and angle of internal friction φ (equal to angle of repose).Vertical, smooth wall; horizontal backfill surface (Rankine conditions).Active state of earth pressure (wall yields sufficiently).Concept / Approach:Rankine active pressure coefficient K_a = tan^2(45° − φ/2) = (1 − sin φ)/(1 + sin φ). Resultant active thrust P_a per metre length isP_a = (1/2) * w * h^2 * K_aacting at h/3 above the base. Substituting K_a gives the expression in terms of φ.Step-by-Step Solution:Compute K_a = tan^2(45° − φ/2).Total thrust: P_a = (w * h^2 / 2) * K_a.Hence P_a = (w * h^2 / 2) * tan^2(45° − φ/2).Verification / Alternative check:Using identity K_a = (1 − sin φ)/(1 + sin φ), P_a = (w h^2 / 2) * (1 − sin φ)/(1 + sin φ), consistent with Rankine theory.Why Other Options Are Wrong:tan^2(45° + φ/2) corresponds to K_p (passive), not active.w h^2 tan φ lacks the correct coefficient and dimensions.(w h^2 / 3) times the ratio mixes location of resultant with magnitude.(w h^2 / 2) ignores soil friction angle entirely.Common Pitfalls:Confusing K_a and K_p signs; forgetting that the resultant acts at h/3; applying Coulomb theory coefficients to Rankine conditions without accounting for wall friction or backface inclination.Final Answer:(w * h^2 / 2) * tan^2(45° − φ/2)

Question 19

Axial extension of a circular bar tapering uniformly from diameter d1 to diameter d2 over length L under axial pull P (E = Young's modulus): Which expression gives the total extension?

Accepted Answer

Introduction / Context:Members with varying cross-section require integration of strain along the length to compute elongation under axial load. A classic result is the extension of a circular bar that tapers linearly from diameter d1 to d2.Given Data / Assumptions:Bar length L; diameters d1 (one end) to d2 (other end), linearly varying.Axial tensile load P; Young's modulus E constant.Small deformations; axial stress uniform over each section.Concept / Approach:For an element at distance x with local diameter d(x), area A(x) = π [d(x)]^2 / 4. Strain at x is P / (E * A(x)). Total extension is the integral of strain over length L. With linear taper, d(x) varies linearly → integration yields a simple closed form.Step-by-Step Solution:Let d(x) = d1 + ( (d2 − d1) / L ) * x.A(x) = π [d(x)]^2 / 4.Incremental extension: du = (P / (E * A(x))) dx.Integrate u = ∫(0→L) (4 P / (π E [d(x)]^2)) dx.For linear d(x), the integral evaluates to u = (4 P L) / (π E d1 d2).Verification / Alternative check:Check limits: if d1 = d2 = d (uniform bar) → u = (4 P L)/(π E d^2) = (P L)/(A E), since A = π d^2/4. This confirms correctness.Why Other Options Are Wrong:Using average area (A_avg) is an approximation; exact integral gives 1/(d1 d2) dependence, not function of (d1 + d2)^2.Forms with d1^2 + d2^2 do not reduce properly for d1 = d2.Factor 2 in the numerator (option E) underestimates extension by half.Common Pitfalls:Arithmetic slips when integrating 1/[d(x)]^2; forgetting to verify the uniform-bar limit; using arithmetic average of areas instead of the correct harmonic-type relation.Final Answer:(4 * P * L) / (π * E * d1 * d2)

Question 20

Beam under tent-shaped load: A simply supported beam carries a load that varies gradually from zero at both supports to w per metre at midspan (symmetric). Is the statement “the bending moment at the centre is w l / 4” correct?

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Introduction / Context:Different load distributions produce different shear and bending moment diagrams. Dimensional consistency and known reference cases help validate formulas. The given statement resembles the classic point-load case at midspan rather than a distributed tent-shaped load.Given Data / Assumptions:Simply supported beam of span l.Load intensity is zero at both supports and increases linearly to w per metre at midspan, then symmetrically decreases.w denotes load intensity (force per unit length).Concept / Approach:For a central point load W on a simply supported beam, M_max at midspan equals W * l / 4 (units of forcelength). For a uniformly distributed load q, M_max = q * l^2 / 8 (forcelength). For a tent-shaped distributed load peaking at midspan, the correct midspan moment involves w * l^2 multiplied by a coefficient distinct from 1/8 and certainly not w * l / 4, which is dimensionally inconsistent if w is per metre.Step-by-Step Solution:Check dimensions: bending moment units should be force * length.Given w is force/length, w * l / 4 has units of force (incorrect). Correct form must include l^2.Therefore, the stated expression cannot be correct for a distributed load described.Verification / Alternative check:Detailed integration of the symmetric triangular intensity gives M_center proportional to w * l^2 with a coefficient less than that for UDL. Regardless, w * l / 4 is invalid dimensionally and numerically.Why Other Options Are Wrong:“Yes” contradicts both dimensional analysis and known beam results.“Only for very small spans” or “Depends on end fixity” are irrelevant; the form remains wrong.“Only if w is total load” would redefine symbols, but the stem clearly defines w as intensity.Common Pitfalls:Memorizing W l / 4 (point load) and misapplying it to distributed loads; ignoring unit consistency; mixing up W (total load) with w (load per unit length).Final Answer:No

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