Q: Elastic elongation under self-weight – Conical bar versus prismatic bar A conical bar and a prismatic (uniform) bar of the same material and length L hang under their own weight. The elongation of the conical bar is how much of the elongation of th…

Introduction / Context: Bars elongate under their self-weight because the material below each cross-section supports the weight of material beneath. The distribution of cross-sectional area changes how stress accumulates along the length, making tapering an effective way to reduce elongation. Given Data / Assumptions: Length L, same material (Young’s modulus E and unit weight γ) for both bars. Prismatic bar has constant area A; conical bar tapers linearly to the tip. Small deformations and linear elasticity. Concept / Approach: Axial strain at a section equals axial stress / E. Stress at a height x from the top equals weight below that section divided by the local area. Integrating strain along the length gives total elongation. For self-weight: δ_prismatic = (γ L^2) / (2 E). For a conical bar: δ_conical = (γ L^2) / (3 E). Step-by-Step Solution: For the prismatic bar: δ_pr = (γ L^2) / (2 E).For the conical bar: δ_co = (γ L^2) / (3 E) (due to area increasing toward the top, lowering stress).Ratio: δ_co / δ_pr = (1/3) / (1/2) = 2/3.Hence the conical bar elongates by two-thirds of the prismatic bar’s elongation. Verification / Alternative check: A taper reduces stress near the top by providing more area where the force is larger, so a smaller overall extension compared to a uniform bar is physically reasonable. Why Other Options Are Wrong: “Equal to” ignores area variation; “half” and “one-third” underestimate; “three-fourths” overestimates relative to the established 2/3. Common Pitfalls: Neglecting to integrate strain properly; using average stress instead of the varying stress distribution; mixing up factors 1/2 and 1/3. Final Answer: two-third

Question 1

In tensile testing of ductile materials, how does ductility vary with the percentage reduction in area measured at fracture?

Accepted Answer

Introduction / Context: Ductility quantifies a material’s ability to undergo plastic deformation before fracture. Common measures include percentage elongation and percentage reduction in area (%RA) of the fractured specimen. Given Data / Assumptions: Ductile material under a standard tensile test. % reduction in area = (A0 − Af) / A0 * 100% at the necked fracture section. Question asks the relation between ductility and %RA. Concept / Approach: A higher percentage reduction in area indicates greater plastic flow and necking before failure, which is the hallmark of higher ductility. Thus, ductility correlates positively with %RA. Step-by-Step Solution: Record initial cross-sectional area A0.After fracture, measure minimum area Af at the neck.Compute %RA = (A0 − Af) / A0 * 100%.Interpretation: larger %RA → more plastic deformation → higher ductility. Verification / Alternative check: Comparing specimens: one with greater necking (smaller Af) displays higher ductility as observed by both % elongation and %RA. Why Other Options Are Wrong: “Decreases” or “remains the same” contradict the definition of %RA as a ductility measure. Common Pitfalls: Confusing %RA with strength; %RA is not a strength metric but a plasticity indicator. Final Answer: Increases

Question 2

Define a principal plane in stress analysis: on a principal plane, the shear stress is what?

Accepted Answer

Introduction / Context: Principal stresses and principal planes are foundational in mechanics of materials. They simplify complex stress states by finding orientations where shear vanishes and only normal stress acts. Given Data / Assumptions: Plane stress at a point in a continuum. Principal planes are the physical planes where normal stress is extremal. We must recall the definition regarding shear stress on such planes. Concept / Approach: By definition, on principal planes, shear stress τ is zero and the normal stress σ reaches principal values (maximum and minimum). This follows from extremising σ_n with respect to plane angle in the stress transformation equations or by Mohr’s circle where principal points lie on the σ-axis (τ = 0). Step-by-Step Solution: Use Mohr’s circle: principal stresses are intersection points with the σ-axis.At these points, τ = 0 by construction.The corresponding physical planes are principal planes. Verification / Alternative check: Differentiating σ_n(θ) with respect to θ shows extrema occur when τ(θ) = 0, confirming the zero-shear condition on principal planes. Why Other Options Are Wrong: “Minimum but not zero” and “maximum” misstate the defining property; principal planes are shear-free. Common Pitfalls: Confusing maximum shear planes (where τ is extreme and σ is intermediate) with principal planes (where τ = 0). Final Answer: Zero

Question 3

Strain measures: The ratio of the change in volume to the original volume of a body under load is called what?

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Introduction / Context: Different strain measures quantify deformation in various directions and modes. Recognising which definition applies avoids errors in bulk modulus or multi-axial stress calculations. Given Data / Assumptions: Small-strain (engineering strain) regime. Volume change ΔV relative to original volume V. Isotropic or anisotropic loading does not change the definition itself. Concept / Approach: Volumetric strain ε_v is defined as ε_v = ΔV / V. Linear (axial) strain is ΔL / L; lateral strain is transverse strain associated with Poisson effects; Poisson’s ratio ν = −(lateral strain)/(linear strain) in uniaxial loading. Step-by-Step Solution: Identify the asked ratio: change in volume to original volume.Map definition: ε_v = ΔV / V.Conclude the correct term: volumetric strain. Verification / Alternative check: Bulk modulus K uses ε_v: p = K * ε_v for hydrostatic loading (small strains), reinforcing the definition. Why Other Options Are Wrong: Linear and lateral strains are one-dimensional measures; Poisson’s ratio is a material constant relating them, not a strain itself. Common Pitfalls: Confusing Poisson’s ratio (dimensionless constant) with a strain measure; mixing linear and volumetric changes. Final Answer: Volumetric strain

Question 4

For a simply supported beam carrying a uniformly distributed load of w per unit length over the entire span, what is the qualitative shape of the shear force diagram?

Accepted Answer

Introduction / Context: Internal shear force and bending moment diagrams are fundamental for beam design. Recognising their shapes for common loadings lets engineers size members quickly and check results by inspection. Given Data / Assumptions: Simply supported beam (pin and roller supports). Uniformly distributed load (UDL) = w per unit length across entire span L. Static, prismatic beam in the usual sign convention (upward reactions positive). Concept / Approach: Shear force V(x) equals the algebraic sum of vertical forces to the left (or right). Under a UDL, dV/dx = −w, so V varies linearly. For symmetry, reactions are equal (wL/2 each). The diagram is a straight line decreasing from +wL/2 at the left support to −wL/2 at the right support—i.e., a right-angled triangle. Step-by-Step Solution: Compute reactions: R_A = R_B = wL/2.Integrate load: dV/dx = −w → V(x) = R_A − w x.Evaluate endpoints: V(0) = +wL/2; V(L) = −wL/2.Plot linear variation → triangular SFD. Verification / Alternative check: The slope of the bending moment diagram equals V; since V is linear, moment is quadratic (parabolic), consistent with standard beam theory. Why Other Options Are Wrong: Two triangles or equilateral forms do not match the single linear ramp imposed by constant w. Common Pitfalls: Misplacing signs at supports; forgetting that UDL produces linear shear and parabolic moment. Final Answer: A single right-angled triangle (linear variation across the span)

Question 5

A closely coiled helical spring is subjected to an axial load. Predominantly, the wire of the spring is under which type of action?

Accepted Answer

Introduction / Context: Helical compression and tension springs are ubiquitous in machines. Understanding how an axial load translates into stresses within the spring wire is vital for sizing and fatigue analysis. Given Data / Assumptions: Closely-coiled helical spring (helix angle small). Axial load applied along the spring axis. Standard thin-wire spring theory (Wahl, curvature effects ignored for first order). Concept / Approach: In a closely-coiled spring, the axial load primarily produces a twisting moment in the wire. Hence, the wire experiences torsion, which induces shear stress. There is a small direct shear component, but torsion dominates the stress state and governs design. Step-by-Step Solution: Model the coil as a curved beam subjected to axial force.Resolve the axial load into a torque about the wire’s centreline.Compute maximum shear using τ_max ≈ 16 W R / (π d^3) with curvature factors as needed.Note direct shear W/(π d^2/4) is secondary compared to torsional shear. Verification / Alternative check: Energy methods and Castigliano’s theorem yield the same deflection when torsional strain energy dominates, confirming torsion as the primary action. Why Other Options Are Wrong: Bending and crushing do not represent the principal action in closely-coiled springs; “direct shear only” ignores the dominant torsion. Common Pitfalls: Neglecting curvature correction (Wahl factor) and stress concentration at the inner fibre in detailed design. Final Answer: Torsion (resulting in shear stress in the wire)

Question 6

Two shafts of the same length (one hollow, one solid) transmit equal torques and have the same maximum shear stress. Which property must be equal for the two shafts?

Accepted Answer

Introduction / Context: Strength in torsion is governed by the relationship between applied torque and maximum shear stress. Distinguishing between stiffness quantities (like J and angle of twist) and strength quantities (like Zp) avoids design errors. Given Data / Assumptions: Equal torque T on both shafts. Same maximum shear stress τ_max on both shafts. Same length; materials may be the same (for strength equality it is not required if τ_max is the same limit). Concept / Approach: The torsion formula for strength is τ_max = T / Zp, where Zp = J / R (R = outer radius). If T and τ_max are equal for both shafts, then Zp must be equal. J alone is insufficient because τ_max also depends on the outer radius. Step-by-Step Solution: Write τ_max1 = T / Zp1 and τ_max2 = T / Zp2.Given τ_max1 = τ_max2 and equal T → Zp1 = Zp2.Conclude that equality of polar section modulus is required. Verification / Alternative check: Angle of twist relates to stiffness: θ = T L / (G J). Equal θ would demand equal J, not required by equal τ and T; hence Zp is the correct equality for strength. Why Other Options Are Wrong: Polar moment J alone ignores the effect of R on stress; diameter equality is unnecessary; angle of twist pertains to stiffness, not strength. Common Pitfalls: Confusing Zp (strength) with J (stiffness) and designing for the wrong criterion. Final Answer: Polar section modulus (Zp = J / R)

Question 7

Riveted joints practice: If a hole of 25 mm diameter is drilled in plates for riveting, what should be the approximate diameter of the rivet shank (nominal rivet) for proper fit and driving?

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Introduction / Context: In structural and boiler riveting practice, the hole is made slightly larger than the rivet shank so the rivet can be inserted and then upset (formed) to fill the hole during driving. Choosing the correct clearance is a standard shop rule. Given Data / Assumptions: Drilled hole diameter d_h = 25 mm. Conventional practice: hole diameter exceeds rivet nominal diameter by about 1.5 mm (typical workshop allowance). Goal: find nominal rivet diameter (shank size before driving). Concept / Approach: The usual clearance c between hole and rivet in hand or power riveting is about 1.0 to 1.5 mm (often 1.5 mm in many textbooks for up to about 25 mm rivets). Therefore, d_r ≈ d_h − c. The rivet expands plastically during upsetting to fill the hole tightly and ensure joint integrity. Step-by-Step Solution: Take clearance c ≈ 1.5 mm for a 25 mm hole.Compute rivet diameter: d_r = d_h − c = 25 − 1.5 = 23.5 mm.Verify that this allows insertion and proper upsetting to fill the hole. Verification / Alternative check: Typical workshop tables specify hole ≈ rivet + 1.5 mm for common sizes; field practice confirms this as a workable allowance. Why Other Options Are Wrong: 23.0 mm: clearance 2.0 mm is excessive, risks poor fit.24.5 mm: clearance 0.5 mm may be too tight for consistent insertion and heating tolerances.25.0 mm: zero clearance; impractical for insertion before upset. Common Pitfalls: Confusing nominal rivet size with hole size; ignoring practical clearance needed for insertion and thermal expansion before closing the head. Final Answer: 23.5 mm

Question 8

Strength of Materials – Simply supported beam with a central point load  Consider a simply supported beam of span L carrying a single concentrated load P exactly at mid-span. State whether the shear force diagram changes sign at the midpoint under t…

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Introduction / Context: In strength of materials, understanding how shear force varies along a beam under specific loads is fundamental. A simply supported beam with a central point load is a classic case used to teach sign conventions and the relationship between loads, shear force, and bending moment. Given Data / Assumptions: Beam is simply supported at two ends. A single point load P acts at mid-span (x = L/2). Beam is prismatic, linearly elastic, and weightless for simplicity. Upward reactions at supports are equal due to symmetry, each = P/2. Concept / Approach: Shear force V(x) is the algebraic sum of vertical forces to the left or right of a section. For a point load, V has a jump (discontinuity) at the load location. With the load at mid-span, the shear changes from positive to negative (or vice versa depending on sign convention) at that exact point. Step-by-Step Solution: Take sections to the left of mid-span: V_left = +P/2 (upward reaction).Take sections immediately to the right of the point load: V_right = +P/2 − P = −P/2.The sudden change from +P/2 to −P/2 at mid-span shows a sign change in V(x).Bending moment is maximum where shear crosses zero; here it occurs at mid-span, consistent with theory. Verification / Alternative check: The bending moment diagram is triangular on each half with peak M_max = (P/2)(L/2) = PL/4 at mid-span. Maximum moment coincides with V = 0, confirming the shear sign change at the center. Why Other Options Are Wrong: “It changes sign only at the supports” is false; supports are reaction points, not zero-shear locations for this load case. “It never changes sign for a point load” contradicts the jump at the load. “Depends on beam material only” is incorrect; material does not affect static equilibrium of internal forces. Common Pitfalls: Mixing sign conventions; forgetting that a concentrated load introduces a shear force jump; assuming zero shear only for distributed loads. Final Answer: Correct

Question 9

Design of riveted joints – Definition of efficiency  In machine design, the efficiency of a riveted joint is defined as which ratio with respect to the original unperforated plate?

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Introduction / Context: Riveted joints are evaluated by comparing the capacity of the perforated, riveted section against the capacity of the original unperforated plate. This comparison leads to a dimensionless efficiency that indicates how much of the base plate strength is retained by the joint. Given Data / Assumptions: Tensile loading is along the joint line. Joint strength is governed by the least of tearing, shearing, and bearing strengths over one pitch length. Solid plate strength is based on full plate section without holes. Concept / Approach: Joint efficiency η measures how effectively the joint transmits load compared to the parent plate. It is defined as a ratio, not a difference. Selecting the minimum failure mode ensures a conservative evaluation. Step-by-Step Solution: Compute solid plate strength: P_solid = plate_thickness * plate_width * allowable_stress.Compute joint strength per pitch: P_joint = minimum of (tearing capacity through net section, shear capacity of rivets, bearing capacity).Define efficiency: η = P_joint / P_solid.Interpretation: 0 < η ≤ 1, with higher values indicating better utilization of plate strength. Verification / Alternative check: Design codes and textbooks consistently define efficiency as a ratio of capacities, confirming the correct option. Why Other Options Are Wrong: (B) inverts the ratio; (C) is a geometric parameter, not efficiency; (D) and (E) restrict to partial modes and ignore the weakest-link principle. Common Pitfalls: Forgetting to use net section for tearing; ignoring bearing limits; comparing per-pitch strengths to the correct reference section. Final Answer: The ratio of the strength of the joint to the strength of the solid (unrivettted) plate

Question 10

Elastic elongation under self-weight – Conical bar versus prismatic bar  A conical bar and a prismatic (uniform) bar of the same material and length L hang under their own weight. The elongation of the conical bar is how much of the elongation of th…

Accepted Answer

Introduction / Context: Bars elongate under their self-weight because the material below each cross-section supports the weight of material beneath. The distribution of cross-sectional area changes how stress accumulates along the length, making tapering an effective way to reduce elongation. Given Data / Assumptions: Length L, same material (Young’s modulus E and unit weight γ) for both bars. Prismatic bar has constant area A; conical bar tapers linearly to the tip. Small deformations and linear elasticity. Concept / Approach: Axial strain at a section equals axial stress / E. Stress at a height x from the top equals weight below that section divided by the local area. Integrating strain along the length gives total elongation. For self-weight: δ_prismatic = (γ L^2) / (2 E). For a conical bar: δ_conical = (γ L^2) / (3 E). Step-by-Step Solution: For the prismatic bar: δ_pr = (γ L^2) / (2 E).For the conical bar: δ_co = (γ L^2) / (3 E) (due to area increasing toward the top, lowering stress).Ratio: δ_co / δ_pr = (1/3) / (1/2) = 2/3.Hence the conical bar elongates by two-thirds of the prismatic bar’s elongation. Verification / Alternative check: A taper reduces stress near the top by providing more area where the force is larger, so a smaller overall extension compared to a uniform bar is physically reasonable. Why Other Options Are Wrong: “Equal to” ignores area variation; “half” and “one-third” underestimate; “three-fourths” overestimates relative to the established 2/3. Common Pitfalls: Neglecting to integrate strain properly; using average stress instead of the varying stress distribution; mixing up factors 1/2 and 1/3. Final Answer: two-third

Question 11

Soil mechanics – Assumptions of Rankine’s theory for active earth pressure  Identify the set of assumptions underlying Rankine’s earth pressure theory for active case acting on a vertical retaining wall.

Accepted Answer

Introduction / Context: Rankine’s theory predicts earth pressure at rest, active, and passive states for cohesionless backfills. For active pressure, the wall yields sufficiently away from the soil to mobilize the minimum lateral stress condition. Given Data / Assumptions: Backfill is idealized as homogeneous and dry (c = 0), with unit weight γ and friction angle φ. Back of the wall is vertical and smooth, implying no wall–soil friction effects. Failure surface is assumed planar for the active case. Concept / Approach: With these assumptions, Rankine derived active earth pressure coefficient Ka = (1 − sin φ) / (1 + sin φ). The active pressure acts at H/3 above the base of the wall for level backfill. The simplifications enable closed-form expressions without wedge equilibrium involving interface friction (as in Coulomb’s theory). Step-by-Step Solution: Adopt cohesionless soil (c = 0) and smooth wall → no interface shear.Assume an extended semi-infinite backfill with level surface → uniform conditions.Postulate planar failure → obtain Ka and the triangular pressure distribution.Compute resultant = (1/2) * Ka * γ * H^2 acting at H/3 above base (if needed). Verification / Alternative check: Coulomb’s theory relaxes the smooth-wall assumption and permits sloping backfills and wall friction; its results reduce toward Rankine’s when interface friction is set to zero and geometry is simplified. Why Other Options Are Wrong: Each of (A), (B), and (C) is correct but incomplete alone; “None of the above” is incorrect because a planar surface is explicitly assumed in Rankine’s formulation. Common Pitfalls: Applying Rankine for cohesive or submerged soils without modification; confusing active and passive signs; using Coulomb coefficients unintentionally. Final Answer: all of the above

Question 12

Beams – Definition of an overhanging beam  An overhanging beam must necessarily extend beyond the supports on both sides. Decide whether this statement is correct.

Accepted Answer

Introduction / Context: An overhanging beam is commonly encountered in structural analysis. Clarifying its definition helps in selecting the proper boundary conditions and internal force diagrams. Given Data / Assumptions: Beam has at least two supports (typically simple supports). An overhang is a portion of the beam that extends beyond a support. Concept / Approach: A beam is termed “overhanging” if it projects beyond a support on one or both sides. The presence of even a single overhang (on one side only) qualifies it as an overhanging beam. Overhanging beams differ from pure simply supported beams and cantilevers in reaction and moment distributions. Step-by-Step Solution: Define: overhanging beam → extension beyond a support.Case 1: Overhang on one side only → still an overhanging beam.Case 2: Overhangs on both sides → also an overhanging beam, but not mandatory.Therefore, the claim that both sides are required is incorrect. Verification / Alternative check: Textbook examples show single-side overhangs producing hogging moments near the support and sagging moments on the span, confirming the classification. Why Other Options Are Wrong: “True only for cantilevers” confuses cantilever (fixed end) with overhangs; “Depends on load position” is irrelevant to the definition; “True only for propped cantilevers” is unrelated. Common Pitfalls: Assuming symmetry is required; mixing the term with cantilevered members fixed at one end. Final Answer: Incorrect

Question 13

Thin spherical pressure vessel – Membrane stress formula  A thin spherical shell of diameter d and wall thickness t is subjected to an internal pressure p. What is the uniform membrane (hoop) stress developed in the shell material?

Accepted Answer

Introduction / Context: Thin-walled pressure vessels are analyzed using membrane theory. For a sphere, the stress is uniform in all tangential directions, making it a preferred shape for withstanding internal pressure. Given Data / Assumptions: Spherical shell, diameter d, thickness t (t ≪ d). Internal gauge pressure p acts uniformly. Material behaves elastically and deformations are small. Concept / Approach: Equilibrium of a hemispherical free body yields the tangential membrane stress σ. For a thin sphere, the resultant of membrane stresses around the great circle balances the internal pressure force on the hemisphere. Step-by-Step Solution: Pressure force on hemisphere = p * projected area = p * (π d^2 / 4).Resultant resisting force = σ * (circumference) * thickness = σ * (π d) * t.Equate forces: σ * (π d) * t = p * (π d^2 / 4).Solve: σ = p d / (4 t). Verification / Alternative check: Compare with thin cylinder (σ_hoop = p d / (2 t)); the sphere carries pressure more efficiently, showing half the hoop stress for the same d, t, p. Why Other Options Are Wrong: (A) and (B) overestimate stress; (D) underestimates; “2 p d / t” is nonphysical for thin-shell equilibrium. Common Pitfalls: Using the cylindrical formula for spheres; forgetting the projected area for pressure force; neglecting thin-wall criteria. Final Answer: p d / (4 t)

Question 14

Thick cylinders – Maximum tangential (hoop) stress vs. internal pressure  In a thick-walled cylinder subjected to internal pressure p_i only, how does the maximum tangential (hoop) stress at the inner surface compare with p_i?

Accepted Answer

Introduction / Context: For thick cylinders, the stress distribution is non-uniform and must be found using Lame’s equations, unlike thin-wall approximations. The hoop stress peaks at the inner surface where the pressure is applied. Given Data / Assumptions: Internal pressure only; no external pressure. Elastic, axisymmetric behavior; long cylinder (plane strain or plane stress assumptions for wall). Inner radius r_i, outer radius r_o. Concept / Approach: Lame’s formula for hoop stress is σ_θ(r) = A + B / r^2. Applying boundary conditions at r = r_i (−p_i radial) and r = r_o (0 radial) gives a maximum σ_θ at r_i. Its magnitude exceeds p_i depending on thickness ratio. Step-by-Step Solution: Use Lame’s relations to determine constants A and B from radial stress boundary conditions.Evaluate σ_θ at r = r_i and compare with p_i.For any finite wall (r_o > r_i), σ_θ(r_i) > p_i; the thicker the wall (smaller r_i/r_o), the larger the amplification factor. Verification / Alternative check: Numerical examples with r_o/r_i > 1 always yield σ_θ(r_i) greater than p_i. The thin-wall limit tends toward σ_θ ≈ p_i * r_i / t, but for thick walls, inner hoop stress is elevated further. Why Other Options Are Wrong: Equal to or less than p_i contradicts Lame’s solution; dependence on Poisson’s ratio is weak for principal stresses; zero at inner surface is impossible under internal pressure. Common Pitfalls: Applying thin-wall formulas when t is not small; ignoring radial stress boundary conditions. Final Answer: greater than the internal pressure

Question 15

Beam theory – Bending moment at mid-span for UDL  A simply supported beam of span l carries a uniformly distributed load of intensity w per unit length over the entire span. What is the bending moment at the center of the beam?

Accepted Answer

Introduction / Context: For basic loading cases, memorizing key formulas helps in quick checks and preliminary sizing. A simply supported beam under a full-span uniformly distributed load (UDL) is among the most common scenarios. Given Data / Assumptions: Span l, UDL = w (force per unit length). Supports are simple (no end moments), beam is prismatic. Linear elastic behavior. Concept / Approach: By symmetry, reactions at supports are R_A = R_B = w l / 2. The bending moment at a distance x from the left is M(x) = R_A x − w x^2 / 2. The maximum occurs where shear V(x) = 0, i.e., at x = l/2. Step-by-Step Solution: Compute reactions: R_A = R_B = w l / 2.Moment at mid-span: M(l/2) = (w l / 2)(l/2) − w(l/2)^2 / 2.Simplify: M(l/2) = w l^2 / 4 − w l^2 / 8 = w l^2 / 8.Hence, the center bending moment equals w l^2 / 8. Verification / Alternative check: The parabolic bending moment diagram for full UDL peaks at the center with the known value w l^2 / 8; area under shear diagram also corroborates this result. Why Other Options Are Wrong: Zero is inconsistent with UDL; w l^2 / 2 and /4 are too high; /12 is not the standard value for this case. Common Pitfalls: Forgetting the 1/2 factor in the distributed load moment term; mixing with the point-load mid-span case P L / 4. Final Answer: w l^2 / 8

Question 16

Energy methods – Definition of proof resilience for springs  The strain energy stored in a spring under the maximum permissible load without permanent set is known as what?

Accepted Answer

Introduction / Context: Energy storage in elastic members is crucial in shock absorption and vibration isolation. Springs are designed to store strain energy up to an elastic limit without plastic deformation. Given Data / Assumptions: Spring is loaded up to its maximum allowable (proof) load in the elastic range. No permanent deformation is permitted at the proof load. Material obeys Hooke’s law up to that load. Concept / Approach: Proof resilience is the maximum strain energy that a material or spring can store per unit volume (for materials) or in total (for springs) without permanent set. For a linear spring: U = (1/2) * F * δ at the elastic limit. Step-by-Step Solution: Identify the proof (maximum elastic) load F_proof.Determine the corresponding deflection δ_proof from the load–deflection relation.Compute stored energy: U_proof = (1/2) * F_proof * δ_proof.Interpretation: This is the spring’s proof resilience. Verification / Alternative check: For material specimens, the area under the stress–strain curve up to elastic limit equals modulus of resilience (per unit volume). For a spring, the analogous total energy at the elastic limit is termed proof resilience. Why Other Options Are Wrong: Impact energy relates to sudden loads, not necessarily elastic limit; proof stress is a stress value; modulus of resilience is per unit volume of material, not total spring energy; toughness covers energy to fracture, including plasticity. Common Pitfalls: Confusing proof resilience with modulus of resilience; exceeding elastic limit in design assumptions. Final Answer: proof resilience

Question 17

Mohr’s circle – Normal stress on an oblique plane (unequal like principal stresses)  Given Mohr’s circle for two unequal and like principal stresses σx and σy acting on orthogonal planes, the normal stress on a plane making angle θ with the minor pr…

Accepted Answer

Introduction / Context: Mohr’s circle graphically relates normal and shear stresses acting on rotated planes. For any plane at a physical angle θ from a reference principal plane, the corresponding point on Mohr’s circle lies at an angle 2θ from the principal stress point. Given Data / Assumptions: Principal stresses σ1 (major) and σ2 (minor) are unequal and of the same sign (like). The circle center C lies at ( (σ1 + σ2)/2 , 0 ) on the normal-stress axis. Point P on the circle represents the stress state on the plane at angle θ; its projection on the σ-axis is Q. Concept / Approach: On Mohr’s circle, the horizontal axis is normal stress (σ), and the vertical axis is shear stress (τ). The normal stress on a plane equals the abscissa of the representative point, i.e., the horizontal projection from the origin to that point’s σ-coordinate. Step-by-Step Solution: Locate center C at σ_avg = (σ1 + σ2)/2.Construct the circle with radius R = (σ1 − σ2)/2.Rotate from the minor principal plane by angle 2θ to find point P.Drop a horizontal projection from P to the σ-axis → this distance from origin is OQ, the normal stress. Verification / Alternative check: Coordinates of P satisfy σ = σ_avg + R cos(2θ) and τ = R sin(2θ). The normal stress is the σ-coordinate only, which is exactly the horizontal projection OQ. Why Other Options Are Wrong: OC is the offset to the circle center, not the stress on a plane; OP mixes σ and τ components; PQ represents shear magnitude; CQ is partial and not referenced to the origin. Common Pitfalls: Confusing physical angle θ with 2θ on the circle; reading radial length instead of horizontal projection for σ; sign conventions for clockwise/counter-clockwise rotations. Final Answer: OQ (the abscissa/projection of the plane point on the normal-stress axis)

Question 18

Tensile test and ductility comparison: If a specimen of material A shows 60% reduction in area at fracture and an identical specimen of material B shows 40% reduction in area, which material is more ductile?

Accepted Answer

Introduction / Context: Ductility in a tensile test is commonly judged by the percentage reduction in area at fracture and/or percentage elongation. A higher percentage reduction in area indicates that the material can undergo more plastic deformation before failure, hence it is more ductile. Given Data / Assumptions: Specimens of equal initial dimensions are tested in tension. Material A: reduction in area = 60%. Material B: reduction in area = 40%. Standard tensile test conditions and accurate measurements are assumed. Concept / Approach: Percentage reduction in area is defined as ((A0 − Af) / A0) * 100%, where A0 is original cross-sectional area and Af is area at the fracture section. Larger values imply greater necking and plastic flow ability, which correspond to higher ductility. Step-by-Step Solution: Compute ductility metric: %RA_A = 60%, %RA_B = 40%.Compare: 60% > 40%.Inference: The material with the higher % reduction in area is more ductile.Conclusion: Material A is more ductile than material B. Verification / Alternative check: One may also check percentage elongation. Typically, a material with higher % reduction in area also shows higher % elongation, reaffirming the same conclusion about ductility ranking. Why Other Options Are Wrong: Material B more ductile: contradicts 40% vs 60% comparison. Equal ductility: data clearly differ. “A is brittle and B is ductile”: opposite of the evidence. “Neither material is ductile”: reduction in area values indicate plasticity in both. Common Pitfalls: Confusing strength with ductility; using ultimate tensile strength alone to judge ductility; ignoring measurement consistency in necked area readings. Final Answer: Material A is more ductile than material B

Question 19

Riveted joints basics: A lap joint is always subjected to which shear condition under load transfer?

Accepted Answer

Introduction / Context: Riveted joints transfer loads through shear in the rivets. Joint configuration dictates whether each rivet is in single or double shear. Understanding this is fundamental in design of plate, boiler, and structural joints. Given Data / Assumptions: Lap joint with two plates overlapping and a single row (or multiple rows) of rivets passing through the overlap. Load is axial, trying to slide one plate relative to the other. Plates remain in a single overlapping plane (no cover plate on the opposite side). Concept / Approach: In a lap joint, each rivet crosses one shear plane at the lapped interface. Therefore, each rivet resists load in single shear. In contrast, a double-strap butt joint places rivets in double shear because the force passes through two shear planes (two strap interfaces). Step-by-Step Solution: Identify joint type: lap joint (one overlap).Count shear planes a rivet crosses: 1 plane at plate interface.Conclusion: Rivets carry load in single shear. Verification / Alternative check: Sketch the free-body diagram of the plates and locate the interface; only one shear plane intersects the rivet for each load path in a lap configuration. Why Other Options Are Wrong: Double shear arises in double-strap butt joints, not in lap joints. Pure bending only or “no load on rivets” is incorrect in a loaded lap joint. Alternating shear conditions do not occur in a single, uniform lap geometry. Common Pitfalls: Confusing lap joints with butt joints; assuming two planes exist because of two plates—only the overlapping interface matters here. Final Answer: Single shear

Question 20

Shear force sign change and bending moment: At a section where the shear force changes sign, what happens to the bending moment?

Accepted Answer

Introduction / Context: Shear force (V) and bending moment (M) are related by beam theory: dM/dx = V and dV/dx = w, where w is the load intensity. Recognizing the qualitative link between the sign of V and extrema of M is key in plotting diagrams quickly and accurately. Given Data / Assumptions: Prismatic beam with standard loading. Shear force diagram (SFD) crosses the zero line (sign change) at a particular section. Continuum mechanics relations apply (beam is linearly elastic within the analysis region). Concept / Approach: If dM/dx = V, then where V = 0, the slope of the bending moment diagram is zero, which indicates a local extremum (maximum or minimum) in M, provided the function is smooth there. Hence a sign change in V broadly marks a peak or valley in M, not necessarily a zero moment value. Step-by-Step Solution: Use dM/dx = V.At the section of interest, V = 0 (sign change implies crossing through zero).Therefore, dM/dx = 0 → M has an extremum at that section. Verification / Alternative check: Plot M for a simply supported beam with UDL: V crosses zero at midspan and M is maximum there. For a cantilever with point load at the tip, V does not change sign in span and M is monotonic—consistent with the rule. Why Other Options Are Wrong: Zero moment occurs at supports of simply supported beams, not necessarily where V changes sign. Infinity/undefined is non-physical here. Constancy of M implies V = 0 everywhere in that region, which is not the case around a sign change. Common Pitfalls: Assuming “zero shear implies zero moment.” The correct inference is an extremum of M, not necessarily M = 0. Final Answer: It attains an extremum (maximum or minimum) at that section

Strength of Materials Questions