Question 1

Elastic energy terms: identify correct statements about strain energy, proof resilience, and modulus of resilience  Consider the standard definitions used in strength of materials. Select the correct group of statements.

Accepted Answer

Definitions Strain energy: total elastic energy stored in a body under load (recoverable).Proof resilience: maximum strain energy up to elastic limit.Modulus of resilience: proof resilience per unit volume. Final Answer All of the above statements are correct.

Question 2

Beam theory: meaning of point of contraflexure (inflection point)  On a bending moment diagram for a beam, identify what occurs at the point of contraflexure.

Accepted Answer

Concept The point of contraflexure (inflection) is where the curvature changes sign; equivalently, the bending moment diagram crosses zero. Final Answer Bending moment changes sign (passes through zero).

Question 3

Solid circular shaft: torque–diameter relation at allowable shear stress  For a solid circular shaft of diameter D, limited by maximum allowable shear stress τ, state the formula for the maximum torque transmitted.

Accepted Answer

Concept/Formula For a solid circular shaft: T = (τ/J)·(J·R) = τ·(J/R).Polar moment J = (π/32)·D⁴, outer radius R = D/2 ⇒ J/R = (π/16)·D³.Thus, T = (π/16)·τ·D³. Final Answer (π/16)·τ·D³.

Question 4

Stability of columns: applicability of Euler's buckling formula  State for which type of column Euler's critical load formula is valid, based on slenderness.

Accepted Answer

Concept Euler's buckling formula assumes elastic instability and is valid for columns with high slenderness ratio (long/slender), where buckling governs before material yield or crushing. Final Answer Long (slender) columns.

Question 5

Plane stress analysis: compute the maximum normal stress (principal stress) for a biaxial state with shear A body is under σ_x = 1200 MPa (tension) and σ_y = 600 MPa (tension) on mutually perpendicular planes. A shear stress τ

Accepted Answer

Given σx = 1200 MPa, σy = 600 MPa, τxy = 400 MPa. Approach Principal stresses for plane stress: σ1,2 = (σx + σy)/2 ± √{ [ (σx − σy)/2 ]² + τxy² }. Step-by-step calculation σavg = (1200 + 600)/2 = 900 MPa.Δ = (σx − σy)/2 = (1200 − 600)/2 = 300 MPa.Radical = √(300² + 400²) = √(90,000 + 160,000) = √250,000 = 500 MPa.σ1 = 900 + 500 = 1400 MPa; σ2 = 900 − 500 = 400 MPa. Verification Mohr’s circle would have center at 900 MPa and radius 500 MPa, consistent with σ1 and σ2 above. Final Answer 1400 MPa.

Question 6

Thick cylinder (Lame’s theory): Maximum tangential (hoop) stress at the inner surface due to internal pressure.  Given: a thick cylindrical shell with inner radius r_i, outer radius r_o, internal pressure p, and negligible external pressure. Choose …

Accepted Answer

Given data Thick cylinder with inner radius r_i and outer radius r_o. Internal pressure = p; external pressure ≈ 0. Concept/Approach Use Lame’s equations for thick cylinders: σ_r = A − B/r^2 and σ_θ = A + B/r^2. Step-by-step derivation Boundary 1 (at r = r_i): σ_r(ri) = −p ⇒ −p = A − B/r_i^2.Boundary 2 (at r = r_o): σ_r(ro) = 0 ⇒ 0 = A − B/r_o^2 ⇒ A = B/r_o^2.Substitute A into Boundary 1: −p = (B/r_o^2) − (B/r_i^2) = B,[1/r_o^2 − 1/r_i^2].Solve for B: B = p, r_i^2 r_o^2 / (r_o^2 − r_i^2).Then A = B/r_o^2 = p, r_i^2 / (r_o^2 − r_i^2).Hoop stress at the inner surface: σ_θ(ri) = A + B/r_i^2 = p(r_o^2 + r_i^2)/(r_o^2 − r_i^2). Common pitfalls Mixing up radial and hoop stresses or sign conventions leads to wrong expressions. Final Answer σ_θ(ri) = p × (r_o^2 + r_i^2) / (r_o^2 − r_i^2)

Question 7

Composite torsional shaft: Equivalent torsional spring constant for two serial segments between rigid ends.  Given two stepped portions with torsional spring constants k₁ and k₂ in series (same torque, angular twists add). Choose the correct express…

Accepted Answer

Given data Two shaft segments in series with torsional spring constants k₁ and k₂. Concept/Approach For springs in series (same torque T), total twist θ = θ₁ + θ₂ = T/k₁ + T/k₂. Step-by-step Define k_eq by T/θ = k_eq ⇒ 1/k_eq = 1/k₁ + 1/k₂.Therefore, k_eq = (k₁ k₂)/(k₁ + k₂). Verification If k₁ = k₂ = k, then k_eq = k/2 (softer than either alone), which is consistent for series combination. Final Answer k_eq = (k₁ k₂)/(k₁ + k₂)

Question 8

Strength of materials (simple bending theory): In the common beam assumptions, “the material is isotropic” means which of the following?

Accepted Answer

Introduction / Context: Classical bending theory relies on simplifying assumptions about material behavior and geometry. One key assumption is isotropy, which impacts how stress and strain relate in different directions when a beam bends. Given Data / Assumptions: Homogeneous, linearly elastic material. Plane sections remain plane and normal to the neutral axis after bending. Small deformations and a straight prismatic member. Concept / Approach: Isotropic materials have identical mechanical properties in all directions at a point. Practically, constants such as Young’s modulus E and shear modulus G are direction-independent. This ensures that the linear stress distribution from bending is valid irrespective of orientation, provided other assumptions hold. Step-by-Step Solution: State bending relation: sigma = My / I for linear elastic isotropic material.Because E is uniform and direction independent, strain compatibility leads to linear stress variation across depth.The neutral axis passes through the centroid for symmetrical sections with isotropic materials.Results such as curvature k = M / (EI) use a single E due to isotropy. Verification / Alternative check: Compare with orthotropic materials (e.g., composites); E differs by direction, violating isotropy and changing bending response. Why Other Options Are Wrong: Normal stress constant in all directions: not a definition of isotropy and not true in bending. Normal stress varies linearly: that is a bending result, not the definition of isotropy. Elastic constants vary linearly: contradicts isotropy; that would be inhomogeneity. Common Pitfalls: Equating isotropy with homogeneity; they are related but distinct—homogeneity means properties are the same at every point, isotropy that they are direction independent at a point. Final Answer: Elastic constants are the same in all directions

Question 9

Welding vs riveted joints: Compared to an equivalent riveted joint, a properly designed welded joint has what relative strength under static loading?

Accepted Answer

Introduction / Context: Permanent joints connect machine or structural elements. Riveted joints were once standard, but welding often replaces them due to weight savings, strength, and fabrication speed. Knowing the comparative strength guides design choices. Given Data / Assumptions: Joints are designed to good practice (adequate weld size, correct electrode/procedure, quality control). Static loading comparison; fatigue and inspection concerns are considered separately. Parent material properties are compatible with the weld process. Concept / Approach: Welding creates near-continuous joints that distribute load across the entire faying surface. Riveting introduces holes that reduce net section and create stress concentrations. Thus, for similar geometry and materials, a welded joint can achieve higher net static strength at lower weight. Step-by-Step Solution: Note that riveted plates have reduced net area A_net = bt − nd*t due to holes.Welded joints maintain gross area and can be sized (throat thickness) to develop parent-metal strength.Under static tension or shear, the effective capacity of a proper weld typically exceeds that of comparable riveted patterns.Result: “More” strength for welded joints under the stated conditions. Verification / Alternative check: Historical transitions in bridges/boilers/shipbuilding from rivets to welding were driven by higher strength-to-weight and lower fabrication time. Why Other Options Are Wrong: Same: ignores net-section loss in riveted joints. Less: only true if welding is defective; not an inherent limitation. Common Pitfalls: Confusing static strength with fatigue performance; weld toes can be fatigue-critical if not designed properly. Final Answer: More

Question 10

Mohr’s circle / stress transformation: A body is subjected to two mutually perpendicular normal stresses of 20 kN/m² (tension) and 10 kN/m² (compression). What is the maximum shear stress?

Accepted Answer

Introduction / Context: When plane stress involves two perpendicular normal stresses with no shear on the reference planes, principal and maximum shear stresses can be computed directly. This is a core application of Mohr’s circle and stress transformation equations. Given Data / Assumptions: Sigma_x = +20 kN/m² (tensile). Sigma_y = −10 kN/m² (compressive). Tau_xy on the reference planes = 0. Concept / Approach: For plane stress with perpendicular normal stresses: maximum shear stress tau_max equals half the difference between the principal stresses. With sigma_1 and sigma_2 as the principal stresses, tau_max = (sigma_1 − sigma_2) / 2. Here, sigma_1 = +20 and sigma_2 = −10. Step-by-Step Solution: Identify principal stresses: sigma_1 = +20, sigma_2 = −10.Compute difference: sigma_1 − sigma_2 = 20 − (−10) = 30 kN/m².Compute maximum shear: tau_max = 30 / 2 = 15 kN/m².Report the result with correct units. Verification / Alternative check: Draw Mohr’s circle: center at (sigma_avg, 0) where sigma_avg = (20 − 10)/2 = 5; radius R = 15. Tau_max equals the radius, confirming 15 kN/m². Why Other Options Are Wrong: 5 or 10 kN/m²: result from taking half the absolute values incorrectly or from using sigma_avg. 20 kN/m²: corresponds to the larger principal stress, not the shear radius. Common Pitfalls: Forgetting the sign of compressive stress; using absolute values without considering direction; mixing up average normal stress with shear magnitude. Final Answer: 15 kN/m²

Question 11

Beam bending (cantilever with end load): When a cantilever beam is loaded at its free end by a downward force, where does the maximum compressive bending stress occur at the fixed support cross-section?

Accepted Answer

Introduction / Context: Bending stress distribution in beams depends on load type and support conditions. For cantilevers, identifying the fibres in tension versus compression is key for safe design and detailing (e.g., reinforcement placement in concrete). Given Data / Assumptions: Prismatic cantilever beam fixed at one end and free at the other. Downward point load applied at the free end. Linear elastic behavior; small deflections. Concept / Approach: Under a downward end load, the cantilever deflects downward, producing a sagging curvature (concave upward). In sagging, the top fibres are in compression and bottom fibres are in tension. Bending stress magnitude at a section is sigma = My / I, largest at extreme fibres (maximum y). Step-by-Step Solution: Compute bending moment at a distance x from the free end: M(x) = Px; maximum at the fixed end where x = L.At the fixed end, extreme fibres carry the largest stress magnitude.With sagging curvature from downward loading, top fibres are compressive, bottom tensile.Therefore the maximum compressive stress is at the top fibre at the fixed support. Verification / Alternative check: Draw a qualitative bending moment diagram: a triangle from zero at free end to maximum at the fixed end; check sign conventions—sagging indicates compression at the top for a horizontal beam with downward load. Why Other Options Are Wrong: Bottom fibre: carries maximum tension, not compression. Neutral axis: zero bending stress by definition. Centre of gravity: coincides with neutral axis for symmetric sections; stress is zero there. Common Pitfalls: Confusing sign conventions for sagging/hogging; assuming the same behaviour as simply supported beams at midspan without considering boundary conditions. Final Answer: Top fibre

Question 12

Design concept: A beam of “uniform strength” is characterised by which of the following conditions along its length under a given bending moment distribution?

Accepted Answer

Introduction / Context: The idea of a “beam of uniform strength” is used to design lightweight members by varying the cross-section so that allowable stress is reached everywhere, fully utilising material while avoiding overdesign. Given Data / Assumptions: Bending stress formula applies: sigma = M*y / I. Material is linearly elastic with allowable stress sigma_allow. Geometry can vary along the length (e.g., variable depth or width). Concept / Approach: Uniform strength means the maximum fibre stress equals the allowable stress at every section. Because bending moment M varies along the span, the section modulus Z = I / y must be varied so that sigma = M / Z = constant (equal to sigma_allow). This yields tapered or cut-out sections that are lighter than prismatic beams. Step-by-Step Solution: Start from sigma = M / Z.Impose sigma = sigma_allow (constant) along the length.Conclude Z(x) = M(x) / sigma_allow: increase Z where M is high; reduce Z where M is low.Translate Z(x) into geometric dimensions (depth/width) for a chosen profile. Verification / Alternative check: For a cantilever with end load, M rises linearly to the fixed end; a triangular (tapered) depth keeping Z proportional to M provides near-uniform stress along the length. Why Other Options Are Wrong: Same cross-section: defines a prismatic beam, not uniform strength under varying moments. Same bending moment: depends on loading, not a design property of the section. Same shear stress: not the principle behind uniform strength; shear varies differently and usually governs near supports in deep webs. Common Pitfalls: Ignoring stress concentrations and local buckling when tapering; confusing section modulus with second moment of area alone. Final Answer: Same bending stress at every section

Question 13

On the standard engineering stress–strain curve for mild steel under tension, what is the nature of the stress at point D (the peak of the curve before necking begins)?

Accepted Answer

Introduction / Context: In materials testing for mechanical engineering, the tensile test produces a characteristic stress–strain curve. Recognising key points on this curve—proportional limit, yield points, ultimate tensile stress, and fracture—is essential for safe design and failure prevention. Given Data / Assumptions: A standard tensile test on a ductile (mild steel–like) specimen. Engineering stress–strain curve is considered (load divided by original area). Point D is the highest point on the curve before the final drop to fracture. Concept / Approach: The curve starts linear (Hooke’s law), then yields (upper/lower yield), then strain hardening increases stress until a maximum is reached (ultimate tensile stress). After this peak, local necking causes a reduction in load-carrying cross-section and the measured engineering stress drops until fracture. Step-by-Step Solution: Identify the linear region ending at proportional limit and elastic limit.Locate yield region (upper and lower yield points) for low-carbon steel.Observe strain hardening region leading to a peak stress value.Recognise that the highest stress reached (point D) is the ultimate tensile stress.Note the subsequent drop due to necking, ending at fracture stress. Verification / Alternative check: The ultimate tensile stress is always the maximum engineering stress recorded in the test; fracture stress is lower for ductile metals due to necking. Why Other Options Are Wrong: Yield point stress: occurs earlier, before strain hardening.Breaking stress: occurs at the end, after necking and load drop.Elastic limit: much earlier in the curve; end of purely elastic behaviour. Common Pitfalls: Confusing ultimate tensile stress with fracture stress; mixing engineering and true stress where the true stress can still rise after necking. Final Answer: Ultimate tensile stress

Question 14

Two shafts of the same material: Shaft A is solid with diameter D; Shaft B is hollow with outer diameter D and inner diameter D/2. Comparing torsional strength (torque capacity at the same allowable shear stress), the hollow shaft is what fraction o…

Accepted Answer

Introduction / Context: Torsion design often favours hollow shafts because material is more effective away from the centre. This question compares the torsional strength (torque for a given maximum shear stress) of a solid versus a hollow circular shaft with the same outer diameter and material. Given Data / Assumptions: Material and allowable shear stress are the same. Outer diameter for both shafts is D; hollow shaft has inner diameter D/2. Strength comparison is based on torque at the same maximum shear stress. Concept / Approach: For circular shafts, torsional strength T is related to polar section modulus Zp: T = τ_max * Zp. For a solid shaft, Zp_solid = (π/16) * D^3. For a hollow shaft, Zp_hollow = J/R = [(π/32) * (D^4 − d^4)] / (D/2). Step-by-Step Solution: Let d = D/2 → d^4 = (D/2)^4 = D^4 / 16.Compute J_hollow = (π/32) * (D^4 − D^4/16) = (π/32) * (15/16) * D^4.Compute Zp_hollow = J_hollow / (D/2) = (π/16) * (15/16) * D^3.Compute Zp_solid = (π/16) * D^3.Strength ratio = Zp_hollow / Zp_solid = 15/16. Verification / Alternative check: Because most torque resistance comes from outer fibres, removing the core barely reduces strength: 15/16 ≈ 0.9375, confirming high efficiency of hollow sections. Why Other Options Are Wrong: 1/16, 1/8, 1/4 drastically underestimate strength; they ignore the outer-radius dominance in torsion. Common Pitfalls: Comparing polar moments J (stiffness-related) instead of polar modulus Zp (strength-related); forgetting to divide J by outer radius for strength. Final Answer: 15/16

Question 15

Mohr’s circle interpretation: The endpoints of any diameter on Mohr’s circle represent stresses on which planes?

Accepted Answer

Introduction / Context: Mohr’s circle is a powerful graphical tool for transforming plane stresses. Correctly reading what points and diameters represent is key to finding stresses on inclined planes without laborious algebra. Given Data / Assumptions: State of plane stress is represented by a Mohr’s circle. A diameter is drawn through the circle with endpoints on the circumference. We relate these endpoints to stresses on real (physical) planes in the body. Concept / Approach: Each point on Mohr’s circle corresponds to the pair (σ_n, τ_nt) acting on some plane through a point. Two points connected by a diameter correspond to two orthogonal (mutually perpendicular) physical planes. Principal stresses are a special case where τ = 0 (points at horizontal intercepts). Step-by-Step Solution: Select any reference plane; find its stress point on Mohr’s circle.Construct the diameter through the circle’s centre to the opposite point.Interpretation: the two endpoints give (σ, τ) for two perpendicular planes in the material.Recognise that rotating the physical plane by θ rotates the Mohr’s radius by 2θ. Verification / Alternative check: Transformations show that stress components on orthogonal planes are located 180° apart on Mohr’s circle, i.e., at endpoints of a diameter. Why Other Options Are Wrong: Principal stresses only: only true at τ = 0 intercepts.Options limited to “45° planes”: too restrictive; a diameter is general. Common Pitfalls: Mistaking principal points for arbitrary diameter endpoints; forgetting the 2θ relationship between physical and Mohr rotations. Final Answer: The normal and shear stresses on two mutually perpendicular physical planes

Question 16

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 17

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 18

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

Accepted Answer

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 19

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 20

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)

Strength of Materials Questions