Question 1

Helical springs – stiffness versus number of turns For a closely-coiled helical spring, the stiffness (k) varies with the number of turns (n) as:

Accepted Answer

Introduction / Context: Helical compression and tension springs are designed to achieve a target stiffness (spring rate). Understanding how geometry affects stiffness enables quick design adjustments. The number of active turns has a direct, simple influence on k. Given Data / Assumptions: Closely-coiled helical spring (helix angle small). Wire diameter d, mean coil diameter D, shear modulus G. n = number of active coils. Concept / Approach: The standard spring rate formula for a closely-coiled helical spring loaded axially is k = (G d^4) / (8 D^3 n). This shows linear inverse dependence on n, holding other parameters constant. Step-by-Step Solution: Start from k = (G d^4) / (8 D^3 n).Treat G, d, D as constants for a given design iteration.Therefore, k ∝ 1 / n. Increasing the number of active turns reduces stiffness. Verification / Alternative check: Doubling n halves k; halving n doubles k. This proportionality matches both analytical derivations and empirical spring catalogs. Why Other Options Are Wrong: Options claiming k ∝ n, n^2, or independent of n contradict the formula; k ∝ 1 / n^2 is also incorrect because the dependence is first power. Common Pitfalls: Forgetting to count only active coils (end coils may be inactive); changing D or d simultaneously and misattributing the effect to n. Final Answer: k is inversely proportional to n

Question 2

Transverse shear – comparing cross-sections of equal area The ratio of maximum shear stress in a rectangular beam to that in a circular beam of the same cross-sectional area is:

Accepted Answer

Introduction / Context: Transverse shear stress distributions depend on cross-sectional shape. For beams of equal area under the same shear force V, maximum shear differs by shape. Knowing these ratios helps in preliminary sizing and in selecting efficient sections. Given Data / Assumptions: One beam has a rectangular cross-section; the other is circular. Both have the same area A and carry the same shear force V. Elastic theory of shear in beams applies; small deformations. Concept / Approach: Maximum shear stress for a rectangle is τ_max(rect) = 1.5 * V / A. For a circle, τ_max(circ) = 4/3 * V / A. Taking the ratio eliminates V and A, leaving a pure number. Step-by-Step Calculation: τ_max(rect) = (3/2) * V / A.τ_max(circ) = (4/3) * V / A.Ratio = τ_max(rect) / τ_max(circ) = (3/2) / (4/3) = (3/2) * (3/4) = 9/8. Verification / Alternative check: Numerical example: choose V = 1 and A = 1. Then τ_max(rect) = 1.5 and τ_max(circ) = 1.333..., giving 1.125 = 9/8, confirming the ratio. Why Other Options Are Wrong: 2/3 and 3/4 are less than 1 and contradict known results; 1 would imply identical maxima; 4/3 in the choices is the circular multiplier itself, not the ratio asked. Common Pitfalls: Comparing shapes with unequal areas; confusing average shear V / A with maximum shear. Final Answer: 9 / 8

Question 3

Beam of triangular section (base horizontal) Where does the maximum transverse shear stress occur?

Accepted Answer

Introduction / Context: Shear stress in beams is not uniform and depends on the first moment of area Q at the level considered. For non-rectangular sections such as triangles, understanding where the maximum shear occurs is important for checking web stresses and potential shear failures. Given Data / Assumptions: Triangular cross-section with base horizontal; width varies linearly with height. Beam subjected to a transverse shear force V. Elastic theory applies; small deformations. Concept / Approach: Shear stress at a level y from a reference is τ = V Q / (I b), where Q is first moment of the area above (or below) the level about the neutral axis, I is second moment of area, and b is width at that level. For a triangle, b varies linearly with y, and Q is a quadratic function of y, so τ becomes a cubic-like function with a single maximum within the depth. Step-by-Step Reasoning (qualitative): 1) At the extreme fibers (base or apex), Q = 0 or b → 0, giving τ = 0 at both extremes.2) Between extremes, Q increases from zero, while b also changes; the ratio Q / b peaks roughly at the mid-depth.3) Detailed derivations show the maximum shear for a triangular section occurs at mid-height (i.e., halfway between base and apex). Verification / Alternative check: Carrying out the full τ(y) = V Q / (I b) derivation with b(y) linear in y and integrating to find Q(y) leads to a stationary point at mid-depth, which is the maximum τ. Why Other Options Are Wrong: Apex and base are extreme fibers where τ = 0; centroid is not the location of maximum τ for a triangle; one-third from base is not the maximum for this shape. Common Pitfalls: Assuming maximum shear always occurs at the neutral axis; for triangles, the neutral axis is at h/3 from the base, but τ_max is at mid-depth, not at the centroidal axis. Final Answer: At mid-height of the section

Question 4

Torsion of circular shafts – shear stress at the centre What is the shear stress at the centre of a solid circular shaft subjected to torsion?

Accepted Answer

Introduction / Context: In torsion of circular shafts, shear stress varies linearly with radius. Knowing where the maximum and minimum shear occur is vital for design against yielding and fatigue. Given Data / Assumptions: Solid circular shaft under torque T. Linear elastic behavior; Saint-Venant torsion theory. r = radial distance from the centre; R = outer radius. Concept / Approach: The torsion formula states τ(r) = T r / J, where J is the polar second moment of area (J = π R^4 / 2 for a solid circle). Shear stress is directly proportional to r, so it increases from the centre to the surface. Step-by-Step Reasoning: At r = 0 (the centre): τ(0) = T * 0 / J = 0.At r = R (outer surface): τ_max = T R / J, which is the maximum value.Therefore, the centre experiences zero shear stress under pure torsion. Verification / Alternative check: Shear strain distribution is γ(r) = r θ / L, also zero at the centre and maximum at the surface. Since τ = G * γ, τ follows the same radial linearity. Why Other Options Are Wrong: “Minimum but non-zero” conflicts with τ(0) = 0; “maximum” and “infinity” contradict the linear relation; “equal to average shear V / A” is a beam shear concept, unrelated to torsion at the centre. Common Pitfalls: Confusing torsional shear distribution with transverse shear in beams; forgetting that τ varies with radius in torsion. Final Answer: Zero

Question 5

Beam theory – shear stress at the neutral axis At the neutral axis of a beam subjected to transverse shear, the shear stress is:

Accepted Answer

Introduction / Context: Understanding the shear stress distribution across a beam depth is essential for web design, particularly in rectangular and I-sections. The neutral axis is the line of zero bending normal stress, but its role in shear distribution is different. Given Data / Assumptions: Beam carrying a transverse shear force V. Cross-sections like rectangles and I-beams with non-zero web thickness at the neutral axis. Elastic, small-deflection conditions. Concept / Approach: The shear stress at a level y is given by τ = V Q / (I b), where Q is the first moment of area about the neutral axis, I is the second moment of area, and b is the width at that level. For common sections, Q is maximum at the neutral axis; consequently, τ is also typically maximum there if b does not spike excessively. Step-by-Step Reasoning: At extreme fibers, Q = 0, hence τ = 0.Moving toward the neutral axis, Q increases, reaching its maximum at the neutral axis.Since b is finite at the neutral axis for rectangular and I-sections (web thickness), τ attains a maximum there. Verification / Alternative check: For a rectangle: τ_max = 1.5 * V / A and occurs at the neutral axis. For an I-beam, the maximum is in the web at the neutral axis region. Why Other Options Are Wrong: Zero or minimum applies at extreme fibers, not at the neutral axis; infinity is non-physical; “equal to bending stress” mixes different stress components—bending normal stress is zero at the neutral axis, not equal to shear stress. Common Pitfalls: Assuming “neutral axis” means all stresses are zero; only bending normal stress is zero there, not shear. Final Answer: Maximum

Question 6

Strength of materials — beam flexure terminology: What is the name given to the product of Young's modulus (E) and the second moment of area (I) of a beam cross-section?

Accepted Answer

Introduction / Context: In strength of materials, the bending stiffness of a prismatic beam is governed by both the material's elastic property and the cross-sectional geometry. The compound quantity that measures this bending stiffness is the product E * I, widely used in deflection and slope calculations. Given Data / Assumptions: E = Young's modulus (material stiffness in tension/compression). I = second moment of area about the neutral axis (geometric stiffness in bending). Small deflection theory (Euler–Bernoulli) is assumed. Concept / Approach: Beam curvature under bending follows M / I = E / R, which rearranges to M = (E * I) / R. The larger the product E * I, the smaller the curvature R for a given bending moment M. Hence E * I quantifies resistance to bending and is termed flexural rigidity. Step-by-Step Solution: Start from bending equation: M / I = E / R.Rearrange: M = (E * I) / R.Interpretation: For fixed M, higher E * I implies larger R (flatter curve) → stiffer in bending.Therefore, E * I is called flexural rigidity. Verification / Alternative check: Deflection formulas for standard cases (e.g., simply supported beam with central load) contain E * I in the denominator, confirming that deflection decreases as E * I increases. Why Other Options Are Wrong: Bulk modulus: volumetric elasticity, unrelated to flexure directly. Torsional rigidity: G * J for twisting, not bending. Modulus of rigidity: the shear modulus G alone, not multiplied by I. Section modulus: geometric term Z = I / y, not multiplied by E. Common Pitfalls: Confusing section modulus Z with second moment I; only E * I represents flexural rigidity. Final Answer: Flexural rigidity

Question 7

Reinforced concrete (RCC) beams — locating the actual neutral axis: State whether the following is correct: “The actual neutral axis in an RCC beam is found by the principle that the moment of areas of the compression and tension zones about the neu…

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Introduction / Context: In reinforced concrete flexural analysis, correctly locating the actual neutral axis (N.A.) is essential to determine stresses and to verify whether the section is under-reinforced, balanced, or over-reinforced. The criterion used must satisfy internal force equilibrium between concrete in compression and steel in tension (and compression steel if present). Given Data / Assumptions: Transformed section or modular ratio approach for elastic analysis. Concrete carries compression; tension is carried by steel (for singly reinforced sections). Plane sections remain plane (linear strain distribution). Concept / Approach: The location of the actual N.A. is obtained from the condition of axial force equilibrium: total compressive force in concrete equals total tensile force in steel (and compression steel if any). This is fundamentally a balance of forces, not a balance of first moments of area. First moments determine centroid locations, but the N.A. in RCC depends on material participation with different moduli, not merely geometrical areas. Step-by-Step Solution: Assume a trial N.A. depth x from extreme compression fibre.Compute compressive force C = ∫(stress_concrete) dA over the compression block (or C = σc_avg * b * x in elastic idealisation).Compute tensile force T = m * Es_contrib = m * σc_equiv * As (transformed steel area), or directly T = As * stress_steel from strain compatibility.Enforce equilibrium: C = T and solve for x → this gives the actual N.A. Verification / Alternative check: For doubly reinforced beams, include compression steel force C_s; still, the governing condition is the equality of internal tensile and compressive forces. Why Other Options Are Wrong: Agree/conditional: Moments of areas equality about N.A. is not the governing condition; force equilibrium is. “Depends on loading only”: The N.A. position depends on material properties and steel ratio, not just external loading pattern. Common Pitfalls: Confusing centroidal axis of a homogeneous section (pure geometry) with N.A. in RCC (composite, different moduli and cracking in tension). Final Answer: Disagree

Question 8

Columns — effective length equivalence: A column of actual length l with both ends fixed is equivalent in buckling strength to a column of what length with one end fixed and the other end free (cantilever)?

Accepted Answer

Introduction / Context: Buckling capacity depends on a column’s effective length, which incorporates end restraints. Converting between end conditions using effective length allows direct comparison of Euler loads for different support cases. Given Data / Assumptions: Euler buckling for long, perfectly straight columns. Effective length factor K: fixed–fixed K = 0.5; fixed–free (cantilever) K = 2.0. Pcr = π^2 * E * I / (K * l)^2. Concept / Approach: Set the Euler load for the fixed–fixed column equal to that of a cantilevered column of unknown length Lc. Equality of Pcr implies equality of effective lengths. Step-by-Step Solution: Fixed–fixed: Le1 = 0.5 * l.Cantilever: Le2 = 2 * Lc.Equate effective lengths: 0.5 * l = 2 * Lc → Lc = l / 4.Therefore, a fixed–fixed column of length l is equivalent to a fixed–free column of length l/4. Verification / Alternative check: Equate Euler loads explicitly: π^2 E I / (0.5 l)^2 = π^2 E I / (2 Lc)^2 → Lc = l/4, confirming the result. Why Other Options Are Wrong: l/8, l/2, l, 2l do not satisfy equality of effective lengths between the two cases. Common Pitfalls: Mixing K and C notations; remember C = K^2 if P = π^2 E I / (C l^2). Final Answer: l/4

Question 9

Thermal stresses in composite members during cooling: If a composite body experiences a fall in temperature, the member having the greater coefficient of linear expansion will be subjected to which type of stress?

Accepted Answer

Introduction / Context: When dissimilar materials are bonded and subjected to temperature changes, incompatible free thermal strains create internal forces. Recognizing the sign of thermal stress is important in bimetallic strips, composite bars, and reinforced concrete. Given Data / Assumptions: Two members are perfectly bonded (no slip), forming a composite bar. Temperature decreases by ΔT. α1 ≠ α2 (coefficients of linear expansion differ). Concept / Approach: Free thermal strain is ε_th = α * ΔT. On cooling (ΔT < 0), a member with larger α desires a larger contraction in magnitude. Bonding prevents differential contraction, so the larger-α member is restrained from contracting as much as it would freely, inducing tensile stress in it; the smaller-α member experiences compressive stress. Step-by-Step Solution: Compute free strains: ε1_free = α1 * ΔT, ε2_free = α2 * ΔT.With α1 > α2 and ΔT < 0, |ε1_free| > |ε2_free|.Compatibility forces prevent extra contraction of member 1 → member 1 is stretched relative to its free state → tensile stress in member with larger α.Member 2 is pushed to contract more than free state → compressive stress. Verification / Alternative check: Energy or force-balance formulations lead to equal and opposite internal forces; signs match the qualitative reasoning above. Why Other Options Are Wrong: Compressive stress in higher-α member is opposite of actual behavior under cooling. Pure shear/zero/bending-only are not consistent with axial composite action. Common Pitfalls: Reversing signs for heating versus cooling; under heating the higher-α member is in compression if constrained. Final Answer: Tensile stress (it wants to contract more but is restrained)

Question 10

Buckling of columns — parameters affecting Euler load: The critical (buckling) load for a given column depends on which set of properties?

Accepted Answer

Introduction / Context: Euler’s formula for long columns shows how geometry, material stiffness, and end conditions jointly govern buckling strength. Understanding each parameter's role is essential in member sizing. Given Data / Assumptions: Pcr = π^2 * E * I / (K * L)^2 (long, straight columns). I = A * k^2, where k is radius of gyration. Least radius of gyration controls the weakest axis buckling. Concept / Approach: Critical load scales linearly with E and I, and inversely with the square of effective length (K * L). Since I involves area A and radius k, area influences Pcr through the second moment of area. Thus multiple properties simultaneously affect buckling. Step-by-Step Solution: Write Euler: Pcr = π^2 * E * I / (K * L)^2.Substitute I = A * k^2 → Pcr ∝ E * A * k^2 / (K^2 * L^2).Identify dependencies: E, A, k (least), and L (with end condition factor K).Therefore, all listed geometric and material factors matter. Verification / Alternative check: Comparing two sections with same area but different k (e.g., solid vs thin-walled tube) shows different Pcr, confirming dependence on k in addition to A. Why Other Options Are Wrong: Single-factor answers ignore key contributors. “Only end load magnitude” is irrelevant; buckling capacity is a property of the member. Common Pitfalls: Equating larger area with higher buckling load without considering how area is distributed (k). Final Answer: All of the above (area through I, length, least radius of gyration, and E)

Question 11

Thin cylindrical shells with riveted joints: For a riveted cylindrical shell of diameter d and plate thickness t under internal pressure p, with joint efficiency η, the hoop (circumferential) stress is given by

Accepted Answer

Introduction / Context: Thin pressure vessels experience two principal membrane stresses: hoop (circumferential) and longitudinal. When a longitudinal riveted joint is present, its efficiency reduces the effective load-carrying thickness against hoop stress. Given Data / Assumptions: Thin cylinder (t ≪ d), uniform internal pressure p. Riveted longitudinal joint with efficiency η (0 < η ≤ 1). Membrane theory (neglecting bending and stress concentrations). Concept / Approach: For a seamless thin cylinder, hoop stress is σh = p * d / (2 * t). With a longitudinal joint of efficiency η, the effective resisting thickness is η * t, so the hoop stress increases to σh = p * d / (2 * η * t). Step-by-Step Solution: Start: σh(seamless) = p * d / (2 * t).Introduce joint efficiency: t_effective = η * t.Hence σh = p * d / (2 * t_effective) = p * d / (2 * t * η).Select the option matching this expression. Verification / Alternative check: Longitudinal stress for a seamless thin cylinder is σL = p * d / (4 * t). The presence of a circumferential joint would affect σL instead; here the longitudinal joint affects hoop stress as shown. Why Other Options Are Wrong: p * d / (4 * t * η) corresponds to longitudinal stress with joint efficiency, not hoop stress. p * d / (t * η) and 2 * p * d / (t * η) overstate the stress by missing the factor 2 or adding an extra 2. p / (2 * t * η * d) has wrong dependence on diameter. Common Pitfalls: Mixing hoop and longitudinal stress formulas; always note the factor 2 difference. Final Answer: σh = p * d / (2 * t * η)

Question 12

Fatigue testing — types of stress-time variations covered: Fatigue tests are carried out under which of the following patterns of fluctuating stress?

Accepted Answer

Introduction / Context: Fatigue is failure under repeated or fluctuating stresses. Materials may fail at stress levels far below static strength when subjected to many cycles. Fatigue testing therefore uses different stress-time histories to represent service conditions. Given Data / Assumptions: Low-cycle and high-cycle fatigue regimes are possible. Mean stress and stress amplitude both influence life (e.g., Goodman, Soderberg diagrams). Common loading patterns: fully reversed, fluctuating with mean tensile/compressive stress, and pulsating (one sign only). Concept / Approach: Fatigue life depends on the combination of stress amplitude and mean stress. Hence testing protocols include multiple patterns: fully reversed (σmax = −σmin), alternating about a nonzero mean (unequal opposite limits), and pulsating (σmin ≥ 0 or σmax ≤ 0). Step-by-Step Solution: Recognize that fatigue damage accumulates with cycles under varied mean stresses.Identify standard test types matching options (a), (b), and (c).Conclude that all listed types are used in fatigue testing. Verification / Alternative check: S–N curves are often generated for fully reversed loading; modified approaches introduce mean stress corrections to cover unequal limits and pulsating loads. Why Other Options Are Wrong: Restricting to only one loading pattern ignores the real variety of service stresses. Common Pitfalls: Assuming fatigue is synonymous only with fully reversed bending; in practice, many components operate with substantial mean stress. Final Answer: All of the above

Question 13

Euler’s formula — end condition constant: In the form p = π^2 * E * I / (C * l^2), the value of C for a column with both ends fixed is

Accepted Answer

Introduction / Context: Euler’s buckling load can be expressed using an end-condition constant C such that Pcr = π^2 * E * I / (C * l^2). The constant C encodes the support restraints via the effective length factor. Given Data / Assumptions: Effective length Le = K * l. Pcr = π^2 * E * I / Le^2 = π^2 * E * I / (K^2 * l^2). Therefore, C = K^2. Concept / Approach: For both ends fixed, K = 0.5. Hence C = K^2 = 0.25 = 1/4. This gives the largest buckling load among common end conditions due to the shortest effective length. Step-by-Step Solution: Identify K for fixed–fixed = 0.5.Compute C = K^2 = (0.5)^2 = 0.25.Thus C = 1/4. Verification / Alternative check: Check other cases: pinned–pinned K = 1 → C = 1; fixed–free K = 2 → C = 4; fixed–pinned K ≈ 0.7 → C ≈ 0.5. These confirm consistency. Why Other Options Are Wrong: 4 corresponds to fixed–free, not fixed–fixed. 1 is for pinned–pinned; 1/2 is not a standard C; 2 is incorrect for any basic case. Common Pitfalls: Mixing effective length K with its square C when switching formula forms. Final Answer: 1/4

Question 14

Close-coiled helical spring under axial load: For a closely-coiled helical spring of mean coil diameter D, wire diameter d, n turns, modulus of rigidity C (G), and axial load W, the deflection δ is

Accepted Answer

Introduction / Context: Close-coiled helical springs store energy primarily by twisting of the wire. The load–deflection relation links geometry (D, d, n) and material shear modulus to the axial compliance. Given Data / Assumptions: Close-coiled (helix angle small), pure torsion of wire dominates. Axial load W causes torque in each coil. Modulus of rigidity C (often denoted G) is constant. Concept / Approach: Each coil behaves like a torsion bar of circular section. Summing angular twist over n coils and converting to axial deflection yields the well-known expression δ = 8 * W * D^3 * n / (C * d^4). Step-by-Step Solution: Torque in wire: T = W * D / 2 ≈ W * (D/2). Often expressed with mean radius R = D/2.Angle of twist per coil: θ = T * l_w / (C * J), with wire polar J = π * d^4 / 32 and l_w ≈ π * D.Axial deflection: δ = n * θ * (pitch angle small) → simplifies to δ = 8 * W * D^3 * n / (C * d^4).Select the matching option. Verification / Alternative check: Dimensional check: δ has units of length; numerator W * D^3 * n divided by C * d^4 (stress per shear-strain scale) yields length. Why Other Options Are Wrong: Other powers of D and d do not arise from torsion theory; they would give incorrect stiffness scaling. Common Pitfalls: Using E instead of C (G); mixing up D (mean diameter) with radius; forgetting the power d^4 in the denominator. Final Answer: δ = 8 * W * D^3 * n / (C * d^4)

Question 15

Materials testing — ductility index by percentage elongation: A tensile test shows 30% elongation for material A and 40% elongation for material B (same gauge length and cross-section). Conclude whether B is more ductile than A.

Accepted Answer

Introduction / Context: Ductility quantifies the extent of plastic deformation a material can sustain before fracture. Common indices are percentage elongation and percentage reduction in area, both measured on standard tensile specimens. Given Data / Assumptions: Two specimens of identical dimensions are tested to fracture. Percentage elongation for A = 30%, for B = 40%. Standard test conditions assumed identical (strain rate, temperature). Concept / Approach: Greater percentage elongation indicates higher ductility because the specimen undergoes more permanent extension prior to fracture. Therefore, with all else equal, material B is more ductile than A. Step-by-Step Solution: Define ductility metric: % elongation = (Lf − L0) / L0 * 100%.Given: 30% for A, 40% for B.Compare: 40% > 30% → B exhibits greater plastic strain capacity.Conclude statement is correct. Verification / Alternative check: For many metals, higher % reduction in area accompanies higher % elongation; both indices generally correlate with ductility. Even if %RA differs, the given elongation already supports the conclusion under identical conditions. Why Other Options Are Wrong: “Incorrect” contradicts the definition of the ductility index used. Requiring %RA or limiting to brittle materials is unnecessary; elongation alone is a valid ductility measure. Common Pitfalls: Comparing elongations from different gauge lengths or non-standard specimens; always compare like with like. Final Answer: Correct

Question 16

Lame’s theory is primarily associated with stress distribution in thick cylindrical shells (pressure vessels with significant wall thickness).

Accepted Answer

Introduction / Context: Lame’s theory provides the fundamental solution for radial and hoop (circumferential) stresses in thick-walled cylinders under internal and/or external pressure. Unlike thin-shell formulas that assume uniform membrane stress through the thickness, thick shells exhibit strong stress gradients that must be captured for safe design. Given Data / Assumptions: Cylinder with inner and outer radii; wall thickness is not negligible. Material assumed homogeneous, isotropic, and linearly elastic. Axisymmetric loading due to internal pressure (and optionally external pressure). Concept / Approach: Lame's equations resolve the stress field as functions of radius r. The general forms are sigma_r = A − B / r^2 and sigma_theta = A + B / r^2, where constants A and B are determined from boundary conditions at the inner and outer surfaces. This captures maximum hoop stress near the inner wall and decreasing magnitude toward the outside surface. Step-by-Step Solution: 1) Assume sigma_r(r) = A − B / r^2 and sigma_theta(r) = A + B / r^2.2) Apply boundary conditions: sigma_r(r_i) = −p_i, sigma_r(r_o) = −p_o.3) Solve for A and B in terms of p_i, p_o, r_i, r_o.4) Evaluate sigma_theta(r) to identify critical (maximum) stresses at r = r_i. Verification / Alternative check: Compare to thin-shell formula sigma_h = p d / (2 t). For small t relative to diameter, Lame’s solution approaches the thin-shell result; for larger thickness, the gradient predicted by Lame is essential. Why Other Options Are Wrong: Thin cylindrical shells: use membrane theory; not Lame's full-field solution.Direct and bending stresses: beam theory, not pressure vessels.'None of these' is incorrect because thick cylinders are explicit domain of Lame's theory. Common Pitfalls: Using thin-wall equations for thick vessels; ignoring inner-wall peak hoop stress; neglecting external pressure effects. Final Answer: thick cylindrical shells

Question 17

In riveted joints, the center-to-center distance between two consecutive rivets in the same row is called the pitch of rivets.

Accepted Answer

Introduction / Context: Riveted and bolted joints are laid out using standard geometric terms. Knowing the correct terminology—pitch, margin, back pitch, and diagonal pitch—is essential for strength checks against tearing, shearing, and bearing. Given Data / Assumptions: Plate joint with one or more rows of rivets. Uniform rivet spacing within a given row. Definitions as used in machine design and fabrication standards. Concept / Approach: The pitch is defined as the center-to-center distance between adjacent rivets along the same row. Other distances measure from rivet center to plate edge (margin), or between adjacent rows (back pitch), or along a diagonal between staggered rows (diagonal pitch). Step-by-Step Solution: Identify the same row of rivets.Measure the linear center distance along that row between two consecutive rivet centers.That measured value is the rivet pitch. Verification / Alternative check: Check a standard joint sketch: symbols p (pitch), m (margin), pb (back pitch), pd (diagonal pitch) are conventionally labeled. Why Other Options Are Wrong: Margin is center-to-near-edge distance.Back pitch is the perpendicular spacing between rows.Diagonal pitch is the center distance between staggered rows measured along a diagonal. Common Pitfalls: Confusing pitch and gauge; using edge distance in place of pitch; measuring across different rows. Final Answer: pitch

Question 18

Define the factor of safety (FoS) used in machine design and structural engineering: choose the correct ratio.

Accepted Answer

Introduction / Context: The factor of safety (FoS) provides a margin between the limiting material strength and the expected working (allowable) stress during service. It accounts for uncertainties in loading, material variability, and analysis assumptions. Given Data / Assumptions: Allowable (working) stress is used for design checks. Limiting strength may be ultimate or yield, depending on design philosophy. Linear elastic behavior assumed up to the allowable. Concept / Approach: By definition, FoS = limiting stress / working stress. For ductile design against yielding, some codes use yield stress; for ultimate strength design, ultimate stress is used. The classical general definition uses ultimate stress in the numerator. Step-by-Step Solution: FoS = sigma_limiting / sigma_workingFor classical definition here: sigma_limiting = ultimate stress.Therefore, FoS = ultimate stress / working stress. Verification / Alternative check: Confirm with design codes: allowable stress = limiting stress / FoS ⇒ rearranging yields FoS as above. Why Other Options Are Wrong: Working/ultimate is the reciprocal.Breaking/ultimate or ultimate/breaking are ill-defined for ductile design and do not match standard FoS definition given. Common Pitfalls: Mixing yield-based and ultimate-based FoS; ignoring different FoS for different failure modes (buckling, fatigue). Final Answer: ultimate stress to working stress

Question 19

Torsional rigidity of a shaft: it is the torque required to produce a twist of one radian per unit length of the shaft. State whether this definition is correct.

Accepted Answer

Introduction / Context: Torsional rigidity quantifies a shaft's resistance to twisting under applied torque. It is fundamental for drive shafts, transmission lines, and rotary machine elements where angular deflection control is important. Given Data / Assumptions: Uniform, prismatic shaft of length L. Linear elastic behavior and Saint-Venant torsion assumption. Polar moment of inertia J depends on cross-section. Concept / Approach: Torsion relation: theta = T L / (G J), where theta is the twist in radians, T is torque, G is shear modulus, and J is polar moment. Rearranging defines torsional rigidity G J as the torque per unit twist per unit length. Step-by-Step Solution: theta = T L / (G J)Twist per unit length: theta / L = T / (G J)Torque to produce unit twist per unit length = G J Verification / Alternative check: Dimensional check: G (stress) * J (length^4) gives torque per radian per unit length. Why Other Options Are Wrong: Depends on cross-section (solid or hollow) through J; definition remains valid for both.Using degrees would require conversion; radians are standard. Common Pitfalls: Confusing torsional rigidity (GJ) with bending rigidity (EI); forgetting that J differs significantly between solid and hollow shafts for the same area. Final Answer: True

Question 20

For a rectangular beam under transverse loading (sagging), where is the maximum tensile stress located across the depth?

Accepted Answer

Introduction / Context: In beam bending, the sign and distribution of normal stress depend on the bending moment sense (sagging or hogging). Correctly locating tensile and compressive zones is vital for reinforcement layout and checking tensile strength. Given Data / Assumptions: Rectangular prismatic beam, linearly elastic. Sagging moment (concave up) due to transverse loads. Small deflection theory; plane sections remain plane. Concept / Approach: Bending stress varies linearly with distance y from the neutral axis: sigma = M y / I. For sagging, the top fibers are in compression (negative y), the bottom fibers in tension (positive y). Maximum magnitudes occur at the extreme fibers farthest from the neutral axis. Step-by-Step Solution: Neutral axis passes through the centroid; sigma = 0 at NA.Magnitude increases linearly with |y|, peaking at the extremes.For sagging: bottom extreme fiber carries maximum tensile stress. Verification / Alternative check: Sign convention check using sagging positive: compression at top, tension at bottom; textbook stress blocks confirm. Why Other Options Are Wrong: Top layer is compressive under sagging.Neutral axis has zero normal stress.Stress is not uniform across section depth. Common Pitfalls: Confusing sagging versus hogging; misplacing neutral axis in non-symmetric sections. Final Answer: bottom layer

Strength of Materials Questions