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

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

Q: 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

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

Q: 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

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)

Q: 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.

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 1

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 2

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…

Accepted Answer

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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

Question 16

Square beam orientation effect: For the same allowable stress, the ratio of the moment of resistance when the square's sides are horizontal to that when its diagonal is horizontal equals what value?

Accepted Answer

Introduction / Context: The moment of resistance of a beam section at a given allowable stress is proportional to its section modulus. For a square, rotating the section by 45 degrees (diagonal horizontal) changes the extreme fiber distance even though the centroidal second moment of area about any axis through the center remains the same. This affects flexural capacity. Given Data / Assumptions: Square cross-section of side a. Linear elastic bending with the same allowable (design) stress in both orientations. Comparison between orientation with sides horizontal versus diagonal horizontal. Concept / Approach: Moment of resistance MR ∝ Z, where Z = I / y_max. For a square, I about a centroidal axis is a^4 / 12 regardless of rotation because I_x = I_y and I_xy = 0. However, y_max (distance from neutral axis to extreme fiber) differs by orientation: a/2 with sides horizontal, and a/√2 with diagonal horizontal. Step-by-Step Solution: Z_sides = (a^4 / 12) / (a / 2) = a^3 / 6Z_diag = (a^4 / 12) / (a / √2) = a^3 * √2 / 12Ratio = Z_sides / Z_diag = (a^3 / 6) / (a^3 * √2 / 12) = 2 / √2 = √2 Verification / Alternative check: Numerical check with a = 100 mm: compute Z for both orientations and take the ratio; the result is approximately 1.414. Why Other Options Are Wrong: 1 or 1/2 assumes no change in y_max or incorrect I.2 overestimates; 1/√2 reverses the true ratio. Common Pitfalls: Assuming I changes with rotation for a square; forgetting that extreme fiber distance changes with orientation, which controls Z. Final Answer: √2

Question 17

Definition of a fixed beam: a beam that is rigidly fixed at both of its ends (no rotation and no translation at supports).

Accepted Answer

Introduction / Context: Support conditions dictate bending moments and deflections in beams. A fixed (built-in) support restrains both rotation and translation, producing end moments even under uniform loading. Given Data / Assumptions: Linear elastic, prismatic beam. Ideal fixed ends with zero slope and zero deflection at supports. Transverse loading. Concept / Approach: A fixed beam has both ends built-in. This contrasts with simply supported beams (pinned/roller) where rotations are free, and cantilevers where one end is fixed and the other is free. Step-by-Step Solution: Identify end conditions: fixity at both ends ⇒ redundant end moments develop.Use fixed-end moment formulas for standard loads to analyze.Deflection is smaller than an equivalent simply supported beam due to added restraint. Verification / Alternative check: Compare slopes: in a fixed beam, slope at ends is zero; in simply supported beam, slope is generally non-zero. Why Other Options Are Wrong: One fixed end only defines a cantilever.Support type does not vary with loading for a given structural detail. Common Pitfalls: Confusing fixed with guided supports; assuming zero moment at ends when fixity exists. Final Answer: both

Question 18

Relationship between modulus of elasticity E, modulus of rigidity C (G), and Poisson’s ratio μ: select the correct formula.

Accepted Answer

Introduction / Context: In isotropic linear elasticity, only two independent elastic constants are required. The commonly used relations connect Young’s modulus E, shear modulus C (also denoted G), bulk modulus K, and Poisson’s ratio μ. Given Data / Assumptions: Material is homogeneous and isotropic. Linear elastic response. Standard definitions of E, C (G), μ. Concept / Approach: The fundamental isotropic elasticity relations include E = 2 C (1 + μ) and E = 3 K (1 − 2 μ). These allow conversion between sets of elastic constants based on measured data. Step-by-Step Solution: Start from generalized Hooke’s law for isotropic solids.Relate normal and shear strains to stresses using E, μ and C.Derive E = 2 C (1 + μ). Verification / Alternative check: Cross-check with E = 3 K (1 − 2 μ) and the identity K = E / 3 (1 − 2 μ) to ensure consistency. Why Other Options Are Wrong: E = 3 C (1 − 2 μ) confuses K with C.C = E (1 − 2 μ) / 2 (1 + μ) is algebraically equivalent to the correct formula, but the question asks for E as a function of C and μ; the simplest correct expression is E = 2 C (1 + μ).Other forms given are incorrect dimensionally or miss the 1 + μ factor. Common Pitfalls: Interchanging K and C; omitting μ; using plastic-range values where these linear relations no longer hold. Final Answer: E = 2 C (1 + μ)

Question 19

Bending moment–shear force relationship: the bending moment on a section is maximum at locations where the shear force is zero (i.e., where the SF diagram changes sign).

Accepted Answer

Introduction / Context: Shear force (SF) and bending moment (BM) diagrams are linked by differential relationships. Identifying where BM attains maxima or minima is critical for section design and reinforcement placement. Given Data / Assumptions: Beam under transverse loading. Linearly elastic behavior; small deflections. Continuous SF and BM functions except at point loads or couples. Concept / Approach: The relationship is dM/dx = V, where M is bending moment and V is shear force. Therefore, extremum of M occurs where dM/dx = 0, i.e., where V = 0. At those points the SF diagram crosses or touches the axis, often changing sign. Step-by-Step Solution: Start from dM/dx = V.Set derivative to zero for extremum: V = 0.Check second derivative dV/dx to distinguish maximum or minimum. Verification / Alternative check: For a simply supported beam with a central point load, SF changes sign at midspan and BM is maximum there. Similarly, under uniformly distributed load, SF passes through zero at midspan where BM peaks. Why Other Options Are Wrong: 'Minimum' or 'maximum' SF does not directly determine BM extremum; the zero condition does.'Changing sign' is often true at the same location, but the defining condition is V = 0. Common Pitfalls: Confusing zero shear with zero moment points; forgetting that point couples create jumps in M without V being zero. Final Answer: zero

Question 20

In strength of materials: when a bar experiences a temperature change but its thermal expansion or contraction is fully prevented, what do we call the stress that develops in the bar?

Accepted Answer

Introduction / Context: Engineers frequently encounter members that heat up or cool down while being restrained by connections or surrounding materials. If a bar tries to expand or contract due to temperature change but is prevented from deforming, internal forces develop. The name and nature of this stress are central to thermal stress analysis in bridges, piping, rails, and machine components. Given Data / Assumptions: Uniform, homogeneous bar. Temperature change occurs uniformly along the member. All free thermal expansion or contraction is fully prevented by rigid supports. Material behaves elastically within the stress range considered. Concept / Approach: A free bar subjected to a temperature rise ΔT would expand by ΔL_free = α * L * ΔT, where α is the coefficient of linear expansion. When expansion is fully restrained, the bar cannot change length, so mechanical stress develops to counter the thermal strain. This internally induced stress due only to restrained thermal strain is termed thermal stress. Step-by-Step Solution: Thermal strain (free) = α * ΔT.If fully restrained, total strain = 0 ⇒ mechanical strain = − α * ΔT.Hooke’s relation gives: sigma = E * (mechanical strain) = − E * α * ΔT.The sign (tension or compression) depends on whether the bar wants to expand (heated) or contract (cooled) but is restrained. Verification / Alternative check: The same conclusion arises from energy methods: the restraint does work against the tendency to thermally deform, storing strain energy associated with the induced thermal stress. Why Other Options Are Wrong: Tensile stress / Compressive stress: These may describe the sign in a given case but do not name the phenomenon; the proper general term is thermal stress.Shear stress: Not produced by purely axial thermal restraint in a straight prismatic bar.Residual stress: A broader term usually related to manufacturing or non-uniform processes; here the cause is temperature change with restraint—specifically thermal stress. Common Pitfalls: Assuming no stress if supports are “flexible”; partial restraint still induces thermal stress. Forgetting that sign depends on whether heating or cooling occurs. Final Answer: thermal stress

Strength of Materials Questions