Q: In brake and clutch design, what is the primary function of the springs installed in the mechanism?

Introduction / Context: Springs are integral to friction brakes and clutches used in automotive and industrial drives. Understanding their primary role ensures correct selection of spring type, rate, and preload for safe and reliable operation. Given Data / Assumptions: Conventional drum/disc brakes and plate/cone clutches. Mechanical or hydraulic actuation with springs providing biasing forces. Focus is on the primary design purpose of the springs. Concept / Approach: In brakes, return or retraction springs pull pads/shoes away when the actuation pressure is removed, preventing drag. In many clutches (e.g., diaphragm or coil-spring plate clutches), springs provide the clamping (engaging) force that presses friction plates together. Thus, the dominant role is to apply a force (either to engage or to return/disengage). Step-by-Step Solution: Identify the needed force in brakes: springs retract shoes when pedal is released.Identify the needed force in clutches: springs clamp plates to transmit torque.Recognise secondary functions (energy storage and shock mitigation) exist but are not the primary design purpose. Verification / Alternative check: Service manuals and clutch/brake catalogs specify spring force ratings and preloads explicitly as the governing design parameter, confirming their primary force-applying role. Why Other Options Are Wrong: Measure forces: Springs do not measure; sensors do.Store strain energy only / absorb shocks only: These are secondary effects, not the main purpose in these mechanisms.Reduce friction coefficient: Springs do not change the material friction property. Common Pitfalls: Confusing damping (shock absorbers) with elastic biasing (springs); assuming springs mainly “absorb shocks” in brakes/clutches rather than provide the required bias or clamping force. Final Answer: To apply forces (engage or return the mechanism)

Q: For a fixed-ended (both ends built-in) beam of length l carrying a central point load W, what is the maximum deflection?

Introduction / Context: Beam deflection formulas are essential in stiffness checks and serviceability design. The fixed-ended beam provides higher stiffness than a simply supported beam because end rotations are restrained, reducing deflection. Given Data / Assumptions: Beam is prismatic with constant E and I. Both ends are fixed (built-in) → zero slope and zero deflection at supports. Central point load W applied at midspan. Concept / Approach: Standard closed-form solutions (e.g., double integration, area-moment, or conjugate beam methods) give the maximum deflection at midspan for this case. Boundary conditions (zero rotations at ends) reduce the curvature distribution compared to a simply supported beam. Step-by-Step Solution: For a simply supported beam with central load: delta_max_ss = W l^3 / (48 E I).For a fixed-ended beam with central load: delta_max_fixed = W l^3 / (192 E I).Ratio: delta_max_fixed / delta_max_ss = (1/192) / (1/48) = 1/4 → fixed ends quarter the deflection.Thus, maximum deflection is W l^3 / (192 E I) at midspan. Verification / Alternative check: Energy methods yield the same expression; numerous beam tables list W l^3 / (192 E I) for a fixed beam with a midspan point load. Why Other Options Are Wrong: W l^3 / (48 E I): Simply supported case.W l^3 / (96 E I) or / (384 E I): Do not match fixed-end boundary conditions for this loading.2 W l^3 / (3 E I): Dimensionally correct but not a standard result; orders of magnitude too large. Common Pitfalls: Confusing fixed-ended and simply supported cases; forgetting that built-in ends reduce rotations and deflection significantly. Final Answer: W l^3 / (192 E I)

Q: For a cantilever beam, the internal shear force and bending moment at the free end are zero. Considering the listed load cases, when is this statement true?

Introduction / Context: Cantilever beams are fixed at one end and free at the other. Understanding boundary conditions—especially that the free end cannot transmit moment or shear—is crucial for drawing correct shear-force and bending-moment diagrams. Given Data / Assumptions: Cantilever of any prismatic or non-prismatic form. Loads may be a point load, uniformly distributed load, or other distributions. Linear elastic beam theory applies for internal force definitions. Concept / Approach: By definition, at the free end of a cantilever, there is no support to develop a reaction shear or reaction moment. Therefore, the internal shear force V and bending moment M evaluated at the very free tip are zero for any applied loading configuration. Step-by-Step Solution: Cut a free-body section at the free end and shrink its length to zero.Since no support exists at the tip, the section cannot develop internal V or M at the end point.Consequently, V_free = 0 and M_free = 0 irrespective of load distribution along the span. Verification / Alternative check: All standard shear and moment diagrams for cantilevers start at zero at the free end and grow towards the fixed end to meet boundary conditions at the built-in support. Why Other Options Are Wrong: Options limiting the zero condition to particular loads (point load or UDL) are incorrect; the zero condition holds universally at the free end.“True only for no loading” is false; even with loads, the free-end values remain zero. Common Pitfalls: Misreading internal shear/moment at a small distance from the tip (which may be nonzero) versus exactly at the tip where both are zero. Final Answer: None of the above (it is always true for the free end of a cantilever)

Q: In Mohr’s circle for plane stress (σ on horizontal axis, τ on vertical axis), the horizontal coordinate of the circle’s centre measured from the Y-axis equals which quantity?

Introduction / Context: Mohr’s circle provides a geometric method to transform stresses on inclined planes. Identifying the centre and radius allows quick determination of principal and shear values. Given Data / Assumptions: Plane stress components σx, σy, and τxy are known. Mohr’s circle is plotted with σ on the horizontal axis and τ on the vertical axis. Centre and radius definitions follow standard conventions. Concept / Approach: The centre C of Mohr’s circle lies midway between the points representing stresses on two orthogonal faces. Therefore, its σ-coordinate is the average of σx and σy, while its τ-coordinate is zero. Step-by-Step Solution: Plot point A: (σx, τxy) and point B: (σy, −τxy).Compute centre: C = midpoint of A and B.C has coordinates: ( (σx + σy)/2 , 0 ).Thus, measured from the Y-axis along +σ, the centre equals the average normal stress. Verification / Alternative check: Analytical stress transformation confirms that principal stresses are at σ = (σx + σy)/2 ± R with τ = 0, reinforcing the centre location. Why Other Options Are Wrong: Sum or difference alone is not the centre coordinate.Maximum shear is the radius, not the centre abscissa.Zero applies only to special cases (σx = −σy and τxy = 0 does not generally make the centre zero). Common Pitfalls: Confusing the centre coordinate with the radius; mixing sign conventions for τ that mirror the circle about the σ-axis. Final Answer: The average normal stress ((σx + σy) / 2)

Q: For a cantilever beam of length l carrying a load that varies linearly from zero at the free end to w per unit length at the fixed end, the shear force diagram (SFD) has what shape?

Introduction / Context: Being able to sketch the qualitative shapes of shear-force and bending-moment diagrams for different loadings is a core skill in structural analysis and machine design. Given Data / Assumptions: Cantilever beam, length l. Linearly varying distributed load: w(x) increases from 0 at free end to w at fixed end. Small-deflection beam theory; sign conventions standard. Concept / Approach: The differential relationships are: dV/dx = −w(x)dM/dx = V(x)With w(x) linear in x, integrating once gives V(x) quadratic in x (a parabola), and integrating again gives M(x) cubic. Step-by-Step Solution: Let w(x) = (w/l) * x (taking x from free end).Integrate: dV/dx = −(w/l) * x → V(x) = −(w/(2l)) * x^2 + C.At free end (x = 0), V(0) = 0 → C = 0.Hence V(x) = −(w/(2l)) * x^2 → a parabola opening downward. Verification / Alternative check: Total shear at the fixed end equals the resultant of the triangular load: W_total = (1/2) * w * l. Evaluating V(l) = −(w/(2l)) * l^2 = −(w l)/2 matches the resultant, confirming the parabolic SFD. Why Other Options Are Wrong: Horizontal/vertical/inclined lines correspond to zero/impulse/constant load distributions, not a linearly varying load.Cubic curve describes the bending moment diagram for this case, not the SFD. Common Pitfalls: Using the wrong origin for x; forgetting that V is the integral of −w(x), hence one degree higher in polynomial order than w(x). Final Answer: Parabolic curve

Q: A body under uniaxial direct tensile stress σ experiences a maximum shear stress equal to what fraction of the maximum normal stress?

Introduction / Context: Relating normal and shear stresses helps predict yielding and failure using criteria such as Tresca and von Mises. In a uniaxial stress state, maximum shear stress occurs on planes at 45° to the load axis. Given Data / Assumptions: Uniaxial tension: principal stresses are σ1 = σ and σ2 = 0 (plane stress). Small deformation linear elasticity. Shear stress sign convention standard. Concept / Approach: Maximum shear stress in plane stress is half the difference of principal stresses: tau_max = (σ1 − σ2) / 2 = (σ − 0) / 2 = σ/2Maximum normal stress equals σ in the loaded direction. Therefore, the ratio is 1/2. Step-by-Step Solution: Identify principal stresses: σ1 = σ, σ2 = 0.Compute tau_max = (σ1 − σ2) / 2 = σ/2.Compare to maximum normal stress σ → fraction = 1/2. Verification / Alternative check: Mohr’s circle has radius R = (σ1 − σ2)/2 = σ/2, which is the maximum shear; the horizontal intercept equals σ, confirming the ratio. Why Other Options Are Wrong: Equal/Twice: Overestimate shear; Two-thirds has no basis for uniaxial stress; Zero is only for hydrostatic stress (σ1 = σ2). Common Pitfalls: Confusing maximum shear for 3D principal differences; mixing engineering and true stress at large strains (not relevant here). Final Answer: One-half of σ

Q: A bolt passes through a tubular sleeve; washers and a nut are tightened to clamp parts together. What is the stress state after tightening?

Introduction / Context: Preloaded bolted joints are ubiquitous. Understanding which member takes tension and which takes compression under tightening is essential for correct sizing, stiffness analysis, and joint integrity. Given Data / Assumptions: A bolt runs through a tube (sleeve) between clamped parts. Nut is tightened to apply preload. Elastic behaviour; no plastic yielding during tightening. Concept / Approach: Tightening the nut stretches the bolt (increasing its tensile force) and shortens the sleeve/stack (putting it into compression). The clamped parts experience compressive contact, generating the clamping force that resists external loads. Step-by-Step Solution: Apply torque to nut → bolt elongates by δ_bolt.Compression of sleeve/parts by δ_sleeve equals bolt elongation in the elastic regime.Bolt carries tensile preload; sleeve carries compressive preload. Verification / Alternative check: Load–deflection models treat bolt and clamped members as springs in series; equilibrium requires equal and opposite internal forces → bolt tension equals sleeve compression. Why Other Options Are Wrong: Both in tension or both in compression violates equilibrium in the absence of external separating forces.Bolt in compression, tube in tension is opposite of actual preload behaviour.“Neither carries load” contradicts the essence of preloading. Common Pitfalls: Assuming only external loads create stresses; forgetting that preload exists even before service loads are applied. Final Answer: Bolt in tension, tube in compression

Q: Define volumetric strain: it is the ratio of which quantities for a deforming body?

Introduction / Context: Strain measures quantify deformation and are central to elasticity, plasticity, and design checks. Volumetric strain is especially important in hydrostatic loading and bulk modulus calculations. Given Data / Assumptions: Small deformation (engineering strain) regime. Volume changes from V to V + ΔV under load. Uniform material properties for definition purposes. Concept / Approach: Volumetric strain ε_v characterises dilatation or compression and is defined as: epsilon_v = ΔV / VThis is analogous to linear strain ε = ΔL / L but extended to three dimensions. Step-by-Step Solution: Measure original volume V.Measure volume change ΔV after loading.Compute ε_v = ΔV / V (dimensionless). Verification / Alternative check: The bulk modulus K relates pressure p to volumetric strain: p = K * ε_vfor small elastic deformations, confirming the definition. Why Other Options Are Wrong: Thickness-based ratios refer to linear or lateral strain, not volumetric.Original volume to change in volume is the reciprocal and not the standard definition.Change in length/original length defines linear strain, not volumetric. Common Pitfalls: Confusing volumetric and linear strains; neglecting that volumetric strain can be nonzero even when linear strains sum to zero in anisotropic cases. Final Answer: Change in volume to the original volume

Q: Which of the following statements about stress, strain, and Hooke’s law is correct within the elastic limit?

Introduction / Context: Basic definitions in strength of materials underpin all later design work. Distinguishing stress from pressure, and knowing the limits of Hooke’s law, prevents fundamental errors in calculations and interpretations. Given Data / Assumptions: Small-strain linear elasticity region (elastic limit not exceeded). Engineering definitions of stress and strain. Isotropic, homogeneous materials for simplicity. Concept / Approach: Hooke’s law states: σ = E * εwithin the elastic limit, i.e., stress is proportional to strain, with E (Young’s modulus) as the constant of proportionality. Strain is dimensionless; stress is force per unit area (not “pressure per unit area,” although numerically both share the same units). Step-by-Step Solution: Define stress: σ = F / A (units of N/m^2).Define strain: ε = ΔL / L (dimensionless).State Hooke’s law: σ ∝ ε up to elastic limit → σ = E ε. Verification / Alternative check: Tensile test data show linear σ–ε relation up to a proportional limit; beyond that, plasticity begins and the relation becomes nonlinear until fracture. Why Other Options Are Wrong: “Stress is pressure per unit area” is a misuse of terms; stress is force per unit area; pressure is a specific isotropic normal stress.“Strain in mm” is incorrect; strain is dimensionless.“Holds up to breaking point” is false; validity ends near the proportional/elastic limit, well before fracture.“Stress equals strain” ignores the modulus E. Common Pitfalls: Conflating units of pressure and stress; assuming linearity beyond yield; mixing true and engineering measures at large strains. Final Answer: Stress is directly proportional to strain within elastic limit

Question 1

In solid mechanics and machine design, thermal (temperature) stress in a fully or partially restrained member depends on which factors? Select the most complete description.

Accepted Answer

Introduction / Context: Thermal stress appears when a component is heated or cooled but is restrained from freely expanding or contracting. This topic is fundamental in mechanical design, civil structures, piping, and pressure vessels where temperature cycles can produce significant stresses even without external loads. Given Data / Assumptions: Uniform temperature change ΔT is imposed. The member experiences some degree of restraint (full or partial). Material properties: elastic modulus E and linear thermal expansion coefficient α are relevant. Concept / Approach: For a fully restrained, homogeneous member, the classic relation for thermal stress is: sigma_thermal = E * α * ΔTPartial restraint scales this result by the degree of fixity. The relation shows direct proportionality to all three: temperature change ΔT, stiffness E, and expansivity α. Step-by-Step Solution: Recognise free thermal strain: epsilon_free = α * ΔT.If fully restrained, mechanical strain equals negative of free thermal strain.Use Hooke’s law: sigma = E * epsilon_mech = E * α * ΔT.For partial restraint, sigma = k * E * α * ΔT, where 0 < k < 1 depends on boundary conditions. Verification / Alternative check: Dimensional reasoning confirms: α (per °C) * ΔT (°C) gives strain; multiplying by E (stress per strain) yields stress, consistent with the formula. Why Other Options Are Wrong: ΔT only / E only / α only: Thermal stress is a product of all three; omitting any factor is incomplete.None of these: Contradicts the standard thermal stress relation. Common Pitfalls: Ignoring restraint (no restraint means no stress); using plastic modulus instead of elastic modulus for elastic-range estimates; forgetting that property values E and α vary with temperature. Final Answer: All of these (ΔT, E, and α)

Question 2

In brake and clutch design, what is the primary function of the springs installed in the mechanism?

Accepted Answer

Introduction / Context: Springs are integral to friction brakes and clutches used in automotive and industrial drives. Understanding their primary role ensures correct selection of spring type, rate, and preload for safe and reliable operation. Given Data / Assumptions: Conventional drum/disc brakes and plate/cone clutches. Mechanical or hydraulic actuation with springs providing biasing forces. Focus is on the primary design purpose of the springs. Concept / Approach: In brakes, return or retraction springs pull pads/shoes away when the actuation pressure is removed, preventing drag. In many clutches (e.g., diaphragm or coil-spring plate clutches), springs provide the clamping (engaging) force that presses friction plates together. Thus, the dominant role is to apply a force (either to engage or to return/disengage). Step-by-Step Solution: Identify the needed force in brakes: springs retract shoes when pedal is released.Identify the needed force in clutches: springs clamp plates to transmit torque.Recognise secondary functions (energy storage and shock mitigation) exist but are not the primary design purpose. Verification / Alternative check: Service manuals and clutch/brake catalogs specify spring force ratings and preloads explicitly as the governing design parameter, confirming their primary force-applying role. Why Other Options Are Wrong: Measure forces: Springs do not measure; sensors do.Store strain energy only / absorb shocks only: These are secondary effects, not the main purpose in these mechanisms.Reduce friction coefficient: Springs do not change the material friction property. Common Pitfalls: Confusing damping (shock absorbers) with elastic biasing (springs); assuming springs mainly “absorb shocks” in brakes/clutches rather than provide the required bias or clamping force. Final Answer: To apply forces (engage or return the mechanism)

Question 3

For a fixed-ended (both ends built-in) beam of length l carrying a central point load W, what is the maximum deflection?

Accepted Answer

Introduction / Context: Beam deflection formulas are essential in stiffness checks and serviceability design. The fixed-ended beam provides higher stiffness than a simply supported beam because end rotations are restrained, reducing deflection. Given Data / Assumptions: Beam is prismatic with constant E and I. Both ends are fixed (built-in) → zero slope and zero deflection at supports. Central point load W applied at midspan. Concept / Approach: Standard closed-form solutions (e.g., double integration, area-moment, or conjugate beam methods) give the maximum deflection at midspan for this case. Boundary conditions (zero rotations at ends) reduce the curvature distribution compared to a simply supported beam. Step-by-Step Solution: For a simply supported beam with central load: delta_max_ss = W l^3 / (48 E I).For a fixed-ended beam with central load: delta_max_fixed = W l^3 / (192 E I).Ratio: delta_max_fixed / delta_max_ss = (1/192) / (1/48) = 1/4 → fixed ends quarter the deflection.Thus, maximum deflection is W l^3 / (192 E I) at midspan. Verification / Alternative check: Energy methods yield the same expression; numerous beam tables list W l^3 / (192 E I) for a fixed beam with a midspan point load. Why Other Options Are Wrong: W l^3 / (48 E I): Simply supported case.W l^3 / (96 E I) or / (384 E I): Do not match fixed-end boundary conditions for this loading.2 W l^3 / (3 E I): Dimensionally correct but not a standard result; orders of magnitude too large. Common Pitfalls: Confusing fixed-ended and simply supported cases; forgetting that built-in ends reduce rotations and deflection significantly. Final Answer: W l^3 / (192 E I)

Question 4

For a cantilever beam, the internal shear force and bending moment at the free end are zero. Considering the listed load cases, when is this statement true?

Accepted Answer

Introduction / Context: Cantilever beams are fixed at one end and free at the other. Understanding boundary conditions—especially that the free end cannot transmit moment or shear—is crucial for drawing correct shear-force and bending-moment diagrams. Given Data / Assumptions: Cantilever of any prismatic or non-prismatic form. Loads may be a point load, uniformly distributed load, or other distributions. Linear elastic beam theory applies for internal force definitions. Concept / Approach: By definition, at the free end of a cantilever, there is no support to develop a reaction shear or reaction moment. Therefore, the internal shear force V and bending moment M evaluated at the very free tip are zero for any applied loading configuration. Step-by-Step Solution: Cut a free-body section at the free end and shrink its length to zero.Since no support exists at the tip, the section cannot develop internal V or M at the end point.Consequently, V_free = 0 and M_free = 0 irrespective of load distribution along the span. Verification / Alternative check: All standard shear and moment diagrams for cantilevers start at zero at the free end and grow towards the fixed end to meet boundary conditions at the built-in support. Why Other Options Are Wrong: Options limiting the zero condition to particular loads (point load or UDL) are incorrect; the zero condition holds universally at the free end.“True only for no loading” is false; even with loads, the free-end values remain zero. Common Pitfalls: Misreading internal shear/moment at a small distance from the tip (which may be nonzero) versus exactly at the tip where both are zero. Final Answer: None of the above (it is always true for the free end of a cantilever)

Question 5

In Mohr’s circle for plane stress (σ on horizontal axis, τ on vertical axis), the horizontal coordinate of the circle’s centre measured from the Y-axis equals which quantity?

Accepted Answer

Introduction / Context: Mohr’s circle provides a geometric method to transform stresses on inclined planes. Identifying the centre and radius allows quick determination of principal and shear values. Given Data / Assumptions: Plane stress components σx, σy, and τxy are known. Mohr’s circle is plotted with σ on the horizontal axis and τ on the vertical axis. Centre and radius definitions follow standard conventions. Concept / Approach: The centre C of Mohr’s circle lies midway between the points representing stresses on two orthogonal faces. Therefore, its σ-coordinate is the average of σx and σy, while its τ-coordinate is zero. Step-by-Step Solution: Plot point A: (σx, τxy) and point B: (σy, −τxy).Compute centre: C = midpoint of A and B.C has coordinates: ( (σx + σy)/2 , 0 ).Thus, measured from the Y-axis along +σ, the centre equals the average normal stress. Verification / Alternative check: Analytical stress transformation confirms that principal stresses are at σ = (σx + σy)/2 ± R with τ = 0, reinforcing the centre location. Why Other Options Are Wrong: Sum or difference alone is not the centre coordinate.Maximum shear is the radius, not the centre abscissa.Zero applies only to special cases (σx = −σy and τxy = 0 does not generally make the centre zero). Common Pitfalls: Confusing the centre coordinate with the radius; mixing sign conventions for τ that mirror the circle about the σ-axis. Final Answer: The average normal stress ((σx + σy) / 2)

Question 6

For a cantilever beam of length l carrying a load that varies linearly from zero at the free end to w per unit length at the fixed end, the shear force diagram (SFD) has what shape?

Accepted Answer

Introduction / Context: Being able to sketch the qualitative shapes of shear-force and bending-moment diagrams for different loadings is a core skill in structural analysis and machine design. Given Data / Assumptions: Cantilever beam, length l. Linearly varying distributed load: w(x) increases from 0 at free end to w at fixed end. Small-deflection beam theory; sign conventions standard. Concept / Approach: The differential relationships are: dV/dx = −w(x)dM/dx = V(x)With w(x) linear in x, integrating once gives V(x) quadratic in x (a parabola), and integrating again gives M(x) cubic. Step-by-Step Solution: Let w(x) = (w/l) * x (taking x from free end).Integrate: dV/dx = −(w/l) * x → V(x) = −(w/(2l)) * x^2 + C.At free end (x = 0), V(0) = 0 → C = 0.Hence V(x) = −(w/(2l)) * x^2 → a parabola opening downward. Verification / Alternative check: Total shear at the fixed end equals the resultant of the triangular load: W_total = (1/2) * w * l. Evaluating V(l) = −(w/(2l)) * l^2 = −(w l)/2 matches the resultant, confirming the parabolic SFD. Why Other Options Are Wrong: Horizontal/vertical/inclined lines correspond to zero/impulse/constant load distributions, not a linearly varying load.Cubic curve describes the bending moment diagram for this case, not the SFD. Common Pitfalls: Using the wrong origin for x; forgetting that V is the integral of −w(x), hence one degree higher in polynomial order than w(x). Final Answer: Parabolic curve

Question 7

A body under uniaxial direct tensile stress σ experiences a maximum shear stress equal to what fraction of the maximum normal stress?

Accepted Answer

Introduction / Context: Relating normal and shear stresses helps predict yielding and failure using criteria such as Tresca and von Mises. In a uniaxial stress state, maximum shear stress occurs on planes at 45° to the load axis. Given Data / Assumptions: Uniaxial tension: principal stresses are σ1 = σ and σ2 = 0 (plane stress). Small deformation linear elasticity. Shear stress sign convention standard. Concept / Approach: Maximum shear stress in plane stress is half the difference of principal stresses: tau_max = (σ1 − σ2) / 2 = (σ − 0) / 2 = σ/2Maximum normal stress equals σ in the loaded direction. Therefore, the ratio is 1/2. Step-by-Step Solution: Identify principal stresses: σ1 = σ, σ2 = 0.Compute tau_max = (σ1 − σ2) / 2 = σ/2.Compare to maximum normal stress σ → fraction = 1/2. Verification / Alternative check: Mohr’s circle has radius R = (σ1 − σ2)/2 = σ/2, which is the maximum shear; the horizontal intercept equals σ, confirming the ratio. Why Other Options Are Wrong: Equal/Twice: Overestimate shear; Two-thirds has no basis for uniaxial stress; Zero is only for hydrostatic stress (σ1 = σ2). Common Pitfalls: Confusing maximum shear for 3D principal differences; mixing engineering and true stress at large strains (not relevant here). Final Answer: One-half of σ

Question 8

A bolt passes through a tubular sleeve; washers and a nut are tightened to clamp parts together. What is the stress state after tightening?

Accepted Answer

Introduction / Context: Preloaded bolted joints are ubiquitous. Understanding which member takes tension and which takes compression under tightening is essential for correct sizing, stiffness analysis, and joint integrity. Given Data / Assumptions: A bolt runs through a tube (sleeve) between clamped parts. Nut is tightened to apply preload. Elastic behaviour; no plastic yielding during tightening. Concept / Approach: Tightening the nut stretches the bolt (increasing its tensile force) and shortens the sleeve/stack (putting it into compression). The clamped parts experience compressive contact, generating the clamping force that resists external loads. Step-by-Step Solution: Apply torque to nut → bolt elongates by δ_bolt.Compression of sleeve/parts by δ_sleeve equals bolt elongation in the elastic regime.Bolt carries tensile preload; sleeve carries compressive preload. Verification / Alternative check: Load–deflection models treat bolt and clamped members as springs in series; equilibrium requires equal and opposite internal forces → bolt tension equals sleeve compression. Why Other Options Are Wrong: Both in tension or both in compression violates equilibrium in the absence of external separating forces.Bolt in compression, tube in tension is opposite of actual preload behaviour.“Neither carries load” contradicts the essence of preloading. Common Pitfalls: Assuming only external loads create stresses; forgetting that preload exists even before service loads are applied. Final Answer: Bolt in tension, tube in compression

Question 9

Define volumetric strain: it is the ratio of which quantities for a deforming body?

Accepted Answer

Introduction / Context: Strain measures quantify deformation and are central to elasticity, plasticity, and design checks. Volumetric strain is especially important in hydrostatic loading and bulk modulus calculations. Given Data / Assumptions: Small deformation (engineering strain) regime. Volume changes from V to V + ΔV under load. Uniform material properties for definition purposes. Concept / Approach: Volumetric strain ε_v characterises dilatation or compression and is defined as: epsilon_v = ΔV / VThis is analogous to linear strain ε = ΔL / L but extended to three dimensions. Step-by-Step Solution: Measure original volume V.Measure volume change ΔV after loading.Compute ε_v = ΔV / V (dimensionless). Verification / Alternative check: The bulk modulus K relates pressure p to volumetric strain: p = K * ε_vfor small elastic deformations, confirming the definition. Why Other Options Are Wrong: Thickness-based ratios refer to linear or lateral strain, not volumetric.Original volume to change in volume is the reciprocal and not the standard definition.Change in length/original length defines linear strain, not volumetric. Common Pitfalls: Confusing volumetric and linear strains; neglecting that volumetric strain can be nonzero even when linear strains sum to zero in anisotropic cases. Final Answer: Change in volume to the original volume

Question 10

Which of the following statements about stress, strain, and Hooke’s law is correct within the elastic limit?

Accepted Answer

Introduction / Context: Basic definitions in strength of materials underpin all later design work. Distinguishing stress from pressure, and knowing the limits of Hooke’s law, prevents fundamental errors in calculations and interpretations. Given Data / Assumptions: Small-strain linear elasticity region (elastic limit not exceeded). Engineering definitions of stress and strain. Isotropic, homogeneous materials for simplicity. Concept / Approach: Hooke’s law states: σ = E * εwithin the elastic limit, i.e., stress is proportional to strain, with E (Young’s modulus) as the constant of proportionality. Strain is dimensionless; stress is force per unit area (not “pressure per unit area,” although numerically both share the same units). Step-by-Step Solution: Define stress: σ = F / A (units of N/m^2).Define strain: ε = ΔL / L (dimensionless).State Hooke’s law: σ ∝ ε up to elastic limit → σ = E ε. Verification / Alternative check: Tensile test data show linear σ–ε relation up to a proportional limit; beyond that, plasticity begins and the relation becomes nonlinear until fracture. Why Other Options Are Wrong: “Stress is pressure per unit area” is a misuse of terms; stress is force per unit area; pressure is a specific isotropic normal stress.“Strain in mm” is incorrect; strain is dimensionless.“Holds up to breaking point” is false; validity ends near the proportional/elastic limit, well before fracture.“Stress equals strain” ignores the modulus E. Common Pitfalls: Conflating units of pressure and stress; assuming linearity beyond yield; mixing true and engineering measures at large strains. Final Answer: Stress is directly proportional to strain within elastic limit

Question 11

Uniaxial tension of a rectangular bar

When a prismatic bar of length l, width b, and thickness t is subjected to an axial tensile pull P, describe how its dimensions change due to the axial stress and Poisson effect.

Accepted Answer

Introduction / Context: In strength of materials, the dimensional response of a bar under uniaxial tension illustrates axial strain and lateral contraction, governed by Young's modulus and Poisson's ratio. This concept is fundamental for predicting deformations in structural members and machine elements. Given Data / Assumptions: Bar is prismatic with initial length l, width b, thickness t. Axial tensile load P acts along the length; material is homogeneous, isotropic, and linearly elastic. Small strains; Hooke's law and Poisson effect are valid. Concept / Approach: Axial normal stress is sigma = P / A, where A = b * t. The axial strain is epsilon_long = sigma / E, producing an increase in length. Due to Poisson effect, lateral strain is epsilon_lat = -nu * epsilon_long in directions transverse to the load, causing width and thickness to decrease when nu > 0 (typical solids). Step-by-Step Solution: Compute axial stress: sigma = P / (b * t).Axial strain: epsilon_long = sigma / E > 0 → length increases.Lateral strains: epsilon_b = -nu * epsilon_long, epsilon_t = -nu * epsilon_long → width and thickness decrease.Resultant dimensional changes: Δl = epsilon_long * l > 0; Δb < 0; Δt < 0. Verification / Alternative check: Energy or compatibility viewpoints lead to the same conclusion: positive axial work elongates the member; volume change is approximately ΔV/V ≈ (1 - 2 nu) * epsilon_long for small strains, supporting lateral contraction for typical nu between 0 and 0.5. Why Other Options Are Wrong: Options (a) and (b) change all three dimensions in the same sense, contradicting Poisson effect. Option (d) suggests compression, not tension. Option (e) ignores lateral contraction observed in virtually all structural metals. Common Pitfalls: Assuming width and thickness stay constant; confusing engineering strain sign conventions; neglecting that Poisson's ratio is positive for most materials. Final Answer: length increases, width and thickness decreases

Question 12

Definition of linear strain

For a member of original length l that undergoes a change in length δl under axial loading, which expression correctly defines engineering (normal) strain?

Accepted Answer

Introduction / Context: Engineering strain is a fundamental measure used throughout solid mechanics and materials testing. It quantifies relative elongation or contraction and underpins stress–strain relationships, modulus determination, and design limits. Given Data / Assumptions: Original (gauge) length = l. Change in length under load = δl (positive in tension, negative in compression). Small deformations where engineering strain is appropriate. Concept / Approach: Engineering (normal) strain is defined as change in length divided by original length. It is a dimensionless quantity that captures relative deformation independent of the absolute size of the specimen. Step-by-Step Solution: Engineering strain: epsilon = (final length − original length)/original length.Given final length = l + δl → epsilon = (l + δl − l)/l.Hence epsilon = δl / l.Units: none (ratio of length to length). Verification / Alternative check: For a bar under uniaxial stress sigma, Hooke's law gives epsilon = sigma / E; integrating strain over length l still yields total elongation δl = epsilon * l, which rearranges to epsilon = δl / l. Why Other Options Are Wrong: (a) inverts the ratio; (c) and (d) have dimensions of length, not strain; (e) invents a nonstandard form with incorrect dimensionality. Common Pitfalls: Confusing engineering strain with true (logarithmic) strain, which is ln(1 + δl/l); mixing sign conventions; using instantaneous length instead of original length for engineering strain. Final Answer: δl/l

Question 13

Power transmitted by a rotating shaft

A shaft rotates at angular speed ω (in rad/s) while transmitting a torque T (in N·m). What is the mechanical power developed?

Accepted Answer

Introduction / Context: Converting rotational motion to power is routine in machine design. Power relates torque and angular speed; getting the correct form and units prevents large sizing errors in shafts, couplings, and motors. Given Data / Assumptions: Torque T in newton-metre (N·m). Angular speed ω in radians per second (rad/s). Rigid shaft; steady transmission (no transients or losses considered). Concept / Approach: Instantaneous mechanical power equals torque multiplied by angular velocity. The radian is dimensionless, so units reduce to watt = N·m/s. No additional 2π factor is used when ω is already in rad/s. Step-by-Step Solution: Definition: Power P = work rate.In rotation, infinitesimal work dW = T * dθ; dividing by dt gives dW/dt = T * dθ/dt.Since dθ/dt = ω, we get P = T * ω.Therefore, the correct power in SI units is P = T * ω watts. Verification / Alternative check: If speed is given in revolutions per minute (N rpm), then ω = 2πN/60 rad/s, yielding P = T * (2πN/60). This explains why a 2π factor appears only when starting from rpm rather than ω. Why Other Options Are Wrong: (b) double counts the 2π when ω is already in rad/s; (c) and (d) invert the relationship; (e) mixes constants applicable to rpm, not to ω. Common Pitfalls: Confusing rpm and rad/s; dropping unit consistency; using T in N·mm without converting to N·m. Final Answer: T * ω watts

Question 14

Units of modulus of elasticity (Young's modulus)

Modulus of elasticity has the same dimensional units as which of the following quantities?

Accepted Answer

Introduction / Context: Understanding units helps cross-check formulas and prevents design mistakes. Young's modulus E quantifies stiffness by relating stress to strain within the elastic range. Given Data / Assumptions: Stress = force/area; strain = dimensionless. Pressure has the same units as normal stress. Modulus of rigidity G relates shear stress to shear strain. Concept / Approach: Because strain is dimensionless, E has the same units as stress: pascal (Pa) = N/m^2. Pressure p is also in pascal. Modulus of rigidity G shares the same units as E because it relates shear stress to shear strain. Step-by-Step Solution: E = σ / ε → units(E) = units(σ)/units(ε).units(σ) = N/m^2; units(ε) = 1 → units(E) = N/m^2.Pressure p has units N/m^2; modulus G has units N/m^2.Therefore, E shares units with stress, pressure, and modulus of rigidity. Verification / Alternative check: Dimensional analysis confirms [E] = [F][L]^-2. Any correct grouping must include only quantities with pascal units, not strain (dimensionless) or force alone (newton). Why Other Options Are Wrong: (a) incorrectly includes strain; (b) includes force (newton) which is not consistent; (c) includes strain and force; (e) includes displacement which has units of length. Common Pitfalls: Confusing stress with force; forgetting that strain is unitless; mixing SI and non-SI units without conversion. Final Answer: stress, pressure, and modulus of rigidity

Question 15

Riveted joints – Crushing (bearing) resistance per pitch length

For a single rivet in a lap or butt joint, the pull required to crush the rivet per pitch length is given by which expression? (Use d = rivet diameter, t = plate thickness, σc = allo…

Accepted Answer

Introduction / Context: Designing riveted joints requires checking multiple failure modes per pitch: tearing of plate, shearing of rivet, and crushing (bearing) of material. The crushing check uses the projected bearing area and the allowable bearing stress. Given Data / Assumptions: Single rivet considered per pitch length. d = rivet diameter; t = plate thickness. σc = allowable bearing (crushing) stress of the weaker bearing pair (plate or rivet). Concept / Approach: The bearing resistance equals allowable bearing stress multiplied by the projected contact area between rivet and plate. For a round rivet bearing on plate thickness t, projected area = d * t. Thus the force to cause bearing failure is d * t * σc. Step-by-Step Solution: Projected bearing area A_b = contact width d * thickness t.Allowable bearing stress = σc.Crushing (bearing) resistance per rivet = A_b * σc = d * t * σc.Compare with other modes to size the joint based on the least capacity. Verification / Alternative check: Dimension check: N = (m) * (m) * (N/m^2) → N, consistent with force. Common textbooks derive the same expression for bearing strength. Why Other Options Are Wrong: (a) uses pitch p and tensile stress σt, which relates to plate tearing not bearing. (c) and (d) use circular area (π/4 * d^2) relevant to shearing or tensile area, not bearing. (e) introduces an extra p, not part of bearing area. Common Pitfalls: Confusing bearing strength with rivet shear or plate tearing; using net-section area instead of projected contact area. Final Answer: d * t * σc

Question 16

Tensile testing – Breaking stress versus ultimate tensile stress

In a standard tensile test on a ductile material, the engineering breaking (fracture) stress is how related to the ultimate tensile stress?

Accepted Answer

Introduction / Context: The engineering stress–strain curve for ductile metals shows a peak (ultimate tensile stress) followed by necking until fracture. Understanding the difference between ultimate and breaking stress is vital for safe design and correct interpretation of material data. Given Data / Assumptions: Engineering stress is calculated using original area. After UTS, localized necking reduces actual area. Material is ductile so necking occurs prior to fracture. Concept / Approach: Ultimate tensile stress (UTS) is the maximum engineering stress attained. After necking starts, load typically drops while the original area is still used to compute engineering stress, so the engineering stress value at fracture (breaking stress) is lower than the UTS. Step-by-Step Solution: Before necking: load rises → stress increases.UTS reached at maximum load divided by original area.Necking: load decreases, original area constant for engineering stress → stress value drops.Breaking stress (at fracture) < UTS for ductile materials. Verification / Alternative check: True stress (based on instantaneous area) may increase during necking, but engineering stress (based on original area) decreases; testing standards and typical curves confirm breaking stress lower than UTS. Why Other Options Are Wrong: (a) equal to is false for ductile materials; (c) greater than contradicts the curve; (d) it can be compared using definitions; (e) is an unnecessary conditional—necking generally occurs in ductile materials. Common Pitfalls: Mixing true stress with engineering stress; misreading the stress–strain curve endpoints. Final Answer: less than

Question 17

Thin cylindrical pressure vessels – Governing stress for design

For thin cylindrical shells under internal pressure, design is primarily based on which stress component?

Accepted Answer

Introduction / Context: Thin-walled cylinders (t ≪ d) such as boilers and tanks are designed against tensile stresses caused by internal pressure. Two principal membrane stresses arise: hoop (circumferential) and longitudinal (axial). Given Data / Assumptions: Uniform internal pressure p; diameter d; thickness t; thin-wall criterion satisfied. Membrane theory with negligible radial stress compared to tangential stresses. Isotropic, linear-elastic material. Concept / Approach: For a thin cylinder, hoop stress σ_h = p d / (2 t) and longitudinal stress σ_l = p d / (4 t). Hoop stress is twice the longitudinal stress; hence it governs the design for maximum tensile stress. Step-by-Step Solution: Derive σ_h by cutting the cylinder along a diametral plane: pressure force balanced by two hoop stress resultants → σ_h = p d / (2 t).Derive σ_l by cutting with a transverse plane through the heads: σ_l = p d / (4 t).Compare magnitudes: σ_h = 2 σ_l → hoop stress governs. Verification / Alternative check: Design codes limit circumferential stress more strictly; thickness sizing formulas are based on hoop stress with allowances for corrosion and joint efficiency. Why Other Options Are Wrong: (b) longitudinal stress is smaller; (c) and (d) use averages not employed in design; (e) radial stress in thin shells is small compared to tangential components. Common Pitfalls: Applying spherical vessel formulas; misusing diameter versus radius; forgetting joint efficiency in welded/riveted shells. Final Answer: hoop stress

Question 18

Simple bending theory – Identify the incorrect statement

Which of the following statements about pure bending of beams is incorrect?

Accepted Answer

Introduction / Context: Elementary beam theory rests on specific assumptions that lead to linear strain variation and corresponding stress distribution. Spotting an incorrect statement tests conceptual clarity about neutral axis behavior and section properties. Given Data / Assumptions: Material is homogeneous and isotropic; Hooke's law valid. Plane sections remain plane; small deflection theory. No shear deformation in the pure bending region. Concept / Approach: Strain varies linearly with distance y from the neutral axis, and stress σ = E * ε = (M * y) / I. At y = 0 (neutral axis), both strain and stress are zero; maximum stress occurs at the extreme fibers where |y| is maximum. Step-by-Step Solution: Bending stress formula: σ = M * y / I.Neutral axis: y = 0 → σ = 0.Extreme fibers: y = ±c → |σ_max| = M * c / I = M / Z, where Z = I / c is section modulus.Hence bending stress is inversely proportional to Z, consistent with option (d). Verification / Alternative check: Plotting the linear stress diagram confirms zero at the neutral axis and maxima at the outermost fibers. Beam design maximizes Z to reduce stresses for a given bending moment M. Why Other Options Are Wrong: Option (c) is wrong because stress at the neutral axis is zero. Options (a), (b), and (d) are standard results. Option (e) is consistent with homogeneous materials. Common Pitfalls: Confusing neutral axis with centroidal axis location for unsymmetrical sections; forgetting linear distribution under elastic bending. Final Answer: At the neutral axis of a beam, the bending stress is maximum.

Question 19

Euler's column theory – End conditions factor C

According to Euler, the critical (crippling) load for a long column is p = π^2 E I / (C l^2). For a column with one end fixed and the other end free, what is the value of C?

Accepted Answer

Introduction / Context: Euler's buckling formula expresses the critical load in terms of column stiffness and effective length. Different end conditions change the effective length through the factor C. Given Data / Assumptions: Long, slender column (Euler region) with modulus E and second moment I. Geometric length = l; one end fixed, other free (cantilever-like). Elastic stability analysis; initial imperfections neglected. Concept / Approach: Effective length L_e captures boundary restraint. For fixed–free, L_e = 2 l. Substituting L_e in Euler's base form p = π^2 E I / L_e^2 gives p = π^2 E I / (4 l^2). Comparing with p = π^2 E I / (C l^2) yields C = 4. Step-by-Step Solution: Identify end condition: fixed–free → L_e = 2 l.Base Euler: p = π^2 E I / L_e^2.Substitute: p = π^2 E I / (2 l)^2 = π^2 E I / (4 l^2).Match forms → C = 4. Verification / Alternative check: Standard effective length factors: pinned–pinned C = 1; fixed–pinned C = 2; fixed–fixed C = 4; fixed–free C = 4 (because L_e = 2 l). Many texts tabulate these values consistently. Why Other Options Are Wrong: (a) and (b) correspond to more restrained cases; (c) is for fixed–pinned; larger C reduces critical load, and fixed–free has the least capacity among common end conditions. Common Pitfalls: Mixing up the two notations: some authors write p = π^2 E I / (K l)^2, where K is the effective length factor (K = 2 for fixed–free). Converting to the given form requires care. Final Answer: 4

Question 20

Beam deflection under central load: A simply supported beam A of length l, breadth b, and depth d carries a central load W. Another identical beam B carries a central load of 2W. How does the midspan deflection of beam B compare with that of beam A?

Accepted Answer

Introduction / Context: For linearly elastic beams, midspan deflection under a central point load is directly proportional to the magnitude of that load when all geometric and material properties remain the same. This question checks recall of proportionality in classic beam-bending relations for a simply supported beam. Given Data / Assumptions: Beam A and Beam B are identical: same length l, breadth b, depth d, modulus of elasticity E, and moment of inertia I. Both beams are simply supported with a central point load. Loads: W on A, and 2W on B. Small deflection theory, linear elasticity, and no shear deformation effects. Concept / Approach: For a simply supported beam with a central point load P, the maximum deflection at midspan is given by delta_max = (P * l^3) / (48 * E * I) Since E, I, and l are unchanged between the two beams, the deflection ratio equals the load ratio. Step-by-Step Solution: For Beam A: delta_A = (W * l^3) / (48 * E * I)For Beam B: delta_B = (2W * l^3) / (48 * E * I)Take the ratio: delta_B / delta_A = (2W) / W = 2Therefore, delta_B = 2 * delta_A, i.e., double. Verification / Alternative check: A quick dimensional check confirms deflection dependence on load P linearly, so doubling P doubles the deflection when geometry and material remain constant. Why Other Options Are Wrong: One-fourth / one-half / four times / one and a half times: These contradict the direct proportionality of deflection to load for this case. Common Pitfalls: Confusing this with uniformly distributed load or cantilever formulae; mixing up how deflection scales with depth (via I) versus with load P. Final Answer: double

Strength of Materials Questions