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

Question 1

Beam of uniform strength: A beam may be designed to have uniform strength (constant maximum fibre stress along its length) by which of the following approaches?

Accepted Answer

Introduction / Context: A beam of uniform strength is shaped so that the allowable bending stress is reached everywhere along the span, reducing weight compared to a prismatic beam. This concept is used in cranes, leaf springs, and tapered structural members. Given Data / Assumptions: Bending stress formula: sigma = M / Z where Z is section modulus. Allowable stress sigma_allow is constant along the beam. Designer may vary cross-sectional dimensions. Concept / Approach: For sigma to be constant while M(x) varies with x, the section modulus must vary proportionally: Z(x) = M(x) / sigma_allow. Since Z depends on both width and depth, either parameter (or both) may be varied to match the bending moment diagram. Step-by-Step Solution: Determine M(x) for the applied loading.Set Z(x) = M(x) / sigma_allow.Relate Z(x) to geometry (e.g., rectangular: Z = b*d^2/6).Choose a practical variation: vary d, vary b, or vary both to achieve the required Z(x). Verification / Alternative check: For a cantilever with end load (M ∝ x), a triangular depth variation provides near-constant stress; other profiles can achieve the same with width variation. Why Other Options Are Wrong: Options a, b, and c each can be correct individually; limiting the designer to only one would be unnecessarily restrictive. Common Pitfalls: Ignoring local stability (buckling of thin webs/flanges) when aggressively tapering members. Final Answer: Any one of the above, depending on design convenience

Question 2

Axial strain calculation: A bar of length L metres extends by l millimetres under a tensile force P. What is the engineering strain produced in the bar?

Accepted Answer

Introduction / Context: Strain is the non-dimensional measure of deformation and is defined as change in length divided by original length. Careful unit handling is essential when lengths are given in different units. Given Data / Assumptions: Original length = L metres. Extension = l millimetres. Small, uniform axial strain (linear elastic range). Concept / Approach: Engineering strain epsilon = delta_L / L_0. Convert all lengths to consistent units before forming the ratio to ensure a dimensionless quantity. Step-by-Step Solution: Convert extension to metres: delta_L = l mm = (l / 1000) m.Original length L_0 = L m.Compute strain: epsilon = (l / 1000) / L = 0.001 * (l / L).Hence epsilon = 0.001 l / L (dimensionless). Verification / Alternative check: Unit check: numerator and denominator both in metres; the ratio is unitless as required for strain. Why Other Options Are Wrong: l / L: ignores unit conversion; would be 1000× too large. 0.1 l / L and 0.01 l / L: incorrect conversion factors. Common Pitfalls: Forgetting to convert millimetres to metres or mixing up engineering strain with percentage strain (%strain = 100 * epsilon). Final Answer: 0.001 l / L

Question 3

Fixed beam with central point load: For a prismatic beam fixed at both ends and loaded by a central point load, where does the maximum vertical deflection occur?

Accepted Answer

Introduction / Context: Serviceability checks require the location and magnitude of maximum deflection. For statically indeterminate beams such as fixed–fixed beams, symmetry provides powerful insight without full analysis. Given Data / Assumptions: Beam is prismatic and fixed at both ends. Single concentrated load acts at midspan. Linear elastic behavior; small deflection assumptions hold. Concept / Approach: Due to geometric and loading symmetry, the slope at midspan is zero and the deflected shape is symmetric about the centreline. The maximum deflection occurs where the slope changes sign, i.e., at the centre under a central load. Step-by-Step Solution: Recognize symmetry: reactions and end moments are equal in magnitude at supports.The elastic curve is symmetric; slope at midspan is zero.The extremum of deflection occurs at zero slope → at the centre.Hence, maximum deflection is at midspan. Verification / Alternative check: Closed-form solution gives y_max at midspan with magnitude y_max = (Pl^3) / (192E*I) for fixed–fixed central load, confirming the location. Why Other Options Are Wrong: At the fixed ends: deflection is zero by boundary condition. At l/3: not the extremum for this loading. None of these: incorrect because the centre is correct. Common Pitfalls: Confusing fixed–fixed with simply supported beams where maximum deflection is also at midspan, but magnitudes differ. Final Answer: At the centre of the beam

Question 4

Plane stress transformation: A body has normal stresses of 1200 MPa (tension) and 600 MPa (tension) on two perpendicular planes, with a shear stress of 400 MPa on the same planes. What is the minimum (least) principal normal stress?

Accepted Answer

Introduction / Context: Principal stresses and maximum shear stresses in plane stress are central to failure theories and design. Given two perpendicular normal stresses and an in-plane shear, we can compute principal values using transformation equations or Mohr’s circle. Given Data / Assumptions: Sigma_x = 1200 MPa (tension). Sigma_y = 600 MPa (tension). Tau_xy = 400 MPa (shear on the same planes). Plane stress; sign convention: tensile positive, shear as given. Concept / Approach: Principal stresses are given by sigma_(1,2) = sigma_avg ± R, where sigma_avg = (sigma_x + sigma_y)/2 and R = sqrt( ((sigma_x − sigma_y)/2)^2 + tau_xy^2 ). The smaller of these is the minimum normal stress. Step-by-Step Solution: Compute average: sigma_avg = (1200 + 600)/2 = 900 MPa.Compute half-difference: (sigma_x − sigma_y)/2 = (600)/2 = 300 MPa.Radius: R = sqrt(300^2 + 400^2) = sqrt(90000 + 160000) = sqrt(250000) = 500 MPa.Principal stresses: sigma_1 = 900 + 500 = 1400 MPa; sigma_2 = 900 − 500 = 400 MPa. Verification / Alternative check: Mohr’s circle center at 900 MPa, radius 500 MPa; intersections at 1400 and 400 MPa confirm the computed values. Why Other Options Are Wrong: 500 MPa and 900 MPa: intermediate values (average or misused components) not principal results. 1400 MPa: this is the maximum, not the minimum principal stress. Common Pitfalls: Sign errors for shear; forgetting that both principal stresses can be tensile when both sigma_x and sigma_y are tensile. Final Answer: 400 MPa

Question 5

Thermal stress in a restrained bar: A circular bar tapers uniformly from diameter d1 at one end to diameter d2 at the other. If its temperature is increased uniformly by t (°C) and the bar is fully restrained against expansion, what is the induced t…

Accepted Answer

Introduction / Context: Thermal stresses arise when thermal expansion or contraction is restrained. Designers must quantify these stresses to prevent cracking or yielding in components subject to temperature changes, such as turbine shafts or bridge members. Given Data / Assumptions: The bar is circular and tapers linearly from diameter d1 to d2. Uniform temperature rise t is applied to the entire bar. Both ends are fully restrained so that net axial extension is zero. Material is homogeneous and isotropic with modulus E and expansion coefficient α. Concept / Approach: If a uniform axial strain α * t would occur freely, but the bar is fully restrained, an equal and opposite mechanical strain must be induced to enforce zero net extension. In uniaxial elasticity, mechanical strain ε_mech = sigma / E. Setting total strain to zero gives sigma / E + α * t = 0, hence sigma = −E * α * t. The magnitude is E * α * t and it is uniform along the bar under full restraint, independent of area variation. Step-by-Step Solution: Free thermal strain (unrestrained): ε_th = α * t.Restraint imposes ε_total = ε_mech + ε_th = 0.Mechanical strain: ε_mech = sigma / E → sigma / E + α * t = 0.Solve for stress: sigma = −E * α * t (compressive for heating); magnitude E * α * t. Verification / Alternative check: A tapered bar has varying stiffness EA(x), but for uniform temperature and full restraint, compatibility requires uniform axial strain (zero total), yielding a uniform stress solution; equilibrium is satisfied without axial load transfer variation. Why Other Options Are Wrong: Area-dependent formulas (options b and d) incorrectly imply geometry dependence for uniform thermal stress under full restraint. sigma = 0: only true if the bar is free to expand (unrestrained), which is not the case here. Common Pitfalls: Confusing partially restrained conditions (which would require integrating EA(x)) with fully restrained ends; forgetting the sign convention (compressive stress under heating). Final Answer: sigma = E * α * t

Question 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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 17

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 18

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 19

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 20

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 = allowa…

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

Strength of Materials Questions