Question 1

Basic definition in mechanics of materials: when a body is pulled by two equal and opposite forces and tends to increase in length, what are the resulting stress and strain called?

Accepted Answer

Introduction / Context: This question checks your understanding of the core terminology in strength of materials: identifying the type of stress and strain under axial pulling (tension). Given Data / Assumptions: A straight member is subjected to two equal and opposite axial pulls. Forces act along the member’s axis and try to elongate it. Material response is within the elastic range for clarity. Concept / Approach: Stress is internal resistance per unit area. When axial forces pull the material outward and it tends to lengthen, the normal stress developed is tensile. The corresponding deformation is an increase in length, i.e., tensile strain (positive linear strain). Step-by-Step Solution: Apply two equal and opposite forces along the axis.Internal resistance develops to balance the external pulls.The bar’s length increases; longitudinal strain is positive (tensile).The stress causing this is tensile stress. Verification / Alternative check: If the same member were pushed (shortened), we would describe the state as compressive stress with compressive strain. Here we explicitly have extension, confirming tensile–tensile. Why Other Options Are Wrong: Compressive pairings contradict the observed extension.Tensile stress with compressive strain is inconsistent with axial pulling.Shear stress–strain applies to tangential loading, not axial pull. Common Pitfalls: Confusing sign conventions for strain; tensile strain is positive elongation, compressive is negative shortening. Final Answer: tensile stress, tensile strain

Question 2

Elastic behavior terminology: the greatest stress that a material can withstand without any permanent set (so that it fully returns to its original shape after unloading) is known as what?

Accepted Answer

Introduction / Context: Engineers must distinguish between several critical stress levels: proportional limit, elastic limit, yield stress, ultimate stress, and breaking stress. Each term captures a distinct point on a stress–strain response curve and has specific design implications. Given Data / Assumptions: Standard tensile test behavior is assumed. Small-scale material nonlinearity may occur near limits. Definitions are based on conventional engineering usage. Concept / Approach: The elastic limit is the maximum stress a material can endure without leaving any residual (permanent) deformation after complete unloading. Below this limit, the strain is fully recoverable. Yield stress marks the onset of significant plastic flow. Ultimate stress is the peak stress prior to necking or fracture. Breaking stress is the stress at fracture. Step-by-Step Solution: Identify the criterion: “no permanent set after unloading.”Map criterion to term: elastic (fully recoverable) behavior persists up to the elastic limit.Therefore, the requested term is elastic limit. Verification / Alternative check: On a stress–strain diagram, unloading from any point below the elastic limit returns along (nearly) the same path to zero strain. Why Other Options Are Wrong: Yield stress: marks plastic flow; unloading leaves a permanent set.Ultimate and breaking stress: occur much later in the test and include plasticity.Proportional limit: where stress is strictly proportional to strain; it may be just below but not above the elastic limit for some materials. Common Pitfalls: Confusing proportional limit with elastic limit; while close, they are not identical for all materials. Final Answer: elastic limit

Question 3

Torsional strength comparison: two solid circular shafts of the same material have diameters D (shaft A) and D/2 (shaft B). The torsional strength (torque capacity at the same allowable shear) of shaft B is what fraction of shaft A?

Accepted Answer

Introduction / Context: Torsional strength of a solid round shaft is proportional to its polar section modulus, which scales as the cube of the diameter. Halving the diameter therefore reduces torque capacity dramatically. Given Data / Assumptions: Shafts are solid, circular, and made of the same material. Allowable maximum shear stress is the same for both. Elastic torsion theory applies. Concept / Approach: Polar section modulus for a solid round: Z_p = J / R = (π d^4 / 32) / (d/2) = π d^3 / 16. Therefore, torque capacity T_allow ∝ Z_p ∝ d^3 for the same allowable shear stress. Step-by-Step Solution: Let d_A = D and d_B = D/2.Strength ratio = (d_B / d_A)^3 = (D/2D)^3 = (1/2)^3 = 1/8.Hence shaft B is one-eighth as strong as shaft A in torsion. Verification / Alternative check: Compute Z_p explicitly for each and take the ratio to confirm 1/8. Why Other Options Are Wrong: One-fourth, one-half: These reflect area or linear scaling, not the correct cubic scaling.Four times: Opposite of the actual reduction; reducing diameter cannot increase strength.Three-eighths: Arbitrary fraction not supported by d^3 scaling. Common Pitfalls: Assuming strength scales with d^2 (area) rather than d^3 (polar section modulus). Final Answer: one-eighth

Question 4

Elastic constants conversion: a material has Young’s modulus E = 125 GPa and Poisson’s ratio ν = 0.25. What is its modulus of rigidity C (shear modulus G)?

Accepted Answer

Introduction / Context: Converting between elastic constants is a routine task in design and analysis. Knowing any two of E, ν, C (G), and K allows you to compute the others for an isotropic material. Given Data / Assumptions: E = 125 GPa. ν = 0.25. Isotropic, linear-elastic behavior. Concept / Approach: Use the identity relating E, ν, and C: C = E / (2 * (1 + ν)). This comes directly from isotropic elasticity theory. Step-by-Step Solution: C = E / (2 * (1 + ν))= 125 / (2 * 1.25) GPa= 125 / 2.5 GPa= 50 GPa. Verification / Alternative check: Cross-check with the inverse formula E = 2C(1 + ν). With C = 50 GPa and ν = 0.25, E = 2 * 50 * 1.25 = 125 GPa, which matches the given value. Why Other Options Are Wrong: 30, 80, 100, 62.5 GPa do not satisfy C = E / (2 * (1 + ν)) for the provided E and ν. Common Pitfalls: Plugging ν into a bulk-modulus relation by mistake, or forgetting the factor (1 + ν) in the denominator. Final Answer: 50 GPa

Question 5

Unit check: since Young’s modulus E is defined as stress divided by strain (and strain is dimensionless), is the unit of E the same as the unit of stress?

Accepted Answer

Introduction / Context: Dimensional consistency is crucial in mechanics. Young’s modulus quantifies stiffness as the ratio of stress to strain and therefore inherits the units of stress. Given Data / Assumptions: E = stress / strain. Strain is dimensionless (change in length / original length). Any coherent unit system (SI, CGS, U.S. customary) is acceptable. Concept / Approach: Because strain has no dimensions, the dimensions of E are exactly those of stress. In SI units, stress is in pascals (Pa = N/m^2), so E is also in pascals. In practice, engineers use MPa or GPa for convenience. Step-by-Step Solution: Define E = sigma / epsilon.epsilon has no units.Therefore, [E] = [sigma] = force / area. Verification / Alternative check: Check typical values: steels have E ≈ 200 GPa, which is a stress unit magnitude, confirming the reasoning. Why Other Options Are Wrong: “False”: contradicts the definition.Conditions like “only in SI” or “only for metals” are irrelevant; the dimensional reasoning is universal. Common Pitfalls: Confusing modulus units with material-dependent numbers; the unit type is always the same as stress, regardless of the material. Final Answer: True

Question 6

Under direct stress, a material elongates in the loading direction and contracts laterally at right angles. What is the name of the strain developed in the directions perpendicular to the applied load?

Accepted Answer

Introduction / Context: Axially loaded members do not deform only in the loading direction. Due to Poisson’s effect, they also undergo deformation in perpendicular directions. Correctly naming these strains is essential for volume-change estimates and multiaxial stress analysis. Given Data / Assumptions: Uniaxial direct stress is applied. Material is isotropic and linearly elastic. Small strains so superposition holds. Concept / Approach: Linear (longitudinal) strain is the change along the loading axis. At the same time, equal and opposite type of strain occurs laterally—this is lateral strain. The ratio of lateral strain to linear strain is Poisson’s ratio ν. Step-by-Step Solution: Apply tensile stress: the member elongates longitudinally (positive linear strain).Simultaneously, it contracts across its thickness/width (negative lateral strain).Therefore, the strain at right angles to the load is termed “lateral strain.” Verification / Alternative check: Volume change can be estimated using linear and lateral strains; these calculations require ν = lateral strain / linear strain. Why Other Options Are Wrong: Linear strain: along the load direction, not perpendicular.Volumetric strain: ratio of change in volume to original volume; a combined effect of all three directional strains.Shear strain: change in angle between material line elements; not applicable to pure uniaxial tension without shear.Plastic strain: refers to permanent deformation, not the directional description. Common Pitfalls: Confusing lateral strain with volumetric strain; the latter aggregates strains in three orthogonal directions. Final Answer: lateral strain

Question 7

Cantilever with uniformly distributed load (UDL): for a cantilever beam of length l carrying a UDL of w per unit length, what is the bending moment at the fixed end?

Accepted Answer

Introduction / Context: Knowing end moments for standard loading cases is essential for rapid design. A classic case is a cantilever with a uniformly distributed load (UDL), where the critical bending moment occurs at the fixed support. Given Data / Assumptions: Cantilever span = l. UDL intensity = w (force per unit length). Static equilibrium; prismatic member. Concept / Approach: For a cantilever, the support must balance the resultant of the distributed load. The resultant of a UDL equals w * l acting at the centroid of the load distribution, which is at l/2 from the support for a full-length UDL. The fixed-end moment equals the moment of this resultant about the support. Step-by-Step Solution: Resultant load R = w * l.Lever arm of R about the fixed support = l / 2.Fixed-end bending moment M_fixed = R * (l / 2) = (w * l) * (l / 2) = w * l^2 / 2. Verification / Alternative check: Derive from the shear diagram (linear from −w l to 0) and integrate to get a quadratic bending moment diagram peaking at w l^2 / 2 at the fixed end. Why Other Options Are Wrong: w l / 4, w l / 2, w l: Missing the essential l^2 dependence of moment.w l^2 / 8: This corresponds to other common cases (e.g., simply supported with central point load) and is not applicable here. Common Pitfalls: Confusing cantilever results with simply supported beam formulas; forgetting the location of the UDL resultant. Final Answer: w l^2 / 2

Question 8

Beam classification: which statement best defines a continuous beam in structural analysis?

Accepted Answer

Introduction / Context: Structural members are classified by their support and loading conditions. Recognizing a continuous beam is important because its analysis involves continuity conditions and distribution of moments across multiple spans. Given Data / Assumptions: Standard structural engineering terminology. Linearly elastic analysis context. Concept / Approach: A continuous beam rests on more than two supports, creating multiple spans that are structurally continuous. This continuity alters moment distribution compared to an isolated, simply supported span and generally reduces maximum positive moments within spans while introducing negative moments over supports. Step-by-Step Solution: Identify the support count: more than two supports ⇒ multi-span member.Apply continuity: rotations and deflections must be compatible at intermediate supports.Conclude that such a member is a continuous beam. Verification / Alternative check: Classical methods (Clapeyron’s theorem of three moments, moment distribution, stiffness method) specifically target continuous beams with more than two supports. Why Other Options Are Wrong: Fixed at both ends: describes an end condition, not necessarily multi-support continuity.Cantilever (fixed–free): is not continuous.Overhanging beams extend beyond a support but may still have only two supports.Simply supported at two ends: by definition, not continuous. Common Pitfalls: Confusing overhanging beams with continuous beams; overhangs are still usually two-support systems with extensions beyond a support. Final Answer: supported on more than two supports

Question 9

Concept of a beam of uniform strength: is it true that, in such a beam, the bending stress at every cross-section is constant and equal to the allowable stress?

Accepted Answer

Introduction / Context: The idea of a beam of uniform strength is a classic design concept where the cross-section is varied along the length to keep the bending stress constant and equal to the permissible value everywhere, thus using material efficiently. Given Data / Assumptions: Beam material follows linear elasticity. Sectional properties (e.g., depth or width) can vary with x along the span. Allowable bending stress is a prescribed limit for the material. Concept / Approach: For bending, sigma = M / Z, where Z is the section modulus. In a beam of uniform strength, Z(x) is tailored so that sigma(x) = sigma_allow for the actual bending moment distribution M(x). This requires the cross-section to be smaller where moments are small and larger where moments are high. Step-by-Step Solution: Start from sigma(x) = M(x) / Z(x).Set sigma(x) = sigma_allow (constant target).Therefore, choose Z(x) = M(x) / sigma_allow so that stress remains uniform.Practical implementations taper depth or width to approximate this requirement. Verification / Alternative check: Check a candidate variable section by computing sigma(x) across the span; if constant and equal to sigma_allow, the design meets the definition. Why Other Options Are Wrong: Stipulations about section type, span length, or varying E are not part of the definition; the key is tailoring Z(x) to M(x). Common Pitfalls: Assuming “uniform strength” means constant cross-section; it means constant stress by varying sectional properties to match the bending moment diagram. Final Answer: True

Question 10

Columns: Effective (equivalent) length for end conditions Statement: For a column with one end fixed and the other end free (cantilever), the equivalent length is half of the actual length.

Accepted Answer

Introduction / Context: In column buckling, the concept of effective or equivalent length converts different end restraints into an equivalent pinned–pinned length for use in Euler's critical load formula. The statement claims that for a column with one end fixed and the other end free, the equivalent length is half of the actual length. Understanding end conditions is essential for safe structural design. Given Data / Assumptions: Slender column subject to axial compression. Elastic behavior with small deflections and prismatic, homogeneous material. End conditions considered: fixed–free (cantilever). Concept / Approach: Euler buckling uses Pcr = (pi^2 * E * I) / (Le^2), where Le is the effective length. The effective length depends on rotational and translational restraint at the ends. Standard K-factors relate Le to the real length l by Le = K * l. Step-by-Step Solution: For pinned–pinned: K = 1.0, Le = l.For fixed–fixed: K = 0.5, Le = 0.5 * l.For fixed–free (cantilever): K = 2.0, Le = 2 * l.For fixed–hinged: K ≈ 0.7, Le ≈ 0.7 * l.Therefore the claim that Le equals half the length for a fixed–free column is incorrect; it should be double the actual length. Verification / Alternative check: The buckled mode shape for a cantilever is a quarter sine wave over 2 * l when mapped to the pinned–pinned reference, justifying K = 2.0 and not 0.5. Why Other Options Are Wrong: Agree: Incorrect because fixed–free gives K = 2.0, not 0.5. Only true for both ends fixed: Half-length applies to fixed–fixed, not fixed–free. Only true for both ends hinged: Hinged–hinged uses Le = l. Only true for fixed–hinged: That case has K ≈ 0.7, not 0.5. Common Pitfalls: Confusing fixed–free with fixed–fixed; memorizing formulas without linking them to end restraints and mode shapes; ignoring that K reflects rotational and translational fixity. Final Answer: Disagree

Question 11

Euler's column theory: critical (crippling) load for a column of length l with one end fixed and the other end hinged Choose the correct expression in terms of E, I, and l.

Accepted Answer

Introduction / Context: Euler's buckling theory provides the elastic critical load at which a slender column becomes unstable. The result depends on the flexural rigidity E * I and the effective length, which varies with end conditions. Here the column has one end fixed and the other end hinged (pinned). Given Data / Assumptions: Column length = l, prismatic, linearly elastic. End conditions: fixed–hinged. Slenderness high enough for Euler theory to apply (no yielding). Concept / Approach: Euler formula: Pcr = (pi^2 * E * I) / (Le^2). The effective length Le = K * l depends on the rotational restraint at ends. For fixed–hinged, K ≈ 0.7 (often taken as 0.7 or 1/sqrt(2)). Using K = 1/sqrt(2) gives Le = l / sqrt(2). Step-by-Step Solution: Le = l / sqrt(2).Pcr = (pi^2 * E * I) / (Le^2).Pcr = (pi^2 * E * I) / ( (l / sqrt(2))^2 ).Pcr = (pi^2 * E * I) / ( l^2 / 2 ).Pcr = 2 * pi^2 * E * I / l^2. Verification / Alternative check: Tabulated K-factors: pinned–pinned K = 1.0, fixed–fixed K = 0.5, fixed–free K = 2.0, fixed–hinged K ≈ 0.7. Substituting K = 0.707 reproduces approximately the same result as 2 * pi^2 * E * I / l^2 when expressed exactly with 1/sqrt(2). Why Other Options Are Wrong: (pi^2 * E * I) / l^2 corresponds to pinned–pinned. 4 * pi^2 * E * I / l^2 corresponds to fixed–fixed. The fractional variants with 1/2 reduce capacity improperly; fixed–hinged has a higher capacity than pinned–pinned, not lower. Common Pitfalls: Treating all end conditions as the same; forgetting that increasing fixity decreases Le and increases Pcr; using inconsistent K values between design references. Final Answer: 2 * pi^2 * E * I / l^2

Question 12

Stress–strain diagram for mild steel: identify point A Select the correct interpretation for point A on the standard mild steel curve.

Accepted Answer

Introduction / Context: A typical engineering stress–strain diagram for mild steel shows distinct regions: linear elastic, yield plateau, strain hardening, ultimate strength, and fracture. Classic labeled diagrams often mark a point A close to the end of the elastic range. Understanding which mechanical property corresponds to point A is vital for deciding safe service stresses. Given Data / Assumptions: Material: mild steel (low carbon steel) tested in tension. Conventional engineering stress–strain representation with clearly visible yield behavior. Point A lies near the end of the initial straight-line portion. Concept / Approach: In the initial straight region, stress is proportional to strain and follows Hooke's law. The end of this region corresponds to the elastic limit or limit of proportionality in classroom diagrams. Many exam conventions use A as the elastic limit, followed by a slight deviation leading to upper and lower yield points. Step-by-Step Solution: Identify the straight-line segment from origin to A.Recognize that unloading anywhere before A returns to near-zero strain, indicating elasticity.Therefore, A corresponds to the maximum stress at which the material still behaves elastically. Verification / Alternative check: Standard textbooks commonly label A as elastic limit, B as upper yield, C as lower yield or vice versa according to the diagram style. While some sources distinguish between elastic limit and limit of proportionality, exam keys usually accept elastic limit for A in mild steel diagrams. Why Other Options Are Wrong: Upper yield point and lower yield point occur immediately after the elastic region. Breaking point is at the far right after necking. Limit of proportionality is sometimes very close to A but, in conventional labeling here, A is taken as the elastic limit. Common Pitfalls: Assuming all diagrams use identical labels; ignoring small differences between elastic limit and proportional limit; confusing yield points with elastic behavior. Final Answer: Elastic limit

Question 13

Cantilever beam of length l with uniformly distributed load w (per unit length): shape of the shear force diagram Choose the correct diagram shape for the entire span.

Accepted Answer

Introduction / Context: Shear force diagrams represent the internal transverse shear variation along a beam. For standard loading and support cases, the diagram shape can be inferred directly from load intensity and boundary conditions. Here, a cantilever carries a uniformly distributed load across its full length. Given Data / Assumptions: Beam type: cantilever, fixed at one end and free at the other. Load: uniformly distributed load w over 0 to l. Linear elastic, prismatic beam. Concept / Approach: Relationship: dV/dx = -w(x). For a constant w, V varies linearly with x. Hence the shear force diagram is a straight line, reaching maximum magnitude at the fixed end and zero at the free end. Step-by-Step Solution: Take origin at the free end; w is constant.Integrate: V(x) = -∫ w dx = -w x + C.Apply boundary: V = 0 at the free end (x = 0) gives C = 0.Therefore V(x) = -w x, a straight line from 0 at free end to -w l at fixed end. Verification / Alternative check: The bending moment diagram is the integral of V and is quadratic (a parabola) peaking at the fixed end. This pairing confirms the linear, triangular SFD. Why Other Options Are Wrong: Isosceles or equilateral triangles imply specific side lengths unrelated to load intensity; the correct feature is right angle due to linear variation ending at zero. Rectangle would require constant V. Parabola corresponds to bending moment, not shear force. Common Pitfalls: Mixing up shear and moment shapes; forgetting sign conventions; assuming constant shear under UDL, which is incorrect. Final Answer: A right-angled triangle

Question 14

Loads on beams: classification Statement: A load acting at a single point on a beam is not called a uniformly distributed load (UDL).

Accepted Answer

Introduction / Context: Beam loading is categorized by how force is spread along the span. A point load is concentrated at a location, while a uniformly distributed load acts continuously with constant intensity per unit length. Clear classification is essential for drawing shear and moment diagrams correctly. Given Data / Assumptions: Beam is prismatic and linearly elastic. Load types under discussion: point load vs uniformly distributed load. Concept / Approach: Definitions govern the classification. A UDL has intensity w (force per unit length) constant along a region. A point load is modeled as acting at an exact position with no length over which it is distributed. Step-by-Step Solution: Identify the load application area.If the load acts over zero length, it is a concentrated load.Since there is no distribution per unit length, it cannot be called a UDL. Verification / Alternative check: In structural analysis, converting a UDL over length a to a resultant W = w * a placed at the centroid is a modeling step; the original load remains a UDL, not a point load. The converse does not apply to a true point load. Why Other Options Are Wrong: Disagree: contradicts the definition. Conditions involving plates, moving loads, or dynamics do not change the classification based on distribution. Common Pitfalls: Confusing the resultant of a UDL with an equivalent point load; mislabeling affects SFD and BMD shapes and magnitudes. Final Answer: Agree

Question 15

Impact energy concepts: proof resilience per unit volume Statement: The proof resilience per unit volume of a material is called the modulus of resilience.

Accepted Answer

Introduction / Context: Designers often need a measure of how much elastic energy a material can absorb before permanent deformation. The terms proof resilience and modulus of resilience are closely related yet used at different scales, and clarity helps in impact and energy storage problems like springs. Given Data / Assumptions: Elastic loading up to yield or proof stress. Material follows Hooke's law within the elastic limit. Concept / Approach: Proof resilience is the maximum strain energy stored in a body up to the elastic limit. When expressed per unit volume, it is called modulus of resilience. For linear elasticity, area under the stress–strain curve up to yield is triangular. Step-by-Step Solution: Strain energy density up to yield = area under curve.For linear elastic region: Uv = (1/2) * sigma_y * epsilon_y.Since epsilon_y = sigma_y / E, Uv = (1/2) * sigma_y * (sigma_y / E).Therefore Uv = (sigma_y^2) / (2 * E), which is the modulus of resilience. Verification / Alternative check: For a spring, energy stored at proof load equals area under load–deflection curve up to the elastic limit; dividing by volume gives the same modulus value. Why Other Options Are Wrong: False: contradicts standard definitions. Constraints about brittleness, temperature, or compression do not redefine the term; it is a general elastic property. Common Pitfalls: Confusing resilience (elastic energy) with toughness (total energy to fracture); mixing up proof load with ultimate load. Final Answer: True

Question 16

Tensile test observation: diameter at the fracture section remains approximately the same as before testing Identify the most likely material behavior.

Accepted Answer

Introduction / Context: In a tensile test, the way a specimen fractures reveals whether it is ductile or brittle. Ductile materials exhibit noticeable necking before fracture, while brittle materials fail suddenly with little or no plastic deformation. The observation given is that the diameter at the fracture remains approximately unchanged, which signals brittle behavior. Given Data / Assumptions: Round bar in uniaxial tension to failure. Diameter near fracture shows no significant necking. Room temperature, quasi-static loading. Concept / Approach: Ductile fracture is preceded by plastic flow and localized necking. Brittle fracture occurs with minimal plasticity and no necking, often along cleavage planes. Cast iron is a classic brittle engineering material in tension due to its graphite morphology and microstructure. Step-by-Step Solution: Check for necking: None observed.Classify behavior: brittle.Among options, cast iron and glass are brittle, but glass is not typically tested as standard round bar in structural labs; cast iron is the conventional exam answer. Verification / Alternative check: Typical fracture surfaces: ductile shows cup-and-cone with reduction of area (mild steel, copper, aluminum); brittle shows granular flat fracture with little contraction (cast iron). Why Other Options Are Wrong: Mild steel, copper, aluminum alloy are ductile and neck noticeably. Glass is brittle but not the usual engineering metal specimen in standard tensile testing contexts for this question. Common Pitfalls: Misinterpreting a slightly rough surface as necking; assuming all metals behave ductilely; ignoring standardized lab examples used in exam problems. Final Answer: Cast iron

Question 17

Thermal stress in a fully restrained bar: proportionality with temperature change Fill the blank: Thermal stress is ______ proportional to the change in temperature (for no expansion allowed).

Accepted Answer

Introduction / Context: Thermal stresses arise when thermal expansion or contraction is restrained. In many practical cases, bars or members cannot freely extend, producing internal stresses that must be considered in design to avoid cracking or yielding. Given Data / Assumptions: Uniform temperature change delta_T. Member is fully restrained from changing length. Material remains within elastic range with modulus E and coefficient of thermal expansion alpha. Concept / Approach: Free thermal strain would be alpha * delta_T. If restraint prevents this strain, an equal and opposite mechanical strain develops to satisfy compatibility. Using Hooke's law, stress equals E times mechanical strain. Step-by-Step Solution: Free strain: epsilon_free = alpha * delta_T.Restrained condition enforces epsilon_total = 0.Mechanical strain: epsilon_mech = - alpha * delta_T.Stress: sigma = E * epsilon_mech = - E * alpha * delta_T.Magnitude: |sigma| = E * alpha * delta_T, which is directly proportional to delta_T. Verification / Alternative check: If partial restraint exists, stress scales with the effective restraint stiffness; still, for a given restraint, sigma increases linearly with delta_T until yielding or creep reduces stress. Why Other Options Are Wrong: Inversely or independent: contradicts the linear relation. Square dependence does not apply in linear elasticity. Saying it depends only on modulus neglects the role of delta_T and alpha. Common Pitfalls: Forgetting sign conventions (tension vs compression); ignoring temperature gradients which cause bending rather than pure axial stress. Final Answer: directly

Question 18

Spring behavior under load: elasticity vs permanent distortion Statement: A spring, when loaded, is permanently distorted and recovers its original shape when the load is removed.

Accepted Answer

Introduction / Context: Springs are energy-storing elements designed to operate primarily within the elastic range so that they return to original geometry after unloading. The statement mixes the terms permanently distorted and recovers, which are mutually exclusive in mechanics. Given Data / Assumptions: Conventional helical spring made from spring steel. Loading within the design range, not exceeding yield. Concept / Approach: Elastic deformation is reversible; plastic deformation is permanent. Properly designed springs operate below yield stress so that deflection is recoverable. Permanent set indicates plasticity, which is not desired in normal service. Step-by-Step Solution: If load is within elastic range: no permanent distortion; full recovery occurs.If load exceeds yield: permanent set occurs and complete recovery does not happen.Therefore the statement claiming permanent distortion yet full recovery is internally inconsistent and wrong for normal operation. Verification / Alternative check: Design formulas for springs use allowable shear stress below yield with safety factors; fatigue design also ensures stresses remain reversible over cycles. Why Other Options Are Wrong: Agree and the conditional variants conflict with elastic design principles. Impact or temperature conditions may change allowable limits, but permanent set still contradicts full recovery. Common Pitfalls: Confusing temporary elastic deflection with permanent set; assuming that any large deflection means plastic deformation. Final Answer: Disagree

Question 19

Neutral axis in bending: stress state at the neutral axis of a beam section Fill the blank with the correct stress condition.

Accepted Answer

Introduction / Context: In pure bending of beams, stresses vary linearly across the section depth. The line where the bending stress changes sign is called the neutral axis. Recognizing the stress condition at this axis is fundamental to section design and to locating tension and compression zones. Given Data / Assumptions: Plane sections remain plane (Bernoulli–Euler assumption). Linear elastic material and small deflections. No axial force superposed, only bending. Concept / Approach: The normal bending stress is sigma = (M * y) / I, where y is measured from the neutral axis. At the neutral axis, y = 0, so sigma = 0. Shear stress distribution is separate and has its own neutral line discussion, but in most standard cases maximum shear occurs at the centroidal axis, not necessarily the same as zero bending stress statement. Step-by-Step Solution: Use sigma = M * y / I.At y = 0 (neutral axis), sigma = 0.Above the neutral axis: one sign (compression); below it: opposite sign (tension), depending on bending sense. Verification / Alternative check: Strain distribution is linear: epsilon = curvature * y; at y = 0, epsilon = 0, hence stress is zero by Hooke's law. Why Other Options Are Wrong: Maximum tensile or compressive occurs at extreme fibers (|y| is maximum). Minimum tensile is not a standard descriptor. Maximum shear does not coincide with zero normal stress fact in bending discussion here. Common Pitfalls: Confusing shear stress distribution with normal bending stress; assuming neutral axis carries some residual normal stress. Final Answer: zero

Question 20

Strength of Materials — Columns and Struts At what load does a slender column just begin to buckle (lose lateral stability) under axial compression?

Accepted Answer

Introduction: Columns fail by buckling long before the material reaches its crushing strength. The precise load at which a perfectly straight, slender column just becomes laterally unstable is a fundamental idea in strength of materials and design of compression members. Given Data / Assumptions: Member: long, slender column. Loading: concentric axial compression. Ideal behavior: linearly elastic, small initial imperfections neglected for definition. Concept / Approach: The onset of buckling is predicted by Euler's theory. The load that makes a column neutrally stable (neither straight nor bent without additional load) is called the critical load. Many texts also use the term crippling load for the same quantity in the elastic range. Buckling load is a generic phrase, but the standard technical term is critical load. Step-by-Step Solution: For an ideal pin–pin column: P_cr = (pi^2 * E * I) / (L^2)For other end conditions: replace L with L_e (effective length).At P = P_cr, any small lateral deflection can persist without additional load, indicating the threshold of instability. Verification / Alternative check: Design codes (e.g., using KL/r) identify the same threshold as the transition from stable shortening to lateral instability. Why Other Options Are Wrong: buckling load: colloquial; not the standard precise term in formulas.crippling load: used in some texts, but critical load is the accepted technical name.any one of these: terminology matters in design; only critical load is unambiguous. Common Pitfalls: Confusing crushing strength with buckling strength; slender columns fail at much lower loads.Ignoring end conditions (effective length) when computing P_cr. Final Answer: critical load

Strength of Materials Questions