Question 1

Section modulus of a beam: The section modulus Z used in flexural design is expressed in terms of the second moment of area I and the distance to the extreme fiber y. What is Z?

Accepted Answer

Introduction / Context: Section modulus Z is a geometric property that links bending moment to maximum normal stress at the outer fiber of a beam. It is widely used to size beams in structural and machine design. Given Data / Assumptions: Linear elastic behavior and Bernoulli–Euler beam theory. Maximum stress occurs at the extreme fiber located a distance y from the neutral axis. Concept / Approach: The flexure formula is σ_max = M * y / I. Rearranging gives M / Z = σ_max where Z = I / y. Thus for a given allowable stress, the required section modulus is M / σ_allow, translating directly into geometry selection. Step-by-Step Solution: Flexure formula: σ = M * y / I.At the extreme fiber: σ_max corresponds to y = y_max.Define Z = I / y_max so that σ_max = M / Z.Therefore, Z = I / y. Verification / Alternative check: For standard shapes (rectangular, circular, I-sections), published Z values equal I / y_max, confirming the definition. Example: for a rectangle b × h, I = b h^3 / 12 and y = h/2, giving Z = b h^2 / 6, a well-known result. Why Other Options Are Wrong: I * y and y / I have incorrect dimensions. M / I is unrelated to Z; it appears in curvature κ = M / (E I). I / y^2 does not match flexure relations for stress at the extreme fiber. Common Pitfalls: Using the wrong y (not the extreme fiber); confusing elastic section modulus with plastic section modulus (Zp), which is different and used in plastic design. Final Answer: I/y

Question 2

Automotive suspension springs: Is it correct to say that the springs in cars are used to store strain energy for ride comfort and load management?

Accepted Answer

Introduction / Context: Vehicle suspension systems use springs (coil, leaf, torsion bar, or air springs) to improve ride comfort, maintain tire contact, and protect the chassis. The fundamental role of a spring is to store and release strain energy when subjected to loads and road irregularities. Given Data / Assumptions: Conventional passive suspension with springs and dampers (shock absorbers). Normal driving conditions with variable road inputs. Concept / Approach: Springs store strain energy U = ∫ F dx, smoothing load transmission from the road. Dampers dissipate energy to control oscillations. Together, they filter road disturbances; the spring’s energy storage is crucial for transient load buffering and for maintaining tire-ground contact. Step-by-Step Solution: Road bump applies an upward force; the spring compresses, storing energy.As the bump passes, the spring releases energy, returning the wheel toward equilibrium.The damper converts part of this kinetic/strain energy to heat, preventing sustained oscillations. Verification / Alternative check: From elementary vibration theory, a sprung mass–spring–damper system exhibits reduced transmissibility in a target frequency range, with the spring’s compliant energy storage central to performance. Why Other Options Are Wrong: “Incorrect” ignores the fundamental mechanical role of a spring. Speed- or vehicle-type-limited statements are false; the principle applies across use cases. “Only static loads” is wrong; dynamic inputs are the main reason for springs. Common Pitfalls: Confusing the roles of spring (stores energy) and damper (dissipates energy); assuming stiffer springs always improve handling without considering comfort and tire grip trade-offs. Final Answer: Correct

Question 3

Cantilever with triangular (linearly varying) load: A cantilever beam of length l carries a load varying from zero at the free end to w per unit length at the fixed end. What is the shear force at the fixed end?

Accepted Answer

Introduction / Context: Linearly varying distributed loads are common in structural members. For cantilevers with a triangular load intensity that is zero at the tip and maximum at the fixed end, the reaction shear and moment at the fixed support are obtained from resultants of the equivalent load diagram. Given Data / Assumptions: Cantilever length = l. Load intensity varies linearly from 0 (free end) to w (fixed end). Static equilibrium; prismatic member. Concept / Approach: The total load equals the area under the load-intensity diagram. For a triangle of base l and height w, the resultant load is (1/2) * w * l, acting at the centroid of the triangle located at a distance l/3 from the high-intensity end (i.e., from the fixed end) or 2l/3 from the free end. The reaction shear at the fixed end equals this resultant (taking the cantilever as a free body). Step-by-Step Solution: Equivalent resultant load: W = (1/2) * w * l.Location: from the fixed end, a distance l/3.Equilibrium of the beam: vertical reaction (shear) at fixed end V_fixed = W = w l / 2. Verification / Alternative check: Integrate the load intensity q(x) = (w / l) * x measured from the free end to the fixed end: ∫_0^l q(x) dx = ∫_0^l (w/l) x dx = (w/l) * (l^2 / 2) = w l / 2, confirming the resultant magnitude equals the support shear. Why Other Options Are Wrong: zero: contradicts presence of a distributed load. w l / 4: underestimates by half. w l and 2 w l: overestimates the triangular resultant. Common Pitfalls: Placing the resultant at l/3 from the free end (incorrect); confusing triangular versus uniform loads (for uniform, resultant would be w l with different symbols). Final Answer: w l / 2

Question 4

Riveting practice: Relative to the rivet shank diameter, how is the plate hole size selected for drilling before riveting?

Accepted Answer

Introduction / Context: In riveted joints, holes are made slightly larger than the rivet shank to facilitate insertion, accommodate manufacturing tolerances, and allow proper rivet expansion during setting. Using the correct hole size is essential for accurate strength calculations and reliable fabrication. Given Data / Assumptions: Conventional hot- or cold-driven rivets in metal plates. Standard allowances of approximately 0.5 mm to 1.5 mm above shank diameter depending on size and practice. Concept / Approach: The hole diameter must exceed the nominal shank diameter to permit insertion and to ensure the rivet upsets properly to fill the hole during driving. Design calculations for net section, rivet shear, and bearing use the hole diameter because it governs the effective areas after drilling/punching and reaming where applicable. Step-by-Step Solution: Select a hole drill size slightly larger than the rivet shank diameter.Account for fit, tolerance, and intended expansion during riveting.Use the hole diameter in strength checks to remain conservative. Verification / Alternative check: Fabrication standards and handbooks specify hole oversize; inspection gauges and reamers are chosen accordingly. Empirical practice aligns with using larger holes to avoid damage during driving and to ensure full seating of the rivet head and shank. Why Other Options Are Wrong: Equal or smaller holes hinder insertion and risk plate damage or misalignment. “Depends only on thickness or material” ignores the primary fit and tolerance requirement; while these affect allowance magnitude, the direction is always “greater than”. Common Pitfalls: Using rivet diameter instead of hole diameter in calculations; neglecting clearance leading to practical assembly issues and residual stresses. Final Answer: greater than

Question 5

Elastic constants relationship: if a material has modulus of elasticity E equal to twice its modulus of rigidity C (i.e., E = 2C), what is the corresponding bulk modulus K for that material?

Accepted Answer

Introduction / Context: Designers often convert between the three fundamental elastic constants for isotropic materials: Young’s modulus E, modulus of rigidity C (also written G), and bulk modulus K. This question checks your fluency in using the standard relations to move from one constant to another. Given Data / Assumptions: E = 2C (i.e., E = 2G). Material is homogeneous, linear-elastic, and isotropic. Standard relationships among E, C, K, and Poisson’s ratio ν apply. Concept / Approach: For isotropic materials, the following formulas hold: E = 2C * (1 + ν) E = 3K * (1 − 2ν) Alternatively, K = E / (3 * (1 − 2ν)) and C = E / (2 * (1 + ν)) Step-by-Step Solution: From E = 2C * (1 + ν) and E = 2C, we get 2C = 2C * (1 + ν).Therefore, 1 + ν = 1 ⇒ ν = 0.Use K = E / (3 * (1 − 2ν)). With ν = 0, K = E / 3.Since E = 2C, substitute to get K = (2C) / 3 = 2C/3. Verification / Alternative check: Using the well-known identity E = 9KC / (3K + C). Set E = 2C and solve for K to confirm K = 2C/3. Why Other Options Are Wrong: 2C, 3C: These ignore the required division by 3 from the E–K–ν relationship.3C/2, C/2: Do not satisfy the isotropic consistency conditions when E = 2C. Common Pitfalls: Forgetting that ν is implicated through E = 2C(1 + ν); if E = 2C, it forces ν = 0, not any arbitrary value. Always solve for ν first when constants are related in this way. Final Answer: 2C/3

Question 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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 17

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 18

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 19

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 20

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

Strength of Materials Questions