Question 1

Elastic energy terminology The capacity of a strained body to do work upon removal of the straining force is called:

Accepted Answer

Introduction / Context: When loads are applied to an elastic body, work is done and stored as recoverable potential energy. The correct terminology distinguishes total stored energy, energy at the elastic limit, and energy to fracture. Using precise terms avoids ambiguity in design calculations, fatigue assessment, and impact problems. Given Data / Assumptions: Body deformed elastically by external force. Upon unloading, energy is released as the body returns toward its original shape. Linear elasticity for simplicity. Concept / Approach: Strain energy is the total energy stored due to deformation, recoverable on unloading. Resilience is the ability to absorb energy elastically; proof resilience is the strain energy at the elastic (proof) limit. Modulus of toughness pertains to the total area under the stress–strain curve up to fracture (energy per unit volume), and impact energy is typically measured in Charpy/Izod tests up to fracture. Step-by-Step Solution: Define strain energy U = ∫ F dx (or = ∫ sigma d epsilon * volume).For linear springs: U = 1/2 * k * x^2; for linear stress–strain: energy density = 1/2 * sigma * epsilon.Upon unloading, recoverable work equals stored strain energy. Verification / Alternative check: Energy conservation and unloading path for linear elasticity show the same triangular area under the force–deflection curve being released. Why Other Options Are Wrong: Resilience describes capacity in the elastic range but is not the general term for total stored energy. Proof resilience refers specifically to the energy at the elastic limit. Impact energy and toughness involve fracture processes, not purely elastic recovery. Common Pitfalls: Interchanging resilience with strain energy; ignoring the distinction between total energy and energy per unit volume (moduli). Final Answer: strain energy

Question 2

Riveted joints: definition of back pitch in zig-zag riveting Statement: The distance between the centres of rivets in adjacent rows of a zig-zag riveted joint is called back pitch.

Accepted Answer

Introduction / Context: In plate-to-plate connections, rivet layout parameters govern strength against tearing, shearing, and bearing failures. Standard terms include pitch (longitudinal spacing), back pitch (transverse spacing between adjacent rows), and diagonal pitch in zig-zag patterns. Knowing these helps compute net section and staggered pitch efficiency. Given Data / Assumptions: Zig-zag riveted joint with multiple rows. Centers of rivets are used for spacing definitions. Concept / Approach: Back pitch is the perpendicular spacing between adjacent rows of rivets. In zig-zag riveting, holes are staggered; the diagonal pitch relates a hole in one row to the nearest in the adjacent row along the joint direction, while back pitch is the transverse distance between rows irrespective of staggering. Step-by-Step Solution: Identify joint direction (longitudinal axis).Measure spacing between row centre lines perpendicular to this axis.This spacing is the back pitch, matching the statement. Verification / Alternative check: Design handbooks and codes define p (pitch), pb (back pitch), and pd (diagonal) consistently; diagrams show back pitch as transverse spacing. Why Other Options Are Wrong: False contradicts standard definition; qualifiers (chain, lap, butt) are unnecessary because the definition of back pitch does not depend on joint type. Common Pitfalls: Confusing diagonal pitch with back pitch; measuring edge distance instead of row-to-row spacing. Final Answer: True

Question 3

Elastic behavior within the elastic limit Statement: If a material is loaded within its elastic limit, it will regain its original shape and size on removal of load.

Accepted Answer

Introduction / Context: The elastic limit separates reversible (elastic) and permanent (plastic) deformation. In engineering design, service stresses are kept below the elastic limit so that components return to their original geometry after unloading and dimensional tolerances are maintained. Given Data / Assumptions: Material behaves linearly up to elastic limit. Loading and unloading are quasi-static, no significant creep. No residual stresses from prior plastic deformation. Concept / Approach: Within the elastic limit, strain is proportional to stress (Hooke’s law) and the unloading path coincides with the loading path. Therefore, when stress is removed, elastic strain vanishes and the body recovers its original dimensions. Step-by-Step Solution: Apply stress sigma less than elastic limit.Induced strain epsilon = sigma / E.Unload: stress returns to zero, and elastic strain recovers to zero.Result: original shape and size are regained. Verification / Alternative check: Tension tests show linear loading–unloading loops below yield with negligible hysteresis; permanent set appears only after yielding. Why Other Options Are Wrong: Material class, temperature (within normal ranges), and loading sense do not negate the definition of elastic recovery. Common Pitfalls: Assuming microplasticity below yield under high cycles; mixing up proportional limit with elastic limit in materials showing slight nonlinearity. Final Answer: Agree

Question 4

Reinforced concrete beams: definition of the critical neutral axis Statement: The critical neutral axis of an RCC beam is located at the centre of gravity of the given section.

Accepted Answer

Introduction / Context: In reinforced concrete (RCC), the position of the neutral axis depends on the transformed section under bending, not merely the geometric centroid. The critical neutral axis separates tension and compression zones and is key to determining whether a section is under-reinforced, balanced, or over-reinforced. Given Data / Assumptions: Transformed section method or limit-state stress block. Concrete carries compression; steel primarily carries tension. Plane sections remain plane; bond is adequate. Concept / Approach: The neutral axis is obtained from force equilibrium: compressive resultant in concrete equals tensile force in steel (plus compression steel, if any). Its position therefore depends on material properties (modular ratio m or design stress block) and reinforcement area and placement, not on the centroid of the gross concrete section. Step-by-Step Solution: Transform steel area to equivalent concrete area using m = Es / Ec (working stress) or use code stress block (limit state).Compute compressive stress block and tensile force in steel.Set compression = tension and solve for neutral axis depth x.Critical neutral axis corresponds to the balanced failure condition (simultaneous concrete crushing and steel yielding), not the geometric centroid. Verification / Alternative check: Code tables provide limiting neutral axis depth (e.g., x_u,lim / d depends on steel grade), confirming dependence on material behavior and reinforcement, not on geometric centroid alone. Why Other Options Are Wrong: Statements limiting to section type or m = 1 are irrelevant; the NA is controlled by equilibrium and compatibility with material laws. Common Pitfalls: Using gross-section centroid for RCC bending; ignoring compression reinforcement and tension steel position in NA computation. Final Answer: False

Question 5

Columns under eccentric load: core for circular sections Limit of eccentricity e for no tensile stress in a circular column of diameter d equals:

Accepted Answer

Introduction / Context: The core (or kern) of a section is the locus of load application points that produce compressive stress everywhere with no tension. For masonry or concrete columns where tension is undesirable, ensuring the load resultant lies within the core prevents cracking. Given Data / Assumptions: Circular cross-section, diameter d. Linear elastic stress distribution (Bernoulli hypothesis). Axial load with eccentricity e about one principal axis. Concept / Approach: For a section with axial load P and moment M = P * e, extreme fiber stress is sigma = (P / A) ± (M * y / I). For no tension, the minimum stress must be ≥ 0. The core radius for a circle is r_core = d / 8, meaning any e ≤ d / 8 keeps the entire section in compression. Step-by-Step Solution: For a circle: A = pi * d^2 / 4; I = pi * d^4 / 64; c = d / 2.Set sigma_min = P / A - M * c / I ≥ 0.Substitute M = P * e and rearrange to e ≤ (I / (A * c)).Compute I / (A * c) = (pi d^4 / 64) / [(pi d^2 / 4) * (d / 2)] = d / 8.Therefore, limit of eccentricity for no tension is e ≤ d / 8. Verification / Alternative check: For a rectangle b × h, core limits are ±b/6 and ±h/6; the circular case giving d/8 is consistent with standard kern values. Why Other Options Are Wrong: d/4 is too large and would cause tensile stress; d/12 and d/16 are conservative but not the theoretical limit; d/6 pertains to rectangles, not circles. Common Pitfalls: Using rectangular core limits for all shapes; forgetting to use I/(A*c) for the specific section geometry. Final Answer: d / 8

Question 6

Applicability of Rankine’s column formula Rankine’s empirical formula is valid for:

Accepted Answer

Introduction / Context: Rankine’s formula blends crushing failure (dominant in short columns) and elastic buckling (dominant in long columns) to estimate critical load. It is widely used because it provides a smooth transition across slenderness ratios where neither pure Euler nor pure crushing alone is adequate. Given Data / Assumptions: Empirical constants derived from tests. Applies to prismatic columns with typical end conditions (via effective length). Material behaves elastically up to failure modes considered. Concept / Approach: Rankine’s formula: P_cr = (sigma_c * A) / [1 + a * (L_e / r)^2], combining material crushing strength sigma_c and slenderness (L_e / r). For very small slenderness, denominator ≈ 1 (crushing). For large slenderness, it trends toward Euler behavior. Step-by-Step Solution: Identify effective length L_e and radius of gyration r.Compute slenderness lambda = L_e / r.Use Rankine formula across lambda values to obtain a single P_cr expression applicable to both regimes. Verification / Alternative check: Limiting cases confirm: lambda → 0 ⇒ crushing; lambda → large ⇒ Euler-like response through 1/(lambda^2) behavior. Why Other Options Are Wrong: Restricting to only short or only long columns ignores the blended nature. End condition exclusivity is handled via L_e, not by the formula’s scope. “Weak columns only” is not a standard category. Common Pitfalls: Using Euler for stocky columns; ignoring effective length factors; misusing crushing stress without stability check. Final Answer: both short and long columns

Question 7

Bending stress–moment relationship in beams Fill the blank: The bending stress in a beam section is __________ the bending moment at that section (other factors constant).

Accepted Answer

Introduction / Context: The flexure formula relates bending stress to internal bending moment and section properties. Understanding proportionality guides selection of sections with appropriate moment of inertia to limit stresses under a given loading. Given Data / Assumptions: Bernoulli–Euler theory: plane sections remain plane. Linear elastic material; small deflections. Stress evaluated at a given fiber distance y from the neutral axis. Concept / Approach: The flexure formula is sigma = M * y / I. For a fixed section (I and y fixed at a particular fiber), sigma varies directly with M. Therefore, as bending moment increases, bending stress increases proportionally at that location. Step-by-Step Solution: Start with sigma = M * y / I.Hold y and I constant for the chosen fiber and section.Conclude sigma ∝ M (direct proportionality). Verification / Alternative check: Dimensional analysis and linearity of constitutive law corroborate the proportional relationship in elastic bending. Why Other Options Are Wrong: Equal/less/more than are not functional relationships; “inversely proportional” contradicts the formula. Common Pitfalls: Comparing stresses at different fibers (varying y); forgetting that changing section (I) alters the proportionality constant. Final Answer: directly proportional to

Question 8

Elastic Constants — Shear Modulus (Modulus of Rigidity) What is shear modulus defined as, in terms of fundamental stress–strain ratios for small deformations?

Accepted Answer

Introduction: Shear modulus, also called modulus of rigidity and denoted by G, quantifies how a material resists shape change under tangential loading. It complements the modulus of elasticity E (normal loading) and bulk modulus K (volumetric loading), forming the core elastic constants used in strength of materials and machine design. Given Data / Assumptions: Small strains and linear elastic behavior. Homogeneous, isotropic material so standard relationships among E, G, K, and Poisson ratio v hold. Pure shear loading leading to angular distortion without significant volume change. Concept / Approach: By definition, shear modulus G is the ratio of shear stress tau to shear strain gamma for small deformations: G = tau / gamma. For isotropic linear elastic solids, G relates to E and v through G = E / (2 * (1 + v)), showing why G is typically a little less than E/2 when v is between 0.25 and 0.35. Step-by-Step Solution: Identify loading mode: tangential force causes layers to slide relative to each other.Measure shear stress: tau = V / A on the considered plane.Measure shear strain: gamma is the change in right angle between orthogonal lines in radians.Apply definition: G = tau / gamma. Verification / Alternative check: Use isotropic relation G = E / (2 * (1 + v)) to cross-check measured values of E and v from tests; the computed G should match tau / gamma from shear tests. Why Other Options Are Wrong: linear stress to linear strain: that is modulus of elasticity E in tension or compression.linear stress to lateral strain: this does not represent a standard elastic constant.volumetric strain to linear strain: this is a nonsensical ratio for defining a basic modulus. Common Pitfalls: Confusing shear modulus with E; note that shear involves angular change, not normal extension.Using large-strain values where linear definitions no longer apply. Final Answer: shear stress to shear strain

Question 9

Cantilever with Uniformly Distributed Load (UDL) For a cantilever of length l carrying a uniformly distributed load w per unit length, what is the bending moment at the free end?

Accepted Answer

Introduction: Bending moment diagrams (BMD) and shear force diagrams (SFD) are foundational tools in beam analysis. Recognizing key values at supports and free ends for standard loads is essential for quick design checks. Given Data / Assumptions: Beam type: cantilever of length l. Loading: uniformly distributed load w per unit length over the entire span. Small deflection, linear elastic behavior; prismatic member. Concept / Approach: For a cantilever with UDL, the shear and moment vary with distance x measured from the free end (or from the fixed end, consistently). At the free end there is no support to provide a resisting couple. Hence, the bending moment must be zero at the free end, and it reaches its maximum magnitude at the fixed end. Step-by-Step Solution: Resultant of UDL: W_total = w * l acting at midspan from the free end.Shear at a section x from the free end: V(x) = w * x.Moment at the same section: M(x) = w * x^2 / 2 (when origin is at the free end).At x = 0 (free end): M(0) = 0, confirming zero bending moment. Verification / Alternative check: Compute the fixed end moment: M_fixed = w * l^2 / 2. The BMD varies quadratically from 0 at the free end to this maximum at the fixed end, consistent with theory. Why Other Options Are Wrong: w l / 4 and w l / 2: nonzero values contradict boundary condition at the free end.w l: also violates the free-end zero-moment condition and has wrong dimensions if units are not balanced. Common Pitfalls: Mixing up shear and moment boundary conditions; remember free end has zero moment and zero reaction.Placing the origin at the fixed end and forgetting M = 0 at the free end by boundary condition. Final Answer: zero

Question 10

Simply Supported Beam with UDL A simply supported beam of span l carries a uniformly distributed load w over the whole span. What are the shear forces at the two supports (beam ends)?

Accepted Answer

Introduction: Support reactions for standard loading cases are frequently used to sketch SFD and BMD quickly. For a symmetric UDL on a simply supported beam, reactions (and hence end shears) are equal and opposite at the two supports. Given Data / Assumptions: Beam: simply supported, span l. Load: UDL of intensity w across entire span. Linear elastic behavior; negligible self-weight beyond w if not included. Concept / Approach: The resultant load equals w * l and acts at midspan. By symmetry, each support carries half the resultant: R_A = R_B = w * l / 2. Shear just to the right of the left support equals +R_A and just to the left of the right support equals -R_B (sign convention). Step-by-Step Solution: Compute resultant: W_total = w * l.Use symmetry: R_A = R_B = W_total / 2 = w * l / 2.Shear at ends: V_left = + w * l / 2, V_right = - w * l / 2. Verification / Alternative check: Area under SFD over the span equals the net change in bending moment, which matches the known parabolic BMD M_max = w * l^2 / 8 at midspan. Why Other Options Are Wrong: zero at both ends: contradicts equilibrium; supports must resist load.w l and - w l: each end carrying full load is impossible for symmetric simply supported case.w l^2 / 2 terms have wrong dimensions for shear force. Common Pitfalls: Confusing shear force with bending moment expressions (note power of l).Sign convention mistakes at the right support. Final Answer: w l / 2 at one end and - w l / 2 at the other end

Question 11

Eccentrically Loaded Struts — Choice of Section For compression members subjected to eccentric axial load, which type of cross-section is generally preferred and why?

Accepted Answer

Introduction: Eccentric compression introduces combined axial stress and bending. The selection of cross-section must therefore consider section modulus, radius of gyration, and material economy to limit maximum compressive stress and reduce buckling risk. Given Data / Assumptions: Member: slender strut/column under axial load with eccentricity e. Material: isotropic, linear elastic. Design aims: high stiffness and strength per unit weight, improved stability (larger radius of gyration) and better distribution away from the neutral axis. Concept / Approach: For the same area, hollow (tubular) sections place more material away from the centroid, increasing both section modulus and radius of gyration k. This reduces bending stress sigma_b = M / Z and increases Euler capacity P_cr proportional to (E * I) / (L_e^2) since I = A * k^2 grows with k when area is away from the center. Step-by-Step Solution: Model combined stress: sigma_max = P / A + M / Z with M = P * e.For given A and e, larger Z reduces M / Z hence lowers peak stress.Hollow sections maximize Z and k for a given mass by shifting material outward.Buckling capacity improves since I = A * k^2 increases with k. Verification / Alternative check: Compare a solid round bar and a tube with the same mass: the tube shows higher I and Z, confirmed by standard section property tables. Why Other Options Are Wrong: solid section: concentrates material near the center, giving smaller Z and k for the same weight.composite section: may be useful but the generic preference for eccentric loading is a hollow section due to efficiency.reinforced section: vague; does not specify geometry that improves Z and k. Common Pitfalls: Ignoring eccentricity in stress checks and relying only on P / A.Forgetting the role of effective length and end conditions on buckling capacity. Final Answer: hollow section

Question 12

Laminated (Leaf) Springs — Maximum Bending Stress For a semi-elliptic leaf spring carrying a central load W with span l and n plates (each of width b and thickness t), what is the expression for the maximum bending stress in the plates?

Accepted Answer

Introduction: Laminated springs distribute bending stresses across several thin plates. Designers need formulas for maximum bending stress and tip deflection to proportion b and t for a target load W and allowable stress, balancing weight and ride comfort in vehicles. Given Data / Assumptions: Semi-elliptic leaf spring of span l with n identical plates. Each plate has width b and thickness t; plates are in contact and share load. Central load W; linear elastic behavior; friction between leaves neglected in stress formula. Concept / Approach: The bending stress in each plate is derived from beam theory using the composite section. The standard result for a semi-elliptic spring under mid-load is sigma_max = (3 * W * l) / (2 * n * b * t^2). Deflection is often y = (3 * W * l^3) / (8 * E * n * b * t^3), showing the strong influence of thickness t. Step-by-Step Solution: Model each plate as a beam; total bending moment is shared by n plates.Use flexure: sigma = M * y / I; for a rectangular plate I_plate = b * t^3 / 12 and y = t / 2.Combine for n plates under central load W on span l to obtain sigma_max.Final result: sigma_max = (3 * W * l) / (2 * n * b * t^2). Verification / Alternative check: Dimensional check: numerator forcelength; denominator arealength^2 yields stress units as required. Why Other Options Are Wrong: (6 * W * l) / (n * b * t^2): incorrect factor; overestimates stress by 4 times.(W * l) / (n * b * t): wrong thickness power; stress scales with 1 / t^2.(3 * W * l) / (8 * n * b * t^3): mixes deflection formula into stress expression with wrong t power. Common Pitfalls: Confusing stress and deflection formulae; note t^2 vs t^3 dependence.Ignoring number of plates n when sizing leaf thickness. Final Answer: sigma_max = (3 * W * l) / (2 * n * b * t^2)

Question 13

Relation between G and E For most engineering materials, how does shear modulus G compare with half the modulus of elasticity E?

Accepted Answer

Introduction: Knowing how elastic constants relate allows engineers to estimate one property from another. The ratio between shear modulus G and modulus of elasticity E depends on Poisson ratio v for isotropic materials. Given Data / Assumptions: Isotropic, homogeneous, linear elastic materials. Typical Poisson ratio v between about 0.25 and 0.35 for metals. Concept / Approach: The fundamental relation is G = E / (2 * (1 + v)). Because v is positive for stable materials, the denominator exceeds 2. Hence G is strictly less than E / 2 for ordinary materials (v > 0). If v were zero, G would equal E / 2, but that is uncommon in practice. Step-by-Step Solution: Start with isotropic relation: G = E / (2 * (1 + v)).For v = 0.3 (typical steel): G = E / (2 * 1.3) = E / 2.6 which is less than E / 2.Thus, for most materials with v between 0.25 and 0.35, G < E / 2. Verification / Alternative check: Cross-check with measured values: for mild steel E ≈ 200 GPa, G ≈ 80 GPa; 80 is less than 100 which equals E / 2. Why Other Options Are Wrong: equal to half: only true if v = 0, rarely the case.more than half: impossible for v > 0 because 1 + v > 1 makes denominator > 2.none of these: incorrect since the correct qualitative relation is known. Common Pitfalls: Forgetting the role of v and assuming a fixed ratio independent of material. Final Answer: less than half

Question 14

Mild Steel — Ultimate Strength in Tension vs Compression For mild steel tested under quasi-static loading, the ultimate tensile strength is generally how related to the ultimate compressive strength?

Accepted Answer

Introduction: Material behavior in tension and compression can differ significantly. Brittle materials like cast iron are much stronger in compression than tension, whereas ductile metals such as mild steel show comparable strength in both modes under standard tests. Given Data / Assumptions: Material: mild steel (ductile, low carbon). Quasi-static tests; effects of buckling or instability not governing the compressive result. Uniform specimens and standard test conditions. Concept / Approach: For ductile metals, yielding is governed by shear-dominated criteria, making tensile and compressive flow behavior similar. Ultimate values measured in laboratory compression (with short, stocky specimens to avoid buckling) are close to tensile ultimate strengths, unlike brittle materials. Step-by-Step Solution: Recognize ductile response: large plastic strains before fracture.Compression test on stubby specimen avoids buckling; stress–strain curve mirrors tensile plastic flow up to similar stress levels.Conclude ultimate strengths are approximately equal for mild steel. Verification / Alternative check: Handbook data typically list UTS for mild steel around 400–550 MPa and compressive ultimate of similar magnitude when buckling is not limiting. Why Other Options Are Wrong: less than: not characteristic of ductile steels; applies more to certain polymers or special conditions.more than: strong compression dominance is typical of brittle materials like cast iron, not mild steel.not applicable: laboratories do define compressive limits with proper specimens. Common Pitfalls: Confusing structural buckling (a stability failure) with material compressive strength; columns may fail early due to buckling, but that is not material ultimate strength. Final Answer: equal to

Question 15

Clockwork Mechanisms — Energy in Watch Springs In a mechanical watch, the coiled spring stores elastic strain energy. This stored energy is released primarily in order to do what?

Accepted Answer

Introduction: Mechanical timepieces rely on energy storage and regulated release. The mainspring or balance spring stores strain energy during winding and releases it gradually to drive the gear train, enabling accurate timekeeping. Given Data / Assumptions: Classic mechanical watch with a wound spring. Energy conversion from elastic potential to mechanical work on the gear train. Escapement mechanism regulates rate of energy release. Concept / Approach: When wound, the spring accumulates strain energy U = integral of torque * d(theta). The escapement meters this energy, allowing the watch to run at a controlled rate rather than releasing it at once. Therefore, the purpose of the stored energy is to power continuous motion of the hands via the gear train. Step-by-Step Solution: Winding increases elastic energy in the spring due to torsional deformation.The spring torque drives the center wheel, third wheel, fourth wheel, and escape wheel.The escapement and balance regulate motion into discrete steps, producing steady hand movement. Verification / Alternative check: Observation: as the spring unwinds, the watch runs; when energy is exhausted, the watch stops, confirming the function. Why Other Options Are Wrong: to stop the watch: the opposite of the function; stopping occurs when energy is depleted.to change the time: time setting is a manual operation via crown and clutch, not driven by stored energy.all of these: includes incorrect purposes, hence wrong. Common Pitfalls: Confusing winding (energy input) with regulation of rate by the escapement. Final Answer: to run the watch

Question 16

Simply Supported Rectangular Beam — Deflection under Central Point Load A rectangular beam of span l, simply supported at both ends, carries a central point load W. Where along the beam does the maximum deflection occur?

Accepted Answer

Introduction: Deflection shapes provide insight into stiffness and serviceability. For statically determinate beams with symmetric loading, the location of maximum deflection can often be inferred from symmetry and slope conditions without solving the full differential equation. Given Data / Assumptions: Simply supported beam, span l. Central point load W. Prismatic beam; linear elastic bending; small deflection theory. Concept / Approach: The loading and boundary conditions are symmetric about midspan, so the deflected shape is symmetric and slope is zero at the center. The point of zero slope is typically an extremum of deflection, giving the location of maximum deflection at the center. Step-by-Step Solution: Write standard result for midspan deflection: y_max = (W * l^3) / (48 * E * I).Symmetry dictates dy/dx = 0 at x = l / 2.Boundary conditions y(0) = y(l) = 0 at the supports; thus the interior extremum occurs at midspan. Verification / Alternative check: Area-moment method or conjugate-beam method both yield the same midspan location and value. Why Other Options Are Wrong: at the ends: deflection is constrained to zero at simple supports.at l / 3: not an extremum for this loading; applies to other cases (e.g., certain distributed loads) only under special conditions.none of these: incorrect since midspan is correct. Common Pitfalls: Confusing the location of maximum moment (also at midspan here) with other load cases where it shifts. Final Answer: at the centre

Question 17

Helical Springs — Effect of Cutting into Two Halves A closely coiled helical spring is cut into two equal halves. How does the stiffness (spring constant) of each resulting spring compare to the original?

Accepted Answer

Introduction: Spring stiffness determines load–deflection behavior. Understanding how stiffness changes when altering the number of active coils helps when adapting springs to new applications by cutting or combining coils. Given Data / Assumptions: Closely coiled helical spring with wire diameter d, mean coil diameter D, and N active turns. Material shear modulus G unchanged by cutting. Ends of the new springs are finished so that the number of active turns halves approximately. Concept / Approach: The stiffness k of a closely coiled helical spring is k = (G * d^4) / (8 * D^3 * N). With all other parameters unchanged, k is inversely proportional to the number of active turns N. Cutting the spring into two halves halves N and thus doubles k. Step-by-Step Solution: Original stiffness: k_0 = (G * d^4) / (8 * D^3 * N).After cutting: N_new = N / 2.New stiffness: k_new = (G * d^4) / (8 * D^3 * (N / 2)) = 2 * k_0. Verification / Alternative check: Series/parallel analogy: two identical springs in series halve stiffness; conversely, halving coil count in a single spring doubles stiffness. Why Other Options Are Wrong: same: would require N unchanged, which is not the case.half: contradicts inverse proportionality to N.one-fourth: would occur only if N increased fourfold, not halved. Common Pitfalls: Forgetting that wire and mean diameters remain the same after cutting, so only N changes. Final Answer: double

Question 18

Modulus of elasticity (Young's modulus) for mild steel — choose the closest standard value. Give your answer in kN/mm^2 (note: 1 kN/mm^2 = 1 GPa).

Accepted Answer

Introduction: Mild steel (low-carbon steel) is a standard structural material. This question tests recall of its elastic stiffness, commonly used in beam, column, shaft, and plate calculations in civil and mechanical engineering. Given Data / Assumptions: Material: mild (low-carbon) steel. Small-strain, linear-elastic range. Units: kN/mm^2 where 1 kN/mm^2 = 1 GPa. Concept / Approach: Young's modulus E links normal stress and normal strain by Hooke's law: σ = Eε. For steels, E is nearly constant across grades and heat treatments within engineering accuracy, typically around 200 to 210 GPa (i.e., 200–210 kN/mm^2). Step-by-Step Solution: Recognize standard value: E_steel ≈ 200–210 GPaConvert understanding of units: 1 GPa = 1 kN/mm^2Select the closest choice: 210 kN/mm^2 Verification / Alternative check: Handbooks and codes often tabulate E_steel = 2.010^5 to 2.1*10^5 N/mm^2, which equals 200–210 kN/mm^2, confirming the chosen answer. Why Other Options Are Wrong: 10, 25, 80, 100 kN/mm^2: Far lower than accepted values for steel; such stiffnesses are more typical of polymers (very low) or some non-ferrous alloys (still higher than listed low values). Common Pitfalls: Confusing strength (e.g., fy) with stiffness (E); mixing up units such as MPa, GPa, and N/mm^2. Remember: E for steels is relatively invariant compared to strength grades. Final Answer: 210 kN/mm^2

Question 19

Steam generation and pressure vessels — evaluate the statement: “A thick pressure vessel is always used for generating steam because it can withstand high pressures.”

Accepted Answer

Introduction: This item checks conceptual understanding of thin vs thick pressure vessel design and the realities of steam generation equipment (boilers and drums). Not every steam generator uses a thick shell; selection depends on diameter, pressure level, and practical fabrication/heat-transfer constraints. Given Data / Assumptions: Comparing “thick” vs “thin” shell criteria. Typical steam drums and boilers operate with diameters where t/D is small. Design must satisfy stress limits and heat-transfer needs. Concept / Approach: Classification often uses t/D thresholds (e.g., thin when t/D is small). Many boiler drums and shells are analyzed with thin-shell formulas because their walls, though substantial in mm, are still “thin” relative to large diameters. Thick cylinders are favored when high pressure and small diameter push t/D into the “thick” regime and stress varies significantly through the wall thickness. Step-by-Step Solution: Identify claim: “always” thick for steam generationCounterpoint: Many boilers use thin-shell design with appropriate thickness, materials, and code safety factorsConclusion: The statement is false because selection depends on t/D and design codes, not the fact of generating steam Verification / Alternative check: Typical water-tube boilers employ drums with large diameters and wall thicknesses small relative to diameter, validating thin-shell assumptions at code-checked thicknesses. Why Other Options Are Wrong: True variants: Overgeneralize; thickness choice is not universal for steam generation. “Thin shells always regardless of pressure”: Also incorrect; very high pressures or small diameters may require thick-cylinder analysis. Common Pitfalls: Equating “withstand high pressure” with “must be thick.” Thickness is set by code equations, diameter, material strength, corrosion allowance, and fabrication—sometimes thin-shell theory remains valid. Final Answer: False

Question 20

Beam classification by supports — name the member: A beam resting on more than two supports over its span is called what?

Accepted Answer

Introduction: Support conditions define common beam types and their analysis methods. Recognizing a beam spanning over multiple supports is foundational for statically indeterminate analysis. Given Data / Assumptions: More than two supports along the beam's length. Standard textbook nomenclature. Concept / Approach: A beam that extends over three or more supports is termed a continuous beam. Internal moments and shears depend on continuity across supports and require compatibility along with equilibrium for solution. Step-by-Step Solution: Identify number of supports: more than twoMatch to standard term: “continuous beam”Conclude correct classification Verification / Alternative check: Indeterminacy check: each additional support beyond two generally introduces indeterminacy requiring slope-deflection, moment distribution, or matrix methods. Why Other Options Are Wrong: Simply supported: exactly two supports without fixity. Fixed beam: both ends fixed, but still two supports. Overhanging: extensions beyond supports, not necessarily >2 supports. Cantilever: fixed at one end only. Common Pitfalls: Confusing “overhanging” with “continuous.” Overhang concerns extension beyond a support, not the number of supports. Final Answer: continuous beam

Strength of Materials Questions