Question 1

Torsion of shafts in parallel:
Two shafts connected in parallel carry a common torque between a pair of rigidly connected flanges. Will the angle of twist of each shaft be the same?

Accepted Answer

Introduction / Context: Multiple shafts in parallel are used to share torque between paths (e.g., multi-shaft drive lines). Understanding compatibility and equilibrium conditions in torsion helps determine torque distribution and shaft sizing. Given Data / Assumptions: Two shafts connected in parallel between common rigid end plates/flanges. Total applied torque T is shared as T = T1 + T2. Linear elastic torsion (Saint-Venant), small angles. Concept / Approach: In parallel, both shafts experience the same end rotation because the end flanges remain plane and rotate together (compatibility). The torque splits according to each shaft’s torsional stiffness Kt = G * J / L. Stiffer shaft carries a larger share, but the twist angle θ is identical for both. Step-by-Step Solution: Angle of twist: θ = T1 * L / (G1 * J1) = T2 * L / (G2 * J2) for common end rotation.Equilibrium: T = T1 + T2.Solve for T1 and T2 using stiffnesses; θ emerges identical for both shafts.Hence, in parallel systems, twist compatibility enforces the same θ. Verification / Alternative check: Electrical analogy: torque ↔ current, angle of twist ↔ voltage, torsional stiffness ↔ conductance. Elements in parallel share the same “voltage” (θ), reinforcing the conclusion. Why Other Options Are Wrong: “No” contradicts compatibility at the rigidly connected ends. Conditions on equal diameter/length are unnecessary; these affect torque split, not equality of θ. “Total torque zero” would trivially make θ = 0, but that does not define the general case. Common Pitfalls: Confusing series vs parallel: in series, torque is the same and twists add; in parallel, twist is the same and torques add. Final Answer: Yes

Question 2

Springs for shock and vibration service:
Which spring type is most commonly used in vehicles and machinery to absorb shocks and vibrations under heavy loads?

Accepted Answer

Introduction / Context: Springs are energy storage elements used for cushioning, vibration isolation, returning mechanisms, and force control. Choosing the right spring type depends on load level, space constraints, stiffness characteristics, and durability requirements. Leaf springs are a classic solution for heavy vehicles and machinery where shock absorption is critical. Given Data / Assumptions: Application requires absorption of shocks and vibration over significant deflections. Robustness and load distribution across multiple leaves are desired. Maintenance and packaging in vehicle suspensions are considered. Concept / Approach: Leaf springs consist of stacked steel leaves forming a semi-elliptic or parabolic profile. Under load, they flex to store strain energy and provide a progressive rate (especially in multi-leaf or parabolic configurations). The broad contact area spreads loads, and interleaf friction contributes to damping—helpful for shock absorption without separate dampers in very old designs (modern systems pair springs with shock absorbers for controlled damping). Step-by-Step Solution: Identify requirement: shock/vibration absorption under heavy loads.Evaluate options: torsion springs store energy in twist but are not typical for heavy-vehicle primary suspension.Conical and disc springs offer high load in small spaces but limited travel and are stiffer with less comfort.Leaf springs provide large travel, load capacity, durability, and inherent damping → best fit. Verification / Alternative check: Common practice in trucks, trailers, and railway rolling stock uses multi-leaf or parabolic leaf springs as primary suspension elements, confirming suitability for shock/vibration service. Why Other Options Are Wrong: Torsion spring: common in door hinges and mechanisms, not heavy suspension. Conical/disc springs: compact high-force solutions but limited deflection and harsher response. Volute spring: used historically in buffers; less common than leaf springs for vehicle suspensions. Common Pitfalls: Assuming any spring isolates vibration equally well; stiffness–mass–damping matching matters. Leaf springs are preferred for heavy-duty shock loads and packaging along the axle. Final Answer: Leaf spring

Question 3

Basic definitions – cantilever support conditions:
A cantilever beam is defined as a beam that is:

Accepted Answer

Introduction / Context: Support conditions control the internal forces and deflections of beams. The cantilever is a fundamental boundary condition found in balconies, crane jibs, and shelf brackets. Recognizing its definition immediately sets the boundary values for shear, moment, slope, and deflection. Given Data / Assumptions: One end of the beam is rigidly fixed (built-in), preventing translation and rotation. The other end is free, with no restraint to translation or rotation. Loads may act along the span or at the free end. Concept / Approach: At the fixed end, reactions include a vertical force and a fixing moment (and potentially a horizontal force in 2D). At the free end, both the shear force and bending moment are zero in the absence of applied end actions. These boundary conditions define the shape of shear and moment diagrams and the characteristic deflection curve of a cantilever. Step-by-Step Solution: State definition: cantilever = one fixed end + one free end.Apply boundary conditions: V_free = 0, M_free = 0; at fixed end, reactions develop as required by loads.Conclude the correct descriptive option matches the cantilever definition. Verification / Alternative check: Compare with simply supported beams (pin and roller): both ends allow rotation and do not provide a fixing moment; that is not a cantilever. Continuous beams over multiple supports also differ. Why Other Options Are Wrong: Fixed–fixed: both ends restrained, not a cantilever. Simply supported or multi-support beams: lack a fixed end; they admit rotations at supports. Overhang with simple supports is not a cantilever because the overhang end is not fixed. Common Pitfalls: Confusing overhanging simply supported beams with cantilevers; only a fixed connection qualifies the member as a cantilever. Final Answer: Fixed at one end and free at the other end

Question 4

Thick-walled cylinder under internal pressure: where along the wall thickness is the hoop (circumferential) stress the greatest—at the inner radius or elsewhere?

Accepted Answer

Introduction / Context: In pressure vessel design, understanding how circumferential (hoop) stress varies through the wall of a thick cylinder is essential for safe sizing and material selection. Thick-cylinder behavior differs from thin-shell formulas because stresses vary significantly from the inside surface to the outside surface. Given Data / Assumptions: Cylindrical vessel with inner radius and outer radius, subjected to internal pressure p only. Isotropic, homogeneous, linearly elastic material. No thermal gradients or external pressure unless stated. Concept / Approach: Lame’s equations govern radial and hoop stresses in thick cylinders. For internal pressure only, the radial stress is most compressive (−p) at the inner surface and approaches zero at the outer surface. The hoop stress is tensile and attains its maximum at the inner surface, decreasing towards the outer surface. Step-by-Step Solution: Use Lame form: sigma_r = A − B/r^2, sigma_theta = A + B/r^2Apply boundary conditions: sigma_r(inner) = −p, sigma_r(outer) = 0Solve for A and B, then evaluate sigma_theta(r).Result: sigma_theta is largest at r = inner radius and reduces with increasing r. Verification / Alternative check: Plot sigma_theta versus r from the inner to outer surface. The curve peaks at the inner wall for an internally pressurized cylinder, matching standard design charts. Why Other Options Are Wrong: False/conditional options conflict with Lame’s solution; dependence on Poisson’s ratio does not overturn the location of maximum hoop stress. Common Pitfalls: Using thin-wall formula sigma_h = p d / (2 t) when D/t is small; neglecting through-thickness stress variation and inner-wall peaking. Final Answer: True

Question 5

In a circular shaft under torsion, how does the shear stress at a point vary with its radial distance from the shaft axis?

Accepted Answer

Introduction / Context: Torsion of circular shafts is a core topic in mechanics of materials. Designers must know how shear stress distributes from the center to the outer surface to ensure safe torque transmission. Given Data / Assumptions: Solid or hollow circular shaft under pure torque T. Linearly elastic, homogeneous, isotropic material. No warping restraint beyond Saint-Venant torsion assumptions. Concept / Approach: The torsion formula gives shear stress at a radius r as tau(r) = T * r / J, where J is the polar moment of inertia of the cross-section. This indicates a linear variation of shear stress with radius—zero at the center and maximum at the outer surface. Step-by-Step Solution: Torsion relation: tau(r) = T * r / JFor a given T and J, tau is proportional to r.At r = 0 (axis), tau = 0; at r = R (outer radius), tau_max = T * R / J. Verification / Alternative check: Mohr’s circle for pure shear at a point yields zero shear at the centerline because r = 0; experimental strain-gage data on shafts confirms linear radial variation. Why Other Options Are Wrong: Inverse or constant dependence contradicts tau = T r / J; saying maximum at center is incorrect as tau(0) = 0. Common Pitfalls: Confusing bending stress distribution (linear with y) with torsional shear; forgetting hollow shaft still has tau proportional to r within its wall. Final Answer: directly proportional to the distance from the axis

Question 6

Stress–strain diagram for a ductile metal: over which labeled portion does stress remain directly proportional to strain (Hooke’s law region)?

Accepted Answer

Introduction / Context: A typical engineering stress–strain curve for mild steel begins with a linear segment where stress is proportional to strain. Correctly identifying this proportional (elastic) region is fundamental to applying Young’s modulus in design. Given Data / Assumptions: Standard tensile test specimen (ductile metal). Diagram points: O (origin), A (proportional limit), followed by yield/ultimate points. Engineering stress and strain used. Concept / Approach: Hooke’s law states sigma = E * epsilon within the proportional limit. On the graph, the straight-line region O–A indicates constant slope equal to E, beyond which nonlinearity and yielding begin. Step-by-Step Solution: Identify linear segment: O–A.Slope of O–A = Young’s modulus E.Therefore, stress ∝ strain in O–A only. Verification / Alternative check: Unload anywhere within O–A: the material returns to original length (primarily elastic behavior). Beyond A, permanent set appears. Why Other Options Are Wrong: Segments including A–C, A–D, D–E include yielding, strain hardening, necking, or plastic flow, where proportionality does not hold. Common Pitfalls: Confusing proportional limit with elastic limit; assuming linearity persists beyond A; mixing true vs engineering curves. Final Answer: from O to A

Question 7

Two closely coiled helical springs A and B are identical except for number of turns. Spring A has twice the number of turns as spring B. How does the stiffness of A compare with B?

Accepted Answer

Introduction / Context: Helical spring stiffness determines load–deflection behavior in suspensions, valves, and precision mechanisms. For closely coiled springs, stiffness depends sensitively on geometry, especially wire diameter, mean coil diameter, and number of active turns. Given Data / Assumptions: Springs A and B have the same wire diameter d, same mean coil diameter D, same material (same shear modulus G). Only difference: number of active turns n_A = 2 n_B. Close-coiled helical spring, small pitch angle. Concept / Approach: Spring stiffness k for a close-coiled helical spring is k = G d^4 / (64 D^3 n). Thus, stiffness is inversely proportional to the number of active turns n when other parameters are fixed. Step-by-Step Solution: k_A = G d^4 / (64 D^3 n_A)k_B = G d^4 / (64 D^3 n_B)n_A = 2 n_B ⇒ k_A / k_B = n_B / n_A = 1 / 2Therefore, k_A = 0.5 * k_B (one-half of B). Verification / Alternative check: Physical reasoning: more coils increase flexibility and reduce stiffness; doubling coils halves stiffness if all else is unchanged. Why Other Options Are Wrong: Fractions like one-eighth or one-sixteenth suggest changes in d or D (since k ∝ d^4 and ∝ 1/D^3), which are not given here. Common Pitfalls: Forgetting which variables most strongly affect k; confusing active turns with total turns; ignoring squared or fourth-power dependencies. Final Answer: one-half of B

Question 8

Identify the stress–strain pair: equal and opposite tangential forces across a resisting section cause the body to tend to slide (shear) on that section. What stress and strain arise?

Accepted Answer

Introduction / Context: Different external loadings generate distinct internal stress and associated strain responses. Recognizing shear scenarios is crucial in connections, rivets, welds, and beam webs where sliding along a plane is critical. Given Data / Assumptions: Two equal and opposite tangential forces acting across a section. Forces act parallel to the area, not normal to it. Small deformations; linear elasticity. Concept / Approach: When loads act tangentially to a surface, the internal response is shear. The related strain measure is change in angle (distortion) between originally orthogonal lines—this is shear strain, usually denoted gamma. Step-by-Step Solution: Identify load direction: tangential ⇒ not normal tension/compression.Internal stress on the resisting plane is shear stress tau.Associated deformation is angular distortion: shear strain gamma. Verification / Alternative check: Mohr’s circle for stress shows pure shear produces equal magnitude principal stresses of opposite signs; strain counterpart displays angular change, not axial extension alone. Why Other Options Are Wrong: Tensile/compressive pairs result from normal forces; shear stress with tensile strain is mismatched. Common Pitfalls: Confusing axial and tangential loadings; assuming all deformations are axial elongations instead of angular distortions under shear. Final Answer: shear stress, shear strain

Question 9

Fillet weld design: Transverse fillet welds (welds placed perpendicular to the direction of applied force) are primarily designed on which strength criterion?

Accepted Answer

Introduction / Context: Fillet welds are ubiquitous in structural and machine design. Their orientation relative to the applied load determines the critical stress at the weld throat used for sizing and verification. Given Data / Assumptions: Transverse fillet weld: weld line is perpendicular to the applied load direction. Static loading; small deformations. Design checked at the effective throat area. Concept / Approach: Transverse fillet welds develop stresses across the throat that act largely in tension due to the load path normal to the weld line. By contrast, longitudinal fillet welds (parallel to the load) are typically designed for shear on the throat area. Step-by-Step Solution: Identify orientation: load perpendicular to weld length ⇒ transverse fillet weld.Critical section: effective throat (typically t = 0.707 * leg size for equal-leg fillet).Dominant stress across throat: tensile stress = P / (throat area). Verification / Alternative check: Standard welding design manuals specify transverse fillet weld allowable loads using tensile stress on the throat, while longitudinal weld checks use shear. Empirical tests corroborate the orientation-dependent failure mode. Why Other Options Are Wrong: Compressive and bending checks are not primary for this orientation; shear is primary for longitudinal, not transverse, fillet welds. Common Pitfalls: Confusing longitudinal and transverse cases; using shear allowables for a transverse throat, leading to unconservative or overconservative results. Final Answer: tensile strength

Question 10

Buckling analysis terminology: The “equivalent length” of a column is defined as the length of a hinged–hinged column of the same material and section that has the same Euler crippling load as the given column with its end conditions. Is this defini…

Accepted Answer

Introduction / Context: In column buckling, end restraints change the effective buckling length and thus the critical load. Using an equivalent length allows all end conditions to be treated with a single Euler formula by adjusting the effective column length. Given Data / Assumptions: Elastic buckling per Euler theory. Same cross-section and material for real and equivalent columns. End conditions captured via an effective length factor K. Concept / Approach: Euler's critical load: P_cr = pi^2 * E * I / (L_e)^2, where L_e = K * l. The equivalent (effective) length L_e is defined so a hypothetical hinged–hinged column of length L_e buckles at the same load as the actual column with given end fixity. Step-by-Step Solution: Identify actual end condition ⇒ determine K (e.g., 0.5 for fixed–fixed, 1.0 for pinned–pinned, 2.0 for fixed–free).Compute L_e = K * l.Use Euler formula with L_e to get the same P_cr as the real column. Verification / Alternative check: Mode shapes for different end conditions show different effective half-wavelengths; the K factors are derived from boundary-condition solutions of the buckling differential equation. Why Other Options Are Wrong: Restriction to short or long columns is incorrect; the definition is general within Euler's elastic range (slender columns). Common Pitfalls: Confusing actual length with effective length; misapplying K, leading to large errors in P_cr. Final Answer: True

Question 11

Thin spherical shell under internal pressure p: is the maximum in-plane shear stress in the shell wall equal to zero?

Accepted Answer

Introduction / Context: Thin spherical pressure vessels (e.g., gas spheres) have equal biaxial tensile stresses on their walls under internal pressure. Determining the maximum shear stress helps in applying yield criteria and comparing with cylindrical shells. Given Data / Assumptions: Thin spherical shell, thickness small relative to diameter. Internal pressure p; negligible bending and radial stress. Uniform biaxial membrane stress state. Concept / Approach: For a thin sphere, principal stresses are equal in all in-surface directions: sigma_1 = sigma_2 = p d / (4 t). Maximum in-plane shear stress equals (sigma_1 − sigma_2) / 2. If sigma_1 = sigma_2, this difference is zero, so the maximum in-plane shear stress vanishes. Step-by-Step Solution: Principal stresses: sigma_1 = sigma_2 = p d / (4 t)Maximum shear: tau_max = (sigma_1 − sigma_2) / 2 = 0Conclusion: No in-plane shear under pure internal pressure for a thin sphere. Verification / Alternative check: Mohr’s circle reduces to a single point at sigma = p d / (4 t) when principal stresses are equal—its radius (shear) is zero. Why Other Options Are Wrong: Non-zero shear would require unequal principal stresses (e.g., cylinders where sigma_h ≠ sigma_l). Common Pitfalls: Applying cylindrical shell results to spheres; forgetting that equal principal stresses imply zero in-plane shear. Final Answer: Yes

Question 12

Effective (equivalent) length for a column with both ends fixed: choose the correct relation between equivalent length L and actual length l.

Accepted Answer

Introduction / Context: End restraint increases the critical buckling load by reducing the effective length of a column. Knowing the correct effective length for each end condition is vital for safe, economical column design. Given Data / Assumptions: Straight, prismatic column under axial compression. Both ends fixed (built-in), elastic buckling per Euler theory. No initial imperfections considered in the basic formula. Concept / Approach: The Euler critical load uses effective length L_e = K * l, where K depends on end conditions. For both ends fixed, K = 0.5. Thus L_e = 0.5 * l. The reduced length raises the critical load to four times that of a pinned–pinned column of the same actual length. Step-by-Step Solution: End condition: fixed–fixed ⇒ K = 0.5Effective length: L = K * l = 0.5 * lEuler load: P_cr = pi^2 * E * I / L^2 = pi^2 * E * I / (0.5 l)^2 = 4 * (pi^2 E I / l^2) Verification / Alternative check: From buckling mode shapes, a fixed–fixed column has half-wavelength equal to l, implying effective pinned length of l/2 for the same mode. Why Other Options Are Wrong: L = l corresponds to pinned–pinned; L = 2l to fixed–free (cantilever); fractional roots do not match standard K-factors. Common Pitfalls: Confusing K-factors among end conditions; misapplying effective length in slenderness ratio and Euler load calculations. Final Answer: L = l/2

Question 13

Poisson’s ratio: identify the correct definition

In strength of materials, Poisson’s ratio (ν) relates the lateral contraction to the axial extension in a uniaxial test. Is the statement “Poisson's ratio is the ratio of linear strain to the volum…

Accepted Answer

Introduction / Context: Understanding Poisson’s ratio is fundamental in solid mechanics. It links how a material thins laterally when stretched longitudinally (or bulges when compressed). Confusing it with volumetric strain leads to serious errors in design and analysis. Given Data / Assumptions: Uniaxial loading of a prismatic specimen within the elastic range. Lateral strain is measured perpendicular to the loading axis; linear (axial) strain is along the load. Small strain, linear elasticity (Hooke’s law) applies. Concept / Approach: Poisson’s ratio ν is defined as the magnitude of lateral strain divided by the linear (axial) strain: ν = −(ε_lateral / ε_axial). The minus sign reflects that lateral strain is typically opposite in sign to axial strain (contraction vs. extension). Volumetric strain is a different quantity equal to the sum of principal strains. Step-by-Step Solution: Define axial strain: ε_axial = ΔL / L.Define lateral strain: ε_lateral = Δd / d (perpendicular to loading).Poisson’s ratio: ν = − ε_lateral / ε_axial.Volumetric strain (small strain): ε_v = ε1 + ε2 + ε3 (sum of principal strains).Therefore ν does not equal “linear strain / volumetric strain” nor its inverse. Verification / Alternative check: For an isotropic linear elastic solid under uniaxial stress σ, the lateral strain equals −ν σ / E, and axial strain equals σ / E. Substituting confirms ν = −ε_lateral / ε_axial, independent of volumetric strain. Why Other Options Are Wrong: “True” is incorrect; ν is not defined using volumetric strain. “True only for incompressible materials” is still false; even for ν ≈ 0.5, the definition remains lateral-to-axial. “True only at very small strains” misstates the definition. “It equals bulk modulus divided by Young’s modulus” is incorrect; elastic constants relate as E = 3K(1 − 2ν) and E = 2G(1 + ν). Common Pitfalls: Mixing up ν with bulk modulus K or with volumetric strain; forgetting the negative sign convention; applying the definition outside the elastic range. Final Answer: False

Question 14

Cantilever with uniformly distributed load (UDL): shear force at the fixed end

A cantilever beam of length l carries a uniformly distributed load of intensity w per unit length over the entire span. What is the shear force at the fixed end?

Accepted Answer

Introduction / Context: Cantilever beams are common in balconies, crane jibs, and machine tool overhangs. For uniform loading, knowing reactions, shear, and moment at the fixed support is essential to size sections and check safety. Given Data / Assumptions: Cantilever length l, fixed at one end and free at the other. Uniformly distributed load w (force per unit length) over the full span. Linear elastic beam theory; static equilibrium. Concept / Approach: The resultant of a UDL of intensity w on length l is a single force W = w * l acting at the centroid of the distribution, located at midspan from the free end (i.e., l/2 from either end). Shear at the fixed support equals this resultant in magnitude but opposite in direction. Step-by-Step Solution: Resultant load: W = w * l.Support shear at fixed end must balance vertical loads: V_fixed + (−W) = 0.Hence V_fixed = W = w * l.(For completeness, fixed-end moment M_fixed = w l^2 / 2.) Verification / Alternative check: Construct shear diagram: starting at fixed end with +w l, shear decreases linearly to zero at the free end in the presence of a constant distributed load w; the intercept at the fixed end is indeed w l. Why Other Options Are Wrong: Zero contradicts equilibrium. w l / 4 and w l / 2 are fractions appropriate to other problems (e.g., reactions for simply supported beams), not a cantilever’s fixed-end shear. 2 w l is an overestimate. Common Pitfalls: Confusing cantilever with simply supported beam; placing the resultant at the wrong location; mixing moment and shear values. Final Answer: w l

Question 15

Name the stress at a local contact between two members

A localized compressive stress developed at the small area of contact between two mating members (such as a pin in a lug or a rivet in a plate) is called:

Accepted Answer

Introduction / Context: Machine and structural joints often transfer load through small contact areas, for example pins, rivets, and bolts in holes. The primary compressive stress in the contact patch must be identified and limited to avoid permanent deformation. Given Data / Assumptions: Two members pressed together over a finite, relatively small area (hole–pin, rivet–plate, key–keyway). Load predominantly compressive and normal to the contact patch. Elastic design intent with allowable bearing stresses. Concept / Approach: The stress of interest is called crushing stress or bearing stress. It represents average compressive stress over the projected contact area (usually hole diameter times plate thickness for design). Step-by-Step Solution: Define projected bearing area A_bearing = d * t (for a plate of thickness t and pin/rivet diameter d).Average bearing stress: σ_bearing = P / A_bearing.Design requires σ_bearing ≤ allowable bearing stress to prevent local yielding or hole elongation. Verification / Alternative check: Observe failure modes in joints: excessive bearing stress causes ovalization of holes and plastic flow around the contact, distinct from shear failure of fasteners or tearing across net sections. Why Other Options Are Wrong: Tensile and bending stresses are not the primary local contact stresses. Shear stress is relevant to fastener shear failure, not the compressive contact between parts. Torsional stress is unrelated to this local phenomenon. Common Pitfalls: Using bolt shear area instead of projected bearing area; ignoring edge distance effects; confusing bearing with contact Hertzian stress in curved contacts—here we refer to average design bearing stress. Final Answer: crushing (bearing) stress

Question 16

Modular ratio of two materials: choose the correct definition

In composite members (e.g., reinforced concrete or bimetal bars), the modular ratio between two materials is defined as the ratio of:

Accepted Answer

Introduction / Context: When different materials act together, designers often transform one material into an equivalent section of another using the modular ratio. This is standard for cracked transformed sections in reinforced concrete and laminated members. Given Data / Assumptions: Two linearly elastic, isotropic materials with moduli E1 and E2. Perfect bond so that strains are compatible at the interface. Axial or bending behavior within elastic range. Concept / Approach: The modular ratio n is defined as n = E1 / E2. It scales the area (or moment of inertia) of one material to an equivalent amount of the other so that identical strains produce proportionate stresses via Hooke’s law. Step-by-Step Solution: Hooke’s law: σ = E * ε.For equal ε in bonded materials, stress ratio equals modulus ratio: σ1 / σ2 = E1 / E2 = n.Transformed section method: replace area A2 of material 2 by n * A2 in material 1’s “units.” Verification / Alternative check: Check bending of a composite section: internal force resultants balance when areas are transformed by n, producing correct stress distribution proportional to E. Why Other Options Are Wrong: Options (a) and (b) describe definitions of E and G individually, not a ratio between two materials. (d) refers to the ratio of shear moduli, not the standard modular ratio used in transformed sections. (e) density is unrelated. Common Pitfalls: Using temperature-dependent or nonlinear moduli without care; mixing tangent and secant moduli; applying wrong n for sustained vs. short-term (as in RC design). Final Answer: their modulus of elasticities (E1 / E2)

Question 17

Definition of pure bending in a beam

A beam section is said to be under “pure bending” when it experiences which combination of internal actions?

Accepted Answer

Introduction / Context: Pure bending is a key idealization used to derive the flexure formula and to understand stress distributions in beams without the complicating influence of shear. Given Data / Assumptions: Straight prismatic beam in the elastic range. Plane sections remain plane (Bernoulli–Euler hypothesis). No torsion; loads cause bending about a principal axis. Concept / Approach: Pure bending occurs in a beam segment where the internal shear force is zero while the bending moment is constant (nonzero). This situation arises in the region between equal and opposite couples or within the constant-moment portion of some loading cases. Step-by-Step Solution: Shear–moment relation: dM/dx = V.If V = 0 → dM/dx = 0 → M = constant.With V = 0, the cross-section experiences only bending; the flexure formula σ = M y / I applies cleanly without shear stress complications. Verification / Alternative check: Example: a beam segment loaded only by two end couples produces a region of constant moment and zero shear—classic pure bending. Why Other Options Are Wrong: Option (a) includes a nonzero shear, which contradicts dM/dx = V = 0. Option (b) with zero M is not bending. Option (d) is trivial and not representative of bending behavior. Option (e) is general bending, not pure. Common Pitfalls: Assuming “no shear” means no transverse stresses at all—note that in reality, Saint-Venant effects near loads exist; pure bending is an interior idealization. Final Answer: Constant bending moment and zero shear force

Question 18

Column classification by slenderness ratio

Columns with slenderness ratio (effective length/radius of gyration) less than about 80 are commonly classified as:

Accepted Answer

Introduction / Context: Column behavior shifts from crushing/yielding to elastic buckling as slenderness increases. Designers classify columns to choose appropriate strength models (yield-based vs. buckling-based). Given Data / Assumptions: Slenderness ratio λ = L_e / r, where L_e is effective length and r is radius of gyration. Reference threshold around λ ≈ 80 (varies with code and material). Axial compression predominant. Concept / Approach: Short columns (low slenderness) fail by material yielding or crushing rather than elastic buckling. As λ increases, columns transition to inelastic and then elastic buckling regimes, termed medium or long columns in various texts. Step-by-Step Solution: If λ <≈ 80 → short: strength governed by compressive yield stress with small buckling influence.Intermediate λ → combined effects; empirical curves used.High λ → long columns governed by Euler buckling. Verification / Alternative check: Compare Euler critical stress σ_cr = π^2 E / (λ^2) with yield stress; the crossover near λ ≈ 80 (for steels) indicates when buckling becomes dominant. Why Other Options Are Wrong: “Long” corresponds to high slenderness; “weak” is not a standard classification term; “medium” refers to the inelastic region; “inelastic columns” describes behavior, not the stated threshold. Common Pitfalls: Using a single universal threshold without checking the material and code; ignoring end conditions in L_e; misusing λ based on total length instead of effective length. Final Answer: short columns

Question 19

Fiber stress state at the lower surface of a sagging beam

For a prismatic beam under typical downward loading causing a sagging (concave-up) bending moment, the lower layer (bottom fiber) of the beam will be in which state?

Accepted Answer

Introduction / Context: Identifying which fibers are in tension or compression under bending is essential for detailing reinforcement, checking brittle materials, and preventing cracking where tension capacity is low. Given Data / Assumptions: Beam experiences sagging bending (positive bending moment by common sign convention). Plane sections remain plane; linear strain profile across depth. Neutral axis passes through the centroid for homogeneous material. Concept / Approach: For sagging (concave-up) bending, the top fibers shorten (compression) and the bottom fibers elongate (tension). The neutral axis is where axial stress is zero; stress varies linearly with distance from that axis. Step-by-Step Solution: Assume positive bending moment M > 0.Strain distribution: ε = −(κ) y, with curvature κ > 0 and y measured positive upward from neutral axis.Top (y > 0) → ε negative (compression); bottom (y < 0) → ε positive (tension).Stress via σ = E ε follows the same sign pattern. Verification / Alternative check: Sketch the deflected shape for a simply supported beam with midspan load: it sags; string analogy shows the underside stretches, confirming tension at the bottom. Why Other Options Are Wrong: “Compression” applies to hogging (negative) bending. “Neither” is only true at the neutral axis, not at an extreme fiber. “Pure shear only” and “torsion” are unrelated to the axial fiber stress caused by bending. Common Pitfalls: Mixing sign conventions; misreading a hogging region (over a fixed support) as sagging; assuming composite sections always have centroidal neutral axes (they may shift). Final Answer: in tension

Question 20

Weight comparison: square vs circular beam with equal bending capacity

A square beam and a circular beam have the same length, the same allowable bending stress, and resist the same bending moment (i.e., same required section modulus). What is th…

Accepted Answer

Introduction / Context: Choosing a cross-section to minimize weight for a target bending capacity is central to efficient design. Here we compare square and circular solid sections sized to deliver the same section modulus (hence same allowable moment at the same allowable stress). Given Data / Assumptions: Equal length and material (same density), so weight ∝ cross-sectional area. Allowable normal bending stress is identical for both sections. Required section modulus Z is the same for both beams. Concept / Approach: For a square of side a: I = a^4 / 12 and c = a/2, hence Z_square = I / c = a^3 / 6. For a circle of diameter D: I = π D^4 / 64 and c = D/2, hence Z_circle = π D^3 / 32. Setting Z_square = Z_circle gives the size ratio, then weight ratio equals area ratio. Step-by-Step Solution: Equate section moduli: a^3 / 6 = π D^3 / 32.Solve for a/D: (a/D)^3 = (3 π) / 16.Therefore a/D = [(3 π) / 16]^(1/3) ≈ 0.838.Areas: A_square = a^2; A_circle = π D^2 / 4.Weight ratio (square : circular) = A_square / A_circle = (a^2) / (π D^2 / 4) = (4 / π) * (a/D)^2 ≈ (4 / π) * 0.838^2 ≈ 0.894.Hence square : circular ≈ 0.894 : 1 ≈ 1 : 1.12, i.e., 1/1.12. Verification / Alternative check: Dimensional and numerical checks confirm the ratio. Using more digits for π barely changes the result; the qualitative conclusion is that the circular section is slightly heavier for the same bending capacity. Why Other Options Are Wrong: 1/2 and 1/1.5 significantly understate the square’s weight. “1” would imply equal weight, which is not the case at equal Z. “2” is far off. Common Pitfalls: Comparing at equal area instead of equal Z; forgetting that weight scales with area when length and density are fixed; using wrong expressions for Z. Final Answer: 1/1.12

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