Question 1

Materials mechanics – definition: What is the ratio of lateral strain to axial strain for a homogeneous, isotropic material called?

Accepted Answer

Introduction / Context: When a prismatic bar is pulled in tension, it elongates in the loading direction and simultaneously contracts in the directions perpendicular to the load. The quantitative link between these two strains is fundamental in strength of materials and helps predict 3D deformation from 1D loading. The name assigned to this ratio is frequently used in civil, mechanical, and materials engineering design. Given Data / Assumptions: Material is homogeneous and isotropic so properties are the same in all directions. Deformation is within the elastic range so linear relations apply. Axial strain means strain along the load; lateral strain means strain perpendicular to the load. Concept / Approach: The property that links the lateral contraction to the axial extension is called Poisson's ratio, commonly denoted by the symbol nu. It is defined as the negative of the ratio of lateral strain to axial strain, but as a magnitude it is often quoted as a positive number between about 0.15 and 0.50 for common engineering materials. It works together with Young's modulus to determine elastic deformation in three dimensions. Step-by-Step Solution: Identify the strains: axial strain = change in length / original length.Lateral strain = change in lateral dimension / original lateral dimension.Use the definition: Poisson's ratio = (lateral strain) / (axial strain) with a minus sign by sign convention.Match the term: the described ratio corresponds to Poisson's ratio. Verification / Alternative check: Typical values: steel about 0.30, concrete about 0.15 to 0.20 (short-term), rubber near 0.5. These known ranges confirm the interpretation of the ratio connecting transverse and axial strains. Why Other Options Are Wrong: Hooke's ratio: not a standard term; Hooke's law relates stress and strain via modulus. Plastic ratio: plasticity refers to permanent deformation, not the elastic strain ratio. Yield ratio: relates strengths, not strains. Modulus ratio: used in composite analysis (e.g., n = Es/Ec), unrelated to lateral/axial strain ratio. Common Pitfalls: Confusing Poisson's ratio with Young's modulus; forgetting the sign convention (lateral strain is negative in tension), even though the reported value is a positive magnitude. Final Answer: Poisson's ratio

Question 2

Springs – limiting capacity: What is the term for the greatest load a spring can sustain without suffering any permanent set (permanent distortion)?

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Introduction / Context: Design of helical and leaf springs requires specifying a safe maximum load so that, after unloading, the spring returns to its original dimensions. The engineering term for the limiting service load that does not cause permanent set is essential for specifying acceptance tests and service ratings. Given Data / Assumptions: Spring is used in the elastic range during service. Permanent set corresponds to plastic deformation. We need the correct terminology used in spring design and testing standards. Concept / Approach: Springs are commonly tested up to a specified maximum force and then unloaded to verify that no permanent deformation remains. That maximum force is called the proof load. It establishes a pass/fail criterion for elastic performance under service conditions. Step-by-Step Solution: Identify the physical requirement: no permanent set after removing the load.Relate to terminology: the greatest load meeting this requirement is the proof load.Differentiate from related terms: proof stress is a stress level; stiffness is a rate of load vs. deflection; proof resilience is energy, not load.Conclude that the correct choice is 'proof load'. Verification / Alternative check: Manufacturing specifications often state: 'Spring shall withstand the specified proof load and recover to within tolerance of free length,' confirming the definition. Why Other Options Are Wrong: Proof stress: a stress value (commonly at a specified offset), not a load. Stiffness: spring rate (force per unit deflection), not a limiting load. Proof resilience: energy stored at the elastic limit, not the force itself. Elastic limit load: descriptive phrase but the standard accepted term in spring testing is 'proof load.' Common Pitfalls: Confusing energy and load terminology; assuming 'stiffness' relates to strength rather than deflection rate; mixing 'proof stress' (material property) with spring system limits. Final Answer: proof load

Question 3

Springs – energy at limiting load: What is the name of the strain energy stored in a spring when it is loaded up to the greatest load that causes no permanent set?

Accepted Answer

Introduction / Context: Besides load capacity, springs are often characterized by how much elastic energy they can store and release without damage. The correct term for the maximum recoverable energy at the elastic limit is frequently used in design reports and acceptance criteria. Given Data / Assumptions: Spring is deflected only within the elastic range (no plastic deformation). The 'greatest load with no permanent set' equals the proof load. Strain energy refers to the recoverable elastic energy stored due to deformation. Concept / Approach: Energy stored up to the elastic limit (or proof point) and fully recoverable upon unloading is called proof resilience. For a linear spring, energy U = 0.5 * P * delta at the proof load, where P is the proof load and delta is the corresponding elastic deflection. Step-by-Step Solution: Recognize that the question asks for an energy term, not a force or a stress.At the limiting elastic load, the spring stores maximum recoverable energy.The proper term for this maximum recoverable energy is 'proof resilience.'Therefore, choose 'proof resilience' as the answer. Verification / Alternative check: Compare with 'modulus of resilience,' which is energy per unit volume for the material; in contrast, proof resilience refers to the total elastic energy stored by the spring at the proof load. Why Other Options Are Wrong: Stiffness: a rate (force per unit deflection), not energy. Proof stress: a stress value at yield/proof point; again, not energy. Proof load: the limiting load itself, not the energy at that load. Modulus of resilience: material energy density, not the spring's total stored energy at proof. Common Pitfalls: Mixing up system-level quantities (load, energy) with material-level properties (stress, modulus of resilience); assuming any energy term at maximum load is 'modulus of resilience' rather than the correct 'proof resilience.' Final Answer: proof resilience

Question 4

Strength of materials – deflection comparison For beams of the same length L and flexural rigidity EI, what is the ratio of the maximum deflection of a simply supported beam with a central point load W to that of a cantilever carrying the same load …

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Introduction / Context: This question assesses recall and application of standard deflection formulas from elementary beam theory. Comparing deflections for common boundary conditions (simply supported vs cantilever) helps engineers quickly estimate stiffness and serviceability under identical loads. Given Data / Assumptions: Both members have the same span L and flexural rigidity EI. Simply supported beam has a single central point load W. Cantilever has a single point load W at the free end. Small-deflection, linear-elastic behavior is assumed; shear deformation is neglected. Concept / Approach: Use the standard closed-form deflection expressions for prismatic beams under classic loads. For a central point load on a simply supported beam, and for an end load on a cantilever, the maximum deflection expressions are well known and depend only on W, L, E, and I. Step-by-Step Solution: Maximum deflection of simply supported beam with central load: delta_ss = WL^3 / (48EI)Maximum deflection of cantilever with end load: delta_cant = WL^3 / (3EI)Form the ratio: R = delta_ss / delta_cantR = (WL^3 / (48EI)) / (WL^3 / (3EI)) = (1/48) / (1/3) = 3/48 = 1/16 Verification / Alternative check: Numerically, set W = 1, L = 1, E*I = 1. Then delta_ss = 1/48 and delta_cant = 1/3. The ratio (1/48)/(1/3) equals 1/16, confirming the result without units. Why Other Options Are Wrong: (1/8) and (1/12) are larger than the correct ratio and would imply the simply supported system deflects relatively more than it truly does; (1/24) is too small; 'None of these' is incorrect because a correct ratio, 1/16, is provided. Common Pitfalls: Mixing the formulas for distributed loads and point loads; forgetting that cantilevers are much more flexible under the same load and span; canceling terms incorrectly when forming the ratio. Final Answer: 1/16

Question 5

Core (kernel) of a circular section – area ratio For a circular section of radius r in no-tension theory, what is the ratio of the total cross-sectional area to the area of its core (kernel)?

Accepted Answer

Introduction / Context: The core (kernel) of a section in masonry or concrete design is the locus within which the resultant compressive load must act to avoid tensile stress anywhere on the section. For circular sections, the core is also circular but reduced in size. Given Data / Assumptions: No-tension material model is assumed (e.g., unreinforced masonry under compression). Section is a full circle of radius r. Core radius for a circle is r/4 (a standard result from elementary strength of materials). Concept / Approach: The area ratio equals the area of the whole section divided by the area of the core. For a circle, areas scale with the square of the radius. Step-by-Step Solution: Total area A = πr^2Core radius rc = r/4, so core area Ac = π(r/4)^2 = πr^2/16Required ratio = A / Ac = (πr^2) / (π*r^2/16) = 16 Verification / Alternative check: Choose r = 1. Then A = π, Ac = π/16. A/Ac = 16. The ratio is independent of r, as expected. Why Other Options Are Wrong: Values 4, 8, and 12 underestimate the reduction in core size; 20 overestimates it. Only 16 matches the known core radius of r/4 for a circular section. Common Pitfalls: Confusing core radius (r/4) with diameter or using r/3 (which applies to a rectangle’s core distance from the edge in a different context). Final Answer: 16

Question 6

Name of the modulus: ratio of shear stress to shear strain What do we call the material constant defined as (shear stress) / (shear strain)?

Accepted Answer

Introduction / Context: Materials exhibit elastic response under small deformations. In shear, the proportionality constant between shear stress and shear strain is fundamental for torsion of shafts, shear deflection of beams, and plate/shell behavior. Given Data / Assumptions: Linear elastic, small-strain behavior (Hooke-type relationship) is assumed. Terminology may vary between texts but standard names exist. Concept / Approach: The constant G = tau / gamma is called the Shear Modulus or Modulus of Rigidity. It is related to Young's Modulus E and Poisson's ratio ν through G = E / (2*(1 + ν)). Young's Modulus (also called Modulus of Elasticity in tension/compression) pertains to normal stress–strain, not shear. Step-by-Step Solution: Identify definition: G = shear stress / shear strain.Recognize synonyms: Modulus of Rigidity = Shear Modulus.Exclude E (Young's Modulus) because it governs normal stress–strain, not shear. Verification / Alternative check: If E = 200 GPa and ν = 0.3 for steel, then G = 200 / (2*(1+0.3)) = 76.9 GPa, which matches typical tabulated values for structural steels. Why Other Options Are Wrong: (c) and (d) refer to axial elasticity; while some authors loosely call E the 'modulus of elasticity,' in this context the required modulus is specifically G, not E. Common Pitfalls: Using E when shear deformation dominates; ignoring the role of Poisson's ratio in relating E and G. Final Answer: Both (a) and (b)

Question 7

Thin cylinders under internal pressure In a thin cylindrical shell carrying a flowing liquid under internal pressure p, what is the ratio of circumferential (hoop) stress to longitudinal stress in the wall?

Accepted Answer

Introduction / Context: Design of thin-walled pressure vessels (pipes, tanks) requires distinguishing between hoop (circumferential) and longitudinal stresses. Knowing their ratio helps prioritize reinforcement and assess failure modes. Given Data / Assumptions: Thin cylinder assumption (wall thickness t is small compared to diameter d). Internal pressure p from the contained fluid. Uniform stress distribution through thickness; end effects ignored for hoop stress derivation. Concept / Approach: Classical thin-cylinder formulas are: Hoop stress, sigma_h = pd / (2t) Longitudinal stress, sigma_l = pd / (4t) Step-by-Step Solution: Write the ratio R = sigma_h / sigma_lR = (pd/(2t)) / (pd/(4t))R = (1/2) / (1/4) = 2 Verification / Alternative check: Use a quick numeric check: let p = 1, d = 1, t = 1. Then sigma_h = 0.5, sigma_l = 0.25, ratio = 2. Why Other Options Are Wrong: 0.5 and 1.5 do not follow from the thin-cylinder relations; 1 would imply equal stresses which is not the case; 'None of these' is incorrect because the correct ratio is exactly 2. Common Pitfalls: Confusing radius with diameter in formulas; using thick-cylinder theory unnecessarily; forgetting that end plates halve the longitudinal stress relative to the hoop stress. Final Answer: 2

Question 8

Overhangs on a uniformly loaded simply supported beam A simply supported beam carries a uniformly distributed load across its span and has two equal overhangs. To make the maximum bending moment as small as possible, what should be the ratio (overha…

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Introduction / Context: Overhangs can be used to reduce peak bending moments under uniform loads by shifting the location and magnitude of positive and negative moments. There exists an optimal overhang ratio that minimizes the maximum absolute bending moment in the member. Given Data / Assumptions: Simply supported interior span with identical overhangs at both ends. Uniformly distributed load of constant intensity over the entire physical length. Prismatic beam; linear elastic behavior. Concept / Approach: The optimal condition is found by equating the magnitude of the maximum positive moment in the span to the magnitude of the maximum negative moment over the supports, then solving for the ratio a/L, where a is each overhang and L is the total physical length. Step-by-Step Solution: Let w be the load per unit length, a the overhang on each end, and L the total length.Write expressions for the reactions and bending moments as functions of a/L.Set the peak negative support moment equal in magnitude to the peak positive span moment.Solve the resulting equation for a/L; the well-known solution is approximately 0.207 (about 20.7% of total length). Verification / Alternative check: Design handbooks and classical beam tables list this optimum ratio near 0.207. A quick energy or influence-line based derivation confirms the same optimum. Why Other Options Are Wrong: Values 0.307, 0.407, 0.508 and 0.107 either overestimate or underestimate the required overhang, yielding unbalanced positive and negative peaks and thus a larger maximum bending moment. Common Pitfalls: Confusing the interior clear span with total length; forgetting that both overhangs must be equal; applying point-load formulas instead of uniform-load relations. Final Answer: 0.207

Question 9

Material behavior at the yield point At the yield point of a metallic test specimen under tension, which behavior best describes the material response?

Accepted Answer

Introduction / Context: The yield point marks the onset of plasticity in ductile metals during a tensile test. Understanding this transition is fundamental to differentiating elastic design limits from plastic behavior and permanent deformation. Given Data / Assumptions: Conventional engineering stress–strain testing of a ductile metal. Elastic behavior precedes yield; beyond yield, plastic strains accumulate. Concept / Approach: Hooke's law (stress proportional to strain) applies in the linear elastic range. At yield, the proportional limit is exceeded and plastic (permanent) strain initiates. On unloading after yielding, only the elastic portion of strain is recovered; the plastic portion remains. Step-by-Step Solution: Before yield: stress = E * strain; unloading follows the same linear slope E.At yield: nonlinearity appears; plastic mechanisms activate (dislocation motion, slip).After yield: on complete unloading, permanent set (plastic strain) is observed; shape is not fully recovered. Verification / Alternative check: Typical mild steel stress–strain curves show a distinct upper and lower yield point; aluminum alloys show a smooth yield transition but still exhibit plastic deformation beyond the 0.2% proof stress. Why Other Options Are Wrong: (a) and (b) would imply purely elastic proportional behavior, which ends at or before yield; (c) contradicts the definition of plastic strain; (e) viscoelasticity is a time-dependent phenomenon not specifically tied to the yield point in metals. Common Pitfalls: Confusing yield point with ultimate strength; assuming full shape recovery after any unloading; ignoring the difference between proportional limit and yield strength. Final Answer: Undergoes plastic deformation

Question 10

Three-hinged arch basics in structural engineering A three-hinged arch is typically provided with hinges at both supports. Where is the third hinge commonly located for analysis and construction convenience?

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Introduction / Context: A three-hinged arch is a classical statically determinate structure used in roofs, bridges, and temporary works. Understanding where the third hinge is placed is fundamental for computing reactions, internal forces, and ensuring thermal and settlement compatibility. Given Data / Assumptions: Supports are hinged at both springings (arch ends). The arch is designed as a three-hinged system. No special geometric irregularities are assumed. Concept / Approach: Adding a third hinge at the crown creates a statically determinate system. The crown hinge divides the arch into two determinate segments, simplifying analysis for horizontal thrust and bending moments. It also relieves the structure of secondary stresses from temperature and support settlements by providing an internal release at the highest point. Step-by-Step Solution: Identify support conditions: two end hinges are present.Add a third hinge: the most effective position is at the crown.Outcome: the structure becomes determinate; analysis uses simple statics on each half. Verification / Alternative check: Standard structural analysis texts model three-hinged arches with the third hinge at the crown specifically to exploit symmetry and facilitate closed-form solutions for thrust and moments. Why Other Options Are Wrong: One-quarter span: uncommon and offers no special analytical advantage.Anywhere along the rib: theoretically possible but not standard and complicates analysis.Only two hinges: that would be a two-hinged (indeterminate) arch, not three-hinged. Common Pitfalls: Confusing two-hinged and three-hinged arches and their determinacy. Assuming the crown hinge is optional; it is essential to the three-hinged definition. Final Answer: at the crown

Question 11

Second moment of area of a triangle about its base For a triangular cross-section with base b and height h, what is the moment of inertia (second moment of area) about its base line?

Accepted Answer

Introduction / Context: The second moment of area (moment of inertia) is a geometric property used to predict a member’s resistance to bending. For standard shapes like triangles, rectangles, and circles, closed-form expressions enable quick section property calculations used in beam design and deflection analysis. Given Data / Assumptions: Cross-section is a triangle with base b and height h. Axis of interest is along the base line of the triangle. Material is homogeneous; the property is purely geometric. Concept / Approach: For a triangle, the centroidal moment of inertia about an axis parallel to the base is I_centroid = (b * h^3) / 36. Using the parallel-axis theorem to shift from centroid to the base line a distance h/3, the base-axis moment of inertia becomes I_base = I_centroid + A * (h/3)^2, where A = (1/2) * b * h. This yields I_base = (b * h^3) / 12. Step-by-Step Solution: Compute area: A = 0.5 * b * h.Use centroidal value: I_centroid = (b * h^3) / 36.Apply parallel-axis theorem: I_base = I_centroid + A * (h/3)^2.I_base = (b * h^3) / 36 + (0.5 * b * h) * (h^2 / 9) = (b * h^3) / 12. Verification / Alternative check: Dimension check: I has units of length^4, satisfied by (b * h^3). Also, compared to the centroidal value, the base-axis inertia must be larger, and 1/12 is larger than 1/36 by a factor of 3, consistent with the shift. Why Other Options Are Wrong: (b * h^3)/36: This is the centroidal inertia, not about the base.Terms with b^3 * h or with h^2 indicate wrong axis or incorrect application of the theorem. Common Pitfalls: Confusing centroidal and base-line axes. Applying the parallel-axis theorem with an incorrect distance (not using h/3). Final Answer: I_base = (b * h^3) / 12

Question 12

Bending and shear stresses across a beam section: identify the correct statements Consider the stress distributions in a prismatic beam under pure bending and shear. Which combination of statements is correct?

Accepted Answer

Introduction / Context: Understanding how bending and shear stresses vary across a beam section is foundational for safe design. Bending stresses vary linearly with distance from the neutral axis, while shear stresses follow shape-dependent distributions (e.g., parabolic in rectangles), peaking at the neutral axis and vanishing at free surfaces. Given Data / Assumptions: (a) The bending stress in a section is zero at its neutral axis and maximum at the outer fibres. (b) The shear stress is zero at the outer fibres and maximum at the neutral axis. (c) The bending stress at the outer fibres is a principal normal stress (in pure bending, shear is zero on the cross-section). (d) The planes of principal stresses are inclined at 45° to the neutral plane. Concept / Approach: In pure bending of a prismatic beam, normal stress varies linearly and no in-plane shear acts on the cross-section, making the normal stress itself principal at that section. Shear stress distribution is zero at the free surfaces and maximum at the neutral axis for common shapes. The statement about principal planes at 45° pertains to a pure shear state, not pure bending. Step-by-Step Solution: Evaluate (a): True—linear distribution from zero at neutral axis to maximum at extreme fibres.Evaluate (b): True—shear stress is highest at the neutral axis and zero at outer surfaces (for rectangular, I-section webs dominate).Evaluate (c): True—on the cross-section in pure bending, shear equals zero, so the normal stress equals a principal normal stress.Evaluate (d): False—45° principal planes occur under pure shear; not applicable to pure bending planes. Verification / Alternative check: Mohr’s circle for pure bending has a point on the normal-stress axis with zero shear, confirming that the normal stress at the section is principal. For a rectangular section in shear, the parabolic distribution confirms zero at the free boundaries and maximum at the neutral axis. Why Other Options Are Wrong: Only (a) and (b): Ignores that (c) is also true in pure bending.Only (a) or All (a)–(d): Either omits valid statements or includes the incorrect (d).None of these: Incorrect since multiple statements are true. Common Pitfalls: Confusing pure shear behavior with bending behavior. Assuming the presence of shear on the cross-section under pure bending. Final Answer: Only (a), (b), and (c) are correct

Question 13

Deflection of a cantilever with an applied end moment A cantilever of length L is subjected to a constant bending moment M at its free end (no transverse load). If the flexural rigidity is EI, what is the vertical deflection of the free end?

Accepted Answer

Introduction / Context: The slope-deflection behavior of a cantilever under an applied end moment is a basic case in elastic beam theory. With pure end moment and no transverse load, the shear force is zero along the span, and curvature is constant, allowing a straightforward integration to obtain slope and deflection. Given Data / Assumptions: Prismatic beam with constant E and I. Cantilever length = L, fixed at one end, free at the other. Applied end moment = M (clockwise or counterclockwise sign as per convention). Concept / Approach: From elastic curve: M(x) = constant = M. Curvature κ = M / (E * I). Then dy' = M / (E * I) under standard sign convention. Integrating the curvature twice with boundary conditions at the fixed end (y = 0 and dy/dx = 0 at x = 0) yields closed-form expressions for slope and deflection. Step-by-Step Solution: Write differential equation: E * I * d^2y/dx^2 = M.Integrate once: E * I * dy/dx = M * x + C1; apply slope at fixed end (x = 0): dy/dx = 0 ⇒ C1 = 0.Integrate again: E * I * y = (M * x^2)/2 + C2; apply deflection at fixed end (x = 0): y = 0 ⇒ C2 = 0.At free end x = L: y_free = (M * L^2) / (2 * E * I). Verification / Alternative check: Deflection varies with L^2, and slope at the free end is θ_free = (M * L) / (E * I). Units: M has forcelength, so y_free has length, confirming dimensional consistency. Why Other Options Are Wrong: (M * L)/(E * I): This is the slope, not the deflection.(M * L^3)/(3 * E * I) and other powers: Correspond to different load cases (e.g., end point load or UDL), not a pure moment. Common Pitfalls: Confusing the pure moment case with an end load or uniformly distributed load. Forgetting the boundary conditions at the fixed end, leading to wrong constants. Final Answer: y_free = (M * L^2) / (2 * E * I)

Question 14

Equal normal and shear components on an inclined plane in uniaxial loading A member is subjected to a single axial force. On a plane inclined at angle θ to the direction of the force, the normal stress component equals the tangential (shear) stress …

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Introduction / Context: Stress transformation in a uniaxially loaded member shows how normal and shear components vary with the orientation of the cut plane. Recognizing special angles simplifies analysis and helps in failure predictions using Mohr’s circle or transformation equations. Given Data / Assumptions: Uniaxial normal stress σ acts along the member axis. Plane is inclined at angle θ to the force direction (equivalently, to the axis of σ). Linear elastic continuum; plane stress transformation applies. Concept / Approach: For uniaxial stress, the normal and shear components on a plane at angle θ to the stress direction are: σ_n = σ * cos^2(θ) and τ = σ * sin(θ) * cos(θ). Equating σ_n and τ gives the characteristic angle where normal and shear magnitudes match. Step-by-Step Solution: Write equality: σ * cos^2(θ) = σ * sin(θ) * cos(θ).Cancel σ and cos(θ) (for θ not equal to 90°): cos(θ) = sin(θ).Therefore, tan(θ) = 1 ⇒ θ = 45°. Verification / Alternative check: Using Mohr’s circle with center at σ/2 and radius σ/2, the plane where σ_n = τ lies at 45° from the stress axis, consistent with the algebraic result. Why Other Options Are Wrong: 30°/60°: These produce unequal components because cos^2(θ) ≠ sin(θ)cos(θ) at those angles.90°: On a plane normal to the stress direction, σ_n = 0 and τ = 0, not equal nonzero values.15°: No special equality occurs at this angle. Common Pitfalls: Mixing the angle from the plane normal vs. plane itself; the formula above uses the angle between the plane and the stress direction. Forgetting that both components scale with σ; equality depends only on angle, not magnitude. Final Answer: 45°

Question 15

In engineering mechanics, a body is said to be in mechanical equilibrium when no net effect of forces or moments tends to change its state. Which statement best describes this condition for a rigid body (e.g., a machine part or structural element)?

Accepted Answer

Introduction / Context: Equilibrium of a rigid body is a foundational concept in engineering mechanics. Designers and analysts check equilibrium before proceeding to deformation, stress, or stability calculations. This question tests your understanding of the complete mathematical condition required for a body to be in mechanical equilibrium. Given Data / Assumptions: Rigid body assumption (deformations are negligible for equilibrium checks). Forces and moments are external resultants acting on the body. Static conditions (no acceleration). Concept / Approach: The equilibrium of a rigid body requires two independent conditions: force equilibrium and moment equilibrium. In vector form, these are written as: ΣF = 0ΣM = 0These must hold simultaneously. If either condition is violated, the body translates or rotates with non-zero acceleration. Step-by-Step Solution: 1) Check translational equilibrium: enforce ΣF_x = 0, ΣF_y = 0, and (in 3D) ΣF_z = 0.2) Check rotational equilibrium: enforce ΣM about any convenient point or axis = 0. For 3D, moments about x, y, z axes must each sum to zero.3) If both are satisfied, no net linear or angular acceleration occurs. The body remains at rest or moves with constant velocity (Newton’s first law). Verification / Alternative check: Using Newton’s second law: ΣF = m * a and ΣM = I * α. For equilibrium, a = 0 and α = 0, which directly yields ΣF = 0 and ΣM = 0. Why Other Options Are Wrong: Moves horizontally at a constant speed only: ignores vertical forces and moments.Moves vertically at a constant speed only: ignores horizontal forces and moments.Only rotates uniformly about its centre of gravity: allows nonzero translation or mismatched force balance.None of these: incorrect because the force-and-moment zero condition is the correct full statement. Common Pitfalls: Forgetting to check moments after balancing forces. Taking moments about a point that hides unknowns without ensuring independence of equations. Assuming rest implies equilibrium without verifying external resultant is zero. Final Answer: The vector sum of all external forces is zero and the algebraic sum of all external moments about any point is zero.

Question 16

In a uniaxial loading case, a plane is taken at an angle θ to the direction of the applied force. What is the ratio of the tangential (shear) stress to the normal stress on that plane?

Accepted Answer

Introduction / Context: Stress transformation on an inclined plane is a classic topic in strength of materials. For a uniaxial direct stress, the normal and shear components on a plane at angle θ have standard relations. This question asks for the ratio of shear to normal stress on that plane. Given Data / Assumptions: Uniaxial direct stress σ acts along a reference direction. An inclined plane is cut at angle θ to this stress direction. Plane stress transformation relations apply. Concept / Approach: For a uniaxial stress σ, the stresses on an inclined plane at angle θ (measured between the plane and the stress direction) are: Normal stress: σ_n = σ * cos^2(θ)Shear stress: τ = σ * sin(θ) * cos(θ)Therefore the desired ratio is τ / σ_n = tan(θ). Step-by-Step Solution: 1) Write σ_n = σ cos^2 θ.2) Write τ = σ sin θ cos θ.3) Compute τ/σ_n = (σ sin θ cos θ) / (σ cos^2 θ) = tan θ. Verification / Alternative check: Using Mohr’s circle, the radius is σ/2 and the coordinates on the circle at angle 2θ from the σ-axis give the same relations, yielding τ/σ_n = tan θ. Why Other Options Are Wrong: sin θ, cos θ, sec θ, cot θ: do not match the derived ratio from the standard transformation equations for uniaxial stress. Common Pitfalls: Confusing the angle definition (plane angle vs. normal angle). Using σ_n = σ cos θ instead of σ cos^2 θ. Dropping the common factor σ when forming the ratio. Final Answer: tan θ

Question 17

Eccentric axial loading on a rectangular section: A bar (or column) with direct compressive load and bending is said to be in combined stress. Which of the following sets of statements about bending and axial effects under eccentric compression is c…

Accepted Answer

Introduction / Context: Members under eccentric axial load experience a combination of direct (uniform) stress and bending stress. Understanding how these two components superpose determines whether any region is in compression or tension. This question asks you to identify the correct statements about combined stress behavior. Given Data / Assumptions: Member is prismatic and behaves elastically. Plane sections remain plane after bending. Axial load is compressive and applied with some eccentricity (causing bending). Concept / Approach: Resultant stress at a fiber is the algebraic sum of direct stress and bending stress contributions. If the bending-induced tensile stress is smaller than the direct compressive stress, the entire cross-section remains in compression. If it equals the axial value, stress at one edge becomes zero. If it exceeds the axial compressive stress, a portion goes into tension. Step-by-Step Solution: (a) The moment of inertia must be taken about the axis about which bending occurs – correct.(b) If tensile stress from bending is less than axial compressive stress, the net stress everywhere is compressive – correct.(c) If tensile stress equals axial compressive stress, the stress at one extreme edge becomes zero, not compressive – statement is incorrect.(d) If tensile stress from bending is greater than axial compressive stress, part of the section goes into net tension – correct. Verification / Alternative check: Use superposition: at an extreme fiber, write σ_edge = σ_axial ± σ_bendingand compare their magnitudes to see sign changes across the section. Why Other Options Are Wrong: Choices listing (c) as correct include a false statement; “All” is therefore wrong. Common Pitfalls: Misinterpreting the sign convention for tensile/compressive stresses. Using the wrong centroidal axis for the second moment of area. Ignoring section edge where stress is most critical. Final Answer: Only (a), (b), and (d) are correct

Question 18

Two-hinged parabolic arch of span l and rise h carries a triangular load that varies from zero at the left support to ω (per unit length) at the right support. What is the horizontal thrust H at the supports?

Accepted Answer

Introduction / Context: Horizontal thrust in a two-hinged parabolic arch depends on the load shape and the arch geometry. Unlike a three-hinged arch, indeterminacy requires using energy or funicular principles to obtain H for non-uniform loading. Here, the loading is triangular from zero to ω across span l with rise h. Given Data / Assumptions: Parabolic two-hinged arch, span = l, rise = h. Distributed load w(x) varies linearly from 0 at left to ω at right: w(x) = (ω/l) * x. Small-slope approximation for the strain energy integral; ds ≈ dx. Concept / Approach: For a two-hinged arch, the horizontal thrust H can be obtained by Castigliano’s theorem or the consistent deformation method: H = (∫ M_0 * y dx) / (∫ y^2 dx)where M_0 is the bending moment of the corresponding simply supported beam under the same load, and y is the arch ordinate measured from the chord line. For a parabolic arch: y = 4h * x * (l - x) / l^2. Step-by-Step Solution: 1) Reactions for the simply supported beam with triangular load: total load W = (1/2) * ω * l, R_left = W/3 = ω l / 6, R_right = 2W/3 = ω l / 3.2) Beam bending moment function: M_0(x) = R_left * x − (ω/(6l)) * x^3.3) Arch ordinate: y(x) = 4h * x * (l − x) / l^2.4) Compute numerator I1 = ∫_0^l M_0(x) * y(x) dx.5) Compute denominator I2 = ∫_0^l y(x)^2 dx.6) Evaluate H = I1 / I2, which simplifies to H = (ω * l^2) / (16 * h). Verification / Alternative check: The expression has correct dimensions of force since ω has units of force/length and l^2 / h is length, yielding force. It also trends correctly: larger h reduces H, larger l or ω increases H. Why Other Options Are Wrong: Values with 8h or 12h correspond to other load cases (e.g., uniform load) and are not valid for a zero-to-ω triangular distribution.Expressions with l or l^3 in the numerator have incorrect dimensional consistency for this case. Common Pitfalls: Using the UDL arch thrust formula H = ω l^2 / (8 h) instead of integrating for the triangular loading. Forgetting the correct centroid and reactions for the triangular load. Using the wrong arch ordinate function. Final Answer: H = (ω * l^2) / (16 * h)

Question 19

Euler–column idealization: What is the equivalent (effective) length for a column of actual length L when both ends are fixed against rotation and translation?

Accepted Answer

Introduction / Context: Effective (equivalent) length is central to Euler buckling and slenderness calculations. End restraints reduce the buckling half-wave, effectively shortening the buckling length compared with the actual length. This question focuses on the standard end condition of a column with both ends fixed. Given Data / Assumptions: Prismatic column of length L. Both ends fully fixed (no rotation, no translation). Elastic buckling per Euler theory. Concept / Approach: For Euler buckling, the critical load is: P_cr = (π^2 * E * I) / (L_e)^2where L_e is the effective length determined by end conditions. Standard effective-length factors K give L_e = K * L. For both ends fixed, K = 0.5, hence L_e = L/2. Step-by-Step Solution: 1) Identify end condition: both ends fixed.2) Use standard K-value table: K = 0.5.3) Compute effective length L_e = K * L = 0.5 * L = L/2. Verification / Alternative check: Compare with other end conditions: both hinged (K = 1 ⇒ L_e = L), one fixed–one free (K = 2 ⇒ L_e = 2L). The fixed–fixed case is the stiffest, so it must have the smallest effective length, L/2, and largest P_cr. Why Other Options Are Wrong: L, 2L, √2 L, 3L/2 correspond to other end conditions, not the fixed–fixed case. Common Pitfalls: Confusing fixed–fixed with hinged–hinged. Using unsupported K-values without checking end fixity details. Forgetting that greater fixity reduces effective length. Final Answer: L / 2

Question 20

A circular bar of diameter d and length l elongates by e under a gradually applied axial load W. If the same bar is simply supported as a beam of span l and carries a central point load W, what is the mid-span deflection in terms of e, l, and d?

Accepted Answer

Introduction / Context: Relating axial deformation to bending deflection through material and geometric properties is a powerful technique. Here, the same load W produces an axial extension in a bar and a mid-span deflection in a simply supported beam made from that bar. Eliminating E and geometric constants links the two responses. Given Data / Assumptions: Uniform circular bar: diameter d, length l. Under axial load W (gradually applied) the extension is e. The same member, simply supported with central load W, deflects at mid-span by δ. Linear elasticity holds. Concept / Approach: Axial extension under load W: e = W * l / (A * E)with A = (π d^2) / 4. Mid-span deflection for a simply supported beam with central load W: δ = W * l^3 / (48 * E * I)with I = (π d^4) / 64. Eliminate W/E using the axial relation to express δ in terms of e, l, d. Step-by-Step Solution: 1) From axial case: W/E = e * A / l.2) Substitute into beam deflection: δ = (e * A / l) * l^3 / (48 * I) = e * A * l^2 / (48 * I).3) Compute A/I for circular section: A/I = [(π d^2 / 4)] / [π d^4 / 64] = 16 / d^2.4) Therefore δ = e * (16 / d^2) * l^2 / 48 = e * l^2 / (3 d^2). Verification / Alternative check: Dimensional check: e has dimension of length; multiplying by (l^2 / d^2) yields length. Trend check: larger diameter reduces deflection, larger span increases deflection – both physically consistent. Why Other Options Are Wrong: Other coefficients (1/6, 2/3, etc.) result from algebraic mistakes in A/I or the beam formula (48EI). Common Pitfalls: Using I for a rectangle instead of a circle. Forgetting that the axial load is gradually applied only affects strain energy, not the static e relation. Mixing up span l and half-span in the beam formula. Final Answer: e * l^2 / (3 d^2)

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