Q: In plastic analysis, identify the shape factor ( Mp / My ) for a rectangular cross-section (keeping symbols and values unchanged).

Introduction / Context: The shape factor S (also written as Mp / My) quantifies the reserve strength beyond first yield in plastic analysis. It depends only on cross-sectional shape, not on material properties, and is central to plastic hinge formation and collapse load calculations. Given Data / Assumptions: Section is rectangular, subjected to bending. My = yield moment (based on elastic stress distribution up to sigma_y). Mp = plastic moment (fully plastic stress block). Concept / Approach: For a rectangle, the elastic section modulus (Z) and plastic section modulus (Zp) have a constant ratio: S = Zp / Z = Mp / My. Classical plastic analysis yields S = 1.5 for a rectangular section because the plastic neutral axis divides the section into two equal areas in full plasticity, doubling the lever arm relative to the elastic centroidal distribution by a factor leading to 1.5 overall. Step-by-Step Solution: 1) My = sigma_y * Z (elastic).2) Mp = sigma_y * Zp (plastic).3) Shape factor S = Mp / My = Zp / Z.4) For a rectangle, Zp / Z = 1.5 → S = 1.5. Verification / Alternative check: Handbooks and steel design texts tabulate S = 1.5 for rectangles, approximately 1.1 for I-sections (compact flanges/webs vary), and 1.7 for diamonds, confirming the standard value for rectangles. Why Other Options Are Wrong: 1.4, 1.6, 1.7: Belong to other shapes or bounds; not the canonical value for a pure rectangular section. Common Pitfalls: Confusing shape factor with safety factor or with section modulus itself. Using a noncompact or perforated section where local buckling alters plastic capacity. Final Answer: 1.5.

Q: Beams built from two or more different materials, rigidly connected to act together as a single unit under bending, are called composite beams.

Introduction / Context: Engineering practice often combines materials (e.g., steel–concrete, timber–steel) to optimize stiffness, strength, weight, or cost. When elements are connected so there is no slip at the interface and they bend together, the assembly behaves as a single member called a composite beam. Given Data / Assumptions: Two or more materials are used. Rigid connection prevents interlayer slip (full composite action). Plane sections remain plane (compatibility across materials via transformed-section method). Concept / Approach: Composite behavior requires shear transfer at the interface (e.g., studs in steel–concrete). Analysis uses modular ratio n = E1/E2 to transform areas and compute stresses/deflections as if the section were homogeneous under the assumption of perfect bond. Step-by-Step Solution: 1) Define objective: different materials to share bending based on stiffness.2) Ensure composite action: provide connectors to prevent slip.3) Use transformed section: A2' = n*A2 to compute neutral axis and section modulus.4) Evaluate stresses with sigma = M*y / I_transformed, respecting each material's allowable limits. Verification / Alternative check: Load tests show reduced deflection versus noncomposite action, confirming stiffness gain from composite behavior. Why Other Options Are Wrong: Compound beams: Term sometimes used for built-up beams of the same material; not specific to multi-material action. Indeterminate/Determinate beams: Describe static determinacy, not material composition. Common Pitfalls: Assuming full composite action without adequate connectors. Ignoring differential shrinkage/creep between materials. Final Answer: Composite beams.

Q: A simply supported rolled-steel joist under a uniformly distributed load shows a maximum midspan deflection of 10 mm and an end slope of 0.002 rad; determine the span length of the joist.

Introduction / Context: For a simply supported beam carrying UDL, classic formulae relate maximum deflection and end slope to load, span, and flexural rigidity. With both deflection and slope known, their ratio eliminates E and I, allowing span determination directly. Given Data / Assumptions: Beam: simply supported, UDL over entire span. Maximum deflection: delta_max = 10 mm = 0.010 m. End slope: theta_end = 0.002 rad. Linear elastic behavior; small deflection theory. Concept / Approach: For UDL w on span L: delta_max = (5*w*L^4)/(384*E*I) and theta_end = (w*L^3)/(24*E*I). Taking the ratio cancels w, E, and I, yielding a simple proportionality between delta_max, theta_end, and L. Step-by-Step Solution: 1) Write ratio: (delta_max)/(theta_end) = [(5*w*L^4)/(384*E*I)] / [(w*L^3)/(24*E*I)] = (5*L/16).2) Solve for L: L = (16/5) * (delta_max / theta_end).3) Substitute values: delta_max = 0.010 m, theta_end = 0.002 → delta_max/theta_end = 5.4) L = (16/5) * 5 = 16 m. Verification / Alternative check: Dimensional consistency: delta/theta has units of length, so L emerges in meters as expected. Cross-check with typical steel spans—16 m is a reasonable magnitude for the given small slope and 10 mm deflection under UDL. Why Other Options Are Wrong: 10, 12, 14, 18 m: Do not satisfy the ratio (5*L/16) = 0.010/0.002 = 5. Common Pitfalls: Using the point-load formula instead of UDL relations. Forgetting to convert 10 mm to meters. Final Answer: 16 m.

Q: Identify the name associated with the total strain energy theory of failure at the elastic limit (choose the classical attribution used in mechanics of materials).

Introduction / Context: Failure theories extend uniaxial yield data to multiaxial stress states. Each classical theory corresponds to a distinct failure criterion, and many are named for the researchers who developed them. Recognizing the correct attributions helps avoid mixing up criteria during design checks. Given Data / Assumptions: Total strain energy theory is referenced. We are at the elastic limit under multiaxial stress. Names provided correspond to well-known criteria. Concept / Approach: Total strain energy theory (sometimes attributed to Beltrami–Haigh) posits that failure occurs when the total elastic strain energy per unit volume reaches the same value as at yielding in simple tension. It differs from the distortional (shear) energy theory of Von Mises and the maximum shear stress (Tresca/Guest) theory. Step-by-Step Solution: 1) Map theories to names: Rankine → maximum principal stress; St. Venant → maximum principal strain; Tresca/Guest → maximum shear stress; Von Mises → distortion (shear) energy; Haigh → total strain energy.2) Select Haigh's theory as the correct attribution for total strain energy. Verification / Alternative check: Mechanics references list Beltrami–Haigh energy criteria under total energy, separate from Von Mises (J2) distortional energy. Why Other Options Are Wrong: Guest/Tresca: Maximum shear stress, not energy-based. St. Venant: Maximum principal strain criterion. Rankine: Maximum normal stress criterion. Von Mises: Distortion (shear) energy, not total energy. Common Pitfalls: Confusing total energy (Haigh) with distortional energy (Von Mises). Assuming all energy-based theories are identical—they are not. Final Answer: Haig's theory.

Q: State the name associated with the shear strain energy (distortion energy) theory for elastic failure under multiaxial stress.

Introduction / Context: Designing ductile materials under multiaxial stress often uses an energy-based failure criterion. The shear (distortion) strain energy theory—also called the Von Mises or Hencky–Von Mises criterion—is widely adopted in metal design because it correlates well with yielding in ductile metals. Given Data / Assumptions: We focus on the distortional component of strain energy as the trigger for yield. Elastic behavior up to yield point is assumed. Concept / Approach: Von Mises theory states that yielding begins when the second deviatoric stress invariant (J2) reaches the same value as in uniaxial tension at yield. Physically, only the distortion energy causes yielding; hydrostatic (volumetric) energy does not. Step-by-Step Solution: 1) Associate names with criteria: Rankine → maximum principal stress; St. Venant → maximum principal strain; Guest/Tresca → maximum shear stress; Haigh → total strain energy; Von Mises → distortion energy.2) Select Von Mises for shear strain (distortion) energy theory. Verification / Alternative check: Modern codes and FEA post-processing commonly report Von Mises equivalent stress for ductile materials, aligning with this theory. Why Other Options Are Wrong: Rankine/St. Venant: Principal stress/strain criteria. Guest or Trecas: Maximum shear stress, not energy-based. Haig: Total strain energy, not distortional-only. Common Pitfalls: Interchanging Tresca and Von Mises; Tresca gives a hexagonal yield surface, Von Mises a circular one in principal stress space. Final Answer: Von Mises.

Q: In a three-hinged arch analysis by funicular/graphical construction, measured distances are AD = 4 cm, DD' = 3 cm, and AD' = 5 cm. Determine the angle between the reaction directions at supports A and B (choose the nearest correct option).

Introduction / Context: Graphical methods for arches and frames use force polygons and funicular constructions. Distances measured on the load line and parallel lines relate to the magnitudes and directions of reactions. This problem uses a simple geometric relation emerging from a right-triangle measurement. Given Data / Assumptions: Measured lengths: AD = 4 cm, DD' = 3 cm, AD' = 5 cm. Standard three-hinged arch force polygon construction is assumed. The angle between reaction directions at A and B is sought. Concept / Approach: The measured segments form a triangle with sides 3, 4, and 5. Recognition of the 3–4–5 Pythagorean triplet indicates a right angle in the associated force polygon. In this classical construction, the right angle corresponds to the angle between the two reaction lines, yielding an orthogonal (90°) relationship. Step-by-Step Solution: 1) Identify triangle with sides AD = 4, DD' = 3, and AD' = 5.2) Check Pythagoras: 4^2 + 3^2 = 16 + 9 = 25 = 5^2 → right triangle.3) Map this right angle in the force polygon to the included angle between reaction vectors.4) Conclude the reaction directions are mutually perpendicular. Verification / Alternative check: Alternate geometric derivations (cosine law on the force triangle) also yield cos(theta) = 0 when the measured lengths satisfy a 3–4–5 relation. Why Other Options Are Wrong: 30°, 45°, 60°: Not supported by the 3–4–5 right-triangle result; these would require different length ratios. Common Pitfalls: Misidentifying which included angle corresponds to the reaction directions in the force polygon. Arithmetic slips when checking Pythagoras. Final Answer: 90°.

Q: For a cantilever beam of span L, what is the ratio of free-end deflections due to the same isolated load P placed at L/3 and at 2L/3 from the fixed end?

Introduction / Context: Deflection of beams under point loads depends on where the load acts. For a cantilever, the same load located nearer to the free end generally causes a larger free-end deflection than the same load applied closer to the fixed end. This question compares free-end deflections for a single cantilever under identical load magnitudes but at two different positions along the span. Given Data / Assumptions: Prismatic cantilever beam with span L, flexural rigidity E*I constant. Point load P applied at a distance a from the fixed end. We compare deflection at the free end for a = L/3 and a = 2L/3. Concept / Approach: For a cantilever with a point load P applied at distance a from the fixed end, the free-end deflection is given by: δ_free = P * a^2 * (3L - a) / (6 * E * I)This arises from integrating the curvature equation M / (E * I) along the span considering the load location. Since P, E, I, and L are the same for both cases, the ratio depends only on the factor a^2 * (3L - a). Step-by-Step Solution: Case 1 (a = L/3): factor = (L/3)^2 * (3L - L/3) = (L^2/9) * (8L/3) = 8L^3/27.Case 2 (a = 2L/3): factor = (2L/3)^2 * (3L - 2L/3) = (4L^2/9) * (7L/3) = 28L^3/27.Ratio δ(L/3) : δ(2L/3) = (8/27) : (28/27) = 8 : 28 = 2 : 7. Verification / Alternative check: Sanity check: load nearer the free end (2L/3) should cause greater free-end deflection; ratio < 1 confirms this expectation. Why Other Options Are Wrong: 7 : 2 and 4 : 1: Invert the true ratio; contradict physical expectation. 1 : 4 and 3 : 8: Do not match the derived factor relationship. Common Pitfalls: Using end-load formula δ = P*L^3/(3*E*I) irrespective of load position; forgetting that only the segment from the load to the free end directly influences δ_free in this relation. Final Answer: 2 : 7

Q: A material may be predicted to fail according to different failure theories; which condition(s) can be a valid failure criterion depending on the material and loading?

Introduction / Context: Different materials fail under different controlling stress/strain states. Classical failure theories provide criteria to assess safety: maximum principal stress (Rankine), maximum principal strain (St. Venant), maximum shear stress (Tresca), and energy-based theories like total strain energy or distortion energy (Beltrami, von Mises). This question checks awareness that multiple criteria may be relevant depending on the material behavior (brittle versus ductile) and design code. Given Data / Assumptions: General 3D stress state possible. Allowable limits σ0 (direct), shear limit, and strain/energy limits are known from material data or codes. We are not restricting ourselves to only one theory. Concept / Approach: Brittle materials (e.g., cast iron): Rankine (maximum normal stress) often governs.Ductile materials (e.g., mild steel): Shear- or energy-based criteria (Tresca or von Mises) are typically appropriate.High strain sensitivity: Maximum principal strain criterion may be used for certain materials/components.Energy viewpoint: Exceeding total allowable strain energy can indicate failure risk under complex states. Step-by-Step Solution: Map each option to a known failure theory.Recognize that design codes allow different criteria per material class and service condition.Therefore, all listed conditions can represent valid failure triggers in the proper context. Verification / Alternative check: Compare predicted safety margins using Rankine vs. Tresca vs. von Mises in a biaxial state; the controlling (lowest margin) criterion dictates design in that scenario. Why Other Options Are Wrong: Any single choice alone ignores other established, code-recognized criteria. Common Pitfalls: Applying a ductile-material criterion to brittle materials (or vice versa) without justification; assuming one universal theory fits all cases. Final Answer: All the above.

Q: A material is said to be perfectly elastic when which of the following statements best describes its behavior under loading and unloading?

Introduction / Context: Elasticity measures a material's ability to recover its original dimensions after the applied load is removed. Perfect elasticity is an idealization used in fundamental mechanics to simplify analysis and define the linear portion of stress–strain behavior. Given Data / Assumptions: Material is homogeneous and isotropic in the ideal model. Loading remains within the elastic limit. No time-dependent effects (creep, relaxation) are considered. Concept / Approach: For a perfectly elastic material, strain is entirely recoverable upon unloading: there is no plastic (permanent) deformation, and the unloading path coincides with the loading path. Hooke's law (stress proportional to strain) typically models this regime for small deformations. Step-by-Step Solution: Define perfect elasticity: zero permanent set after full unloading.Relate to stress–strain curve: linear reversible behavior below yield point.Choose the option that explicitly states complete recovery. Verification / Alternative check: In tensile tests, if the specimen returns exactly to its original gauge length after removal of load within the elastic range, the behavior matches perfect elasticity. Why Other Options Are Wrong: Partial recovery: Indicates viscoelastic or elastoplastic behavior. No recovery: Represents plastic or damaged state. None of these: Incorrect because a precise correct statement exists. Common Pitfalls: Confusing perfect elasticity with high elasticity; believing linearity is required at all strains (the idealization usually concerns small strains). Final Answer: It regains its original shape completely upon removal of load (no permanent set).

Q: Stress is a measure of internal force per unit area. Which of the following expresses valid SI-consistent units for stress?

Introduction / Context: Stress (σ or τ) quantifies internal forces distributed over a cross-sectional area. Its SI unit is the Pascal, defined as 1 N/m^2, but engineering practice commonly uses scaled units such as N/mm^2 or N/cm^2 for convenience with typical magnitudes. Given Data / Assumptions: Stress = Force / Area. Force unit: Newton (N). Area units: m^2, cm^2, mm^2 are all valid metric areas. Concept / Approach: Any combination of Newtons divided by a squared length unit yields a pressure/stress unit. Conversions are straightforward: 1 N/mm^2 = 10^6 N/m^2 (MPa), and 1 N/cm^2 = 10^4 N/m^2 (0.1 MPa). Therefore, N/m^2, N/cm^2, and N/mm^2 are all valid stress units; the Pascal is the SI base (N/m^2). Step-by-Step Solution: Write the definition: stress = force / area.Identify SI base: N/m^2 = Pa.Acknowledge engineering scales: N/mm^2 (MPa) and N/cm^2 (0.1 MPa).Therefore, all listed unit forms are acceptable expressions of stress. Verification / Alternative check: Perform conversions: 1 N/mm^2 = (1 N)/(10^-6 m^2) = 10^6 N/m^2.1 N/cm^2 = (1 N)/(10^-4 m^2) = 10^4 N/m^2.Consistency confirms validity. Why Other Options Are Wrong: None of these: Incorrect because all three are legitimate unit expressions. Common Pitfalls: Believing only N/m^2 is correct; confusing kgf/cm^2 with N/cm^2; mixing mass and force units. Final Answer: All of the above (unit-consistent pressure measures)

Question 1

For a circular shaft subjected simultaneously to a bending moment M and a torque T, determine the ratio of maximum bending normal stress to maximum torsional shear stress (express the ratio in terms of M and T without changing the given symbols).

Accepted Answer

Introduction / Context: When a circular shaft is subjected to combined loading—bending moment (M) and torque (T)—two peak stresses occur at the outer surface: maximum bending normal stress and maximum torsional shear stress. This item tests your ability to recall the fundamental stress expressions for circular shafts and to form a clean ratio independent of section dimensions. Given Data / Assumptions: Shaft is circular and prismatic (constant diameter). Loading consists of a bending moment M and a torque T acting simultaneously. Elastic theory (Hooke's law) and small deformations. We seek (maximum bending stress) / (maximum shear stress). Concept / Approach: For a solid circular shaft of diameter d: bending stress at outer fiber is sigma_max = M*c/I and torsional shear stress at outer fiber is tau_max = T*c/J. For a circle, c = d/2, I = (pi*d^4)/64, and J = (pi*d^4)/32. Substituting and simplifying eliminates the geometric terms, leaving a simple ratio in M and T only. Step-by-Step Solution: 1) sigma_max = (M * c) / I = (M * (d/2)) / (pi*d^4/64) = 32*M / (pi*d^3).2) tau_max = (T * c) / J = (T * (d/2)) / (pi*d^4/32) = 16*T / (pi*d^3).3) Form the ratio: (sigma_max) / (tau_max) = (32*M / (pi*d^3)) / (16*T / (pi*d^3)) = 2M / T. Verification / Alternative check: The pi and d^3 terms cancel, confirming that the ratio is independent of diameter for a solid circular shaft. The same ratio 2M/T also results for a thin-walled circular tube because the linear dependence on c and section properties cancels similarly when forming the ratio. Why Other Options Are Wrong: M / T: Misses the factor 2 arising from I and J of a circular section. M / (2T): Incorrect scaling; would imply sigma_max < tau_max for equal M and T, which contradicts the derived expressions. T / (2M): Inverted ratio, not consistent with bending vs torsion formulae. Common Pitfalls: Using noncircular I and J values; the result depends on circular section properties. Forgetting that J for a solid circle is pi*d^4/32, not /64. Final Answer: 2M / T.

Question 2

In plastic analysis, identify the shape factor ( Mp / My ) for a rectangular cross-section (keeping symbols and values unchanged).

Accepted Answer

Introduction / Context: The shape factor S (also written as Mp / My) quantifies the reserve strength beyond first yield in plastic analysis. It depends only on cross-sectional shape, not on material properties, and is central to plastic hinge formation and collapse load calculations. Given Data / Assumptions: Section is rectangular, subjected to bending. My = yield moment (based on elastic stress distribution up to sigma_y). Mp = plastic moment (fully plastic stress block). Concept / Approach: For a rectangle, the elastic section modulus (Z) and plastic section modulus (Zp) have a constant ratio: S = Zp / Z = Mp / My. Classical plastic analysis yields S = 1.5 for a rectangular section because the plastic neutral axis divides the section into two equal areas in full plasticity, doubling the lever arm relative to the elastic centroidal distribution by a factor leading to 1.5 overall. Step-by-Step Solution: 1) My = sigma_y * Z (elastic).2) Mp = sigma_y * Zp (plastic).3) Shape factor S = Mp / My = Zp / Z.4) For a rectangle, Zp / Z = 1.5 → S = 1.5. Verification / Alternative check: Handbooks and steel design texts tabulate S = 1.5 for rectangles, approximately 1.1 for I-sections (compact flanges/webs vary), and 1.7 for diamonds, confirming the standard value for rectangles. Why Other Options Are Wrong: 1.4, 1.6, 1.7: Belong to other shapes or bounds; not the canonical value for a pure rectangular section. Common Pitfalls: Confusing shape factor with safety factor or with section modulus itself. Using a noncompact or perforated section where local buckling alters plastic capacity. Final Answer: 1.5.

Question 3

Select the correct set of assumptions in the elementary (Euler–Bernoulli) theory of bending of beams used in structural analysis and design.

Accepted Answer

Introduction / Context: The elementary beam theory underpins bending stress and deflection formulae used widely in civil and mechanical engineering. Knowing its assumptions is essential to understand applicability and limitations when dealing with real materials and sections. Given Data / Assumptions: We work within linear elastic theory (Hooke's law). Small deflections, plane sections remain plane (implied in classical theory). Statements list material homogeneity, isotropy, equal E in tension/compression, and layer behavior in bending. Concept / Approach: Euler–Bernoulli bending assumes a linear strain distribution across depth and proportional stress via a single modulus E. Homogeneous and isotropic material behavior ensures uniform response. Taking E the same in tension and compression yields symmetric stress blocks. The phrasing 'each layer acts independently to expand or contract' reflects that layers are free to undergo compatible axial strains dictated by curvature without through-thickness shear slip in the ideal model. Step-by-Step Solution: 1) Adopt homogeneity → uniform composition/properties.2) Adopt isotropy → properties identical in all directions.3) Use a single E for both tension and compression → linear, symmetric response.4) Permit layers to experience compatible axial strains set by curvature → linear strain profile.5) Conclude the complete set of listed assumptions holds in the elementary theory. Verification / Alternative check: Standard texts list: plane sections remain plane, material homogeneous and isotropic, stresses within elastic limit, and E same in tension/compression. The listed statements align with that framework. Why Other Options Are Wrong: Picking only A–D individually omits the completeness of the assumption set. Common Pitfalls: Confusing Euler–Bernoulli with Timoshenko theory (which includes shear deformation). Applying elementary bending to composite/anisotropic materials without modification. Final Answer: all the above.

Question 4

For a simply supported rectangular beam of span L carrying a central load W, determine the length of the elasto–plastic zone of the plastic hinge that forms at midspan (answer in terms of L).

Accepted Answer

Introduction / Context: As bending moment increases under a central point load, yielding starts at the extreme fibers near midspan and spreads until a plastic hinge forms. The length of the yielded (elasto–plastic) region depends on the shape factor of the section and the moment diagram around midspan. Given Data / Assumptions: Simply supported rectangular beam, span L. Central load W → triangular bending-moment diagram peaking at midspan. Rectangular section with shape factor S = Mp / My = 1.5. Linear elastic-perfectly plastic behavior; small deflection. Concept / Approach: Let z measure distance from midspan. For a central point load, near midspan M(z) decreases linearly with slope equal to the shear V = W/2. The elasto–plastic zone is where M(z) ≥ My. At collapse, M(0) = Mp = S*My. The half-length of the yielded zone z_y satisfies M(z_y) = My, which allows solving for z_y in terms of S and L, and then doubling to get the full zone length. Step-by-Step Solution: 1) Maximum moment at midspan: M_max = W*L/4.2) At collapse for a rectangle: M_max = Mp = S*My with S = 1.5.3) Near midspan, M(z) = M_max - (W/2)*|z| (linear drop with shear W/2).4) Define z_y by M(z_y) = My ⇒ S*My - (W/2)*z_y = My ⇒ (S - 1)*My = (W/2)*z_y.5) Also, My = M_max / S = (W*L/4)/S.6) Substitute to get z_y = 2*(S - 1)*My / W = 2*(S - 1)*(W*L/4S)/W = (S - 1)*L/(2S).7) Full elasto–plastic length = 2*z_y = (S - 1)*L/S = (1.5 - 1)*L/1.5 = L/3. Verification / Alternative check: Dimensional check shows a length proportional to L. Using S = 1.5 reproduces classic results for rectangular sections under a central point load. Why Other Options Are Wrong: L/6, L/4: Underestimate the spread implied by S = 1.5. L/2: Overestimates; would imply too large a yielded region for a rectangular section. Common Pitfalls: Confusing the plastic hinge length with the entire constant-moment region (there is none here). Forgetting the role of the shape factor in determining spread of yield. Final Answer: L / 3.

Question 5

Beams built from two or more different materials, rigidly connected to act together as a single unit under bending, are called composite beams.

Accepted Answer

Introduction / Context: Engineering practice often combines materials (e.g., steel–concrete, timber–steel) to optimize stiffness, strength, weight, or cost. When elements are connected so there is no slip at the interface and they bend together, the assembly behaves as a single member called a composite beam. Given Data / Assumptions: Two or more materials are used. Rigid connection prevents interlayer slip (full composite action). Plane sections remain plane (compatibility across materials via transformed-section method). Concept / Approach: Composite behavior requires shear transfer at the interface (e.g., studs in steel–concrete). Analysis uses modular ratio n = E1/E2 to transform areas and compute stresses/deflections as if the section were homogeneous under the assumption of perfect bond. Step-by-Step Solution: 1) Define objective: different materials to share bending based on stiffness.2) Ensure composite action: provide connectors to prevent slip.3) Use transformed section: A2' = n*A2 to compute neutral axis and section modulus.4) Evaluate stresses with sigma = M*y / I_transformed, respecting each material's allowable limits. Verification / Alternative check: Load tests show reduced deflection versus noncomposite action, confirming stiffness gain from composite behavior. Why Other Options Are Wrong: Compound beams: Term sometimes used for built-up beams of the same material; not specific to multi-material action. Indeterminate/Determinate beams: Describe static determinacy, not material composition. Common Pitfalls: Assuming full composite action without adequate connectors. Ignoring differential shrinkage/creep between materials. Final Answer: Composite beams.

Question 6

A simply supported rolled-steel joist under a uniformly distributed load shows a maximum midspan deflection of 10 mm and an end slope of 0.002 rad; determine the span length of the joist.

Accepted Answer

Introduction / Context: For a simply supported beam carrying UDL, classic formulae relate maximum deflection and end slope to load, span, and flexural rigidity. With both deflection and slope known, their ratio eliminates E and I, allowing span determination directly. Given Data / Assumptions: Beam: simply supported, UDL over entire span. Maximum deflection: delta_max = 10 mm = 0.010 m. End slope: theta_end = 0.002 rad. Linear elastic behavior; small deflection theory. Concept / Approach: For UDL w on span L: delta_max = (5*w*L^4)/(384*E*I) and theta_end = (w*L^3)/(24*E*I). Taking the ratio cancels w, E, and I, yielding a simple proportionality between delta_max, theta_end, and L. Step-by-Step Solution: 1) Write ratio: (delta_max)/(theta_end) = [(5*w*L^4)/(384*E*I)] / [(w*L^3)/(24*E*I)] = (5*L/16).2) Solve for L: L = (16/5) * (delta_max / theta_end).3) Substitute values: delta_max = 0.010 m, theta_end = 0.002 → delta_max/theta_end = 5.4) L = (16/5) * 5 = 16 m. Verification / Alternative check: Dimensional consistency: delta/theta has units of length, so L emerges in meters as expected. Cross-check with typical steel spans—16 m is a reasonable magnitude for the given small slope and 10 mm deflection under UDL. Why Other Options Are Wrong: 10, 12, 14, 18 m: Do not satisfy the ratio (5*L/16) = 0.010/0.002 = 5. Common Pitfalls: Using the point-load formula instead of UDL relations. Forgetting to convert 10 mm to meters. Final Answer: 16 m.

Question 7

Identify the name associated with the total strain energy theory of failure at the elastic limit (choose the classical attribution used in mechanics of materials).

Accepted Answer

Introduction / Context: Failure theories extend uniaxial yield data to multiaxial stress states. Each classical theory corresponds to a distinct failure criterion, and many are named for the researchers who developed them. Recognizing the correct attributions helps avoid mixing up criteria during design checks. Given Data / Assumptions: Total strain energy theory is referenced. We are at the elastic limit under multiaxial stress. Names provided correspond to well-known criteria. Concept / Approach: Total strain energy theory (sometimes attributed to Beltrami–Haigh) posits that failure occurs when the total elastic strain energy per unit volume reaches the same value as at yielding in simple tension. It differs from the distortional (shear) energy theory of Von Mises and the maximum shear stress (Tresca/Guest) theory. Step-by-Step Solution: 1) Map theories to names: Rankine → maximum principal stress; St. Venant → maximum principal strain; Tresca/Guest → maximum shear stress; Von Mises → distortion (shear) energy; Haigh → total strain energy.2) Select Haigh's theory as the correct attribution for total strain energy. Verification / Alternative check: Mechanics references list Beltrami–Haigh energy criteria under total energy, separate from Von Mises (J2) distortional energy. Why Other Options Are Wrong: Guest/Tresca: Maximum shear stress, not energy-based. St. Venant: Maximum principal strain criterion. Rankine: Maximum normal stress criterion. Von Mises: Distortion (shear) energy, not total energy. Common Pitfalls: Confusing total energy (Haigh) with distortional energy (Von Mises). Assuming all energy-based theories are identical—they are not. Final Answer: Haig's theory.

Question 8

State the name associated with the shear strain energy (distortion energy) theory for elastic failure under multiaxial stress.

Accepted Answer

Introduction / Context: Designing ductile materials under multiaxial stress often uses an energy-based failure criterion. The shear (distortion) strain energy theory—also called the Von Mises or Hencky–Von Mises criterion—is widely adopted in metal design because it correlates well with yielding in ductile metals. Given Data / Assumptions: We focus on the distortional component of strain energy as the trigger for yield. Elastic behavior up to yield point is assumed. Concept / Approach: Von Mises theory states that yielding begins when the second deviatoric stress invariant (J2) reaches the same value as in uniaxial tension at yield. Physically, only the distortion energy causes yielding; hydrostatic (volumetric) energy does not. Step-by-Step Solution: 1) Associate names with criteria: Rankine → maximum principal stress; St. Venant → maximum principal strain; Guest/Tresca → maximum shear stress; Haigh → total strain energy; Von Mises → distortion energy.2) Select Von Mises for shear strain (distortion) energy theory. Verification / Alternative check: Modern codes and FEA post-processing commonly report Von Mises equivalent stress for ductile materials, aligning with this theory. Why Other Options Are Wrong: Rankine/St. Venant: Principal stress/strain criteria. Guest or Trecas: Maximum shear stress, not energy-based. Haig: Total strain energy, not distortional-only. Common Pitfalls: Interchanging Tresca and Von Mises; Tresca gives a hexagonal yield surface, Von Mises a circular one in principal stress space. Final Answer: Von Mises.

Question 9

In a three-hinged arch analysis by funicular/graphical construction, measured distances are AD = 4 cm, DD' = 3 cm, and AD' = 5 cm. Determine the angle between the reaction directions at supports A and B (choose the nearest correct option).

Accepted Answer

Introduction / Context: Graphical methods for arches and frames use force polygons and funicular constructions. Distances measured on the load line and parallel lines relate to the magnitudes and directions of reactions. This problem uses a simple geometric relation emerging from a right-triangle measurement. Given Data / Assumptions: Measured lengths: AD = 4 cm, DD' = 3 cm, AD' = 5 cm. Standard three-hinged arch force polygon construction is assumed. The angle between reaction directions at A and B is sought. Concept / Approach: The measured segments form a triangle with sides 3, 4, and 5. Recognition of the 3–4–5 Pythagorean triplet indicates a right angle in the associated force polygon. In this classical construction, the right angle corresponds to the angle between the two reaction lines, yielding an orthogonal (90°) relationship. Step-by-Step Solution: 1) Identify triangle with sides AD = 4, DD' = 3, and AD' = 5.2) Check Pythagoras: 4^2 + 3^2 = 16 + 9 = 25 = 5^2 → right triangle.3) Map this right angle in the force polygon to the included angle between reaction vectors.4) Conclude the reaction directions are mutually perpendicular. Verification / Alternative check: Alternate geometric derivations (cosine law on the force triangle) also yield cos(theta) = 0 when the measured lengths satisfy a 3–4–5 relation. Why Other Options Are Wrong: 30°, 45°, 60°: Not supported by the 3–4–5 right-triangle result; these would require different length ratios. Common Pitfalls: Misidentifying which included angle corresponds to the reaction directions in the force polygon. Arithmetic slips when checking Pythagoras. Final Answer: 90°.

Question 10

In strength of materials, identify the incorrect proportionality in the torsional resistance of a circular shaft (choose the statement that is NOT directly proportional to torsional resistance).

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Introduction / Context: Torsion of circular shafts is a core topic in strength of materials. The torque–twist relation for a prismatic circular shaft is T / J = G * θ / L, where T is torque, J is the polar moment of inertia, G is modulus of rigidity, θ is angle of twist, and L is length. This question asks which proportionality statement is incorrect when discussing what torsional resistance (torque for a given twist) depends upon. Given Data / Assumptions: Torsion of a homogeneous, circular shaft. Elastic (Hookean) behavior, Saint-Venant torsion. Symbols: G (shear modulus), θ (angle of twist), L (length), J (polar moment of inertia). Concept / Approach: From T = (G * J / L) * θ, torsional resistance grows with G, J, and θ, and decreases with L. Importantly, the relevant geometric property is the polar moment of inertia J, not the area (second) moment of inertia I used for bending about an axis. Confusing I (bending stiffness) with J (torsional stiffness) is a common mistake. Step-by-Step Solution: Start with T = (G * J / L) * θ.Identify proportionalities: T ∝ G, T ∝ J, T ∝ θ, and T ∝ (1 / L).Note that “moment of inertia of the shaft section (I)” refers to bending, not torsion.Therefore, the statement linking torsional resistance directly to I is incorrect; it should be J. Verification / Alternative check: Check limiting cases: doubling G or J doubles T for the same θ and L; doubling L halves T; using I in place of J fails dimensional and physical consistency for torsion. Why Other Options Are Wrong: Modulus of rigidity (G): Correct proportionality—higher G increases torsional resistance. Angle of twist (θ): For a given shaft, larger θ requires proportionally larger T. Reciprocal of length (1 / L): Longer shafts twist more easily; resistance varies inversely with L. Polar moment of inertia (J): Correct geometric measure for torsion. Common Pitfalls: Mixing up I (bending) and J (torsion); assuming length does not influence torsional stiffness; forgetting that G—not E—controls torsional response. Final Answer: Moment of inertia of the shaft section (second moment, I)

Question 11

For a cantilever beam of span L, what is the ratio of free-end deflections due to the same isolated load P placed at L/3 and at 2L/3 from the fixed end?

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Introduction / Context: Deflection of beams under point loads depends on where the load acts. For a cantilever, the same load located nearer to the free end generally causes a larger free-end deflection than the same load applied closer to the fixed end. This question compares free-end deflections for a single cantilever under identical load magnitudes but at two different positions along the span. Given Data / Assumptions: Prismatic cantilever beam with span L, flexural rigidity E*I constant. Point load P applied at a distance a from the fixed end. We compare deflection at the free end for a = L/3 and a = 2L/3. Concept / Approach: For a cantilever with a point load P applied at distance a from the fixed end, the free-end deflection is given by: δ_free = P * a^2 * (3L - a) / (6 * E * I)This arises from integrating the curvature equation M / (E * I) along the span considering the load location. Since P, E, I, and L are the same for both cases, the ratio depends only on the factor a^2 * (3L - a). Step-by-Step Solution: Case 1 (a = L/3): factor = (L/3)^2 * (3L - L/3) = (L^2/9) * (8L/3) = 8L^3/27.Case 2 (a = 2L/3): factor = (2L/3)^2 * (3L - 2L/3) = (4L^2/9) * (7L/3) = 28L^3/27.Ratio δ(L/3) : δ(2L/3) = (8/27) : (28/27) = 8 : 28 = 2 : 7. Verification / Alternative check: Sanity check: load nearer the free end (2L/3) should cause greater free-end deflection; ratio < 1 confirms this expectation. Why Other Options Are Wrong: 7 : 2 and 4 : 1: Invert the true ratio; contradict physical expectation. 1 : 4 and 3 : 8: Do not match the derived factor relationship. Common Pitfalls: Using end-load formula δ = P*L^3/(3*E*I) irrespective of load position; forgetting that only the segment from the load to the free end directly influences δ_free in this relation. Final Answer: 2 : 7

Question 12

A material may be predicted to fail according to different failure theories; which condition(s) can be a valid failure criterion depending on the material and loading?

Accepted Answer

Introduction / Context: Different materials fail under different controlling stress/strain states. Classical failure theories provide criteria to assess safety: maximum principal stress (Rankine), maximum principal strain (St. Venant), maximum shear stress (Tresca), and energy-based theories like total strain energy or distortion energy (Beltrami, von Mises). This question checks awareness that multiple criteria may be relevant depending on the material behavior (brittle versus ductile) and design code. Given Data / Assumptions: General 3D stress state possible. Allowable limits σ0 (direct), shear limit, and strain/energy limits are known from material data or codes. We are not restricting ourselves to only one theory. Concept / Approach: Brittle materials (e.g., cast iron): Rankine (maximum normal stress) often governs.Ductile materials (e.g., mild steel): Shear- or energy-based criteria (Tresca or von Mises) are typically appropriate.High strain sensitivity: Maximum principal strain criterion may be used for certain materials/components.Energy viewpoint: Exceeding total allowable strain energy can indicate failure risk under complex states. Step-by-Step Solution: Map each option to a known failure theory.Recognize that design codes allow different criteria per material class and service condition.Therefore, all listed conditions can represent valid failure triggers in the proper context. Verification / Alternative check: Compare predicted safety margins using Rankine vs. Tresca vs. von Mises in a biaxial state; the controlling (lowest margin) criterion dictates design in that scenario. Why Other Options Are Wrong: Any single choice alone ignores other established, code-recognized criteria. Common Pitfalls: Applying a ductile-material criterion to brittle materials (or vice versa) without justification; assuming one universal theory fits all cases. Final Answer: All the above.

Question 13

A material is said to be perfectly elastic when which of the following statements best describes its behavior under loading and unloading?

Accepted Answer

Introduction / Context: Elasticity measures a material's ability to recover its original dimensions after the applied load is removed. Perfect elasticity is an idealization used in fundamental mechanics to simplify analysis and define the linear portion of stress–strain behavior. Given Data / Assumptions: Material is homogeneous and isotropic in the ideal model. Loading remains within the elastic limit. No time-dependent effects (creep, relaxation) are considered. Concept / Approach: For a perfectly elastic material, strain is entirely recoverable upon unloading: there is no plastic (permanent) deformation, and the unloading path coincides with the loading path. Hooke's law (stress proportional to strain) typically models this regime for small deformations. Step-by-Step Solution: Define perfect elasticity: zero permanent set after full unloading.Relate to stress–strain curve: linear reversible behavior below yield point.Choose the option that explicitly states complete recovery. Verification / Alternative check: In tensile tests, if the specimen returns exactly to its original gauge length after removal of load within the elastic range, the behavior matches perfect elasticity. Why Other Options Are Wrong: Partial recovery: Indicates viscoelastic or elastoplastic behavior. No recovery: Represents plastic or damaged state. None of these: Incorrect because a precise correct statement exists. Common Pitfalls: Confusing perfect elasticity with high elasticity; believing linearity is required at all strains (the idealization usually concerns small strains). Final Answer: It regains its original shape completely upon removal of load (no permanent set).

Question 14

Stress is a measure of internal force per unit area. Which of the following expresses valid SI-consistent units for stress?

Accepted Answer

Introduction / Context: Stress (σ or τ) quantifies internal forces distributed over a cross-sectional area. Its SI unit is the Pascal, defined as 1 N/m^2, but engineering practice commonly uses scaled units such as N/mm^2 or N/cm^2 for convenience with typical magnitudes. Given Data / Assumptions: Stress = Force / Area. Force unit: Newton (N). Area units: m^2, cm^2, mm^2 are all valid metric areas. Concept / Approach: Any combination of Newtons divided by a squared length unit yields a pressure/stress unit. Conversions are straightforward: 1 N/mm^2 = 10^6 N/m^2 (MPa), and 1 N/cm^2 = 10^4 N/m^2 (0.1 MPa). Therefore, N/m^2, N/cm^2, and N/mm^2 are all valid stress units; the Pascal is the SI base (N/m^2). Step-by-Step Solution: Write the definition: stress = force / area.Identify SI base: N/m^2 = Pa.Acknowledge engineering scales: N/mm^2 (MPa) and N/cm^2 (0.1 MPa).Therefore, all listed unit forms are acceptable expressions of stress. Verification / Alternative check: Perform conversions: 1 N/mm^2 = (1 N)/(10^-6 m^2) = 10^6 N/m^2.1 N/cm^2 = (1 N)/(10^-4 m^2) = 10^4 N/m^2.Consistency confirms validity. Why Other Options Are Wrong: None of these: Incorrect because all three are legitimate unit expressions. Common Pitfalls: Believing only N/m^2 is correct; confusing kgf/cm^2 with N/cm^2; mixing mass and force units. Final Answer: All of the above (unit-consistent pressure measures)

Question 15

Select the correct statements about the shear centre of open sections and its relation to bending loads.

Accepted Answer

Introduction / Context: The shear centre is the point in the cross-section through which a transverse load must act to produce bending without twist. Its location depends on the geometry and symmetry of the section and is critical for thin-walled open sections in structural design. Given Data / Assumptions: Open thin-walled sections (I, channel, angle) are considered. Linear elastic behavior; small deformations. Loads may be eccentric to centroid, inducing twist. Concept / Approach: For sections with two axes of symmetry (e.g., ideal I-sections), the shear flows are symmetric and the shear centre lies at the centroid. For singly-symmetric or unsymmetric open sections (channels, angles), the shear centre is offset from the centroid to balance shear flows and prevent torsion. Applying loads through the shear centre prevents coupled bending–twisting response. Step-by-Step Solution: Define shear centre: point of zero twist under transverse loading.I-sections: double symmetry → shear centre at centroid.Channels: asymmetry → shear centre shifts outside or away from centroidal location.Design implication: place loads through shear centre to avoid torsional effects. Verification / Alternative check: Standard tables show shear-centre locations: I-beams (at centroid), channels (outside the web–flange junction line), angles (outside the section). Why Other Options Are Wrong: Any single statement alone is correct but incomplete; the comprehensive answer is that all apply. Common Pitfalls: Equating shear centre with centroid for all sections; ignoring twist when loads are applied off the shear centre in open sections. Final Answer: All the above.

Question 16

For a cross-section of area A with least centroidal second moment of area I, what is the radius of gyration about that centroidal axis?

Accepted Answer

Introduction / Context: The radius of gyration characterizes how area is distributed about an axis and is widely used in column buckling and section classification. It provides a compact measure linking area and second moment of area for stability checks. Given Data / Assumptions: Cross-sectional area = A. Least centroidal second moment of area = I (about the governing axis). We seek the radius of gyration r about that centroidal axis. Concept / Approach: By definition, I = A * r^2, so r = sqrt(I / A). For buckling, the least radius of gyration controls Euler-type critical loads since it corresponds to the weakest bending axis. Step-by-Step Solution: Start from definition: I = A * r^2.Rearrange: r^2 = I / A.Take square root: r = sqrt(I / A). Verification / Alternative check: Dimensional check: [I] = L^4, [A] = L^2 → I/A = L^2 → sqrt yields L, consistent for a length measure. Why Other Options Are Wrong: r = I / A and r = A / I: Wrong dimension (length vs. length squared or inverse). r = sqrt(A / I): Inverse of the correct relationship. r = I * A: Nondimensional nonsense for r. Common Pitfalls: Confusing section modulus (I / y) with radius of gyration; forgetting that r is a geometric length derived from I and A. Final Answer: r = sqrt(I / A)

Question 17

A simply supported beam of length a carries a linearly varying load from zero at one end to w at the other end. Determine the maximum bending moment in terms of w and a (do not change symbols).

Accepted Answer

Introduction / Context: The beam is simply supported and subjected to a triangular (linearly varying) load that is zero at one end and reaches w at the other end over span a. This classic load case appears in civil/mechanical design when wind pressure, soil pressure, or linearly distributed weights act on members. The task is to compute the location and magnitude of the maximum bending moment (M_max). Given Data / Assumptions: Support conditions: simply supported at both ends. Load intensity: q(x) varies linearly from 0 to w over length a. Elastic, small-deflection theory; prismatic beam. We use x measured from the zero-load end. Concept / Approach: For a triangular load q(x) = (w/a)*x. The total load is W_total = (1/2)*w*a, acting at 2a/3 from the zero-load end. Compute reactions, write shear V(x) = R_A − ∫0^x q(s) ds, set V = 0 to find where bending moment is extremal, then integrate V to get M(x) and evaluate at the critical x. Step-by-Step Solution: 1) Resultant load: W_total = (1/2)*w*a at x̄ = 2a/3 from the zero-load end.2) Reactions: R_A = W_total*(distance to right support)/a = (1/2*w*a)*(a/3)/a = w*a/6; R_B = W_total − R_A = w*a/3.3) Shear: V(x) = R_A − ∫0^x (w/a)*s ds = w*a/6 − (w/(2a))*x^2.4) Critical point: set V = 0 → w cancels → a/6 = x^2/(2a) → x = a/√3 from the zero-load end.5) Moment from V: M(x) = ∫ V dx = (w*a/6)*x − (w/(2a))*x^3/3 = (w*a*x)/6 − (w*x^3)/(6a).6) Evaluate at x = a/√3: M_max = (w*a^2)/(6√3) − (w*a^2)/(18√3) = w*a^2/(9√3). Verification / Alternative check: Units: w has units of force/length, multiplying by a^2 yields force*length, consistent with bending moment. Also, symmetry of the triangular load about the heavier end justifies the maximum occurring closer to that end at x = a/√3 from zero-load end. Why Other Options Are Wrong: w a^2/8 and w a^2/12: Typical of uniform loads or different patterns, not this triangular case. w a^2/(6√3): This is the intermediate term before subtracting the cubic contribution; it overestimates M_max. Common Pitfalls: Placing the resultant at the centroid wrongly (a/3 instead of 2a/3) from the zero-load end. Using uniform load formulae for a non-uniform (triangular) load. Final Answer: w a^2 / (9√3).

Question 18

Three two-hinged semicircular arches A, B, and C (radii 5 m, 7.5 m, and 10 m, respectively) each carry an identical point load W at the crown. What is the ratio of the horizontal thrusts at the supports (H_A : H_B : H_C)?

Accepted Answer

Introduction / Context: Two-hinged semicircular arches of geometrically similar shape under the same type and position of load exhibit horizontal thrusts that scale with a characteristic length. With a crown load on a semicircular arch, the thrust depends on the span–rise geometry; for a semicircle, span = 2R and rise = R, giving a constant span/rise ratio but a thrust that scales linearly with size (radius) for identical load placement and type. Given Data / Assumptions: Arches are semicircular, radii: R_A = 5 m, R_B = 7.5 m, R_C = 10 m. Identical load W at the crown for each arch. Similar geometry and material behavior; standard two-hinged analysis. Concept / Approach: For similar arches under the same nondimensional loading position (crown), the horizontal thrust H scales with a characteristic dimension (here, R) when the load magnitude W is the same. Therefore, H ∝ R for the crown point load case in similar semicircular arches, and the ratio of H values follows the ratio of radii. Step-by-Step Solution: 1) Recognize geometric similarity: semicircular shape with identical load position.2) For the given loading, adopt proportionality H ∝ R (same W).3) Compute ratio using radii: 5 : 7.5 : 10 → divide by 5 → 1 : 1.5 : 2.4) Map to options: 1 : 1 1/2 : 2. Verification / Alternative check: Dimensionless analysis with forces normalized by W and lengths by R shows thrusts scale with R for similar arches under identical load types and positions, confirming the 1 : 1.5 : 2 ratio. Why Other Options Are Wrong: 2 : 1 1/2 : 1 and 1 : 1 : 2: Do not follow the linear scaling with radius. None of these: Incorrect because 1 : 1 1/2 : 2 fits exactly. Common Pitfalls: Assuming H is independent of size for similar shapes—this neglects the geometric scaling. Confusing span scaling with rise scaling; for a semicircle both scale with R. Final Answer: 1 : 1 1/2 : 2.

Question 19

For an isotropic elastic material with Young's modulus E, shear modulus N (rigidity), bulk modulus K, and Poisson's ratio ν = 1/m, identify the correct relationship(s) among E, N, K, and m.

Accepted Answer

Introduction / Context: In linear elasticity for isotropic materials, the four elastic constants (E, N, K, ν) are interrelated; any two determine the others. Many exam questions test recognition of these canonical identities, including notational variants where ν = 1/m. Given Data / Assumptions: E: Young's modulus; N: modulus of rigidity (shear modulus G); K: bulk modulus. Poisson's ratio ν = 1/m. Material is homogeneous and isotropic; small-strain elasticity applies. Concept / Approach: Standard relationships are: E = 2G(1 + ν), E = 3K(1 − 2ν), and E = 9KG/(3K + G). Substituting G = N and ν = 1/m yields the given forms. These identities are dimensionally consistent and widely tabulated. Step-by-Step Solution: 1) From shear relation: E = 2N(1 + ν) → using ν = 1/m → E = 2N(1 + 1/m).2) From volumetric relation: E = 3K(1 − 2ν) → E = 3K(1 − 2/m).3) Combined relation: E = 9 K N / (3K + N) (independent of ν explicitly). Verification / Alternative check: Pick E, ν and compute N and K using the first two formulas; then back-substitute into the third to confirm equality. All three are consistent for isotropic materials (ν between −1 and 0.5 for stability). Why Other Options Are Wrong: Each of A–C is correct; omitting any would be incomplete. Common Pitfalls: Confusing m with ν (remember ν = 1/m here). Using E = 3K(1 − ν) (incorrect; the factor is 1 − 2ν). Final Answer: All of the above.

Question 20

A steel bar (area A_s, modulus E_s, coefficient α_s) and a copper bar (area A_c, modulus E_c, coefficient α_c) of equal initial length L are rigidly connected in parallel and subjected to a uniform temperature rise of t°. Find the common extension δ…

Accepted Answer

Introduction / Context: Compound (composite) bars of dissimilar materials bonded in parallel develop internal self-equilibrating forces under temperature change because each material tends to expand by a different free amount. Compatibility enforces a common extension, while equilibrium requires net axial force to vanish when there is no external axial load. Given Data / Assumptions: Bars in parallel: same initial length L, rigidly connected at ends. Material data: A_s, E_s, α_s for steel; A_c, E_c, α_c for copper. Uniform temperature rise t°, no external axial load. Elastic, small-strain behavior. Concept / Approach: Let ε be the common strain, δ = ε*L. Free thermal strains would be α_s t and α_c t. Actual stresses: σ_s = E_s(ε − α_s t), σ_c = E_c(ε − α_c t). Force equilibrium (no external load) gives σ_s A_s + σ_c A_c = 0. Solve for ε, then δ. Step-by-Step Solution: 1) Write stresses: σ_s = E_s(ε − α_s t), σ_c = E_c(ε − α_c t).2) Equilibrium: E_s A_s (ε − α_s t) + E_c A_c (ε − α_c t) = 0.3) Factor ε: ε (E_s A_s + E_c A_c) = t (E_s A_s α_s + E_c A_c α_c).4) Solve: ε = t (E_s A_s α_s + E_c A_c α_c) / (E_s A_s + E_c A_c).5) Extension: δ = ε * L = L t * (α_s E_s A_s + α_c E_c A_c) / (E_s A_s + E_c A_c). Verification / Alternative check: Special case A_s E_s ≫ A_c E_c → δ ≈ L t α_s (steel dominates); conversely if A_c E_c ≫ A_s E_s → δ ≈ L t α_c (copper dominates). This limiting behavior validates the formula. Why Other Options Are Wrong: Arithmetic averages (A, D) ignore stiffness weighting. Option B ignores areas; stiffness depends on E*A, not just E. Common Pitfalls: Forgetting that the net axial force must be zero (no external axial load). Using α-weighted averages without E*A weighting. Final Answer: δ = L t * (α_s E_s A_s + α_c E_c A_c) / (E_s A_s + E_c A_c).

Theory of Structures Questions