Question 1

Point of contraflexure (inflection point) occurs in which beams under usual loading?

Accepted Answer

Introduction / Context: The point of contraflexure (inflection) is where the bending moment changes sign, and the curvature of the beam changes from sagging to hogging or vice versa. Recognizing beam types that exhibit this is essential for reinforcement and section design. Given Data / Assumptions: Usual static load cases (point loads, UDL, UVL). Beam behavior is linear elastic. We consider typical “textbook” loadings. Concept / Approach: Simply supported beams under common symmetrical loads have positive bending moment within span and zero at supports—no sign change within the span. Cantilevers typically have one sign of bending throughout. Overhanging beams can have positive midspan and negative near supports, creating a sign change. Fixed beams often have negative moments at supports and positive toward midspan, producing two inflection points. Step-by-Step Solution: Identify beam type and support moments.Plot qualitative BMD to observe sign regions.Locate where M(x) = 0 within the span; that is the contraflexure point. Verification / Alternative check: For a fixed–fixed beam with UDL, analytical M(x) shows negative at ends, positive at midspan, guaranteeing two zeros within the span. Why Other Options Are Wrong: Cantilevers and simply supported beams (typical loadings) lack an internal sign change. “No real beam” is false; “all beams” is overgeneralized. Common Pitfalls: Confusing zero moment at supports of simply supported beams with an internal inflection point; forgetting that overhangs and fixity modify end moments. Final Answer: Overhanging and fixed beams (where bending moment changes sign)

Question 2

Deflection of a cantilever with uniformly distributed load: Maximum deflection in terms of W = w * l

Accepted Answer

Introduction / Context: Deflection formulas are central to serviceability checks. For a cantilever of length l with a uniformly distributed load w (per unit length), maximum deflection occurs at the free end. The problem defines W = w * l for convenience, and asks for the deflection in that form. Given Data / Assumptions: Cantilever length = l. UDL intensity = w, total load W = w * l. Young’s modulus = E, second moment of area = I. Small deflection theory, prismatic beam. Concept / Approach: Standard result for a cantilever with UDL: y_max = w * l^4 / (8 * E * I). Substituting W = w * l gives y_max = (W * l^3) / (8 * E * I). This is obtained by integrating beam differential equations or using tabulated results. Step-by-Step Solution: Start with y_max = w * l^4 / (8 * E * I).Given W = w * l → w = W / l.Substitute: y_max = (W / l) * l^4 / (8 * E * I) = W * l^3 / (8 * E * I).Hence, the correct expression is W * l^3 / (8 * E * I). Verification / Alternative check: Dimensional check: numerator has force * length^3; denominator E * I has force * length^2; result is length, as required. Why Other Options Are Wrong: W * l^3 / (3 * E * I) is the cantilever tip load case factor; the 48 denominators belong to simply supported beam formulae; W * l^2 / (8 * E * I) has wrong power of l. Common Pitfalls: Mixing up the 8 vs 3 vs 48 factors; forgetting to substitute W = w * l correctly; using simply supported results for cantilever problems. Final Answer: W * l^3 / (8 * E * I)

Question 3

Strength of materials – Polar modulus for a solid circular shaft
For a solid shaft of diameter D subjected to torsion, what is the polar section modulus (also called polar modulus) expressed in terms of D?

Accepted Answer

Introduction / Context: The polar section modulus is a geometry property used in torsion of circular shafts. It links the applied torque to the maximum shear stress at the outer surface and is essential for sizing shafts to carry torque safely. Given Data / Assumptions: Solid circular shaft with diameter D. Material is homogeneous and isotropic; elastic torsion theory applies. Saint-Venant torsion (no warping restraint). Concept / Approach: The torsion relation is T / J = τ / R = G * θ / L, where J is the polar moment of inertia and R is outer radius. The polar section modulus Zp is defined as J / R. For a solid circle, J depends on D^4, while Zp scales with D^3. Step-by-Step Solution: For a solid circular section: J = (π/32) * D^4.Outer radius: R = D / 2.Polar section modulus: Zp = J / R.Zp = ((π/32) * D^4) / (D/2) = (π/16) * D^3.Therefore, the correct expression is (π/16) * D^3. Verification / Alternative check: Dimensional check: J has units of length^4; dividing by R (length) gives length^3, which matches Zp. This is consistent with torsion formulas used in machine design textbooks. Why Other Options Are Wrong: (π/32) * D^3 and (π/64) * D^3 underestimate Zp by factors of 2 and 4 respectively; (π/8) * D^3 overestimates by 2. The option with D^4 is dimensionally incorrect for a modulus. Common Pitfalls: Confusing J (polar moment) with Zp (section modulus) or forgetting to divide by R. Another common mistake is using area moment of inertia I instead of polar J for torsion. Final Answer: (π/16) * D^3

Question 4

Strength of materials – Polar moment of inertia of a hollow circular shaft
For a hollow shaft with outer diameter D and inner diameter d, what is the polar moment of inertia J?

Accepted Answer

Introduction / Context: The polar moment of inertia J quantifies a circular section’s resistance to torsion. For hollow shafts, material is distributed away from the center to increase torsional stiffness for a given weight. Given Data / Assumptions: Hollow circular shaft, outside diameter D, inside diameter d. Linear elastic torsion; uniform material. Concept / Approach: For circular sections, J equals the sum of second moments about two orthogonal axes through the centroid. Using the standard result for rings, J is the difference between the solid outer and inner polar moments. Step-by-Step Solution: Solid circle polar moment: J_solid = (π/32) * D^4.Inner void polar moment: J_void = (π/32) * d^4.Hollow section: J = J_solid − J_void = (π/32) * (D^4 − d^4). Verification / Alternative check: Boundary cases: if d = 0, J reduces to (π/32) * D^4 (solid case). If d → D, J → 0 as expected. Why Other Options Are Wrong: (π/64) * (D^4 − d^4) is off by a factor of 1/2. Expressions with D^2 or D^3 are dimensionally incorrect. Using a plus sign increases J erroneously. Common Pitfalls: Mixing polar moment J with area moment I, and forgetting that the hollow result is the outer minus inner solid contribution. Final Answer: J = (π/32) * (D^4 − d^4)

Question 5

Materials property – Poisson’s ratio of cast iron
Within what approximate range does Poisson’s ratio for cast iron typically lie?

Accepted Answer

Introduction / Context: Poisson’s ratio ν relates lateral strain to axial strain under uniaxial loading. It is a key elastic constant that influences deformation behavior and is used in stress–strain relations and deflection analysis. Given Data / Assumptions: Material: cast iron (commonly gray cast iron). Elastic, small-strain regime. Concept / Approach: Typical ν for metals ranges from about 0.25 to 0.35 for ductile steels and aluminums. Cast irons are more brittle and display lower Poisson’s ratio, commonly near 0.22–0.25 depending on grade and microstructure. Step-by-Step Solution: Identify standard handbook range for cast iron: ν ≈ 0.21 to 0.26.Compare with options and select the correct interval. Verification / Alternative check: Cross-check with engineering handbooks: gray cast iron often cited around ν = 0.23; nodular/ductile iron can be slightly higher but still within or near the stated band. Why Other Options Are Wrong: 0.33–0.37 and 0.31–0.34 are more typical for steels/aluminum alloys. 0.32–0.42 is too high for cast iron. 0.12–0.18 is too low for common engineering cast irons. Common Pitfalls: Using a generic ν = 0.3 for all metals leads to noticeable errors in deflection predictions for cast iron components. Final Answer: 0.21 to 0.26

Question 6

Riveted joints – Material choice for rivets
Are rivets generally manufactured from ductile materials to permit upsetting and reliable formation of the head during riveting?

Accepted Answer

Introduction / Context: Riveted joints rely on forming heads by deforming the rivet shank. The material of the rivet must withstand plastic deformation without cracking, ensuring a tight joint and adequate shear/tension capacity. Given Data / Assumptions: Common practice in structural and boiler riveting. Hot and cold riveting both used depending on application. Concept / Approach: Ductile materials such as mild steel, wrought iron, aluminum, or copper are preferred because they can be upset to form the head and fill the hole, improving load transfer and minimizing stress concentrations. Step-by-Step Explanation: During riveting, the shank is inserted and a head is formed by plastic deformation.Ductility prevents brittle fracture during upsetting and service vibrations.Common specifications call for rivet materials with adequate elongation and toughness. Verification / Alternative check: Standards and handbooks list typical rivet materials (e.g., low-carbon steels and nonferrous ductile alloys), confirming the practice. Why Other Options Are Wrong: Limiting ductility to only hot or only cold riveting is incorrect; both benefit from ductile rivets. Plate material choice does not negate the need for rivet ductility. Common Pitfalls: Using overly hard or brittle fasteners can lead to cracking during heading or service, compromising joint integrity. Final Answer: Yes

Question 7

Uniaxial tension – Resultant (traction) on an inclined plane
A body is subjected to a direct tensile stress σ. On a plane inclined at an angle θ to the normal cross-section, what is the magnitude of the resultant stress (traction) on that plane?

Accepted Answer

Introduction / Context: In strength of materials, the traction vector on an inclined plane has a normal component (normal stress) and a tangential component (shear stress). For uniaxial tension, their combination yields a simple expression for the resultant magnitude. Given Data / Assumptions: Uniform direct tensile stress σ along one axis. A plane is taken at angle θ to the normal cross-section (i.e., to the plane perpendicular to the load). Continuum mechanics relations for plane stress apply. Concept / Approach: Resolve traction into normal and shear components using standard transformation equations: σ_n = σ cos^2 θ and τ = σ sin θ cos θ. The resultant traction magnitude is the vector sum of these components. Step-by-Step Solution: Normal stress on plane: σ_n = σ cos^2 θ.Shear stress on plane: τ = σ sin θ * cos θ = 0.5 * σ * sin 2θ.Resultant: t = sqrt(σ_n^2 + τ^2).t = sqrt(σ^2 cos^4 θ + σ^2 sin^2 θ cos^2 θ) = σ * sqrt(cos^4 θ + sin^2 θ cos^2 θ).Factor cos^2 θ: t = σ * cos θ * sqrt(cos^2 θ + sin^2 θ) = σ * cos θ. Verification / Alternative check: At θ = 0°, the plane is normal to the load, so t = σ, and σ cos 0° = σ. At θ = 90°, the plane is parallel to load, so t = 0, matching σ cos 90° = 0. Why Other Options Are Wrong: σ sin θ understates traction and is zero at θ = 0°, which is incorrect. σ sin 2θ and σ cos 2θ relate to transformed shear/normal components, not the resultant magnitude. Common Pitfalls: Confusing θ as the angle with the axis instead of the normal section, or mixing up individual components with the resultant. Final Answer: σ cos θ

Question 8

Bending theory – Neutral axis in a rectangular beam under transverse load
When a rectangular beam is transversely loaded, is the bending (normal) stress zero at the neutral axis?

Accepted Answer

Introduction / Context: In pure bending theory for prismatic beams, the neutral axis (NA) passes through the centroid and separates tension and compression zones. Understanding stress distribution relative to the NA is foundational in beam design. Given Data / Assumptions: Rectangular, homogeneous, prismatic beam. Elastic, small deflections; Bernoulli–Euler assumptions. Transverse loading causing bending and shear. Concept / Approach: The bending (normal) stress varies linearly with distance y from the NA: σ = (M/I) * y. At y = 0 (on the NA), σ = 0. While shear stress may be maximum near the NA, the normal bending stress is zero exactly at the NA. Step-by-Step Explanation: Identify NA at the centroid for a symmetric rectangular section.Bending stress distribution is linear: tension on one side, compression on the other, zero at NA.Shear stress distribution is parabolic and peaks at the NA but is a different stress component. Verification / Alternative check: Plot σ(y) = (M/I) * y; at y = 0, σ = 0. Superimpose shear distribution to see that shear does not alter the zero bending stress at NA. Why Other Options Are Wrong: Stating false unless shear is zero confuses normal and shear stresses. The condition holds for general transverse loading within elastic theory, not just for cantilevers or “symmetric loading only”. Common Pitfalls: Mixing the concepts: maximum shear occurs at the NA, but bending stress there is still zero. Students often conflate these two stress types. Final Answer: True

Question 9

Buckling of compression members – Direction of buckling tendency
Compression members (columns/struts) tend to buckle preferentially in which direction?

Accepted Answer

Introduction / Context: Column stability depends on flexural stiffness EI about potential buckling axes. The effective slenderness ratio determines the critical Euler load; weaker bending axes buckle first. Given Data / Assumptions: Prismatic, initially straight column under axial compression. Elastic buckling (Euler regime) with appropriate end conditions. Concept / Approach: The radius of gyration r = sqrt(I/A). For a given axial load and member, the slenderness ratio λ = L_e / r governs buckling. Smaller r means larger λ and lower critical load, so buckling occurs about the axis with least r (and least I). Step-by-Step Explanation: Compute r_x and r_y for principal axes.Compare slenderness ratios λ_x and λ_y; the larger one indicates the weaker direction.Expect buckling about the axis with least r (i.e., least flexural stiffness). Verification / Alternative check: Euler load P_cr = π^2 * E * I / (L_e^2). For the axis with smaller I, P_cr is smaller; initial imperfections trigger buckling there first. Why Other Options Are Wrong: Buckling is not along the load axis (axial shortening). “Any direction” ignores anisotropy of section stiffness. Minimum area alone is not decisive unless it correlates with least I and least r. Greatest I would be the strongest direction, not the weakest. Common Pitfalls: Ignoring end-restraint effects on L_e; overlooking weak-axis bracing requirements in design. Final Answer: In the direction of least radius of gyration

Question 10

Euler’s column theory – Comparing end conditions
According to Euler’s theory, how does the critical (crippling) load of a column of length l fixed at both ends compare with that of an identical column hinged at both ends?

Accepted Answer

Introduction / Context: Euler’s elastic buckling load depends on the effective length L_e determined by end restraints. Stronger end fixity reduces L_e and increases the critical load dramatically. Given Data / Assumptions: Same material (E), moment of inertia (I), and clear length l. One column is fixed–fixed; the other is hinged–hinged. Elastic Euler buckling applies (long, slender columns). Concept / Approach: Euler formula: P_cr = π^2 * E * I / (L_e^2). For hinged–hinged, L_e = l. For fixed–fixed, L_e = l / 2. The ratio of critical loads follows from the square of the effective length ratio. Step-by-Step Solution: Hinged–hinged: P_cr,HH = π^2 * E * I / l^2.Fixed–fixed: P_cr,FF = π^2 * E * I / (l/2)^2 = π^2 * E * I / (l^2/4) = 4 * π^2 * E * I / l^2.Ratio: P_cr,FF / P_cr,HH = 4. Verification / Alternative check: Effective length factors K: K_HH = 1.0, K_FF = 0.5. Since P_cr ∝ 1/K^2, the factor is 1/(0.5^2) = 4. Why Other Options Are Wrong: Two times or eight times contradict the 1/L_e^2 relationship. “Equal” ignores end restraint effects. Common Pitfalls: Confusing clear length with effective length; misapplying K factors for mixed end conditions (e.g., fixed–free or fixed–hinged). Final Answer: Four times

Question 11

Strength of materials – Buckling: The slenderness ratio of a column is defined as the ratio of the effective length of the column to its least radius of gyration (l_e / k_min). Identify the correct definition.

Accepted Answer

Introduction / Context: The slenderness ratio is a fundamental measure in column buckling used to judge whether a column will fail by crushing (short column) or by elastic buckling (long/medium column). A correct understanding of this ratio helps engineers select suitable design formulas such as Euler's or Rankine–Gordon. Given Data / Assumptions: The column has an effective length l_e determined by end conditions. The least radius of gyration is k_min = sqrt(I_min / A), where I_min is the minimum second moment of area and A is the cross-sectional area. Material is homogeneous and straight; initial imperfections are neglected for the definition. Concept / Approach: The slenderness ratio, commonly denoted by λ, compares geometric slenderness to sectional stiffness: λ = l_e / k_min. A larger λ indicates a more slender member and a higher propensity for elastic buckling at lower loads. Step-by-Step Solution: Identify effective length from end conditions (e.g., pinned–pinned l_e = l; fixed–free l_e = 2l).Compute k_min from the axis with the smallest I: k_min = sqrt(I_min / A).Form the ratio: slenderness ratio λ = l_e / k_min.Use λ to choose design criteria: low λ → crushing/yield; high λ → Euler buckling. Verification / Alternative check: Dimensional analysis: l_e has units of length and k_min has units of length, so λ is dimensionless, validating the ratio. Why Other Options Are Wrong: Area to k_min: incorrect dimensional form and not used in buckling criteria. k_min to area: also dimensionally inconsistent with a pure ratio used in codes. k_min to length: reciprocal of the correct ratio and not standard. Common Pitfalls: Using the radius of gyration about the wrong axis; always use the least (weak) axis value to capture the critical buckling mode. Final Answer: Length of column to least radius of gyration

Question 12

Euler's column theory: For columns of equal material and length l, the crippling load for a column with one end fixed and the other end free is __________ compared to a similar column hinged at both ends. Choose the correct comparison.

Accepted Answer

Introduction / Context: Euler's buckling theory relates the critical (crippling) load to the boundary conditions via the effective length. Knowing how end restraint changes buckling capacity is essential for safe column design. Given Data / Assumptions: Same column length l, material (E), and section (I). Case 1: One end fixed and the other free. Case 2: Hinged (pinned) at both ends. Elastic buckling (Euler region). Concept / Approach: Euler load P_cr = (pi^2 * E * I) / (l_e^2), where l_e is the effective length. For pinned–pinned: l_e = l. For fixed–free: l_e = 2l. Larger l_e reduces P_cr quadratically. Step-by-Step Solution: Pinned–pinned: P_cr,pp = (pi^2 * E * I) / (l^2).Fixed–free: P_cr,ff = (pi^2 * E * I) / (2l)^2 = (pi^2 * E * I) / (4l^2).Ratio: P_cr,ff / P_cr,pp = 1/4 → fixed–free carries one-quarter of the pinned–pinned capacity.Therefore, P_cr for fixed–free is less than that for pinned–pinned. Verification / Alternative check: Effective length factors K: K = 1.0 (pinned–pinned), K = 2.0 (fixed–free). Since P_cr ∝ 1/K^2, doubling K quarters the load, consistent with the result. Why Other Options Are Wrong: Equal to/More than: contradict the effective-length dependence where a cantilevered column is the weakest restraint case. Common Pitfalls: Confusing fixed–free with fixed–fixed (the latter has higher capacity, K = 0.5). Final Answer: Less than

Question 13

Deflection of beams: For a cantilever beam of length l carrying a uniformly distributed load w per unit length, the location of maximum deflection is where?

Accepted Answer

Introduction / Context: Determining where the maximum deflection occurs helps in serviceability checks and control of deflections in structures. For common loadings on standard supports, locations are well known and can be derived from beam theory. Given Data / Assumptions: Cantilever of length l, fixed at one end and free at the other. Uniformly distributed load w over the entire span. Linear elastic behavior, small deflections. Concept / Approach: Deflection curves y(x) are obtained by integrating the moment–curvature relationship: EI * d^2y/dx^2 = M(x). For a UDL on a cantilever, the slope and deflection increase monotonically toward the free end, so maximum deflection occurs at the free tip. Step-by-Step Solution: Bending moment at distance x from the free end: M(x) = wx^2/2 (taking origin at free end) or equivalent form from fixed end.Integrate twice and apply boundary conditions (y = 0 and dy/dx = 0 at the fixed end).Resulting deflection function shows y increases with x toward the free end.Hence, the maximum deflection is at the free end. Verification / Alternative check: Closed-form maximum deflection magnitude: y_max = wl^4 / (8E*I) for UDL on a cantilever, which occurs at the free end. Why Other Options Are Wrong: Fixed end or midspan: deflection there is smaller; at the fixed end it is zero by boundary condition. l/3 from fixed end: not the correct extremum for UDL on cantilever. Common Pitfalls: Confusing locations for different load cases (e.g., concentrated end load versus UDL); mixing up simply supported and cantilever cases. Final Answer: At the free end

Question 14

Riveted joints: Rivets are generally specified in practice by which primary dimension?

Accepted Answer

Introduction / Context: Standard practice for specifying fasteners ensures correct selection and interchangeability. For rivets, the shank diameter governs hole size, pitch, and joint efficiency and is therefore the primary specification. Given Data / Assumptions: Conventional hot or cold-driven rivets used in plate joints. Holes are drilled/punched to suit the rivet shank diameter plus clearance. Other dimensions (length, head size) follow standards once the diameter is set. Concept / Approach: The rivet's shank diameter determines net-section area of plates and the shear capacity of the rivet itself. Designers first choose diameter to meet shear and bearing requirements; length is then chosen to accommodate grip length and head formation. Step-by-Step Solution: Select rivet diameter d from strength checks (shear of rivet, bearing on plate, net tension of plate).Choose hole diameter typically d_h ≈ d + clearance (manufacturing standard).Select rivet length to suit total grip plus allowance for forming head.Finalize head type per standard (e.g., snap, pan, counter-sunk). Verification / Alternative check: Joint design tables list capacities primarily by shank diameter, confirming it is the controlling specification in catalogs and standards. Why Other Options Are Wrong: Plate thickness: used to pick length, not the primary rivet designation. Overall length and head diameter: derived after diameter selection. Common Pitfalls: Specifying length without confirming grip range leads to improper set and reduced strength. Final Answer: Shank diameter

Question 15

Types of riveted joints: When one plate overlaps the other and the two plates are riveted together with two rows of rivets, the joint is termed as what?

Accepted Answer

Introduction / Context: Riveted joints are classified by plate arrangement (lap or butt) and by the number of rivet rows and covers. Differentiating these types is essential for computing efficiency and detailing pitch and edge distances. Given Data / Assumptions: Two plates overlap directly (lap arrangement). There are two rows of rivets across the joint. No cover plates are mentioned, so it is not a butt joint. Concept / Approach: A lap joint uses overlapping plates connected by rivets. When there are two rows of rivets, it is designated as a double riveted lap joint. Butt joints, by contrast, use cover plates (single or double) and do not rely on plate overlap. Step-by-Step Solution: Identify geometry: plates overlapping → lap joint.Count rows: two rows → double riveted.Therefore, the correct name is double riveted lap joint.Exclude butt joint options, which require cover plates. Verification / Alternative check: Textbook nomenclature lists lap joints (single- or double-riveted) versus butt joints (single- or double-strap/cover). Why Other Options Are Wrong: Single riveted lap joint: only one row of rivets. Double riveted butt joints: require one or two cover plates with aligned or staggered rows. Common Pitfalls: Confusing number of rows with number of covers; “double” in lap joints refers to rows, not straps. Final Answer: Double riveted lap joint

Question 16

Beam of uniform strength: A beam may be designed to have uniform strength (constant maximum fibre stress along its length) by which of the following approaches?

Accepted Answer

Introduction / Context: A beam of uniform strength is shaped so that the allowable bending stress is reached everywhere along the span, reducing weight compared to a prismatic beam. This concept is used in cranes, leaf springs, and tapered structural members. Given Data / Assumptions: Bending stress formula: sigma = M / Z where Z is section modulus. Allowable stress sigma_allow is constant along the beam. Designer may vary cross-sectional dimensions. Concept / Approach: For sigma to be constant while M(x) varies with x, the section modulus must vary proportionally: Z(x) = M(x) / sigma_allow. Since Z depends on both width and depth, either parameter (or both) may be varied to match the bending moment diagram. Step-by-Step Solution: Determine M(x) for the applied loading.Set Z(x) = M(x) / sigma_allow.Relate Z(x) to geometry (e.g., rectangular: Z = b*d^2/6).Choose a practical variation: vary d, vary b, or vary both to achieve the required Z(x). Verification / Alternative check: For a cantilever with end load (M ∝ x), a triangular depth variation provides near-constant stress; other profiles can achieve the same with width variation. Why Other Options Are Wrong: Options a, b, and c each can be correct individually; limiting the designer to only one would be unnecessarily restrictive. Common Pitfalls: Ignoring local stability (buckling of thin webs/flanges) when aggressively tapering members. Final Answer: Any one of the above, depending on design convenience

Question 17

Axial strain calculation: A bar of length L metres extends by l millimetres under a tensile force P. What is the engineering strain produced in the bar?

Accepted Answer

Introduction / Context: Strain is the non-dimensional measure of deformation and is defined as change in length divided by original length. Careful unit handling is essential when lengths are given in different units. Given Data / Assumptions: Original length = L metres. Extension = l millimetres. Small, uniform axial strain (linear elastic range). Concept / Approach: Engineering strain epsilon = delta_L / L_0. Convert all lengths to consistent units before forming the ratio to ensure a dimensionless quantity. Step-by-Step Solution: Convert extension to metres: delta_L = l mm = (l / 1000) m.Original length L_0 = L m.Compute strain: epsilon = (l / 1000) / L = 0.001 * (l / L).Hence epsilon = 0.001 l / L (dimensionless). Verification / Alternative check: Unit check: numerator and denominator both in metres; the ratio is unitless as required for strain. Why Other Options Are Wrong: l / L: ignores unit conversion; would be 1000× too large. 0.1 l / L and 0.01 l / L: incorrect conversion factors. Common Pitfalls: Forgetting to convert millimetres to metres or mixing up engineering strain with percentage strain (%strain = 100 * epsilon). Final Answer: 0.001 l / L

Question 18

Fixed beam with central point load: For a prismatic beam fixed at both ends and loaded by a central point load, where does the maximum vertical deflection occur?

Accepted Answer

Introduction / Context: Serviceability checks require the location and magnitude of maximum deflection. For statically indeterminate beams such as fixed–fixed beams, symmetry provides powerful insight without full analysis. Given Data / Assumptions: Beam is prismatic and fixed at both ends. Single concentrated load acts at midspan. Linear elastic behavior; small deflection assumptions hold. Concept / Approach: Due to geometric and loading symmetry, the slope at midspan is zero and the deflected shape is symmetric about the centreline. The maximum deflection occurs where the slope changes sign, i.e., at the centre under a central load. Step-by-Step Solution: Recognize symmetry: reactions and end moments are equal in magnitude at supports.The elastic curve is symmetric; slope at midspan is zero.The extremum of deflection occurs at zero slope → at the centre.Hence, maximum deflection is at midspan. Verification / Alternative check: Closed-form solution gives y_max at midspan with magnitude y_max = (Pl^3) / (192E*I) for fixed–fixed central load, confirming the location. Why Other Options Are Wrong: At the fixed ends: deflection is zero by boundary condition. At l/3: not the extremum for this loading. None of these: incorrect because the centre is correct. Common Pitfalls: Confusing fixed–fixed with simply supported beams where maximum deflection is also at midspan, but magnitudes differ. Final Answer: At the centre of the beam

Question 19

Plane stress transformation: A body has normal stresses of 1200 MPa (tension) and 600 MPa (tension) on two perpendicular planes, with a shear stress of 400 MPa on the same planes. What is the minimum (least) principal normal stress?

Accepted Answer

Introduction / Context: Principal stresses and maximum shear stresses in plane stress are central to failure theories and design. Given two perpendicular normal stresses and an in-plane shear, we can compute principal values using transformation equations or Mohr’s circle. Given Data / Assumptions: Sigma_x = 1200 MPa (tension). Sigma_y = 600 MPa (tension). Tau_xy = 400 MPa (shear on the same planes). Plane stress; sign convention: tensile positive, shear as given. Concept / Approach: Principal stresses are given by sigma_(1,2) = sigma_avg ± R, where sigma_avg = (sigma_x + sigma_y)/2 and R = sqrt( ((sigma_x − sigma_y)/2)^2 + tau_xy^2 ). The smaller of these is the minimum normal stress. Step-by-Step Solution: Compute average: sigma_avg = (1200 + 600)/2 = 900 MPa.Compute half-difference: (sigma_x − sigma_y)/2 = (600)/2 = 300 MPa.Radius: R = sqrt(300^2 + 400^2) = sqrt(90000 + 160000) = sqrt(250000) = 500 MPa.Principal stresses: sigma_1 = 900 + 500 = 1400 MPa; sigma_2 = 900 − 500 = 400 MPa. Verification / Alternative check: Mohr’s circle center at 900 MPa, radius 500 MPa; intersections at 1400 and 400 MPa confirm the computed values. Why Other Options Are Wrong: 500 MPa and 900 MPa: intermediate values (average or misused components) not principal results. 1400 MPa: this is the maximum, not the minimum principal stress. Common Pitfalls: Sign errors for shear; forgetting that both principal stresses can be tensile when both sigma_x and sigma_y are tensile. Final Answer: 400 MPa

Question 20

Thermal stress in a restrained bar: A circular bar tapers uniformly from diameter d1 at one end to diameter d2 at the other. If its temperature is increased uniformly by t (°C) and the bar is fully restrained against expansion, what is the induced t…

Accepted Answer

Introduction / Context: Thermal stresses arise when thermal expansion or contraction is restrained. Designers must quantify these stresses to prevent cracking or yielding in components subject to temperature changes, such as turbine shafts or bridge members. Given Data / Assumptions: The bar is circular and tapers linearly from diameter d1 to d2. Uniform temperature rise t is applied to the entire bar. Both ends are fully restrained so that net axial extension is zero. Material is homogeneous and isotropic with modulus E and expansion coefficient α. Concept / Approach: If a uniform axial strain α * t would occur freely, but the bar is fully restrained, an equal and opposite mechanical strain must be induced to enforce zero net extension. In uniaxial elasticity, mechanical strain ε_mech = sigma / E. Setting total strain to zero gives sigma / E + α * t = 0, hence sigma = −E * α * t. The magnitude is E * α * t and it is uniform along the bar under full restraint, independent of area variation. Step-by-Step Solution: Free thermal strain (unrestrained): ε_th = α * t.Restraint imposes ε_total = ε_mech + ε_th = 0.Mechanical strain: ε_mech = sigma / E → sigma / E + α * t = 0.Solve for stress: sigma = −E * α * t (compressive for heating); magnitude E * α * t. Verification / Alternative check: A tapered bar has varying stiffness EA(x), but for uniform temperature and full restraint, compatibility requires uniform axial strain (zero total), yielding a uniform stress solution; equilibrium is satisfied without axial load transfer variation. Why Other Options Are Wrong: Area-dependent formulas (options b and d) incorrectly imply geometry dependence for uniform thermal stress under full restraint. sigma = 0: only true if the bar is free to expand (unrestrained), which is not the case here. Common Pitfalls: Confusing partially restrained conditions (which would require integrating EA(x)) with fully restrained ends; forgetting the sign convention (compressive stress under heating). Final Answer: sigma = E * α * t

Strength of Materials Questions