Question 1

Euler buckling — permissible crippling load for a circular column (hinged ends) A prismatic column has a uniform circular cross-section of diameter d = 7.5 cm and effective length L = 5 m (both ends hinged). The material has modulus of elasticity E …

Accepted Answer

Introduction / Context: Long, slender columns fail by elastic instability (buckling). For ideal columns with hinged ends and within the Euler regime, the critical load depends on the modulus of elasticity, the least second moment of area, and the effective length. Correct unit handling and section properties are essential to avoid large numerical errors. Given Data / Assumptions: Diameter d = 7.5 cm; circular cross-section. Effective length L = 5 m = 500 cm; hinged–hinged ends. Elastic modulus E = 2100 tonnes/cm^2 (use consistent tonne–cm units). Euler formula applies (column is slender; no inelastic effects). Concept / Approach: Euler critical load for pinned–pinned column: P_cr = (pi^2 * E * I) / (L_eff^2). For a circle, I = (pi * d^4) / 64. Use centimetre units throughout to remain consistent with E and express the final load in tonnes. Step-by-Step Solution: Compute second moment: I = (pi * d^4) / 64 = pi * (7.5^4) / 64 ≈ 155.32 cm^4.Effective length L_eff = 500 cm (hinged–hinged).Apply Euler: P_cr = (pi^2 * 2100 * 155.32) / (500^2) tonnes ≈ 12.88 tonnes.Round suitably: 12.88 tonnes (matches option). Verification / Alternative check: Slenderness ratio λ = L/r where r = sqrt(I/A). For d = 7.5 cm, r ≈ 7.5/4 = 1.875 cm; λ ≈ 500 / 1.875 ≈ 267 > 100, confirming Euler applicability for a long column. Why Other Options Are Wrong: 128.8 t and 288.0 t: an order-of-magnitude too high, typical of unit mistakes (metres vs. centimetres) or squaring length incorrectly. 1.288 t: too small, from dividing by an extra power of 10. None of these: unnecessary since 12.88 t is correct. Common Pitfalls: Mixing SI and gravitational units; forgetting to square the effective length; using wrong end conditions (which change L_eff). Also, using radius r incorrectly for a circle (r = d/4 for the radius of gyration is only approximate; the exact value is d/4 for thin rings; for solid circles r = d/4 * sqrt(2) is incorrect—always compute r = sqrt(I/A)). Final Answer: 12.88 tonnes

Question 2

Definition check — bending moment at a section of a beam Which statement correctly defines the bending moment at a given cross-section of a beam in terms of the forces acting on the beam?

Accepted Answer

Introduction / Context: Bending moment (BM) is a fundamental concept in structural analysis. It represents the tendency of external forces to cause rotation about a point on the beam's axis, producing curvature. Correctly defining BM is essential for drawing shear force and bending moment diagrams, sizing members, and checking stresses using flexure formulas. Given Data / Assumptions: Beam loaded by general external loads (point loads, UDLs, etc.). Sign convention is consistent (e.g., sagging positive). Section of interest cuts the beam at some location x. Concept / Approach: At a given section, the internal bending moment equals the algebraic sum of the moments of all external forces to one side of the section about that section. The word algebraic emphasizes that sign conventions must be used (e.g., counterclockwise positive), unlike a mere arithmetic (unsigned) summation. Shear force is the algebraic sum of lateral forces, whereas bending moment is the algebraic sum of their moments. Step-by-Step Solution: Cut the beam at the section of interest and consider free body of either side.Sum the moments of all external forces about the cut section using the chosen sign convention.The algebraic result equals the internal bending moment at that section. Verification / Alternative check: In differential form, dM/dx = V, where V is shear force. This reinforces that moments must be treated algebraically, consistent with sign. Why Other Options Are Wrong: Arithmetic sum (without signs) ignores direction, yielding incorrect results. Sum of forces (not moments) defines shear, not bending moment. “None of these” is incorrect because option (c) is the standard definition. Resultant shear force is a different internal action and not a definition of bending moment. Common Pitfalls: Mixing sign conventions between left and right free bodies; forgetting to include reactions; using absolute values instead of algebraic sums. Final Answer: Algebraic sum of the moments of all the forces on either side of the section

Question 3

Welding practice — butt joint type most commonly used in fabrication Among the following butt weld preparations used in steel fabrication, which one is most commonly used in practice for general plate thicknesses?

Accepted Answer

Introduction / Context: Butt joints are widely used to join plates and rolled sections in bridges, buildings, tanks, and pipelines. The edge preparation affects accessibility, weld metal volume, distortion control, and economy. Selecting a common, versatile joint preparation ensures practicality for shop and site welding with standard equipment. Given Data / Assumptions: General structural fabrication context (mild or low-alloy steels). Range of moderate plate thicknesses where both cost and ease of execution matter. Common manual or semi-automatic welding processes (SMAW, GMAW, FCAW). Concept / Approach: The single V-butt joint is the most commonly used butt joint preparation. It provides a straightforward groove accessible from one side, requires moderate beveling, and balances penetration with reasonable weld metal volume. For thicker plates, double V can be preferable to reduce total weld metal and balance shrinkage, but in everyday fabrication, single V prevails for simplicity and productivity. Step-by-Step Solution: Identify typical shop practice: plates of moderate thickness are often prepped as single V grooves.Consider alternatives: double V or U reduce weld metal but require access to both sides and more complex preparation.Conclude that the most common, general-purpose choice is the single V-butt joint. Verification / Alternative check: Standard welding handbooks and codes (e.g., AWS D1.1) show single V as a default detail for a wide range of thicknesses, with prequalified groove angles and root openings for typical processes. Why Other Options Are Wrong: Double V-butt: common for thick sections but not the most common overall due to two-sided access and increased prep. Double U-butt: efficient for very thick plates but requires specialized machining and higher prep cost. Single inverted V: uncommon; gravity and access issues make it less practical as a standard choice. Square butt with no prep: limited to thin plates where full penetration is achievable; not a general solution. Common Pitfalls: Assuming the most economical for very thick plate (double V or U) is the most common in all situations; conflating “possible” with “prevalent.” Final Answer: Single V-butt joint

Question 4

Three-hinged arch in structural analysis In classical structural analysis and bridge engineering, how is a three-hinged arch classified with respect to static determinacy? Provide the most accurate description used in design and exam standards.

Accepted Answer

Introduction / Context: A three-hinged arch is a fundamental form used in bridge and roof structures. Knowing whether it is statically determinate or indeterminate is critical because it dictates the analysis method, sensitivity to support settlements, and thermal effects. Given Data / Assumptions: An arch with hinges at both springings (supports) and one hinge at the crown (third hinge). Loads are arbitrary, supports are ideal pins, and the arch is a rigid body between hinges. Linear elastic behavior is assumed for internal force interpretation. Concept / Approach: Each pin (hinge) releases a moment. With hinges at both ends and at the crown, the arch has just enough internal releases to remove redundancy. Therefore, all reaction components can be obtained using only the three static equilibrium equations in two dimensions: ΣFx = 0, ΣFy = 0, and ΣM = 0. Step-by-Step Solution: Count unknown reactions: two at each springing (Ax, Ay and Bx, By) → four unknowns.Add the internal crown hinge condition: the bending moment at the crown hinge is zero → this gives an additional relationship (used via sectional equilibrium).Use global and sectional equilibrium to solve for all unknowns without invoking compatibility → structure is determinate. Verification / Alternative check: Three-hinged arches are immune to thermal stress buildup and support settlement–induced indeterminacy because the crown hinge accommodates rotations, allowing solution purely by statics. Why Other Options Are Wrong: Statically indeterminate requires more unknowns than equilibrium equations; not the case here. Geometrically unstable is false because three hinges provide stability with proper geometry. Structurally sound but indeterminate contradicts determinate nature. Common Pitfalls: Confusing two-hinged arches (indeterminate to degree 1) with three-hinged arches (determinate). Also, misusing the crown hinge condition. Final Answer: statically determinate

Question 5

Parabolic three-hinged arch under uniform load For a three-hinged parabolic arch carrying a uniformly distributed load over the entire span, what internal action does every section primarily resist?

Accepted Answer

Introduction / Context: Funicular shapes transmit loads primarily through axial action. A parabolic three-hinged arch is the classic example: when subjected to a uniform load across its span, it closely follows the funicular (cable) polygon for that load. Given Data / Assumptions: Arch geometry is parabolic. Loading is uniformly distributed across the entire span. Three hinges: at both springings and at the crown. Concept / Approach: For a parabolic arch under a full-span uniform load, the bending moment at every section vanishes because the arch axis coincides with the thrust line (funicular line). The line of compression stays within the depth, so internal forces reduce to normal thrust (compression) with negligible bending. Shear may be present at supports in global equilibrium, but the key sectional action is axial compression. Step-by-Step Solution: Identify funicular shape for UDL → parabola.Match arch axis to funicular → bending moment M ≈ 0 throughout.Internal force left is axial compression (normal thrust) at each section. Verification / Alternative check: Using conventional arch equations shows M(x) = 0 when the arch profile equals the funicular; influence lines also confirm zero bending for uniform loading in the ideal case. Why Other Options Are Wrong: Tensile force is incorrect because arches work in compression. Shear force is not the defining action at each section for this ideal case. Bending moment is specifically minimized to zero for the funicular arch. Common Pitfalls: Assuming any arch always has bending; geometry–load matching can eliminate bending ideally. Final Answer: compressive force

Question 6

Location of the third hinge in a three-hinged arch In a standard three-hinged arch used for bridges and long-span roofs, where is the third (internal) hinge generally placed for analysis and construction convenience?

Accepted Answer

Introduction / Context: The position of hinges controls force flow and computational simplicity. A crown hinge provides a clear geometric reference and simplifies bending checks at midspan. Given Data / Assumptions: Hinges exist at both springings (supports). One internal hinge is placed along the arch rib. Symmetric spans and loading are common in practice. Concept / Approach: Placing the third hinge at the crown (highest point) divides the arch into two determinate segments and ensures zero bending moment at that location. It also enhances tolerance to temperature variations and settlements by allowing relative rotation. Step-by-Step Solution: Model arch with hinges at supports A and B.Insert the third hinge at the crown → internal moment there is zero.Use sectional equilibrium to solve reactions without compatibility equations. Verification / Alternative check: Influence lines for a three-hinged arch show a zero-moment line passing through the crown hinge, confirming the benefit of that location. Why Other Options Are Wrong: Off-crown hinge locations complicate analysis and are atypical in standard three-hinged arch configurations. Common Pitfalls: Confusing three-hinged with two-hinged arches, where no crown hinge exists and indeterminacy arises. Final Answer: crown of the arch

Question 7

Shear in a three-hinged arch For a three-hinged arch carrying general loads, where along the span is the shear force commonly at its maximum value?

Accepted Answer

Introduction / Context: Shear distribution in arches resembles that in beams: near supports, reactions and geometry combine to produce larger shear values. Recognizing this helps in sizing section depth and detailing near springings. Given Data / Assumptions: Three-hinged arch with hinges at both springings and at the crown. Arbitrary static loading (dead plus live). Linear elastic response and standard sign conventions. Concept / Approach: Support reactions contribute directly to shear near the springings. For typical symmetric or unsymmetric loads, the shear envelopes peak close to supports. The crown hinge enforces zero moment at midspan but does not create maximum shear there. Step-by-Step Solution: Compute reactions from statics.Construct shear diagram; highest magnitudes usually occur adjacent to supports.Confirm that the crown region shows lower shear compared to springings under standard loading. Verification / Alternative check: Influence lines for shear in an arch show peak ordinates near supports for moving loads. Why Other Options Are Wrong: Maximum at crown is uncommon; quarter points may see peaks for special load cases but not generally; varies only with slope is incorrect—shear is load and reaction driven. Common Pitfalls: Assuming crown hinge influences shear maxima; it primarily affects moment distribution. Final Answer: maximum at springings (supports)

Question 8

Moving point load on a simply supported beam For a single moving concentrated load traveling across a simply supported beam, where does the resulting bending moment attain its maximum value?

Accepted Answer

Introduction / Context: Moving load analysis is essential for bridges and cranes. Two classic results exist: (1) the absolute maximum bending moment for a single point load occurs at midspan; (2) the maximum bending moment at a given section occurs when the load is directly over that section. The question asks the general location where the maximum occurs for the moving load itself. Given Data / Assumptions: Beam: simply supported, span L. Load: single moving concentrated load W. Linear elastic behavior; no secondary effects. Concept / Approach: For a given position x of the load, the bending moment at that same section is M = R_left * x, and the influence line for bending moment at a section peaks when the load is at the section. Therefore, as the load travels, the bending moment is most intense directly under the load. The absolute maximum over the entire span occurs when the load is at midspan, which still satisfies “under the load.” Step-by-Step Solution: Write bending moment at a section z: M_z = IL_M(z) * W.Influence line IL_M(z) attains its peak when the load is at section z.Hence, for the moving load, the maximum at any location is under the load; the global maximum is at midspan when the load is there. Verification / Alternative check: Setting derivative dM/dx = 0 for a single point load yields x = L/2 as the position for absolute maximum; this still coincides spatially with the load location. Why Other Options Are Wrong: At the supports bending moment is zero for simple supports. At the mid span is true for the absolute maximum location but misses the general “under the load” principle. Anywhere on the beam is vague and incorrect. Common Pitfalls: Confusing “absolute maximum over the span” with “location of maximum at any instant.” The correct unifying statement is that the maximum occurs under the load. Final Answer: under the load

Question 9

Eccentric loading and weld groups For a weld group resisting an in-plane moment about its centroid, which part of the weld offers the greatest resistance to rotation?

Accepted Answer

Introduction / Context: Weld groups under eccentric loads experience a combination of direct shear and torsion (or bending) about the group centroid. The distribution of secondary shear (or bending) due to the moment depends on each weld element’s distance from the centroid. Given Data / Assumptions: Weld group in the plane (e.g., fillet weld rectangle). Resultant includes a moment M about the centroid. Linear elastic distribution proportional to radius from centroid. Concept / Approach: The torsional (or rotational) component of stress is proportional to the distance r from the centroidal axis: q_secondary ∝ M * r / J (for torsion-like distribution). Hence, the weld segments farthest from the centroid carry the largest share of rotational resistance. Step-by-Step Solution: Compute centroid of weld group and polar moment J (or equivalent).Secondary shear due to moment: q = M * r / J.Largest r → largest q → elements most distant resist most rotation. Verification / Alternative check: Design handbooks and codes distribute eccentric shear to welds in proportion to distance from centroid, confirming the principle. Why Other Options Are Wrong: Least distant and centrally located have smaller r and thus smaller contribution. Either end is ambiguous and not necessarily the farthest location from the centroid. Common Pitfalls: Ignoring the direct shear component; total weld demand is vector sum of direct and secondary components. Final Answer: most distant

Question 10

Failure stress of a simply supported cast iron bar under center load A cast iron bar of square section 2.5 cm × 2.5 cm and span 1.0 m is simply supported. It fails when a 100 kg weight is applied at midspan. Determine the maximum bending stress at f…

Accepted Answer

Introduction / Context: This is a classic flexure problem for a simply supported member under a central point load. It tests the ability to compute section modulus and relate bending moment to extreme fiber stress. Given Data / Assumptions: Cross-section: square, b = d = 2.5 cm. Span L = 1.0 m = 100 cm. Central point load W = 100 kgf (kilogram-force), using kg/cm^2 units. Material behaves linearly up to failure for stress calculation. Concept / Approach: For a simply supported beam with a central load, maximum bending moment occurs at midspan and equals M_max = W * L / 4. The extreme fiber stress follows flexure formula: σ = M / Z, where Z is the section modulus of a rectangle, Z = b * d^2 / 6 (about the strong axis). Step-by-Step Solution: Compute M_max = W * L / 4 = 100 * 100 / 4 = 2500 kgf·cm.Compute section modulus Z = b * d^2 / 6 = 2.5 * (2.5)^2 / 6 = 15.625 / 6 = 2.6042 cm^3 (approx).Compute stress σ = M / Z = 2500 / 2.6042 ≈ 960 kg/cm^2.Therefore, the failure stress is about 960 kg/cm^2. Verification / Alternative check: Using more precise arithmetic yields the same rounded value (≈ 959.9 kg/cm^2). The answer aligns with typical cast iron flexural strength ranges. Why Other Options Are Wrong: 980 and 1000 kg/cm^2 are overestimates from rounding errors; 1200 kg/cm^2 is significantly high for the given geometry and load. Common Pitfalls: Mixing N and kgf units; using I instead of Z directly; forgetting to convert L to centimeters when working in kg/cm^2. Final Answer: 960 kg/cm^2

Question 11

General internal actions in an arch Under arbitrary loading and realistic boundary conditions, what internal actions can an arch rib develop at a section?

Accepted Answer

Introduction / Context: Unlike a cable, which ideally carries pure tension, a rigid arch rib can carry a combination of internal actions. Understanding these actions is vital for sizing the rib and checking serviceability and ultimate limit states. Given Data / Assumptions: Rigid arch rib (steel, concrete, or masonry) with finite depth. Arbitrary loading (dead plus live plus wind, etc.). Hinged or fixed boundary conditions typical of arch structures. Concept / Approach: Equilibrium of any free body cut from the arch requires three internal resultants in plane: normal thrust (axial force), shear force, and bending moment. Special geometries (e.g., a parabolic three-hinged arch under full UDL) may reduce bending moment to near zero, but the general case admits all three. Step-by-Step Solution: Isolate a segment of the arch and apply ΣFx = 0, ΣFy = 0, ΣM = 0.Solutions require internal axial (thrust), shear, and moment to satisfy equilibrium.Hence, the complete set of actions is present in general. Verification / Alternative check: Design codes for arches require checks for axial compression, shear, and bending (including interaction diagrams), confirming the presence of all actions. Why Other Options Are Wrong: Pairs of actions listed in other options are incomplete subsets; they omit at least one necessary component for general equilibrium. Common Pitfalls: Assuming an arch always follows a funicular line; in practice, load changes and construction tolerances introduce bending. Final Answer: thrust, shear force and bending moment

Question 12

Points of contraflexure for a simply supported beam under UDL A simply supported beam carries a uniformly distributed load over the entire span. How many points of contraflexure (sign change of bending moment within the span) occur?

Accepted Answer

Introduction / Context: Points of contraflexure are locations where the bending moment changes sign, i.e., crosses zero within the span. Identifying them is important for reinforcement detailing and structural diagnostics. Given Data / Assumptions: Beam is simply supported. Uniformly distributed load (UDL) across the full span. No additional couples or varying boundary conditions. Concept / Approach: For a simply supported beam under full-span UDL, the bending moment diagram is a single sagging parabola: zero at each support and maximum positive at midspan. There is no internal sign change of bending moment between the supports, therefore no internal point of contraflexure. Step-by-Step Solution: Write M(x) for UDL on simply supported beam: M(x) = R_left * x − w * x^2 / 2 (for 0 ≤ x ≤ L).With symmetry, M(0)=0, M(L)=0, and M(x)>0 for 0Hence, there is no x in (0, L) for which M(x)=0 → no contraflexure within the span. Verification / Alternative check: Plotting the parabolic bending moment confirms a single positive dome; contraflexure points arise in beams with couples, overhangs, or nonuniform loading, not in this basic case. Why Other Options Are Wrong: Options 1, 2, or 3 imply internal zero crossings which do not occur under pure UDL on a simple span. Common Pitfalls: Counting the supports as contraflexure points; by convention, points of contraflexure are internal to the span, not at the simple supports where M is trivially zero. Final Answer: 0

Question 13

Plane pin-jointed trusses and frames – redundancy check using member–joint count Let n be the number of members and j be the number of joints in a planar pin-jointed frame (truss). The frame contains redundant (statistically indeterminate) members w…

Accepted Answer

Introduction / Context: For planar pin-jointed trusses, a quick determinacy check compares the number of members and joints. This avoids lengthy calculations and flags whether additional compatibility equations (beyond static equilibrium) are required. Given Data / Assumptions: Planar (2D) truss with ideal pin joints and two-force members. Proper external support reactions so the overall structure is stable. No internal mechanisms (i.e., geometry prevents collapse under small loads). Concept / Approach: The basic relation for a statically determinate, stable planar truss is n = 2j − 3. If the actual number of members n exceeds this value, there are more unknown member forces than can be solved by the three global equilibrium equations and joint-by-joint equilibrium; such a truss is statically indeterminate (redundant). If n is less, the truss is a mechanism (unstable) unless additional bracing exists. Step-by-Step Solution: Determinacy criterion (planar truss): n_determinant = 2j − 3.Redundancy condition: if n − (2j − 3) > 0 → extra (redundant) members.Therefore, redundancy ⇔ n > 2j − 3. Verification / Alternative check: Try a triangular truss: j = 3, 2j − 3 = 3; a triangle with n = 4 is indeterminate, aligning with the rule. Why Other Options Are Wrong: n = 2j − 3: determinate (no redundancy). n = 3j − 2: unrelated to 2D pin-jointed determinacy. n < 2j − 3 and n < j − 2: imply a mechanism/instability, not redundancy. Common Pitfalls: Forgetting that stability also depends on geometry; a truss may satisfy n = 2j − 3 yet be unstable if poorly arranged. Final Answer: n > 2j − 3

Question 14

Materials science – name for materials with identical elastic properties in all directions In mechanics of materials, what do we call a material that exhibits the same elastic properties (such as modulus and Poisson’s ratio) in every direction?

Accepted Answer

Introduction / Context: Describing material symmetry is essential for selecting correct constitutive models. Engineers must distinguish isotropy, orthotropy, and anisotropy because directional dependence affects stress–strain predictions and design. Given Data / Assumptions: Elastic range behavior only. No microstructural texture that would cause directional properties. Small deformations (linear elasticity context). Concept / Approach: An isotropic material has elastic properties that are identical in all directions. A homogeneous material has properties uniform from point to point (spatially), but they can still be direction-dependent. Brittle and hard are failure/strength descriptors, not symmetry classes. Orthotropic has three mutually orthogonal planes of material symmetry (direction-dependent). Step-by-Step Solution: Map definition: same elastic response for any loading direction → isotropic.Reject others based on meaning (not symmetry or not direction-independent). Verification / Alternative check: Common isotropic examples: fully annealed metals, glass, and many polymers (approximately). Wood and composites are orthotropic/anisotropic. Why Other Options Are Wrong: Brittle/Hard: describe failure or hardness, not elastic isotropy. Homogeneous: uniform in space, not necessarily same in all directions. Orthotropic: direction-dependent along principal directions. Common Pitfalls: Confusing homogeneity with isotropy; a material can be homogeneous yet anisotropic. Final Answer: Isotropic

Question 15

Riveted joints – Unwin’s empirical formula for rivet diameter According to Unwin’s rule, the recommended rivet diameter d (mm) for a plate of thickness t (mm) is approximately:

Accepted Answer

Introduction / Context: Sizing rivets for structural and boiler plate work often uses empirical relations that balance plate tearing and rivet shear. Unwin’s formula provides a quick estimate of rivet diameter from plate thickness. Given Data / Assumptions: Plate thickness t is in millimetres. Rivet diameter d is desired in millimetres. General-purpose structural riveting (lap/butt joints) without special constraints. Concept / Approach: Unwin’s rule of thumb states that the rivet diameter should scale with the square root of plate thickness: d ≈ k * √t with k close to 6 (many texts quote 6 or 6.05). This ensures practical proportions and adequate capacity without oversizing. Step-by-Step Solution: Select the baseline Unwin coefficient: k ≈ 6.Write the relation: d = 6 √t (mm, mm).Alternative close variant: d = 6.05 √t (option b) is essentially the same in practice; option (a) is the classic rounded form. Verification / Alternative check: Checking typical plates (e.g., t = 16 mm): d = 6 √16 = 24 mm, which aligns with common practice. Why Other Options Are Wrong: d = 5 √t underestimates diameter for many standards. d = 1.5 t + 4 grows linearly with t and becomes excessive for thick plates. d = 0.5 t ignores the square-root dependence and oversizes thin plates while undersizing thick ones. Common Pitfalls: Forgetting unit consistency (t in mm), and using the formula outside its empirical range without checking code requirements. Final Answer: d = 6 √t

Question 16

Maxwell’s reciprocal theorem – applicability in structural analysis Maxwell’s reciprocal theorem (Betti’s theorem) can be applied to which type of structures?

Accepted Answer

Introduction / Context: Maxwell’s reciprocal theorem underpins many displacement methods and influence-line concepts. It states that the displacement at point A due to a unit load at B equals the displacement at point B due to a unit load at A—under appropriate conditions. Given Data / Assumptions: Small displacements; linear geometry. Material behaves linearly elastically (superposition holds). Structures may be statically determinate or indeterminate but within elastic range. Concept / Approach: The reciprocity result (also Betti’s theorem) derives from symmetry of the stiffness matrix in linear elasticity. Plastic or nonlinear behavior breaks superposition and the theorem generally does not apply. Step-by-Step Solution: Identify requirement: linear elastic behavior → symmetric stiffness.Conclude applicability: theorem valid for linear elastic structures regardless of geometric symmetry. Verification / Alternative check: Matrix methods: for linear elastic systems, k = k^T; thus, displacement influence coefficients are reciprocal. Why Other Options Are Wrong: Plastic/nonlinear systems violate linear superposition. Symmetry of geometry is not required; material linearity is the key. “All the above” cannot be true because (a) and (e) are incorrect contexts. Common Pitfalls: Applying reciprocity across different boundary conditions or after yielding; ensure all loads remain within the elastic range. Final Answer: Linear elastic structures

Question 17

Plastic, brittle, and rigid idealizations – compare plastic zones near cracks/notches Which statement about the size of the plastic zone (or plastic region) is correct for idealized material behaviors?

Accepted Answer

Introduction / Context: In fracture and yielding, the extent of plasticity ahead of stress raisers controls ductility and energy absorption. Materials are often idealized as ductile, brittle, or rigid to discuss limiting behavior. Given Data / Assumptions: Ductile = significant plastic deformation before fracture. Brittle = negligible plastic deformation before fracture. Rigid = no deformation until failure (idealization). Concept / Approach: Ductile materials undergo yielding in a finite region near stress concentrations, creating a relatively large plastic zone. Brittle materials exhibit little to no yielding, so the plastic zone is tiny or absent. A perfectly rigid material is an abstraction with zero strain up to failure, hence no plastic zone. Step-by-Step Solution: Map each idealization to expected plastic region size.Ductile → large plastic zone; brittle → negligible; rigid → none.Therefore all three individual statements are correct, making the combined option correct. Verification / Alternative check: Fracture mechanics indicates higher fracture toughness and larger plastic zones for ductile metals versus glass/ceramics (brittle). Why Other Options Are Wrong: Any single choice omits portions of the correct set. “None” contradicts well-established definitions. Common Pitfalls: Assuming “no plastic zone” means “no yielding ever”; ductile/brittle are spectrums and depend on temperature and strain rate. Final Answer: All of the above

Question 18

Shear centre locations – identify the single incorrect statement Consider the following statements about the shear centre (elastic center) of open thin-walled sections. Which one is incorrect?

Accepted Answer

Introduction / Context: The shear centre (elastic centre) is crucial when transverse shear is applied. Loading away from this point induces twist in open sections. Designers must know where it lies for channels, angles, tees, and I-sections. Given Data / Assumptions: Open, thin-walled prismatic sections. Shear flow derived from elementary bending theory. No warping restraint and Saint-Venant torsion assumptions. Concept / Approach: By definition, the shear centre is where the resultant shear force must act to cause bending without twist. For doubly-symmetric sections (e.g., I or rectangular), it coincides with the centroid. For singly-symmetric sections (channel, tee), it lies on the axis of symmetry but generally not at the centroid. For unequal or angle sections (no symmetry axes coinciding with shear), it lies outside the material, often along the angle bisector. Step-by-Step Solution: (a) Matches the definition → correct.(b) Channels: shear centre outside web on symmetry line → correct.(c) Equal angles: shear centre outside along leg bisector → correct.(d) T-sections: only one symmetry axis; shear centre lies on that axis but not necessarily at the centroid → statement is incorrect. Verification / Alternative check: Handbooks list e-values for channels and tees showing offsets from centroid; for tees, the shear-centre offset depends on flange/web proportions. Why Other Options Are Wrong: (a)–(c) are standard results for open sections. (e) cannot be chosen because one specific incorrect statement exists, namely (d). Common Pitfalls: Assuming “centroid = shear centre”; this is true only for doubly symmetric sections or closed thin-walled sections with uniform thickness. Final Answer: For T-sections, the shear centre always coincides exactly with the centroid of the section

Question 19

Tensile test on a ductile material – compare true and nominal (engineering) stresses at fracture During a standard tension test on a ductile metal specimen, which statement is correct regarding stresses computed at the instant of fracture?

Accepted Answer

Introduction / Context: In tensile testing, two stress definitions are common: engineering (nominal) stress based on original area and true stress based on instantaneous area. After necking, these values diverge significantly. Given Data / Assumptions: Ductile material showing clear yield, strain hardening, necking, and final fracture. Ultimate stress (engineering) occurs at maximum load P_max divided by original area A_0. True stress at fracture uses the instantaneous minimum area A_f at fracture. Concept / Approach: Once necking begins, the load may drop, but area reduces faster; consequently, true stress σ_true = P/A_inst can continue to rise even as P decreases. Engineering stress σ_eng = P/A_0 ignores area reduction and therefore underestimates the final local stress state. Step-by-Step Solution: Before necking: σ_true ≈ σ_eng (areas similar).At maximum load: σ_ult = P_max/A_0 (engineering definition).At fracture: A has shrunk to A_f ≪ A_0; therefore σ_true,fracture = P_fracture/A_f is typically higher than σ_ult, while σ_eng,fracture = P_fracture/A_0 is lower.Hence, the correct comparison is: σ_true at fracture > σ_ult (engineering). Verification / Alternative check: Consistent with Considère criterion and necking behavior; true-stress–true-strain curves show continued hardening up to fracture in many ductile metals. Why Other Options Are Wrong: (a) Engineering stress at fracture is usually less than the ultimate (since load has dropped). (c) and (e) contradict the different areas used. (d) is invalid because (b) is correct. Common Pitfalls: Comparing loads instead of stresses; forgetting the strong reduction in area after necking. Final Answer: True stress at fracture is higher than the ultimate engineering stress

Question 20

Members of uniform strength – definition in axial loading If, under axial loading, the stress at every cross-section of a bar is equal to the allowable (working) stress, the bar is referred to as a:

Accepted Answer

Introduction / Context: Bars can be shaped so that the maximum permissible stress is fully utilized along the entire length. This is important in weight-critical designs (e.g., ties, columns, connecting rods). Given Data / Assumptions: Axial (tension/compression) loading only. Allowable working stress is the same everywhere. Bar geometry may vary to achieve uniform stress. Concept / Approach: A body of equal (uniform) strength is one in which the cross-sectional area A(x) is adjusted along the length so that σ(x) = P/A(x) equals the allowable stress everywhere. For constant load P, this implies A(x) is constant; for members with self-weight or varying load, A(x) varies to keep σ constant. Step-by-Step Solution: Define target: σ_allow = P/A(x) for all x.Solve for area profile: A(x) = P/σ_allow; if P varies with x (e.g., self-weight), A(x) changes accordingly.Name the concept: body of equal (uniform) strength. Verification / Alternative check: Design texts show tapered bars under self-weight as classic examples where A increases toward the fixed end to maintain constant stress. Why Other Options Are Wrong: Equal section does not guarantee equal stress unless load is constant and the bar is prismatic. “Body of equal (incomplete term)” is not a proper descriptor. “Mechanism” is unrelated. Common Pitfalls: Confusing equal strength with equal section; the former is a stress condition, the latter is geometric. Final Answer: Body of equal strength (uniform strength)

Strength of Materials Questions