Q: Structural Analysis – Conditions Common to Elastic and Plastic Methods Which condition must be satisfied by both elastic analysis and plastic (limit/upper-bound) analysis of indeterminate structures?

Introduction / Context: Structural analysis can be performed using elastic methods (stiffness/slope-deflection) or plastic (limit state) methods. Although their assumptions differ, certain fundamental requirements must be satisfied by any admissible solution to ensure physical realism and safety. Given Data / Assumptions: Indeterminate structure under static loading. Elastic analysis: stresses/strains within elastic range, compatibility enforced. Plastic analysis: considers yield and collapse mechanisms, may relax elastic compatibility locally. Concept / Approach: Equilibrium of forces and moments is a universal requirement. Elastic analysis also enforces compatibility and constitutive relations; plastic limit analysis replaces compatibility with yield and mechanism conditions while still maintaining global and member-wise equilibrium. Step-by-Step Solution: 1) Identify universal requirement: force and moment equilibrium must hold for the entire structure and its parts.2) Elastic approach adds compatibility (unique displacements) and Hooke's law.3) Plastic approach adds yield criteria and mechanism conditions but may not enforce elastic-range compatibility everywhere.4) Therefore, equilibrium is common to both methods. Verification / Alternative check: Lower- and upper-bound theorems of plasticity explicitly require equilibrium (statically admissible stress field) or mechanism with yield, reinforcing its universality. Why Other Options Are Wrong: Yield and mechanism conditions: specific to plastic analysis. Compatibility: enforced in elastic analysis; not strictly required in plastic upper-bound solutions. Common Pitfalls: Assuming compatibility is necessary in all methods; overlooking that limit analysis targets collapse load rather than elastic displacements. Final Answer: Equilibrium

Q: Highway Design – When Overtaking Sight Distance (OSD) Is Not Feasible On stretches where the required OSD cannot be provided, which minimum sight distance criterion should be adopted to maintain safety under mixed traffic?

Introduction / Context: Overtaking sight distance (OSD) ensures that a vehicle can overtake safely on two-lane highways. When geometric or environmental constraints prevent providing full OSD, standards prescribe a fallback ensuring a base level of safety for drivers who must at least be able to stop safely if a hazard appears. Given Data / Assumptions: Mixed traffic, two-lane highways. OSD not achievable due to terrain, right-of-way, or environmental constraints. Need a conservative, implementable alternative. Concept / Approach: The recommended fallback is Intermediate Sight Distance (ISD), defined as twice the Stopping Sight Distance (SSD). ISD offers better safety than SSD alone by allowing drivers more distance to react and adjust speed in the absence of full OSD. Step-by-Step Solution: 1) Determine SSD for the design speed and conditions.2) Set ISD = 2 * SSD when OSD cannot be provided.3) Implement warning/markings as necessary to discourage overtaking where ISD replaces OSD. Verification / Alternative check: Design manuals and standard practice list ISD = 2 * SSD as the fallback where OSD is infeasible, balancing safety and feasibility. Why Other Options Are Wrong: Half or one third of OSD: Insufficient safety margin. Three times SSD: More conservative than typical prescriptions and often impractical. Common Pitfalls: Failing to post appropriate signage; not adjusting for gradients or friction conditions when computing SSD. Final Answer: At least twice the stopping sight distance (Intermediate Sight Distance)

Q: Plastic Analysis – Yield Stress from Plastic Modulus and Plastic Moment A section has plastic modulus Zp = 4.8 × 10^−4 m^3. The shape factor is 1.2. If the plastic moment capacity is Mp = 120 kN·m, what is the yield stress of the material?

Introduction / Context: In plastic analysis, the fully plastic moment Mp is related to the yield stress σy and the plastic section modulus Zp by Mp = σy * Zp. Knowing Zp and Mp allows direct computation of σy, independent of the elastic modulus or shape factor (the latter relates Zp to the elastic modulus Ze but is not needed once Zp is given). Given Data / Assumptions: Plastic modulus: Zp = 4.8 × 10^−4 m^3. Plastic moment capacity: Mp = 120 kN·m = 120,000 N·m. Small-strain plasticity assumptions; uniform yield stress σy across the plastic blocks. Concept / Approach: Use Mp = σy * Zp. Units must be consistent: N·m for moment and m^3 for modulus give N/m^2 (Pa) for stress. Convert the result to MPa for convenience. Step-by-Step Solution: 1) Write σy = Mp / Zp.2) Substitute: σy = 120,000 / (4.8 × 10^−4) N/m^2.3) Compute: σy = 2.5 × 10^8 N/m^2.4) Convert: 2.5 × 10^8 N/m^2 = 250 MPa. Verification / Alternative check: The provided shape factor (Zp/Ze = 1.2) is a cross-check; it is not required here since Zp and Mp are already known. Why Other Options Are Wrong: 100 MPa, 240 MPa, 300 MPa: Do not satisfy Mp = σy * Zp for the given values. Common Pitfalls: Mixing units (kN·m with mm^3); using elastic modulus Ze instead of Zp when Mp is given; forgetting to convert to MPa. Final Answer: 250 MPa

Q: Hydrology – What Is an Isochrone on a Basin Map? Choose the correct definition of an isochrone plotted on a catchment (basin) map used for runoff and hydrograph studies.

Introduction / Context: Isochrones are key to transforming spatial rainfall into temporal runoff at the basin outlet. By delineating lines of equal travel time, engineers visualize how different parts of the basin contribute to the hydrograph at successive times after a rainfall event. Given Data / Assumptions: Catchment with defined outlet and drainage paths. Overland and channel flow velocities estimated or mapped. Focus on surface runoff travel time, not rainfall amount or duration. Concept / Approach: An isochrone connects points that deliver runoff to the outlet after the same elapsed time since effective rainfall begins. Stacking the areas between successive isochrones and multiplying by an effective rainfall rate yields the unit hydrograph by time-area method. Step-by-Step Solution: 1) Compute/estimate travel times from each basin point to the outlet.2) Draw lines connecting points with equal travel time: these are isochrones.3) Use time-area histogram (areas between isochrones) to derive the outflow hydrograph. Verification / Alternative check: Results can be checked against observed hydrographs; calibration adjusts velocities and losses to align modeled peaks and timing. Why Other Options Are Wrong: Equal rainfall duration/depth: describe isohyetal analysis (isohyets), not isochrones. Equal distance: ignores hydraulic velocity differences; distance is not time. Common Pitfalls: Assuming constant velocity everywhere; neglecting channelized flow that shortens travel time relative to overland flow. Final Answer: Joining points having equal time of travel of surface runoff to the catchment outlet

Q: Highway Geometrics – Superelevation for High Speed with Mixed Traffic Limits Determine the design rate of superelevation for a horizontal curve of radius R = 450 m under mixed traffic conditions at a speed of 125 km/h. Apply practical limits for mix…

Introduction / Context: Superelevation (e) counters the lateral acceleration experienced by vehicles on curves. The theoretical rate needed for a given speed and radius follows from equilibrium of forces, but practical design for mixed traffic imposes an upper cap to accommodate slow-moving vehicles, heavy vehicles, and safety margins in rain. Given Data / Assumptions: Design speed v = 125 km/h. Curve radius R = 450 m. Mixed traffic conditions with a capped maximum e. Concept / Approach: The theoretical no-friction superelevation is e_theory = v^2 / (225 R) (v in km/h, R in m). Compute e_theory and then apply the practical limit for mixed traffic. If e_theory exceeds the cap, adopt the capped value. Step-by-Step Solution: 1) Compute e_theory = v^2 / (225 R) = 125^2 / (225 * 450) ≈ 0.154.2) Mixed traffic design caps e at about 0.07 (7%).3) Since 0.154 > 0.07, adopt e = 0.07. Verification / Alternative check: Design charts similarly show that for very high speeds on modest radii, the required theoretical e exceeds the cap; designers therefore use the maximum permissible e and manage the remainder with speed control, signage, or increased radius. Why Other Options Are Wrong: 1.0: Impossible; would imply a 100% cross slope. 0.05: Below the needed maximum; less conservative than the cap. 0.154: Theoretical value without practical cap; not used for mixed traffic. Common Pitfalls: Using e_theory directly without checking the allowable maximum; ignoring speed management measures when e is capped. Final Answer: 0.07

Question 1

Impact Loading on a Vertical Rod – How to Reduce Maximum Axial Stress A vertical rod PQ of length L is fixed at its top end P and has a flange at the bottom end Q. A weight W is dropped vertically from a height h (with h < L) onto the flange. Which …

Accepted Answer

Introduction / Context: When a weight drops onto a member, part of the potential energy W * h is converted into strain energy in the rod. The resulting peak stress is higher than for a static load and depends on the member's stiffness and energy absorption capacity. Designers must recognize which parameters reduce the maximum impact stress to improve safety and durability. Given Data / Assumptions: Rod: prismatic, length L, cross-sectional area A, modulus E. Impact: weight W dropped from height h (< L) onto a rigidly attached flange at the free end. Material remains within elastic range for the discussion. Concept / Approach: For elastic impact, peak stress approximately follows from equating strain energy to drop energy, giving a peak stress that increases with stiffness and decreases with greater compliance. A representative relation shows peak stress increasing with E and decreasing with L and A. Thus, making the member more compliant (larger L and/or larger A for energy capacity per unit stress) reduces the maximum stress developed. Step-by-Step Solution: 1) Drop energy: U = W * h (neglecting local losses).2) Elastic strain energy in an axially loaded bar: U = ∫(σ^2 / (2E)) * (A dz) = (σ_max^2 * A * L) / (2E) for a uniform bar (conceptual form).3) Higher L or A raises the energy capacity for a given peak stress, thereby lowering σ_max required to absorb U.4) Conversely, larger E raises stiffness, reducing deflection and increasing σ_max for the same U.5) Therefore, to reduce peak stress, increase L (and/or A), or choose a lower E material; among the options, increasing L is correct. Verification / Alternative check: Energy methods or exact elastic-impact formulas with static extension terms both show σ_max decreasing with increasing L and A, and increasing with E. Why Other Options Are Wrong: Decreasing L: Makes the bar stiffer; increases σ_max. Decreasing A: Increases stress for the same force; also lowers energy capacity. Increasing E: Stiffer bar yields higher σ_max for the same drop energy. Common Pitfalls: Confusing static and impact loading; ignoring that energy balance, not just force equilibrium, governs peak stress under impact. Final Answer: Increasing the length of the rod

Question 2

Structural Analysis – Conditions Common to Elastic and Plastic Methods Which condition must be satisfied by both elastic analysis and plastic (limit/upper-bound) analysis of indeterminate structures?

Accepted Answer

Introduction / Context: Structural analysis can be performed using elastic methods (stiffness/slope-deflection) or plastic (limit state) methods. Although their assumptions differ, certain fundamental requirements must be satisfied by any admissible solution to ensure physical realism and safety. Given Data / Assumptions: Indeterminate structure under static loading. Elastic analysis: stresses/strains within elastic range, compatibility enforced. Plastic analysis: considers yield and collapse mechanisms, may relax elastic compatibility locally. Concept / Approach: Equilibrium of forces and moments is a universal requirement. Elastic analysis also enforces compatibility and constitutive relations; plastic limit analysis replaces compatibility with yield and mechanism conditions while still maintaining global and member-wise equilibrium. Step-by-Step Solution: 1) Identify universal requirement: force and moment equilibrium must hold for the entire structure and its parts.2) Elastic approach adds compatibility (unique displacements) and Hooke's law.3) Plastic approach adds yield criteria and mechanism conditions but may not enforce elastic-range compatibility everywhere.4) Therefore, equilibrium is common to both methods. Verification / Alternative check: Lower- and upper-bound theorems of plasticity explicitly require equilibrium (statically admissible stress field) or mechanism with yield, reinforcing its universality. Why Other Options Are Wrong: Yield and mechanism conditions: specific to plastic analysis. Compatibility: enforced in elastic analysis; not strictly required in plastic upper-bound solutions. Common Pitfalls: Assuming compatibility is necessary in all methods; overlooking that limit analysis targets collapse load rather than elastic displacements. Final Answer: Equilibrium

Question 3

Highway Design – When Overtaking Sight Distance (OSD) Is Not Feasible On stretches where the required OSD cannot be provided, which minimum sight distance criterion should be adopted to maintain safety under mixed traffic?

Accepted Answer

Introduction / Context: Overtaking sight distance (OSD) ensures that a vehicle can overtake safely on two-lane highways. When geometric or environmental constraints prevent providing full OSD, standards prescribe a fallback ensuring a base level of safety for drivers who must at least be able to stop safely if a hazard appears. Given Data / Assumptions: Mixed traffic, two-lane highways. OSD not achievable due to terrain, right-of-way, or environmental constraints. Need a conservative, implementable alternative. Concept / Approach: The recommended fallback is Intermediate Sight Distance (ISD), defined as twice the Stopping Sight Distance (SSD). ISD offers better safety than SSD alone by allowing drivers more distance to react and adjust speed in the absence of full OSD. Step-by-Step Solution: 1) Determine SSD for the design speed and conditions.2) Set ISD = 2 * SSD when OSD cannot be provided.3) Implement warning/markings as necessary to discourage overtaking where ISD replaces OSD. Verification / Alternative check: Design manuals and standard practice list ISD = 2 * SSD as the fallback where OSD is infeasible, balancing safety and feasibility. Why Other Options Are Wrong: Half or one third of OSD: Insufficient safety margin. Three times SSD: More conservative than typical prescriptions and often impractical. Common Pitfalls: Failing to post appropriate signage; not adjusting for gradients or friction conditions when computing SSD. Final Answer: At least twice the stopping sight distance (Intermediate Sight Distance)

Question 4

Plastic Analysis – Yield Stress from Plastic Modulus and Plastic Moment A section has plastic modulus Zp = 4.8 × 10^−4 m^3. The shape factor is 1.2. If the plastic moment capacity is Mp = 120 kN·m, what is the yield stress of the material?

Accepted Answer

Introduction / Context: In plastic analysis, the fully plastic moment Mp is related to the yield stress σy and the plastic section modulus Zp by Mp = σy * Zp. Knowing Zp and Mp allows direct computation of σy, independent of the elastic modulus or shape factor (the latter relates Zp to the elastic modulus Ze but is not needed once Zp is given). Given Data / Assumptions: Plastic modulus: Zp = 4.8 × 10^−4 m^3. Plastic moment capacity: Mp = 120 kN·m = 120,000 N·m. Small-strain plasticity assumptions; uniform yield stress σy across the plastic blocks. Concept / Approach: Use Mp = σy * Zp. Units must be consistent: N·m for moment and m^3 for modulus give N/m^2 (Pa) for stress. Convert the result to MPa for convenience. Step-by-Step Solution: 1) Write σy = Mp / Zp.2) Substitute: σy = 120,000 / (4.8 × 10^−4) N/m^2.3) Compute: σy = 2.5 × 10^8 N/m^2.4) Convert: 2.5 × 10^8 N/m^2 = 250 MPa. Verification / Alternative check: The provided shape factor (Zp/Ze = 1.2) is a cross-check; it is not required here since Zp and Mp are already known. Why Other Options Are Wrong: 100 MPa, 240 MPa, 300 MPa: Do not satisfy Mp = σy * Zp for the given values. Common Pitfalls: Mixing units (kN·m with mm^3); using elastic modulus Ze instead of Zp when Mp is given; forgetting to convert to MPa. Final Answer: 250 MPa

Question 5

Hydrology – What Is an Isochrone on a Basin Map? Choose the correct definition of an isochrone plotted on a catchment (basin) map used for runoff and hydrograph studies.

Accepted Answer

Introduction / Context: Isochrones are key to transforming spatial rainfall into temporal runoff at the basin outlet. By delineating lines of equal travel time, engineers visualize how different parts of the basin contribute to the hydrograph at successive times after a rainfall event. Given Data / Assumptions: Catchment with defined outlet and drainage paths. Overland and channel flow velocities estimated or mapped. Focus on surface runoff travel time, not rainfall amount or duration. Concept / Approach: An isochrone connects points that deliver runoff to the outlet after the same elapsed time since effective rainfall begins. Stacking the areas between successive isochrones and multiplying by an effective rainfall rate yields the unit hydrograph by time-area method. Step-by-Step Solution: 1) Compute/estimate travel times from each basin point to the outlet.2) Draw lines connecting points with equal travel time: these are isochrones.3) Use time-area histogram (areas between isochrones) to derive the outflow hydrograph. Verification / Alternative check: Results can be checked against observed hydrographs; calibration adjusts velocities and losses to align modeled peaks and timing. Why Other Options Are Wrong: Equal rainfall duration/depth: describe isohyetal analysis (isohyets), not isochrones. Equal distance: ignores hydraulic velocity differences; distance is not time. Common Pitfalls: Assuming constant velocity everywhere; neglecting channelized flow that shortens travel time relative to overland flow. Final Answer: Joining points having equal time of travel of surface runoff to the catchment outlet

Question 6

Highway Geometrics – Superelevation for High Speed with Mixed Traffic Limits Determine the design rate of superelevation for a horizontal curve of radius R = 450 m under mixed traffic conditions at a speed of 125 km/h. Apply practical limits for mix…

Accepted Answer

Introduction / Context: Superelevation (e) counters the lateral acceleration experienced by vehicles on curves. The theoretical rate needed for a given speed and radius follows from equilibrium of forces, but practical design for mixed traffic imposes an upper cap to accommodate slow-moving vehicles, heavy vehicles, and safety margins in rain. Given Data / Assumptions: Design speed v = 125 km/h. Curve radius R = 450 m. Mixed traffic conditions with a capped maximum e. Concept / Approach: The theoretical no-friction superelevation is e_theory = v^2 / (225 R) (v in km/h, R in m). Compute e_theory and then apply the practical limit for mixed traffic. If e_theory exceeds the cap, adopt the capped value. Step-by-Step Solution: 1) Compute e_theory = v^2 / (225 R) = 125^2 / (225 * 450) ≈ 0.154.2) Mixed traffic design caps e at about 0.07 (7%).3) Since 0.154 > 0.07, adopt e = 0.07. Verification / Alternative check: Design charts similarly show that for very high speeds on modest radii, the required theoretical e exceeds the cap; designers therefore use the maximum permissible e and manage the remainder with speed control, signage, or increased radius. Why Other Options Are Wrong: 1.0: Impossible; would imply a 100% cross slope. 0.05: Below the needed maximum; less conservative than the cap. 0.154: Theoretical value without practical cap; not used for mixed traffic. Common Pitfalls: Using e_theory directly without checking the allowable maximum; ignoring speed management measures when e is capped. Final Answer: 0.07

Question 7

Superelevation with Friction – Compute e for Given Speed and Radius A vehicle travels at 60 km/h on a horizontal curve of radius R = 128 m. The pavement's side-friction factor is f = 0.15. What superelevation e should be provided (use e + f = v^2 / …

Accepted Answer

Introduction / Context: On horizontal curves, part of the lateral acceleration is balanced by roadway banking (superelevation e) and the remainder by side friction f between tires and pavement. Design aims to keep both within safe limits. Given Data / Assumptions: Speed v = 60 km/h. Radius R = 128 m. Side-friction factor f = 0.15. Design relation: e + f = v^2 / (127 R). Concept / Approach: Compute the right-hand side from speed and radius, then subtract the allowable friction contribution to find the required superelevation e. Step-by-Step Solution: 1) Evaluate v^2 / (127 R) = 60^2 / (127 * 128).2) 60^2 = 3600; denominator = 16256; ratio ≈ 0.22146.3) e = 0.22146 − f = 0.22146 − 0.15 ≈ 0.07146.4) Provide e ≈ 0.07 (7%). Verification / Alternative check: Checking with design tables for v = 60 km/h and R = 125–130 m yields similar e values around 7%, confirming the calculation. Why Other Options Are Wrong: 0.05: Too low, would overload friction. 0.09 or 0.11: Higher than required and may be uncomfortable or drain poorly. Common Pitfalls: Confusing the constants 127 vs 225; forgetting to subtract f from the total required e + f; rounding e to an impractically fine precision (use 7%). Final Answer: e = 0.07

Question 8

Water and Wastewater Treatment – What is flocculation? Choose the statement that best describes the process of flocculation in environmental engineering practice.

Accepted Answer

Introduction / Context: Flocculation is a key unit process in water and wastewater treatment. It bridges coagulation and clarification by gently mixing water so that micro-flocs grow into settleable flocs, improving sedimentation, dissolved air flotation, and filtration performance. Given Data / Assumptions: We are considering the generic definition of flocculation as used in treatment trains. Gravity separation (settling or flotation) follows flocculation. Coagulant addition may precede flocculation to destabilize particles. Concept / Approach: Fine colloids are stable due to surface charges. Coagulation neutralizes these charges, and flocculation provides controlled hydraulic energy (characterized by velocity gradient G and Gt) to promote inter-particle collisions, forming larger, denser flocs that are more readily removed by gravity processes. Step-by-Step Solution: 1) Destabilize colloids via coagulants (e.g., alum, ferric salts, polymers).2) Apply gentle mixing → collisions between micro-flocs increase.3) Flocs grow to sizes with higher settling velocities per Stokes’ law.4) Downstream units (clarifiers/DAF/filters) remove flocs efficiently. Verification / Alternative check: Jar tests optimize coagulant dose and flocculation gradients; improved turbidity removal and filter run length confirm proper floc formation. Why Other Options Are Wrong: (a) describes one application (algae removal) rather than the definition; it can be true in context but is not the generic meaning. (c) is incorrect because (a) alone does not define the process. (d) is false since (b) correctly states the definition. Common Pitfalls: Overmixing (breaking flocs), insufficient Gt, poor coagulant dosing, or skipping alkalinity control, all of which reduce floc size and settleability. Final Answer: A process that promotes aggregation of fine particles into larger flocs to enhance gravity removal

Question 9

Tacheometry – Horizontal distance by stadia formula (line of collimation horizontal) f = 200 mm, distance from instrument vertical axis to object-glass centre = 100 mm, stadia hair spacing i = 4 mm. Staff readings: top = 1.000 m, middle = 2.000 m, b…

Accepted Answer

Introduction / Context: Tacheometry employs a stadia diaphragm to quickly measure horizontal distance without chaining. When the line of sight is horizontal, the classic formula D = k * s + C applies, where k = f / i and C = f + d (additive constant from instrument geometry). Given Data / Assumptions: Focal length f = 200 mm = 0.2 m. Distance from instrument vertical axis to objective centre d = 100 mm = 0.1 m. Stadia hair interval i = 4 mm = 0.004 m. Staff readings: top = 1.000 m, middle = 2.000 m, bottom = 3.000 m (collimation horizontal). Concept / Approach: For a horizontal line of collimation, the horizontal distance equals the stadia constant times the intercept plus the additive constant. The intercept s is the difference between top and bottom readings in metres. Step-by-Step Solution: 1) Compute intercept: s = top − bottom = 1.000 − 3.000 = −2.000 m → use magnitude 2.000 m.2) k = f / i = 0.2 / 0.004 = 50.3) C = f + d = 0.2 + 0.1 = 0.3 m.4) Horizontal distance D = k * s + C = 50 * 2.0 + 0.3 = 100.3 m. Verification / Alternative check: A quick ratio check shows that with k = 50, every 0.02 m intercept corresponds to 1 m distance; an intercept of 2 m gives ~100 m, and the additive constant adds 0.3 m → 100.3 m. Why Other Options Are Wrong: 103.0, 150.0, and 153.0 m ignore the computed constants or misuse the intercept sign/units. Common Pitfalls: Confusing mm and m for f and i, using the middle reading incorrectly, or forgetting to add C. Final Answer: 100.3

Question 10

Steel Beams – Allowable average shear stress in an unstiffened web (Fe 250 grade steel)

Accepted Answer

Introduction / Context: In working-stress design tradition, Indian steel codes specify permissible average shear stress values for webs of rolled beams and plate girders. For unstiffened webs in Fe 250 steel, a conservative allowable shear controls web sizing and prevents shear buckling and yielding under service loads. Given Data / Assumptions: Material: structural steel grade 250 N/mm2 (Fe 250). Member: beam with unstiffened web region. Design basis: permissible/allowable stress format. Concept / Approach: The web primarily resists transverse shear. Average web shear stress τ_avg = V / (b_w * t_w) is checked against the code-specified allowable value. Traditional values adopt about 100 N/mm2 for Fe 250 to maintain adequate safety against shear yielding and instability for typical web slenderness limits. Step-by-Step Solution: 1) Compute factored/service shear V as per design method.2) Evaluate average shear in web: τ_avg = V / (b_w * t_w).3) Ensure τ_avg ≤ 100 N/mm2 for unstiffened web (Fe 250) in allowable stress design practice. Verification / Alternative check: Design tables and classic examples align with the 100 N/mm2 benchmark; higher-grade steels or limit-state methods use different formulas and factors but this question targets the traditional allowable value. Why Other Options Are Wrong: 250 N/mm2 equals nominal yield strength, not an allowable; 165 or 150 N/mm2 are higher than standard permissible shear for unstiffened webs in Fe 250 under WSD. Common Pitfalls: Confusing allowable stresses with limit-state design shear capacities, or using web area of the whole section instead of the clear web area. Final Answer: 100 N/mm2

Question 11

Pavement Design (IRC method) – Design traffic in msa for a 2-lane single carriageway (15-year life) Two-way CVPD at opening = 2000; vehicle damage factor = 3.0; lane distribution factor = 0.75; annual growth = 7.5%. Compute cumulative standard axles…

Accepted Answer

Introduction / Context: IRC pavement design requires estimating cumulative standard axles (N) over the design life using initial traffic, growth rate, lane distribution, and vehicle damage factor (VDF). Correct interpretation of the distribution factors is critical to avoid under- or over-design. Given Data / Assumptions: Two-way commercial vehicles at opening, P = 2000 CVPD. Lane distribution factor (design lane) = 0.75 for a 2-lane single carriageway. VDF = 3.0. Annual growth r = 7.5% for n = 15 years. Concept / Approach: Design traffic N (standard axles) is computed by N = 365 * A * F * [[(1 + r)^n − 1] / r], where A is the initial CVPD in the design lane, F is the vehicle damage factor. With the data provided, a common interpretation uses A = P * 0.75 for the design lane on a 2-lane single carriageway. Step-by-Step Solution: 1) Initial design-lane traffic: A = 2000 * 0.75 = 1500 CVPD.2) Growth factor: G = ((1.075)^15 − 1) / 0.075 ≈ 26.0.3) Compute N = 365 * 1500 * 3.0 * 26.0.4) N ≈ 365 * 1500 = 547,500; 547,500 * 3.0 = 1,642,500; 1,642,500 * 26.0 ≈ 42,705,000 ≈ 42.9 × 10^6 standard axles. Verification / Alternative check: Using the per-direction split (0.5) and a 0.75 lane factor would halve A; however, the presence of an explicit 0.75 design-lane factor in the problem statement indicates A is to be taken as P × 0.75, matching the given answer band. Why Other Options Are Wrong: 22.6, 10.1, and 5.3 msa arise from misapplying growth or lane/directional factors. Common Pitfalls: Double-counting both directional and lane factors, or using monthly/annual traffic instead of daily CVPD in the IRC formula. Final Answer: 42.9 × 10^6

Question 12

Strength of Materials – Proof resilience of mild steel (maximum strain energy density without permanent set) E = 2 × 10^5 MPa, yield stress σ_y = 250 MPa. Compute the maximum strain energy per unit volume (Nmm/mm3).

Accepted Answer

Introduction / Context: Proof resilience is the maximum strain energy stored per unit volume in a material without causing permanent deformation. For linearly elastic behavior up to yield, it depends only on the yield stress and Young’s modulus. Given Data / Assumptions: Material: mild steel. Young’s modulus E = 2 × 10^5 MPa. Yield stress σ_y = 250 MPa. Elastic behavior up to σ_y. Concept / Approach: Strain energy density u for linear elasticity under uniaxial stress is u = ∫0^ε σ dε = 0.5 * σ * ε. At yield, ε_y = σ_y / E, so u_max = 0.5 * σ_y * (σ_y / E) = σ_y^2 / (2E). Units of MPa convert directly to N/mm^2, which equal Nmm/mm^3 for energy density. Step-by-Step Solution: 1) Compute ε_y = σ_y / E = 250 / (2 × 10^5) = 0.00125.2) u_max = 0.5 * σ_y * ε_y = 0.5 * 250 * 0.00125 = 0.15625 Nmm/mm3.3) Rounded value = 0.156 Nmm/mm3. Verification / Alternative check: Direct formula check: u = σ_y^2 / (2E) = 62,500 / 400,000 = 0.15625 Nmm/mm3, consistent with the step calculation. Why Other Options Are Wrong: 1.56, 15.6, and 156 are 10×, 100×, and 1000× larger, stemming from unit mistakes (e.g., mixing MPa with N/mm2 or omitting the 1/2 factor). Common Pitfalls: Confusing energy per unit volume units, or using ultimate stress instead of yield for proof resilience. Final Answer: 0.156 Nmm/mm3

Question 13

Soil Mechanics – Earth pressure state when a retaining wall moves away from the backfill Identify the correct term for the pressure condition on the wall.

Accepted Answer

Introduction / Context: Earth pressure on retaining structures depends on wall movement. Small outward movements reduce lateral stress to a minimum state known as “active” pressure; inward movements raise it toward “passive.” Correctly identifying the state is essential for safe design. Given Data / Assumptions: Retaining wall supporting granular backfill. Wall moves away from the soil mass (outward). Drainage adequate so that pressures are effective (total versus pore pressure distinguished). Concept / Approach: With outward wall movement, the soil mass tends to dilate and expand laterally, allowing shear along a failure surface to mobilize. The lateral stress decreases until the active state is reached, characterized by Rankine or Coulomb active pressure coefficients (Ka). Step-by-Step Solution: 1) Recognize movement direction: wall moves away from backfill.2) Mobilization: soil reaches limiting equilibrium with reduced lateral stress.3) Identify state: “active” earth pressure (minimum lateral thrust for given vertical stress). Verification / Alternative check: Conversely, when the wall is pushed into the backfill, lateral stress increases to the “passive” state (Kp), confirming that outward movement corresponds to “active.” Why Other Options Are Wrong: Passive requires inward movement; swelling pressure concerns expansive clays; pore pressure is a hydraulic component, not the overall earth pressure state. Common Pitfalls: Ignoring drainage and water table effects, or confusing active/passive with at-rest (K0) conditions when the wall does not move. Final Answer: Active earth pressure

Question 14

Clay Consistency – Toughness index (It) of clayey soils Select the correct expression.

Accepted Answer

Introduction / Context: The toughness index It measures the resistance of a clay to deformation near the plastic limit by relating plasticity to the rate of change of consistency. It is useful in differentiating clays of similar plasticity but different workability. Given Data / Assumptions: Plasticity index PI = LL − PL. Flow index IF from the slope of the flow curve in the liquid limit test. All limits expressed in percentage terms consistently. Concept / Approach: Toughness index is defined as It = PI / IF. A higher It indicates a tougher clay exhibiting greater resistance to shearing near the plastic limit region. Step-by-Step Solution: 1) Determine PI from liquid and plastic limits.2) Determine IF from flow curve of LL test.3) Compute It = PI / IF and interpret: larger It → tougher clay. Verification / Alternative check: Comparing clays with equal PI but different IF demonstrates that lower IF (steeper consistency change) increases It, reflecting greater toughness. Why Other Options Are Wrong: Options (b), (c), and (d) mix unrelated indices; only PI/IF is the standard definition of toughness index. Common Pitfalls: Confusing toughness index with liquidity index or consistency index, or using inconsistent units for limits. Final Answer: Plasticity index / Flow index

Question 15

Fluid Mechanics – Lift on a body in a flowing fluid Identify the correct description of the lift force acting on an immersed body.

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Introduction / Context: When a body is immersed in a moving fluid, it experiences hydrodynamic forces due to pressure and viscous effects. These forces are decomposed into drag (parallel to the approach flow) and lift (normal to the approach flow). Understanding the lift definition is fundamental for aerodynamics and hydraulic structures. Given Data / Assumptions: Steady external flow around a body. Approach velocity defines the reference direction for drag/lift components. Buoyancy (hydrostatic) considered separately from dynamic forces. Concept / Approach: Lift is the resultant dynamic force component normal to the free-stream direction, generated by asymmetric pressure distribution and circulation around the body. Drag is the parallel component. Buoyancy acts vertically upward independent of flow and is not “lift” in this context. Step-by-Step Solution: 1) Resolve total hydrodynamic force into orthogonal components relative to the flow direction.2) Identify the component normal to the approach velocity → this is lift.3) Recognize contributing mechanisms: angle of attack, camber, boundary layer behavior, and circulation. Verification / Alternative check: For symmetrical bodies at zero angle of attack, lift ≈ 0; tilting or cambering the body produces a non-zero pressure difference and hence lift, confirming the definition. Why Other Options Are Wrong: (a) Buoyancy is hydrostatic, not part of dynamic lift. (b) Lift direction need not oppose gravity; it is defined normal to flow, which may be upward, downward, or lateral. (c) Wake contributes to drag and unsteady forces; lift is not caused “only” by the wake. Common Pitfalls: Confusing aerodynamic lift with buoyancy, or defining lift relative to vertical instead of the approach velocity direction. Final Answer: The dynamic fluid-force component normal to the approach velocity

Question 16

Shallow foundations on cohesionless soil — compare ultimate bearing capacities Two footings, one circular and one square, rest on the surface of a purely cohesionless soil. The diameter of the circular footing equals the side length of the square fo…

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Introduction / Context: Ultimate bearing capacity for shallow foundations in cohesionless (c = 0) soil at ground surface depends strongly on footing shape. Terzaghi's general bearing capacity expression uses shape factors that differ for circular and square footings. This question checks whether you recall the relative magnitudes of the shape coefficients for the N_gamma term, which governs bearing for c = 0 at D_f = 0. Given Data / Assumptions: Soil is purely cohesionless; c = 0 and φ > 0. Footings are at the ground surface; surcharge term q = 0. Diameter of circular footing = side of square footing = B. Ultimate capacity governed by the N_gamma term. Concept / Approach: For D_f = 0 and c = 0, qult ≈ 0.5 * γ * B * N_gamma * s_gamma, where s_gamma is the shape factor. Commonly used values: s_gamma ≈ 0.4 for a square footing and ≈ 0.3 for a circular footing (strip footing baseline is 0.5). Therefore the ratio (circular / square) reduces to 0.3 / 0.4. Step-by-Step Solution: 1) Write qult (square) = 0.4 * γ * B * N_gamma.2) Write qult (circular) = 0.3 * γ * B * N_gamma.3) Take ratio = (0.3)/(0.4) = 0.75.4) Express as a fraction → 3/4. Verification / Alternative check: As shape becomes more compact (circular), the effective contribution of the N_gamma term decreases compared to square of the same breadth; many design handbooks list exactly these multipliers. Why Other Options Are Wrong: 4/3 and 1.3 incorrectly imply the circular footing has a larger capacity than the square for the same breadth. 1.0 ignores shape effects. Common Pitfalls: Confusing area-based effects with shape factors; applying strip footing coefficients to square or circular footings. Final Answer: 3/4

Question 17

Combined action in fasteners — interaction limit for shear and tension Rivets or bolts subjected simultaneously to design shear stress (tau_vf,cal) and design axial tensile stress (sigma_tf,cal) must satisfy the interaction requirement. The permissi…

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Introduction / Context: Steel connections often subject a single fastener to both shear and tension due to eccentricities or prying forces. Codes control this by an interaction check so that the combined utilization in shear and tension does not exceed the allowable envelope. Given Data / Assumptions: A fastener experiences concurrent shear and tension. Allowable design stresses are tau_vf (shear) and sigma_tf (tension). Interaction is evaluated using a code-specified expression. Concept / Approach: A common interaction form is of the type (tau_vf,cal / tau_vf) + (sigma_tf,cal / sigma_tf) ≤ 1.0 or a similar squared-sum form bounded by 1.0. Regardless of the exact expression used, the non-dimensional combined utilization must not exceed unity for safety. Step-by-Step Solution: 1) Compute the demand ratios in shear and in tension with respect to their allowable values.2) Substitute into the prescribed interaction formula.3) Check that the evaluated interaction ≤ 1.0.4) If > 1.0, redesign by increasing capacity or reducing demand. Verification / Alternative check: Design examples in steel manuals consistently use unity as the limiting combined utilization threshold. Why Other Options Are Wrong: 1.2, 1.4, and 1.8 would permit overstress; interaction checks are capped at unity by definition. Common Pitfalls: Ignoring prying action in tension; double-counting or omitting load combinations; using nominal instead of design allowables. Final Answer: 1.0

Question 18

Field compaction — best soil type for vibratory rollers Vibration-based compaction is most effective for which of the following soil conditions?

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Introduction / Context: Different compaction equipment suits different soil types because the mechanism of densification varies. Vibratory rollers induce particle rearrangement by reducing intergranular friction momentarily, which favors free-draining granular soils. Given Data / Assumptions: Equipment: vibratory roller. Target: achieve high relative density efficiently. Soil choices include granular and cohesive categories. Concept / Approach: Granular soils (sands and gravels) compact primarily by particle rearrangement. Vibration is highly effective, especially for well-graded dry or slightly moist sands where pore pressures do not build up. Cohesive soils (silts/clays), particularly plastic or highly compressible silts, respond better to kneading/sheepsfoot rollers due to their low permeability and structure. Step-by-Step Solution: 1) Identify mechanism: vibration reduces interparticle friction and mobilizes rearrangement.2) Match soil: well graded dry sand has broad size distribution and drains freely.3) Expect rapid density gains and high compaction energy transfer with vibratory rollers on sands.4) Therefore, select well graded dry sand as the best match. Verification / Alternative check: Compaction charts in field manuals rank vibratory rollers as most suitable for sands/gravels and least for clays/silts. Why Other Options Are Wrong: Moist silty sand may respond, but silt fraction dampens vibration; clays and high-compressibility silts require kneading and have drainage issues. Common Pitfalls: Over-wetting sands, which causes pumping; using vibratory rollers on saturated silts leading to instability. Final Answer: well graded dry sand

Question 19

Buckling of a cantilevered circular electric pole An electric pole of length L and constant diameter is erected in the ground and behaves as a cantilever. The effective length for buckling is:

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Introduction / Context: Effective length represents the distance between inflection points of the buckled shape. For column design, the effective length coefficient K depends on end restraints. A free-top, fixed-base electric pole approximates a classic cantilever boundary condition. Given Data / Assumptions: Pole with one end fixed in ground, other end free. Uniform diameter and material along length. Euler buckling framework is applicable. Concept / Approach: For a cantilever, the buckling half-wavelength spans from a point of zero moment (free end) to a maximum moment at the fixed end. The effective length factor is K = 2.0, so LE = K * L. Step-by-Step Solution: 1) Identify end conditions: fixed–free.2) Use K = 2.0 for cantilever.3) Compute effective length LE = 2.0 * L.4) Select 2.00 L as the correct value. Verification / Alternative check: Euler load P_cr = pi^2 * E * I / (K * L)^2; using K = 2 correctly predicts the lowest critical load for a cantilever. Why Other Options Are Wrong: 0.80 L, 1.20 L, and 1.50 L correspond to other end conditions (e.g., fixed–fixed, fixed–pinned); they do not model a cantilevered pole. Common Pitfalls: Assuming partial fixity or soil flexibility without justification; for embedded poles, p–y effects may modify K, but the textbook answer for a cantilever is K = 2.0. Final Answer: 2.00 L

Question 20

Roundabout capacity by weaving formula (IRC concept) Given: average entry width e = 8.4 m, weaving width w = 14 m, weaving length l = 35 m, crossing traffic = 1000 PCU/h, total traffic on weaving section = 2000 PCU/h. Estimate the capacity of the ro…

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Introduction / Context: At unsignalized roundabouts, the practical capacity of the weaving section can be estimated using the IRC-style empirical weaving formula. It accounts for geometry (entry width, weaving width, length) and traffic composition (weaving proportion). Given Data / Assumptions: Average entry width e = 8.4 m. Weaving width w = 14 m. Weaving length l = 35 m. Total traffic on weaving section = 2000 PCU/h. Crossing (weaving) traffic = 1000 PCU/h → proportion p = 1000/2000 = 0.5. Concept / Approach: IRC weaving capacity formula (one common form): Q = 280 * w * (1 + e/w) * (1 - p/3) / (1 + w/l), where Q is in PCU/h, w and e in metres, l in metres, and p is the proportion of weaving traffic. Step-by-Step Solution: 1) Compute e/w = 8.4 / 14 = 0.6 → (1 + e/w) = 1.6.2) Compute w/l = 14 / 35 = 0.4 → denominator (1 + w/l) = 1.4.3) Compute (1 - p/3) with p = 0.5 → 1 - 0.5/3 = 1 - 0.1667 = 0.8333.4) Q = 280 * 14 * 1.6 * 0.8333 / 1.4 ≈ 3730 PCU/h → nearest option 3700. Verification / Alternative check: Magnitude consistency: larger w, longer l, and lower p increase capacity; here values yield capacity in the mid-3,000s PCU/h, which is reasonable for the geometry. Why Other Options Are Wrong: 3300 underestimates the computed value; 4500 and 5200 are too high for the given geometry and weaving proportion. Common Pitfalls: Using wrong p (must be weaving proportion); mixing up units; rounding too early. Final Answer: 3700

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