Question 1

Prestressing systems identification in building construction: The system that uses high-tensile alloy steel bars (silico-manganese bars) as prestressing tendons, tightened by nuts and anchor plates, is known as which of the following?

Accepted Answer

Introduction / Context: Prestressed concrete can be executed using several proprietary or classical systems. Each system is distinguished by the type of tendon (wire, strand, or bar), the anchorage arrangement, and the stressing method. Recognizing these systems is a common exam objective in civil engineering and structural design. Given Data / Assumptions: The tendon material in question is a high-tensile alloy steel bar, often described as silico-manganese steel. The question asks for the name of the system that specifically employs such bars with mechanical anchorage (nuts/anchor plates). No numerical calculations are required; this is a factual classification problem. Concept / Approach: Classic systems include: Freyssinet (multiple high-tensile wires with conical wedges), Magnel–Blaton (grouped wires with sandwich plates), C.C.L. systems (strands/wires with wedge anchors), and Lee–McCall (high-tensile alloy steel bars stressed by nuts against anchor plates). Matching the tendon type to the system identifies the correct choice. Step-by-Step Solution: Identify tendon: high-tensile alloy steel bars (not wires/strands).Map to system: bar tendons with nut-and-plate anchorage correspond to Lee–McCall.Confirm: other systems listed mainly use wires/strands with wedge grip anchors.Select answer: Lee–McCall system. Verification / Alternative check: Reference descriptions of historical prestressing methods consistently attribute bar-tendon, nut-anchored techniques to Lee–McCall, whereas Freyssinet and Magnel–Blaton are wire/strand-based systems. Why Other Options Are Wrong: Freyssinet system: uses multiple wires with conical wedge anchorages, not alloy bars. Magnel–Blaton system: sandwich plate anchor with grouped wires; not alloy bars. C.C.L. standard system: wedge-grip anchors for strands/wires; not a bar-with-nut system. Common Pitfalls: Confusing bar tendons with strand/wire systems, or assuming anchor plates imply any system. The key identifier here is the bar tendon stressed by a nut. Final Answer: Lee–McCall system

Question 2

Prestress losses by creep and shrinkage (working-stress viewpoint): If C is the creep coefficient of concrete, f is the original concrete stress at tendon level, m is modular ratio, E is the Young’s modulus of steel, and e is the shrinkage strain of…

Accepted Answer

Introduction / Context: In prestressed concrete, time-dependent effects like creep and shrinkage of concrete reduce the effective prestress in steel. Analytical formulae express the change in steel stress in terms of concrete parameters and modular ratio. Understanding the signs and relative contributions of creep and shrinkage is fundamental in estimating long-term losses. Given Data / Assumptions: C = creep coefficient of concrete referred to the initial concrete stress at tendon level. f = initial concrete stress (compression) at tendon level due to prestress and loading. m = modular ratio = Es / Ec. E = Young’s modulus of steel; e = free shrinkage strain of concrete (shortening). Working-stress treatment, linear elastic superposition, no secondary effects considered. Concept / Approach: Creep under sustained compression increases concrete strain beyond the instantaneous elastic value, causing steel to relax (loss of steel stress). Shrinkage is a free uniform contraction of concrete; compatibility through bond imposes shortening on steel, again reducing steel stress. Sign convention: additional concrete shortening implies a reduction in steel stress (negative contribution for shrinkage term). Step-by-Step Solution: Instantaneous steel stress linked to concrete stress: m f (compatibility/transformations).Creep addition to concrete strain proportional to (C - 1), hence steel stress change term involves (C - 1) m f.Shrinkage causes a direct steel stress change of magnitude E * e but as a loss (opposes initial prestress), so with negative sign.Combined expression: Δσs = (C - 1) m f - e E. Verification / Alternative check: Both mechanisms reduce effective steel stress: creep term depends on existing concrete stress (through m f), and shrinkage term depends on the free strain e multiplied by Es. The signs therefore align with a net loss in steel stress. Why Other Options Are Wrong: (1 - C) m f ± e E: Reverses creep sign; would imply gain if C > 1. (C - 1) m f + e E: Assigns a positive sign to shrinkage; contradicts loss nature. Common Pitfalls: Mixing up whether the expression represents “loss” (drop in steel stress) or “remaining stress”; always check sign conventions. Also, ensure e refers to concrete shrinkage strain (shortening), not thermal expansion. Final Answer: (C - 1) m f - e E

Question 3

Nominal shear stress in an R.C.C. beam: The maximum nominal shear stress in the concrete web for design is commonly evaluated as which of the following expressions, where Vu is the design shear force, b is the breadth (web width), and d is the effec…

Accepted Answer

Introduction / Context: Shear design in reinforced concrete beams starts by computing the nominal shear stress in the concrete web. This value is compared with permissible concrete shear stress and shear capacity of stirrups to ensure adequate resistance against diagonal tension cracking and web shear failure. Given Data / Assumptions: Vu = factored (design) shear force at the section under consideration. b = breadth of web (for flanged beams, the web width). d = effective depth to tension steel. Nominal shear stress symbol τv used for checks against τc and limits. Concept / Approach: Nominal shear stress is the shear force per unit area of the web resisting shear, approximated by b * d. This assumes the shear is primarily carried by the concrete web around the centroidal region and provides a consistent basis for code checks and stirrup design. Step-by-Step Solution: Compute Vu at the section (factored).Compute resisting area ≈ b * d.Evaluate τv = Vu / (b * d).Compare τv with code-permitted concrete shear stress; design stirrups if τv exceeds concrete capacity. Verification / Alternative check: This simple rectangular-block model aligns with standard code procedures, serving as the basis for minimum shear reinforcement and detailed stirrup design. Why Other Options Are Wrong: Vu/(bb) or Vu/(dd): use incorrect resisting areas. 2Vu/(bd): applies an unjustified factor, over-estimating stress. Common Pitfalls: Using flange width instead of web width for T/L beams; forgetting to use factored shear; neglecting the effect of axial force or prestress on shear checks. Final Answer: τv = Vu / (b * d)

Question 4

Pile load test acceptance criterion: During a routine pile load test, the platform is loaded up to 1.5 times the design load. After unloading, the net settlement (gross settlement minus rebound) per tonne of test load shall not exceed which value fo…

Accepted Answer

Introduction / Context: Pile load tests verify capacity and serviceability. A simple acceptance criterion uses the net settlement per unit of applied test load to judge whether the pile exhibits acceptable stiffness and creep characteristics. Limiting net settlement helps control serviceability issues in the supported structure. Given Data / Assumptions: Pile is loaded to 1.5 times the design (working) load. Settlement during loading and rebound on unloading are measured. Net settlement per tonne of test load is assessed for acceptability. Concept / Approach: By subtracting elastic rebound from total settlement, the residual (plastic and time-dependent) component is isolated. A lower allowable net settlement per tonne indicates a stiffer pile–soil system. Traditional acceptance guidelines adopt a relatively stringent limit around 10 mm per tonne of test load for routine acceptance, especially in granular soils or where settlement control is critical. Step-by-Step Solution: Conduct incremental loading up to 1.5 × design load.Record total settlement at each stage; then unload to observe rebound.Compute net settlement = total settlement − rebound.Normalize by test load (tonnes) to obtain mm/tonne and check against the limit. Verification / Alternative check: Comparing with established practice and many project specifications, the 10 mm per tonne criterion is commonly adopted for routine compression tests; stricter limits may be specified for sensitive structures. Why Other Options Are Wrong: 15–30 mm: These larger limits risk excessive long-term settlement and are generally less conservative for routine acceptance. Common Pitfalls: Confusing total settlement with net; ignoring time-dependent settlement during maintained load; not accounting for test load increments when normalizing. Final Answer: 10 mm

Question 5

Tendon profile for uniformly distributed load: For a prestressed rectangular beam carrying a uniformly distributed load over its span, which tendon profile best counterbalances the sagging bending moment diagram for economy and serviceability?

Accepted Answer

Introduction / Context: In prestressed concrete, tendon profiles are shaped so the line of action of prestress (pressure line) counteracts the bending moments from external loads. For a uniformly distributed load (UDL), the bending moment diagram is parabolic, so an appropriately parabolic tendon profile is most efficient. Given Data / Assumptions: Rectangular beam section under UDL. Objective is to balance moments to minimize tensile stresses and deflection. Standard bonded tendon behavior; no special end blocks considered. Concept / Approach: The pressure line locus should mirror the external moment curve for best balancing. A UDL causes maximum sagging at midspan and zero at supports. Hence the tendon should have maximum eccentricity at midspan and minimal eccentricity near supports. Geometrically, this is a parabolic profile that is concave up (convexity upward), placing the tendon lowest at midspan to produce a counteracting upward moment. Step-by-Step Solution: Identify moment shape from UDL: parabolic sagging, peak at midspan.Select tendon profile that gives maximum downward eccentricity at midspan: parabolic concave up.Confirm anchor elevations: tendon rises toward supports to reduce eccentricity near zero-moment zones.Conclude: a parabolic tendon with convexity upward is appropriate. Verification / Alternative check: Balancing load method shows that a parabolic tendon can be proportioned so that prestress induces an equivalent upward uniform load, effectively offsetting the UDL over the span. Why Other Options Are Wrong: Straight profiles cannot match the parabolic moment curve efficiently. Parabolic with convexity downward creates the opposite eccentricity trend, increasing tension at midspan. Common Pitfalls: Confusing “convex up” with “convex down.” Remember: for UDL, tendon is lowest at midspan and higher at supports. Final Answer: Parabolic with convexity upward

Question 6

Span-to-depth limit for two-way simply supported slabs: What is the maximum basic ratio of span to effective depth commonly adopted for slabs that are simply supported and span in two directions under normal service conditions?

Accepted Answer

Introduction / Context: Control of deflection is essential in slab design. Codes and design guides specify span-to-depth limits as a quick serviceability check prior to detailed deflection calculations. Two-way slabs supported on all four edges typically permit a higher span/depth ratio than one-way slabs due to two-directional stiffness. Given Data / Assumptions: Simply supported slab spanning in two directions. Normal serviceability, typical finishes and imposed loads. Basic (unmodified) ratio prior to applying adjustment factors for tension steel, drop panels, etc. Concept / Approach: Empirical limits (basic l/d ratios) are provided for preliminary sizing. For two-way simply supported slabs, a basic limit of about 35 is widely used. Adjustments may be applied later based on percentage of tension reinforcement, concrete grade, and other parameters. Step-by-Step Solution: Identify slab system: two-way action on simple supports.Adopt the standard basic ratio: l/d ≈ 35.Apply modification factors in detailed design if needed; ensure deflections are within service limits. Verification / Alternative check: Design examples confirm that two-way systems can span longer per unit depth than one-way due to bidirectional stiffness. The figure of 35 is commonly referenced as a starting point. Why Other Options Are Wrong: 25 or 30: conservative for two-way slabs; more typical for beams/one-way slabs. 40: may be excessive without favorable modification factors. 15: unrealistically low for standard RC slabs. Common Pitfalls: Assuming basic limits are final; always check deflection especially with long spans, high service loads, or low reinforcement ratios. Final Answer: 35

Question 7

Working-stress “moment of resistance factor” for M150 (1:2:4) concrete: For concrete of nominal grade M150 (mix 1:2:4), which of the following is the typical moment of resistance factor used in classic working-stress design tables?

Accepted Answer

Introduction / Context: Older working-stress design (WSD) methods tabulate a “moment of resistance factor” for common concrete mixes, enabling rapid preliminary sizing of plain and reinforced members. Although limit-state design is prevalent today, many exam questions still reference these WSD values. Given Data / Assumptions: Nominal concrete grade M150 corresponding approximately to mix 1:2:4. Use within classic WSD tables for preliminary/educational calculations. No limit-state partial safety factors are applied here. Concept / Approach: These tabulated factors encapsulate allowable stresses and section properties to simplify hand calculations. For M150 concrete, a commonly cited factor is about 7.5 (in appropriate consistent units as used in those tables), enabling quick estimation of section capacity before detailed checks. Step-by-Step Solution: Identify grade: M150 (1:2:4).Consult/WSD convention: standard factor ≈ 7.50 for this grade.Select value: 7.50. Verification / Alternative check: Numerous traditional textbooks list the factor 7.5 for M150; it aligns with the allowable stress set used in WSD tables. Why Other Options Are Wrong: 0.87: Not a WSD moment factor; it is typically the steel stress factor used in LSM (0.87fy). 5.80 or 8.50: Values associated with different mix strengths/assumptions, not the standard for M150. Common Pitfalls: Applying WSD factors to limit-state design without conversion; always be consistent with the design philosophy and units. Final Answer: 7.50

Question 8

Purpose of lateral (transverse) reinforcement in R.C.C. columns: Which statement best captures a correct role of lateral reinforcement such as ties or spirals provided in reinforced concrete columns?

Accepted Answer

Introduction / Context: Lateral reinforcement (ties or spirals) in R.C.C. columns provides confinement to the core concrete and restrains the main longitudinal bars. Proper detailing of ties is crucial for ductility, shear resistance, and to prevent bar instability under compression. Given Data / Assumptions: Conventional reinforced concrete building columns. Standard tie detailing as per common codes. Focus on the primary structural role of lateral reinforcement. Concept / Approach: Under axial load and bending, longitudinal bars can buckle between tie nodes if not adequately restrained. Ties maintain bar spacing, prevent outward buckling, provide confinement (enhancing concrete strength/ductility), and assist in shear resistance via truss action around the core. Step-by-Step Solution: Identify the principal function: restrain longitudinal bars and confine concrete.Recognize secondary benefits: improved shear performance, delay of cover spalling.Check minimum diameters: many codes require ≥ 6 mm or ≥ 8 mm depending on steel type and bar size; 5 mm is usually below code minima.Therefore, select the clearly correct core statement (option a). Verification / Alternative check: Design standards emphasize prevention of bar buckling as the primary motivation for ties; confinement and shear capacity are well-documented additional advantages. Why Other Options Are Wrong: b and c: Partially true as secondary effects, but the question asks for a correct role; (a) is the clearest primary function. d: “Not less than 5 mm” conflicts with common minima (often 6 mm or 8 mm), hence is incorrect as a universal rule. e: Since (d) is incorrect, “All the above” is also incorrect. Common Pitfalls: Assuming minimal tie diameter can be as low as 5 mm; ignoring code-specific minima and spacing rules can compromise column ductility and strength. Final Answer: Lateral reinforcement in R.C.C. columns is provided to prevent the longitudinal reinforcement from buckling

Question 9

Reinforced concrete stairs (longitudinally spanning): If the flights span longitudinally and are cast monolithically with their landings (i.e., the slab is continuous over landings), the maximum bending moment per metre width is taken as which of th…

Accepted Answer

Introduction / Context: This question tests the structural design convention for stairs that span longitudinally and are cast integrally with landings. When landings act as supports, the stair slab behaves like a continuous slab, which reduces the midspan bending moment compared to a simply supported case. Given Data / Assumptions: Stair slab spans longitudinally between landings/walls. Flights and landings are cast monolithically (continuity provided). Uniformly distributed load per metre width = w. Clear span between effective supports = l. Concept / Approach: For simply supported slabs, the design midspan moment is w l^2 / 8. For continuous slabs with adequate end fixity or continuity, recognized coefficients reduce the midspan moment. A common conservative coefficient for a continuous slab/flight over landings is w l^2 / 10 for the maximum positive moment in the span, acknowledging continuity without performing a full frame analysis. Step-by-Step Solution: Model the stair slab as a continuous member over supports at landings.Adopt the standard moment coefficient for a continuous slab: M_max,span ≈ w l^2 / 10.Use this value per metre width for reinforcement design in the flight. Verification / Alternative check: Where more accuracy is needed, a continuous-beam analysis (e.g., moment distribution) can be done; the resulting span moments will typically be close to or lower than the w l^2 / 10 coefficient, confirming the conservatism of the rule-of-thumb. Why Other Options Are Wrong: w l^2 / 8: Suits simply supported behavior, unconservative for continuous action.w l^2 / 12 or /16: Too small for general continuous stair design without a full analysis.w l^2 / 2: Corresponds to cantilever case; not applicable here. Common Pitfalls: Designing as simply supported despite monolithic landings (overestimates steel at midspan and underestimates negative moments at supports). Ignoring effective span definition and support conditions at landings. Final Answer: w l^2 / 10

Question 10

Rebar detailing in tension zones: Lapped splices in tensile reinforcement are generally avoided for bars larger than which nominal diameter, as per common RC detailing practice (e.g., IS 456 recommendations)?

Accepted Answer

Introduction / Context: Lapped splices transfer force via bond between overlapping bars. In high-tension zones with large bar diameters, long lap lengths and heavy congestion can compromise constructability and quality. Design codes therefore discourage laps for large-diameter tension bars and recommend welded or mechanical splices instead. Given Data / Assumptions: Tension reinforcement (not compression). Nominal bar diameters following standard metric sizes. General practice aligned with IS 456 recommendations. Concept / Approach: For tension bars greater than a code-specified cutoff diameter, lap splicing is generally not preferred. For IS 456, this practical limit is 36 mm for tension bars. Beyond this, welding or mechanical couplers are recommended to ensure reliable force transfer and reduce congestion. Step-by-Step Reasoning: Identify that the context is tensile laps, not compression.Recall the recommended maximum bar diameter for lapping in tension (≈ 36 mm).Conclude that for bars larger than 36 mm, avoid lap splicing and adopt welded/mechanical splices. Verification / Alternative check: Project specifications often reiterate this practice because couplers improve site handling, reduce reinforcement congestion, and enhance quality control in highly stressed regions. Why Other Options Are Wrong: 18/24/30/32 mm: Commonly lapped in practice; no general code prohibition for these diameters. Common Pitfalls: Using long laps for 40 mm bars in tension zones, causing congestion and poor compaction. Not verifying the splice location (avoid peak moment regions). Final Answer: 36 mm diameter

Question 11

Two-way slabs with corners held down (IS 456 design strips): According to IS 456, a slab spanning in two directions with corners held down is divided, in each direction, into a middle strip and two edge strips. What is the width of the middle strip …

Accepted Answer

Introduction / Context: For two-way slabs, bending moments are distributed between a central “middle strip” and adjacent “edge strips.” IS 456 provides conventional strip widths to simplify distribution of design moments and reinforcement, especially when corners are held down to prevent uplift. Given Data / Assumptions: Two-way slab. Corners held down. Use of IS 456 strip method. Concept / Approach: IS 456 divides the slab, in each direction, into a central middle strip of three-fourths of the panel width and two edge strips (one on each side) making up the remaining one-fourth (i.e., one-eighth each). This aids moment partitioning and bar placing without a full elastic analysis. Step-by-Step Reasoning: In any one direction, total width = 1.00.Middle strip width = 3/4 of the panel width.Edge strips = 1/8 on each side (total 1/4). Verification / Alternative check: Using these conventional widths, the detailing becomes straightforward: heavier steel in the middle strip (where moments are higher) and reduced reinforcement in edge strips, consistent with two-way action. Why Other Options Are Wrong: Half, two-third, four-fifth, three-fifth: Not the standard fractions used in IS 456 for corners-held-down slabs. Common Pitfalls: Confusing corners “held down” vs “free to lift,” which affects moment coefficients and detailing. Misallocating steel between middle and edge strips. Final Answer: three-fourth of the width of the slab

Question 12

Minimum thickness criteria for flat slabs: Which of the following thickness rules apply to flat slabs under normal design practice (choose the most comprehensive statement)?

Accepted Answer

Introduction / Context: Flat slabs are two-way slabs supported directly by columns, sometimes with column heads and drops. Thickness must satisfy both a minimum absolute value for robustness and span-to-depth limits for deflection control, with different limits for interior and edge panels and for the presence/absence of drops. Given Data / Assumptions: General building-type flat slabs. Common rules of thumb and code-based span/depth limits. Concept / Approach: A typical minimum absolute thickness is about 125–130 mm; 13 cm (130 mm) is a common conservative minimum in practice. Additionally, deflection control is checked via effective span-to-depth ratios such as L/32 or L/36, with specific values depending on whether the panel is interior or at the edge and whether drops are present. Thus, multiple criteria simultaneously apply; the governing requirement is the most stringent among them. Step-by-Step Reasoning: Check absolute minimum thickness (≈ 13 cm as a practical floor).Apply span/depth ratios (e.g., L/32, L/36) depending on panel location and detailing.Select the largest thickness demanded by any controlling criterion. Verification / Alternative check: Even after meeting these limits, perform serviceability checks (immediate and long-term deflection). Shear (punching) near columns must also be verified, which can require further local thickening or shear reinforcement regardless of the basic slab thickness. Why Other Options Are Wrong: Each single option represents only one criterion. In practice, designers must comply with all applicable checks, hence the comprehensive answer is the inclusive choice. Common Pitfalls: Using only an absolute minimum without deflection verification. Ignoring changes between interior and edge panels or the presence of drops. Final Answer: all the above

Question 13

Position of neutral axis in a T-beam: Depending on flange and web proportions and loading, where can the neutral axis (NA) lie in a reinforced concrete T-beam section?

Accepted Answer

Introduction / Context: The neutral axis location in a T-beam depends on the relative stiffness of the flange and web and the magnitude of compressive force to be carried at ultimate or working conditions. Understanding possible NA positions is central to deciding whether the flange is fully or partially effective. Given Data / Assumptions: T-beam comprised of slab flange and web (rib). Bending causing compression on the slab (top) side. Concept / Approach: If compressive force can be accommodated entirely within the flange, the NA lies within the flange. When compressive demand increases, the NA can drop to the flange–web junction or even into the web (below the slab), indicating that both flange and part of the web are participating in compression. Step-by-Step Reasoning: Small compression demand → NA within flange (flange alone resists).Moderate demand → NA near the flange bottom.High demand → NA inside the web below slab level. Verification / Alternative check: In design, trial NA positions are assumed and section capacity is computed. The final position is the one that balances internal compression and tension to match the applied bending moment. Why Other Options Are Wrong: Each isolated location can occur; therefore the inclusive option correctly captures all possible NA positions. Common Pitfalls: Assuming NA is always within flange; not true for heavy moments or narrow flanges. Ignoring effective flange width limitations per code. Final Answer: all the above

Question 14

Minimum clear spacing for horizontal parallel bars (same diameter): When horizontal reinforcement bars of the same size are placed side-by-side in one layer, the minimum clear spacing should not be less than…

Accepted Answer

Introduction / Context: Minimum bar spacing requirements ensure adequate concrete flow, proper compaction, and reliable bond. Insufficient spacing can cause honeycombing, poor aggregate accommodation, and reduced development of reinforcement strength. Given Data / Assumptions: Bars are of equal diameter placed in the same layer. We are addressing minimum clear spacing (not center-to-center spacing). Concept / Approach: A common rule in RC codes is that clear spacing between parallel bars shall be at least the bar diameter, while also not being less than a value related to the maximum aggregate size (e.g., 5 mm more than nominal maximum aggregate size) and a small absolute minimum (e.g., 25 mm) where applicable. For same-size bars, the bar-diameter criterion is the primary rule cited in such objective tests. Step-by-Step Reasoning: Identify the governing minimum: clear spacing ≥ bar diameter.Check aggregate size requirement separately during detailed design. Verification / Alternative check: Site experience confirms that spacing less than one diameter leads to poor concrete flow and increased risk of voids, which is unacceptable for durability and bond. Why Other Options Are Wrong: Values such as 2.5d, 3d, etc., may be used in specific detailing or bundled bars, but the basic minimum for same-diameter parallel bars in a layer is one diameter. Common Pitfalls: Confusing clear spacing with centre-to-centre spacing. Neglecting aggregate size checks that can govern in some mixes. Final Answer: one diameter

Question 15

Reducing load-induced stress in an RC section: Which practical detailing change most effectively reduces the stress in the section under a given load, without altering member dimensions?

Accepted Answer

Introduction / Context: For a given member size and external load, detailing choices affect stress distribution, crack width control, and bond performance. Using many smaller bars instead of a few large bars spreads the reinforcement more uniformly and improves bond, which reduces peak steel stress and crack widths for the same total steel area. Given Data / Assumptions: Member geometry unchanged. Total steel area can be maintained while changing bar sizes and count. Load and boundary conditions unchanged. Concept / Approach: Stress in steel ≈ internal force / steel area. With more, smaller-diameter bars, the steel is distributed closer to the tensile face, increasing the effective lever arm slightly and providing better crack control through more uniform strain distribution and greater perimeter for bond. This combination reduces peak stress demand in individual bars and improves serviceability. Step-by-Step Reasoning: Keep total As approximately constant but distribute it among more bars.Better distribution → improved crack control and reduced peak bar stress.Often a small increase in effective lever arm results due to placement flexibility. Verification / Alternative check: Design guides for crack control favor smaller bars at closer spacing because maximum crack width is proportional to bar spacing and steel strain. Thus, swapping a few large bars for many small bars is beneficial for reducing stress concentrations and crack width. Why Other Options Are Wrong: Decreasing lever arm increases internal force demand in steel for equilibrium.Increasing perimeter alone (option b) is a partial explanation; option c operationalizes it effectively.Using larger bars (option d) concentrates force and worsens crack control. Common Pitfalls: Overcrowding bars; maintain minimum spacing and cover. Ignoring development length requirements when changing bar sizes. Final Answer: replacing larger bars by a greater number of smaller bars

Question 16

Circular slab under uniformly distributed load: If a simply supported circular slab of radius R carries a total uniformly distributed load W (i.e., load intensity w = W / (π R^2)), what is the maximum radial bending moment at the centre of the slab …

Accepted Answer

Introduction / Context: For circular slabs supported along the circumference and subjected to a uniformly distributed load, classic plate theory gives closed-form coefficients for bending moments. The peak radial moment occurs at the centre; knowing this coefficient is essential for reinforcement design in the radial direction. Given Data / Assumptions: Circular slab of radius R. Simply supported at the edge. Total uniformly distributed load W over the area (so w = W / (π R^2)). Concept / Approach: For a simply supported circular slab under UDL w (force per unit area), the maximum radial moment at the centre is given by M_r,max = (3 w R^2) / 16 per unit width. Substituting w = W / (π R^2) simplifies the expression in terms of total load W. Step-by-Step Solution: Start with M_r,max = (3 w R^2) / 16.Use w = W / (π R^2).Then M_r,max = (3 / 16) * (W / (π R^2)) * R^2 = 3 W / (16 π). Verification / Alternative check: Dimensional consistency: W has dimensions of force; dividing by π gives force; the coefficient 3/16 is dimensionless; the result is bending moment per unit width, which is in force–length units once the per-unit-width interpretation is acknowledged from plate theory conventions. Why Other Options Are Wrong: W/π and W/(2π): Miss the correct coefficient; overestimate moment.W/(8π): Undercounts by a factor of 3/2.W/(4π): Also incorrect coefficient. Common Pitfalls: Confusing simply supported with clamped edges (different coefficients). Mixing up total load W and intensity w. Final Answer: 3 W / (16 π)

Question 17

Continuous floor beams supporting slabs: For reinforced concrete floor beams carrying slab loads, what is a typical recommended ratio of end span to intermediate span lengths in a multi-span continuous system?

Accepted Answer

Introduction / Context: In continuous beam-and-slab systems, the end spans often differ from interior spans to achieve favorable moment distributions, deflection control, and practical grid layouts. A conventional proportion is used in many building grids as a guide when exact architectural constraints permit. Given Data / Assumptions: Reinforced concrete floor beams continuous over several supports. Span ratio guideline sought, not a rigid code requirement. Concept / Approach: Keeping end spans a bit shorter than interior spans reduces negative support moments at the exterior support and balances sagging and hogging moments along the beam. A commonly adopted ratio is about 0.8 of the interior span for the end span, improving overall economy and serviceability. Step-by-Step Reasoning: Set interior span length = L.Choose end span ≈ 0.8 L for improved moment distribution.Verify deflections and reinforcement demands with standard analysis. Verification / Alternative check: Analytical checks using standard continuous-beam coefficients show that a slightly shorter end span reduces exterior negative moments and often leads to better utilization of section capacity. Why Other Options Are Wrong: 0.7 or 0.6: End span becomes too short relative to interior spans, which may be impractical architecturally.0.9: Very close to interior span; moment balancing benefit is reduced. Common Pitfalls: Assuming a fixed mandatory ratio; this is a guideline subject to architectural grid and analysis results. Final Answer: 0.8

Question 18

Stirrups spacing along a rectangular RC beam: How should the spacing of shear stirrups generally vary along the length of a simply supported rectangular beam under uniformly distributed load?

Accepted Answer

Introduction / Context: Shear force is highest near the supports and lowest at midspan for a simply supported beam with uniformly distributed load. Stirrups resist shear; therefore, their spacing should reflect the shear force diagram rather than remain constant along the length. Given Data / Assumptions: Simply supported beam. Uniformly distributed load. Design according to standard RC practice. Concept / Approach: Nominal shear stress v = V / (b d) is largest near supports because V is maximum there. To maintain capacity and economy, stirrup spacing is closer (smaller) where shear is high and can be relaxed (larger spacing) where shear is lower, i.e., towards midspan. Step-by-Step Reasoning: Plot shear diagram: V_max at supports; V = 0 at midspan.Detail stirrups with minimum spacing near supports, increasing spacing towards the centre.Comply with code minimum and maximum spacing limits in all regions. Verification / Alternative check: Running a shear check at critical sections confirms that the most demanding spacing is near supports. Midspan checks usually govern minimum code stirrup provisions only. Why Other Options Are Wrong: Constant spacing ignores varying shear.Decreasing spacing towards centre or increasing at ends contradicts the shear distribution.Option e is not a recognized pattern for UDL on simple spans. Common Pitfalls: Forgetting special checks near point loads or openings, where local shear may spike. Violating code maximum spacing, especially near supports. Final Answer: increased at the centre of the beam

Question 19

In prestressed concrete design, which material combination is most appropriate for achieving the intended precompression and crack control in members?

Accepted Answer

Introduction / Context: Prestressed concrete relies on precompression introduced by tensioned steel so that service loads do not cause tensile cracking. Selecting the correct combination of concrete and steel grades is fundamental to achieve adequate transfer of prestress, limit losses, and ensure durability under service conditions. Given Data / Assumptions: Member is prestressed prior to (pretensioning) or after (post-tensioning) casting. Goal is to keep concrete largely in compression under service load. Materials must resist high initial stresses and long-term losses. Concept / Approach: High-strength concrete provides higher compressive capacity, improved bond, reduced creep and shrinkage (thereby reducing losses), and better cracking resistance. High tensile steel (prestressing strands or wires) sustains high initial stress with low relaxation to produce the required precompression. The synergy of both is the foundation of prestressed systems. Step-by-Step Solution: Choose concrete with characteristic strength adequate for transfer and service (often ≥ 40 MPa) to resist compressive stress and reduce time-dependent losses.Use high tensile, low relaxation steel to anchor the prestress effectively while limiting long-term stress reduction.Detail anchorage, duct profile, and cover to maintain intended force and durability. Verification / Alternative check: Service stress checks confirm that with high-strength concrete and high tensile steel, tensile stresses remain limited or negative (compression) across critical sections, validating material choice. Why Other Options Are Wrong: Low- or high-strength concrete alone cannot provide prestress; steel is needed. Low-strength concrete increases creep/shrinkage losses. Low tensile steel cannot carry required prestress safely. Common Pitfalls: Underestimating relaxation of steel, using concrete that has not reached transfer strength, and neglecting environmental exposure leading to durability issues. Final Answer: High-strength concrete with high tensile steel.

Question 20

Ribbed (joist) slab detailing: which set of limits is appropriate for safe and code-aligned design of a ribbed slab system?

Accepted Answer

Introduction / Context: Ribbed (joist) slabs achieve economy by concentrating concrete in ribs while using a thin topping slab for distribution. Proper geometric limits on rib spacing, rib width, and overall depth relative to rib breadth are required to ensure stiffness, shear capacity, fire cover, and good load distribution. Given Data / Assumptions: Monolithic ribbed slab with regular ribs and a thin topping slab. Need typical detailing limits widely used in building codes and practice. Focus is on spacing, minimum rib width, and overall depth cap. Concept / Approach: Minimum rib width ensures adequate space for reinforcement and cover, reducing congestion and improving fire resistance. Limiting clear spacing controls two-way action in the topping and prevents excessive transverse bending. Limiting overall depth relative to rib breadth maintains a realistic rib slenderness and avoids brittle behavior. Step-by-Step Solution: Adopt rib width ≥ 75 mm to allow main bars plus cover and to provide concrete confinement.Keep clear spacing ≤ about 450 mm so the topping acts as an effective flange between ribs.Restrict overall depth to ≤ 4 × rib breadth so ribs remain stocky, controlling shear and deflection. Verification / Alternative check: Check shear at the rib–slab junction and distribution steel in the topping slab; deflection calculations corroborate that these geometric limits lead to serviceable deformations. Why Other Options Are Wrong: Options b and d: rib widths too small and overall depth too restrictive for practical reinforcement and capacity. Option c: spacing excessively large and depth overly slender, harming distribution and shear. Option e: unlimited spacing is unsafe; topping alone cannot maintain action. Common Pitfalls: Ignoring topping reinforcement, using too-narrow ribs causing congestion, and exceeding spacing that the topping can bridge. Final Answer: Clear spacing between ribs ≤ 450 mm; rib width ≥ 75 mm; overall depth of slab ≤ 4 × rib breadth.

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