Q: Dead-load allowance in wall footing estimates: As a rule of thumb in preliminary design, the self-weight of a foundation is taken as approximately what percentage of the supported wall weight?

Introduction / Context: Preliminary sizing and load estimation often rely on practical thumb rules before detailed analysis. One such rule is to approximate the self-weight of footings and pedestals as a percentage of the wall or column load to account for dead load in foundation design and settlement checks. Given Data / Assumptions: Common masonry or concrete wall resting on a spread footing. Preliminary stage estimate; detailed design will refine self-weight. Typical densities and footing proportions assumed. Concept / Approach: Experience indicates that the self-weight of a footing is a modest fraction of the superimposed load. A frequently used allowance is around 10% of the supported load for initial calculations, sufficient for sizing area and checking bearing pressure prior to exact quantity computations. Step-by-Step Solution: Assume wall load W_w.Estimate foundation self-weight W_f ≈ 0.10 * W_w.Use W_total = W_w + W_f for preliminary bearing and settlement checks. Verification / Alternative check: Detailed weight can be computed later from dimensions and density. Preliminary designs using 10% typically converge to final solutions with minor adjustments. Why Other Options Are Wrong: 5%–7% may understate the footing mass for common proportions; 12%–15% is conservative and may overestimate, reducing economy. The commonly cited mid-range value is 10%. Common Pitfalls: Persisting with the rule of thumb into final design; ignoring pedestals, grade beams, or backfill loads; not adjusting for unusually thick or deep footings. Final Answer: 10% of wall weight

Q: Ribbed (waffle) slab reinforcement: What is the maximum diameter generally permitted for bars used in the slab component of a ribbed slab system, considering serviceability and detailing practice?

Introduction / Context: Ribbed (waffle) slabs consist of thin topping slabs cast monolithically over closely spaced ribs. While ribs may contain larger longitudinal bars, the slab (topping) reinforcement is governed by cover, spacing, crack control, and practical placement. Excessively large diameters are unsuitable for thin slabs due to spacing and serviceability constraints. Given Data / Assumptions: Typical ribbed slab with a relatively thin topping slab. Standard cover and bar spacing limits for crack control apply. Question refers to bars in the slab portion, not the main rib longitudinal bars. Concept / Approach: Codes limit bar diameters as a function of slab thickness and spacing. For thin toppings, practical bar diameters are small (commonly 8–12 mm). Larger diameters complicate placement, risk inadequate cover, and impair crack dispersion, which relies on multiple smaller bars at closer spacing. Step-by-Step Solution: Identify the element: slab (topping) of a ribbed system, not ribs.Apply typical detailing practice: max bar dia ≈ 12 mm for thin slabs.Select 12 mm as the appropriate ceiling from the given choices. Verification / Alternative check: Design manuals and detailing guides commonly show 8, 10, or 12 mm top meshes for waffle slabs; ribs carry larger bars as required by flexure. Why Other Options Are Wrong: 6 mm may be too light for structural slab reinforcement beyond distribution; 20–24 mm diameters are generally reserved for ribs or beams and are impractical in thin slabs. Common Pitfalls: Confusing the slab mesh with rib reinforcement; violating minimum spacing/cover; using large bars that create honeycombing or poor consolidation around reinforcement. Final Answer: 12 mm

Q: In reinforced concrete design, a T-beam behaves like a rectangular (flanged) beam of width equal to its flange only when the neutral axis lies where? Choose the condition under which the compression block remains entirely within the slab flange.

Introduction / Context: T-beams arise when a slab and its supporting beam act monolithically. For flexural design, whether the section behaves as a simple rectangular section or a true T-section depends on the position of the neutral axis (N.A.) relative to the flange thickness. Given Data / Assumptions: Monolithic slab and beam (R.C.C.). Flange in compression, web (rib) below. Working stress or limit state concepts both recognize compression block extent. Concept / Approach: If the N.A. falls within the slab thickness, the entire compression zone is in the flange. The web is then largely unstressed in compression; the section can be analysed as a rectangular beam of width equal to the effective flange width. If the N.A. lies below the flange, compression spreads into the web and T-beam analysis is required. Step-by-Step Solution: Identify requirement for rectangular behaviour: compression block contained in the flange.This occurs when the neutral axis depth is less than or equal to the effective flange thickness.Therefore, the N.A. must remain within the flange. Verification / Alternative check: Design codes treat two cases: N.A. in flange (rectangular compression block) and N.A. in web (T-section). In the first case, breadth = effective flange width, depth = neutral axis depth within flange. Why Other Options Are Wrong: Remains below the slab: Implies N.A. in the web; T-analysis needed. Coincides with geometric centre: Not a criterion for behaviour; depends on reinforcement and loads. None of these: Incorrect, because there is a correct condition. Common Pitfalls: Assuming T-beam action always; ignoring effective flange width limits; misplacing N.A. without checking strain compatibility. Final Answer: remains within the flange

Q: For bar bending schedules in reinforced concrete, the internal bend radius when forming a standard hook should not be less than what multiple of the bar diameter?

Introduction / Context: Minimum bend radii are specified to avoid cracking of concrete and bar damage during bending. Hooks provide anchorage in tension zones, and an adequate internal radius limits strain hardening and micro-cracking of the bar. Given Data / Assumptions: Conventional reinforcing steel bending for standard hooks. General guideline for minimum internal bend radius. No special high-strength bar exception is invoked. Concept / Approach: Practical detailing rules cap the sharpness of bends by specifying a minimum internal radius as a multiple of bar diameter d. A commonly adopted minimum for standard hooks in mild steel is about 2d, ensuring the bar is bent without excessive damage and providing reliable anchorage. Step-by-Step Solution: Identify minimum internal bend radius requirement for hooks.Adopt the common detailing guideline: r(min) ≈ 2d for standard hooks in typical bars.Select the corresponding option. Verification / Alternative check: Detailing manuals provide bend diameter or radius limits; the 2d internal radius (equivalently bend diameter 4d) is a widely used baseline for plain bars. (Some codes specify larger radii for deformed/high-strength bars.) Why Other Options Are Wrong: 3d, 4d, 5d: More conservative than the common baseline and may be required for certain bar grades, but not the typical minimum for standard hooks cited in many exam contexts. None of these: Incorrect because a standard minimum exists. Common Pitfalls: Confusing internal radius with bend diameter; using too small a radius causing bar distress; not checking code-specific exceptions for high-strength deformed bars or seismic hooks. Final Answer: twice the diameter

Question 1

In reinforced concrete detailing, the lap length for tension reinforcement shall not be less than bar_dia * (actual_tension / (4 * permissible_average_bond_stress)) if this calculated value is greater than which minimum benchmark (expressed in bar d…

Accepted Answer

Introduction / Context: Lap splices ensure continuity of reinforcing bars where a bar is terminated and another begins. In tension zones, adequate lap length is critical to transfer stress safely through bond with concrete and avoid premature slip or cracking. Given Data / Assumptions: The expression for lap length in tension: L_calc = bar_dia * (actual_tension / (4 * permissible_average_bond_stress)). We must compare L_calc with a minimum benchmark expressed in bar diameters. Concept / Approach: Codes and standard practice stipulate a minimum lap length in tension to prevent short splices. Even when the calculated value is small, a floor value in terms of bar diameters is enforced. A widely adopted benchmark is 30 bar diameters for tension laps (subject to additional detailing requirements, cover, confinement, and bar type). Step-by-Step Solution: Compute L_calc from the given formula.Compare L_calc to the minimum benchmark.If L_calc exceeds the benchmark, use L_calc; otherwise, not less than the benchmark is permitted. Verification / Alternative check: Typical detailing guides and many code interpretations adopt 30φ as the minimum for tension splices in ordinary conditions, with adjustments for bar type (plain vs deformed), confinement, and concrete class where applicable. Why Other Options Are Wrong: 18φ and 24φ: too short for tension splices under most design scenarios. 36φ: conservative but not the standard threshold referred to in this formulation. Common Pitfalls: Confusing development length with lap length, and neglecting reductions or increases due to hooks, confinement, or poor bond conditions. Always verify with current code provisions for the project jurisdiction. Final Answer: 30 bar diameters.

Question 2

For a vertical retaining wall of height h supporting a hydrostatic or earth-pressure distribution that varies linearly with depth, the resultant total pressure acts parallel to the free surface and at what distance above the base?

Accepted Answer

Introduction / Context: Retaining walls resist lateral pressures from fluids or soils. When the pressure distribution is triangular (zero at the top and maximum at the base), determining the line of action of the resultant is essential for stability checks against overturning and sliding. Given Data / Assumptions: Wall height = h. Pressure varies linearly from zero at the top to a maximum at the base. Resultant acts parallel to the free surface (i.e., horizontal for hydrostatic/active pressure). Concept / Approach: A linearly varying pressure distribution forms a triangle. The resultant of a triangular load acts through its centroid, located at one-third of the height from the base (i.e., h/3 above the base). The magnitude equals the area of the triangle (0.5 * base_pressure * h), and its line of action is crucial for overturning moment calculations. Step-by-Step Solution: Identify pressure diagram: triangular with zero at top and maximum at base.Compute resultant location: centroid of triangle is at h/3 from the base.Direction: parallel to the free surface (typically horizontal). Verification / Alternative check: For hydrostatics, an identical conclusion arises from integrating pressure p = gamma * z over depth. For earth pressure with linear distribution, the same centroidal property holds, confirming h/3 from the base. Why Other Options Are Wrong: h/4 and h/2: not the centroid of a triangle. 2h/3: corresponds to the location from the top, not from the base. Common Pitfalls: Mixing the location measured from the top versus from the base; forgetting that the direction is parallel to the free surface, not inclined like the wall backface in some cases. Final Answer: h/3.

Question 3

In a simply supported reinforced concrete slab, distribution (secondary) reinforcement is provided primarily to handle which effects and functions?

Accepted Answer

Introduction / Context: Reinforced concrete slabs have main (primary) reinforcement in the direction of the principal bending. Distribution (secondary) reinforcement runs orthogonally and serves multiple serviceability functions beyond mere strength in the main span direction. Given Data / Assumptions: Simply supported slab with predominant one-way action. Ordinary room temperature variations and drying shrinkage are present. Local loads may occur due to point or line contacts (e.g., partitions, wheel loads). Concept / Approach: Distribution steel limits crack widths and spacing caused by temperature gradients and drying shrinkage; it also helps distribute local wheel or impact loads over a wider area, reducing stress concentrations and improving serviceability. Though not sized for primary flexure, it is indispensable for crack control and load sharing. Step-by-Step Solution: Identify the main causes of secondary cracking: temperature and shrinkage.Recognize load spread: orthogonal bars help spread wheel/point loads to main reinforcement.Therefore, its role encompasses load distribution plus restraint of temperature and shrinkage strains. Verification / Alternative check: Crack control provisions in design standards set minimum percentages for distribution steel based on bar size, spacing, and exposure. Field experience confirms slabs with inadequate secondary steel exhibit wider crack patterns and local punching or rutting near concentrated loads. Why Other Options Are Wrong: Each single-cause option is incomplete. The correct role is multi-faceted, including both environmental strain effects and load distribution. Common Pitfalls: Reducing distribution steel as a cost-saving measure often leads to surface cracking, durability problems, and poor performance under localized loads. Final Answer: All the above (load distribution as well as temperature and shrinkage effects).

Question 4

For an inaccessible roof in building design (i.e., not intended for regular occupancy), what live load should be considered for structural design calculations?

Accepted Answer

Introduction / Context: Structural design must account for appropriate live loads as per occupancy and usage. Roofs that are not intended for regular foot traffic (inaccessible roofs) still require a nominal live load to cover occasional access for inspection and maintenance. Given Data / Assumptions: Roof is designated inaccessible (no regular occupancy or assembly use). Loads are to be taken as uniformly distributed for design. Local code conventions equivalent to 0.75 kN/m2 are recognized as 75 kg/m2. Concept / Approach: Design standards typically prescribe a minimum live load for inaccessible roofs to ensure safety during maintenance. While accessible or landscaped roofs can require higher loads, the baseline for inaccessible roofs is significantly lower but non-zero. Step-by-Step Solution: Identify roof category: inaccessible.Apply prescribed nominal live load: 75 kg/m2.Ensure combinations with wind, seismic, and ponding checks as applicable. Verification / Alternative check: Conversion check: 75 kg/m2 ≈ 0.75 kN/m2, which aligns with common building code prescriptions for inaccessible roofs. This value provides a safety buffer for occasional maintenance loads without overdesigning the structure. Why Other Options Are Wrong: Nil: unsafe, ignores maintenance access. 150 kg/cm2: incorrect unit and unrealistically large. 200 kg/m2: applicable to more heavily used roofs, not typical for inaccessible types. Common Pitfalls: Confusing accessible terraces with inaccessible roofs; using incorrect units (kg/cm2 rather than kg/m2) leading to gross errors in design forces. Final Answer: 75 kg/m2.

Question 5

In flexural design, “steel beam theory” refers to an approach applied to which situation among the following?

Accepted Answer

Introduction / Context: When transforming reinforced concrete sections for analysis, simplified theories are sometimes used for special cases. “Steel beam theory” is one such simplification historically used for doubly reinforced beams where concrete in compression is neglected and compression is assumed to be carried entirely by compression steel. Given Data / Assumptions: Doubly reinforced RC beam with both tension and compression steel. Concrete compressive contribution ignored for the idealization. Linear elastic assumptions for transformed steel areas within service range. Concept / Approach: By neglecting the compressive resistance of concrete, the section is treated as a “steel-like” flanged member where a couple is formed by tension and compression steel. This simplifies calculations of stress–strain compatibility and moment capacity for conceptual understanding or preliminary checks, though modern code design prefers more accurate methods acknowledging concrete compression. Step-by-Step Solution: Identify the context demanding compression steel (e.g., limited depth, high bending).Apply “steel beam theory”: assume compression is resisted by compression steel only.Compute internal couple from equal and opposite steel forces times lever arm. Verification / Alternative check: Comparing with conventional RC theory (concrete in compression + tension steel), steel beam theory gives conservative/approximate insights but should be cross-checked against full rectangular stress block approaches per current codes. Why Other Options Are Wrong: (a) Concerns steel-only members, not the specific RC simplification. (b) Composite behavior is different and includes concrete compression explicitly. (d) Relates to shear design criteria, unrelated to the flexural idealization named here. Common Pitfalls: Using steel beam theory as a final design method without verifying code-compliant stress blocks, ductility, and detailing; confusing the concept with composite beam theory. Final Answer: Doubly reinforced concrete beams while neglecting compressive stress in concrete (compression carried by steel).

Question 6

In building construction and staircase design, what is the correct term for the horizontal part of a step on which the foot is placed?

Accepted Answer

Introduction / Context: Understanding the terminology of stair components is fundamental in architecture and civil engineering. Accurate terms support correct detailing, safety compliance, and communication on drawings and at site. This question focuses on the name of the horizontal surface that a user steps on while ascending or descending a staircase. Given Data / Assumptions: The question refers to a conventional staircase used in buildings. Standard architectural nomenclature is assumed. No special stair types (spiral, alternating tread, ship ladder) are implied; the terms remain consistent across common stair typologies. Concept / Approach: Each step has two primary dimensions: the vertical dimension (rise/riser) and the horizontal dimension (tread). Additional terms such as flight, landing, and winder describe groupings or special shapes of steps. By matching definitions to the options, we isolate the correct term for the horizontal surface where foot contact occurs during use. Step-by-Step Solution: Identify the element asked: “horizontal part of a step”.Recall standard definitions: tread = horizontal part; riser = vertical face; flight = series of steps between landings; landing = resting platform; winder = a step that is wedge-shaped used at turns.Therefore, the horizontal part is the tread. Verification / Alternative check: Building codes and handbooks (e.g., NBC/IBC handbooks) consistently define “tread” as the horizontal component of a step. Detailing conventions on drawings label the tread depth and riser height for ergonomic and code compliance checks. Why Other Options Are Wrong: Rise denotes the vertical height of a step, not the horizontal surface. Flight is a series of steps between landings. Winder is a wedge-shaped step used to turn a stair. Landing is a flat platform, not the individual horizontal part of a step. Common Pitfalls: Confusing “tread” with “landing,” and “rise” with “riser.” The “riser” is the vertical plate; the “rise” is the dimension. The “tread” is the surface and also a dimension (tread depth). Final Answer: Tread

Question 7

Reinforced concrete column under axial load: A 30 cm diameter R.C.C. column is reinforced with six 12 mm bars placed symmetrically. If it carries 40,000 kg axially, what is the working compressive stress using the transformed (modular ratio) area me…

Accepted Answer

Introduction / Context: Axially loaded reinforced concrete (R.C.C.) columns are commonly designed under the working stress approach using a transformed area. Steel is converted to an equivalent concrete area by the modular ratio so that a uniform compressive stress can be computed for service load checks. This question applies that concept to a circular column with symmetrically placed bars. Given Data / Assumptions: Column diameter = 30 cm (radius = 15 cm). Number of bars = 6; diameter of each bar = 12 mm = 1.2 cm. Axial load = 40,000 kg (service load). Use modular ratio method; take a typical working-stress modular ratio m ≈ 18 (standard textbook value). Uniform axial compression, no eccentricity considered. Concept / Approach: Working-stress design converts steel area to an equivalent concrete area using Ae = Ac + m * As. The service compressive stress is then load / Ae. This accounts for higher stiffness of steel so that the combined section shares load realistically under small strains. Step-by-Step Solution: Gross concrete area Ag = π * 15^2 = 706.858 cm².Area of one 12 mm bar = π * (0.6)^2 = 1.131 cm²; for 6 bars, As = 6 * 1.131 = 6.786 cm².Net concrete area Ac ≈ Ag − As = 700.072 cm².Transformed (equivalent) area Ae = Ac + m * As = 700.072 + 18 * 6.786 ≈ 700.072 + 122.148 ≈ 822.220 cm².Working compressive stress σ = Load / Ae = 40,000 / 822.220 ≈ 48.6 kg/cm² (rounds to about 49–50 kg/cm²). Verification / Alternative check: Using gross area only gives 56.6 kg/cm², which ignores the stiffness contribution of steel. The transformed area approach gives a lower, more realistic stress, matching the textbook answer choices. Why Other Options Are Wrong: 100, 175, and 250 kg/cm² are too high for the given load and section. 56.6 kg/cm² uses gross area only and neglects modular ratio; the problem context implies transformed area. Common Pitfalls: Forgetting to subtract steel area from concrete before adding m * As; using mm² vs cm² inconsistently; ignoring eccentricity when present in practice. Final Answer: 49.9 kg/cm²

Question 8

Foundation slab profiling for uniform soil pressure: How should the thickness of a footing vary from the column face toward the edge to maintain practical and economical stress distribution?

Accepted Answer

Introduction / Context: Footings are designed to safely spread column loads onto soil while controlling bending, shear, and punching. Although the soil pressure is governed by load and plan area, the slab’s thickness profile strongly influences internal forces and economy. Textbook practice uses thicker concrete near the column (where bending/shear are highest) tapering toward the edge. Given Data / Assumptions: Isolated spread footing under vertical load and uniform bearing capacity. No unusual edge loads or moments beyond typical soil reactions. Conventional reinforced concrete detailing and code-compliant cover. Concept / Approach: Bending moment and shear are maximum near the column face and reduce toward the edge. A variable depth section satisfies strength where needed and saves material elsewhere. While soil pressure distribution is a function of plan geometry and load, the footing thickness taper improves structural efficiency and constructability without compromising bearing performance. Step-by-Step Solution: Recognize internal design actions peak under/near the column.Provide greater depth where bending and shear are highest.Reduce thickness toward the edge to maintain economy while meeting code for minimum thickness and cover. Verification / Alternative check: Design charts and code clauses for one-way shear, punching shear, and bending confirm higher required effective depth near the column, permitting a slope (haunch) or step down toward the edges. Why Other Options Are Wrong: Uniform thickness wastes concrete or may be unsafe near the column; increasing thickness toward the edge is structurally illogical; “zero” thickness at the edge is impossible; arbitrary variation is poor engineering practice. Common Pitfalls: Ignoring minimum cover and edge thickness; not checking punching shear at the column face; over-simplifying soil pressure as affected by thickness profile. Final Answer: Decreased gradually toward the edge

Question 9

R.C.C. beam detailing: Which statement about the behavior/placement of tensile reinforcement in a rectangular beam is INCORRECT?

Accepted Answer

Introduction / Context: Proper reinforcement detailing ensures that reinforced concrete beams safely resist flexure and shear. Traditional detailing may use bent-up bars and stirrups for shear; however, the direction and purpose of bending must be correct and consistent with the flow of tensile forces in the member. Given Data / Assumptions: Rectangular, singly reinforced beam under gravity loading. Working stress or limit state concepts for anchorage, curtailment, and shear resistance apply. Standard code-based bond and anchorage provisions are observed. Concept / Approach: Tension reinforcement is placed near the tension face (bottom for sagging midspan). When bending moment decreases toward supports, bars may be curtailed subject to anchorage and shear checks. Bent-up bars are traditionally “bent up” toward the compression zone to intersect potential shear cracks, contributing to shear resistance along with stirrups. The phrase “bent down” for shear is incorrect in this context. Step-by-Step Solution: Identify the role of tensile bars: resist flexure at the tension face.At support approach, reduce bending → permitted bar curtailment with anchorage.For shear, bars are bent up (not down) to cross inclined cracks; stirrups provide the principal shear reinforcement in modern practice. Verification / Alternative check: Historic detailing practice and many codes illustrate bent-up bars sloping upward toward supports; modern design often relies primarily on vertical stirrups with continuous bottom bars for ductility and constructability. Why Other Options Are Wrong: (a), (b), and (d) reflect correct practice; (e) is an acceptable added note that negative moments at supports can require top bars. The incorrect statement is (c) because “bent down” contradicts the established detailing direction for shear. Common Pitfalls: Premature bar curtailment without sufficient anchorage length; inadequate stirrups; misinterpreting diagrams and bending directions. Final Answer: Tensile bars are bent down at selected locations to act as shear reinforcement

Question 10

Purpose of shear reinforcement in R.C.C.: Shear reinforcement is primarily provided to resist which internal action produced by shear?

Accepted Answer

Introduction / Context: Concrete is strong in compression and weak in tension. Under shear, inclined tensile stresses (diagonal tension) develop, causing potential cracking. Shear reinforcement (stirrups/bent-up bars) is used to intercept these cracks and carry tensile forces safely, enhancing ductility and preventing brittle web shear failure in beams. Given Data / Assumptions: Reinforced concrete beam under typical gravity loading. Shear forces cause inclined principal stresses. Standard code provisions for shear design are assumed. Concept / Approach: Shear in beams leads to diagonal tension (tensile principal stress on planes approximately at 45 degrees). Concrete alone cannot resist this tension beyond a small limit (diagonal tension capacity). Shear reinforcement functions as tension ties across these potential crack planes, while concrete and aggregate interlock provide residual shear transfer. Step-by-Step Solution: Recognize that “vertical shear” is a resultant; the critical failure mode is tension along inclined planes.Provide stirrups (vertical or inclined) to act across the crack as tension members.Ensure spacing and size satisfy code: V_us ≥ V_u − V_c with adequate anchorage. Verification / Alternative check: Design formulas decompose shear resistance into concrete contribution and steel contribution; the steel leg force is proportional to A_sv * f_y * (d / s) for vertical stirrups, directly addressing diagonal tension. Why Other Options Are Wrong: Horizontal shear relates to composite action along interfaces; diagonal compression is typically carried by concrete. “Vertical shear” wording misses the mechanism of crack control; torsion requires special closed stirrups and longitudinal reinforcement beyond ordinary shear design. Common Pitfalls: Underestimating shear near supports; exceeding maximum stirrup spacing; inadequate hooks/anchorage for stirrups. Final Answer: Diagonal tension

Question 11

Doubly reinforced beam—role of critical neutral axis: How do maximum stresses develop relative to the position of the actual neutral axis compared to the critical neutral axis?

Accepted Answer

Introduction / Context: In flexural design, the position of the neutral axis (NA) relative to a critical (balanced) depth governs the failure mode. For doubly reinforced beams, compression steel supplements the concrete when the section is shallow or moments are high. The comparison of actual NA with the critical NA indicates whether concrete crushes first or steel yields first, guiding ductility and safety considerations. Given Data / Assumptions: Balanced (critical) NA corresponds to simultaneous attainment of limiting concrete strain and steel yield at ultimate. Over-reinforced: NA deeper (greater depth) than critical; concrete crushes first. Under-reinforced: NA shallower than critical; tensile steel yields first, providing ductile warning. Concept / Approach: The distribution of strains is linear in flexure. The strain at the extreme compression fiber and at the steel layers depends on NA depth. Positioning relative to critical depth dictates which material hits its limit first. Doubly reinforced beams allow redistribution: compression steel participates when concrete’s compression block is insufficient. Step-by-Step Solution: Compare actual NA depth to critical depth.If actual > critical → compression zone larger → concrete crushes first.If actual < critical → tension steel strain higher → steel yields before concrete crushes.If actual = critical → simultaneous limiting strains in concrete and steel. Verification / Alternative check: Ultimate limit state strain diagrams and standard code equations for balanced reinforcement confirm these qualitative outcomes independent of specific numerical values. Why Other Options Are Wrong: Each of (a), (b), and (c) states a correct condition; hence the combined correct choice is (d) “All of the above statements are correct”. Common Pitfalls: Confusing “greater NA depth” with safer behavior; over-reinforced sections are less ductile. Neglecting compression steel contribution in analysis of doubly reinforced sections. Final Answer: All of the above statements are correct

Question 12

Dead-load allowance in wall footing estimates: As a rule of thumb in preliminary design, the self-weight of a foundation is taken as approximately what percentage of the supported wall weight?

Accepted Answer

Introduction / Context: Preliminary sizing and load estimation often rely on practical thumb rules before detailed analysis. One such rule is to approximate the self-weight of footings and pedestals as a percentage of the wall or column load to account for dead load in foundation design and settlement checks. Given Data / Assumptions: Common masonry or concrete wall resting on a spread footing. Preliminary stage estimate; detailed design will refine self-weight. Typical densities and footing proportions assumed. Concept / Approach: Experience indicates that the self-weight of a footing is a modest fraction of the superimposed load. A frequently used allowance is around 10% of the supported load for initial calculations, sufficient for sizing area and checking bearing pressure prior to exact quantity computations. Step-by-Step Solution: Assume wall load W_w.Estimate foundation self-weight W_f ≈ 0.10 * W_w.Use W_total = W_w + W_f for preliminary bearing and settlement checks. Verification / Alternative check: Detailed weight can be computed later from dimensions and density. Preliminary designs using 10% typically converge to final solutions with minor adjustments. Why Other Options Are Wrong: 5%–7% may understate the footing mass for common proportions; 12%–15% is conservative and may overestimate, reducing economy. The commonly cited mid-range value is 10%. Common Pitfalls: Persisting with the rule of thumb into final design; ignoring pedestals, grade beams, or backfill loads; not adjusting for unusually thick or deep footings. Final Answer: 10% of wall weight

Question 13

Ribbed (waffle) slab reinforcement: What is the maximum diameter generally permitted for bars used in the slab component of a ribbed slab system, considering serviceability and detailing practice?

Accepted Answer

Introduction / Context: Ribbed (waffle) slabs consist of thin topping slabs cast monolithically over closely spaced ribs. While ribs may contain larger longitudinal bars, the slab (topping) reinforcement is governed by cover, spacing, crack control, and practical placement. Excessively large diameters are unsuitable for thin slabs due to spacing and serviceability constraints. Given Data / Assumptions: Typical ribbed slab with a relatively thin topping slab. Standard cover and bar spacing limits for crack control apply. Question refers to bars in the slab portion, not the main rib longitudinal bars. Concept / Approach: Codes limit bar diameters as a function of slab thickness and spacing. For thin toppings, practical bar diameters are small (commonly 8–12 mm). Larger diameters complicate placement, risk inadequate cover, and impair crack dispersion, which relies on multiple smaller bars at closer spacing. Step-by-Step Solution: Identify the element: slab (topping) of a ribbed system, not ribs.Apply typical detailing practice: max bar dia ≈ 12 mm for thin slabs.Select 12 mm as the appropriate ceiling from the given choices. Verification / Alternative check: Design manuals and detailing guides commonly show 8, 10, or 12 mm top meshes for waffle slabs; ribs carry larger bars as required by flexure. Why Other Options Are Wrong: 6 mm may be too light for structural slab reinforcement beyond distribution; 20–24 mm diameters are generally reserved for ribs or beams and are impractical in thin slabs. Common Pitfalls: Confusing the slab mesh with rib reinforcement; violating minimum spacing/cover; using large bars that create honeycombing or poor consolidation around reinforcement. Final Answer: 12 mm

Question 14

In reinforced concrete design, a T-beam behaves like a rectangular (flanged) beam of width equal to its flange only when the neutral axis lies where? Choose the condition under which the compression block remains entirely within the slab flange.

Accepted Answer

Introduction / Context: T-beams arise when a slab and its supporting beam act monolithically. For flexural design, whether the section behaves as a simple rectangular section or a true T-section depends on the position of the neutral axis (N.A.) relative to the flange thickness. Given Data / Assumptions: Monolithic slab and beam (R.C.C.). Flange in compression, web (rib) below. Working stress or limit state concepts both recognize compression block extent. Concept / Approach: If the N.A. falls within the slab thickness, the entire compression zone is in the flange. The web is then largely unstressed in compression; the section can be analysed as a rectangular beam of width equal to the effective flange width. If the N.A. lies below the flange, compression spreads into the web and T-beam analysis is required. Step-by-Step Solution: Identify requirement for rectangular behaviour: compression block contained in the flange.This occurs when the neutral axis depth is less than or equal to the effective flange thickness.Therefore, the N.A. must remain within the flange. Verification / Alternative check: Design codes treat two cases: N.A. in flange (rectangular compression block) and N.A. in web (T-section). In the first case, breadth = effective flange width, depth = neutral axis depth within flange. Why Other Options Are Wrong: Remains below the slab: Implies N.A. in the web; T-analysis needed. Coincides with geometric centre: Not a criterion for behaviour; depends on reinforcement and loads. None of these: Incorrect, because there is a correct condition. Common Pitfalls: Assuming T-beam action always; ignoring effective flange width limits; misplacing N.A. without checking strain compatibility. Final Answer: remains within the flange

Question 15

For bar bending schedules in reinforced concrete, the internal bend radius when forming a standard hook should not be less than what multiple of the bar diameter?

Accepted Answer

Introduction / Context: Minimum bend radii are specified to avoid cracking of concrete and bar damage during bending. Hooks provide anchorage in tension zones, and an adequate internal radius limits strain hardening and micro-cracking of the bar. Given Data / Assumptions: Conventional reinforcing steel bending for standard hooks. General guideline for minimum internal bend radius. No special high-strength bar exception is invoked. Concept / Approach: Practical detailing rules cap the sharpness of bends by specifying a minimum internal radius as a multiple of bar diameter d. A commonly adopted minimum for standard hooks in mild steel is about 2d, ensuring the bar is bent without excessive damage and providing reliable anchorage. Step-by-Step Solution: Identify minimum internal bend radius requirement for hooks.Adopt the common detailing guideline: r(min) ≈ 2d for standard hooks in typical bars.Select the corresponding option. Verification / Alternative check: Detailing manuals provide bend diameter or radius limits; the 2d internal radius (equivalently bend diameter 4d) is a widely used baseline for plain bars. (Some codes specify larger radii for deformed/high-strength bars.) Why Other Options Are Wrong: 3d, 4d, 5d: More conservative than the common baseline and may be required for certain bar grades, but not the typical minimum for standard hooks cited in many exam contexts. None of these: Incorrect because a standard minimum exists. Common Pitfalls: Confusing internal radius with bend diameter; using too small a radius causing bar distress; not checking code-specific exceptions for high-strength deformed bars or seismic hooks. Final Answer: twice the diameter

Question 16

In reinforcement detailing, the straight extension beyond the end of a standard hook (i.e., the length of the straight portion past the bend) shall be at least how many times the bar diameter?

Accepted Answer

Introduction / Context: Hooks and bends provide mechanical anchorage for tension bars. However, a hook is not effective unless it is followed by a minimum straight length that develops the steel and ensures proper force transfer to the concrete. Given Data / Assumptions: Standard hook at the end of a tension bar. Ordinary R.C.C. detailing for anchorage. Concept / Approach: Detailing rules prescribe a minimum straight length beyond the bend (tail). A widely practiced minimum is 4d (four times bar diameter), which, along with the bend, provides adequate development against pullout and splitting forces. Step-by-Step Solution: Recognize the requirement for a minimum tail beyond the hook.Adopt the commonly specified value of 4d for standard hooks.Select the correct multiple from the options. Verification / Alternative check: Bar bending schedules and detailing handbooks list the tail beyond 90°/135° hooks as not less than 4d (subject to code-specific refinements for seismic detailing). Why Other Options Are Wrong: 2d or 3d: Often insufficient to mobilize full anchorage capacity. 5d or 7d: Conservative and may be used in special cases, but exceed the standard minimum requirement referenced here. Common Pitfalls: Confusing development length with hook tail length; ignoring bar coating effects (e.g., epoxy) that may require longer development. Final Answer: four times the diameter

Question 17

For ribbed (joist) slabs in building construction, the in-situ topping (wearing/finishing slab cast over the ribs) generally has what thickness range?

Accepted Answer

Introduction / Context: Ribbed slabs consist of closely spaced ribs with a thin topping slab forming the flange. The topping provides distribution of loads, diaphragm action, and a surface for finishes and services. Given Data / Assumptions: Conventional building floors with ribbed slab systems. Normal live loads and finishing requirements. Concept / Approach: A topping that is too thin may crack and fail to distribute loads; one that is too thick adds dead load without significant benefit. Practice typically adopts about 50–80 mm (5–8 cm) thickness for a good balance of performance and weight. Step-by-Step Solution: Identify typical topping thickness used in ribbed slabs.Select the 5–8 cm range as most common in building practice.Confirm that this thickness allows cover to reinforcement, conduits, and finishing layers. Verification / Alternative check: Structural detailing manuals and manufacturer guidance for joist systems commonly specify a minimum topping thickness in the 50–80 mm range. Why Other Options Are Wrong: 3–5 cm: Often inadequate for crack control and embedment of services. 8–10 cm or 12–18 cm: Heavier than necessary for typical ribbed slabs, increasing dead load. Common Pitfalls: Forgetting to include screed/finish thickness in dead load; using too thin a topping over deep rib spacing causing localized punching or cracking. Final Answer: 5 cm to 8 cm

Question 18

In a singly reinforced concrete beam, the effective depth d is measured from which reference to which location?

Accepted Answer

Introduction / Context: Effective depth is a key geometric parameter in flexural design because it determines the internal lever arm between compression and tension resultants. Accurate d directly affects moment capacity calculations. Given Data / Assumptions: Singly reinforced rectangular or flanged section. Compression at the top (compression edge). Tension steel placed near the soffit. Concept / Approach: By definition, effective depth d is the distance from the extreme compression fibre (compression edge) to the centroid of the tensile reinforcement. It is not measured to the neutral axis or to the free tension edge; what matters is the actual steel centroid since that is where the tensile force acts. Step-by-Step Solution: Locate compression edge (usually top surface for positive bending).Locate the centroid of tension steel (distance from soffit equals cover + 0.5bar diameter for a single layer).Measure d between these two points. Verification / Alternative check: Design charts and code equations for moment resistance use lever arm z ≈ jd, where d is as defined above. Any mis-measurement leads to incorrect steel area estimates. Why Other Options Are Wrong: Tensile edge: Not necessarily coincident with steel centroid. Neutral axis: Moves with loading/material; not the definition of d. Longitudinal central axis: A geometric convenience, not relevant for resisting couple. Common Pitfalls: Ignoring multiple steel layers (centroid shifts); failing to account for bar diameter and clear cover in computing d. Final Answer: tensile reinforcement (centroid of tension steel)

Question 19

In working-stress analysis of reinforced concrete, the expression [A + (m − 1)Asc] representing the equivalent concrete area of an R.C.C. section corresponds to which method of analysis?

Accepted Answer

Introduction / Context: Before limit-state design became universal, working-stress (elastic) analysis used transformed sections so that concrete and steel could be analysed with a single material by scaling areas using a modular ratio m. Given Data / Assumptions: A = gross concrete area of the section under consideration. Asc = area of steel in compression (if any). m = modular ratio = Es / Ec. Concept / Approach: The modular ratio method converts steel area to an equivalent concrete area: steel in compression is multiplied by m to reflect its higher stiffness compared to concrete. The transformed (equivalent) concrete area becomes A + (m − 1)Asc because the steel already displaces its own area in concrete (hence m − 1). Step-by-Step Solution: Start with steel in compression Asc.Transform to concrete: equivalent = m*Asc.Add to concrete area but subtract the displaced concrete area Asc → net addition (m − 1)Asc.Total equivalent area = A + (m − 1)Asc. Verification / Alternative check: Similar transformations are used to compute transformed second moment of area for elastic stress checks at service loads. Why Other Options Are Wrong: Load factor/ultimate load methods belong to plastic or limit-state philosophy; they do not use this transformed-area expression. None of these: Incorrect since the modular ratio method explicitly uses it. Common Pitfalls: Forgetting that tension steel is ignored in concrete (cracked) tension zone; using the wrong m value; mixing service and ultimate stress checks. Final Answer: modular ratio method

Question 20

Horizontal shear stress distribution in a beam: pick the incorrect proportionality statement for the intensity of horizontal shear at a given layer.

Accepted Answer

Introduction / Context: In beam theory, the horizontal (longitudinal) shear stress at a layer is given by the well-known formula τ = VQ / (Ib). Understanding how τ varies with different geometric and loading quantities helps in web/flange design and shear connection detailing. Given Data / Assumptions: Prismatic beam, small deflection, linear elastic behaviour. V = transverse shear force at the section. Q = first moment of area of the portion about the neutral axis for the layer considered. I = second moment of area about the neutral axis for the whole section. b = width at the layer where τ is evaluated. Concept / Approach: From τ = VQ / (Ib): τ is directly proportional to V and Q, and inversely proportional to I and b. Since Q = Aȳ (area above/below the layer times the distance of its C.G. from the N.A.), τ is directly proportional to both A (of the considered part) and ȳ. Step-by-Step Solution: Recognize direct proportionality: τ ∝ V and τ ∝ Q.Note that Q = Aȳ → τ ∝ A and τ ∝ ȳ.Recognize inverse proportionality: τ ∝ 1/I and τ ∝ 1/b.Thus, any statement claiming direct proportionality to I is incorrect. Verification / Alternative check: Compute τ for a rectangular section: τmax = 1.5V/(bh); here I is in the denominator, confirming inverse dependence. Why Other Options Are Wrong: Shear force: τ increases with V (directly proportional). Area of the section (portion considered via Q): τ increases with A of the part in Q. Distance of C.G. from N.A.: τ increases with ȳ via Q. Width of beam: τ is inversely proportional to b, not directly; the option did not claim direct proportionality to width. Common Pitfalls: Confusing total area with the partial area in Q; ignoring the layer width variation in flanged sections; misinterpreting I's effect. Final Answer: moment of the beam section about its neutral axis

RCC Structures Design Questions