Q: Axially loaded R.C.C. short column: Given Ac (area of concrete), Asc (area of longitudinal steel), A (gross section area), m (modular ratio), and σc (permissible compressive stress in concrete), which expression for the axial load-carrying strength …

Introduction / Context: In reinforced concrete (R.C.C.) columns designed by the working stress method, the axial load capacity is obtained by summing the contributions of concrete and steel at their permissible stresses. This question checks whether you recognize the equivalent algebraic forms of the same strength expression when expressed in terms of Ac, Asc, A, and the modular ratio m. Given Data / Assumptions: Ac = area of concrete only. Asc = area of longitudinal steel. A = gross area = Ac + Asc. σc = permissible compressive stress in concrete. m = modular ratio (Es/Ec under working stress concepts). Short, concentrically loaded column; secondary effects ignored. Concept / Approach: Under concentric compression, steel shares the load in proportion to its modulus difference relative to concrete, hence the use of modular ratio m. The strength (permissible axial load) equals the sum of concrete's share plus steel's share converted into an equivalent concrete stress block using m. Step-by-Step Solution: Start with P = σc Ac + (m σc) Asc.Since A = Ac + Asc, Ac = A − Asc → P = σc (A − Asc) + m σc Asc.Factor σc: P = σc [A + (m − 1) Asc]. Verification / Alternative check: All three expressions reduce to the same value when you substitute Ac = A − Asc, proving their equivalence algebraically and dimensionally. Why Other Options Are Wrong: Options (a), (b), and (c) are each correct individually; the only fully correct choice that captures this is “All of the above.” “None of these” is obviously incorrect. Common Pitfalls: Forgetting that A = Ac + Asc; mixing permissible stresses with design strengths; applying the expression to columns with eccentricity without modifications. Final Answer: All of the above

Q: Continuous floor slab supported on beams: What ratio of end-span length to intermediate-span length is typically adopted for analysis and detailing assumptions?

Introduction / Context: For continuous slabs supported on beams, simplified span assumptions are used when applying moment-coefficient methods or when distributing loads and reinforcement. End spans are commonly taken slightly shorter (or treated differently) than interior spans to reflect support conditions and cracking patterns. This question asks for the typical ratio used in many design guides and exam problems. Given Data / Assumptions: Normal continuous one-way slab strips supported on beams. Regular framing; no large stiffness discontinuities. Rule-of-thumb span ratio for preliminary design and moment coefficients. Concept / Approach: Because end spans have only one interior support, their negative moment capacity and continuity differ from interior spans. Many textbooks adopt end-span length ≈ 0.8 of interior-span length as a practical simplifying assumption to apply standard coefficient tables consistently and to keep deflections and cracking comparable across spans. Step-by-Step Solution: Identify end span and interior span conditions.Apply the common assumption: L_end ≈ 0.8 * L_interior.Use this ratio in moment-coefficient tables and reinforcement detailing. Verification / Alternative check: Comparative analyses (elastic frame or FEM) show end spans often develop somewhat different moment envelopes; the 0.8 ratio is a conservative simplification used in many curricula. Why Other Options Are Wrong: 0.6 and 0.7 make the end span overly short relative to interior spans, affecting continuity assumptions; 0.9 is closer to equal spans but does not reflect the commonly adopted simplification for teaching-level design. Common Pitfalls: Treating end and interior spans as identical without checking support fixity; ignoring serviceability differences; not validating assumptions when significant stiffness changes occur. Final Answer: 0.8

Q: Proportioning an R.C.C. beam section: A common preliminary thumb rule keeps the ratio of breadth b to effective depth d approximately equal to what value?

Introduction / Context: At the preliminary stage of reinforced concrete beam design, dimensional ratios are used to obtain a practical section before detailed checks. The breadth b to effective depth d ratio affects shear capacity, bar placement, and deflection control. Many textbooks provide a simple starting ratio for ordinary beams carrying typical floor loads. Given Data / Assumptions: Ordinary reinforced concrete beam, simply supported or continuous. Serviceability and constructability considerations typical of building work. Working-stress or limit-state preliminary sizing intent. Concept / Approach: Keeping b about one quarter of d (b/d ≈ 0.25) often balances shear web width, bar congestion, and economy, while allowing sufficient depth to control deflections. Final proportions must be verified by bending, shear, deflection, and detailing checks, but 0.25 is a standard rule-of-thumb starting point in many courses and older handbooks. Step-by-Step Solution: Choose d from span/overall-depth limits.Set b ≈ 0.25 d initially.Iterate after detailed moment, shear, and serviceability checks. Verification / Alternative check: Trial designs using b/d = 0.25 commonly meet shear and reinforcement spacing requirements; adjustments are made if serviceability or bar spacing dictates a wider web. Why Other Options Are Wrong: 0.50, 0.70, and 0.75 produce excessively wide beams for a given depth in typical building floors, which may be uneconomical or create beam-slab conflicts. “None of these” is incorrect because 0.25 is the widely taught starting value. Common Pitfalls: Freezing preliminary ratios without later verification; ignoring minimum web width for bar placement and shear links; not coordinating with architectural beam widths. Final Answer: 0.25

Q: Combined footing for two equal column loads with restricted overall width: Which footing shape is generally adopted to satisfy centroidal alignment and uniform soil pressure?

Introduction / Context: When two columns of equal load share a combined footing and site width is limited (e.g., property line constraint), the footing must still position the resultant soil reaction under the combined centroid of column loads. For equal loads at reasonable spacing, a rectangular combined footing is typically suitable and economical. Given Data / Assumptions: Two columns with equal axial loads. Restricted overall footing width (e.g., due to boundary line). Uniform allowable soil bearing capacity. Eccentricity control needed so resultant passes through centroid. Concept / Approach: For equal loads and symmetric geometry, the pressure distribution is uniform when the footing centroid coincides with the load resultant. A rectangular footing centered beneath the two columns satisfies this condition if its length spans between columns and is centered along the line joining them. Trapezoidal shapes are usually adopted when loads are unequal or when a property line prevents centering the footing; triangular shapes do not suit two equal point loads well. Step-by-Step Solution: Establish column loads: P and P.Resultant = 2P at the midpoint between columns.Choose a rectangular plan area centered on the midpoint to give uniform soil pressure q = (2P)/A within allowable limits. Verification / Alternative check: Basic statics confirms that a symmetric rectangular area gives no net moment about the centroid line, maintaining uniform bearing. A trapezoid is warranted primarily when P1 ≠ P2 or centering is not feasible. Why Other Options Are Wrong: Square is only possible if spacing and allowable pressure make it viable; triangular and circular are inappropriate for two equal, spaced point loads; trapezoidal is for unequal loads or boundary constraints causing eccentricity. Common Pitfalls: Using a trapezoid unnecessarily; ignoring footing self-weight and overburden; failing to check punching shear near columns. Final Answer: Rectangular

Question 1

T-beam with neutral axis below the slab (flange in compression): which relationship between section variables holds for equilibrium of internal forces?

Accepted Answer

Introduction / Context: In T-beams where the neutral axis lies below the flange (slab), the flange is wholly in compression. Force equilibrium between compression in the effective flange thickness and tension in steel leads to a specific relationship linking geometric and material terms. Given Data / Assumptions: B = effective flange width. ds = thickness of slab (flange). n = depth of neutral axis from the top surface. d = effective depth to tension steel centroid. At = area of tensile steel; m = modular ratio (Es/Ec). Neutral axis lies below the slab, so the compressive block within slab is rectangular. Concept / Approach: With NA below the flange, compression is confined to slab thickness ds, giving a compression force equal to stress block intensity multiplied by area. In transformed section analysis (working stress method), compressive force in concrete equals the tension in transformed steel: C = T. The lever arm is not needed for this force equilibrium statement—only force balance is used to derive the presented relation. Step-by-Step Solution: Transformed area of steel = m * At at NA depth n.Compression in slab portion = B * ds * (n - ds) in transformed terms consistent with linear strain assumption.Equate compression to transformed steel tension to get B * ds * (n - ds) = m * At * (d - n). Verification / Alternative check: Strain diagram confirms linear distribution; centroid of compression block lies at mid-thickness of ds when NA is below slab, conforming to the adopted expression for transformed equilibrium. Why Other Options Are Wrong: Options a, b, c: misuse distances (d + n) or invert terms; dimensional and physical inconsistency. Option e: incorrect because a valid relation exists. Common Pitfalls: Confusing n with ds location, and mixing ultimate-limit-state rectangular stress blocks with working-stress transformed analysis without consistency. Final Answer: B * ds * (n - ds) = m * At * (d - n).

Question 2

Prestressed rectangular member with a concentric tendon: If a tendon is prestressed by force P through the centroidal longitudinal axis of a prism with area A, what is the uniform compressive stress induced in the concrete immediately after transfer…

Accepted Answer

Introduction / Context: In concentric prestressing, a tendon passes through the centroid. The prestressing force is therefore applied along the centroidal axis, causing a uniform precompression across the section with no bending moment at transfer. Given Data / Assumptions: Rectangular prismatic member with cross-sectional area A. Tendon stressed to force P and anchored concentrically. Self-weight and time-dependent losses neglected at the instant of transfer. Concept / Approach: Direct stress = Force / Area. With a concentric force, there is no eccentricity; hence no additional bending stress. The entire section is uniformly compressed by magnitude P/A. Step-by-Step Solution: Identify area A and prestress force P.Compute uniform compressive stress: sigma_c = P / A.Confirm eccentricity e = 0, so bending stress M/Z = 0. Verification / Alternative check: Stress distribution is rectangular and uniform; strain compatibility ensures uniform shortening. Any deviation from concentricity would add bending: sigma = P/A ± M/Z; with M = Pe = 0 here. Why Other Options Are Wrong: P/(2A) and 2P/A are incorrect scalings. PA and A/P are dimensionally wrong for stress. Common Pitfalls: Ignoring eccentricity in non-concentric layouts, or forgetting to adjust for immediate losses in real designs; however, the ideal concentric case remains P/A. Final Answer: P / A.

Question 3

Why are prestressed concrete members preferred in many structural applications compared to ordinary reinforced concrete members?

Accepted Answer

Introduction / Context: Prestressing introduces deliberate precompression to counteract tensile stresses from service loads. This approach enhances serviceability and often reduces structural depth, weight, and cracking compared to conventional reinforced concrete. Given Data / Assumptions: Use of high tensile steel and high-strength concrete. Prestress suitably profiled to balance loads and control deflection. Focus on service-level performance (crack control and deflection) and shear behavior. Concept / Approach: Precompression offsets part or all of the tensile stress envelope that would otherwise crack concrete. With proper web reinforcement and tendon profiling (sag), shear capacity improves via clamping action and inclined tendon components, alleviating the need for oversized sections. Step-by-Step Solution: Introduce prestress to maintain compressive stresses across critical zones at service.Utilize tendon profiles to generate counter-moments and beneficial shear components.Detail to minimize losses and maintain long-term compression, controlling shrinkage and temperature cracking. Verification / Alternative check: Service stress checks show reduced or eliminated tensile stresses; deflection checks indicate improved stiffness. Shear verifications confirm enhanced capacity due to prestress and transverse reinforcement. Why Other Options Are Wrong: Each of options a, b, and c accurately states a recognized advantage. Hence the comprehensive choice “All the above” is correct, while “None of these” is false. Common Pitfalls: Overlooking relaxation and other losses, inadequate anchorage/fretting in post-tensioning ducts, and insufficient cover or grouting that compromise durability. Final Answer: All the above.

Question 4

Span-to-overall-depth limit for one-way slabs: For a simply supported slab spanning in one direction, what is the basic maximum span/depth ratio used for preliminary design (before modification factors)?

Accepted Answer

Introduction / Context: Span-to-depth limits provide a rapid serviceability check to control deflections without full calculation. For one-way slabs, codes give basic ratios depending on support conditions that are later adjusted by modification factors for tension steel and compression reinforcement. Given Data / Assumptions: One-way slab behavior, simply supported at ends. Basic ratio sought prior to applying modification factors. Normal reinforcement and concrete grades assumed. Concept / Approach: Simply supported slabs are the most flexible among common support cases and therefore have the most restrictive (lowest) basic span/depth ratio. Adopting the basic value streamlines sizing, with further refinement via percentage steel, stress levels, and finishes. Step-by-Step Solution: Identify slab as simply supported, one-way.Adopt the basic limit span/depth = 20 for preliminary sizing.Apply modification factors later if needed to raise or lower this limit based on reinforcement and stress. Verification / Alternative check: Comparing continuous slabs (often 26) or cantilevers (often 7) confirms that simply supported cases fall at 20, aligning with common code tables. Why Other Options Are Wrong: Higher values such as 25, 30, or 35 correspond to stiffer support or modified conditions and risk excessive deflection for simple supports. 15 is overly conservative and uneconomical in many ordinary cases. Common Pitfalls: Using beam limits for slabs, ignoring modification factors, or not checking detailed deflection when finishes/partitions are sensitive. Final Answer: 20.

Question 5

Maximum spacing (pitch) of main tension reinforcement in slabs: The pitch of main bars in a reinforced concrete slab should not exceed which multiple of the slab’s effective depth (subject to absolute spacing limits)?

Accepted Answer

Introduction / Context: Bar spacing rules in slabs ensure crack control, load distribution, and proper bond. For the main tension reinforcement, both a relative limit (as a multiple of effective depth) and an absolute limit (a maximum in millimeters) are applied in design practice. Given Data / Assumptions: Reinforced concrete slab in flexure with main tension steel. Effective depth d is known. Absolute spacing limit (such as 300 mm) also applies, but question asks the multiple of d. Concept / Approach: To control flexural cracking and ensure adequate distribution, the maximum spacing of main bars is commonly limited to 3d or an absolute cap (whichever is smaller). This balances crack width restraint with constructability and bar economy. Step-by-Step Solution: Determine effective depth d of the slab.Limit main bar spacing to ≤ 3d and also ≤ the absolute cap (often 300 mm).Provide distribution bars orthogonal to the main reinforcement per code minimums. Verification / Alternative check: Crack control checks (bar diameter, spacing, stress) confirm that limiting pitch to 3d is compatible with acceptable service crack widths and load distribution. Why Other Options Are Wrong: 4d and 5d exceed typical limits and can lead to wide cracks. 2d or equal to d are overly conservative and uneconomical for most slabs. Common Pitfalls: Forgetting the absolute cap (e.g., 300 mm), using too-large bars at wide spacing, or not providing adequate distribution reinforcement. Final Answer: Three times the effective depth.

Question 6

Definition of a short reinforced concrete column: A column may be treated as a short column if its slenderness ratio (effective length / least lateral dimension) is less than which threshold?

Accepted Answer

Introduction / Context: Column classification into short or slender governs the design approach. Short columns primarily fail by material crushing when overloaded, while slender columns are dominated by instability (buckling) and require second-order effects to be considered. Given Data / Assumptions: Slenderness ratio defined as effective length divided by least lateral dimension. Ordinary building column with braced conditions. Code-aligned threshold used for classification. Concept / Approach: If the slenderness ratio is below a small threshold, deflection magnification and secondary moments are minimal and crushing governs. A commonly used boundary for reinforced concrete is 12, below which columns are treated as short for design purposes. Step-by-Step Solution: Compute effective length Le based on end restraints and bracing.Obtain least lateral dimension b of the column.Evaluate slenderness ratio SR = Le / b; if SR < 12, treat as short column. Verification / Alternative check: Comparing with typical limits—higher limits indicate growing slenderness and buckling sensitivity—confirms that 12 is a conservative boundary for many reinforced concrete design guides. Why Other Options Are Wrong: 18, 24, 30, 40: these values permit too much slenderness; second-order effects can no longer be neglected safely. Common Pitfalls: Using unsupported length instead of effective length, neglecting sway sensitivity, or using the larger lateral dimension rather than the least. Final Answer: 12.

Question 7

Losses in prestressing after transfer: Which list correctly identifies common sources of prestress loss in practice (excluding friction/anchorage for simplicity)?

Accepted Answer

Introduction / Context: Prestress in concrete reduces over time due to short-term and long-term mechanisms. Understanding these losses is crucial to ensure that the effective prestress at service remains adequate to control cracking and deflection. Given Data / Assumptions: Losses considered after transfer to the concrete. Friction and anchorage slip are omitted here by the question. Material properties correspond to typical prestressed concrete practice. Concept / Approach: The main loss components are: (1) elastic shortening of concrete at transfer (especially significant in pretensioned members), (2) creep of concrete under sustained compression, (3) shrinkage of concrete as moisture is lost, and (4) relaxation of prestressing steel (reduction of steel stress under constant strain). Together, they can be a substantial percentage of the initial prestress. Step-by-Step Solution: Quantify immediate elastic loss using concrete elastic modulus and transformed-section relations.Estimate creep loss based on sustained stress and creep coefficients over time.Estimate shrinkage loss using shrinkage strain models and steel modulus.Account for steel relaxation per manufacturer or code curves (low-relaxation strands reduce this loss). Verification / Alternative check: Comparing code-based estimates (short-term plus long-term) against long-term monitoring validates that these four factors dominate loss predictions in practice. Why Other Options Are Wrong: Options b and c: single-mechanism lists are incomplete. Option d: creep of steel is not a recognized dominant mechanism; relaxation is the appropriate steel-related loss. Option e: incorrect because real losses exist and must be considered. Common Pitfalls: Double-counting elastic losses in post-tensioned members, ignoring environmental humidity on shrinkage, and assuming zero relaxation for all steel types. Final Answer: Shrinkage of concrete, elastic shortening of concrete, creep of concrete, relaxation of prestressing steel.

Question 8

Axially loaded R.C.C. short column: Given Ac (area of concrete), Asc (area of longitudinal steel), A (gross section area), m (modular ratio), and σc (permissible compressive stress in concrete), which expression for the axial load-carrying strength …

Accepted Answer

Introduction / Context: In reinforced concrete (R.C.C.) columns designed by the working stress method, the axial load capacity is obtained by summing the contributions of concrete and steel at their permissible stresses. This question checks whether you recognize the equivalent algebraic forms of the same strength expression when expressed in terms of Ac, Asc, A, and the modular ratio m. Given Data / Assumptions: Ac = area of concrete only. Asc = area of longitudinal steel. A = gross area = Ac + Asc. σc = permissible compressive stress in concrete. m = modular ratio (Es/Ec under working stress concepts). Short, concentrically loaded column; secondary effects ignored. Concept / Approach: Under concentric compression, steel shares the load in proportion to its modulus difference relative to concrete, hence the use of modular ratio m. The strength (permissible axial load) equals the sum of concrete's share plus steel's share converted into an equivalent concrete stress block using m. Step-by-Step Solution: Start with P = σc Ac + (m σc) Asc.Since A = Ac + Asc, Ac = A − Asc → P = σc (A − Asc) + m σc Asc.Factor σc: P = σc [A + (m − 1) Asc]. Verification / Alternative check: All three expressions reduce to the same value when you substitute Ac = A − Asc, proving their equivalence algebraically and dimensionally. Why Other Options Are Wrong: Options (a), (b), and (c) are each correct individually; the only fully correct choice that captures this is “All of the above.” “None of these” is obviously incorrect. Common Pitfalls: Forgetting that A = Ac + Asc; mixing permissible stresses with design strengths; applying the expression to columns with eccentricity without modifications. Final Answer: All of the above

Question 9

Continuous floor slab supported on beams: What ratio of end-span length to intermediate-span length is typically adopted for analysis and detailing assumptions?

Accepted Answer

Introduction / Context: For continuous slabs supported on beams, simplified span assumptions are used when applying moment-coefficient methods or when distributing loads and reinforcement. End spans are commonly taken slightly shorter (or treated differently) than interior spans to reflect support conditions and cracking patterns. This question asks for the typical ratio used in many design guides and exam problems. Given Data / Assumptions: Normal continuous one-way slab strips supported on beams. Regular framing; no large stiffness discontinuities. Rule-of-thumb span ratio for preliminary design and moment coefficients. Concept / Approach: Because end spans have only one interior support, their negative moment capacity and continuity differ from interior spans. Many textbooks adopt end-span length ≈ 0.8 of interior-span length as a practical simplifying assumption to apply standard coefficient tables consistently and to keep deflections and cracking comparable across spans. Step-by-Step Solution: Identify end span and interior span conditions.Apply the common assumption: L_end ≈ 0.8 * L_interior.Use this ratio in moment-coefficient tables and reinforcement detailing. Verification / Alternative check: Comparative analyses (elastic frame or FEM) show end spans often develop somewhat different moment envelopes; the 0.8 ratio is a conservative simplification used in many curricula. Why Other Options Are Wrong: 0.6 and 0.7 make the end span overly short relative to interior spans, affecting continuity assumptions; 0.9 is closer to equal spans but does not reflect the commonly adopted simplification for teaching-level design. Common Pitfalls: Treating end and interior spans as identical without checking support fixity; ignoring serviceability differences; not validating assumptions when significant stiffness changes occur. Final Answer: 0.8

Question 10

Proportioning an R.C.C. beam section: A common preliminary thumb rule keeps the ratio of breadth b to effective depth d approximately equal to what value?

Accepted Answer

Introduction / Context: At the preliminary stage of reinforced concrete beam design, dimensional ratios are used to obtain a practical section before detailed checks. The breadth b to effective depth d ratio affects shear capacity, bar placement, and deflection control. Many textbooks provide a simple starting ratio for ordinary beams carrying typical floor loads. Given Data / Assumptions: Ordinary reinforced concrete beam, simply supported or continuous. Serviceability and constructability considerations typical of building work. Working-stress or limit-state preliminary sizing intent. Concept / Approach: Keeping b about one quarter of d (b/d ≈ 0.25) often balances shear web width, bar congestion, and economy, while allowing sufficient depth to control deflections. Final proportions must be verified by bending, shear, deflection, and detailing checks, but 0.25 is a standard rule-of-thumb starting point in many courses and older handbooks. Step-by-Step Solution: Choose d from span/overall-depth limits.Set b ≈ 0.25 d initially.Iterate after detailed moment, shear, and serviceability checks. Verification / Alternative check: Trial designs using b/d = 0.25 commonly meet shear and reinforcement spacing requirements; adjustments are made if serviceability or bar spacing dictates a wider web. Why Other Options Are Wrong: 0.50, 0.70, and 0.75 produce excessively wide beams for a given depth in typical building floors, which may be uneconomical or create beam-slab conflicts. “None of these” is incorrect because 0.25 is the widely taught starting value. Common Pitfalls: Freezing preliminary ratios without later verification; ignoring minimum web width for bar placement and shear links; not coordinating with architectural beam widths. Final Answer: 0.25

Question 11

Combined footing for two equal column loads with restricted overall width: Which footing shape is generally adopted to satisfy centroidal alignment and uniform soil pressure?

Accepted Answer

Introduction / Context: When two columns of equal load share a combined footing and site width is limited (e.g., property line constraint), the footing must still position the resultant soil reaction under the combined centroid of column loads. For equal loads at reasonable spacing, a rectangular combined footing is typically suitable and economical. Given Data / Assumptions: Two columns with equal axial loads. Restricted overall footing width (e.g., due to boundary line). Uniform allowable soil bearing capacity. Eccentricity control needed so resultant passes through centroid. Concept / Approach: For equal loads and symmetric geometry, the pressure distribution is uniform when the footing centroid coincides with the load resultant. A rectangular footing centered beneath the two columns satisfies this condition if its length spans between columns and is centered along the line joining them. Trapezoidal shapes are usually adopted when loads are unequal or when a property line prevents centering the footing; triangular shapes do not suit two equal point loads well. Step-by-Step Solution: Establish column loads: P and P.Resultant = 2P at the midpoint between columns.Choose a rectangular plan area centered on the midpoint to give uniform soil pressure q = (2P)/A within allowable limits. Verification / Alternative check: Basic statics confirms that a symmetric rectangular area gives no net moment about the centroid line, maintaining uniform bearing. A trapezoid is warranted primarily when P1 ≠ P2 or centering is not feasible. Why Other Options Are Wrong: Square is only possible if spacing and allowable pressure make it viable; triangular and circular are inappropriate for two equal, spaced point loads; trapezoidal is for unequal loads or boundary constraints causing eccentricity. Common Pitfalls: Using a trapezoid unnecessarily; ignoring footing self-weight and overburden; failing to check punching shear near columns. Final Answer: Rectangular

Question 12

Self-weight of footing in foundation design: How should it be treated in pressure and sizing calculations?

Accepted Answer

Introduction / Context: In shallow foundation design, the base pressure equals the total downward load divided by the plan area. The total downward load includes the superstructure load transferred at the column base plus the self-weight of the footing and the weight of soil backfill over the footing. Recognizing these components is essential to size the footing correctly and to compute base pressure for bearing and settlement checks. Given Data / Assumptions: Isolated or combined shallow footing on soil with allowable bearing capacity. Dead load of footing concrete and overburden soil are present. Service-level bearing pressure design by classical approach. Concept / Approach: Upward soil reaction must balance the total downward load. Therefore, footing self-weight and backfill are included both in pressure computations and in determining the required area A = (P_struct + W_footing + W_soil) / q_allowable. Neglecting these terms underestimates base pressure and can undersize the footing. Step-by-Step Solution: Write total downward W_total = P + W_f + W_s.Compute base pressure q = W_total / A.Determine A such that q ≤ q_allowable → A ≥ W_total / q_allowable. Verification / Alternative check: Iterative design often assumes A, estimates W_f, recomputes q, and adjusts A until convergence. This confirms the inclusion of self-weight is necessary in both pressure and area sizing. Why Other Options Are Wrong: (a) and (c) omit significant downward loads, leading to unconservative designs. (d) combines a correct statement with an incorrect one. (e) is too restrictive; self-weight matters for pressure and strength, not only settlement. Common Pitfalls: Ignoring self-weight; forgetting water table effects on buoyant weight; failing to include footing weight when checking sliding and overturning for retaining structures. Final Answer: Also considered for calculating the upward pressure on the footing

Question 13

Effect of over-reinforcing a singly reinforced concrete beam: By how much (approximately) can the moment of resistance be increased beyond the balanced condition?

Accepted Answer

Introduction / Context: In working-stress-era beam design, a “balanced” section is one where concrete reaches its permissible compressive stress simultaneously as steel reaches its permissible tensile stress. Adding more tensile steel (over-reinforcing) without altering dimensions does not proportionally increase the moment capacity because the limiting concrete stress controls. This question checks the typical upper bound often cited in basic texts. Given Data / Assumptions: Singly reinforced rectangular section under bending. Working stress method assumptions and permissible stresses. Concrete compression is governing at balanced or over-reinforced states. Concept / Approach: Beyond the balanced level, additional tension steel carries little extra moment because the compression block of concrete governs the capacity. Textbooks commonly note that over-reinforcement may raise the moment of resistance modestly, typically not exceeding about 25% over the balanced value, after which increasing depth or making the beam doubly reinforced is more effective. Step-by-Step Solution: Identify balanced moment M_bal from stress-block relations.Add tension steel beyond balanced: concrete stress still controls → limited gain.Rule-of-thumb upper increment ≈ 25% for over-reinforced singly reinforced sections. Verification / Alternative check: Comparison of M values from basic rectangular stress block relations confirms diminishing returns of adding tensile steel without increasing depth or adding compression steel. Why Other Options Are Wrong: 10–20% are conservative but do not reflect the common upper bound; 40% is too high for the stated assumptions. Common Pitfalls: Pursuing heavy tension reinforcement to gain moment instead of increasing depth; ignoring ductility and failure mode concerns with over-reinforced sections. Final Answer: 25%

Question 14

Retaining wall safety against sliding: When the factor of safety is found < 1.5, the downward projection at the heel provided to mobilize passive resistance is called what?

Accepted Answer

Introduction / Context: Retaining walls must be checked for overturning, sliding, and bearing. If the sliding check gives a factor of safety below the target (often 1.5 at working loads), additional measures are taken to increase resistance. One common detail is a shear key projecting downward from the base slab, typically under the heel, to mobilize passive earth pressure and increase the sliding capacity. Given Data / Assumptions: Gravity or cantilever retaining wall with base slab (toe and heel). Sliding check shows FS < 1.5. Foundation soil can develop passive pressure in front of a key. Concept / Approach: A shear key increases resistance by engaging passive pressure on its front face, adding to frictional resistance at the base. It is distinct from a cut-off wall used primarily for seepage control and different from ribs used for stiffening thick members. Step-by-Step Solution: Compute sliding FS = resisting / driving forces.If FS < 1.5, provide a key sized by passive pressure and shear transfer.Recompute FS including passive resistance on key face to verify adequacy. Verification / Alternative check: Alternative measures include increasing base width, adding anchors, or roughening the base. The shear key is a practical detail when plan dimensions are constrained. Why Other Options Are Wrong: (b) A cut-off wall addresses seepage, not primarily sliding. (c) A rib is not the standard term or function for this purpose. (d) “All” is incorrect because functions differ. Common Pitfalls: Overestimating passive pressure without accounting for excavation/compaction; placing the key on the wrong side; ignoring global stability or bearing after adding the key. Final Answer: A key (shear key)

Question 15

Deflection control criterion: For a simply supported reinforced concrete beam under normal conditions, the span-to-overall-depth ratio should not exceed what value to satisfy basic stiffness limits?

Accepted Answer

Introduction / Context: Serviceability (deflection) control is essential in beam design to prevent cracking, ponding, and functional issues. Many codes present basic span-to-depth ratios as starting limits for stiffness, which may be modified for tension steel percentage and stress levels. This question recalls the basic limit for simply supported beams in traditional teaching practice. Given Data / Assumptions: Simply supported beam. Normal service conditions and materials. Overall depth D used in the ratio; basic (unmodified) limit considered. Concept / Approach: A common baseline limit is L/D ≤ 20 for simply supported beams. Continuous beams often allow higher limits due to continuity (e.g., 26–28 baseline), while cantilevers are more restrictive (e.g., 7–10). Modifications are later applied based on steel stress and compression reinforcement, but the unmodified benchmark is 20 for simply supported members. Step-by-Step Solution: Identify support condition: simply supported.Apply basic stiffness rule: L/D ≤ 20.Adjust if code-specific modification factors are used. Verification / Alternative check: Elastic deflection calculations with typical spans and reinforcement confirm that L/D around 20 is a practical starting point to meet serviceability in ordinary floors. Why Other Options Are Wrong: 10 and 15 are overly conservative; 25 or 30 are too slender for a baseline limit without favorable modification factors and may lead to excessive deflection. Common Pitfalls: Confusing effective depth d with overall depth D in the ratio; applying continuous-beam limits to simply supported beams; skipping modification checks at high service stresses. Final Answer: 20

Question 16

Plain concrete (unreinforced) columns: Up to what limit of slenderness may plain concrete columns be used, expressed as the ratio of unsupported length to least lateral dimension?

Accepted Answer

Introduction / Context: Plain concrete has negligible tension capacity and limited ductility. Therefore, code provisions restrict its use for columns (pedestals) to very stocky members with small slenderness. This question asks for the typical slenderness limit stated as the ratio of unsupported length to the least lateral dimension of the section. Given Data / Assumptions: Plain concrete column or pedestal under axial load. Unsupported length measured between restraints. Least lateral dimension is the smaller cross-sectional dimension. Concept / Approach: To avoid instability and to keep eccentricity effects minimal, the unsupported length should be very short relative to the cross-section size. A commonly cited restriction is that plain concrete columns should not have unsupported length exceeding three times the least lateral dimension; beyond this, reinforcement must be provided to ensure ductility and stability. Step-by-Step Solution: Identify least lateral dimension b_min.Check slenderness ratio: L_unsupported / b_min ≤ 3 for plain concrete columns.If the ratio exceeds 3, design as an R.C.C. column with minimum steel and ties. Verification / Alternative check: Euler-type instability and accidental eccentricities become critical as slenderness grows; providing reinforcement and ties addresses these concerns for taller columns. Why Other Options Are Wrong: 2 is overly restrictive and uneconomical; 4–6 are too slender for plain concrete and risk instability and brittle failures. Common Pitfalls: Ignoring minimum eccentricity; misidentifying the unsupported length; assuming plain concrete can be used for tall piers without reinforcement. Final Answer: Three times

Question 17

In reinforced cement concrete (RCC) design using the working stress method, what is the commonly adopted allowable (permissible) tensile stress for mild steel used in stirrups (shear reinforcement)? Choose the closest standard value in kg/cm².

Accepted Answer

Introduction / Context: In RCC design under the traditional working stress method, permissible (allowable) stresses are specified for different materials. For mild steel (plain bars) used as stirrups or ties, designers use a standard allowable tensile stress to proportion shear reinforcement safely while controlling service-level behavior. Given Data / Assumptions: Material: mild steel stirrups (typical yield strength of plain MS ≈ 250 MPa). Design philosophy: working stress method with permissible stresses. Unit system: kg/cm² as used in many legacy RCC problems. Concept / Approach: Permissible stress for mild steel in tension under working stress design has been conventionally taken around 1400 kg/cm². This value reflects historical safety factors applied to the characteristic strength to keep service-stress levels within elastic range and to limit crack widths under service loads. Step-by-Step Solution: Identify the steel type: mild steel (not high-strength deformed bars).Use standard permissible tensile stress for mild steel in shear reinforcement design = about 1400 kg/cm².Select the matching option. Verification / Alternative check: Legacy handbooks and many exam-oriented RCC texts list 1400 kg/cm² as the permissible tensile stress for mild steel in tension for working stress calculations, especially for stirrups and ties. Modern limit-state codes use different approaches (factored capacities), but the question context is clearly working stress in kg/cm². Why Other Options Are Wrong: 190 kg/cm² and 230 kg/cm²: Far too low; they do not reflect conventional permissible levels for mild steel tension in working stress design. 260 kg/cm²: Still much lower than practice and would result in uneconomical, over-conservative reinforcement. Common Pitfalls: Mixing up permissible stresses of mild steel with those of high strength bars; confusing limit-state design parameters with working stress permissible values; converting MPa to kg/cm² incorrectly (1 MPa ≈ 10.197 kg/mm² = 101.97 kg/cm²). Final Answer: 1400 kg/cm²

Question 18

Staircase design — minimum headroom clearance: What is the minimum head room that must be provided over a stair (measured vertically from the nosing line to the soffit above)?

Accepted Answer

Introduction / Context: Head room over stairs ensures safe passage without the risk of head injury. Building practices and many codes prescribe a minimum vertical clearance measured from the line joining the nosings of treads to the underside of the soffit or any obstruction above. Given Data / Assumptions: Conventional residential or institutional stair in RCC/masonry. Vertical headroom measured along the stair flight. Standard practice values expressed in centimeters. Concept / Approach: Common practice and many code-based guides adopt about 2.1 m as the minimum headroom for comfort and safety. This allows users to traverse the stair without having to stoop, accommodates taller individuals, and permits finishes/services below the slab or floor framing without infringing on clearance. Step-by-Step Solution: Identify the metric minimum: approximately 2.1 m = 210 cm.Check listed options and select 210 cm as the standard minimum.Note that larger values (e.g., 230 cm) are acceptable but not the minimum requirement. Verification / Alternative check: Architectural data handbooks and model codes internationally tend to cluster the minimum headroom around 2.0–2.1 m, with 2.1 m widely adopted for comfortable use and to reduce maintenance clashes with services. Why Other Options Are Wrong: 200 cm / 205 cm: Often considered marginal or below many standard recommendations, potentially uncomfortable. 230 cm: Permissible but not the stated minimum. Common Pitfalls: Measuring from the tread rather than the nosing line; forgetting finishes/plaster thickness that can reduce actual clearance below the required minimum; ignoring soffit drops or duct runs. Final Answer: 210 cm

Question 19

Shear safety check of an RCC beam carrying UDL: A simply supported RCC beam of span 6 m is 30 cm wide. The lever arm (j·d) is 55 cm. It carries a uniformly distributed load of 12 t/m (including self-weight). If the allowable shear stress in concrete…

Accepted Answer

Introduction / Context: Shear design in RCC beams requires checking the nominal shear stress against permissible limits. If the nominal shear exceeds the maximum allowable shear capacity of concrete (even with shear reinforcement), the section must be revised (e.g., increased depth or width) for safety. Given Data / Assumptions: Beam span L = 6 m (simply supported). Uniformly distributed load w = 12 t/m = 12,000 kg/m (includes self-weight). Width b = 30 cm; lever arm j·d = 55 cm (given directly). Allowable shear stress in concrete τ_allow = 5 kg/cm² (working stress context). Concept / Approach: For a simply supported beam under UDL, the maximum shear force occurs at the supports and equals V = wL/2. The nominal shear stress is approximated as τ_v = V / (b * j·d) when the effective lever arm j·d is given. Compare τ_v with the allowable value. If τ_v far exceeds code limits (and likely the maximum permissible with reinforcement), the section must be revised. Step-by-Step Solution: Compute support shear: V = wL/2 = 12,000*6/2 = 36,000 kg.Nominal shear stress: τ_v = V / (b * j·d) = 36,000 / (30 * 55) = 36,000 / 1,650 = 21.82 kg/cm² (approx.).Compare with allowable: 21.82 kg/cm² » 5 kg/cm² → greatly exceeds limit.Conclusion: Even with shear reinforcement, such a high nominal shear typically surpasses maximum permissible limits; revise section. Verification / Alternative check: In classic working stress design tables, maximum shear stresses that concrete can sustain (even with stirrups) are much lower than ~22 kg/cm². Increasing depth (thus j·d and b·j·d) significantly reduces τ_v, offering a viable revision path. Why Other Options Are Wrong: Safe in shear / with stirrups / with stirrups and inclined bars: The computed τ_v is far beyond permissible; these are inadequate remedies for such an exceedance. Common Pitfalls: Using d instead of j·d; forgetting to convert t/m to kg/m; checking shear at midspan instead of at supports; assuming stirrups can compensate unlimitedly for undersized sections. Final Answer: needs revision of section

Question 20

Flat slab column head (capital) sizing guideline: For a flat slab supported on columns, the diameter of the column head (capital) is generally taken as which fraction of the slab span length (rule-of-thumb)?

Accepted Answer

Introduction / Context: In flat slab construction, a column head (capital) or drop panel is often provided to reduce punching shear and to enhance negative moment capacity over columns. Preliminary sizing rules help during conceptual design before detailed analysis. Given Data / Assumptions: Flat slab (beamless) system with columns and possible capitals. Preliminary sizing question seeking a proportion to span length. Concept / Approach: A typical rule-of-thumb sets the capital diameter on the order of one-quarter of the span length to effectively spread the load path and reduce punching shear stress around column perimeters. Exact sizes are later refined by punching shear checks and moment distribution requirements. Step-by-Step Solution: Identify target: provide a capital large enough to reduce shear stresses.Adopt preliminary proportion: D_cap ≈ 0.25 × span.Select the corresponding option. Verification / Alternative check: Many design aids suggest 0.2–0.3 of span as a starting point, later adjusted after performing punching shear and serviceability checks. This aligns well with 0.25 of span. Why Other Options Are Wrong: Proportions to column diameter or fixed 4–5 cm increases are too small to influence punching shear effectively in most practical spans. “None of these” is unnecessary because a standard preliminary ratio exists. Common Pitfalls: Relying solely on thumb rules without verifying punching shear; forgetting the role of drop panels vs. column capitals; overlooking edge/corner column differences that may need larger capitals. Final Answer: 0.25 times the span length

RCC Structures Design Questions