Q: Development (embedment) length — for M 150 concrete with average bond conditions (average bending stress 6 kg/cm²), the required embedment length for a bar of diameter d as per IS 456 is approximately:

Introduction / Context: Development length (also called embedment length) ensures that reinforcing bars can safely transfer stress to the surrounding concrete without bond failure. In working-stress/older tabulated approaches used alongside M 150 concrete (nominal 1:2:4), the development length is often quoted as a multiple of the bar diameter d under average bond conditions. This question assesses recognition of that typical multiple for design memory work. Given Data / Assumptions: Concrete grade: M 150 (nominal mix context). Average bond conditions and average bending stress 6 kg/cm² stated. Bar in tension; diameter = d (goal is Ld expressed as a multiple of d). Concept / Approach: General bond design uses the relationship Ld = (phi * sigma_s) / (4 * tau_bd), where phi is bar diameter, sigma_s is steel stress to be developed, and tau_bd is design bond stress (dependent on concrete grade and bar type). For M 150 concrete and typical working-stress levels historically used in many references, the tabulated or memory value corresponding to average conditions is approximately 58 d for tension development in plain/deformed reinforcement at the indicated stress level. Step-by-Step Solution: Identify the code-era/tabulated multiple for M 150 under average bond → around 58 d.Match to options → select 58 d. Verification / Alternative check (if short method exists): A more rigorous derivation would require tau_bd from tables and sigma_s from permissible stress for the steel grade, but for this exam-style prompt, the remembered multiple suffices. Why Other Options Are Wrong: 28 d, 38 d, and 48 d are insufficient for the specified stress level with M 150; 95 d is unnecessarily high for the stated conditions. Common Pitfalls (misconceptions, mistakes): Using development lengths from higher concrete grades; forgetting that tension development length is longer than for compression; overlooking modifications for deformed bars and hooks where applicable. Final Answer: 58 d

Q: Purpose of secondary (distribution) steel — transverse reinforcement placed at right angles to the main reinforcement in slabs and other members primarily serves to:

Introduction / Context: Reinforced concrete slabs and walls include two orthogonal reinforcement systems: main (flexural) bars in the primary span direction and secondary (distribution) bars transverse to them. Understanding why we provide transverse steel is fundamental to good crack control and serviceability performance. Given Data / Assumptions: Member type: slab or similar thin R.C.C. element. Transverse bars are orthogonal to main bars. Serviceability limit states (crack width control) and durability are important objectives. Concept / Approach: Transverse (distribution) reinforcement equalizes strain fields and restrains crack widths, particularly those caused by temperature variations, drying shrinkage, and local stress concentrations near loads and supports. It also assists in distributing point or line loads to a wider area, reducing peak stresses in concrete. Hence, its functions span load distribution and restraint against temperature- and shrinkage-induced tensile strains. Step-by-Step Solution: Recognize typical slab behavior → bidirectional cracking risk if only one set of bars exists.Provide transverse steel → controls cracks from temperature and shrinkage.Transverse steel also spreads localized effects → improves load distribution. Verification / Alternative check (if short method exists): Design standards specify minimum transverse reinforcement percentages precisely to address these serviceability concerns and achieve a more uniform crack pattern with smaller widths. Why Other Options Are Wrong: Options A, B, and C are each true but incomplete alone. Only “all of the above” captures the complete purpose of distribution steel. Common Pitfalls (misconceptions, mistakes): Thinking transverse steel is optional because it contributes little to primary moment capacity; underestimating temperature and shrinkage effects in exposed slabs. Final Answer: all of the above

Q: Design of a simply supported R.C.C. beam subjected to uniformly distributed load (U.D.L.) — the governing bending moment location for flexural design is at:

Introduction / Context: In beam design, correctly identifying the critical bending moment location is the first step. For a simply supported beam under a uniformly distributed load (U.D.L.), the maximum positive bending moment occurs at midspan, which then governs the main flexural reinforcement design for sagging moments. Given Data / Assumptions: Support condition: simply supported (no end fixity). Loading: uniformly distributed load across the span. Objective: find the section where bending moment is maximum. Concept / Approach: For a simply supported beam with U.D.L., the shear force diagram is linear and the bending moment diagram is parabolic with its peak at the center. The maximum sagging moment value is wL^2/8 at midspan (where w is load per unit length and L the span). Therefore, the design for main tension reinforcement is typically controlled by the midspan section. Step-by-Step Solution: Write the standard results: R_A = R_B = wL/2.Compute bending moment function: M(x) = R_A x − w x^2 / 2.Differentiate to find extremum: dM/dx = 0 → x = L/2 (midspan).Evaluate M(L/2) = wL^2 / 8 → maximum sagging moment at midspan. Verification / Alternative check (if short method exists): Draw the bending moment diagram: a concave-down parabola peaking at midspan. This visual confirmation matches the analytic result. Why Other Options Are Wrong: Supports (A) carry zero bending moment for a pin support in the ideal model; “every section” (C) is not meaningful; quarter-span (D) is not where maximum occurs; near free edges (E) is irrelevant to simply supported beams. Common Pitfalls (misconceptions, mistakes): Confusing shear maxima near supports with moment maxima; forgetting that fixed or continuous beams shift critical sections toward supports for hogging moments. Final Answer: mid span

Q: Stair terminology — in one flight of stairs, the number of treads equals which relationship relative to the number of risers?

Introduction / Context: Correct stair terminology is vital for setting out safe and comfortable staircases. Two core terms are riser (vertical step height) and tread (horizontal going). Understanding their count relationship helps in layout and quantity estimation for architectural and structural drawings. Given Data / Assumptions: A single flight of stairs runs from one floor level to a landing or to the next floor. There is a starting floor at the bottom and a landing/floor at the top. The top level is reached by stepping onto the landing, not onto an extra tread beyond the last riser. Concept / Approach: In a typical flight, the number of risers equals the number of vertical increments needed to reach the next level. Treads are the horizontal surfaces between risers. Since the upper landing/floor serves as the final horizontal surface, the number of treads is one fewer than the number of risers in that flight. Step-by-Step Solution: Count total vertical increments → this gives the riser count.Recognize that the top landing provides the last horizontal surface.Therefore, treads = risers − 1. Verification / Alternative check (if short method exists): Create a quick sketch: label risers and intervening treads; you will observe one fewer tread than risers from floor to landing. Why Other Options Are Wrong: “Risers plus one” double counts the landing; “risers in the flight” ignores the landing; “none of these” is incorrect because a standard relation exists. Common Pitfalls (misconceptions, mistakes): Confusing stair runs that include an additional platform or winders; forgetting that the first step may start at floor level (no tread before the first riser). Final Answer: risers minus one

Q: Flange width limitation for T-beams — the effective flange width should be restricted to which of the following (select the least governing limit)?

Introduction / Context: In T-beams, part of the slab acts compositely with the beam web (rib) as the flange under compression. To avoid overestimating the contribution of slab concrete, design codes limit the effective flange width using practical geometric rules. Designers must select the least (most restrictive) of several limits to be safe. Given Data / Assumptions: Monolithic slab-beam construction forming T-section under sagging moments. Effective span, spacing of adjacent beams, and slab thickness are known design parameters. Standard effective flange width rules apply. Concept / Approach: Effective flange width accounts for shear lag and non-uniform stress distribution. The beam cannot mobilize the entire slab width equally; thus, codes limit the assumed flange width by span-based and spacing-based limits as well as a local width tied to web breadth plus a multiple of slab thickness. The controlling value is the smallest among the specified limits. Step-by-Step Solution: List limits: b_f ≤ L/3; b_f ≤ spacing of beams; b_f ≤ b_w + 12 * t_s.Compute each for the given geometry in actual design.Adopt the least → ensures conservative and safe design. Verification / Alternative check (if short method exists): Shear lag theory and test data underpin these limits; using the least value prevents unconservative flange participation assumptions. Why Other Options Are Wrong: Any single limit (A–C) alone may be unconservative; “Any of the above” (E) ignores the requirement to take the least. Common Pitfalls (misconceptions, mistakes): Adding limits instead of taking the minimum; confusing effective flange width with total slab width between beams. Final Answer: Least of the above

Question 1

Axially loaded short R.C.C. column: A 20 cm × 20 cm column is reinforced with four bars totaling 20 cm² of steel. With permissible stresses of 40 kg/cm² in concrete and 300 kg/cm² in steel, what is the maximum safe axial load?

Accepted Answer

none of these

Question 2

Flat slab terminology in building design: The locally thickened portion of a flat slab provided over a supporting column to reduce negative moments and punching shear is called what?

Accepted Answer

Introduction / Context: Flat slab systems distribute loads directly to columns without deep beams. To control negative moments around column supports and to enhance punching shear capacity, designers commonly use local thickening or flaring near columns. Knowing the correct terminology is essential for detailing and communication. Given Data / Assumptions: Flat slab with column supports. Local thickness increase of the slab over the column panel. No drop beams; two-way action prevails. Concept / Approach: The slab thickening over the column is called a drop panel. It reduces peak negative bending moments and increases the slab's punching shear capacity by increasing effective depth and critical perimeter. Related terms include capital or column head, which typically describe a flared enlargement of the column itself rather than a thickening of the slab plate. Step-by-Step Solution: Identify element: thickened slab zone above a column.Apply correct term: ‘‘drop panel’’ is the slab thickening; a ‘‘capital/column head’’ is a flaring or mushrooming of the column.Function: mitigate high negative moments and improve punching resistance. Verification / Alternative check: Design standards for flat slabs (direct design or equivalent frame methods) explicitly reference drop panels and column capitals, distinguishing their geometry and structural roles. Why Other Options Are Wrong: Capital / column head: These are column enlargements, not slab thickenings. None of these: Incorrect because the established term is ‘‘drop panel’’. Haunch: More typical for beam-slab interfaces or precast connections. Common Pitfalls: Confusing column capitals with drop panels; both may be used together in ‘‘mushroom’’ slabs but are distinct features. Final Answer: drop panel

Question 3

Minimum concrete strength requirement: The minimum 28-day cube compressive strength of concrete typically specified for prestressed concrete members is closest to which value?

Accepted Answer

Introduction / Context: Prestressed concrete relies on a high-quality concrete matrix to limit prestress losses, control cracking, and safely transfer high compressive stresses from tendons/strands. Therefore, minimum characteristic strengths are higher than for ordinary reinforced concrete. Given Data / Assumptions: Question refers to standard practice minimum 28-day cube strength for prestressed members. Working-stress era units: kg/cm² (1 MPa ≈ 10.2 kg/cm²). Concept / Approach: Common practice specifies concrete not less than about M40 (≈ 40 MPa) for prestressed members, especially pretensioned elements, to ensure adequate bond and early-age transfer. In kg/cm², 40 MPa ≈ 400 kg/cm², which aligns with the typical minimum strength cited in design guides and examinations. Step-by-Step Solution: Convert 40 MPa to kg/cm²: 40 * 10.2 ≈ 408 kg/cm² → reported as ≈ 400 kg/cm².Choose the nearest listed value representing standard minimum for PSC: 400 kg/cm². Verification / Alternative check: Many codes allow higher grades (e.g., M45, M50) and stipulate minimum transfer strengths at the time of prestress application (often ≥ 70% of f_ck). The exam-style minimum for the 28-day cube strength typically rounds to 400 kg/cm². Why Other Options Are Wrong: 50–250 kg/cm²: Too low for prestressed members; would risk bond failure and excessive losses. 350 kg/cm²: Used in some reinforced concrete, but below the common PSC baseline of ≈400 kg/cm². Common Pitfalls: Confusing minimum 28-day strength with minimum transfer strength (at jacking/release), which is an early-age requirement. Final Answer: 400 kg/cm²

Question 4

Circular slab under uniformly distributed load (UDL): The ratio of maximum negative radial moment to maximum positive radial moment in a circular slab is approximately equal to which value?

Accepted Answer

2

Question 5

Retaining wall detailing thumb rule: The thickness of the base slab of a cantilever retaining wall is generally provided as which proportion of the stem width at the base?

Accepted Answer

Introduction / Context: In preliminary proportioning of cantilever retaining walls, designers use practical thumb rules for member thicknesses before refined analysis. The base slab must be thick enough to resist bending and shear from soil/overburden reactions while remaining economical. Given Data / Assumptions: Conventional cast-in-place cantilever retaining wall. Proportioning at concept stage; soil and surcharge loads will be checked later. Stem thickness at the base is known (controls stress flow into the base slab near the heel/toe). Concept / Approach: Preliminary dimensions commonly adopt a base slab thickness around one-fourth of the stem thickness at the base. This provides a reasonable effective depth against flexure near heel and toe and ensures shear capacity prior to detailed design with load combinations and soil pressures. Step-by-Step Solution: Establish the relation between base slab thickness and stem thickness at base.Use the rule of thumb: t_base ≈ (1/4) * (stem width at base).Proceed to validate by structural analysis (bending, shear, punching near key sections) and serviceability checks (crack control). Verification / Alternative check: Iterative sizing with bending moment envelopes under active/passive pressures typically results in base slab thicknesses close to this rule, adjusted for material strengths and exposure conditions. Why Other Options Are Wrong: 1/2 or equal or twice: Typically excessive and uneconomical for normal walls. 1/3: Sometimes used conservatively, but 1/4 is a more standard starting point in many handbooks. Common Pitfalls: Using thumb rules without subsequent verification of heel/toe bending and shear, sliding, overturning, and bearing pressure under factored loads. Final Answer: one fourth of the width of the stem at the bottom

Question 6

Foundation support basics: A foundation transfers load to which supporting medium(s)? Choose the most accurate statement.

Accepted Answer

Introduction / Context: Foundations are structural elements that transmit loads from the superstructure to the ground. Clarity in terminology helps avoid design misunderstandings: ‘‘subgrade’’ and ‘‘foundation soil’’ are commonly used to denote the supporting ground strata beneath the foundation element. Given Data / Assumptions: General building foundation (shallow or deep). Soil or rock as the load-bearing medium. Standard geotechnical terminology. Concept / Approach: The foundation element (footing, mat, pile cap, etc.) distributes loads to the underlying geomaterial. That supporting medium is the ‘‘subgrade’’ or ‘‘foundation soil/rock.’’ Saying a foundation ‘‘rests on the base of the foundation’’ is tautological and not physically meaningful; the foundation itself bears on the soil or rock. Step-by-Step Solution: Identify the actual load-carrying medium: soil/rock beneath the foundation.Match terminology: ‘‘subgrade’’ and ‘‘foundation soil’’ both refer to that supporting medium.Therefore, the accurate choice is ‘‘both (b) and (c).’’ Verification / Alternative check: Soil–structure interaction models and bearing capacity equations (e.g., Terzaghi, Meyerhof) explicitly treat the underlying soil/rock as the resisting medium. Why Other Options Are Wrong: Base of the foundation: Not a supporting medium; it is the foundation's underside surface. Bedrock only: Not all foundations bear on rock; many bear on competent soils. Common Pitfalls: Confusing structural parts (footing base) with geotechnical support (soil/subgrade). Final Answer: both (b) and (c)

Question 7

Definition threshold for deep beams (continuous members): A continuous reinforced concrete beam shall be deemed a ‘‘deep beam’’ if the ratio of effective span (L_eff) to overall depth (D) is which of the following?

Accepted Answer

Introduction / Context: Deep beams are members in which plane sections do not remain plane; arching and strut-and-tie action dominate over classical beam theory. Codes therefore define deep beams by geometric limits that flag when special analysis and detailing rules are required. Given Data / Assumptions: Continuous beam (not a simple one-way slab strip). Effective span L_eff and overall depth D are known. Concept / Approach: For continuous beams, a typical code-style threshold is L_eff / D < 2 to classify a member as a deep beam. Such members need refined design approaches (e.g., strut-and-tie modeling), web reinforcement provisions, and special detailing at supports/openings due to nonlinear stress fields. Step-by-Step Solution: Compute slenderness ratio SR = L_eff / D.If SR < 2 → treat as deep beam; else → ordinary beam theory generally applies (subject to other checks).Therefore, select ‘‘less than 2’’ as the classification criterion. Verification / Alternative check: Design guides align with the same order-of-magnitude limit (with slightly different wording for simply supported vs. continuous members). The intent is to capture members where shear span is short and arching prevails. Why Other Options Are Wrong: 2.0 or 2.5 without the inequality does not state the classification boundary. less than 2.5: Too lenient; would misclassify many ordinary beams as deep. greater than 3: Opposite of deep-beam behavior; that is relatively slender. Common Pitfalls: Using span/effective depth (L/d) rather than span/overall depth (L/D) without checking the specific code clause and member category (simply supported vs. continuous). Final Answer: less than 2

Question 8

Reinforced concrete design — a reinforced concrete (R.C.C.) column is classified as a long column if its slenderness ratio exceeds which limiting value?

Accepted Answer

Introduction / Context: Classifying a reinforced concrete (R.C.C.) column as short or long is essential because slender (long) columns experience additional bending due to buckling effects (P–Δ), which lowers their axial load capacity compared with short columns. The classification is made using a non-dimensional slenderness ratio, and selecting the correct limit is a basic design knowledge check in building design and civil engineering exams. Given Data / Assumptions: The term slenderness ratio refers to effective length divided by an appropriate sectional dimension parameter. For routine conceptual questions, a common cut-off used for R.C.C. columns is taken as 40. The intent is conceptual classification (not a full code check of effective length factors, end conditions, or biaxial bending). Concept / Approach: Slenderness ratio is a measure of susceptibility to buckling. In reinforced concrete, when the ratio rises above a customary limit, second-order effects become significant and design must allow for magnified moments or a reduced axial capacity. A widely used educational threshold is 40 for a column to be treated as “long,” whereas “short” columns are below this limit and governed primarily by material strength and first-order bending effects. Step-by-Step Solution: Identify the classification criterion → compare slenderness ratio with a recognized threshold.For R.C.C. conceptual classification → use 40 as the limit separating short and long columns.Hence, if the slenderness ratio is greater than 40, the column is treated as long. Verification / Alternative check (if short method exists): In practice, designers also check code-specific rules (effective length factors, end restraints) and may use more conservative thresholds depending on the situation. The conceptual value of 40 aligns with many standard references and examination conventions. Why Other Options Are Wrong: 30 and 35 are too low for the usual R.C.C. classification; 50 or 60 are overly high and may under-predict slenderness effects, risking unsafe assumptions. Common Pitfalls (misconceptions, mistakes): Confusing concrete column limits with structural steel limits; forgetting to use effective length (not clear length) and the least lateral dimension or appropriate radius of gyration in actual design checks. Final Answer: 40

Question 9

Development (embedment) length — for M 150 concrete with average bond conditions (average bending stress 6 kg/cm²), the required embedment length for a bar of diameter d as per IS 456 is approximately:

Accepted Answer

Introduction / Context: Development length (also called embedment length) ensures that reinforcing bars can safely transfer stress to the surrounding concrete without bond failure. In working-stress/older tabulated approaches used alongside M 150 concrete (nominal 1:2:4), the development length is often quoted as a multiple of the bar diameter d under average bond conditions. This question assesses recognition of that typical multiple for design memory work. Given Data / Assumptions: Concrete grade: M 150 (nominal mix context). Average bond conditions and average bending stress 6 kg/cm² stated. Bar in tension; diameter = d (goal is Ld expressed as a multiple of d). Concept / Approach: General bond design uses the relationship Ld = (phi * sigma_s) / (4 * tau_bd), where phi is bar diameter, sigma_s is steel stress to be developed, and tau_bd is design bond stress (dependent on concrete grade and bar type). For M 150 concrete and typical working-stress levels historically used in many references, the tabulated or memory value corresponding to average conditions is approximately 58 d for tension development in plain/deformed reinforcement at the indicated stress level. Step-by-Step Solution: Identify the code-era/tabulated multiple for M 150 under average bond → around 58 d.Match to options → select 58 d. Verification / Alternative check (if short method exists): A more rigorous derivation would require tau_bd from tables and sigma_s from permissible stress for the steel grade, but for this exam-style prompt, the remembered multiple suffices. Why Other Options Are Wrong: 28 d, 38 d, and 48 d are insufficient for the specified stress level with M 150; 95 d is unnecessarily high for the stated conditions. Common Pitfalls (misconceptions, mistakes): Using development lengths from higher concrete grades; forgetting that tension development length is longer than for compression; overlooking modifications for deformed bars and hooks where applicable. Final Answer: 58 d

Question 10

Purpose of secondary (distribution) steel — transverse reinforcement placed at right angles to the main reinforcement in slabs and other members primarily serves to:

Accepted Answer

Introduction / Context: Reinforced concrete slabs and walls include two orthogonal reinforcement systems: main (flexural) bars in the primary span direction and secondary (distribution) bars transverse to them. Understanding why we provide transverse steel is fundamental to good crack control and serviceability performance. Given Data / Assumptions: Member type: slab or similar thin R.C.C. element. Transverse bars are orthogonal to main bars. Serviceability limit states (crack width control) and durability are important objectives. Concept / Approach: Transverse (distribution) reinforcement equalizes strain fields and restrains crack widths, particularly those caused by temperature variations, drying shrinkage, and local stress concentrations near loads and supports. It also assists in distributing point or line loads to a wider area, reducing peak stresses in concrete. Hence, its functions span load distribution and restraint against temperature- and shrinkage-induced tensile strains. Step-by-Step Solution: Recognize typical slab behavior → bidirectional cracking risk if only one set of bars exists.Provide transverse steel → controls cracks from temperature and shrinkage.Transverse steel also spreads localized effects → improves load distribution. Verification / Alternative check (if short method exists): Design standards specify minimum transverse reinforcement percentages precisely to address these serviceability concerns and achieve a more uniform crack pattern with smaller widths. Why Other Options Are Wrong: Options A, B, and C are each true but incomplete alone. Only “all of the above” captures the complete purpose of distribution steel. Common Pitfalls (misconceptions, mistakes): Thinking transverse steel is optional because it contributes little to primary moment capacity; underestimating temperature and shrinkage effects in exposed slabs. Final Answer: all of the above

Question 11

Design of a simply supported R.C.C. beam subjected to uniformly distributed load (U.D.L.) — the governing bending moment location for flexural design is at:

Accepted Answer

Introduction / Context: In beam design, correctly identifying the critical bending moment location is the first step. For a simply supported beam under a uniformly distributed load (U.D.L.), the maximum positive bending moment occurs at midspan, which then governs the main flexural reinforcement design for sagging moments. Given Data / Assumptions: Support condition: simply supported (no end fixity). Loading: uniformly distributed load across the span. Objective: find the section where bending moment is maximum. Concept / Approach: For a simply supported beam with U.D.L., the shear force diagram is linear and the bending moment diagram is parabolic with its peak at the center. The maximum sagging moment value is wL^2/8 at midspan (where w is load per unit length and L the span). Therefore, the design for main tension reinforcement is typically controlled by the midspan section. Step-by-Step Solution: Write the standard results: R_A = R_B = wL/2.Compute bending moment function: M(x) = R_A x − w x^2 / 2.Differentiate to find extremum: dM/dx = 0 → x = L/2 (midspan).Evaluate M(L/2) = wL^2 / 8 → maximum sagging moment at midspan. Verification / Alternative check (if short method exists): Draw the bending moment diagram: a concave-down parabola peaking at midspan. This visual confirmation matches the analytic result. Why Other Options Are Wrong: Supports (A) carry zero bending moment for a pin support in the ideal model; “every section” (C) is not meaningful; quarter-span (D) is not where maximum occurs; near free edges (E) is irrelevant to simply supported beams. Common Pitfalls (misconceptions, mistakes): Confusing shear maxima near supports with moment maxima; forgetting that fixed or continuous beams shift critical sections toward supports for hogging moments. Final Answer: mid span

Question 12

Stair terminology — in one flight of stairs, the number of treads equals which relationship relative to the number of risers?

Accepted Answer

Introduction / Context: Correct stair terminology is vital for setting out safe and comfortable staircases. Two core terms are riser (vertical step height) and tread (horizontal going). Understanding their count relationship helps in layout and quantity estimation for architectural and structural drawings. Given Data / Assumptions: A single flight of stairs runs from one floor level to a landing or to the next floor. There is a starting floor at the bottom and a landing/floor at the top. The top level is reached by stepping onto the landing, not onto an extra tread beyond the last riser. Concept / Approach: In a typical flight, the number of risers equals the number of vertical increments needed to reach the next level. Treads are the horizontal surfaces between risers. Since the upper landing/floor serves as the final horizontal surface, the number of treads is one fewer than the number of risers in that flight. Step-by-Step Solution: Count total vertical increments → this gives the riser count.Recognize that the top landing provides the last horizontal surface.Therefore, treads = risers − 1. Verification / Alternative check (if short method exists): Create a quick sketch: label risers and intervening treads; you will observe one fewer tread than risers from floor to landing. Why Other Options Are Wrong: “Risers plus one” double counts the landing; “risers in the flight” ignores the landing; “none of these” is incorrect because a standard relation exists. Common Pitfalls (misconceptions, mistakes): Confusing stair runs that include an additional platform or winders; forgetting that the first step may start at floor level (no tread before the first riser). Final Answer: risers minus one

Question 13

Prestressing methods — which statement best describes the current use-case and economy of post-tensioning systems in concrete construction?

Accepted Answer

Introduction / Context: Prestressed concrete enhances span capacity and serviceability by introducing a pre-compression that counteracts tensile stresses from external loads. There are two principal methods: pre-tensioning (tendons stressed before casting) and post-tensioning (tendons stressed after concrete hardens). Selecting the right method impacts economy and constructability, especially for long spans. Given Data / Assumptions: Modern bridges, parking structures, long-span floors, and segmental construction often exploit post-tensioning. Economy is judged across materials, formwork, construction speed, and life-cycle performance. Question focuses on broad practical adoption rather than a niche case. Concept / Approach: Post-tensioning uses ducts cast into the concrete. After curing, tendons are stressed and anchored, allowing long spans with thinner sections and reduced cracking/deflection. For large spans, savings from reduced self-weight, fewer supports, and faster erection typically outweigh tendon and anchorage costs. Consequently, it is a mainstream solution today for bridges and large floor plates. Step-by-Step Solution: Recognize span length as a key driver → larger spans benefit more from prestress.Post-tensioning is versatile → segmental, cast-in-place, and slab band systems.Therefore, the economical, commonly adopted approach for large spans is post-tensioning. Verification / Alternative check (if short method exists): Industry practice and numerous case studies show cost/weight reductions and performance improvements using post-tensioning for spans beyond conventional reinforced solutions. Why Other Options Are Wrong: It is not obsolete (A) nor generally uneconomical (B); option D is inapplicable; option E is factually incorrect. Common Pitfalls (misconceptions, mistakes): Assuming prestress is only for bridges; overlooking post-tensioned slabs in buildings and parking structures. Final Answer: Post-tensioning is economical for large spans and is widely adopted nowadays.

Question 14

Flexural design basis — the quantity of main reinforcement in a slab should primarily be based on which structural demand?

Accepted Answer

Introduction / Context: In reinforced concrete slabs, the principal steel (main reinforcement) is placed in the direction of the largest bending moments to resist tensile stresses. Correctly identifying the governing internal action ensures adequate strength and serviceability without uneconomical over-reinforcement. Given Data / Assumptions: Slab is designed primarily for flexure in one or two directions depending on support and aspect ratio. Shear and torsion checks are also required but usually do not dictate the main bar quantity in typical floor slabs. Question asks what governs the amount of main reinforcement. Concept / Approach: Determination of main steel area As is based on the design bending moment envelope. Codes also require minimum temperature and shrinkage reinforcement orthogonal to main bars, but the main bar area itself must satisfy the maximum design flexural demand to meet ultimate limit state requirements. Step-by-Step Solution: Establish design loading and support conditions → find bending moment distribution.Identify peak (maximum) design moment(s) in the governing direction(s).Compute required As to resist those maximum bending moments with appropriate material strengths and safety factors. Verification / Alternative check (if short method exists): Check that As also exceeds minimum crack-control steel; verify deflection and crack width criteria. Shear check is performed, but in ordinary slabs shear rarely controls bar quantity (shear is often satisfied without shear reinforcement). Why Other Options Are Wrong: “Minimum bending moment” would under-design; “maximum shear” or “minimum shear” do not set main bar quantity for flexure; torsion may affect edge strips or corners but does not generally dictate the main bar quantity across the slab panel. Common Pitfalls (misconceptions, mistakes): Confusing distribution (secondary) steel with main steel; ignoring local peak moments near openings or discontinuities. Final Answer: maximum bending moment

Question 15

Pile selection — which is a comprehensive advantage of using a concrete pile instead of a timber pile?

Accepted Answer

Introduction / Context: Pile foundations transfer loads to deeper soils/rock or mobilize skin friction when near-surface soils are weak. Material choice (timber, steel, reinforced or prestressed concrete) affects durability, capacity, and constructability. This question contrasts concrete and timber piles in terms of practical advantages. Given Data / Assumptions: Concrete pile refers to cast-in-situ or precast (reinforced or prestressed) concrete. Timber piles are susceptible to biological decay and insect attack above permanent groundwater level. Design must consider capacity, durability, and installation limits. Concept / Approach: Concrete piles offer superior durability in aggressive environments (with proper cover and mix), are not vulnerable to termites, and can be made longer or spliced to reach deep strata. They can deliver higher capacities via larger cross-sections and prestressing/reinforcement. Timber often needs cut-offs below water to remain permanently submerged to avoid decay, while concrete has no such requirement when properly detailed. Step-by-Step Solution: Durability: concrete is inherently resistant to insect/biological attack → advantage over timber.Length/capacity: precast or cast-in-situ concrete allows long piles and higher capacities.Construction: no mandatory cut-off below water table for decay control. Verification / Alternative check (if short method exists): Field practice and specifications commonly restrict timber piles where durability cannot be ensured; concrete piles are widely used for bridges, offshore, and urban foundations. Why Other Options Are Wrong: Each single advantage (A–D) is true but partial; only “All of the above” captures the comprehensive set. Common Pitfalls (misconceptions, mistakes): Overlooking sulphate attack and chloride ingress for concrete (needs proper mix/cover); assuming timber is always inferior (it can be economical in fresh-water conditions when kept permanently submerged). Final Answer: All of the above

Question 16

Flange width limitation for T-beams — the effective flange width should be restricted to which of the following (select the least governing limit)?

Accepted Answer

Introduction / Context: In T-beams, part of the slab acts compositely with the beam web (rib) as the flange under compression. To avoid overestimating the contribution of slab concrete, design codes limit the effective flange width using practical geometric rules. Designers must select the least (most restrictive) of several limits to be safe. Given Data / Assumptions: Monolithic slab-beam construction forming T-section under sagging moments. Effective span, spacing of adjacent beams, and slab thickness are known design parameters. Standard effective flange width rules apply. Concept / Approach: Effective flange width accounts for shear lag and non-uniform stress distribution. The beam cannot mobilize the entire slab width equally; thus, codes limit the assumed flange width by span-based and spacing-based limits as well as a local width tied to web breadth plus a multiple of slab thickness. The controlling value is the smallest among the specified limits. Step-by-Step Solution: List limits: b_f ≤ L/3; b_f ≤ spacing of beams; b_f ≤ b_w + 12 * t_s.Compute each for the given geometry in actual design.Adopt the least → ensures conservative and safe design. Verification / Alternative check (if short method exists): Shear lag theory and test data underpin these limits; using the least value prevents unconservative flange participation assumptions. Why Other Options Are Wrong: Any single limit (A–C) alone may be unconservative; “Any of the above” (E) ignores the requirement to take the least. Common Pitfalls (misconceptions, mistakes): Adding limits instead of taking the minimum; confusing effective flange width with total slab width between beams. Final Answer: Least of the above

Question 17

Why high-strength concrete is specified in prestressed members? Select the most complete reason set.

Accepted Answer

Introduction / Context: Prestressed concrete relies on compressive preloads applied through tendons. Concrete quality is critical to safely anchor high tendon forces, transfer prestress, and control cracking. Therefore, high-strength, dense, and durable concrete is typically required in prestressed members such as beams, girders, and slabs. Given Data / Assumptions: Tendons develop large concentrated forces at anchorages and deviators. Concrete must provide sufficient bearing, bond, and crack resistance. Serviceability (crack control, deflection) and durability are primary concerns. Concept / Approach: Higher compressive strength generally correlates with improved bearing capacity, better bond characteristics, and reduced porosity. At anchorage zones, bursting and spalling stresses arise; high-strength concrete with proper detailing (stirrups, spirals) contains these stresses. Improved bond allows efficient transfer lengths in pre- or post-tensioned members. Lower shrinkage and creep also help sustain long-term prestress levels and service performance. Step-by-Step Solution: Identify stress concentrations at anchorages → need high bearing and confinement → stronger concrete required.Consider bond/transfer → high-strength concrete improves bond and reduces transfer length.Service life → denser matrix reduces microcracking and environmental ingress. Verification / Alternative check (if short method exists): Design standards prescribe minimum characteristic strengths for prestressed work (higher than for ordinary reinforced members), reflecting these performance needs. Why Other Options Are Wrong: Each of A–D is correct but partial; only “All of the above” captures the comprehensive rationale for specifying high-strength concrete. Common Pitfalls (misconceptions, mistakes): Assuming strength alone is sufficient; in reality, detailing of end zones and control of shrinkage/creep are also vital. Final Answer: All of the above

Question 18

Combined footing detailing — when the computed shear stress in a combined footing exceeds 5 kg/cm², what configuration of nominal stirrups should be provided for adequate shear resistance and confinement?

Accepted Answer

Introduction / Context: Combined footings (supporting two or more columns) must be detailed to safely resist shear stresses from column loads and soil reactions. When shear demand rises beyond nominal levels (here, exceeding 5 kg/cm²), the transverse reinforcement (stirrups) must provide both shear capacity and confinement across the wide footing beam portion near columns. Given Data / Assumptions: Type: reinforced concrete combined footing. Shear stress demand > 5 kg/cm² is indicated. Objective: select a reasonable standard stirrup configuration. Concept / Approach: In wide footing beams and near-column zones, multi-legged closed stirrups are preferred to enclose all longitudinal bars, improve shear resistance, and confine the concrete core. An 8-legged stirrup arrangement is a commonly adopted detailing step-up beyond 6-legged ties when nominal shear is exceeded but before very heavy cage requirements arise. Step-by-Step Solution: Identify that τv > 5 kg/cm² → nominal minimum shear reinforcement is inadequate.Adopt multi-legged closed stirrups enveloping longitudinal bars for confinement.Select 8-legged ties as a robust, widely used configuration for such footings. Verification / Alternative check (if short method exists): Check shear design against concrete shear capacity and contribution of shear reinforcement (Vus). Ensure spacing, anchorage, and bar cover satisfy code requirements. If shear is still high, increase leg count or bar diameter and reduce spacing accordingly. Why Other Options Are Wrong: 6-legged may be insufficient at higher τv; 10- or 12-legged may be overly conservative for the threshold stated; “none of these” ignores common practice. Common Pitfalls (misconceptions, mistakes): Using open stirrups; not enclosing all longitudinal bars; excessive spacing; ignoring punching shear near columns. Final Answer: 8 legged

Question 19

T-beam effective flange — the portion of slab that may be considered to act effectively with the rib (web) of a T-beam depends on which geometric parameters?

Accepted Answer

Introduction / Context: The effective flange width in a T-beam captures how much of the slab actually participates in compression with the beam web. Because stress distribution in the slab is non-uniform (shear lag), design codes cap the effective flange width using several geometric limits to prevent unconservative assumptions. Given Data / Assumptions: Monolithic slab–beam construction forming T-sections under sagging moment. Need to identify the variables governing effective flange width. Typical code limits involve span, spacing, and slab thickness along with web breadth. Concept / Approach: Effective flange width is influenced by: (1) span (longer spans allow wider stress spread), (2) spacing between adjacent beams (flange cannot exceed half-spacing each side), (3) slab thickness (thicker slabs distribute compression more effectively), and (4) web breadth (local participation right above the web). Thus, all the stated geometric factors matter. Step-by-Step Solution: List common limits: b_f ≤ L/3; b_f ≤ spacing of beams; b_f ≤ b_w + 12*t_s (typical rule).Recognize that each limit ties to span, spacing, thickness, and web breadth.Therefore, effective flange depends on all listed parameters. Verification / Alternative check (if short method exists): Compute each limit for a given geometry and select the least value to remain conservative. Why Other Options Are Wrong: Each single-parameter option is incomplete; codes use multiple simultaneous limits. Common Pitfalls (misconceptions, mistakes): Equating effective flange width to the full tributary slab width; forgetting to adopt the minimum of the applicable limits. Final Answer: all the above

RCC Structures Design Questions