Q: Mechanical properties — typical minima for steel materials: Select the correct set of statements regarding common steel grades used in structural practice.

Introduction / Context: Structural design relies on characteristic mechanical properties such as tensile strength, yield stress, and ductility (elongation). Familiarity with typical minima for different steel products helps in selecting materials and checking compliance with specifications. Given Data / Assumptions: Traditional units kg/mm^2 used (1 kg/mm^2 ≈ 9.81 MPa). Properties quoted are typical minima for specification compliance. Elongation is referred to a proportional gauge length 5.65√A. Concept / Approach: Cast steels generally have lower yield stress relative to tensile strength; mild steels exhibit balanced strength and ductility; high tensile steels provide higher strength levels. The statements list representative baseline values consistent with many older specifications and handbooks. Step-by-Step Solution: Step 1: Check steel castings: 41 kg/mm^2 ultimate and 20.5 kg/mm^2 yield are standard minima.Step 2: Check mild steel: about 44 kg/mm^2 ultimate with ≈ 23% elongation at 5.65√A is typical.Step 3: Check high tensile steel: ultimate around 58 kg/mm^2 is a common minimum threshold.Step 4: All three statements are consistent; hence “All of the above”. Verification / Alternative check: Converting to SI: 41 kg/mm^2 ≈ 402 MPa, 44 kg/mm^2 ≈ 431 MPa, 58 kg/mm^2 ≈ 569 MPa, aligning with expected ranges for these products. Why Other Options Are Wrong: Any single-statement option omits other true statements; “None” is incorrect because all are valid. Common Pitfalls: Confusing yield stress with ultimate tensile strength. Ignoring ductility criteria like elongation at specified gauge length. Final Answer: All of the above

Q: Effective flange width of a T-beam in reinforced concrete design Given: b = actual (available) width of the slab/flange, Dt = thickness of flange, bw = breadth (web width) of the rib/web, l0 = distance between points of zero moment (effective span s…

Introduction: In reinforced concrete floor systems with monolithic slabs and ribs, T-beams are analyzed by replacing the slab over the web with an 'effective' flange width that realistically participates in flexure. This question checks your understanding of how codes cap that width using geometry and span parameters so bending stresses are evaluated correctly. Given Data / Assumptions: b = actual slab width available as flange. Dt = thickness of the slab (flange thickness). bw = web (rib) width. l0 = distance between points of contraflexure (often approximated by the effective span for simple spans). Concept / Approach: The flange cannot be taken arbitrarily wide. Codes cap the effective flange width by several limits. When spacing to adjacent webs is not explicitly provided, the controlling limit is usually the lesser of the actual available slab width b and a span- and thickness-dependent expression: bw + (l0 / 6) + 6 * Dt. Taking the lesser prevents overestimation of compression block and moment capacity. Step-by-Step Solution: 1) Start with the code limit tied to span: beff1 = bw + (l0 / 6) + 6 * Dt.2) Recognize that the actual flange cannot exceed its physical availability: beff2 = b.3) Adopt the effective flange width: beff = lesser of { beff1 , beff2 }. Verification / Alternative check: If center-to-center spacing of beams were known, it would provide another upper bound. In its absence, using the lesser of b and bw + (l0/6) + 6Dt remains conservative and code-consistent. Why Other Options Are Wrong: Option B ignores mechanics and code caps. Option C exaggerates width with l0/3 and 12Dt. Option D underestimates the span/thickness influence. Option E incorrectly takes the greater, which is non-conservative. Common Pitfalls: Confusing l0/6 with l0/12 (edge beams) and forgetting the physical cap b are common errors. Final Answer: Effective flange width = lesser of { b , bw + (l0 / 6) + 6 * Dt }

Q: Reinforcement (steel bars) used in concrete work must be clean and free from contaminants State what the bars must be free from before placement to ensure proper bond and durability.

Introduction: Clean reinforcement is essential for achieving proper bond between steel and concrete. Contaminants can reduce bond strength, increase slip, and compromise durability. This question checks basic site workmanship requirements. Given Data / Assumptions: Reinforcement is carbon steel bars intended for reinforced concrete. Surface condition affects bond and corrosion risk. Concept / Approach: Standards require bars to be free from loose mill scale, loose rust, paint, oil, mud, or any deleterious coating that prevents adhesion. A thin, tight rust film is typically acceptable, but loose flaky rust is not. Step-by-Step Solution: 1) Identify typical contaminants: loose scale, loose rust, paint and oil.2) Check code guidance: remove anything that can act as a bond breaker.3) Conclude that bars should be free from all listed contaminants. Verification / Alternative check: Pull-out tests and site inspections confirm that clean, roughened metal surfaces provide better bond. Why Other Options Are Wrong: Choosing only one contaminant underestimates the requirement; all listed items are harmful to bond if present. Common Pitfalls: Leaving oil from cutting/greasing on bars or painting bars for identification without removing paint lines in bond zones. Final Answer: all the above

Q: Concrete mixing time for uniform colour and consistency when using ordinary Portland cement (OPC) Select a suitable total mixing time in a standard batch mixer to obtain uniform paste coating and colour.

Introduction: Mixing time influences homogeneity, workability, and strength. Insufficient mixing causes colour streaks and honeycombing; excessive mixing can increase temperature and cause segregation. The question seeks a practical range for OPC concrete in typical drum/batch mixers. Given Data / Assumptions: Ordinary Portland cement. Conventional on-site tilting drum or pan mixer. Normal slump concrete, standard aggregate grading. Concept / Approach: Standards and good practice suggest a minimum time to coat all particles with paste and achieve uniform colour. For common on-site mixers, the total mixing time after all materials are in the drum is typically around 1.5 to 2 minutes. Step-by-Step Solution: 1) Load sequence ensures part mixing during charging, but uniformity requires post-charging mixing.2) Empirical studies show diminishing returns beyond ~2 minutes in standard mixers.3) Therefore, select 1.5 to 2.0 minutes for uniform colour and consistency. Verification / Alternative check: Perform a field uniformity check: visual colour uniformity and absence of cement balls; slump tests between successive batches should be similar. Why Other Options Are Wrong: 0.5–1.0 minute is too short; 1.0–1.5 minute may still be borderline; 2.0–3.0 minutes can be acceptable but often unnecessary and may risk temperature rise; a blanket rule of >3 minutes is not mandatory. Common Pitfalls: Starting the stopwatch before all materials are in; not accounting for mixer performance; ignoring admixture manufacturer guidance. Final Answer: 1.5 to 2.0 minutes

Question 1

Minimum design eccentricity for reinforced concrete columns: Columns shall be designed for a minimum eccentricity equal to 1/30 of the unsupported length, subject to a minimum of how much (whichever is greater)?

Accepted Answer

Introduction / Context: In reinforced concrete column design, codes prescribe a minimum design eccentricity to account for inevitable construction tolerances, misalignments, and second-order effects. This ensures that columns are not designed for unrealistically perfect concentric compression. Given Data / Assumptions: Minimum eccentricity e_min = L_u / 30, where L_u is the unsupported length of the column. A floor minimum absolute value is also prescribed. Concept / Approach: The design must consider the larger of the length-based eccentricity and a fixed minimum value. This prevents e from becoming too small for short columns, where L_u/30 can be less than a practical lower bound. Step-by-Step Solution: Step 1: Compute e_1 = L_u / 30.Step 2: Compare e_1 with the absolute minimum recommended by code.Step 3: Use e = max(L_u / 30, 20 mm). Therefore, the minimum is 20 mm. Verification / Alternative check: Design handbooks and IS-based provisions consistently adopt 20 mm as the absolute minimum eccentricity floor for RC columns. Why Other Options Are Wrong: 5 mm/10 mm/15 mm: Too small; do not satisfy the prescribed absolute minimum. 25 mm: Exceeds the standard minimum and would be unnecessarily conservative for many cases. Common Pitfalls: Forgetting to take the maximum of L_u/30 and the absolute minimum. Using center-to-center story height instead of unsupported length. Final Answer: 20 mm

Question 2

Selecting workability for shallow sections with vibration: For concreting shallow members that will be compacted using mechanical vibration, which workability level is most appropriate?

Accepted Answer

Introduction / Context: Workability selection depends on member geometry and compaction method. When efficient vibration is available, mixes need not be very flowable; lower slumps reduce segregation and bleeding while still enabling full compaction. Given Data / Assumptions: Shallow sections (short placement depth, relatively simple reinforcement). Mechanical vibration is used for compaction. Objective is adequate compaction with minimal segregation. Concept / Approach: With vibration, a low to moderate slump suffices. Very high workability is unnecessary and can trigger segregation, especially in shallow members where confinement is limited. Step-by-Step Solution: Step 1: Match workability to compaction method: vibration allows stiffer mixes.Step 2: Prefer lower slump ranges (often termed “low”) to control bleeding/segregation.Step 3: Select “Low” as most appropriate for shallow sections with vibration. Verification / Alternative check: Handbooks list slump ranges: for vibrated RC, low slump is commonly recommended for slabs and shallow beams, provided placement is not congested. Why Other Options Are Wrong: Very low: Risk of honeycombing if vibration/placement is not excellent. Medium: Acceptable in some cases, but “low” better targets vibration efficiency and segregation control. High/Very high: Unnecessary with vibration; increase segregation risk and water demand. Common Pitfalls: Choosing high slump “for ease” and then suffering edge bleeding and coarse aggregate settlement. Ignoring reinforcement congestion; if very congested, reassess to medium slump or use plasticizers. Final Answer: Low (stiff but placeable with vibration)

Question 3

Riveted connections — nominal diameters and hole allowances: Which of the following statements about rivet diameter and rivet hole diameter (for steelwork) are correct?

Accepted Answer

Introduction / Context: Riveted steel connections (as treated in classic design texts and old codes) define nominal diameters and hole allowances to ensure proper driving and account for expansion/shrinkage during installation. Understanding these definitions is key for strength checks in shear and bearing. Given Data / Assumptions: Nominal rivet diameter refers to the shank before driving. Standard oversize allowances for drilling/punching of holes are used. Design checks use gross rivet area based on the nominal shank diameter. Concept / Approach: Hole diameters exceed nominal shank diameters to permit insertion and proper forming of the head. Shear area uses the nominal shank area, whereas net section of connected plates accounts for the hole diameter when evaluating plate capacity. Step-by-Step Solution: Step 1: Identify nominal rivet diameter: the pre-driven shank diameter (true).Step 2: Apply standard hole allowances: d + 1.5 mm for d < 25 mm; d + 2.0 mm for d ≥ 25 mm (true).Step 3: Recall that rivet shear area uses the nominal diameter (gross rivet area), not the hole diameter (true).Step 4: Therefore, all listed statements are correct. Verification / Alternative check: Classical steel design references and previous IS/BS practice show the same allowances and area definitions for rivets. Why Other Options Are Wrong: Any single statement taken alone misses the complete picture; the comprehensive, correct choice is “All of the above”. Common Pitfalls: Using hole diameter to compute rivet shear area (should use nominal shank diameter). Forgetting to increase hole diameter in plate net section calculations. Final Answer: All of the above

Question 4

Mechanical properties — typical minima for steel materials: Select the correct set of statements regarding common steel grades used in structural practice.

Accepted Answer

Introduction / Context: Structural design relies on characteristic mechanical properties such as tensile strength, yield stress, and ductility (elongation). Familiarity with typical minima for different steel products helps in selecting materials and checking compliance with specifications. Given Data / Assumptions: Traditional units kg/mm^2 used (1 kg/mm^2 ≈ 9.81 MPa). Properties quoted are typical minima for specification compliance. Elongation is referred to a proportional gauge length 5.65√A. Concept / Approach: Cast steels generally have lower yield stress relative to tensile strength; mild steels exhibit balanced strength and ductility; high tensile steels provide higher strength levels. The statements list representative baseline values consistent with many older specifications and handbooks. Step-by-Step Solution: Step 1: Check steel castings: 41 kg/mm^2 ultimate and 20.5 kg/mm^2 yield are standard minima.Step 2: Check mild steel: about 44 kg/mm^2 ultimate with ≈ 23% elongation at 5.65√A is typical.Step 3: Check high tensile steel: ultimate around 58 kg/mm^2 is a common minimum threshold.Step 4: All three statements are consistent; hence “All of the above”. Verification / Alternative check: Converting to SI: 41 kg/mm^2 ≈ 402 MPa, 44 kg/mm^2 ≈ 431 MPa, 58 kg/mm^2 ≈ 569 MPa, aligning with expected ranges for these products. Why Other Options Are Wrong: Any single-statement option omits other true statements; “None” is incorrect because all are valid. Common Pitfalls: Confusing yield stress with ultimate tensile strength. Ignoring ductility criteria like elongation at specified gauge length. Final Answer: All of the above

Question 5

Effective flange width of a T-beam in reinforced concrete design Given: b = actual (available) width of the slab/flange, Dt = thickness of flange, bw = breadth (web width) of the rib/web, l0 = distance between points of zero moment (effective span s…

Accepted Answer

Introduction: In reinforced concrete floor systems with monolithic slabs and ribs, T-beams are analyzed by replacing the slab over the web with an 'effective' flange width that realistically participates in flexure. This question checks your understanding of how codes cap that width using geometry and span parameters so bending stresses are evaluated correctly. Given Data / Assumptions: b = actual slab width available as flange. Dt = thickness of the slab (flange thickness). bw = web (rib) width. l0 = distance between points of contraflexure (often approximated by the effective span for simple spans). Concept / Approach: The flange cannot be taken arbitrarily wide. Codes cap the effective flange width by several limits. When spacing to adjacent webs is not explicitly provided, the controlling limit is usually the lesser of the actual available slab width b and a span- and thickness-dependent expression: bw + (l0 / 6) + 6 * Dt. Taking the lesser prevents overestimation of compression block and moment capacity. Step-by-Step Solution: 1) Start with the code limit tied to span: beff1 = bw + (l0 / 6) + 6 * Dt.2) Recognize that the actual flange cannot exceed its physical availability: beff2 = b.3) Adopt the effective flange width: beff = lesser of { beff1 , beff2 }. Verification / Alternative check: If center-to-center spacing of beams were known, it would provide another upper bound. In its absence, using the lesser of b and bw + (l0/6) + 6Dt remains conservative and code-consistent. Why Other Options Are Wrong: Option B ignores mechanics and code caps. Option C exaggerates width with l0/3 and 12Dt. Option D underestimates the span/thickness influence. Option E incorrectly takes the greater, which is non-conservative. Common Pitfalls: Confusing l0/6 with l0/12 (edge beams) and forgetting the physical cap b are common errors. Final Answer: Effective flange width = lesser of { b , bw + (l0 / 6) + 6 * Dt }

Question 6

Reinforcement (steel bars) used in concrete work must be clean and free from contaminants State what the bars must be free from before placement to ensure proper bond and durability.

Accepted Answer

Introduction: Clean reinforcement is essential for achieving proper bond between steel and concrete. Contaminants can reduce bond strength, increase slip, and compromise durability. This question checks basic site workmanship requirements. Given Data / Assumptions: Reinforcement is carbon steel bars intended for reinforced concrete. Surface condition affects bond and corrosion risk. Concept / Approach: Standards require bars to be free from loose mill scale, loose rust, paint, oil, mud, or any deleterious coating that prevents adhesion. A thin, tight rust film is typically acceptable, but loose flaky rust is not. Step-by-Step Solution: 1) Identify typical contaminants: loose scale, loose rust, paint and oil.2) Check code guidance: remove anything that can act as a bond breaker.3) Conclude that bars should be free from all listed contaminants. Verification / Alternative check: Pull-out tests and site inspections confirm that clean, roughened metal surfaces provide better bond. Why Other Options Are Wrong: Choosing only one contaminant underestimates the requirement; all listed items are harmful to bond if present. Common Pitfalls: Leaving oil from cutting/greasing on bars or painting bars for identification without removing paint lines in bond zones. Final Answer: all the above

Question 7

Concrete mixing time for uniform colour and consistency when using ordinary Portland cement (OPC) Select a suitable total mixing time in a standard batch mixer to obtain uniform paste coating and colour.

Accepted Answer

Introduction: Mixing time influences homogeneity, workability, and strength. Insufficient mixing causes colour streaks and honeycombing; excessive mixing can increase temperature and cause segregation. The question seeks a practical range for OPC concrete in typical drum/batch mixers. Given Data / Assumptions: Ordinary Portland cement. Conventional on-site tilting drum or pan mixer. Normal slump concrete, standard aggregate grading. Concept / Approach: Standards and good practice suggest a minimum time to coat all particles with paste and achieve uniform colour. For common on-site mixers, the total mixing time after all materials are in the drum is typically around 1.5 to 2 minutes. Step-by-Step Solution: 1) Load sequence ensures part mixing during charging, but uniformity requires post-charging mixing.2) Empirical studies show diminishing returns beyond ~2 minutes in standard mixers.3) Therefore, select 1.5 to 2.0 minutes for uniform colour and consistency. Verification / Alternative check: Perform a field uniformity check: visual colour uniformity and absence of cement balls; slump tests between successive batches should be similar. Why Other Options Are Wrong: 0.5–1.0 minute is too short; 1.0–1.5 minute may still be borderline; 2.0–3.0 minutes can be acceptable but often unnecessary and may risk temperature rise; a blanket rule of >3 minutes is not mandatory. Common Pitfalls: Starting the stopwatch before all materials are in; not accounting for mixer performance; ignoring admixture manufacturer guidance. Final Answer: 1.5 to 2.0 minutes

Question 8

Slab with main reinforcement parallel to a T-beam (rib): required transverse steel at mid-span State the minimum transverse reinforcement (as a percentage of the main reinforcement area) to be provided at mid-span when the main bars run parallel to …

Accepted Answer

Introduction: When slab main bars run parallel to supporting T-beams, adequate transverse steel is needed to control cracking, distribute loads to ribs, and resist shrinkage/temperature effects. Codes prescribe a minimum proportion relative to the main reinforcement at mid-span. Given Data / Assumptions: Main bars are parallel to the T-beam direction. Transverse bars act as distribution and secondary reinforcement. Normal room exposure with typical slab thickness. Concept / Approach: The transverse steel should not be arbitrarily reduced because load transfer to ribs and crack control depend on a minimum amount. A widely used rule is to provide at least half of the main steel area at mid-span in the transverse direction. Step-by-Step Solution: 1) Identify the controlling requirement for distribution steel where main bars are parallel to ribs.2) Adopt minimum transverse steel = 0.5 * (main steel at mid-span).3) Provide appropriate spacing to satisfy bar spacing and crack width limits. Verification / Alternative check: Check crack control and serviceability: spacing and bar diameter should satisfy service crack widths under working loads. Why Other Options Are Wrong: 40% can under-reinforce for distribution; 60% and 70% are conservative but not required as a minimum; 30% is inadequate for load distribution between ribs. Common Pitfalls: Providing the same transverse steel at supports as mid-span without considering negative moments; using large diameters at wide spacing, which is less effective for crack control. Final Answer: 50%

Question 9

For design calculations, the modulus of elasticity of steel (Es) is generally assumed as Choose the most appropriate standard value and unit.

Accepted Answer

Introduction: The elastic modulus of reinforcing steel is a fundamental property used in deflection, crack width, and composite action checks. Designers typically adopt a standard value suitable for most carbon steel rebars. Given Data / Assumptions: Carbon steel reinforcement used in normal temperature ranges. Linear elastic behavior assumed up to yield for serviceability checks. Concept / Approach: For structural design, Es is taken as approximately 200,000 N/mm², which is equivalently 200 kN/mm². This value aligns with material test data and code recommendations. Step-by-Step Solution: 1) Recognize that 1 kN/mm² = 1000 N/mm².2) Standard Es = 200,000 N/mm² = 200 kN/mm².3) Select the exact matching option. Verification / Alternative check: Material certificates for Fe415/Fe500 typically report Es near 200 GPa, confirming the chosen value. Why Other Options Are Wrong: 225, 250, 275, and 150 kN/mm² do not reflect the standard modulus used in design and would misstate stiffness. Common Pitfalls: Confusing yield strength (e.g., 500 MPa) with modulus; mixing units of GPa, N/mm², and kN/mm². Final Answer: 200 kN/mm²

Question 10

Span-to-depth ratios for preliminary sizing of beams up to 10 m span Select the correct set of safe span/depth ratios for cantilever, simply supported, and continuous beams.

Accepted Answer

Introduction: Span/depth ratios help in preliminary sizing to meet deflection control before detailed analysis. For spans up to about 10 m, common guideline values exist for different support conditions. Given Data / Assumptions: Ordinary reinforced concrete beams. Typical service loads and normal exposure. Preliminary sizing stage (to be refined later). Concept / Approach: Cantilevers deflect more for the same depth, so they require smaller span/depth ratios. Continuous beams are stiffer than simply supported and can use a larger ratio. Step-by-Step Solution: 1) Recall guide values: cantilever ≈ 7, simply supported ≈ 20, continuous ≈ 26.2) Compare options with these targets.3) Choose the exact matching set. Verification / Alternative check: Final designs should include modification factors for tension reinforcement percentage, compression steel, and flanged sections. Why Other Options Are Wrong: Other sets either over- or under-estimate stiffness for the given support system. Common Pitfalls: Treating these ratios as code limits rather than starting points; ignoring long-term creep and shrinkage effects. Final Answer: Cantilever 7; Simply supported 20; Continuous 26

Question 11

Laced compression members: recommended inclination of lacing bars with the member axis Choose the suitable range for the angle that lacing bars make with the longitudinal axis.

Accepted Answer

Introduction: Lacing systems connect components of a built-up compression member so that they act together. The angle of lacing bars ensures efficient shear transfer and stability while keeping bar forces reasonable. Given Data / Assumptions: Built-up steel compression member with double lacing. Lacing bars primarily resist shear and provide lateral restraint. Concept / Approach: Design recommendations limit the inclination so that the axial and transverse components of bar forces are balanced. A commonly adopted range is 30° to 60° with the member axis. Step-by-Step Solution: 1) Select a bar angle that avoids very flat lacing (inefficient) and very steep lacing (large forces).2) The practical window 30°–60° satisfies both strength and detailing needs.3) Choose the option exactly matching this window. Verification / Alternative check: Check slenderness of lacing and node strength; connection design must also satisfy shear flow. Why Other Options Are Wrong: 20°–50° admits too-flat lacing at the lower end; 40°–70° and 50°–80° skew the range upward; 15°–30° is excessively flat. Common Pitfalls: Ignoring out-of-plane buckling of lacing, inadequate end tie plates, and mismatched single vs double lacing assumptions. Final Answer: 30° to 60°

Question 12

Concrete tensile strength relationships: flexural (modulus of rupture) and split tensile strength If fck is the characteristic compressive strength of concrete, the flexural/split tensile strength may be expressed as f = a * sqrt(fck). Identify the …

Accepted Answer

Introduction: Concrete is weak in tension. Codes provide empirical relationships linking tensile properties to compressive strength for preliminary design and serviceability checks. This question targets the coefficient used for flexural tensile strength (modulus of rupture). Given Data / Assumptions: Characteristic cube strength fck. Relationship of the form f = a * sqrt(fck). Concept / Approach: For flexural strength (modulus of rupture), the recommended coefficient is about 0.7. For split tensile strength, a lower coefficient around 0.56 is often used. Recognizing which tensile property is referenced is key. Step-by-Step Solution: 1) Identify property: flexural tensile strength (modulus of rupture).2) Use empirical formula fcr ≈ 0.7 * sqrt(fck).3) Match with choices and select 0.7. Verification / Alternative check: Back-check against test data or code annex values; split tensile is lower (~0.56 * sqrt(fck)), confirming that 0.7 corresponds to flexural strength. Why Other Options Are Wrong: 0.4, 0.5, and 0.6 are too low for flexural strength; 0.8 is unconservative. Common Pitfalls: Confusing split tensile with modulus of rupture; using fck in MPa but forgetting the square-root relationship which is sensitive to units. Final Answer: 0.7

Question 13

Minimum reinforcement in slabs in each direction Specify the minimum steel area in either direction as a percentage of the gross concrete section of the slab.

Accepted Answer

Introduction: Minimum reinforcement controls cracking from shrinkage and temperature and ensures ductility even when applied loads are low. The requirement applies in each orthogonal direction of a reinforced concrete slab. Given Data / Assumptions: Reinforced concrete slab with deformed (or mild) bars. Percentage calculated on gross concrete cross section. Concept / Approach: Typical code provisions specify minimum percentages. For slabs with mild steel, 0.15% is a common minimum; for high strength deformed bars, 0.12% may be permitted. The question asks for a conservative minimum figure suitable across cases. Step-by-Step Solution: 1) Identify minimum in either direction of slab.2) Adopt 0.15% as a safe minimum applicable to mild steel and acceptable as a conservative choice more generally.3) Ensure spacing and bar diameters also satisfy crack control. Verification / Alternative check: Check against bar spacing limits and cover; sometimes 0.12% is permitted with HYSD, but 0.15% remains a safe baseline. Why Other Options Are Wrong: 0.75% is excessive for minimum steel; 0.10% and 0.12% can be insufficient or code-restricted depending on bar type; 0.20% is not the standard minimum. Common Pitfalls: Providing minimum percentage but violating maximum spacing; using large diameter bars at very wide spacing, which is less effective for crack control. Final Answer: 0.15%

Question 14

Rectangular steel column base plate (slab base) thickness for uniform bearing Let w be the uniform bearing pressure on concrete (force per unit area) and m be the projection of the base plate beyond the column face. Determine the plate thickness t f…

Accepted Answer

Introduction: In steel design, a slab base (base plate) spreads column load to concrete. The plate projects beyond the column face and behaves like a cantilevered strip under uniform bearing pressure. Correct thickness ensures bending stresses remain within the permissible value. Given Data / Assumptions: w = uniform bearing pressure on concrete under the plate. m = projection of plate beyond the column face measured to the edge. sigma_b = permissible bending stress in the plate steel. Strip theory: plate projection behaves as a cantilever of unit width. Concept / Approach: For a cantilever strip of unit width loaded by uniform pressure w over length m, the resultant load is w * m and the bending moment at the fixed edge is M = w * m^2 / 2 per unit width. The section modulus of a plate strip of unit width is Z = t^2 / 6. Step-by-Step Solution: 1) Bending stress: sigma = M / Z = (w * m^2 / 2) / (t^2 / 6) = 3 * w * m^2 / t^2.2) Limit sigma to sigma_b: 3 * w * m^2 / t^2 = sigma_b.3) Solve for thickness: t = m * sqrt( 3 * w / sigma_b ). Verification / Alternative check: Design thickness should also satisfy local checks near anchor bolts and consider biaxial projections if m differs on two sides; adopt the larger required t. Why Other Options Are Wrong: Options A and B underestimate thickness (unconservative). Option D overestimates bending effect excessively. Option E is dimensionally inconsistent for plate thickness. Common Pitfalls: Using gross plate width instead of projection m; neglecting grout stiffness; ignoring the weaker projection side when projections differ. Final Answer: t = m * sqrt( 3 * w / sigma_b )

Question 15

Steel Design – Critical Compression Stress in Bending (Sections Symmetrical about x–x Axis) In the standard formula for the critical compression stress Cs used for lateral–torsional behaviour, the shear modulus G is conventionally taken as a proport…

Accepted Answer

Introduction / Context: In steel design, the critical compression stress in bending for members susceptible to lateral–torsional buckling depends on elastic constants. Two key constants are the Young's modulus E and the shear modulus G. Designers often need a reliable value of G, expressed as a fraction of E, to evaluate buckling resistance for sections symmetrical about the x–x axis. Given Data / Assumptions: Material: structural steel with Poisson's ratio approximately 0.3. Relationship between elastic constants: G = E / (2 * (1 + nu)). Design practice uses rounded proportions of E for G. Concept / Approach: The shear modulus G links shear strain to shear stress and appears in lateral–torsional buckling expressions. Using nu around 0.3, one can estimate G as a consistent fraction of E for routine design without detailed material testing. Step-by-Step Solution: 1) Start from G = E / (2 * (1 + nu)).2) Take nu = 0.3 for structural steel.3) Compute denominator: 2 * (1 + 0.3) = 2 * 1.3 = 2.6.4) Therefore, G = E / 2.6 ≈ 0.3846 * E.5) Rounded for design convenience, G is taken as approximately 0.4 E. Verification / Alternative check: If nu varies slightly (for example 0.28 to 0.31), G changes only marginally, remaining very close to 0.38 E to 0.39 E. Thus 0.4 E is a safe, conventional approximation. Why Other Options Are Wrong: 0.2 E: Far below the physical value implied by nu ≈ 0.3, overly conservative and incorrect. 0.3 E: Still too low relative to E / 2.6 ≈ 0.385 E. 0.5 E: Much higher than realistic shear modulus for steel, unconservative. Common Pitfalls: Confusing G with E, or assuming G equals E / 3 by rule of thumb without checking nu. Always use the relation G = E / (2 * (1 + nu)). Final Answer: 0.4 E

Question 16

Reinforced Concrete – Minimum Diameter of Longitudinal Bars in a Column According to standard RCC practice (IS guidelines), what is the minimum bar diameter permitted for longitudinal reinforcement in a column?

Accepted Answer

Introduction / Context: Longitudinal reinforcement in columns must satisfy minimum detailing rules to ensure ductility, durability, and constructability. A key rule is the minimum diameter for longitudinal bars, which helps avoid handling damage and excessive corrosion risk. Given Data / Assumptions: Member type: RCC column subjected to axial load with bending. Code context: Indian practice per IS 456 detailing provisions. Objective: Identify the minimum permissible diameter for column longitudinal bars. Concept / Approach: Minimum diameters are set to ensure adequate stiffness of individual bars, proper bond, and practical placement with ties, cover, and lap splices. Too small a diameter compromises robustness and corrosion allowance. Step-by-Step Solution: 1) For columns, codes specify minimum bar size for longitudinal reinforcement.2) IS 456 practice recognizes 12 mm as the minimum longitudinal bar diameter in columns.3) Bars smaller than 12 mm are allowed for stirrups/ties, not for main longitudinal bars. Verification / Alternative check: Reviewing typical schedules confirms that column main bars are rarely smaller than 12 mm due to lapping and buckling considerations during construction and service. Why Other Options Are Wrong: 6 mm and 8 mm: Suitable for ties or links, not for main column bars. 10 mm: Common for secondary uses; still below the minimum for column longitudinal bars. 20 mm: Allowed but not the minimum; using it increases steel without being mandatory. Common Pitfalls: Confusing minimum longitudinal bar size with minimum tie diameter. Ties may be 6 mm or 8 mm depending on bar sizes and spacing, but main bars must be larger. Final Answer: 12 mm

Question 17

Durability of RCC in Marine Exposure For reinforced concrete directly in sea water or exposed to severe marine conditions, which minimum concrete grade should be used for durability and strength?

Accepted Answer

Introduction / Context: Marine environments impose severe durability demands on reinforced concrete due to chloride ingress, wetting and drying cycles, and sulphate action. Codes mandate a minimum concrete grade to limit permeability and protect reinforcement. Given Data / Assumptions: Exposure: structures in sea water or directly exposed to sea spray and saline atmosphere. Performance target: lower permeability and adequate cover to resist corrosion. Concrete grades considered: M10, M15, M20, M30. Concept / Approach: Higher characteristic strength generally corresponds to a denser matrix, reduced capillary porosity, and improved durability. Codes for severe exposure prescribe not just cover and cement type, but also minimum grade. Step-by-Step Solution: 1) Identify exposure class as severe to very severe for sea water.2) For RCC in such exposure, minimum grade stipulated is M30.3) Grades below M30 may be acceptable for plain concrete elements with additional measures, but not typically for RCC in direct marine exposure. Verification / Alternative check: Durability tables in standard practice align with M30 or higher along with adequate cover, low water–cement ratio, and suitable cementitious materials to resist chlorides. Why Other Options Are Wrong: M10 and M15: Intended for lean or non-structural uses; too porous for marine RCC. M20: Common for general RCC but inadequate for severe chloride exposure without special measures. Common Pitfalls: Assuming strength alone ensures durability. In practice, cover, w/c ratio, curing, and supplementary cementitious materials also matter, but the minimum grade M30 remains a baseline requirement. Final Answer: M 30

Question 18

Elastic Modulus of Concrete – Empirical Coefficient in Ec Relation If fck is the characteristic cube strength of concrete in N/mm^2, the short-term modulus of elasticity is expressed as Ec = a * sqrt(fck). What is the empirical constant a used in co…

Accepted Answer

Introduction / Context: The static modulus of elasticity Ec of concrete correlates with compressive strength. Empirical formulas of the form Ec = a * sqrt(fck) are widely used for initial sizing and serviceability checks such as deflection. Given Data / Assumptions: Strength term: fck in N/mm^2 (MPa). Empirical model: Ec = a * sqrt(fck). Objective: identify coefficient a used in common Indian practice for short-term Ec. Concept / Approach: Different standards suggest slightly different a values to reflect aggregate type and testing method. The value 5700 is frequently adopted in competitive exams and many Indian references for short-term Ec estimation with normal-weight concrete. Step-by-Step Solution: 1) Recognize Ec depends on aggregate stiffness and mix; a single coefficient generalizes typical materials.2) Among the options, 5700 is the accepted empirical constant for Ec in many Indian design aids for normal-weight mixes.3) Thus Ec = 5700 * sqrt(fck) provides a reasonable short-term estimate. Verification / Alternative check: Comparisons with measured Ec values for common fck (for example fck = 25 to 40) show that 5700 * sqrt(fck) predicts within practical bounds for preliminary design when proper curing and quality are assumed. Why Other Options Are Wrong: 5900, 5800: Slightly higher than typical; may overestimate stiffness and underpredict deflection. 5600, 5500: Slightly lower; conservative but not the commonly cited coefficient for short-term Ec in normal-weight concrete. Common Pitfalls: Using Ec directly from cube strength without considering long-term effects. For serviceability, creep and cracking reduce effective stiffness, so designers often adopt reduced Ec or moment of inertia in checks. Final Answer: 5700

Question 19

Aggregate Nominal Size for RCC – Commonly Accepted Choice Considering cover, spacing of bars, and workability in typical reinforced concrete members, which nominal maximum size of coarse aggregate is generally acceptable for RCC?

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Introduction / Context: Choosing an appropriate nominal maximum size of aggregate affects workability, segregation resistance, and the ability to place concrete around reinforcement. For general RCC work, a balanced choice is preferred. Given Data / Assumptions: Typical RCC elements such as slabs, beams, and columns. Normal reinforcement congestion and standard covers. Conventional vibration during placing. Concept / Approach: The aggregate size must be smaller than clear cover and the minimum clear spacing between bars to avoid blocking. A too-large size impedes flow around bars, while too small a size may increase water demand for a given workability. Step-by-Step Solution: 1) Check bar spacing and cover constraints; maximum size should not exceed these geometric limits.2) For most routine RCC, 20 mm works well, balancing pumpability and strength.3) Hence, 20 mm is commonly accepted for general use. Verification / Alternative check: Many standard specifications and SOR schedules list 20 mm nominal size as default for RCC unless special conditions call for 10 mm or 12.5 mm for congested zones, or 25 mm for massive pours. Why Other Options Are Wrong: 10 mm, 15 mm: Useful in thin sections or heavy congestion but not the general default. 25 mm: May be acceptable in lightly reinforced, thicker members but not universally adopted. 40 mm: Typically for mass concrete, not for regular RCC with close bar spacing. Common Pitfalls: Forgetting to coordinate aggregate size with pump line, cover, and spacing near laps or anchorage regions where congestion increases locally. Final Answer: 20 mm

Question 20

Minimum Horizontal Spacing Between Parallel Main Reinforcing Bars For two parallel main reinforcement bars placed side by side, the minimum clear horizontal distance should be chosen as which of the following?

Accepted Answer

Introduction / Context: Adequate spacing between reinforcing bars ensures proper aggregate flow, minimizes honeycombing, and allows compaction around bars. Codes prescribe a minimum clear spacing that depends on bar size and aggregate size. Given Data / Assumptions: Two parallel main bars in the same layer. Nominal maximum size of coarse aggregate known from mix design. Objective: choose the controlling minimum clear spacing. Concept / Approach: The minimum clear spacing should prevent aggregate blocking and allow adequate grout to surround bars. Hence, spacing is tied to bar diameter and aggregate size, and the controlling rule is to adopt the greatest of specified checks. Step-by-Step Solution: 1) If bars are equal in diameter, one check equals that diameter.2) If bars are unequal, use the larger bar diameter as another check.3) A separate check is 5 mm more than the nominal maximum aggregate size.4) Adopt the greatest of these values to guarantee placement quality. Verification / Alternative check: Concrete placement trials consistently show that respecting the largest controlling spacing delivers fewer voids and better consolidation around reinforcement. Why Other Options Are Wrong: Options a, b, c singly may be non-conservative when the other parameter governs. Only option d captures the intent to take the largest controlling value. Common Pitfalls: Using bar diameter alone and ignoring aggregate size, which can cause blocking and honeycombing in congested regions. Final Answer: Greatest value of the above

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