Question 1

Transverse reinforcement (ties/spirals) pitch limits in reinforced concrete columns What is the code-based limiting pitch for transverse reinforcement of a column?

Accepted Answer

Introduction: Transverse reinforcement (ties or spirals) in columns restrains longitudinal bars against buckling and confines the core concrete, improving ductility. The maximum pitch (spacing) is limited to ensure effective confinement and to prevent bar instability between ties. Given Data / Assumptions: Reinforced concrete column with properly detailed ties/spirals. Normal seismic demand is assumed (non-special confining reinforcement). Standard bar sizes and cover are provided. Concept / Approach: Codes specify that the tie pitch shall not exceed certain limits tied to the member size and bar diameters. The governing maximum pitch is the least of specific expressions so that confinement is effective regardless of bar sizes or column dimension. Step-by-Step Solution: 1) Compute limit based on member size: s1 = least lateral dimension.2) Compute limit based on longitudinal bars: s2 = 16 * smallest longitudinal bar diameter.3) Compute limit based on tie diameter: s3 = 48 * tie diameter.4) Adopt smax = least(s1, s2, s3). Verification / Alternative check: Detailing examples and experimental data show that exceeding these limits reduces confinement efficiency and increases risk of longitudinal bar buckling in compression zones. Why Other Options Are Wrong: Options A, B, and C give only individual caps; the governing requirement is the least of them. Option E introduces a non-standard multiplier (twenty), which is not the conventional limit. Common Pitfalls: Using a single limit and ignoring the other caps; not reducing pitch near beam-column joints or where special confining reinforcement is required. Final Answer: least value of the above

Question 2

Steel Design – Maximum Pitch of Rivets in Members For both compression and tension members in riveted steelwork, what overall limit should the maximum pitch of rivets not exceed (choose the governing upper bound commonly adopted in practice)?

Accepted Answer

Introduction / Context: Riveted (or bolted) built-up members require limits on the spacing, or pitch, of fasteners to ensure adequate force transfer between connected elements and to prevent plates from buckling away from each other under axial force. The question checks knowledge of the customary upper bound for maximum rivet pitch in tension and compression members in steel design. Given Data / Assumptions: Member type: built-up steel members subject to axial tension or compression. Symbols: t denotes the thickness of the thinner connected plate. Design intent: ensure shear lag and plate buckling are controlled by suitable fastener pitch limits. Concept / Approach: Fastener pitch is limited by two kinds of criteria: a dimensional limit proportional to plate thickness (multiples of t) and an absolute dimensional cap in millimetres. The governing value is obtained by taking the smaller of the two limits so that both thin and thick plates remain adequately stitched together over the member length. Step-by-Step Solution: 1) Recognize that pitch cannot be allowed to grow indefinitely; otherwise, plates may separate locally or carry load non-uniformly.2) A commonly adopted envelope limit is 32 t or 300 mm, whichever is less, for general riveted members.3) Therefore the correct choice is the combined condition that references both the thickness-based and the absolute millimetre limit. Verification / Alternative check: Connection detailing handbooks, classic steel design texts, and long-standing specifications use similar dual limits to keep plate panels adequately stitched. While specific codes can impose additional limits for special cases, the 32 t/300 mm envelope is a widely taught maximum cap for general members. Why Other Options Are Wrong: 32 t (alone): Misses the millimetre cap; can be too large in thick plates. 200 mm: Too stringent as a general upper cap in many traditional riveted cases; the taught envelope uses 300 mm. 40 t: Too permissive for thicker plates and not a standard taught value. 300 mm (alone): Misses the thickness-based proportionality that controls thin plates. Common Pitfalls: Applying only a single limit and ignoring the combined “lesser of two” rule; overlooking that special members (e.g., cover plates, gastight seams) may attract tighter limits. Final Answer: Both 32 t and 300 mm (use the lesser)

Question 3

Permissible Axial Tension – Rolled I-Beams and Channels (Working Stress Context) As per classic Indian working-stress practice for rolled I-beams and channels, what permissible axial tensile stress is commonly adopted (kg/cm^2)?

Accepted Answer

Introduction / Context: Before the widespread adoption of limit-state design, many Indian practices used working-stress design for steel. Under this method, permissible (allowable) stresses were specified as fractions of material yield stress. This question checks basic recall of the typical permissible axial tensile stress for rolled I-beams and channels in that system, expressed in kg/cm^2. Given Data / Assumptions: Material: mild structural steel with yield stress around 2500 kg/cm^2. Design philosophy: working-stress design (allowable stress approach). Member: rolled I-beam or channel subjected to axial tension. Concept / Approach: Working-stress design limits the service stress to a specified fraction of yield to provide an implicit factor of safety. A widely taught rule of thumb is 0.6 * fy for axial tension in mild steel under working-stress assumptions, which for fy ≈ 2500 kg/cm^2 yields 1500 kg/cm^2. Step-by-Step Solution: 1) Take fy ≈ 2500 kg/cm^2 (equivalent to ~250 MPa).2) Permissible tension (working-stress) ≈ 0.6 * fy.3) Compute 0.6 * 2500 = 1500 kg/cm^2.4) Therefore 1500 kg/cm^2 is the conventional permissible axial tensile stress. Verification / Alternative check: Historic handbooks and solved examples in working-stress steel design frequently use 1500 kg/cm^2 for axial tension in mild steel members, aligning with the 0.6 fy rule of thumb. Why Other Options Are Wrong: 1800 kg/cm^2: Represents 0.72 fy; too high for the conventional allowable stress. 1400 kg/cm^2 and 1420 kg/cm^2: Slightly lower than the standard 1500 and not the commonly taught benchmark value. Common Pitfalls: Confusing working-stress values with limit-state design stresses; mixing units between MPa and kg/cm^2; forgetting that permissible stresses vary with material grade and philosophy. Final Answer: 1500 kg/cm^2

Question 4

RCC Columns – Minimum Diameter of Lateral Ties What is the minimum diameter permitted for lateral ties (column ties) in reinforced concrete columns under standard detailing practice?

Accepted Answer

Introduction / Context: Lateral ties in columns confine the core concrete, restrain longitudinal bars from buckling, and maintain reinforcement geometry during casting. Codes specify a minimum diameter to ensure adequate stiffness and robustness of ties, especially during vibration and handling. Given Data / Assumptions: Member: reinforced concrete column. Role of ties: confinement, bar restraint, and geometry control. Standard practice: non-seismic, ordinary detailing context. Concept / Approach: The minimum tie diameter must be sufficient to avoid yielding or distortion during placement and service. Standard practice sets an absolute lower bound along with a conditional rule that ties should not be less than one-fourth of the largest longitudinal bar diameter; however, the absolute minimum remains a fixed value in millimetres. Step-by-Step Solution: 1) Identify that ties must be stiff enough to restrain the main bars.2) Apply the basic detailing rule: tie dia ≥ 6 mm and also ≥ one-fourth of the largest longitudinal bar dia.3) Therefore, the absolute minimum permissible diameter is 6 mm. Verification / Alternative check: Typical bar bending schedules and standard RCC detailing textbooks prescribe 6 mm as the minimum tie diameter in ordinary conditions, while seismic detailing often increases this (for example to 8 mm) in special confining regions. Why Other Options Are Wrong: 4 mm and 5 mm: Too slender for reliable confinement. 10 mm: Allowed, but not the minimum; used in heavily loaded columns or seismic regions. Common Pitfalls: Confusing seismic provisions with general detailing; ignoring the one-fourth largest bar rule that can govern when main bars are large. Final Answer: 6 mm

Question 5

Formwork Tolerances – Columns and Beams Within what tolerance band are the dimensions of formwork for columns and beams typically permitted (negative to positive)?

Accepted Answer

Introduction / Context: Formwork tolerances control the dimensional accuracy of reinforced concrete members. Excessive deviation can alter cover, stiffness, and load-carrying capacity, as well as affect architectural alignment. The question tests awareness of the typical tolerance window used for columns and beams. Given Data / Assumptions: Member types: cast-in-situ columns and beams. Context: general building construction with competent formwork. Measure: dimensional tolerance range (negative to positive). Concept / Approach: Practical tolerances strike a balance between constructability and performance. A modest negative tolerance ensures sections are not undersized significantly (which would reduce capacity), while a positive tolerance compensates for buildability and stripping effects without creating excessive oversizing. Step-by-Step Solution: 1) Identify common tolerance bands used in site QA/QC checklists.2) For columns and beams, a widely used band is −6 mm to +12 mm.3) This range protects structural performance while remaining achievable with quality formwork. Verification / Alternative check: Project specifications often echo the −6/+12 mm tolerance for typical member dimensions unless special finishes or precast standards demand tighter control. Why Other Options Are Wrong: −3/+6 and −4/+8: Too tight for routine site conditions; may be reserved for special finishes. −8/+12: The negative side is overly permissive, risking section undersizing. Common Pitfalls: Measuring tolerances at incorrect reference points; neglecting that cover tolerances and alignment tolerances are checked separately. Final Answer: - 6 mm to + 12 mm

Question 6

Shear Reinforcement – Maximum Spacing Rule The maximum spacing of vertical shear reinforcement (stirrups) should not exceed 0.75 d, subject to what absolute limit (whichever is less)?

Accepted Answer

Introduction / Context: Stirrups resist diagonal tension cracking and confine the concrete core in beams. Codes provide a dual limit on the maximum spacing of shear reinforcement: a depth-related fraction and an absolute millimetre cap, with the governing value being the lesser of the two. Given Data / Assumptions: Member: reinforced concrete beam with vertical stirrups. Symbol: d is the effective depth. Objective: determine the absolute spacing cap paired with 0.75 d. Concept / Approach: Crack control and shear transfer mechanisms require sufficient stirrup frequency near potential failure planes. The absolute cap ensures a minimum confinement even in deep beams where 0.75 d could otherwise be quite large. Step-by-Step Solution: 1) Identify the dual limit: s_max ≤ 0.75 d and s_max ≤ s_cap.2) Standard practice sets s_cap = 300 mm.3) Apply the governing rule: use the lesser of 0.75 d and 300 mm. Verification / Alternative check: QA/QC checklists, bar bending schedules, and site inspections commonly reference “0.75 d or 300 mm, whichever is less” as the maximum stirrup spacing for beams without special seismic requirements. Why Other Options Are Wrong: 350, 400, 450, 500 mm: Too large; may allow wide crack spacing and inadequate shear restraint. Common Pitfalls: Forgetting that closer spacing is required near supports where shear demand is higher; confusing maximum spacing with minimum spacing or with confinement requirements in seismic zones. Final Answer: 300 mm

Question 7

Composite Construction – Provisions for Cased Columns Which statements correctly describe standard provisions for steel sections encased in reinforced concrete (cased columns)?

Accepted Answer

Introduction / Context: Cased columns combine a structural steel core with reinforced concrete encasement to enhance fire resistance, stiffness, and durability. Proper detailing ensures composite action and corrosion protection while maintaining constructability on site. Given Data / Assumptions: Steel core: rolled I-section or channels back-to-back, optionally with flange plates. Concrete encasement: dense concrete with suitable aggregate size and target strength. Cover and dimensional limits: minimum covers and overall size caps for practical site work. Concept / Approach: Encasement improves fire rating and protects the steel against corrosion while allowing the composite section to share load. Minimum cover and concrete quality ensure durability and adequate confinement, and dimensional limits keep transportation and shuttering practical. Step-by-Step Solution: 1) Section choice: single I or paired channels are standard for cased columns; flange plates may be added for capacity.2) Size control: overall dimensions are kept within practical limits such as around 750 × 450 mm to ease encasement and shuttering.3) Concrete specification: use dense concrete with ~10 mm aggregate and characteristic strength ≥ 160 kg/cm^2 at 28 days for durability.4) Cover: ensure not less than 50 mm concrete cover to the steel on all exposed faces for fire and corrosion performance. Verification / Alternative check: Composite design guides and standard detailing manuals present similar practical detailing rules that align with the above statements. Why Other Options Are Wrong: Options a–d each reflect correct aspects; none are individually incorrect, therefore the combined correct choice is “All the above”. Common Pitfalls: Painting the steel before encasement (may impair bond), inadequate cover in aggressive exposure, or oversize steel cores that complicate reinforcement placement and concrete compaction. Final Answer: All the above

Question 8

Roof Trusses – Slenderness of Tie Members with Stress Reversal For a roof-truss tie member that may experience reversal of stress (tension to compression), what is the limiting slenderness ratio (effective length / least radius of gyration)?

Accepted Answer

Introduction / Context: Slenderness ratio limits control buckling risk in compression and govern vibration and serviceability in tension members. If a nominal “tie” can be subjected to stress reversal due to wind, erection, or temperature effects, codes restrict its slenderness as if limited compression might occur. Given Data / Assumptions: Member: roof-truss tie with possible tension-compression reversal. Definition: slenderness = effective length / least radius of gyration. Design intent: ensure stability under accidental or occasional compression. Concept / Approach: Members that can see compression must satisfy a lower upper bound on slenderness to provide robustness. Ties with potential reversal therefore adopt limits closer to those for light compression members rather than very high ratios typical of pure tension bars. Step-by-Step Solution: 1) Recognize that stress reversal requires a conservative slenderness cap.2) A commonly taught limit for such tie members is 180.3) Thus, the ratio (effective length / least radius of gyration) should not exceed 180. Verification / Alternative check: Classic steel design references list 180 for tension members liable to reversal, while members always in tension may be allowed higher values (e.g., up to ~400). Why Other Options Are Wrong: 200 or 240: Higher than the commonly adopted cap for reversal cases; less conservative. 350: Suitable for pure tension elements in some contexts, not for reversal. Common Pitfalls: Assuming a tie never experiences compression; neglecting erection and wind suction scenarios; using the high tension-only limits indiscriminately. Final Answer: 180

Question 9

Load Testing of Structural Components – 24-Hour Test Level When load-testing an individual part of a structure for 24 hours, the applied load equals the full dead load plus what multiple of the imposed (live) load?

Accepted Answer

Introduction / Context: Load tests verify that constructed members satisfy service performance by limiting residual deflection and damage under a sustained test load. Codes specify the magnitude and duration of such tests to provide a uniform basis for acceptance. Given Data / Assumptions: Test duration: 24 hours. Loading composition: full dead load plus a multiple of imposed load. Acceptance intent: check deflection recovery and absence of distress. Concept / Approach: The test load must be high enough to exercise the structure while remaining within safe limits. A commonly adopted magnitude for the live component is 1.25 times the imposed load, applied in combination with full dead load, to provide a meaningful performance check of stiffness and serviceability. Step-by-Step Solution: 1) Identify required combination: full dead load + k * imposed load.2) Select k = 1.25 as the widely taught factor.3) Maintain the load for 24 hours; record deflections and inspect for cracks. Verification / Alternative check: Many acceptance criteria require that residual deflection after unloading not exceed a small fraction of the maximum test deflection, confirming elastic recovery at the 1.25 live-load level. Why Other Options Are Wrong: 1.0 imposed load: Less demanding; may not reveal stiffness deficiencies. 1.15 imposed load: Still conservative but below the commonly prescribed level. 1.3 imposed load: More severe than customary 24-hour test requirements in many building codes. Common Pitfalls: Not maintaining the load for the full duration; measuring deflection from an unstable datum; neglecting temperature effects during long tests. Final Answer: 1.25 times imposed load

Question 10

Beam Mechanics – Purlin under Uniformly Distributed Load For a purlin of span L subjected to a uniformly distributed load ω (per unit length), what is the maximum bending moment assumed for a simply supported condition?

Accepted Answer

Introduction / Context: Purlins span between roof trusses or rafters and carry roofing loads idealized as uniform line loads. Determining the maximum bending moment for a given support condition is a fundamental step in sizing the purlin section and its connections. Given Data / Assumptions: Span: L (simply supported). Loading: uniform line load ω (force per unit length along the span). Objective: compute the maximum bending moment magnitude for design. Concept / Approach: For a simply supported beam under uniform load, reactions are equal at supports and the bending moment diagram is a parabola with a peak at midspan. The standard closed-form result for the midspan moment is ω L^2 / 8, widely used in preliminary and detailed design. Step-by-Step Solution: 1) Write support reactions: R_A = R_B = ω L / 2.2) Bending moment at distance x from the left: M(x) = R_A x − ω x^2 / 2.3) Set dM/dx = 0 ⇒ R_A − ω x = 0 ⇒ x = L / 2 (midspan).4) Evaluate M_max = M(L/2) = (ω L / 2) * (L/2) − ω (L/2)^2 / 2 = ω L^2 / 8. Verification / Alternative check: Area-under-shear-diagram or energy methods yield the same value. Standard tables of beam formulas list ω L^2 / 8 for this case, confirming the result. Why Other Options Are Wrong: ω L^2 / 12 and ω L^2 / 16: Too low; these belong to other support/load cases. ω L^2 / 10: Not a standard exact result for classical boundary conditions. Common Pitfalls: Confusing simply supported with continuous purlins (which reduce peak moment due to continuity), or neglecting load components from roof slope and cladding self-weight. Final Answer: ω L^2 / 8

Question 11

Concrete – Water Quality Acceptance Criterion for Mix Water (IS practice) Compare 28-day cube strengths made with available site water vs distilled water; the site-water concrete should achieve at least this percentage of the distilled-water strengt…

Accepted Answer

Introduction / Context: Quality of mixing water directly affects setting, strength, and durability of concrete. Indian practice allows the use of non-distilled (available) water if it does not reduce compressive strength of standard cubes beyond a small tolerance when compared with cubes made using distilled water. This question tests the acceptance threshold used in field quality control. Given Data / Assumptions: Two sets of at least three 15 cm cubes are prepared identically except for water type. One set uses available site water; the other uses distilled water. Cubes are cured and tested at 28 days under standard conditions. Decision is based on average strengths of the two sets. Concept / Approach: The acceptance criterion compares average 28-day strengths. If the cubes made with available water achieve a sufficiently high percentage of the strength of cubes made with distilled water, the water is deemed acceptable for mixing concrete. The commonly adopted benchmark is 90% or more, reflecting a small allowable difference due to impurities that do not materially impair hydration or long-term performance. Step-by-Step Solution: 1) Cast ≥ 3 cubes with available water and ≥ 3 cubes with distilled water under otherwise identical procedures.2) Cure all cubes identically and test at 28 days to obtain average strengths f_site and f_distilled.3) Compute percentage = (f_site / f_distilled) * 100.4) Accept the site water if percentage ≥ 90; otherwise, reject for mixing or treat accordingly. Verification / Alternative check: Field experience shows potable water and many safe non-potable sources typically meet or exceed 90% of the distilled-water strength; markedly lower values signal harmful contaminants (e.g., excessive organics, sugars, acids, or salts) that retard hydration or create porosity. Why Other Options Are Wrong: 70% and 80% are too lenient and could allow strength loss with serious durability risks. 95% is overly strict and may reject otherwise acceptable water without practical benefit. Common Pitfalls: Not controlling other variables (w/c ratio, cement, curing), using too few specimens, or comparing individual cubes rather than averages. Final Answer: 90%

Question 12

Steel Girders – Lateral Restraint at Supports For a simply supported girder whose compression flange is fully restrained against lateral movement at both ends, the effective laterally unsupported length is taken as:

Accepted Answer

Introduction / Context: Lateral–torsional buckling (LTB) limits the bending strength of beams when the compression flange is not adequately restrained. Codes provide effective length factors that convert the clear distance between restraints into an equivalent laterally unsupported length for design. This item checks the factor used when the compression flange of a simply supported girder is fully restrained laterally at both ends. Given Data / Assumptions: Simply supported girder. Compression flange laterally restrained at both supports (no lateral movement). Standard code-based effective length factors apply. Span is the distance between supports/points of zero moment for a typical UDL case. Concept / Approach: End restraint to the compression flange reduces the effective laterally unsupported length because the flange cannot sway at the bearings. Empirical/analytical studies reflected in code tables yield a reduction factor from the full span to about 0.85 × span for such end conditions in simply supported girders. Step-by-Step Solution: 1) Identify end condition: compression flange restrained laterally at both supports.2) Select effective length factor from standard guidance for simply supported beams with lateral end restraint.3) Compute effective length = 0.85 * span for LTB checks of the compression flange. Verification / Alternative check: In the limiting case of no restraint, effective length would be close to the full span. With full lateral restraint and additional warping/rotation restraint, even lower factors may apply in special cases; 0.85 × span is the commonly adopted value for end lateral restraint only. Why Other Options Are Wrong: span assumes no beneficial restraint; 0.7 × span and 0.5 × span are too low for the stated boundary condition; they would overestimate capacity. Common Pitfalls: Confusing lateral restraint (preventing sideways movement of compression flange) with full rotational fixity, and misapplying factors intended for continuous beams or braced points along the span. Final Answer: 0.85 × span

Question 13

Weld / Combined Stresses – Equivalent (resultant) stress for combined bending tension and shear Given bending tensile stress fbt and shear stress fs at a point, choose the correct equivalent stress fe used for design checks.

Accepted Answer

Introduction / Context: Connections and welds often experience simultaneous normal stress (from axial force or bending) and shear stress (from transverse loads or torsion). Design practice uses an equivalent stress to ensure the combined state does not exceed the allowable for the material and detail. This question probes the commonly used expression for combining bending tension and shear, especially for fillet weld throats and similar details. Given Data / Assumptions: Normal (tensile) stress due to bending at the checked point = fbt. Shear stress at the same point = fs. Design per traditional Indian steel practice (working stress/allowable stress format) for weld/connection checks. Concept / Approach: A conservative and widely used interaction is based on the maximum shear (or distortion energy) theory adapted for weld design, leading to an equivalent resultant stress of the form fe = sqrt(fbt^2 + 3 * fs^2). The factor 3 reflects the relative contribution of shear in the combined state at the weld throat metal compared with pure normal stress. Step-by-Step Solution: 1) Determine fbt from section forces (M) and geometry (Z) at the critical fibre/throat.2) Determine fs from V / (area effective) or from combined torsion/longitudinal effects as appropriate.3) Compute fe using fe = sqrt(fbt^2 + 3 * fs^2).4) Check fe ≤ permissible/allowable stress for the detail and electrode/base metal used. Verification / Alternative check: Where code permits the principal stress approach, results are similar in magnitude for typical ratios of fbt to fs. The chosen interaction has long-standing acceptance in weld design tables and examples. Why Other Options Are Wrong: sqrt(fbt^2 + fs^2) underestimates the influence of shear; linear addition fbt + fs lacks physical basis; the principal stress expression (fbt/2 + sqrt((fbt/2)^2 + fs^2)) gives a principal stress, not the weld-throat equivalent used in allowable-stress weld design. Common Pitfalls: Mixing limit-state factors with allowable-stress values, or evaluating stresses at mismatched locations (e.g., extreme fibre vs weld throat centreline). Final Answer: fe = sqrt(fbt^2 + 3 * fs^2)

Question 14

Corrosion Protection – Minimum steel thickness when directly exposed to weather and not accessible for cleaning/repainting

Accepted Answer

Introduction / Context: Atmospheric corrosion rate depends strongly on exposure and maintenance accessibility. Where members are exposed to weather and cannot be cleaned or repainted easily, a thicker section is mandated to provide corrosion allowance and preserve structural capacity over time. This question recalls the minimum thickness requirement for such inaccessible members. Given Data / Assumptions: Structural steel member directly exposed to weather. Member is not accessible for routine maintenance. Design per traditional IS steel practice and detailing guidelines. Concept / Approach: Minimum plate/section thickness limits aim to ensure that expected corrosion loss during the service life does not reduce the net thickness below safe levels. When access is poor, the code stipulates a higher minimum thickness than for accessible members, commonly 8 mm, to provide durability without excessive reliance on maintenance. Step-by-Step Solution: 1) Classify exposure: weather-exposed and inaccessible → adopt higher minimum.2) Select minimum thickness = 8 mm for such members.3) Detail with sealed edges, drip plates, and avoid crevices to further limit corrosion. Verification / Alternative check: Accessible members (regularly maintainable) may be permitted at 6 mm, but this does not apply here. The 8 mm limit aligns with corrosion allowance needs in practical Indian conditions. Why Other Options Are Wrong: 4 mm is too thin for outdoor exposure; 6–7 mm suit better-maintained situations; 8 mm is the accepted minimum for inaccessible, weather-exposed details. Common Pitfalls: Overlooking water traps, not sealing lap edges, or assuming paint alone will offset chronic inaccessibility. Final Answer: 8 mm

Question 15

Shear Reinforcement (IS 456) – Minimum area of vertical stirrup legs Given beam width b, stirrup spacing x, and characteristic strength of stirrup steel fc (yield strength), the minimum total cross-sectional area of stirrup legs Asv should satisfy:

Accepted Answer

Introduction / Context: Minimum shear reinforcement in reinforced concrete beams prevents sudden diagonal tension failure, controls crack widths, and provides ductility even when computed shear demand is low. IS 456 prescribes a minimum stirrup area based on beam width, stirrup spacing, and steel strength. This question asks for the correct expression using the given symbols. Given Data / Assumptions: b = beam width (mm). x = stirrup spacing along the beam (mm). fc = characteristic/yield strength of stirrup steel (N/mm2). Vertical two-legged (or multi-legged) stirrups crossing the shear plane. Concept / Approach: IS 456 gives a minimum shear reinforcement requirement in the form Asv/s ≥ 0.4 b / 0.87 fy for vertical stirrups, where fy is the yield strength of stirrup steel. Rearranging for total stirrup area within spacing x (i.e., s = x) gives Asv ≥ (0.4 b x) / (0.87 fy). Mapping fy to the symbol fc yields the requested expression. Step-by-Step Solution: 1) Start with IS 456 minimum: Asv / s ≥ 0.4 * b / (0.87 * fy).2) Substitute s = x and fy = fc (as defined): Asv ≥ (0.4 * b * x) / (0.87 * fc).3) Ensure Asv is the sum of all stirrup legs crossing the shear plane (e.g., two legs for a typical closed stirrup). Verification / Alternative check: Try b = 230 mm, x = 200 mm, fc = 415 N/mm2 → Asv ≥ (0.4230200)/(0.87*415) ≈ 51 mm2 per spacing; a 2-legged 8 mm bar (area ≈ 100 mm2) exceeds this, satisfying the minimum. Why Other Options Are Wrong: (a) misses the 0.87 factor; (c) and (d) are dimensionally and conceptually incorrect for minimum shear steel; only (b) matches the codal rearrangement. Common Pitfalls: Using bar area of a single leg instead of total legs, ignoring hooks/bends for detailing, or confusing design shear with minimum requirement. Final Answer: Asv ≥ (0.4 * b * x) / (0.87 * fc)

Question 16

Steel Column Bases – Minimum plan dimension relative to column size The minimum length of the side or diameter of a column base shall not be less than (d is the minimum diameter/side of the column):

Accepted Answer

Introduction / Context: Base plates spread column loads to concrete foundations, limiting bearing pressure and providing room for anchorages and edge distances. Codes set minimum plate dimensions in relation to column size to ensure adequate edge cover and fabrication tolerances. This item recalls the commonly used minimum expression for plan dimension. Given Data / Assumptions: Steel column with minimum across-flat/diameter dimension d. Base plate bearing on concrete with grout layer. Need minimum plan dimension to accommodate load diffusion and detailing. Concept / Approach: The formula 1.6 (d + 75) mm embeds a basic margin (75 mm) around the column and a scale factor (1.6) to ensure practical edge distances, anchor position, and bearing area. It functions as a baseline before detailed bearing and bending checks determine the final size. Step-by-Step Solution: 1) Identify d (smallest side/diameter of the column section).2) Compute side length = 1.6 * (d + 75) mm.3) Verify against bearing stress and plate bending; increase size if required by bearing/bending checks. Verification / Alternative check: Trial sizing by bearing pressure and cantilever strip method typically confirms this minimum is a realistic starting point for most columns. Why Other Options Are Wrong: d + 75 is too small for anchors/edge cover; 1.2 (d − 75) is dimensionally suspect and may be unsafe; 1.5 (d + 75) is slightly smaller than recommended minimum for general practice. Common Pitfalls: Ignoring grout holes/anchor bolt edge distances or not checking for biaxial load cases after preliminary sizing. Final Answer: 1.6 (d + 75) mm

Question 17

Nominal Mix Concrete M20 – Recommended water per 50 kg bag of cement (for normal workability with standard aggregate size)

Accepted Answer

Introduction / Context: Water content governs workability and, through the water–cement ratio, strongly influences strength and durability. For nominal mixes like M20 (1:1.5:3 by volume, typical), guidance tables relate target workability to water per 50 kg of cement for standard maximum aggregate size. This question asks for that recommended value used in field batching. Given Data / Assumptions: Nominal mix concrete M20. Standard maximum aggregate size (e.g., 20 mm) and normal workability. Good compaction and curing are assumed. Concept / Approach: Typical guidance indicates about 30 litres of water per 50 kg bag for normal workability with 20 mm aggregate. Final adjustments are made for measured aggregate moisture/surface water and for specific workability requirements (slump). Keeping water low while satisfying placement needs maintains a reasonable w/c ratio for M20 strength levels. Step-by-Step Solution: 1) Start from code/table value for M20 with standard aggregate → approximately 30 L per 50 kg cement.2) Adjust at site for sand moisture and required slump using trial mixes, avoiding unnecessary water addition.3) Ensure curing to achieve target strength at 28 days. Verification / Alternative check: Many mix-design examples show that 28–32 L is typical for 20 mm aggregate; 30 L sits at the centre of this band and is widely adopted in nominal work. Why Other Options Are Wrong: 60 L would produce extremely high w/c; 45 L and 34 L are excessive for M20 with standard aggregate; 30 L aligns with accepted practice. Common Pitfalls: Ignoring moisture corrections, using water to solve placement issues instead of improving compaction or using plasticizers, or skipping curing. Final Answer: 30 litres

Question 18

Concrete Cover – Correct detailing statements for columns, beams, and slabs Identify the most comprehensive correct statement about nominal cover requirements for longitudinal reinforcement in typical members.

Accepted Answer

Introduction / Context: Nominal cover protects reinforcement against corrosion, ensures bond, and provides fire resistance. Codes specify minimum covers that depend on member type, exposure, and bar size. This question consolidates typical minimums used in day-to-day detailing for columns, beams, and slabs under normal exposure. Given Data / Assumptions: Normal exposure conditions unless specified otherwise. Longitudinal bars considered for columns, beams, and slabs. Nominal cover measured to the surface of reinforcement. Concept / Approach: For columns, 40 mm (or bar diameter, whichever is greater) is a common minimum; beams often adopt 25 mm (or bar diameter, whichever is greater) at sides/soffit; ends/anchorage zones in beams require added allowance (e.g., 2 × dia or 25 mm minimum); slabs, being thinner, may use 15–20 mm depending on exposure and diameter, with the rule of diameter or a small fixed minimum, whichever governs. Step-by-Step Solution: 1) Check column rule → adopt ≥ max(dia, 40 mm).2) Check beam sides/soffit → adopt ≥ max(dia, 25 mm).3) Check end zones/anchorage → provide ≥ 2 × dia or 25 mm.4) Check slab tension face → adopt ≥ max(dia, 15 mm), increasing for severe exposure as required. Verification / Alternative check: Detailing manuals and standard schedules align with these values as baseline minima; higher covers may be required for durability classes or fire ratings. Why Other Options Are Wrong: Each of A–D states a valid rule in isolation; the most complete/correct consolidated choice is “All of the above”. Common Pitfalls: Not distinguishing cover for different exposure classes, forgetting increased cover for large bars or marine environments, or measuring cover to tie wires instead of bar surface. Final Answer: All of the above

Question 19

Thermal Properties of Steel – Linear coefficient of thermal expansion (per °C per unit length) as per IS 800 (traditional value)

Accepted Answer

Introduction / Context: Temperature changes cause steel members to expand or contract. The linear coefficient of thermal expansion α (per °C) is needed for computing thermal strains, movements, expansion-joint gaps, and secondary forces in restrained systems. IS steel design commonly uses a standard value for α used in everyday calculations. Given Data / Assumptions: Carbon structural steel within typical service temperature range. Uniform temperature change across member length. Linear elastic behavior assumed for strain computation. Concept / Approach: Thermal strain = α * ΔT. For structural steel, α is taken approximately as 12 × 10^-6 per °C, written in decimal form as 0.000012 per °C per unit length. This value is sufficiently accurate for most civil engineering applications and aligns with IS 800 practice. Step-by-Step Solution: 1) Adopt α = 12 × 10^-6 / °C for structural steel.2) Compute free expansion = α * L * ΔT; introduce expansion joints/restraints accordingly.3) If restrained, estimate induced thermal force using compatibility and member stiffness. Verification / Alternative check: Materials handbooks list α for steel between about 11–13 × 10^-6 / °C; using 12 × 10^-6 / °C (0.000012) is standard and within this band. Why Other Options Are Wrong: 0.000011 and 0.000013 are plausible but not the standard value typically adopted; 0.000014 is too high for common structural steels at ambient temperatures. Common Pitfalls: Ignoring temperature gradients through depth (leading to curvature), or omitting allowance for joint movements in long roofs/bridges. Final Answer: 0.000012

Question 20

Elastic Properties – Young's modulus E used in column allowable-compression formula (units: kg/cm2)

Accepted Answer

Introduction / Context: Column buckling and allowable compressive stresses depend on the elastic modulus E of steel. Indian steel design in older working-stress formulations expresses E in kg/cm2, and a standardized value is used for routine calculations. This question asks for that numerical value in traditional units. Given Data / Assumptions: Structural steel with typical elastic modulus near 200 GPa. Allowable-stress column formulas require E in kg/cm2. 1 GPa ≈ 10^4 kg/mm2 = 10^6 kg/cm2 / 100. Concept / Approach: Taking E ≈ 200–205 GPa for structural steel corresponds to about 2.0–2.05 × 10^6 kg/cm2. A commonly adopted value in Indian practice is ≈ 2.04 × 10^6 kg/cm2; in tabulations this appears as 2,044,000 kg/cm2 (2044000 kg/cm2). Step-by-Step Solution: 1) Start with E ≈ 200–205 GPa.2) Convert: 1 GPa ≈ 10^4 kg/mm2; 1 kg/mm2 = 100 kg/cm2 → E ≈ 2.04 × 10^6 kg/cm2.3) Choose the nearest tabulated value → 2,044,000 kg/cm2. Verification / Alternative check: Using 2.04 × 10^6 kg/cm2 yields column buckling predictions consistent with code examples and design aids; small variations within 200–205 GPa do not materially affect preliminary sizing. Why Other Options Are Wrong: 2,000,000 is slightly low; 2,045,000 and 2,047,000 are close but 2,044,000 is the standard rounded figure commonly cited in traditional tables. Common Pitfalls: Mixing SI units (MPa, GPa) with gravitational units (kg/cm2) mid-calculation, or using inconsistent E values across checks. Final Answer: 2044000 kg/cm2

Structural Design Specifications Questions