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Steel Structures MCQ Practice Set — 50 Questions with Answers

50 exam-oriented Steel Structures multiple-choice questions with the correct answer and a clear explanation for each. Frequently asked in AE Level Civil Engineering, JE Level Civil Engineering. Solve the full set below for free — no login required.

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  1. 1
    Steel StructuresHARD

    The 'residual stresses' in hot-rolled steel sections arise from:

    AApplied loads during fabrication
    BNon-uniform cooling after hot rolling — thick zones cool slowly (tension), thin zones cool fast (compression)
    CChemical treatment of steel
    DWelding of reinforcement to web

    Answer: B. Non-uniform cooling after hot rolling — thick zones cool slowly (tension), thin zones cool fast (compression)

    Explanation: Residual stresses: locked-in stresses from non-uniform cooling after hot rolling. Flange tips (thin, cool fast) → residual compression. Flange-web junction (thick, cool slowly) → residual tension. Typical values: ±100–150 MPa. Effect on columns: premature yielding at tips reduces column curve below Euler; this is why column curves (a, b, c, d in IS 800) depend on section geometry and residual stress level.

  2. 2
    Steel StructuresHARD

    The 'effective slenderness ratio' for a single angle compression member (IS 800) connected through one leg is:

    AL/r_min × 2.0 (more conservative)
    B0.85 × (L/r_min) — gusset provides partial end restraint to single-leg-connected angle compression member
    CSame as fully restrained strut: 0.5×L/r
    DL/r_vv without any reduction

    Answer: B. 0.85 × (L/r_min) — gusset provides partial end restraint to single-leg-connected angle compression member

    Explanation: IS 800 Cl. 7.5.2.2: for angle sections connected through one leg only, effective slenderness ratio = 0.85×(L/r_min) for equal angles or 0.85×L/r_vv for unequal angles connected through shorter leg. r_min or r_vv = minimum radius of gyration. The 0.85 factor accounts for end restraint from gusset connection. If connected through both legs: standard (L/r_min). For double angles back-to-back, use 0.85×L/r_min.

  3. 3
    Steel StructuresHARD

    The 'tension field action' in a plate girder web after buckling means:

    AThe entire web resists only compressive stresses
    BPost-buckling diagonal tension strips carry additional shear — web acts like a diagonal tension field after shear buckling
    CThe web is replaced by stiffeners only
    DFlanges carry all shear and web carries none

    Answer: B. Post-buckling diagonal tension strips carry additional shear — web acts like a diagonal tension field after shear buckling

    Explanation: Tension field action: after the web buckles in shear (diagonal compression buckles, diagonal tension survives). The web carries additional shear beyond elastic shear buckling strength by acting like diagonal tension strips (truss analogy). Flanges and vertical stiffeners act as chord/posts of this truss. IS 800 allows design using tension field where the web to flange connection can resist the tension field forces.

  4. 4
    Steel StructuresMEDIUM

    'Fatigue failure' in steel structures occurs under:

    AA single static overload event
    BRepeated cyclic loading at below-yield stress — crack initiates at notches/welds and propagates per S-N curve
    COnly in plastic range (fy exceeded)
    DCreep at high temperature

    Answer: B. Repeated cyclic loading at below-yield stress — crack initiates at notches/welds and propagates per S-N curve

    Explanation: Fatigue: failure under repeated cyclic loading at stresses BELOW static yield strength. Crack initiates at stress concentration (weld toe, notch, hole). Propagates with each cycle. S-N curve (stress range vs. cycles to failure): no endurance limit for steel in corrosive environment. IS 800 Cl. 13 provides fatigue design: stress ranges and detail categories. Critical for: crane girders, bridges, offshore structures.

  5. 5
    Steel StructuresMEDIUM

    Bolts in 'slip-critical' (HSFG) connections resist load by:

    AShear in bolt shank (bearing type)
    BFriction between contact surfaces (clamping force × slip factor) — no bolt shank shear at service load level
    COnly bolt tension (no friction used)
    DAdhesive bonding between connected parts

    Answer: B. Friction between contact surfaces (clamping force × slip factor) — no bolt shank shear at service load level

    Explanation: HSFG (High Strength Friction Grip) bolts: (1) Tightened to proof load (creating high clamping force); (2) Friction between contact surfaces transfers shear load (under service loads, no slip occurs); (3) If slip occurs: bolts then bear. Advantages: no slip at service → good for fatigue, vibration, reversal. Design: resistance = μ × N_b (μ = slip factor ≈ 0.45 for clean-blasted surface, N_b = bolt tension). IS 4000 covers HSFG.

  6. 6
    Steel StructuresMEDIUM

    The 'portal frame analysis' approximation for multi-storey frames under horizontal load assumes inflection points at:

    ASupports only (zero moment at base and top)
    BMid-height of all columns AND midspan of all beams — makes each sub-frame statically determinate for lateral load analysis
    COnly at beam-column joints
    DNo inflection points — full fixed frame analysis only

    Answer: B. Mid-height of all columns AND midspan of all beams — makes each sub-frame statically determinate for lateral load analysis

    Explanation: Portal method (lateral loads): assumes inflection points (zero moment, zero rotation) at: (1) Midheight of each column; (2) Midspan of each beam. This makes the frame statically determinate for each storey. Shear distribution: interior columns carry twice the shear of exterior columns. Portal method is appropriate for low-rise frames (H/W < 4). Cantilever method for tall frames: column axial forces proportional to distance from centroid.

  7. 7
    Steel StructuresMEDIUM

    In a 'composite beam' (steel + RC slab), 'shear connectors' are provided to:

    AOnly aesthetic — the slab always acts compositely
    BTransfer horizontal interface shear — forcing slab and steel beam to deflect together without slip, enabling composite bending action
    CProvide fire resistance to steel
    DAnchor the beam to the foundation

    Answer: B. Transfer horizontal interface shear — forcing slab and steel beam to deflect together without slip, enabling composite bending action

    Explanation: Composite steel-concrete beam: steel beam + RC slab act together if interface is connected. Shear connectors (stud connectors, IS 11384): transfer horizontal shear at steel-concrete interface → slab acts with steel in compression → neutral axis rises → reduced depth needed. Without connectors: no composite action (slip at interface). Partial composite: 50–75% shear connection for economy. Effective flange width: IS 11384 limits slab acting width.

  8. 8
    Steel StructuresHARD

    The 'wind-induced vibration' of a long-span cable-stayed bridge deck is controlled by:

    AOnly by increasing deck self-weight
    BStreamlined deck section (aerofoil), guide vanes/fairing, and tuned mass dampers (TMD) — prevent flutter and VIV
    CPainting the deck with anti-rust paint
    DOnly by using shorter spans

    Answer: B. Streamlined deck section (aerofoil), guide vanes/fairing, and tuned mass dampers (TMD) — prevent flutter and VIV

    Explanation: Wind effects on bridges: (1) Flutter (self-excited aeroelastic instability — coupled bending + torsion → catastrophic, Tacoma Narrows); (2) Vortex-induced vibrations (VIV) — lock-in at critical wind speed; (3) Buffeting (turbulent wind). Control: (1) Aerodynamic deck shape (streamlined box girder); (2) Fairing and guide vanes; (3) Tuned mass dampers (TMD); (4) Increased torsional stiffness (π-shape deck section). Wind tunnel testing mandatory for long-span bridges.

  9. 9
    Steel StructuresHARD

    'Vierendeel truss' differs from regular truss in that its members resist:

    AOnly axial forces (no moments like Pratt/Warren)
    BBending + shear + axial — rigid joints, no diagonals; shear transferred by chord/vertical bending (frame action)
    COnly torsion from eccentric loading
    DOnly prestress from cable system

    Answer: B. Bending + shear + axial — rigid joints, no diagonals; shear transferred by chord/vertical bending (frame action)

    Explanation: Vierendeel truss (frame truss): NO diagonals — only top chord, bottom chord, and vertical members with rigid joints. Members carry: bending + shear + axial (NOT axial only like Pratt/Warren trusses). Panel shear transferred by bending of chords and verticals (like a multi-storey frame). Heavier than diagonal trusses for same span. Used when: diagonal members cannot be accommodated (airport concourses, pedestrian bridges needing open web for views, Vierendeel facade).

  10. 10
    Steel StructuresHARD

    The 'crane girder' (overhead travelling crane beam) is designed for:

    AOnly dead load of the crane itself
    BFatigue (repeated cycles), biaxial bending (vertical + lateral surge), local web bearing, and impact factor (25%)
    COnly horizontal wind loads on the building
    DStatic wheel load with no impact or fatigue

    Answer: B. Fatigue (repeated cycles), biaxial bending (vertical + lateral surge), local web bearing, and impact factor (25%)

    Explanation: Crane girder: supports overhead crane wheels. Special considerations: (1) Fatigue loading (repeated wheel passes — IS 807 classifies cranes M1 to M8, IS 800 Cl. 13); (2) Biaxial bending (vertical wheel load → vertical bending; lateral surge force from crane → lateral bending on top flange); (3) Local web crippling under concentrated wheel load; (4) Impact factor: typically 25% added to static wheel load. Worst combination: maximum vertical + lateral surge. Dynamic analysis needed for heavy cranes.

  11. 11
    Steel StructuresMEDIUM

    'Fire protection' for structural steel (IS 800 Cl. 16) is needed because:

    ASteel corrodes rapidly in fire
    BSteel strength drops to 60% at 500°C — fire protection maintains structural integrity until evacuation; required FRR by IS 800 Cl. 16
    CSteel expands too much without protection
    DFire protection only improves aesthetics

    Answer: B. Steel strength drops to 60% at 500°C — fire protection maintains structural integrity until evacuation; required FRR by IS 800 Cl. 16

    Explanation: Steel at elevated temperatures: yield strength fy drops to 60% at 500°C and 23% at 700°C (IS 800 Table 21). Structural failure when steel temperature reaches critical temperature (typically 550°C for members at design load ratio). Protection methods: (1) Intumescent coating (expands and chars at heat → insulating layer); (2) Board/spray fire protection (gypsum boards, vermiculite spray); (3) Concrete encasement; (4) Water-filling hollow sections. Fire resistance rating (FRR) depends on protection thickness and material.

  12. 12
    Steel StructuresMEDIUM

    A 'castellated beam' is fabricated by:

    AAdding plates to existing I-section
    BCutting web in zigzag pattern, offsetting halves, rewelding — increases depth by 50–60% with same weight; web openings for services
    CCompressing solid rectangular billets at high pressure
    DUsing two channels connected back-to-back

    Answer: B. Cutting web in zigzag pattern, offsetting halves, rewelding — increases depth by 50–60% with same weight; web openings for services

    Explanation: Castellated beam: I-section beam with web cut in a zigzag pattern and the two halves welded back together after offsetting — creates hexagonal/cellular openings in web. Result: increased section depth (50–60% taller than original) → higher moment of inertia, higher bending stiffness → reduced deflection. Web openings: allow services (ducts, pipes) to pass through — reduces floor-to-floor height in buildings. Weight same as original I-section. Used for: long span roofs, floors where services integration needed.

  13. 13
    Steel StructuresMEDIUM

    The 'bending stress' in a plate girder flange is highest at:

    AThe neutral axis (zero stress there)
    BExtreme fibre of flange (farthest from neutral axis) where y is maximum: σ = M×(D/2)/I
    CAt the web-flange junction only
    DAt mid-flange width, not at the tip

    Answer: B. Extreme fibre of flange (farthest from neutral axis) where y is maximum: σ = M×(D/2)/I

    Explanation: Plate girder: I-section fabricated from plates (flanges + web). Bending stress σ = M×y/I. Bending stress in top/bottom flange: uniform across flange width in simple theory (but reduced by shear lag for wide flanges). Maximum bending stress at extreme fibre of flange = M×(D/2)/Ixx. Compressive flange: susceptible to lateral torsional buckling. Tension flange: fatigue (weld between flange and web — IS 1024/IS 800 Cl. 13). Fillet weld joins flange to web: carries horizontal shear flow q = VAȳ/I.

  14. 14
    Original practiceMEDIUM

    A steel tension member carries 100 kN on gross area 1500 mm². Average tensile stress is

    A133.33 N/mm²
    B116.67 N/mm²
    C33.33 N/mm²
    D66.67 N/mm²

    Answer: D. 66.67 N/mm²

    Explanation: Average direct (axial) stress sigma = P/A, where P = axial force and A = cross-sectional area. Assumes the load acts through the centroid and stress is uniform across the section (valid far from the load application point, per Saint-Venant's principle). Units: N/mm^2 (MPa). Stress=P/A=100×1000/1500=66.67 N/mm².

  15. 15
    Original practiceMEDIUM

    A steel plate 100 mm wide and 12 mm thick has 2 bolt hole(s) of 20 mm diameter across the critical section. Net area is

    A960 mm²
    B360 mm²
    C720 mm²
    D1200 mm²

    Answer: C. 720 mm²

    Explanation: Net sectional area = (b - n*d_h) * t, where b = plate width, n = number of bolt holes in the critical section, d_h = hole diameter (bolt diameter + 2 mm for drilled holes per IS 800), t = plate thickness. Net area is used to check net section rupture (tensile failure through bolt holes): capacity = 0.9 * An * fu / gamma_m1. Net area=(b-nd)t=(100-2×20)×12=720 mm².

  16. 16
    PYQ/PYQ-PatternMEDIUM

    A steel compression member has effective length 2500 mm and least radius of gyration 20 mm. Slenderness ratio is

    A250
    B125
    C50000
    D62

    Answer: B. 125

    Explanation: Slenderness ratio lambda = Le / r, where Le = effective length (depends on end conditions: both ends pinned Le=L; one fixed one free Le=2L; etc.) and r = least radius of gyration = sqrt(I_min/A). IS 800:2007 limits: lambda <= 180 for compression members, <= 400 for tension. Higher lambda means greater tendency to buckle under compression (Euler critical load = pi^2*E*I/Le^2). Slenderness ratio = Le/r = 2500/20=125.

  17. 17
    Bolt SpacingMEDIUM

    As per IS 800:2007, the minimum pitch (centre-to-centre distance) between bolts in the direction of load is:

    A2.0 x bolt diameter
    B2.5 x bolt diameter
    C3.0 x bolt diameter
    D1.5 x bolt diameter

    Answer: B. 2.5 x bolt diameter

    Explanation: As per IS 800:2007 Clause 10.2.2, the minimum pitch between bolts shall not be less than 2.5 times the nominal diameter of the bolt.

  18. 18
    Steel StructuresMEDIUM

    A compression member tends to buckle in the direction of

    AAxis of load
    BPerpendicular to the axis of load
    CMinimum cross section
    DLeast radius of gyration

    Answer: D. Least radius of gyration

    Explanation: Buckling occurs about the axis with the least radius of gyration, which corresponds to the direction of least stiffness.

  19. 19
    IS 800 BasicsMEDIUM

    As per IS 800:2007 (LSM for steel structures), partial safety factor for material γm0 for yielding check:

    A1.0
    B1.1
    C1.5
    D1.25

    Answer: B. 1.1

    Explanation: IS 800:2007: γm0 = 1.10 (resistance governed by yielding, e.g. tension member); γm1 = 1.25 (resistance governed by buckling). γmb = 1.25 (bolts). γmf = 1.25 (welds — partial).

  20. 20
    Roof Trusses SteelMEDIUM

    Minimum roof slope for corrugated GI sheet roofing (IS standards):

    A1:12 (5°)
    B1:3 (18°)
    C1:6 (9°)
    D1:20 (3°)

    Answer: C. 1:6 (9°)

    Explanation: GI corrugated sheet: minimum slope 1:6 (≈9°). Asbestos cement sheet: 1:5 (≈11°). Lower slopes risk water ponding and leakage at laps. End laps should increase with decreasing slope.

  21. 21
    Steel StructuresMEDIUM

    'Cold-formed steel' sections compared to hot-rolled (IS 811) have advantages in:

    AVery thick walls and high self-weight
    BCustom cross-section shapes, high stiffness-to-weight ratio, pre-galvanized, and punched holes for services — using thin sheets
    CBetter for very heavy structural applications > 50t columns
    DNo advantage over hot-rolled in any application

    Answer: B. Custom cross-section shapes, high stiffness-to-weight ratio, pre-galvanized, and punched holes for services — using thin sheets

    Explanation: Cold-formed sections: manufactured by bending (roll-forming/press-braking) thin sheets (0.5–8 mm) at room temperature. Advantages: (1) Wide variety of custom cross-sections; (2) Thin walls → high section modulus/weight ratio; (3) Galvanized (corrosion-resistant); (4) Pre-punched service holes; (5) Light weight (easier erection). Disadvantages: local buckling (thin walls), b/t limits in IS 811, susceptible to distortional buckling. Used in: purlins, girts, light steel framing.

  22. 22
    Steel StructuresHARD

    The 'intermediate transverse web stiffeners' in a plate girder (IS 800) are provided to:

    AReduce self-weight of girder
    BIncrease shear capacity by permitting tension field action and preventing web shear buckling — spaced at 1–2× web depth
    CProvide composite action with slab
    DImprove thermal expansion accommodation

    Answer: B. Increase shear capacity by permitting tension field action and preventing web shear buckling — spaced at 1–2× web depth

    Explanation: Plate girder web: slender web (d/tw > 67ε, IS 800) has reduced shear capacity — web buckles diagonally. Intermediate transverse stiffeners: (1) Increase shear capacity (post-buckling tension field action — diagonal tension in buckled web panels); (2) Prevent web buckling. Stiffener spacing: typically 1–2× web depth. Stiffener design: acts as column with web strip. Bearing stiffeners at ends/load points: transfer concentrated loads.

  23. 23
    Steel StructuresHARD

    The 'bulk storage silo' (bin) for granular materials is designed for:

    AOnly external wind load (no material pressure)
    BGranular material pressure (Janssen theory for deep bins), eccentric fill/discharge, and flow patterns (mass/funnel)
    CSame as water tank design
    DOnly the weight of stored material on the floor

    Answer: B. Granular material pressure (Janssen theory for deep bins), eccentric fill/discharge, and flow patterns (mass/funnel)

    Explanation: Silo design: Janssen''s theory (1895) for deep bins — granular material develops wall friction → horizontal pressure p_h = γ/K_μ × (1−e^(-K_μ z/R)) where K=lateral pressure ratio, μ=wall friction, R=hydraulic radius of plan. For shallow bins: Rankine. Pressures depend on: eccentric filling/discharge (causes unsymmetric loads → bending + hoop), mass flow vs funnel flow. IS 875 and Eurocode EN 1991-4 cover silo loads.

  24. 24
    Steel StructuresMEDIUM

    The 'slenderness ratio' limit for compression members in IS 800 is:

    AAlways 120 for all members
    B180 for primary compression; 250 for wind/seismic; 350 for tension with occasional compression per IS 800
    C50 for all members
    DNo limit specified in IS 800

    Answer: B. 180 for primary compression; 250 for wind/seismic; 350 for tension with occasional compression per IS 800

    Explanation: IS 800 Cl. 7.3.3: maximum slenderness ratio (KL/r) for compression members: (1) Members carrying load from dead + live (primary compression): 180; (2) Members carrying wind/seismic + gravity: 250; (3) Tension members used as compression in load reversal: 350; (4) Members normally in tension (sag rods etc.): 400. Higher limit → more slender allowed for light loading.

  25. 25
    Steel StructuresMEDIUM

    'Purlins' in a roof truss system are designed as:

    ACompression members (columns)
    BBiaxial bending members — gravity load resolved along roof slope and normal to it, plus wind in/out of plane
    CTension-only rods
    DOnly for resisting horizontal wind load

    Answer: B. Biaxial bending members — gravity load resolved along roof slope and normal to it, plus wind in/out of plane

    Explanation: Purlins: secondary members spanning between roof trusses, carrying roof sheets/cladding. Subjected to biaxial bending because: (1) Major axis bending from vertical (gravity) loads resolved along and perpendicular to slope; (2) Wind suction/pressure. IS 800: Z-section or C-section purlins common. Designed as beam in biaxial bending; can use IS 800 method for bending about both axes.

  26. 26
    Concrete Technology, RCC and Steel DesignHARD

    Slender compression elements in steel sections are prone to:

    AShrinkage cracking
    BChloride attack
    CLocal buckling
    DPlastic settlement

    Answer: C. Local buckling

    Explanation: Thin plates can buckle locally before the whole member fails.

  27. 27
    Concrete Technology, RCC and Steel DesignMEDIUM

    Lug angles in steel tension members are provided to:

    APrevent soil settlement
    BMeasure deflection
    CIncrease concrete cover
    DReduce length of connection

    Answer: D. Reduce length of connection

    Explanation: Lug angles connect outstanding legs and improve efficiency.

  28. 28
    Concrete Technology, RCC and Steel DesignHARD

    Limiting neutral axis depth ratio xu,max/d for Fe415 steel is commonly taken as:

    A0.48
    B0.46
    C0.87
    D0.53

    Answer: A. 0.48

    Explanation: For Fe415, limiting xu/d is 0.48.

  29. 29
    Plastic SectionMEDIUM

    Which of the following about cross-section classification is correct?

    Aplastic sections are never compact
    Ba plastic section buckles before yield at every fibre
    Cclassification is unrelated to width-thickness ratio
    Da plastic section can develop and sustain plastic moment with rotation capacity

    Answer: D. a plastic section can develop and sustain plastic moment with rotation capacity

    Explanation: Section class controls local buckling and rotation capacity. Correct option: D.

  30. 30
    Base PlateMEDIUM

    A steel column base plate transfers the column load to the concrete pedestal. The base plate area is determined by:

    ABearing strength of concrete only
    BColumn cross-sectional area
    CRequired area = P / (bearing strength of concrete)
    DWeld capacity

    Answer: C. Required area = P / (bearing strength of concrete)

    Explanation: Required base plate area A = P / f_bearing, where P = factored column load and f_bearing = 0.45 x f_ck (IS 456 bearing stress) or as per IS 800. The plate is then designed for the upward pressure on the projection beyond the column footprint.

  31. 31
    Stiffener DesignMEDIUM

    Load bearing stiffeners at supports of plate girders are designed as:

    ASimple tension members
    BCompression members (columns) with an effective section including stiffener plus a portion of the web
    CPure bending members
    DMembers in combined bending and torsion

    Answer: B. Compression members (columns) with an effective section including stiffener plus a portion of the web

    Explanation: Bearing stiffeners carry concentrated reaction/load by acting as columns. IS 800 clause on plate girders: bearing stiffener + 20 x tw of web on each side (for interior stiffeners) acts as a short compression column.

  32. 32
    Beam DesignMEDIUM

    The plastic section modulus Z_p for a symmetric I-section is used when the section class is:

    ASlender
    BSemi-compact
    CCompact or Plastic (for plastic moment M_p = f_y x Z_p)
    DElastic only

    Answer: C. Compact or Plastic (for plastic moment M_p = f_y x Z_p)

    Explanation: For Plastic and Compact sections (IS 800), design moment capacity = f_y x Z_p / gamma_m0. For Semi-compact: f_y x Z_e / gamma_m0. For Slender: effective section considering local buckling.

  33. 33
    Steel Grade IS 2062MEDIUM

    As per IS 2062, the grade Fe410 structural steel has a minimum yield strength of:

    A250 MPa
    B410 MPa
    C250 MPa (for plates up to 20 mm)
    D350 MPa

    Answer: C. 250 MPa (for plates up to 20 mm)

    Explanation: IS 2062 steel grades: Fe 410 has minimum yield strength 250 MPa (for thickness up to 20 mm), 240 MPa (20-40 mm), 230 MPa (>40 mm). The number 410 refers to minimum UTS (410 MPa), not yield strength.

  34. 34
    GeneralMEDIUM

    In plastic analysis of steel structures, collapse occurs when

    Afirst fibre reaches proportional limit
    Bsufficient plastic hinges form a mechanism
    Cdeflection becomes exactly zero
    Dall bolts are removed

    Answer: B. sufficient plastic hinges form a mechanism

    Explanation: A collapse mechanism forms after the required number of plastic hinges develops.

  35. 35
    GeneralMEDIUM

    A gusset plate in a truss joint is used to

    Aprovide road camber
    Breduce span length to zero
    Cconnect multiple members at a joint
    Dact as concrete cover

    Answer: C. connect multiple members at a joint

    Explanation: Gusset plates collect and transfer forces among connected members.

  36. 36
    GeneralMEDIUM

    High strength friction grip bolts transfer load primarily by

    Aadhesion of paint
    Bbearing of bolt shank only after slip
    Cfriction between connected plates
    Dweld fusion

    Answer: C. friction between connected plates

    Explanation: HSFG bolts are pre-tensioned and resist slip by friction.

  37. 37
    RCC, Steel and Timber StructuresHARD

    For mild steel, the yield plateau is seen clearly in:

    Asewer profile
    Bcompaction curve
    Chydrograph
    Dstress-strain curve

    Answer: D. stress-strain curve

    Explanation: Mild steel shows distinct upper and lower yield points in a tensile test.

  38. 38
    Theory of Structures and DesignHARD

    For a double-angle laced compression member, lacing bars are generally inclined to the member axis at:

    A80 to 90 degrees
    B10 to 20 degrees
    C20 to 30 degrees
    D40 to 70 degrees

    Answer: D. 40 to 70 degrees

    Explanation: Steel design practice keeps lacing inclination between about 40 and 70 degrees to the member axis.

  39. 39
    Theory of Structures and DesignHARD

    Block shear failure in a tension member involves:

    Aonly local buckling of outstanding leg
    Btension on one plane and shear on another connected plane
    Cpure compression only
    Donly weld throat crushing

    Answer: B. tension on one plane and shear on another connected plane

    Explanation: Block shear is a combined rupture/yielding mode along a block bounded by tension and shear planes near the connection.

  40. 40
    Theory of Structures and DesignHARD

    For a fillet weld, the effective throat thickness is approximately:

    A1.0 times weld size
    B0.7 times weld size
    C0.5 times weld size
    D1.414 times weld size

    Answer: B. 0.7 times weld size

    Explanation: For a standard 45 degree fillet weld, effective throat thickness is 0.7 times the weld leg size.

  41. 41
    Theory of Structures and DesignHARD

    The limiting moment coefficient Mu,lim/(fck b d^2) for Fe415 steel is nearest to:

    A0.138
    B0.133
    C0.149
    D0.111

    Answer: A. 0.138

    Explanation: Using xu,max/d = 0.48 and Mu = 0.36 fck b xu(d - 0.42xu), the coefficient is about 0.138 for Fe415.

  42. 42
    Concrete Technology, RCC and Steel DesignHARD

    Block shear failure in steel tension member involves combination of:

    APure torsion only
    BPure compression only
    CConcrete crushing only
    DShear along one path and tension along another

    Answer: D. Shear along one path and tension along another

    Explanation: A block of material tears out through shear and tension planes.

  43. 43
    Bolt Edge DistanceMEDIUM

    Minimum edge distance for a bolted steel connection is specified mainly to reduce the risk of:

    APlate tearing or splitting near the hole
    BElastic shortening of the whole member
    CExcessive concrete creep
    DHydraulic jump formation

    Answer: A. Plate tearing or splitting near the hole

    Explanation: Adequate edge distance prevents local tearing/bearing failure around bolt holes.

  44. 44
    Concrete Technology, RCC and Steel DesignMEDIUM

    HSFG bolts transfer load mainly by:

    AFriction between connected plates
    BBearing of bolt shank only
    CTimber dowel action
    DConcrete bond

    Answer: A. Friction between connected plates

    Explanation: High-strength friction grip bolts rely on clamping force and friction.

  45. 45
    Steel StructuresMEDIUM

    Battens shall be designed to carry bending moment and shear force arising from transverse shear force V which is ____ % of total axial load on column

    A25
    B2.5
    C2
    D3

    Answer: B. 2.5

    Explanation: Battens are designed for transverse shear generally taken as 2.5% of the axial load

  46. 46
    Steel StructuresMEDIUM

    In a roof truss, which of the following supports the roofing material?

    ATie beam
    BBase plate
    CPurlins
    DGusset plate

    Answer: C. Purlins

    Explanation: Purlins run between the trusses and directly carry the roofing sheets.

  47. 47
    Steel StructuresMEDIUM

    The thickness of a gusset plate attached to a tension member, as per code, should not be less than

    A6 mm
    B8 mm
    C12 mm
    D16 mm

    Answer: A. 6 mm

    Explanation: A minimum gusset-plate thickness of about 6 mm is commonly specified. [Self-solved — please verify.]

  48. 48
    Steel StructuresMEDIUM

    The diameter of a rivet hole is made larger than the rivet diameter by 2 mm for rivet diameters

    Aup to 12 mm
    Bup to 22 mm
    Cup to 15 mm
    Dexceeding 25 mm

    Answer: D. exceeding 25 mm

    Explanation: As per IS 800, the hole is 1.5 mm larger for rivets up to 25 mm and 2.0 mm larger for rivets exceeding 25 mm.

  49. 49
    Steel StructuresMEDIUM

    The density of steel used in structural members is taken as

    A1 gm/mm3
    B6.4 gm/mm3
    C7.85 gm/cc
    D13.6 gm/cc

    Answer: C. 7.85 gm/cc

    Explanation: Structural steel has a density of about 7.85 g/cc (7850 kg/m3).

  50. 50
    Steel StructuresMEDIUM

    The yield stress of a bolt of grade 4.6 is

    A400 MPa
    B600 MPa
    C420 MPa
    D240 MPa

    Answer: D. 240 MPa

    Explanation: For grade 4.6: fu = 4 x 100 = 400 MPa and fy = 0.6 x 400 = 240 MPa.

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