Question 1

According to traditional working-stress design values, the permissible bending stress for rolled steel I-beams and channels is closest to which value (units: kg/cm^2)?

Accepted Answer

Introduction / Context: Before modern limit-state design became prevalent, steel members were sized using working-stress design (WSD) with permissible (allowable) stresses. For structural steel of typical yield strength, a standard allowable bending stress was adopted to ensure a safety margin against yielding and serviceability concerns. Given Data / Assumptions: Rolled I-beams and channels in bending. Permissible stress values expressed in kg/cm^2. Traditional WSD context (not LRFD/limit-state). Concept / Approach: For mild steel with yield stress around 2500 kg/cm^2 (≈ 250 MPa), a typical WSD allowable bending stress is about 0.66 * Fy, giving ≈ 0.66 * 2500 ≈ 1650 kg/cm^2. This value was widely used for beams subject to bending about the major axis under service loads. Step-by-Step Solution: Take Fy ≈ 2500 kg/cm^2 as a representative yield stress for mild steel.Compute 0.66 * Fy ≈ 1650 kg/cm^2.Select the option that matches this conventional permissible bending stress. Verification / Alternative check: Older steel design handbooks list allowable bending stresses near 1650 kg/cm^2 for bending, with different values for tension, compression, and shear reflecting safety factors and instability checks. Why Other Options Are Wrong: 1500 or 1575 kg/cm^2: Conservative relative to the classic 0.66Fy benchmark. 945 kg/cm^2: In the range of typical allowable shear, not bending. 1875 kg/cm^2: Too high for the classic WSD bending allowance for Fy ≈ 2500 kg/cm^2. Common Pitfalls: Mixing units (MPa vs kg/cm^2) and misapplying conversion (1 MPa ≈ 10.197 kg/cm^2). Final Answer: 1650 kg/cm2.

Question 2

Riveting terminology check: which one of the following statements is incorrect regarding nominal diameter, gross diameter, gross area, and hole size of rivets?

Accepted Answer

Introduction / Context: Correct terminology for rivets and bolt holes is essential for net-section and bearing calculations. Misusing terms like nominal diameter, hole diameter, and gross area can lead to incorrect deductions and unsafe designs. Given Data / Assumptions: Nominal diameter refers to the shank size before driving. Holes are drilled/punched slightly larger than nominal to allow fit-up. Gross area of a rivet refers to its shank area, not the hole area. Concept / Approach: Nominal diameter is the specified shank diameter of the rivet. The rivet hole is larger by a clearance (commonly ≈ 1.5 mm for standard sizes). The gross diameter of the rivet is often taken as the nominal shank diameter (while the gross diameter of the hole is the hole size). The gross area of a rivet is the cross-sectional area of the rivet itself (based on nominal diameter), whereas the plate's net area is reduced by the area of the hole, not by the rivet's gross area. Step-by-Step Solution: Review each statement against standard definitions.Identify that calling the rivet's gross area equal to the hole area is wrong—the hole belongs to the plate, not the rivet.Hence, option stating 'gross area of a rivet equals hole area' is incorrect. Verification / Alternative check: Design examples compute plate net area by subtracting n * (hole diameter) * thickness, while rivet strength checks use rivet shank area (based on nominal diameter) for shear/tension capacities. Why Other Options Are Wrong: Option (a) and (d) align with standard practice. Option (b) is acceptable wording when explicitly referring to 'gross diameter of the hole'; the key incorrect statement remains the one equating rivet gross area to hole area. 'None of these' is incorrect since one statement is indeed wrong. Common Pitfalls: Using hole area for rivet strength or rivet area for plate net-area deduction—these are distinct checks. Final Answer: The gross area of a rivet is the cross-sectional area of the rivet hole.

Question 3

In working-stress design of rolled steel beams, the typical permissible average shear stress (units: kg/cm^2) is approximately which of the following values?

Accepted Answer

Introduction / Context: Permissible shear stress governs the design of web thickness and stiffening in beams subjected to significant shear near supports. Traditional WSD adopts an allowable average shear stress derived as a fraction of yield stress to maintain adequate safety. Given Data / Assumptions: Mild steel yield stress ≈ 2500 kg/cm^2 (≈ 250 MPa). Permissible shear often taken around 0.4 * Fy in older practice. Values given are typical exam figures. Concept / Approach: Using τ_allow ≈ 0.4 * Fy yields ≈ 0.4 * 2500 = 1000 kg/cm^2. Many handbooks round to a nearby conventional value such as 945 kg/cm^2 for rolled sections, reflecting further safety considerations and standardization across tables. Step-by-Step Solution: Adopt Fy ≈ 2500 kg/cm^2 for mild steel.Compute 0.4 * Fy ≈ 1000 kg/cm^2.Select the closest standard permissible value used for beams: 945 kg/cm^2. Verification / Alternative check: Older design tables and example problems commonly reference 945 kg/cm^2 for shear in rolled beams, while bending uses ~1650 kg/cm^2, maintaining consistent proportionality to Fy. Why Other Options Are Wrong: 845 and 1025 kg/cm^2: Not the conventional tabulated value for rolled beams. 1500 and 1875 kg/cm^2: Too high for shear; closer to bending/tension allowances in WSD. Common Pitfalls: Confusing average shear (V/A_web) with maximum parabolic shear in rectangular webs; WSD limits are typically for average shear in design checks. Final Answer: 945 kg/cm2.

Question 4

Dead loads in structural design: which statement is correct regarding their nature and treatment in sizing members?

Accepted Answer

Introduction / Context: Accurate characterization of loads is critical for safe and economical design. Dead loads represent permanent, stationary actions on a structure and are the first inputs estimated when sizing beams, columns, and foundations. Given Data / Assumptions: Dead load comprises self-weight and other permanently fixed items. Dead load is typically constant in time and space for a given configuration. Designers determine dead load early using unit weights of materials. Concept / Approach: Dead loads are permanent: structural self-weight, floor finishes, roofing, partitions that are fixed, cladding, and permanently attached equipment. They do not migrate spatially and their magnitude is predictable from quantities and densities. In contrast, live loads vary with occupancy and usage, both in magnitude and location. Step-by-Step Solution: List dead-load components (self-weight + fixed attachments).Use unit weights (e.g., concrete, steel, masonry) and dimensions to estimate magnitudes at the outset.Recognize that dead load classification excludes position-varying loads. Verification / Alternative check: Codes define dead loads as permanent, immovable loads; live loads are moveable and variable. Estimation tables confirm designers compute dead loads at concept stage to select preliminary sizes and then refine as detailing progresses. Why Other Options Are Wrong: Dead loads change position and vary: describes live loads, not dead. Dead loads are known at the beginning: generally true, but option (a) is the clearer definition; the question asks for a correct statement, and (a) most directly captures the definition. None: incorrect because (a) is correct. Common Pitfalls: Classifying removable partitions or equipment as dead load when they may be treated as live or superimposed variable load by code. Final Answer: Dead load includes self-weight of the structure plus superimposed loads permanently attached.

Question 5

Within the elastic limit, the ratio of shearing stress to shearing strain is called what?

Accepted Answer

Introduction / Context: Different moduli describe material stiffness under distinct modes of deformation. In shear, the proportionality between shear stress and shear strain in the elastic range defines the shear modulus, a key parameter for torsion, shear deformation, and plate/beam analyses. Given Data / Assumptions: Linear elastic behavior is assumed. Shear stress (tau) and shear strain (gamma) are considered. Isotropic material relations may link E, G, and nu. Concept / Approach: The shear modulus (G), also called modulus of rigidity, is defined by G = tau / gamma. For isotropic materials, G relates to Young’s modulus E and Poisson’s ratio nu through E = 2G(1 + nu). Bulk modulus K is used for volumetric compression under hydrostatic stress, while tangent modulus is a local slope in nonlinear regimes. Step-by-Step Solution: Identify the deformation mode: shear.Apply Hooke’s law in shear: tau = G * gamma.Select the modulus corresponding to shear stiffness: G. Verification / Alternative check: Typical values: structural steel G ≈ 80,000 MPa when E ≈ 200,000 MPa and nu ≈ 0.3, matching the isotropic relation. Why Other Options Are Wrong: Modulus of elasticity (E): pertains to normal stress–strain. Bulk modulus (K): volumetric stiffness under hydrostatic loading. Tangent modulus: used for inelastic analysis. All the above: only one correct definition applies here. Common Pitfalls: Using E in torsion or shear deflection formulas instead of G, leading to significant error. Final Answer: Shear modulus of elasticity.

Question 6

For a tension member with rivet or bolt holes, the net sectional area is computed as which of the following (relative to gross area)?

Accepted Answer

Introduction / Context: The presence of fastener holes reduces the effective area that resists tension. Correctly computing net area is essential for capacity checks and to prevent brittle net-section rupture across the line of holes. Given Data / Assumptions: Member contains one or more lines of holes for bolts/rivets. Hole diameter used for deduction is the nominal hole size, not the fastener size. Staggered patterns may require additional adjustments. Concept / Approach: Net area A_n = A_g - n * d_hole * t for a straight path through n holes (single plate), where A_g is gross area, d_hole is hole diameter, and t is plate thickness. For staggered lines, a zig-zag correction increases A_n due to longer failure path. The principle remains: subtract the material removed by holes from the gross area along the critical section. Step-by-Step Solution: Identify the critical failure path (straight or zig-zag).Sum the total hole width crossing the path and multiply by thickness to obtain area removed.Compute A_n by subtracting the hole area from A_g, applying any stagger correction if required. Verification / Alternative check: Design examples and codes adopt identical logic; capacity in tension is based on the lesser of gross yielding and net-section rupture, the latter using the net area just computed. Why Other Options Are Wrong: Adding, dividing, or multiplying by hole area has no physical basis for section loss. None of these: incorrect because the subtraction method is established practice. Common Pitfalls: Using fastener nominal diameter instead of hole diameter (which is larger due to clearance). Neglecting stagger correction when applicable. Final Answer: Gross area minus the area of the rivet holes.

Question 7

Slenderness calculation: a stanchion has unsupported length 4 metres and least radius of gyration r = 5 cm. What is its slenderness ratio L/r?

Accepted Answer

Introduction / Context: Slenderness ratio is a key parameter for column stability and determines which buckling curve or formula applies. It is the ratio of the member's effective length to the least radius of gyration of its cross-section, expressed in consistent units. Given Data / Assumptions: Unsupported (effective) length L = 4 m. Least radius of gyration r = 5 cm. No end-restraint modification is specified; interpret L as given. Concept / Approach: The slenderness ratio λ is computed as L/r. Units must be consistent; convert metres to centimetres when r is in centimetres. Higher λ indicates greater susceptibility to buckling and lower allowable compressive stress for a given material and section. Step-by-Step Solution: Convert L to centimetres: L = 4 m = 400 cm.Compute λ = L / r = 400 / 5.Obtain λ = 80. Verification / Alternative check: A quick magnitude check: an 80:1 length-to-radius ratio is a moderate slenderness typical of rolled columns; it lies within ranges where inelastic buckling considerations may be relevant depending on the steel grade and code curve used. Why Other Options Are Wrong: 60, 70, 90, 100: Do not match the straightforward calculation with correct unit conversion. Common Pitfalls: Forgetting to convert L to the same unit system as r. Confusing least radius of gyration with least radius of curvature or with section depth. Final Answer: 80.

Question 8

Definition check: a beam is most appropriately defined as a structural member primarily subjected to which type of loading?

Accepted Answer

Introduction / Context: Clear definitions of structural members help assign the correct analysis models and design checks. A beam is distinguished from truss and column members by the dominant type of loading and the resulting internal actions it resists. Given Data / Assumptions: Member is part of a frame or floor system. Loads typically act perpendicular to the member's longitudinal axis. Primary internal effects are bending and shear. Concept / Approach: A beam primarily resists transverse loads (perpendicular to its axis), developing bending moments and shear forces. Axial forces may be present (e.g., from frame action or temperature), but they are secondary in the classical beam idealization. Purely axial members (tension/compression) are categorized as ties or struts/columns, while beam-columns handle significant combined axial and bending actions. Step-by-Step Solution: Identify the dominant loading orientation: perpendicular to the member axis.Recall that this leads to bending and shear as primary internal forces.Select the option that names transverse loading as the defining characteristic. Verification / Alternative check: Beam theory (Euler–Bernoulli, Timoshenko) is built on transverse loading assumptions; design checks focus on flexure, shear, deflection, and sometimes lateral-torsional buckling due to bending about the strong axis. Why Other Options Are Wrong: Axial loading: describes ties/struts/columns. Axial and transverse: may describe a beam-column, not a pure beam definition. None: incorrect because beams are classically defined by transverse loading. Common Pitfalls: Misclassifying members with small axial force as beam-columns; if axial is minor, beam design still governs. Final Answer: Transverse loading.

Question 9

For a column of height L that is fixed in position and in direction at both its top and bottom (fixed–fixed end conditions), what is its effective length used for buckling checks?

Accepted Answer

Introduction / Context: In column buckling theory for steel and reinforced concrete members, the concept of effective length converts different end restraints into an equivalent pin–pin length for use in slenderness and Euler load calculations. When both ends of a column are fixed in position and restrained against rotation, the restraining stiffness greatly increases the buckling capacity, which is reflected by a shorter effective length than the actual height. Given Data / Assumptions: Actual column height = L. Top end: fixed in position (no lateral translation) and direction (no rotation). Bottom end: fixed in position and direction as well. Member is straight, prismatic, and concentrically loaded until elastic buckling. Concept / Approach: Effective length L_eff is the notional length for which a pin–pin column would buckle in the same load as the real column with restrained ends. Classical end-condition factors (K-factors) provide L_eff = K * L. For fixed–fixed, K = 0.5, so L_eff = 0.5 * L. This smaller L_eff raises critical load and reduces slenderness ratio, λ = L_eff / r. Step-by-Step Solution: Identify end condition: fixed–fixed.Use K-factor: K = 0.5.Compute effective length: L_eff = K * L = 0.5 * L.Apply in design: slenderness λ = (0.5 * L) / r and P_cr = (pi^2 * E * I) / (L_eff^2). Verification / Alternative check: Relative comparison: for pin–pin, K = 1.0; for fixed–free (cantilever), K = 2.0; for fixed–pin, K ≈ 0.7. The fixed–fixed case appropriately gives the smallest K and thus the largest buckling load for a given L and section. Why Other Options Are Wrong: L or “L (same as actual length)”: corresponds to pin–pin, not fixed–fixed. 2L: corresponds to a cantilever (fixed–free), which is the least stiff case. Common Pitfalls: Confusing actual length with effective length. Mixing up end-condition descriptions (fixed vs pinned vs guided). Final Answer: (1/2)*L

Question 10

A compression member of actual length L is held in position and restrained in direction (fixed) at one end, and at the other end it is effectively restrained in direction but not held in position. What effective length should be used for buckling de…

Accepted Answer

Introduction / Context: Effective length captures the influence of boundary restraints on the buckling strength of columns. Real-world columns seldom have identical end restraints, so design tables present K-factors for various combinations of translation and rotation restraints at ends. Given Data / Assumptions: Member length = L. End A: held in position (no lateral translation) and restrained in direction (no rotation) → fixed. End B: effectively restrained in direction (no rotation) but not held in position (translation free) → rotationally fixed but laterally free (sway permitted). Elastic buckling; concentric axial compression. Concept / Approach: For the mixed case where one end is fully fixed and the other end cannot rotate but is free to translate (sway), standard alignment charts give a relatively large K due to the permitted sway mode. The widely used value for this end pair is K = 1.5, hence L_eff = 1.5 L. This reflects a much lower buckling capacity than fixed–fixed and even lower than pin–pin in many alignment conditions because sway increases the effective length. Step-by-Step Solution: Identify end A (fixed) and end B (rotation restrained, translation free).Select corresponding K-factor from end-condition table: K ≈ 1.5.Compute effective length: L_eff = K * L = 1.5 L.Use in design checks: λ = L_eff / r and P_cr = (pi^2 * E * I) / (L_eff^2). Verification / Alternative check: Compare with common cases: fixed–fixed (0.5L), fixed–pin (≈0.7L), pin–pin (L), fixed–free (2L). The given arrangement reasonably lies between pin–pin and cantilever, thus 1.5L is consistent. Why Other Options Are Wrong: L, 0.85L, 0.67L: correspond to stiffer end-restraint combinations (no sway or partial fixity). 2L: is for a cantilever (fixed–free) and is too large here. Common Pitfalls: Misreading “held in position” (translation restraint) vs “restrained in direction” (rotation restraint). Assuming the same K for non-sway and sway frames. Final Answer: 1.5 L

Question 11

Structural steelwork – common use of rolled Tee (T) sections in plate and tank construction In steel fabrication practice, rolled Tee-sections are commonly used:

Accepted Answer

Introduction / Context: Rolled Tees are versatile secondary members. In tank and plate construction, they are convenient for connecting and stiffening plates along seams and corners, often paired with flats for bolting or welding. Given Data / Assumptions: Fabrication of rectangular steel tanks and platework. Need for edge stiffeners and attachment members. Concept / Approach: Tee-sections provide flange and stem surfaces ideal for attaching plates and distributing local loads. While Tees can be part of other members, their frequent use as stiffeners and connectors in plate assemblies is a well-established fabrication practice. Step-by-Step Solution: Identify the context (tank/plate construction) → Tees used with flats to connect/stiffen plates.Eliminate options implying primary heavy columns or specialized tension-only roles that are not typical Tee applications. Verification / Alternative check: Shop drawings for oil/water tanks often show T-stiffeners along walls and roofs, combined with flats and angles for plate connections. Why Other Options Are Wrong: (a) Columns prefer doubly symmetric sections (e.g., I-sections) for buckling resistance. (c) Tees are not limited to tension-only applications. (d) Cable hangers typically use round bars/ropes or tailored fittings, not standard Tees. (e) Incorrect because a common correct usage exists. Common Pitfalls: Assuming any rolled section is equally suited to columns; section class and symmetry matter for stability. Final Answer: With flat strips to connect plates in steel rectangular tanks (as stiffeners/attachments)

Question 12

Riveted joint detailing – minimum pitch requirement relative to rivet diameter For a riveted joint with rivet-hole diameter d, the minimum pitch (center-to-center spacing of holes) should not be less than:

Accepted Answer

Introduction / Context: Pitch limits in riveted (and bolted) joints ensure plate integrity, prevent tear-out, and allow proper load transfer between fasteners. Codes specify minimum spacing as multiples of fastener diameter. Given Data / Assumptions: Traditional structural steel riveted joint. Minimum pitch must avoid hole coalescence and splitting while allowing tool access. Concept / Approach: Empirical and code-based rules of thumb set the minimum pitch at about 2.5 times the hole diameter for rivets, balancing plate net-section strength and practical fabrication. Closer spacing risks weakening the plate due to reduced ligament width. Step-by-Step Solution: Compare options with standard detailing rules → select 2.5 d.Recognize that larger pitches may be used for other considerations (seal, load distribution), but 2.5 d is the minimum threshold. Verification / Alternative check: Classic steel design handbooks and bridge specifications cite 2.5 d as the lower bound for rivet pitch in most configurations. Why Other Options Are Wrong: d to 2 d provide inadequate ligament and risk tearing. 1.25 d and 1.5 d are far below accepted practice. Common Pitfalls: Confusing minimum pitch with edge distance requirements; both must be satisfied independently. Final Answer: 2.5 d

Question 13

Indian Standards context for steel design: Which Indian Standard (IS) code specifically deals with the design and construction of steel structures used in buildings and allied works?

Accepted Answer

Introduction / Context: Structural design in India is governed by a suite of Indian Standards that separately address materials, loading, and special actions such as seismic effects. Knowing which code applies to steel is foundational for safe and compliant design. Given Data / Assumptions: We are selecting the Indian Standard that provides the core provisions for structural steel design. Other listed standards may be related to loads, concrete, or seismic actions, but are not the principal steel design code. Concept / Approach: IS 800 is the primary code for general construction in steel. It prescribes design philosophy, material properties, member design, connection design, and detailing requirements. IS 875 specifies design loads (dead, imposed, wind, etc.), which are then used in conjunction with IS 800 for steel buildings. IS 456 and IS 1343 pertain to concrete. IS 1893 covers seismic actions and is applied alongside material codes. Step-by-Step Solution: Identify the need: a code devoted to steel structural design.List candidates: IS 875 (loads), IS 800 (steel), IS 456 (RCC), IS 1893 (seismic), IS 1343 (PSC).Select IS 800 as the correct steel code. Verification / Alternative check: Typical design workflow: obtain loads from IS 875, seismic forces from IS 1893, then design and detail steel members and connections per IS 800. Why Other Options Are Wrong: IS 875 and IS 1893 provide actions, not steel member design; IS 456 and IS 1343 are for concrete systems. Common Pitfalls: Confusing load codes with material design codes; mixing concrete and steel provisions. Final Answer: IS 800 (General Construction in Steel)

Question 14

Working stress approach (legacy practice): What is the typical permissible bearing stress in steel (for bearing of fasteners on connected steel) used in traditional working-stress design problems?

Accepted Answer

Introduction / Context: Before the widespread adoption of limit-state design, many exam problems and legacy references used working-stress values for riveted or bolted connections in steel. One such property is the permissible bearing stress of steel under fastener bearing. Given Data / Assumptions: Old working-stress method context for steel connections. Permissible bearing stress refers to bearing of rivet/bolt on the connected plate (or vice versa) in mild steel. Concept / Approach: Permissible bearing stress historically adopted in many Indian university and competitive exam problems is approximately 1890 kg/cm2 for standard steel quality and typical shop-driven rivets/bolts. These values ensured adequate safety under service loads, accounting for uncertainties of materials and construction. Step-by-Step Solution: Recognize the question refers to working-stress (permissible stress) era, not limit-state.Recall the standard tabulated value for bearing stress ~ 1890 kg/cm2.Select the nearest exact choice from the options. Verification / Alternative check: Compare with shear/stress tables used alongside permissible tensile and shear stresses; bearing typically exceeded shear and was matched to plate thickness constraints to avoid tear-out and block shear. Why Other Options Are Wrong: 1500 kg/cm2 is on the low side for bearing; 2025–2340 kg/cm2 are higher than typical legacy permissible values; 2250 kg/cm2 appears in some contexts but not as the common teaching value for basic mild-steel bearing in standard problems. Common Pitfalls: Mixing up limit-state design strengths with permissible stresses; misapplying bearing on bolt vs. plate. Final Answer: 1890 kg/cm2

Question 15

Riveted connections terminology: In steelwork practice, the “gross diameter” of a rivet refers to the diameter of which item?

Accepted Answer

Introduction / Context: Precise terminology avoids mistakes in connection design and checking. In design of riveted joints, engineers distinguish between rivet diameter, hole diameter, gross and net areas of connected plates. Given Data / Assumptions: “Gross diameter of a rivet” is asked, not “gross diameter of the hole”. Nominal sizes are the basis for rivet designation before installation. Concept / Approach: Gross diameter of a rivet is the nominal shank diameter of the rivet before it is driven. The rivet hole is usually slightly larger (clearance allowance) than the rivet diameter. After driving, the shank upsets and fills the hole, but design checks still rely on standard nominal definitions. Step-by-Step Solution: Identify what “gross diameter of rivet” means in practice.Separate it from “hole diameter,” which is larger to permit insertion.Choose the cold, pre-driven shank diameter as the intended definition. Verification / Alternative check: Connection design tables: rivet shear and bearing capacities are tabulated against nominal rivet diameters, while plate net area uses hole diameter when checking tearing. Why Other Options Are Wrong: Driven rivet diameter is not a standard reference; the hole diameter corresponds to another definition; heated state is transient and not used for nominal sizing. Common Pitfalls: Using hole diameter in shear capacity of rivet; using rivet diameter instead of hole diameter when computing member net area. Final Answer: The cold rivet shank before driving (nominal rivet size)

Question 16

Strength of a riveted lap joint (working-stress method): The safe (governing) strength of a riveted lap joint is taken as which of the following?

Accepted Answer

Introduction / Context: Riveted joints can fail in multiple modes. A safe design recognizes the weakest (governing) mode under the applied load and limits the joint capacity accordingly. Given Data / Assumptions: Lap joint with standard rivets and plates. Working-stress design framework. Concept / Approach: The joint fails if any of its critical capacities is exceeded. Three primary capacities are checked: (1) shear of rivets, (2) bearing of rivet/plate, and (3) tearing of the plate across the net section. The safe joint strength is the smallest (least) of these capacities because that mode will govern first. Step-by-Step Solution: Compute rivet shear capacity based on number of shear planes.Compute bearing capacity using permissible bearing stress and plate (or rivet) thickness.Compute plate net-section capacity using net width and permissible tensile stress.Adopt the least value as the joint strength. Verification / Alternative check: Compare efficiencies of the joint (joint strength divided by member strength) to ensure acceptable performance. Why Other Options Are Wrong: Selecting any single mode (shear or bearing or tearing) ignores other potential lower capacities; picking the greatest would be unsafe. Common Pitfalls: Using gross rather than net section in tearing checks; forgetting double shear in cover-plate joints; miscounting effective rivets. Final Answer: Least of (a), (b), and (c)

Question 17

“Simple design” (simple construction) assumption for steel frames: Which statement best reflects the principal assumption used in the simple method of steel framework design?

Accepted Answer

Introduction / Context: Structural steel frames can be analyzed and designed under different connection assumptions: simple (pinned), semi-rigid, or continuous (rigid). The “simple design” method is common in building frames with shear connections. Given Data / Assumptions: The design method assumes negligible moment transfer at beam–column connections. Beams carry gravity loads primarily by bending as simply supported members. Columns are designed for axial load with allowances for minimum eccentricity. Concept / Approach: In simple construction, connections are detailed to transmit shear and axial force but not significant end moment, so the global analysis treats beams as simply supported. However, design still checks columns for code-specified minimum eccentricities and second-order effects as required. Hence, all three statements are consistent with simple design practice. Step-by-Step Solution: Identify connection idealization: pinned (no significant moment).Model beams as simply supported for gravity analysis.Design compression members considering eccentricities and stability. Verification / Alternative check: Connection details such as fin plates and flexible end plates exemplify simple shear connections; moment frames would contradict the “simple” assumption. Why Other Options Are Wrong: Choosing only one or two statements would omit integral parts of the simple design philosophy. Common Pitfalls: Assuming zero moment entirely and ignoring frame stability bracing; neglecting minimum eccentricity provisions for columns. Final Answer: All the above

Question 18

Grillage footing for two unequal column loads: For a steel grillage footing supporting two unequal column reactions, which of the following statements are followed to achieve uniform pressure and practical detailing?

Accepted Answer

Introduction / Context: Grillage footings distribute heavy column loads to soil using tiers of steel beams and base plates. When adjacent columns have unequal reactions, the footing layout must ensure uniform soil pressure and manageable bending in the grillage beams. Given Data / Assumptions: Two column loads of different magnitudes act on a common grillage base. Objective: maintain uniform soil pressure and efficient beam design. Concept / Approach: The resultant of column loads should pass through the centroid of the base area to avoid net overturning or differential soil pressure. A trapezoidal footprint is a practical response to unequal loads, aligning area with reaction demand. Lower-tier beam projections (cantilevers beyond the column faces) are adjusted to balance bending under each column and between them. Step-by-Step Solution: Compute resultant R of the two column loads and locate it.Size the base such that centroid coincides with line of action of R.Adopt trapezoidal plan area to match pressure distribution needs.Select beam sizes/projections so moments under columns are compatible and economical. Verification / Alternative check: Check soil pressure uniformity (q = R / A) and verify no uplift; ensure beam bending and shear capacities are adequate with serviceability checks. Why Other Options Are Wrong: Individually, each statement addresses only part of the solution; the combined approach is required, hence “All the above”. Common Pitfalls: Ignoring the location of the resultant; using a rectangular base without checking pressure eccentricity; neglecting differential bending under unequal loads. Final Answer: All the above

Question 19

Contact pressure concept: When a load is transferred through one surface to another surface in direct contact (for example, a pin against a plate), the relevant stress is called what?

Accepted Answer

Introduction / Context: Mechanical and structural joints often transmit force through surface contact—pins in lugs, bolts against plates, or rivets in holes. The design must ensure that local surface pressure does not crush the material. Given Data / Assumptions: Load is transferred via surfaces in contact (no significant tension across the interface). The stress of interest is localized surface pressure. Concept / Approach: The appropriate term for stress due to contact pressure is bearing stress. While it is compressive in nature, calling it simply “compressive stress” is too general and misses the contact-specific meaning. “Working stress” is a design-level stress, not a type, and “shearing stress” acts parallel to area, not normal contact pressure. Step-by-Step Solution: Identify the load path: surface-to-surface contact (e.g., bolt shank against hole wall).Define the stress as force divided by the projected bearing area (often t * d for plate thickness t and fastener diameter d).Name the stress: bearing stress. Verification / Alternative check: Design checks for connections compare bearing stress to permissible/limit values and also verify tear-out and net-section tension, confirming that “bearing” is the intended limit state. Why Other Options Are Wrong: Tension and shear are different stress modes; “working stress” is not a stress type; generic compression ignores the contact/area definition used in joint design. Common Pitfalls: Using nominal fastener area instead of projected bearing area; overlooking edge distance and end distance limits to prevent tear-out. Final Answer: Bearing stress

Question 20

Compression members using angle sections: A compression member built from steel angle sections can practically be configured as which of the following?

Accepted Answer

Introduction / Context: Angle sections are versatile for compression members in trusses, bracing, and light columns. Their use spans single-angle struts to double-angle and continuous built-up members. Given Data / Assumptions: Structural steel angles may be used singly or paired. End connections and effective lengths vary with application. Concept / Approach: Single-angle struts are common in roof trusses; double-angle struts improve symmetry and reduce eccentricity. Where angles are continuous in frames (e.g., as columns or posts), they become continuous members. All such configurations are recognized in design practice with corresponding slenderness, effective length, and connection detailing rules. Step-by-Step Solution: List practical configurations: single-angle, double-angle, continuous.Check end conditions and connection eccentricity (gusset connection often eccentric for single angles).Conclude that all listed configurations are feasible and used. Verification / Alternative check: Design guides and codes include capacity formulas and effective length factors for different angle arrangements and restraint conditions. Why Other Options Are Wrong: Excluding any configuration would be incorrect because all are standard in practice. Common Pitfalls: Ignoring eccentricity in single-angle struts; inadequate lacing/tying in built-up columns; overlooking local buckling. Final Answer: All the above

Steel Structure Design Questions