Question 1

Plate girder web slenderness control: the maximum permissible clear panel dimension of a web of thickness t (between stiffeners) is limited to what multiple of t?

Accepted Answer

Introduction / Context:Thin webs in plate girders must be stiffened at intervals to prevent shear buckling. Traditional exam problems reference legacy limits by relating panel clear dimension directly to web thickness t.Given Data / Assumptions:Web thickness = t.Clear panel dimension measured between adjacent transverse stiffeners.Mild-steel plate girder under shear.Concept / Approach:When the web panel becomes too slender, elastic shear buckling occurs. A prescriptive limit ties the maximum clear dimension to a multiple of t. A commonly cited limit in older practice for mild steel is approximately 270 t, beyond which transverse stiffeners must be provided or spacing reduced.Step-by-Step Solution:Identify web thickness t.Set stiffener spacing such that clear panel dimension ≤ 270 * t.If design demands larger spacing, increase t or add longitudinal stiffeners and verify buckling resistance by calculation.Verification / Alternative check:Modern codes use rational shear buckling models with limits based on yield strength and panel aspect ratio; this legacy 270 t rule is a quick check aligning with exam-style questions.Why Other Options Are Wrong:Smaller multiples (180 t, 220 t, 230 t, 260 t) are more conservative than this traditional limit and not the typical maximum stated in such questions.Common Pitfalls:Confusing clear dimension along depth versus along length; ignoring end/bearing stiffener requirements; overlooking weld-induced distortions affecting slender panels.Final Answer:270 t

Question 2

Spacing rule for vertical (transverse) stiffeners in a plate girder: if d is the clear distance between flange angles (web depth), the spacing should be limited to what range?

Accepted Answer

Introduction / Context:Transverse stiffeners subdivide the plate girder web into panels and must be spaced to control shear buckling while avoiding unnecessarily close placement that complicates fabrication.Given Data / Assumptions:d = clear distance between flange angles (effective web depth).Standard plate girder proportions and mild steel.Concept / Approach:A common exam rule-of-thumb limits the maximum spacing to about 1.5 d to prevent excessive panel slenderness, while a minimum spacing of roughly 0.33 d avoids crowding and allows practical connections and access.Step-by-Step Solution:Compute 1.5 d: set as maximum spacing between vertical stiffeners.Compute 0.33 d: set as minimum practical spacing.Detail stiffeners at or within these bounds and verify web panel shear buckling capacity.Verification / Alternative check:Check web panel aspect ratio and use shear buckling equations for confirmation; adjust spacing or web thickness accordingly.Why Other Options Are Wrong:Smaller maxima like d may be overly restrictive; larger maxima like 2.0 d or 2.5 d risk slender panels; different minima may not reflect customary practice.Common Pitfalls:Measuring d as overall depth rather than clear depth; omitting end/bearing stiffeners at supports; failing to align stiffeners with load application points.Final Answer:Not greater than 1.5 d and not less than 0.33 d

Question 3

Building construction – lintels: For a masonry wall bearing on a lintel, the load above is assumed to disperse and act as a uniformly distributed load on the lintel when the height of wall above the lintel is up to what multiple of the effective spa…

Accepted Answer

Introduction / Context:Lintels support masonry over openings such as doors and windows. In civil engineering design, the actual load from the masonry above a lintel disperses into the span. For practical analysis and safe detailing, we often convert that dispersed load into an equivalent uniformly distributed load (UDL) when sufficient masonry height is available to spread the load evenly across the lintel span.Given Data / Assumptions:Solid masonry above the lintel without large openings immediately above.Standard load dispersion through masonry at roughly 45 degrees (or similar empirical spread).We seek the limiting height above which the load on the lintel may be idealized as UDL.Concept / Approach:As height of masonry above the lintel increases, arching and load spread reduce load concentration near the jambs. Traditional building practice and many exam standards adopt a rule: if the height of masonry above the lintel is at least about 1.25 times the effective span, the load acting on the lintel can reasonably be treated as a UDL for design purposes. Below this height, a more triangular or trapezoidal pressure distribution may be considered.Step-by-Step Solution:Identify the dispersion concept: load spreads through masonry to the lintel.Recall the commonly accepted limit for UDL idealization.Select the closest codal/empirical value: 1.25 × effective span.Verification / Alternative check:Many design handbooks use the 1.25 span height as a conservative threshold. If the height is very large compared to the span, the UDL idealization is even more accurate; if much less, a varying load should be assumed.Why Other Options Are Wrong:Equal to span or 1.50×, 2.0×, 2.5×: these values are either less consistent with common practice (span) or overly conservative and not the usual threshold used in preliminary lintel design.Common Pitfalls:Ignoring heavy concentrated loads like beams bearing above the lintel.Assuming UDL when height is very small; triangular loading would be more appropriate.Final Answer:1.25 times the effective span.

Question 4

Steel roof trusses: What is the maximum permissible slenderness ratio for a member that normally acts as a tie (primarily in tension) in a roof truss, considering occasional stress reversal under wind?

Accepted Answer

Introduction / Context:Slenderness ratio (effective length divided by least radius of gyration) limits help ensure that steel members do not buckle under compressive effects. In roof trusses, some members normally act as ties (in tension), but under wind or erection conditions they may experience limited compression, so design standards allow higher slenderness than for primary compression struts.Given Data / Assumptions:Member is a truss tie, primarily in tension under gravity loads.Occasional load cases (wind, temperature, erection) may cause minor compression.We seek the maximum permitted slenderness consistent with common practice and exam standards.Concept / Approach:Members always in compression typically have tighter limits (for example around 180). For ties in roof trusses, many traditional guidelines permit higher slenderness, recognizing that the member is usually in tension. A commonly cited maximum for “members normally acting as ties” is 350, which balances economy with stability under occasional reversal.Step-by-Step Solution:Classify the member: tie (tension) with possible minor compression.Recall typical slenderness allowance for such members: up to 350.Choose 350 as the correct value.Verification / Alternative check:Historic working-stress references and many exam-centric summaries list 350 for ties, while main compression members stay near 180. This differentiation reflects the lower buckling risk for predominantly tensioned elements.Why Other Options Are Wrong:180–300: typical for compression struts or more restrictive serviceability cases; unnecessarily low for ties.Common Pitfalls:Applying the 180 limit meant for compression columns to truss ties.Ignoring end conditions when computing effective length.Final Answer:350.

Question 5

Wind engineering basics: If P is the wind pressure (kg/cm²) and v is the wind speed (km/h), which proportional relationship correctly expresses the dependence of pressure on velocity, using a constant of proportionality K?

Accepted Answer

Introduction / Context:Wind load calculations start from the fundamental idea that dynamic pressure exerted by wind increases with the square of the wind speed. This principle underlies building codes and structural wind design for cladding, frames, towers, and roofs.Given Data / Assumptions:P is wind pressure in kg/cm² (any consistent pressure unit works).v is wind speed in km/h (or any consistent speed unit).K is a lumped constant accounting for air density and unit conversions.Concept / Approach:Dynamic pressure q is proportional to 0.5 * rho_air * V^2. When expressed in practical units, constants and conversion factors are absorbed into K. Thus, pressure varies as the square of speed: P ∝ v^2. This is why doubling wind speed roughly quadruples the pressure, a key design insight for extreme wind events.Step-by-Step Solution:Recognize physical relationship: P ∝ V^2.Introduce a proportionality constant K to accommodate units.Write P = K v^2.Verification / Alternative check:Checking dimensions confirms a constant K must convert velocity-squared into pressure. Empirical code equations are consistent with this quadratic dependence.Why Other Options Are Wrong:P = K/v^2 or P = K v: contradicted by fluid dynamics; do not reflect energy considerations.v = K/P^2, P = K √v: mathematically inconsistent with dynamic pressure theory.Common Pitfalls:Using linear relationships that underestimate loads at high wind speeds.Mixing units without adjusting K appropriately.Final Answer:P = K v^2.

Question 6

Riveted joints – efficiency: How is the efficiency of a riveted joint defined, relative to the original (unperforated) plate strength?

Accepted Answer

Introduction / Context:Riveted joints reduce the net area of plates due to holes and introduce additional failure modes. To compare a riveted joint to the uncut (solid) plate, designers use the term efficiency, indicating how much of the original plate strength is retained after riveting.Given Data / Assumptions:Riveted plate with one or more rows of rivets.Possible failure modes: plate tearing across holes, rivet shear, and bearing (crushing) of rivet/plate.Efficiency is a ratio, not a stress value.Concept / Approach:Compute joint strength for each failure mode and take the least of these as the critical joint strength. Divide this least joint strength by the strength of the solid plate (without holes) across the same width to obtain efficiency, typically expressed as a percentage. This provides a direct measure of capacity retained after introducing fastener holes.Step-by-Step Solution:Determine net-section plate tearing strength.Determine rivet shear strength.Determine bearing strength of rivets/plate.Choose the least of the above = joint strength.Efficiency = (least joint strength) / (solid plate strength).Verification / Alternative check:Across textbooks, this definition is uniform. The efficiency cannot exceed 100% because holes inevitably reduce the solid plate capacity in the joint region.Why Other Options Are Wrong:Greatest joint strength over solid plate (option b) overestimates efficiency.Ratios that invert numerator/denominator (option c) are not meaningful measures of retained strength.“All of the above” is incorrect because only one definition is valid.Common Pitfalls:Forgetting to use net width for plate tearing.Ignoring stagger effect on net area when rivets are zig-zag.Final Answer:Least strength of the riveted joint divided by the strength of the solid plate.

Question 7

Serviceability limits – deflection: For a simply supported steel beam in building work, what is a commonly adopted maximum permitted deflection under service load, expressed as a fraction of the span?

Accepted Answer

Introduction / Context:Serviceability criteria limit deflections to prevent cracking of finishes, ponding, and user discomfort. For simply supported beams in general building applications, historical working-stress practice specifies an allowable deflection as a fraction of span, depending on usage and finishes.Given Data / Assumptions:Simply supported steel beam in a typical building.Service load combination considered (not ultimate).No unusually sensitive finishes beyond normal plaster/ceilings.Concept / Approach:Commonly adopted limits are in the range L/240 to L/400 depending on occupancy and finishes. A widely cited exam-standard value for general steel floor beams is L/325. More stringent limits (e.g., L/360 or L/400) may be used for vibration control or brittle finishes; less stringent values may be acceptable where finishes are not critical.Step-by-Step Solution:Identify member and serviceability limit category.Recall typical rule-of-thumb for general buildings: L/325.Select 1/325 of the span as the permissible deflection.Verification / Alternative check:Many older codes and handbooks list values around L/325 for general building beams; local codes may specify different values for specific occupancies, but L/325 remains a standard exam reference figure.Why Other Options Are Wrong:1/350, 1/375, 1/400: more restrictive than the commonly quoted general limit; they may be used in special cases but are not the standard value sought here.“None of these” is incorrect because a valid value is provided.Common Pitfalls:Confusing live-load-only limits with total-load limits.Applying the same limit to cantilevers (typically more stringent, e.g., L/150).Final Answer:1/325 of the span.

Question 8

Terminology in steel structures: A structural member that carries tensile force along its longitudinal axis is commonly called what?

Accepted Answer

Introduction / Context:Steel frameworks contain elements that are primarily in tension or compression. Correct terminology helps in communication, specification, and code application. Members designed to resist axial tension are used in trusses, bracing systems, and hangers.Given Data / Assumptions:The member carries axial tensile force parallel to its length.No significant bending is intended in the basic definition.Standard structural engineering terminology is used.Concept / Approach:Common interchangeable terms include tie, tie member, and tension member. In truss nomenclature, “tie” denotes a member resisting tension. Design checks for such members focus on gross and net section strength, slenderness for serviceability, and connection detailing to avoid net-section fracture or block shear.Step-by-Step Solution:Identify the action: axial tension along the member.Match to accepted names: tie, tie member, tension member.Conclude that all listed terms correctly describe the member.Verification / Alternative check:Textbooks and codes use these terms almost synonymously in the context of trusses and bracing. Drawings often label long tensile elements as “ties”.Why Other Options Are Wrong:“None of these” is wrong because all listed terms are valid.Common Pitfalls:Ignoring net-area reductions at bolt/rivet holes when sizing tension members.Overlooking minimum thickness and slenderness limits even for members in tension.Final Answer:all the above.

Question 9

Steel design – naming the beam outside a wall up to the floor level above it In building construction terminology, what is the beam provided along the outside face of a wall, supporting the floor immediately above that wall called?

Accepted Answer

Introduction / Context:Correct terminology is essential in structural engineering drawings and site communication. The beam that runs along the outside face of a wall and supports the edge of a floor slab or floor beams above is commonly used in framed buildings and masonry-plus-frame construction. Knowing this term helps avoid confusion with roof members (rafters, purlins) and openings (lintels).Given Data / Assumptions:Member location: outside of a wall, at or just below the floor it supports.Function: supports the outer edge of the floor system and transfers loads to columns/walls.No numerical calculation is required; this is a definitions/terminology question.Concept / Approach:The beam at the perimeter of a building that supports the floor slab or secondary joists alongside an external wall is termed a spandrel beam. It picks up floor loads and façade loads, and often serves as the structural line for cladding supports. Do not confuse it with a lintel, which sits over wall openings, or roof elements like rafters and purlins which belong to roofing systems.Step-by-Step Solution:Identify the beam’s position: outside the wall and at the level supporting the floor above → a perimeter beam.Cross-check typical terms: rafter/purlin are roof elements; lintel is over an opening; “spandrel beam” matches perimeter floor-support function.Select the correct terminology: spandrel beam.Verification / Alternative check:Look at architectural-structural coordination: spandrel panels and façade systems often connect to spandrel beams at each floor line, confirming the placement and function implied by the question.Why Other Options Are Wrong:Rafter supports roof slopes, not floors. Purlin supports roof sheeting on rafters. Lintel spans over wall openings like doors/windows. Header beam is a generic term and does not specifically denote the outside wall-floor beam in multistorey buildings.Common Pitfalls:Assuming any beam near an opening or wall is a lintel; mixing roof and floor terminology; overlooking that “spandrel” also refers to the façade panel region between floor lines.Final Answer:Spandrel beam

Question 10

Web area definition in a rolled steel beam (I- or H-section) How is the gross cross-sectional area of the web of a steel beam defined for basic calculations?

Accepted Answer

Introduction / Context:In preliminary steel design and section property checks, we often separate the web and flange contributions. Knowing the correct expression for the gross web area helps in assessing shear capacity, weld sizing to web, and estimating self-weight and material usage accurately.Given Data / Assumptions:Rolled I- or H-section (beam) with a single web of uniform thickness over the clear depth between fillets.“Depth of the beam” refers to the overall section depth typically used for a gross area estimate.Flanges are considered separately from the web.Concept / Approach:The web is the vertical plate joining the two flanges. To a first approximation, its gross area is the overall depth multiplied by the nominal web thickness (neglecting minor fillet adjustments). This convention aligns with handbook calculations where the web area is treated independently from flange areas.Step-by-Step Solution:Identify web geometry → a rectangle of thickness t_w and depth approximately equal to the beam depth d.Compute area A_w ≈ d * t_w.Select the option that states “depth of beam × web thickness”.Verification / Alternative check:Standard steel tables partition total area into web area plus two flange areas, where the web portion is shown as d * t_w to a close approximation.Why Other Options Are Wrong:Flange width × web thickness ignores depth; sum with flange width double counts and is not an area; half depth × thickness underestimates; flange area minus web area is not a definition.Common Pitfalls:Confusing clear web depth (between fillets) with overall depth; mixing flange and web contributions; overlooking taper in some rolled sections (still negligible for basic estimates).Final Answer:Depth of the beam multiplied by its web thickness

Question 11

Shear resistance in rolled steel beams – which element takes most of the shear? In a typical rolled I- or H-beam under vertical loads, which part primarily resists the shear force?

Accepted Answer

Introduction / Context:Understanding how internal forces distribute within an I- or H-section helps in safe and economical design. While flanges carry most of the bending moment (tension and compression), the web is the main element resisting vertical shear in beams.Given Data / Assumptions:Standard rolled steel I- or H-beam under vertical gravity loading.Plane sections remain plane (elementary beam theory).No unusual torsion or complex load paths.Concept / Approach:From elementary beam theory, bending stress distribution is maximum at flanges and zero at the neutral axis. Shear stress distribution is parabolic with a maximum at the neutral axis, which lies in the web region. Since the web occupies most of the area near the neutral axis, it resists the bulk of the shear force, while flanges contribute little to vertical shear.Step-by-Step Solution:Identify the dominant shear stress region → around the neutral axis (web).Recall shear stress distribution in an I-section → nearly all vertical shear carried by the web.Hence, the primary shear-resisting component is the web.Verification / Alternative check:Design codes allow shear capacity checks using the web area (or effective web area) times an allowable shear stress, reflecting the web-dominant mechanism.Why Other Options Are Wrong:Flanges only is incorrect since flanges mainly resist bending. Equal measure contradicts the well-known distribution. Shear negligible is unsafe. Flange-web junction only is not a recognized design assumption.Common Pitfalls:Assuming large flange area also means large shear capacity; overlooking the parabolic shear stress profile that peaks in the web.Final Answer:Web only

Question 12

Classification of rolled steel beam sections as per Indian Standards According to Indian Standards practice, rolled steel beam sections are grouped into how many series?

Accepted Answer

Introduction / Context:Rolled steel beam sections in India have historically been categorized into standard series (for example, ISMB, ISJB, ISLB, ISWB in older practice). Recognizing the count of series helps engineers navigate steel tables and select appropriate sections during preliminary design.Given Data / Assumptions:Conventional/classical IS practice regarding rolled beam series.Question tests recognition of the standard grouping rather than exact nomenclature updates.Concept / Approach:The Indian Standards historically provided four principal series of hot-rolled beams differing by depth-to-weight characteristics: medium, junior, light, and wide flange (beam). Though modern editions have updated designations, the examination concept references the classic four series.Step-by-Step Solution:Recall IS beam families commonly listed in handbooks/tables.Identify the traditional grouping count → four series.Choose the option indicating four series.Verification / Alternative check:Cross-check old steel tables used academically: they present four beam series under Indian Standards, affirming the conventional answer.Why Other Options Are Wrong:Two or three understate the variety; five or six exceed the classical categorization and cause confusion with other rolled shapes (channels, angles).Common Pitfalls:Mixing current proprietary/market designations with classical IS series counts; confusing beams with channels or universal columns.Final Answer:Four series

Question 13

Definition of equivalent axial load for columns with eccentricity When a column is loaded eccentrically, the “equivalent axial load” is defined as the load that produces a stress equal to which reference stress?

Accepted Answer

Introduction / Context:Eccentric loading in columns induces combined axial compression and bending, creating a nonuniform stress distribution. To compare such a case with pure axial compression, designers sometimes use the concept of an “equivalent axial load”.Given Data / Assumptions:Eccentricity about one or more axes generates bending moments.Equivalent axial load is a conceptual tool for comparison or simplified checks.We are concerned with the peak compressive stress, which governs safety.Concept / Approach:Under an eccentric load, the extreme fibre sees the largest compressive stress (or compression plus tension if eccentricity is large). The “equivalent axial load” is defined so that, if it acted concentrically, it would produce the same maximum compressive stress as in the real eccentric case.Step-by-Step Solution:Compute eccentric case maximum stress: σ_max = P/A ± M/Z (largest compression).Define P_eq such that σ_max = P_eq / A.Therefore, P_eq corresponds to the maximum stress produced by the eccentric load.Verification / Alternative check:Equivalency is meaningful only if stresses match; using σ_max ensures conservative comparison for crushing or yield checks.Why Other Options Are Wrong:Most compressed fibre only is a location, not a definition; pure bending stress ignores axial part; average stress underestimates peak; shear stress is irrelevant for axial equivalence.Common Pitfalls:Misinterpreting “equivalent” as equal average stress; neglecting biaxial bending when eccentric in two directions.Final Answer:Maximum stress produced by the eccentric load

Question 14

Laced built-up columns – transverse shear to be resisted by lacing bars (IS practice) According to Indian Standard working-stress design practice, lacing bars in a built-up column are designed to resist transverse shear equal to what percentage of t…

Accepted Answer

Introduction / Context:Laced built-up columns use lacing systems to tie individual components so they act as a single unit. The lacing not only keeps components in position but also resists a share of transverse shear induced by load eccentricities, initial imperfections, and secondary effects.Given Data / Assumptions:Working-stress design style (traditional IS practice).Lacing bars are arranged in a single lacing system (typical exam assumption).We seek the code-recommended percentage of axial load to be taken as transverse shear for lacing design.Concept / Approach:Codes provide a simplified shear allowance that lacing must carry to ensure integrity against secondary shear effects. Traditional IS provisions take this nominal shear as a fixed small percentage of the member axial force, commonly 2.5% for single lacing (with slightly different values sometimes adopted for double lacing or battened systems).Step-by-Step Solution:Identify column type → laced built-up member.Apply traditional IS rule-of-thumb → V_lacing = 0.025 * P_axial (2.5%).Design lacing for this shear while checking slenderness of lacing bars and connection details.Verification / Alternative check:Classic Indian steel design texts and older IS 800 guidance list 2.5% as the nominal shear for single lacing systems, corroborating the value.Why Other Options Are Wrong:1.0% and 2.0% may underdesign; 3.0% or 4.0% are conservative beyond typical code guidance for single lacing, potentially uneconomical without justification.Common Pitfalls:Forgetting to check slenderness of lacing bars; omitting additional checks for double lacing or batten plates where recommended shear allowances differ; ignoring out-of-plane buckling of laced members.Final Answer:2.5% of the axial load

Question 15

Riveted connections (IS practice) – permissible shear: According to Indian Standards for structural steelwork, what is the recommended working shear stress on the gross (shank) area of a rivet?

Accepted Answer

Introduction / Context:In steel design using riveted connections (still encountered in assessment of existing bridges and buildings), permissible or working stresses are used to size rivets and to verify legacy details. One frequently tested value is the allowable shear stress on the gross shank area of a rivet as per traditional Indian Standards practice.Given Data / Assumptions:Structural steel rivets in shear.Working stress (allowable stress) design context.Shear is evaluated on the gross shank area (not reduced for threads since rivets are headed shanks).Concept / Approach:Codes specify different allowable stresses for rivets in shear and in bearing. Typical legacy values used in exam problems are about 1025 kg/cm2 for rivet shear (on gross area) and a much higher value for bearing (crushing) on connected plates or rivets. Recognizing which value corresponds to which limit state is essential to choosing the correct option.Step-by-Step Solution:Identify the limit state asked: shear on gross rivet area.Recall standard IS-era values: shear allowance ≈ 1025 kg/cm2 (shop-riveted), distinctly lower than bearing allowance.Select the option that matches this commonly adopted working shear stress: 1025 kg/cm2.Verification / Alternative check:Many solved examples in classic Indian steel design texts adopt about 1000–1025 kg/cm2 for rivet shear, while bearing stresses are commonly taken around 2360 kg/cm2. The presence of both numbers among the options is a cue to avoid mixing up shear and bearing values.Why Other Options Are Wrong:785 kg/cm2: too low for the standard working shear value; may apply to special cases or field constraints.2360 kg/cm2: this aligns with bearing (crushing) allowance, not shear.None of these: incorrect because 1025 kg/cm2 is the expected answer in this setting.Common Pitfalls:Confusing allowable shear with allowable bearing.Using bolt thread-area logic for rivets (rivets are not threaded in the shear plane).Final Answer:1025 kg/cm2

Question 16

Combined footing (two equal column loads): For an economical design of the longitudinal combined footing beam (supporting two equal column loads), the end projections are proportioned so that the bending moment under each column equals which referen…

Accepted Answer

Introduction / Context:Combined footings for two columns are designed to achieve uniform soil pressure and economic use of concrete and steel. When the two column loads are equal and the footing is rectangular, the longitudinal strip acts like a beam on an elastic foundation. A practical proportioning guideline is to balance the bending moment under the columns with the maximum span moment to reduce peak steel demand and achieve a more uniform reinforcement layout.Given Data / Assumptions:Two equal column loads supported by a single rectangular combined footing.Uniform allowable soil pressure targeted through appropriate footing dimensions.Longitudinal action idealized as a beam with overhangs (end projections).Concept / Approach:With equal column loads and uniform soil pressure, the shear and moment diagrams can be shaped by adjusting the end projections (cantilevers) beyond the outer columns. An economical proportion commonly adopted is to make the sagging bending moment at midspan essentially equal to the hogging (negative) moment at the column faces. This balances steel requirements and minimizes peak reinforcement disparity.Step-by-Step Solution:Assume uniform soil pressure under the footing area.Choose overhang lengths so that the resultant soil reaction under the overhangs counterbalances outer-span shears.Iterate until the hogging moment under each column approximately equals the largest sagging moment at midspan.This yields an efficient and symmetrical bending profile and reinforcement layout.Verification / Alternative check:Trial proportioning followed by soil-pressure and section-capacity checks confirms that reinforcement near the columns and at midspan can be kept similar, reducing congestion at the columns and avoiding unnecessarily deep sections at the span centre.Why Other Options Are Wrong:Half the midspan moment: would leave the column zones under-reinforced compared to midspan.Twice the midspan moment: overconservative at columns, wasting material.None of these: not correct because matching column-face and midspan moments is a well-known economical criterion.Common Pitfalls:Proportioning only by geometry without checking uniform soil pressure and resultant location.Ignoring shear near columns when maximizing overhangs.Final Answer:the bending moment at the centre (midspan) of the combined beam

Question 17

Slenderness limit (compression members): What is the maximum permissible slenderness ratio for compression members that carry dead load and imposed (superimposed) load according to standard steel design practice?

Accepted Answer

Introduction / Context:Slenderness ratio limits are prescribed to ensure compression members have adequate stiffness against buckling and to limit serviceability issues. For columns carrying sustained gravity loads (dead plus imposed), the most widely cited limit in traditional Indian practice is 180, frequently examined in design quizzes.Given Data / Assumptions:Member is a compression element of a building frame.Loads include dead load plus superimposed (imposed) load.Slenderness ratio is defined as effective length divided by least radius of gyration.Concept / Approach:To prevent excessive elastic buckling and secondary effects, codes cap the slenderness ratio for different functions: bracing, compression with stress reversal, tension members, and compression under permanent loads. For gravity-carrying columns, the limit often adopted in problems is 180.Step-by-Step Solution:Identify the functional class: compression member carrying dead and imposed load.Recall the conventional limit: slenderness ratio ≤ 180.Select the correct option accordingly.Verification / Alternative check:Design handbooks tabulate different limits (e.g., 180, 250, 350) for different roles. The 180 value aligns with members in sustained compression rather than bracing or tension-only components.Why Other Options Are Wrong:250, 350: used for bracing or members not primarily in sustained compressive stress.80 or 75: too restrictive for ordinary columns; usually for special stiff elements or serviceability-critical parts.Common Pitfalls:Using the wrong effective length because of misidentified end restraints, which changes the computed slenderness.Forgetting to use the least radius of gyration (weak axis) in the ratio.Final Answer:180

Question 18

Thin cylinder hoop stress – water main thickness: A water main of internal diameter 80 cm carries a pressure of 10 kg/cm2. If the permissible circumferential (hoop) stress in the metal is limited to 200 kg/cm2, what is the minimum required thickness…

Accepted Answer

Introduction / Context:Water mains are commonly designed using thin-cylinder theory for hoop stress. The wall thickness must be sufficient so that the circumferential stress due to internal pressure does not exceed the allowable stress in the metal. This is a direct substitution problem widely used in exams.Given Data / Assumptions:Internal pressure p = 10 kg/cm2.Internal diameter D = 80 cm.Allowable hoop stress σ_allow = 200 kg/cm2.Thin cylinder approximation; neglect joint efficiency and corrosion allowance for this calculation step.Concept / Approach:For a thin cylinder, hoop stress σ_h = p * D / (2 * t). Solving for thickness t gives t = p * D / (2 * σ_allow). This is adequate for preliminary sizing; refined formulas include a small addition of pressure in the denominator, but the thin-wall formula is standard in objective problems.Step-by-Step Solution:Write: t = p * D / (2 * σ_allow).Insert values: t = 10 * 80 / (2 * 200).Compute numerator: 10 * 80 = 800.Compute denominator: 2 * 200 = 400.Thus t = 800 / 400 = 2.0 cm.Verification / Alternative check:A refined expression t = p * D / (2 * σ_allow + p) gives 800 / (400 + 10) ≈ 1.95 cm. Rounding and provision for allowances would still make a 2.0 cm plate the practical choice among the options.Why Other Options Are Wrong:1.0 or 1.5 cm: would over-stress the metal above the 200 kg/cm2 limit.2.5 or 3.0 cm: overly conservative for the stated limits; increase cost and weight without need.Common Pitfalls:Using radius instead of diameter in the thin-cylinder formula.Mixing units (e.g., MPa versus kg/cm2).Final Answer:2.0 cm

Question 19

Butt joints – double-cover types: Which of the following are main types of double-cover butt joints used in riveted steel construction?

Accepted Answer

Introduction / Context:Riveted butt joints connect two plates end-to-end using one or two cover plates. Double-cover butt joints are popular because they produce near-symmetric force transfer. The arrangement of rivets (chain or zig-zag) and the governing limit state (shear vs bearing) define common types referenced in exams.Given Data / Assumptions:End-to-end plate connection using double cover plates.Rivets used as shear connectors and bearing elements.Terminology: chain vs zig-zag patterns relate to rivet pitch and transverse staggering.Concept / Approach:A double-cover joint has two cover plates (one on each side of the main plates). The rivet patterns may be chain (aligned) or zig-zag (staggered), and design checks include rivet shear, plate/rivet bearing, net-section tension, and block shear. All listed types are recognized variants of double-cover butt joints.Step-by-Step Solution:Identify the joint as a butt joint with two cover plates.Recognize that both chain and zig-zag rivet patterns are used.Note that rivets are primarily designed for shear and bearing; hence “shear-riveted” is a valid descriptive type.Therefore, all listed options are applicable.Verification / Alternative check:Standard steelwork texts illustrate double-cover joints with either chain or zig-zag patterns; both are structurally acceptable when pitch, edge distances, and net-section checks are satisfied.Why Other Options Are Wrong:Each individual option is valid; thus the comprehensive correct choice is “all the above.”Common Pitfalls:Overlooking net-section reduction due to stagger and multiple rows.Ignoring block shear paths in end connections with short end distances.Final Answer:all the above

Question 20

Channel section – outstanding (projection) length in compression: For a compression member made from a single channel, the outstanding length is measured how?

Accepted Answer

Introduction / Context:In built-up compression members using channels, the free (outstanding) leg is prone to local buckling. Codes limit the projection-to-thickness ratio, so one must measure the outstanding length correctly to check local buckling criteria and detail transverse connections.Given Data / Assumptions:Member uses a single channel or channels with one leg outstanding.Fasteners are placed along a line connecting the channel web/flange to a gusset or companion member.Outstanding length is required for outstand slenderness checks.Concept / Approach:The outstanding width for local buckling is the free projection measured from the free edge up to the first line of restraint (i.e., the first row of fasteners). This dimension is used in the width-to-thickness (b/t) check of the outstanding element to ensure it remains within the “compact” or “semi-compact” limits, depending on the code basis.Step-by-Step Solution:Locate the free edge of the channel flange (outstand).Identify the first line of restraint provided by rivets/bolts or welds.Measure the perpendicular distance between these two: that is the outstanding length.Verification / Alternative check:Design examples consistently define outstand b as the free projection to the first line of connection. This aligns with local buckling formulas using b/t ratios for plate elements.Why Other Options Are Wrong:Half or full nominal width: do not necessarily correspond to the restrained line; connections may not be centered.None of these: incorrect because a standard definition exists.Common Pitfalls:Measuring to the web centreline instead of the first fastener line.Ignoring the effect of fillets or returns; measure the flat projection.Final Answer:from the free edge of the outstanding leg to the first row of rivets/bolts connecting it to the gusset or companion member

Steel Structure Design Questions

Questions

Practice smarter