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Correct ANSWER is (a) ≤ 200εw

The best explanation: For serviceability CRITERIA, when only transverse stiffeners are provided, the depth-to-thickness ratio of web should be ≤ 200 εw, when 3d ≥ c ≥d and for c > 3d, the web is considered as unstiffened, where c is clear distance between stiffeners and d is depth of web, εw = √(250/fyw), fyw is yield stress of web.

The correct answer is (C) ≤90ε

Best explanation: When transverse stiffeners are not provided, the depth-to-thickness ratio of WEB connected to FLANGES ALONG one longitudinal edge only is ≤ 90ε and when web connected to flanges along both longitudinal edges is ≤ 200ε to meet the serviceability criteria.

The correct ANSWER is (a) 0.4kN/m^2

The explanation is: The live LOAD for roof truss should not be LESS than 0.4kN/m^2. For roof slopes ≤ 10^o and access provided, the live load to be TAKEN is 1.5kN/m^2 of PLAN area. For roof slopes > 10^o and access is not provided , the live load to be taken is 0.75kN/m^2 of plan area.

Correct answer is (c) DEAD load + earthquake load

The best I can explain: Earthquake loads are not significant for roof trusses because of the SMALL self weights. The following load combinations can be CONSIDERED : (i) Dead load + snow load, (ii) Dead load + partial/full live load, (iii) dead load + live load + internal POSITIVE air pressure, (iv) dead load + live load + internal SUCTION air pressure, (v) dead load + live load + wind load.

Right OPTION is (d) cost of TRUSSES should be equal to twice the cost of PURLINS PLUS cost of roof coverings

The best explanation: For economic spacing of roof trusses, the cost of trusses should be equal to twice the cost of purlins plus cost of roof coverings. This equation is used for checking the spacing of trusses and not for design of trusses.

Correct option is (b) cost of purlins and cost of roof covering

For EXPLANATION: The economic spacing of the truss is the spacing that makes the overall cost of trusses, purlins, roof coverings, columns, etc. the minimum. It depends upon the relative cost of trusses, purlins, roof coverings, spacing of columns, etc. If the spacing is large, the cost of these trusses PER UNIT area decreases but the cost of purlin INCREASES. But if the spacing of trusses is SMALL, the cost of trusses per unit area increases. Roof coverings cost more if the spacing of trusses is large.

Right option is (a) L/500

For explanation I would say: The vertical deflection of gantry girder where the cranes are manually operated should not exceed L/500, where L is the SPAN of gantry girder. The maximum vertical deflection allowed for a gantry girder where the cranes are travelling overhead and operated ELECTRICALLY upto 500kN is L/750 and operated electrically over 500kN is L/1000. When gantry GIRDERS carry moving LOADS such as charging cars, the deflection should not exceed L/600.

Correct answer is (b) Nsc < 5×10^6 [(27/γmft)/γmt]^3

The explanation: Fatigue effect for light and medium DUTY cranes need not be checked if normal and shear DESIGN STRESS ranges F ≤ (27/γmft) or if actual NUMBER of stress cycles, Nsc < 5×10^6 [(27/γmft)/γmt]^3, where f = actual fatigue stress range, γmft = partial safety factor for strength, γmf = partial safety factor for material = 1.10.

Correct choice is (b) crane crab is closest to GANTRY girder

Explanation: The maximum wheel load is obtained when crane crab is closest to gantry girder. The crab in such position on the crane girder GIVES maximum reaction on the gantry girder. The VERTICAL reaction of crane girder is transferred through its TWO wheels on to the gantry girder. Therefore, the maximum wheel load is half of this reaction. This maximum wheel load is then increased for impact and used for design of gantry girder.

Correct answer is (a) 10% of weight of crab and weight lifted on the crane

Best EXPLANATION: For gantry girders CARRYING electrically OPERATED overhead travelling cranes, the lateral forces are increased by 10% of weight of crab for impact allowance and weight lifted on the crane. The vertical forces can be increased by 25% of MAXIMUM static wheel LOAD.

The correct option is (C) 10% of weight of crab and weight lifted on the crane

Easy EXPLANATION: For gantry GIRDERS carrying HAND operated cranes, the VERTICAL forces are increased by 10% of maximum static wheel loadfor impact allowance. The lateral forces can be increased by 5% of weight of crab and weight lifted on the crane.

Right answer is (c) W1 = [Wt(Lc-L1)]/(2Lc)

The explanation is: Since trolley moves on the crane girder along the span of truss, its weight is transferred to the crane WHEELS as the AXLE load and finally to gantry girder. The wheel load transferred from trolley to gantry girder is given by W1 = [Wt(Lc-L1)]/(2Lc), where W1 is load of each wheel on gantry girder, Wt is weight of trolley or crab CAR, Lc is DISTANCE between gantry giders, L1 is distance between centre of GRAVITY of trolley and gantry.

The correct option is (d) wind loads

To explain I would say: The loads considered in the design of gantry girders are vertical loads or GRAVITY loads, LONGITUDINAL loads, lateral loads and IMPACT loads. The vertical force is the reaction from crane girder, acting vertically downward. The longitudinal thrust is DUE to starting and stopping of crane acting in longitudinal direction. The lateral thrust is due to starting and stopping of the crab acting horizontally normal to the gantry girder.

The correct option is (b) It is subjected to impact load

Explanation: Gantry girder are different from beams in buildings. It is GENERALLY laterally unsupported except at the COLUMNS. It is subjected to impact load. It must be analysed for unsymmetrical bending because of LATERAL THRUST from the starting and stopping of the crab. It is subjected to longitudinal load due to starting and stopping of crane BRIDGE itself. They are always simply supported.

Correct answer is (c) Is ≥ 0.34αsD^3Tcf

Explanation: When bearing stiffeners are required to provide torsional restraint at the support of the beams, the second moment of area of the stiffener section about center line of the WEB should be such that Is ≥ 0.34αsD^3Tcf , where αs = 0.006 for LLT/ry ≤50, 0.3/( LLT/ry) for 50< LLT/ry≤100, 30/( LLT/ry)^2 for LLT/ry ≥100, D = overall depth of beam at the support, TCF = MAXIMUM thickness of compression flange in the span under consideration, KL = LATERALLY unsupported effective length of compression flange of beam, ry = RADIUS of gyration of the beam about minor axis.

The correct option is (B) tw^2/5bs

Best EXPLANATION: Intermediate TRANSVERSE stiffeners not subjected to external loading should be connected to the web so as to withstand a shear between each component of the stiffener and the web not less than tw^2/5bs, where tw is thickness of web, bs is OUTSTAND WIDTH of stiffener.

The correct option is (d) first HORIZONTAL stiffener is provide at neutral axis

Explanation: Longitudinal STIFFENERS are also called horizontal stiffeners. They increase buckling resistance considerably as compared to transverse stiffeners when the web is subjected to buckling. They consist of angle section for riveted/bolted plate girder and PLANE section for WELDED plate girder and are provided in the compression zone of the web. The first horizontal stiffener is provide at one-fifth of distance from compression flange to tension flange. If required ANOTHER stiffener is provided at the neutral axis.

Right OPTION is (a) [(Fq-Fs)/Fqd]+(Fs/Fsd)+(Mq/Myq) < 1

Explanation: Stiffeners subjected to EXTERNAL loads and moments should meet the conditions of load carrying stiffeners. In addition, they should satisfy the following interaction equation: [(Fq-Fs)/Fqd]+(Fs/Fsd)+(Mq/Myq) < 1, where Fq is stiffener force, Fqd is design resistance of INTERMEDIATE stiffeners CORRESPONDING to buckling about axis parallel to the web, Fs is external load or reaction at stiffener, Fsd is design resistance of load carrying stiffener corresponding to buckling about axis parallel to the web, Md is MOMENT on the stiffener due to eccentrically applied load and transverse load, Myq is yield moment capacity of stiffener based on its elastic modulus about its centroidal axis parallel to the web.

Right option is (b) Is ≥ 0.75dtw^2

To explain: TRANSVERSE stiffeners not subjected to external loads or MOMENTS should have second MOMENT of area Is about centreline of the web, if stiffeners are on both SIDES of the web and about face of the web, if STIFFENER is on only one side of the web such that Is ≥ 0.75dtw^2 for c/d ≥ √2 and Is ≥ 1.5dtw^2/c^2 for c/d < √2.

Correct option is (d) 0.5 times the length of stiffener

Best explanation: The EFFECTIVE length of intermediate TRANSVERSE stiffener is taken as 0.7 times the length of stiffener. The intermediate transverse stiffener is provided MAINLY to improve shear buckling resistance of the WEB.

The correct choice is (c) 40 tw

To elaborate: The effective length of WEB on each side of centreline of STIFFENERS is limited to 20 times the web thickness, i.e. 40tw for interior stiffeners and 20tw for end stiffeners . The effective section is the full area or core area of STIFFENER together with effective length of web on each side of centreline of stiffeners.

The CORRECT choice is (a) length which cannot deform appreciably in bending

Explanation: The stiff bearing length of any element B1 is that length which cannot deform appreciably in bending. To determine b1, the DISPERSION of load through a steel bearing element should be TAKEN as 45˚ through SOLID material, such as bearing plates, flange plates, etc.

Correct choice is (b) preclude any crushing of web

To explain I would SAY: The function of BEARING stiffener is to preclude any crushing of web at locations of heavy concentrated loads. Thus, they transfer heavy REACTIONS or concentrated loads to the full DEPTH of web. They are placed in pairs on the web of plate GIRDERS at unframed girder ends and where required for concentrated loads.

The correct option is (a) MFR = 0.25bftf^2fyf {1-[Nf/( bftffyf/γm0)^2]}

The best explanation: The reduced plastic moment capacity of RESPECTIVE FLANGE plate is calculated after accounting for axial force in flange, due to overall bending and any external axial force in the cross section. It is GIVEN by Mfr = 0.25bftf^2fyf {1-[Nf/( bftffyf/γm0)^2]}, where BF and tf are width and thickness of flange respectively, fyf is yield stress of flange and Nf is the axial force in flange.

Right option is (b) WTF = d cosφ + (c-sc-st)sinφ

Easy explanation: The width of tension field in the tension field ACTION method is given by wtf = d cosφ + (c-sc-st)sinφ, where d is depth of beam, φ = inclination of tension field = tan^-1(d/c), c is spacing between stiffeners, sc and STARE anchorage LENGTHS of tension field along the compression and tension flanges respectively and depends on reduced plastic moment capacity, inclination of tension field, thickness of web and yield stress of web.

Right OPTION is (d) tan^-1(d/c)

For explanation: The INCLINATION of TENSION field is given by φ = tan^-1(d/c), where d is depth of web and c is spacing between stiffeners. The change in angle of inclination of tension field affects the width and YIELD strength of tension field.

The correct answer is (a) 1.5 τb sin2φ

The EXPLANATION: The VALUE of Ψ in the FV for nominal shear strength according to tension field method is given by Ψ =1.5 τb sin2φ, where τb is buckling strength or shear stress corresponding to web buckling, φ is inclination of tension field which DEPENDS on DEPTH of web and spacing of stiffeners.

Right answer is (d) Avτb + 0.9wtftwfvsinφ

Best explanation: In TENSION field method, the NOMINAL shear strength is given by VN = Avτb + 0.9wtftfvsinφ, where AV is area of web, τb is buckling strength or shear stress corresponding to web buckling, fv is yield strength of tension field which depends on inclination of tension field, wtf is the width of tension field, tw is width of web.

The CORRECT choice is (d) it MAY be used for webs with intermediate stiffeners

Explanation: The tension field method, based on the post-shear buckling strength, may be used for webs with intermediate transverse stiffeners at SUPPORTS, provided the panels ADJACENT to the panel angle tension field action or the END posts provide anchorage for the tension field and is c/d>1.0.

Correct choice is (b) 4.0 + [5.35/(c/d)^2].

Easiest EXPLANATION: The value of KV in the elastic critical shear STRESS equation is given by kv = 5.35 when transverse stiffeners are provided only at supports, kv = 4.0 + [5.35/(c/d)^2] for c/d < 1.0, kv = 5.35 + [4/(c/d)^2] for c/d ≥ 1.0, where c and d are spacing of transverse stiffeners and DEPTH of web respectively.

The CORRECT option is (c) kvπ^2E/[12(1-μ^2)(d/tw)^2].

To elaborate: The elastic CRITICAL shear stress of the web is GIVEN by τcr,e = kvπ^2E/[12(1-μ^2)(d/tw)^2], where E is elastic modulus, μ is Poisson’s ratio, kv is CONSTANT which depends on spacing of transverse stiffeners and depth of web.

Right answer is (a) √(fyw/(√3τcr,e))

Explanation: The VALUE of non-dimensional WEB slenderness ratio in the NOMINAL shear strength equation ACCORDING to simple post-critical METHOD is given by λw =√(fyw/(√3τcr,e)), where fyw is yield strength of web, τcr,e is elastic critical shear stress of the web.

Right choice is (b) Avτb

To explain I would SAY: Simple post-critical method based on the shear buckling strength can be used for web of I-section girders, with or without intermediate transverse stiffeners, provided that web has transverse stiffeners at the SUPPORTS. The NOMINAL shear strength is given by Vn=Avτb, where AV = area of web, τb = shear stress corresponding to web buckling.

The correct CHOICE is (b) web will yield under shear before buckling

Best explanation: When d/tw ratio is SUFFICIENTLY low, the elastic CRITICAL STRESS INCREASES above the value of yield shear stress and the web will yield under shear before buckling.

Right option is (c) before onset of buckling strength and post buckling strength

The explanation is: The shear CAPACITY of web comprises of strength before onset of buckling strength and post buckling strength. strength before onset of buckling is CONTRIBUTED because of elastic BEHAVIOUR wherein STRESSES are entirely elastic and the only requirement for the stiffeners is to keep the web FLAT.

Correct choice is (a) less

To elaborate: The flange cover PLATES should not be thicker than the flange angles in riveted/bolted connections. Hence, more than one plate may be required. It is PREFERRED that all cover plates should have same THICKNESS. If they are of different thickness, then outer plates should not be thicker THIN the inner plates.

The correct option is (c) when moment resisting capacity has to be INCREASED

The explanation: When moment resisting capacity of the plate GIRDER has to be increased, flange COVER plates are provided over flange angles. The moment of INERTIA and consequently the moment resisting capacity of the girder is increased considerably as the flange cover plates are at the GREATEST distance from neutral axis.

The correct choice is (a) when LARGE number of CONNECTORS are required to connect flange angle to web

For explanation I would say: When shear is heavy, large number of connectors are required to connect flange angle to web. In such case, EQUAL angles may be preferred, if unequal angles are used, the longer LEG MSY be placed parallel to the web.

Correct answer is (d) unequal ANGLE with LONG leg horizontal

To elaborate: Unequal angles with long leg horizontal are PREFERRED. It is because the moment of INERTIA of the section will be more and a large length will be AVAILABLE for making the connection with the flange plate.

Correct CHOICE is (B) 1/3^rd of calculated flange area

Explanation: Flanges of riveted/ bolted plate girder consists of pair of angles with or without cover plates. The flange angles should form as LARGE part of the area of the flange as practicable and PREFERABLY not less than 1/3rd of the calculated flange area to keep the centre of gravity of flange within back of angles and not in the flange cover plate else stability is affected.

The correct choice is (d) C < 1.5d

The explanation is: To avoid buckling of COMPRESSION FLANGE into web, d/tw ≤ 345 εf^2 for c ≥ 1.5d, d/tw < 345 εffor c < 1.5d when only transverse stiffeners are provided, where c is clear DISTANCE between stiffeners, d is depth of web, εf = √(250/FYF), fyf is yield stress of compression flange.

The correct OPTION is (b) ≤ 345 εf^2

The best I can explain: To avoid buckling of COMPRESSION flange into web when transverse stiffeners are not PROVIDED, d/tw ≤ 345 εf^2, where d is depth of web, εf = √(250/fyf), fyf is YIELD stress of compression flange.