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Right option is (C) span/300

The BEST I can explain: The maximum DEFLECTION in other buildings (other than industrial buildings) for FLOOR subjected to live load and elements not susceptible to cracking is span/300.

The correct ANSWER is (a) span/150

The BEST explanation: The maximum DEFLECTION in industrial building for purlins and girts subjected to LIVE load/wind load for elastic cladding is span/150 and for brittle cladding is span/180.

Right answer is (C) increasing depth of beam

To EXPLAIN I would say: Deflection can be reduced by increasing depth of beam, reducing the span, providing GREATER end RESTRAINTS or by other MEANS such as providing camber.

The correct option is (c) FCRIP = (b+2n1)fywt

To explain I would say: The crippling strength of web at interior point where concentrated LOAD is acting is given by Fcrip = (b+2n1)fywt, where b+2n1 is length obtained by DISPERSION through flange, t is thickness of web, fyw is design YIELD strength of web.

The correct choice is (B) provide stiffeners which bear against flanges at load points

For EXPLANATION I would SAY: Web crippling can be PREVENTED by spreading load over large PORTION of flange. The other remedy is provide stiffeners which bear against flanges at load points and are connected to web to transfer force to it gradually. The other remedy is to make the web thicker.

Correct option is (d) Fcrip = (b+n1)fywt

To explain I would say: The crippling strength of WEB at supports is given by Fcrip = (b+n1)fywt, where b+n1 is LENGTH obtained by DISPERSION through flange, t is thickness of web, FYW is design YIELD strength of web.

Correct CHOICE is (b) 1:2.5

For explanation I WOULD say: The angle of dispersion of load for web crippling is assumed to be 1:2.5 .With REFERENCE to this, bearing length is CALCULATED.

Correct answer is (d) at NEUTRAL axis

To elaborate: The MAXIMUM diagonal compression in plate girder simply SUPPORTED OCCURS at neutral axis . It will be inclined at 45˚ to the neutral axis.

Correct answer is (c) web near portion of stress CONCENTRATION tends to FOLD over flange

The explanation: Webs of rolled SECTION are subjected to LARGE amount of stresses just below CONCENTRATED loads and above reactions from support. Stress concentration occurs at junction of web and flange. As a result, large bearing stresses are developed below concentrated load. Consequently, the web near portion of stress concentration tends to fold over flange. This type of local buckling phenomenon is called crippling or crimpling of web.

The correct option is (a) Fwb =Btwfcd

To elaborate: Web BUCKLING strength at support of simply supported plate girder is given by Fwb =Btwfcd , where B is length of stiff portion of bearing PLUS additional length given by dispersion at 45˚to the level of neutral axis, tw is thickness of web, fcd is allowable COMPRESSIVE stress corresponding to assumed web strut according to buckling curve C.

The correct choice is (b) 2d

To explain: The effective depth when top FLANGES are not RESTRAINED against rotation and lateral DEFLECTION is 2d, where d is depth of web. Bottom FLANGE is assumed to be restrained against lateral deflection and rotation. for top flanges, the end restraints and effective depth of the web are to be considered.

Correct option is (a) 2d/3

For EXPLANATION I would say: The effective depth when TOP FLANGES are restrained against LATERAL deflection but not against rotation is 2d/3, where d is depth of web. Bottom flange is assumed to be restrained against lateral deflection and rotation. for top flanges, the end restraints and effective depth of the web are to be considered.

Right option is (B) web in ROLLED section behaves like a column when placed under CONCENTRATED loads

Best EXPLANATION: The web in rolled section behaves like a column when placed under concentrated loads. The web is QUITE thin and is therefore, subjected to buckling.

The correct choice is (a) when web thickness is large in plate GIRDERS

The explanation is: Shear determines design of beam when depth of beam section is SMALL and LOADED uniformly, when large concentrated LOADS are placed near beam supports, when two members are rigidly connected together with their WEBS lying in same plane, when web thickness is small in plate girders.

Right option is (c) FLANGES RESIST very small portion of shear

The explanation: For an I-section, flanges resist very small portion of shear and a SIGNIFICANT portion is resisted by WEB. The maximum and average shear for I-section are almost same.

The correct CHOICE is (b) τ = VAy/Izt

Best explanation: The longitudinal shear is given by τ = VAy/Izt, where V is vertical shear force at section under consideration, Iz is moment of inertia of entire section about zz-axis, neutral axis, Ay is STATIC moment of area of cross section, t is thickness of portion of cross section at which longitudinal shear is OBTAINED.

Right option is (d) parabolically with DEPTH

Easiest EXPLANATION: The shear stress DISTRIBUTION of I-section varies parabolically with depth with MAXIMUM occurring at the neutral axis.

Correct option is (a) beam is short

Easy EXPLANATION: SHEAR forces will GOVERN the DESIGN of beam if beams are short and are heavily loaded (HEAVY concentrated load) or deeply coped.

Correct choice is (c) ∑biti^3/3

The BEST EXPLANATION: ST. Venant’s constant is GIVEN by It = ∑biti3/3. For OPEN section (e.g. I -section) : It = 2bft3f/3 + bft3w/3.

The CORRECT answer is (b) (1-βf)βf Iy h^2f

The explanation: Warping constant in elastic CRITICAL MOMENT is given by Iw = (1-βf)βf Iy h^2f, where βf is ratio of moment of inertia of compression flange to sum of MOMENTS of inertia of compression and tension flanges, Iy = moment of inertia about minor axis, hf = centre-to-centre DISTANCE between flanges.

The correct choice is (d) Mcr= βbZp FCR,b

To elaborate: The ELASTIC critical moment is GIVEN by Mcr= √{[π^2EIy/ L^2MLT][ GIt + (π^2EIw/L^2LT)]} = βbZp fcr,b , IY = moment of inertia about minor axis, Iw =warping constant, It = St. Venant’s constant, G = Shear MODULUS.

Right option is (c) λLT ≤ 1.2 √(Zefy/Mcr)

To explain I would say: The NON- DIMENSIONAL slenderness ratio is given by λLT = √(βbZpfy/Mcr). The check for it is given by λLT ≤ 1.2 √(Zefy/Mcr), where ZE = elastic section modulus, Mcris elastic critical MOMENT.

The CORRECT answer is (d) φLT= 0.5 [ 1 + αLT (λLT – 0.2) + λ^2LT].

EASIEST explanation: The value of φLT in bending stress reduction FACTOR is given by φLT= 0.5 [ 1 + αLT (λLT – 0.2) + λ^2LT], where αLTis IMPERFECTION factor, λLT is non dimensional slenderness ratio.

The correct OPTION is (b) 0.21

The explanation is: Imperfection FACTOR for rolled section is 0.21. The imperfection factor takes into account all the RELEVANT defects in real structure when considering buckling, GEOMETRIC imperfections, eccentricity of applied loads and RESIDUAL stresses. It depends on the buckling curve.

Right answer is (d) 0.49

The BEST explanation: Imperfection factor for welded section is 0.49. The imperfection factor depends on the buckling CURVE and takes into account all the relevant defects in REAL STRUCTURE when considering buckling, geometric imperfections, eccentricity of applied LOADS and residual stresses.

Right OPTION is (b) XLT fy /fy

Explanation: The VALUE of DESIGN bending compressive stress fbd is XLT fy /fy, where XLT is bending stress reduction factor to account for lateral TORSIONAL buckling, fy is yield stress, fy is partial safety factor for material (=1.10).

Right OPTION is (a) Ze/ZP

The explanation is: The value of βb in the equation of design bending strength of laterally unsupported beams for semi-compact sections is Ze/Zp, where Ze is ELASTIC section MODULUS, Zp is plastic section modulus.

Right choice is (d) XLT = 1/{φLT + (φ^2LT – λ^2LT)0.5}

The best explanation: The bending stress reduction FACTOR to account for lateral BUCKLING is GIVEN by XLT = 1/{φLT + (φ^2LT – λ^2LT)0.5}, where φLT depends upon imperfection factor and non dimensional slenderness ratio, λLT is non dimensional slenderness ratio.

Right choice is (d) Md= βbZpfbd

Easy explanation: The bending strength of LATERALLY UNSUPPORTED beams is given by Md= βbZpfbd, where βb is a constant, Zp is plastic section modulus, FBD is DESIGN bending COMPRESSIVE stress.

Right choice is (c) lateral torsional buckling

The best EXPLANATION: Beams with major AXIS bending and compression FLANGE not restrained against lateral bending (or INADEQUATE lateral support) FAIL by lateral torsional buckling before attaining their bending strength.

The CORRECT choice is (a) λLT ≤ 0.4

Explanation: The EFFECT of lateral-torsional BUCKLING need not be CONSIDERED when λLT ≤ 0.4, where λLT is the non dimensional slenderness ratio for lateral torsional buckling.

The CORRECT ANSWER is (B) point load causes more shear lag than uniform load

The best I can EXPLAIN: The resultant stress DISTRIBUTION across flanges is non-uniform and is called shear lag. Point load causes more shear lag than uniform load.

Right CHOICE is (a) Ze/Zp

The EXPLANATION: The value of βb in the equation of design bending strength for semi-compact section is GIVEN by βb = Ze/Zp , where Ze, Zp are ELASTIC and PLASTIC moduli of the cross section.

Correct choice is (b) βbZpfy / γm0

To explain: The DESIGN bending strength of beams when V ≤ 0.6Vd is given by Md = βbZpfy / γm0 , where βb is a constant, Zp = PLASTIC SECTION modulus of cross section, FY is yiled stress of MATERIAL, γm0 = 1.1,partial safety factor.

The correct ANSWER is (b) material of beam

For explanation: The IMPORTANT aspects which need to be CONSIDERED for beam design are moments, shears, DEFLECTION, crippling, buckling, and LATERAL support.

The CORRECT option is (c) its flexural stiffness should be 6EIb/S

Easy explanation: TORSIONAL bracing ATTACHED to TOP flange should BEND in reverse curvature and its flexural stiffness should be 6EIb/S, where S is spacing between girders.

The CORRECT OPTION is (a) yielding of section starts at lower moments

Explanation: Due to presence of residual STRESSES, yielding of section starts at lower moments. Then with INCREASE in moment, yielding spreads through the cross section.