89 + Interview Questions in Plastic And Local Buckling Behaviour Of Steel in Design Of Steel Structures Page 1 InterviewSolution

1.	Flexural torsional buckling cannot occur in ________(a) unsymmetrical members(b) cross section with one axis of symmetry(c) cross section with no axis of symmetry(d) doubly symmetric membersThis question was posed to me in an online quiz.The above asked question is from Behaviour and Ultimate Strength of Plates and Possible Failure Modes topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct answer is (d) doubly symmetric members To explain: Flexural TORSIONAL buckling is a combination of flexural and torsion buckling. The member bends and TWISTS simultaneously. It can occur only with open sections that have unsymmetrical cross section – both with one axis of symmetry(eg: channels, DOUBLE angled shapes) and those with no axis of symmetry (eg: unequal leg single angles). SINCE the shear centre and centroid coincide with each other, doubly symmetric or point symmetric open sections are not subjected to flexural torsional buckling. Close sections are also IMMUNE to flexural torsional buckling.

Discussion

2.	Which of the following is not a solution for torsional buckling?(a) increasing length of members subjected to torsion(b) by careful arrangement of members(c) by providing bracing to prevent lateral movement and twisting(d) box section fabricated by adding welding side plates to ISHB sectionsThe question was asked in an interview for job.The doubt is from Behaviour and Ultimate Strength of Plates and Possible Failure Modes topic in division Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct option is (a) increasing length of MEMBERS subjected to TORSION For explanation: Torsional buckling can be prevented by careful arrangement of members, by providing BRACING to prevent lateral movement and TWISTING. In situations where torsion is expected either a box SECTION fabricated by adding welding side plates to ISHB sections or by shortening box section fabricated by adding welding side plates to ISHB sections becomes the solution.

Discussion

3.	Which of the following is true about torsional buckling?(a) failure occurs by bending about shear centre in longitudinal axis(b) failure occurs when torsional rigidity of member is greater than bending rigidity(c) standard hot rolled shapes are not susceptible to torsional buckling(d) it cannot occur with doubly symmetric cross sectionThe question was posed to me in class test.The origin of the question is Behaviour and Ultimate Strength of Plates and Possible Failure Modes topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct choice is (c) standard hot ROLLED shapes are not susceptible to torsional buckling The best I can explain: In torsional buckling, failure occurs by twisting about shear centre in longitudinal axis. It occurs when torsional rigidity of member is APPRECIABLY smaller than bending rigidity. It can OCCUR only with DOUBLY symmetric cross section with very slender cross SECTIONAL elements. Standard hot rolled shapes are not susceptible to torsional buckling.

Discussion

4.	What is overall flexural buckling?(a) failure occurs by excessive deflection in plane of weaker principal axis(b) failure occurs by excessive deflection in plane of stronger principal axis(c) failure occurs by twisting of member(d) failure caused by seismic loadThe question was posed to me by my school teacher while I was bunking the class.This interesting question is from Behaviour and Ultimate Strength of Plates and Possible Failure Modes topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct answer is (a) failure occurs by EXCESSIVE deflection in plane of weaker PRINCIPAL AXIS The explanation: Overall flexural BUCKLING is failure which occurs by excessive deflection caused by bending or flexure, about axis corresponding to weaker principal axis(minor) – one with smallest RADIUS of gyration, largest slenderness ratio.

Discussion

5.	Which of the following is true about local buckling?(a) failure occurs by twisting of one or more individual elements of member(b) failure occurs by buckling of one or more individual elements of member(c) failure occurs by both buckling and twisting of one or more individual elements of member(d) cannot be prevented by selecting suitable width-to-thickness ratio of elementsI had been asked this question in my homework.I'd like to ask this question from Behaviour and Ultimate Strength of Plates and Possible Failure Modes topic in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» CORRECT answer is (b) failure occurs by buckling of one or more individual elements of MEMBER Explanation: Local buckling is failure which occurs by buckling of one or more individual elements of member. It can be prevented by SELECTING suitable width-to-thickness ratio of elements.

Discussion

6.	The effective width of cold-formed steel sections is given by(a) be/b = (fcr/fy)[1-0.22√(fcr/fy)].(b) be/b = (fcr/fy)[1+0.22√(fcr/fy)].(c) be/b = (fy/fcr)[1-0.22√(fcr/fy)].(d) be/b = (fcr/fy)[1+0.22√(fy/fcr)].This question was posed to me in an interview for job.I want to ask this question from Behaviour and Ultimate Strength of Plates and Possible Failure Modes in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct OPTION is (a) be/b = (fcr/fy)[1-0.22√(fcr/fy)]. Best EXPLANATION: The effective width based on TESTS on cold-formed steel sections is given by be/b = (fcr/fy)[1-0.22√(fcr/fy)]. This formula was first adopted in AISC specification for light gauge cold-formed sections.

Discussion

7.	The effective width of hot-rolled and welded plates is given by(a) be/b = α √(fy/fcr)(b) be/b = α √(fcr x fy)(c) be/b = α √(fcr +fy)(d) be/b = α √(fcr/fy)I had been asked this question in class test.Query is from Behaviour and Ultimate Strength of Plates and Possible Failure Modes topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct CHOICE is (d) be/b = α √(fcr/fy) Easy explanation: The effective width of hot-rolled and welded plates is given by be/b = α √(fcr/fy), where α is the PARAMETER that INDICATES the inclusions of influence of initial curvatures and residual STRESS. GENERALLY α=0.65.

Discussion

8.	What is apparent modulus of elasticity?(a) ratio of average strain carried by plate to average stress(b) ratio of average stress carried by plate to average strain(c) product of average strain carried by plate to average stress(d) product of average stress carried by plate to average strainThis question was posed to me in an interview for internship.I want to ask this question from Behaviour and Ultimate Strength of Plates and Possible Failure Modes topic in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct OPTION is (b) ratio of AVERAGE stress carried by plate to average strain To explain I would say: Elastic post-buckling stiffness is measured in terms of apparent modulus of elasticity, E. It is the ratio of average stress carried by plate to average strain. In most of the cases, the value of E is in the range 0.408 – 0.5 E and MAY be approximately TAKEN as 0.5E.

Discussion

9.	Which of the following statement is true?(a) plate cannot carry loads higher than elastic critical load(b) plate cannot carry loads lesser than elastic critical load(c) secondary path for a plate is unstable(d) secondary path for a plate is stableI had been asked this question during an online interview.The query is from Behaviour and Ultimate Strength of Plates and Possible Failure Modes topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct OPTION is (d) secondary path for a plate is STABLE Explanation: The secondary path shows that plate can carry loads HIGHER than elastic CRITICAL load. The secondary path for a plate is stable.

Discussion

10.	Which of the following is true about secondary path?(a) lateral displacement increases indefinitely at constant load(b) lateral displacement decreases indefinitely at constant load(c) lateral displacement remains same at constant load(d) lateral displacement increases indefinitely and decreases at constant loadThis question was addressed to me by my college professor while I was bunking the class.The doubt is from Behaviour and Ultimate Strength of Plates and Possible Failure Modes topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» CORRECT option is (a) LATERAL displacement increases indefinitely at constant load The explanation: When the axial load reaches the Euler bucking load, the lateral displacement increases indefinitely at constant load. This is CALLED secondary PATH, which BIFURCATES from fundamental path at the buckling load.

Discussion

11.	What is the fundamental path in graph?(a) line along load axis up to P > Pcr(b) line along load axis up to P < Pcr(c) line along load axis up to P = Pcr(d) line along load axis up to P ≥ PcrThis question was addressed to me during an interview for a job.My doubt stems from Behaviour and Ultimate Strength of Plates and Possible Failure Modes in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Right OPTION is (C) line along LOAD axis up to P = Pcr Easy explanation: If axial load verses LATERAL displacement is plotted, we GET a line along the load axis up to P=Pcr, this is called fundamental path.

Discussion

12.	Which of the following statement is correct?(a) Internal elements are elements attached along both longitudinal edges to other elements(b) Outstanding elements are elements attached along both longitudinal edges to other elements(c) Outstanding elements are elements which are free along both the edges(d) Internal elements are elements which are free along both the edgesI had been asked this question in quiz.My doubt is from Cross Sectional Classification topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct OPTION is (a) Internal elements are elements attached ALONG both longitudinal edges to other elements For EXPLANATION I would say: Internal elements are elements attached along both longitudinal edges to other elements or to longitudinal stiffeners connected at suitable intervals to transverse stiffeners connected at suitable intervals totransverse stiffeners (e.g. web of I-SECTIONS and flange and web of box sections). Outstanding elements are attached along only one of the longitudinal edges to an adjacent element, the other edge being free to displace out of plan (e.g. flange overhang of an I-section, stem of T-section, LEG of angle section).

Discussion

13.	Which of the following is correct regarding class I section?(a) They are not fully effective under pure compression(b) They are capable of reaching and maintaining full plastic moment in bending(c) They are not capable of reaching and maintaining full plastic moment in bending(d) They does not exhibit sufficient ductilityThe question was posed to me by my school principal while I was bunking the class.Question is from Cross Sectional Classification topic in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct choice is (b) They are capable of REACHING and MAINTAINING full plastic moment in BENDING To explain: Class I sections are fully effective under pure compression, and capable of reaching and maintaining full plastic moment in bending and hence used in plastic design. These sections will exhibit sufficient DUCTILITY.

Discussion

14.	What is a slender section?(a) cross section which can develop plastic moment resistance(b) cross section which can resist seismic force(c) cross section in which elastically calculated stress in extreme compression fibre can reach yield strength(d) cross section in which local buckling will occur before yield stressI have been asked this question in unit test.Asked question is from Cross Sectional Classification in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct choice is (d) cross section in which LOCAL BUCKLING will occur before yield stress The BEST explanation: Slender or class IV SECTIONS are cross section in which local buckling occurs even before the yield stress is attained in one or more parts of the cross section.

Discussion

15.	What is a semi-compact section?(a) cross section which can develop plastic moment resistance(b) cross section which can resist seismic force(c) cross section in which elastically calculated stress in extreme compression fibre can reach yield strength(d) cross section which can develop plastic hingesThe question was posed to me by my college director while I was bunking the class.Question is from Cross Sectional Classification topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct choice is (c) cross section in which elastically calculated stress in EXTREME compression fibre can REACH YIELD strength Easy explanation: Semi-compact or class III sections are cross sections in which elastically calculated stress in extreme compression fibre of STEEL can reach yield strength.

Discussion

16.	What is a compact section?(a) cross section which can develop plastic moment resistance(b) cross section which can resist seismic force(c) cross section in which buckling can occur(d) cross section which can develop plastic hingesThe question was posed to me during an interview for a job.This intriguing question originated from Cross Sectional Classification topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct CHOICE is (a) CROSS section which can develop plastic moment resistance Explanation: Compact or class II sections are cross sections which can develop plastic moment resistance, but have INADEQUATE plastic hinge rotation CAPACITY because of LOCAL buckling.

Discussion

17.	What is a plastic section?(a) cross section which can develop plastic moment(b) cross section which can resist seismic force(c) cross section in which buckling can occur(d) cross section which can develop plastic hingesThe question was asked during an online interview.My question is from Cross Sectional Classification in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Right ANSWER is (d) cross SECTION which can develop plastic hinges The best EXPLANATION: Plastic or class I sections are cross sections which can develop plastic hinges and have a rotation capacity required for failure of structure by formation of plastic mechanism.

Discussion

18.	Which of the following factor is considered for classification of cross section?(a) location where member is used(b) width-to-thickness ratio(c) length of member(d) seismic forceThis question was posed to me in an international level competition.The question is from Cross Sectional Classification in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The CORRECT answer is (b) width-to-thickness RATIO The best I can explain: CROSS section are classified into four behavioural groups depending uponthe material yield strength, width-to-thickness ratio of INDIVIDUAL components (e.g. webs and flanges) within the cross section, and the loading arrangement.

Discussion

19.	Which of following statement is correct?(a) elastic buckling stress may be decreased by using longitudinal stiffeners(b) elastic buckling stress may be decreased by using intermediate stiffeners(c) elastic buckling stress may be increased by using intermediatetransverse stiffeners(d) elastic buckling stress is not affected by intermediate or longitudinal stiffenersI have been asked this question in an interview.I would like to ask this question from Local Buckling of Plates topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Right choice is (c) elastic buckling STRESS may be increased by using intermediatetransverse STIFFENERS The best I can explain: The elastic buckling stress may be increased by using intermediate TRANSVERSE stiffeners (which will decrease the aspect ratio L/d, thus INCREASING the value of buckling coefficient), or by using longitudinal stiffeners to decrease the depth-thickness ratio.

Discussion

20.	The elastic buckling stress for thin flat plate of length L, depth d and thickness t simply supported along four edges and loaded by bending stress distribution is given by(a) fcr= π^2E/k[12(1-μ^2)(d/t)^2].(b) fcr= π^2E/k[12(1+μ^2)(d/t)^2].(c) fcr= kπ^2E/[12(1+μ^2)(d/t)^2].(d) fcr= kπ^2E/[12(1-μ^2)(d/t)^2].The question was asked in a national level competition.The query is from Local Buckling of Plates in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct OPTION is (d) fcr= kπ^2E/[12(1-μ^2)(d/t)^2]. The best I can explain: The elastic buckling stress for thin flat plate of length L, depth d and thickness t simply supported ALONG all four edges and loaded by bending stress distribution, which varies linearly across its width is given by fcr= kπ^2E/[12(1-μ^2)(d/t)^2], where buckling coefficient k DEPENDS on aspect RATIO L/d and the NUMBER of buckles along the plate.

Discussion

21.	The elastic buckling stress of thin flat plate of length L, depth d and thickness t simply supported along four edges and loaded by shear stresses distributed uniformly along its edges is given by(a) fcr = kπ^2E / [12(1-μ^2)(d/t)^2].(b) fcr = kπ^2E / [12(1+μ^2)(d/t)^2].(c) fcr = kπ^2E / [12(1-μ^2)(d/t)].(d) fcr = kπ^2E / [12(1+μ^2)(d/t)].I got this question during an interview.I'd like to ask this question from Local Buckling of Plates topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct option is (a) fcr = kπ^2E / [12(1-μ^2)(d/t)^2]. Easy explanation: The elastic buckling stress fcr of thin flat plate of length L, depth d and thickness t simply SUPPORTED ALONG all four edges and loaded by SHEAR STRESSES distributed uniformly along its edges is given by fcr = kπ2E / [12(1-μ^2)(d/t)^2], where buckling coefficient can be approximated byk=5.35 + 4(d/L)^2, when L ≥ d and k = 5.35(d/L)2 + 4, when L ≤ d.

Discussion

22.	The buckling coefficient for thin flat plate free along one longitudinal edge is given by(a) k = 0.425 + (b/a)(b) k = 0.425 + (b/a)^2(c) k = 0.425 + (a/b)^2(d) k = 0.425 – (b/a)^2This question was addressed to me in a job interview.The above asked question is from Local Buckling of Plates in division Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Right option is (b) k = 0.425 + (b/a)^2 The BEST EXPLANATION: For a thin plate simply supported along both TRANSVERSE edged and one longitudinal edge and free along the other longitudinal edge, the BUCKLING coefficient can be approximated by k = 0.425 + (b/a)^2.

Discussion

23.	The buckling stress fcr varies _____(a) inversely as plate slenderness or width-to-thickness ratio(b) directly as plate slenderness or width-to-thickness ratio(c) inversely as square of plate slenderness or width-to-thickness ratio(d) directly as square of plate slenderness or width-to-thickness ratioThe question was asked in an internship interview.Enquiry is from Local Buckling of Plates topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» RIGHT answer is (c) INVERSELY as square of plate slenderness or width-to-thickness ratio Explanation: The buckling stress fcr varies inversely as square of plate slenderness or width-to-thickness ratio,√(FY /fcr) = (b/t)√{(fy / E)[12(1-μ^2)/(π^2k)]} .

Discussion

24.	Lowest value of buckling coefficient for simply supported plates is _____(a) 4.0(b) 2.0(c) 5.0(d) 3.0I have been asked this question in homework.This question is from Local Buckling of Plates topic in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» CORRECT answer is (a) 4.0 The BEST I can explain: The lowest value of buckling COEFFICIENT for simply supported PLATES is 4.0. The buckilng STRESS depends upon buckling coefficient.

Discussion

25.	Which of the following statement is correct?(a) stiffened elements are supported along one edge perpendicular to axial stress(b) un-stiffened elements are supported along one edge perpendicular to axial stress(c) stiffened elements are supported along one edge parallel to axial stress(d) un-stiffened elements are supported along one edge parallel to axial stressThe question was posed to me in examination.This intriguing question originated from Local Buckling of Plates in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct choice is (d) un-stiffened elements are SUPPORTED along one edge parallel to axial stress Best explanation: Unstiffened elements are supported along one edge parallel to axial stress (eg : legs of single angles, flanges of BEAMS, and stems of T-section). Stiffened elements are supported along both the edges parallel to axial stress (eg: flanges of square and rectangular hollow sections, PERFORATED cover plates, and webs of I-sections and channel sections).

Discussion

26.	The critical stress of infinite plate having width b and thickness t loaded by compressive forces acting on simply supported sides is given by(a) (kπ^2E)/ [12(1-μ^2)(b/t)].(b) (kπ^2E)/ [12(1-μ^2)(b/t)^2].(c) (kπ^2E)/ [12(1+μ^2)(b/t)].(d) (kπ^2E)/ [12(1+μ^2)(b/t)^2].I had been asked this question in an interview.This question is from Local Buckling of Plates topic in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» CORRECT answer is (b) (kπ^2E)/ [12(1-μ^2)(b/t)^2]. To explain: The critical stress of infinite plate having width b and THICKNESS t loaded by compressive forces acting on SIMPLY supported sides is given by fcr = (kπ^2E)/ [12(1-μ^2)(b/t)^2], where μ is Poisson’s ratio of material, b/t is width-to-thickness ratio of plate, k is buckling coefficient and E is Young’s modulus of rigidity of material. The value of coefficient k depends on constraints along non-loaded EDGES of plate.

Discussion

27.	Buckling occurs to members or elements mainly subjected to ________(a) seismic forces(b) tensile forces(c) compressive forces(d) shear forcesI got this question by my school teacher while I was bunking the class.This question is from Local Buckling of Plates topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct choice is (c) compressive forces The EXPLANATION is: Buckling MAY be defined as structural behavior in which mode of deformation develops in direction or plane perpendicular to that of BENDING which produces it. Such deformation changes rapidly with increase in magnitude of applied loading. It occurs MAINLY members or ELEMENTS that are subjected to compressive forces.

Discussion

28.	Loss of web capacity is given by ___(a) Zpw / (fy – fw)(b) Zpw (fy – fw)(c) Zpw (fy + fw)(d) Zpw /(fy + fw)The question was asked during an interview for a job.Asked question is from Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Right answer is (b) Zpw (FY – FW) To EXPLAIN: LOSS of web capacity is GIVEN by Zpw (fy – fw)= Zpw fy { 1-√[1-(τw/τy)2] }.

Discussion

29.	At full plasticity, the stress in web is given by(a) fw = fy√[1+(τw/τy)^2 ].(b) fw = fy√[(τw/τy)^2 ].(c) fw = fy√[1-(τw/τy)^2 ].(d) fw = fy√[1+2(τw/τy)^2 ].I have been asked this question in an interview for job.My doubt is from Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct choice is (c) fw = fy√[1-(τw/τy)^2 ]. The explanation: When full plasticity is produced, the STRESS in WEB will be equal to fw, fw =√(fy^2 -3τw^2 ) = fy√[1-(w/τy)^2 ]where τy is SHEAR STRENGTHS when yield is in pure shear.

Discussion

30.	Yield in pure shear occurs when ______(a) 0.58 fy(b) 1.58 fy(c) 2.8 fy(d) 3.5 fyI had been asked this question during a job interview.Asked question is from Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct choice is (a) 0.58 FY For EXPLANATION: Yield in PURE SHEAR occurs when τy = fy/√3 = 0.58 fy.

Discussion

31.	When a member is subjected to uniaxial tensile or compressive stress in presence of shear stressτ , yield occurs when ___(a) fy^2 = f^2 – 3 τ(b) fy^2 = f^2 + 3 τ^2(c) fy^2 = f^2 – 3 τ^2(d) fy^2 = f^2 + 3 τThe question was asked by my school teacher while I was bunking the class.This interesting question is from Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct CHOICE is (b) fy^2 = f^2 + 3 τ^2 The best EXPLANATION: When a member is subjected to uniaxial tensile or compressive STRESS in presence of shear stressτ, yield occurs when fy^2 = f^2 + 3 τ^2.

Discussion

32.	Which of the following relation is correct for I- section of width b and depth d subjected to axial force N together with moment M?(a) (N/Np) – (1/1.18)(M/Mp) ≤ 1, when N/Np > 0.15(b) (N/Np) – (1/1.18)(M/Mp) ≤ 1, when N/Np < 0.15(c) (N/Np) + (1/1.18)(M/Mp) ≤ 1, when N/Np < 0.15(d) (N/Np) + (1/1.18)(M/Mp) ≤ 1, when N/Np > 0.15The question was asked during an online exam.This intriguing question originated from Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct answer is (d) (N/Np) + (1/1.18)(M/Mp) ≤ 1, when N/Np > 0.15 To elaborate: For I- SECTION SUBJECTED to AXIAL force N TOGETHER with moment M, (N/Np) + (1/1.18)(M/Mp) ≤ 1, when N/Np > 0.15 M = Mp, when N/Np < 0.15 .

Discussion

33.	Which of the following relation is correct for rectangular section of width b and depth d subjected to axial force N together with moment M?(a) (Mpr/Mp) + (N/Np)^2 = 1(b) (Mpr/Mp) – (N/Np)^2 = 1(c) (Mpr/Mp) + (N/Np) = 1(d) (Mpr/Mp) – (N/Np) = 1The question was asked in homework.Origin of the question is Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» CORRECT option is (a) (Mpr/Mp) + (N/Np)^2 = 1 To explain: For a RECTANGULAR SECTION of width b and depth d subjected to AXIAL force N together with moment M, (Mpr/Mp) + (N/Np)2 = 1, where Mpr is moment with axial force, Mp is moment with axial force, Np is axial force without any moment.

Discussion

34.	The presence of axial equation implies that _________(a) sum of tension forces is always zero(b) sum of compression forces is always zero(c) sum of tension and compression forces is not zero(d) sum of tension and compression forces is zeroThis question was posed to me in an internship interview.I would like to ask this question from Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The CORRECT choice is (c) sum of tension and compression forces is not zero The best I can explain: The presence of AXIAL equation IMPLIES that sum of tension and compression forces is not zero and hence following equation is USED : ∫AfydA – P = 0, where P is the axial force.

Discussion

35.	Which of the following statement is true?(a) combined mechanism is combination of elementary mechanism(b) elementary mechanism is combination of combined mechanism(c) combined mechanism is not combination of elementary mechanism(d) elementary mechanism is combination of elementary and combined mechanismI got this question during an interview.My question is based upon Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct option is (a) combined mechanism is combination of elementary mechanism To EXPLAIN I would say: The POSSIBLE mechanisms can be classified into two TYPES : elementary and combined mechanism. Elementary mechanism is independent of each other, combined mechanism is linear combination of elementary mechanisms.

Discussion

36.	Which of the following relation is correct for pin based frames?(a) Mp =γL(wL^2/8)[1-k+(1+k)^0.5].(b) Mp =γL(wL^2/8)[1+k+(1-k)^0.5].(c) Mp =γL(wL^2/8)[1+k-(1+k)^0.5].(d) Mp =γL(wL^2/8)[1+k+(1+k)^0.5].I got this question in exam.My question is based upon Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct option is (d) Mp =γL(wL^2/8)[1+k+(1+k)^0.5]. Easiest explanation: When gravity load governs the design, a good ESTIMATE of REQUIRED section may be obtained by using following FORMULAE : For pin based frames : Mp =γL(wL^2/8)[1+k+(1+k)^0.5] For fixed=-base FRAME : Mp =γL(wL^2/8)[1/[1+0.5k+(1+k)^0.5]], where γL = 1.7 global load FACTOR, k = h2/h1.

Discussion

37.	Which method is used when mechanism is applied to structures with sloping members?(a) method of instantaneous centre(b) method of centre(c) method of seismic centre(d) method of metacentreI got this question by my school principal while I was bunking the class.The origin of the question is Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Right choice is (a) method of instantaneous centre The BEST EXPLANATION: When MECHANISM is applied to structures with sloping members, the determination of DISPLACEMENTS in the direction of applied forces is required. It is done USING method of instantaneous centre or centre-of-rotation technique.

Discussion

38.	If order of indeterminacy is r, then minimum number of plastic hinges required for total collapse is _______(a) r-1(b) r(c) r+1(d) r+2The question was asked in an international level competition.Origin of the question is Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity in division Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct option is (c) r+1 For EXPLANATION I would SAY: If order of indeterminacy is r, then MINIMUM number of plastic hinges required for total COLLAPSE is r+1.

Discussion

39.	Single bay portal frames with fixed bases have _______(a) two redundancies(b) three redundancies(c) four redundancies(d) zero redundanciesI had been asked this question during an online exam.This is a very interesting question from Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct CHOICE is (b) THREE redundancies Easiest explanation: Single bay portal FRAMES with fixed bases have three redundancies and require four hinges to PRODUCE a MECHANISM.

Discussion

40.	What is the value of load factor for I-section when fbc= 0.66fy and mean value of v is 0.14?(a) 1.727(b) 2.7(c) 1.56(d) 3.98I had been asked this question in an online interview.My doubt stems from Theorem of Plastic Collapse & Methods of Plastic Analysis topic in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer»

Discussion

41.	Which of the following is load factor for simply supported beam with central point load?(a) (fyfbc)v(b) (fbc/ fy)v(c) (fy/fbc)v(d) (fy + fbc)vI got this question during an online exam.The above asked question is from Theorem of Plastic Collapse & Methods of Plastic Analysis in division Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Right answer is (c) (fy/fbc)V Easiest explanation: For a simply supported beam with CENTRAL point load, M = WL/4 = fbcZe, MP = WcL/4 = fyvZe, where v is SHAPE factor F = Mp/M = Wc/W = (fy/fbc)v.

Discussion

42.	Which of the following is true about kinematic theorem?(a) it represents lower limit to true ultimate load(b) it represents plastic load(c) it has small factor of safety(d) it satisfies equilibrium and yield conditionsThe question was asked in quiz.Question is from Theorem of Plastic Collapse & Methods of Plastic Analysis in division Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct option is (c) it has small factor of safety Best explanation: Load computed on basis of this MECHANISM will always be greater than or at LEAST equal to true ultimate load. Hence, kinematic method represents an UPPER limit to the true ultimate load and has a SMALLER factor of safety. This theorem satisfies equilibrium and continuity conditions.

Discussion

43.	Which of the following relation between load factor, collapse load(Wc) and working load (W)(a) F = Wc / W(b) F = W / Wc(c) F = Wc W(d) F = Wc + WI had been asked this question in class test.The query is from Theorem of Plastic Collapse & Methods of Plastic Analysis topic in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» CORRECT answer is (a) F = WC / W The best explanation: Load factor is ratio of COLLAPSE load to working load. It is REPRESENTED by, F = Wc / W.

Discussion

44.	In static method of analysis, moment at any section is _______ plastic moment capacity.(a) greater than(b) two times(c) less than(d) three timesThe question was asked in an interview for internship.This intriguing question comes from Theorem of Plastic Collapse & Methods of Plastic Analysis in division Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Correct OPTION is (c) less than The EXPLANATION is: In STATIC method of analysis, an equilibrium moment diagram is OBTAINED such that the moment at any section is less than or equal to the PLASTIC moment capacity.

Discussion

45.	The number of independent mechanism is related to number of possible plastic hinge locations by ________(a) n = h * r(b) n = h / r(c) n = h + r(d) n = h – rThe question was asked in a national level competition.I want to ask this question from Theorem of Plastic Collapse & Methods of Plastic Analysis in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct choice is (d) n = H – R To explain I would say: The number of INDEPENDENT mechanism (n) is RELATED to number of possible PLASTIC hinge locations (h) and number of degree of redundancy (r) of the frame by equation n = h-r.

Discussion

46.	Which of the following is true about kinematic analysis?(a) virtual work equations are not used to determine collapse load(b) virtual work equations are used to determine collapse load(c) equilibrium condition is assumed(d) plasticity condition is assumedThe question was asked during an interview.I need to ask this question from Theorem of Plastic Collapse & Methods of Plastic Analysis topic in division Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct option is (b) virtual work equations are USED to determine collapse LOAD To ELABORATE: In kinematic method, a MECHANISM is assumed and virtual work equations are used to determinethe collapse load.

Discussion

47.	Load is called as correct collapse load when(a) static theorem is not satisfied(b) kinematic theorem is not satisfied(c) only static theorem is satisfied(d) both static and kinematic theorem are satisfiedThis question was posed to me in semester exam.This interesting question is from Theorem of Plastic Collapse & Methods of Plastic Analysis in section Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct option is (d) both STATIC and kinematic THEOREM are satisfied To explain I would say: LOAD that satisfies both static and kinematic theorem at the same time is CALLED correct COLLAPSE load.

Discussion

48.	Which of the following condition is true for uniqueness theorem?(a) load must be greater than collapse load(b) load must be less than collapse load(c) load must be equal to collapse load(d) load cannot be related to collapse loadI had been asked this question in an interview for internship.My question is from Theorem of Plastic Collapse & Methods of Plastic Analysis topic in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The CORRECT option is (C) load must be equal to collapse load For explanation: For given frame and loading, if at least one safe and statically admissible bending moment DISTRIBUTION can be found, and if in this distributionbending moment is equal to fully plastic moment at sufficient cross sections to cause failure of frame as a mechanism DUE to rotation of plastic hinges, then CORRESPONDING load will be equal to collapse load.

Discussion

49.	Which of the following condition is true for kinematic theorem?(a) load must be greater than collapse load(b) load must be less than collapse load(c) load must be not equal to collapse load(d) load cannot be related to collapse loadI had been asked this question by my college director while I was bunking the class.My query is from Theorem of Plastic Collapse & Methods of Plastic Analysis topic in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» Right OPTION is (a) LOAD MUST be greater than collapse load Explanation: For a given frame subjected to a set of loads P, the value of P which is found to correspond to any ASSUMED mechanism, must be greater than or equal to the collapse load Pu.

Discussion

50.	Which of the following is true about static theorem?(a) it represents upper limit to true ultimate load(b) it represents plastic load(c) it has minimum factor of safety(d) it satisfies equilibrium and yield conditionsI got this question by my school principal while I was bunking the class.My question comes from Theorem of Plastic Collapse & Methods of Plastic Analysis in division Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures
Answer» The correct option is (d) it SATISFIES equilibrium and yield CONDITIONS The explanation is: The static method represents the lower limit to the true ultimate LOAD and has maximum factor of SAFETY. The THEOREM satisfies equilibrium and yield conditions.

Discussion

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Lowest value of buckling coefficient for simply supported plates is _____(a) 4.0(b) 2.0(c) 5.0(d) 3.0I have been asked this question in homework.This question is from Local Buckling of Plates topic in portion Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures