Multiple choice question for engineering
Set 1
1. Which of the following assumptions is correct for plastic design?
a) material obeys Hooke’s law before the stress reaches fy
b) yield stress and modulus of elasticity does not have same value in compression and tension
c) material is homogenous and isotropic in both elastic and plastic states.
d) material is not sufficiently ductile to permit large rotations
Answer
Answer: c [Reason:] The material obeys Hooke’s law till the stress reaches fy. The yield stress and modulus of elasticity have the same value in compression and tension. The material is homogeneous and isotropic in both elastic and plastic states. The material is assumed to be sufficiently ductile to permit large rotations of section to take place.
2. Which of the following assumptions is not correct for plastic design?
a) plastic hinge rotations are small compared with elastic deformations so all the rotations are concentrated at plastic hinges
b) segments between plastic hinges are rigid
c) influence of normal and shear forces on plastic moments is not considered
d) plane section remains plane after bending and the effect of shear is neglected
Answer
Answer: a [Reason:] The plastic hinge rotations are large compared with elastic deformations so all the rotations are concentrated at plastic hinges. The segments between plastic hinges are rigid. The influence of normal and shear forces on plastic moments is not considered. The plane section remains plane even after bending and the effect of shear is neglected.
3. Which of the following are the conditions that stress strain characteristics of steel should obey to ensure plastic moment redistribution?
a) yield plateau should be less than 6 times the yield strain
b) ratio of ultimate tensile stress to yield stress should be less than 1.2
c) steel should not exhibit strain-hardening capacity
d) elongation on standard gauge length should be more than 15%
Answer
Answer: d [Reason:] The conditions that stress strain characteristics of steel should obey to ensure plastic moment redistribution are: (i) yield plateau(horizontal portion of stress strain curve) should be greater than 6 times the yield strain, (ii) ratio of ultimate tensile stress to yield stress should be more than 1.2, (iii) elongation on standard gauge length should be more than 15%, (iv) steel should exhibit strain-hardening capacity.
4. Which of the following conditions are true for using plastic method of analysis as per IS 800?
a) members shall not be hot-rolled or fabricated using hot-plates
b) yield stress of steel should not be greater than 450MPa
c) cross section should be unsymmetrical about its axis perpendicular to axis of plastic hinge rotation
d) cross section of members not containing plastic hinges should be ‘plastic’ and those members containing plastic hinges should be ‘compact’
Answer
Answer: b [Reason:] The following are conditions for using plastic method of analysis as per IS 800: (i) yield stress of steel should not be greater than 450MPa, (ii) members shall be hot-rolled or fabricated using hot-plates, (iii) cross section should be symmetrical about its axis perpendicular to the axis of plastic hinge rotation, (iv) cross section of members not containing plastic hinges should be ‘compact’ and those members containing plastic hinges should be ‘plastic’.
5. Which of the following is true regarding plastic design methods?
a) design needs to satisfy elastic strain compatibility conditions
b) different factor of safety for all parts of the structure
c) saving of material over elastic methods resulting in lighter structures
d) design is effected by temperature changes, settlement of support, etc
Answer
Answer: c [Reason:] The following are the advantages of plastic design methods: (i) realisation of uniform and realistic factor of safety for all parts of the structure , (ii)simplified analytical procedure and rapidity of obtaining moments since no need to satisfy elastic strain compatibility conditions, (iii) no effect due to temperature changes, settlement of support, etc, (iv) saving of material over elastic methods resulting in lighter structures.
6. Which of the following is true regarding plastic design methods?
a) difficult to design for fatigue
b) more saving in column design
c) lateral bracing requirements are less stringent than for elastic design
d) moments produced by different loading conditions can be added together
Answer
Answer: a [Reason:] (i) There is little saving in column design, (ii) lateral bracing requirements are more stringent than for elastic design, (iii) difficult to design for fatigue,(iv) moments produced by different loading conditions needs to be calculated separately(cannot be added together) and the largest plastic moment is selected.
7. What is plastic hinge?
a) zone of bending due to flexure in a structural member
b) zone of yielding due to flexure in a structural member
c) zone of non-yielding due to flexure in a structural member
d) zone of yielding due to twisting in a structural member
Answer
Answer: b [Reason:] Plastic hinge is a zone of yielding due to flexure in a structural member. It is used to describe a deformation of a section when plastic bending occurs.
8. Plastic hinge behaves like a ______
a) friction mechanical hinge except that there is always a fixed moment constraint
b) frictionless mechanical hinge except that there is no fixed moment constraint
c) friction mechanical hinge except that there is no fixed moment constraint
d) frictionless mechanical hinge except that there is always a fixed moment constraint
Answer
Answer: d [Reason:] Plastic hinge behaves like a frictionless mechanical hinge except that there is always a fixed moment constraint which is equal to plastic moment capacity of the section.
9. Which of the following is true about hinged length?
a) value of moment adjacent to yield zone is more than yield moment up to hinged length of structural member
b) value of moment adjacent to yield zone is less than yield moment up to hinged length of structural member
c) value of moment adjacent to yield zone is half the yield moment up to hinged length of structural member
d) value of moment adjacent to yield zone is equal to yield moment up to hinged length of structural member
Answer
Answer: a [Reason:] The value of moment adjacent to yield zone is more than yield moment up to a certain length of structural member. This length is called hinged length.
10. Hinged length depends upon
a) weight of member
b) type of connection
c) type of loading
d) number of bolts used in connection
Answer
Answer: c [Reason:] The hinged length depends upon the type of loading and the geometry of cross section of structural member.
11. What is the hinged length for simply supported rectangular beam of span L with central concentrated load?
a) L/√2
b) 2L
c) L/2
d) L/3
Answer
Answer: d [Reason:] For simply supported rectangular beam with central concentrated load, the hinged length is equal to one-third of the span.
12. What is the hinged length for simply supported rectangular beam of span L with uniformly distributed load?
a) L/√3
b) L/√2
c) L/2
d) L/5
Answer
Answer: a [Reason:] For simply supported rectangular beam with uniformly distributed load, the hinged length is equal to span/√3.
Set 2
1. The design action effects for design basis earthquake loads is obtained by _____________
a) Elastic Analysis
b) Plastic Analysis
c) Advanced Analysis
d) Dynamic Analysis
Answer
Answer: a [Reason:] The design action effects for design basis earthquake loads is obtained by elastic analysis. The maximum credible earthquake load can be assumed to correspond to load at which significant plastic hinges are formed and shall be obtained by plastic analysis.
2. What is non-sway frame?
a) transverse displacement of one end of member relative to other end is not effectively prevented
b) longitudinal displacement of one end of member relative to other end is not effectively prevented
c) transverse displacement of one end of member relative to other end is effectively prevented
d) longitudinal displacement of one end of member relative to other end is effectively prevented
Answer
Answer: c [Reason:] In non-sway frame, transverse displacement of one end of member relative to other end is effectively prevented. Example : in triangulated frames and trusses, in plane stiffeners is provided by bracings or by shear walls, etc.
3. What is sway frame?
a) longitudinal displacement of one end of member relative to other end is not effectively prevented
b) transverse displacement of one end of member relative to other end is effectively prevented
c) longitudinal displacement of one end of member relative to other end is effectively prevented
d) transverse displacement of one end of member relative to other end is not effectively prevented
Answer
Answer: d [Reason:] In sway frame, transverse displacement of one end of member relative to other end is not effectively prevented. Such members and frames occur in structure which depend on flexural action of members to resist lateral loads and sway.
4. Which of the following is true about rigid construction?
a) connections between members at their junction does not have sufficient rigidity
b) connections between members at their junction have sufficient rigidity
c) members are not connected
d) connection between members at junction will not resist any moment
Answer
Answer: b [Reason:] In rigid construction, connections between members at their junction have sufficient rigidity to hold angles between members connected at joint, unchanged under load.
5. In simple construction, connection between members at their junction will __________
a) resist moment
b) resist force
c) not resist force
d) not resist moment
Answer
Answer: d [Reason:] In simple construction, connection between members at their junction will not resist any appreciable moment and shall be assumed to be hinged.
6. What percent of factored load is notional horizontal force?
a) 0.5%
b) 1%
c) 10%
d) 20%
Answer
Answer: a [Reason:] Notional horizontal force = 0.5% of factored dead load + vertical imposed load at that level. Notional horizontal force are applied to analyse a frame subjected to gravity loads, considering sway stability of frame.
7. In first order elastic analysis, equilibrium is expressed in terms of ___________
a) geometry of deformed structure
b) geometry of undeformed structure
c) geometry of both deformed and undeformed structure
d) geometry of any structure
Answer
Answer: b [Reason:] In first order elastic analysis, equilibrium is expressed in terms of geometry of undeformed structure. This assumption is valid when elastic displacement are small compared to dimensions of structure.
8. Which of the following is not an assumption of first order elastic analysis of rigid jointed frame?
a) materials behave linearly
b) yielding effects can be ignored
c) member instability effects cannot be ignored
d) frame instability effects can be ignored
Answer
Answer: c [Reason:] The following are the assumptions of first order elastic analysis of rigid jointed frame : (i) materials behave linearly, (ii) yielding effects can be ignored, (iii) member instability effects such as those caused by axial compression can ignored, (iv) frame behaves linearly frame instability effects such as those caused by moments due to horizontal frame deflection, etc can be ignored.
9. Basic objective of second order elastic analysis is :
a) to calculate external force equilibrium in deformed geometry of structure
b) to calculate internal force equilibrium in deformed geometry of structure
c) to calculate external force equilibrium in undeformed geometry of structure
d) to calculate internal force equilibrium in undeformed geometry of structure
Answer
Answer: b [Reason:] Basic objective of second order elastic analysis is to calculate internal force equilibrium in deformed geometry of structure. In this P-Δ effect is also considered.
10. First order analysis yields bending moment equal to those for a _____
a) cantilever beam
b) continuous beam
c) fixed beam
d) simply supported beam
Answer
Answer: d [Reason:] First order analysis yields bending moment equal to those for a simply supported beam since material and frame are assumed to behave linearly and member instability effects and frame instability effects can be ignored.
11. When plastic analysis is used, the yield stress of grade of steel used shall not exceed _____
a) 250 MPa
b) 500 MPa
c) 450 MPa
d) 800 MPa
Answer
Answer: c [Reason:] When plastic analysis is used, the yield stress of grade of steel used shall not exceed 450 MPa and the stress strain distribution of steel shall not be significantly different from those obtained for steels complying with IS 2062.
Set 3
1. 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 ≥ Pcr
Answer
Answer: c [Reason:] 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.
2. 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 load
Answer
Answer: a [Reason:] 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.
3. 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 stable
Answer
Answer: d [Reason:] The secondary path shows that plate can carry loads higher than elastic critical load. The secondary path for a plate is stable.
4. 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 strain
Answer
Answer: b [Reason:] 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.
5. 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)
Answer
Answer: d [Reason:] 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.
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)].
Answer
Answer: a [Reason:] 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.
7. 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 elements
Answer
Answer: b [Reason:] 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.
8. What is squash load?
a) yield stress + area of cross section
b) yield stress – area of cross section
c) yield stress / area of cross section
d) yield stress x area of cross section
Answer
Answer: d [Reason:] When length of column is relatively small and its component elements are prevented from local buckling, then column will be able to attain its full strength or squash load (squash load = yield stress x area of cross section).
9. 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 load
Answer
Answer: a [Reason:] 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.
10. 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 section
Answer
Answer: c [Reason:] 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.
11. 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 sections
Answer
Answer: a [Reason:] 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.
12. 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 members
Answer
Answer: d [Reason:] 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.
Set 4
1. Single bay portal frames with fixed bases have _______
a) two redundancies
b) three redundancies
c) four redundancies
d) zero redundancies
Answer
Answer: b [Reason:] Single bay portal frames with fixed bases have three redundancies and require four hinges to produce a mechanism.
2. 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+2
Answer
Answer: c [Reason:] If order of indeterminacy is r, then minimum number of plastic hinges required for total collapse is r+1.
3. 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 metacentre
Answer
Answer: a [Reason:] 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.
4. Which of the following relation is correct for pin based frames?
a) Mp =γL(wL2/8)[1-k+(1+k)0.5].
b) Mp =γL(wL2/8)[1+k+(1-k)0.5].
c) Mp =γL(wL2/8)[1+k-(1+k)0.5].
d) Mp =γL(wL2/8)[1+k+(1+k)0.5].
Answer
Answer: d [Reason:] 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(wL2/8)[1+k+(1+k)0.5]
For fixed=-base frame : Mp =γL(wL2/8)[1/[1+0.5k+(1+k)0.5]], where γL = 1.7 global load factor, k = h2/h1.
5. 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 mechanism
Answer
Answer: a [Reason:] 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.
6. 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 zero
Answer
Answer: c [Reason:] The presence of axial equation implies that sum of tension and compression forces is not zero and hence following equation is used : ∫A fydA – P = 0, where P is the axial force.
7. 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) = 1
Answer
Answer: a [Reason:] 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.
8. 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 > 0.15
Answer
Answer: d [Reason:] 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 .
9. When a member is subjected to uniaxial tensile or compressive stress in presence of shear stress τ , yield occurs when ___
a) fy2 = f2 – 3 τ
b) fy2 = f2 + 3 τ2
c) fy2 = f2 – 3 τ2
d) fy2 = f2 + 3 τ
Answer
Answer: b [Reason:] When a member is subjected to uniaxial tensile or compressive stress in presence of shear stress τ, yield occurs when fy2 = f2 + 3 τ2.
10. Yield in pure shear occurs when ______
a) 0.58 fy
b) 1.58 fy
c) 2.8 fy
d) 3.5 fy
Answer
Answer: a [Reason:] Yield in pure shear occurs when τy = fy/√3 = 0.58 fy.
11. 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 ].
Answer
Answer: c [Reason:] When full plasticity is produced, the stress in web will be equal to fw, fw =√(fy2 -3τw2 ) = fy√[1-(w/τy)2 ] where τy is shear strengths when yield is in pure shear.
12. Loss of web capacity is given by ___
a) Zpw / (fy – fw)
b) Zpw (fy – fw)
c) Zpw (fy + fw)
d) Zpw /(fy + fw)
Answer
Answer: b [Reason:] Loss of web capacity is given by Zpw (fy – fw) = Zpw fy { 1-√[1-(τw/τy)2] }.
Set 5
1. What are pin connections?
a) structural members connected by bolts
b) structural members connected by cylindrical pins
c) structural members connected by bolts and pins
d) structural members connected by welding
Answer
Answer: b [Reason:] When two structural members are connected by means of cylindrical-shaped pins, the connection is known as pinned connection. It resists horizontal and vertical movement, but not moment.
2. Pin connections are provided when _______ required.
a) hinge joint
b) fixed joint
c) irrotational joint
d) rigid joint
Answer
Answer: a [Reason:] Pin connections are provided when hinged joints are required , where zero moments or free rotation is desired and horizontal and vertical movement are not desired.
3. Pins used for the connection _________
a) does not affect secondary stresses
b) increase secondary stresses
c) reduce secondary stresses
d) doubles secondary stresses
Answer
Answer: c [Reason:] Pins used for the connection reduce secondary stresses. It serves the same purpose as shank of bolt.
4. Forces acting on pin are ______ those on bolt
a) less than
d) equal to
c) half the force
d) greater than
Answer
Answer: d [Reason:] Since only one pin is present in the connection, forces acting on pin are greater than those on bolt.
5. In which of the following cases pin connections are not used?
a) truss bridge girders
b) hinged arches
c) tall buildings
d) diagonal bracing connection
Answer
Answer: c [Reason:] Pin connections are used in following cases : (i) truss bridge girders, (ii) hinged arches, (iii)tie rod connection in water tanks, (iv)as diagonal bracing connections in beams and columns, (v)chain-link cables suspension bridges.
6. Shear capacity of pin when rotation is allowed is given by
a) 0.5fypA
b) 0.6fypA
c) 0.7fypA
d) 0.8fypA
Answer
Answer: a [Reason:] Shear capacity of pin is given by (i) 0.5fypA, when rotation is required, (ii) 0.6fypA, when rotation is not required, where fyp=design strength of pin, A = cross sectional area of pin.
7. Bearing capacity of pin when rotation is not allowed is given by
a) 0.8fypdt
b) 0.6fypdt
c) 0.7fypdt
d) 1.5fypdt
Answer
Answer: d [Reason:] Bearing capacity of pin given by (i) 1.5fypA, when rotation is not required, (ii) 0.8fypdt, when rotation is not required, where fyp=design strength of pin.
8. Moment capcity of pin when rotation is not allowed is given by
a) 0.8fypZ
b) 0.6fypZ
c) 1.5fypZ
d) 2.0fypZ
Answer
Answer: c [Reason:] Moment capacity of pin given by (i) 1.5fypZ, when rotation is not required, (ii) 1.0fypdt, when rotation is not required, where fyp=design strength of pin, Z=section modulus of the pin.
9. Members joined by pin connections are separated some distance _____
a) to allow friction
b) to allow for bolt heads
c) to allow bending
d) to allow to be removed
Answer
Answer: b [Reason:] Members joined by pin connections are separated some distance (i) to prevent friction, (ii) to allow for bolt heads, if the members are built up, (iii) to facilitate painting.
10. Design of pin connections is primarily governed by
a) shear
b) bending
c) flexure
d) friction
Answer
Answer: c [Reason:] Large bending moments are generated since members joined by pin connections are separated some distance. So the pin diameter is generally governed by flexure.
