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Multiple choice question for engineering

Set 1

1. What is slenderness ratio of a tension member?
a) ratio of its least radius of gyration to its unsupported length
b) ratio of its unsupported length to its least radius of gyration
c) ratio of its maximum radius of gyration to its unsupported length
d) ratio of its unsupported length to its maximum radius of gyration

View Answer

Answer: b [Reason:] Slenderness ratio of tension member is ratio of its unsupported length to its least radius of gyration. This limiting slenderness ratio is required in order to prevent undesirable lateral movement or excessive vibration.

2. What is the maximum effective slenderness ratio for a tension member in which stress reversal occurs?
a) 180
b) 200
c) 280
d) 300

View Answer

Answer: a [Reason:] The maximum effective slenderness ratio for a tension member in which stress reversal occurs due to loads other than wind or seismic forces is 180.

3. What is the maximum effective slenderness ratio for a member subjected to compressive forces resulting only from combination of wind/earthquake actions?
a) 180
b) 200
c) 340
d) 250

View Answer

Answer: d [Reason:] The maximum effective slenderness ratio for a member subjected to compressive forces resulting only from combination of wind or earthquake actions, such that the deformation of such member does not adversely affect stresses in any part of structure is 250.

4. What is the maximum effective slenderness ratio for a member normally acting as a tie in roof truss or a bracing member?
a) 180
b) 200
c) 350
d) 400

View Answer

Answer: c [Reason:] The maximum effective slenderness ratio for a member normally acting as a tie in roof truss or a bracing member, which is not considered when subject to stress reversal resulting from action of wind or earthquake forces is 350.

5. What is the maximum effective slenderness ratio for members always in tension?
a) 400
b) 200
c) 350
d) 150

View Answer

Answer: a [Reason:] The maximum effective slenderness ratio for members always in tension other than pre-tensioned members is 400.

6. The limits specified for slenderness ratio are not
a) applicable to cables
b) applicable to angle sections
c) applicable to built-up sections
d) applicable to circular sections

View Answer

Answer: a [Reason:] The limits specified for slenderness ratio in the IS code are not applicable to cables. They are applicable to angle sections, built-up sections, circular sections.

7. The displacement of tension member under service load is given by
a) PLEAg
b) PLE/Ag
c) PL/EAg
d) P/LEAg

View Answer

Answer: c [Reason:] The displacement, that is increase in length of tension member, under service load is given by Δ = PL/EAg, where Δ = Elongation of member in mm, P= unfactored axial load in N, L = length of member in mm, E = elastic modulus = 2×105MPa, Ag = gross cross sectional area of member in mm2.

8. What is gross section yielding?
a) considerable deformation of the member in longitudinal direction may take place before it fractures, making the structure unserviceable
b) considerable deformation of the member in longitudinal direction may take place before it fractures, making the structure serviceable
c) considerable deformation of the member in lateral direction may take place before it fractures, making the structure unserviceable
d) considerable deformation of the member in lateral direction may take place before it fractures, making the structure serviceable

View Answer

Answer: a [Reason:] Tension member without bolt holes can resist loads up to ultimate load without failure. But such a member will deform in longitudinal direction considerably(10-15% of its original length)before fracture and the structure becomes unserviceable.

9. What is net section rupture failure?
a) rupture of member when the cross section reaches yield stress
b) rupture of member when the cross section reaches ultimate stress
c) rupture of member when the cross section reaches less value than yield stress
d) rupture of member when the cross section is reaches very less value than ultimate stress

View Answer

Answer: b [Reason:] The point adjacent to hole reaches yield stress first when tension member with hole is loaded statically. The stress at that point remains constant and each fibre away from hole progressively reaches yield stress on further loading. With increasing load, deformations continue until finally rupture of member occurs when entire net cross section of member reaches ultimate stress.

10. The tensile stress adjacent to hole will be ____________
a) about five times the average stress on the net area
b) about half the average stress on the net area
c) equal to average stress on the net area
d) about two to three times the average stress on the net area

View Answer

Answer: d [Reason:] From the theory of elasticity, the tensile stress adjacent to hole will be about two to three times the average stress on the net area, depending upon the ratio of diameter of hole to the width of plate normal to direction of stress.

11. What is stress concentration factor?
a) ratio of average stress to maximum elastic stress
b) product of average stress and maximum elastic stress
c) ratio of maximum elastic stress to average stress
d) twice the average stress

View Answer

Answer: c [Reason:] The ratio of maximum elastic stress to average stress (fmax/favg)is called as stress concentration factor. It becomes very significant when repeated applications of load may lead to fatigue failure or where there is possibility of brittle fracture of tension member under dynamic load. It may minimised by providing suitable joint and member details.

12. What is block shear failure?
a) failure of fasteners occurs along path involving tension on one plane and shear on perpendicular plane along fasteners
b) failure of member occurs along path involving tension on one plane and shear on perpendicular plane along fasteners
c) failure of member occurs along path involving tension on one plane and shear on parallel plane along fasteners
d) failure of fasteners occurs along path involving tension on one plane and shear on parallel plane along fasteners

View Answer

Answer: b [Reason:] Failure of member occurs along path that involves (i) tension on one plane and (ii) shear on perpendicular plane along fasteners in block shear failure mode.

13. The possibility of block shear failure increases by
a) larger connection length
b) increasing the number of bolts per connection
c) with use of low strength bolts
d) with use of high bearing strength material

View Answer

Answer: d [Reason:] The block shear failure becomes a possible mode of failure when material bearing strength and bolt shear strength are higher. When high bearing strength of material and high strength bolts are used, only few bolts are required in connection. Decreasing number of bolts per connection results in smaller connection length, but the possibility of block shear failure increases.

Set 2

1. Which of the following equation is correct for bolt subjected to combined shear and tension?
a) (Vsb/Vdb)2 + (Tsb/Tdb)2 ≤ 1
b) (Vsb/Vdb)2 + (Tsb/Tdb)2 ≥ 1
c) (Vsb/Vdb) + (Tsb/Tdb) ≤ 1
d) (Vsb/Vdb) + (Tsb/Tdb) ≥ 1

View Answer

Answer: a [Reason:] Bolt required to satisfy both shear and tension at the same time should satisfy (Vsb/Vdb)2 + (Tsb/Tdb)2 ≤ 1 , where Vsb= factored shear force, Vdb = design shear capacity, Tsb = factored tensile force, Tdb= design tensile capacity.

2. Shear Capacity of HSFG bolts is
a) μfnekhFo
b) μfnekhFoγmf
c) μfnekhoγmf
d) μfnekhFomf

View Answer

Answer: d [Reason:] Shear Capacity of HSFG bolts is μfnekhFomf, where μf = coefficient of friction(0.55), ne = number of frictional interfaces offering frictional resistance to slip, kh = 1 for fasteners in clearance holes, 0.85 for fasteners in over sized and short slotted holes, γmf = 1.1 (slip resistance designed at service load), 1.25 (slip resistance designed at ultimate load), Fo = minimum bolt tension = Anbf0 , where Anb = net area of bolt, f0 = 0.7fub , fub = ultimate tensile stress of bolt.

3. The maximum number of bolts of diameter 25mm that can be accomodated in one row in 200mm wide flat are:
a) 2
b) 3
c) 4
d) 5

View Answer

Answer: b [Reason:] Minimum end distance = 2.5×25 = 62.5mm Number of bolts that can be accommodated = (200-2×62.5)/25 = 3 bolts.

4. Calculate strength in shear of 16mm diameter of bolt of grade 4.6 for lap joint
a) 50 kN
b) 40 kN
c) 29 kN
d) 59 kN

View Answer

Answer: c [Reason:] Bolts will be in single shear. Diameter of bolt = 16mm. Net area = 0.78x(π/4)x162=156.83mm2. Strength of bolt in shear = Anbfub/(√3 x 1.25) = 156.83x400x10-3/1.25x√3 = 28.97kN.

5. What is the value of kb in nominal bearing strength for a bolt of 20mm diameter of grade 4.6?
a) 0.5
b) 1
c) 0.97
d) 2

View Answer

Answer: a [Reason:] diameter of bolt = 20mm, diameter of hole = 20+2 =22mm e=1.5×22=33mm, p=2.5×20=50mm e/3d0 = 33/(3×22) = 0.5, p/3d0 -0.25 = 50/(3×22) -0.25=0.5, fub /fb = 400/410=0.975 kb = minimum of (e/3d0 , p/3d0 -0.25, fub /fb, 1) = 0.5.

6. Calculate bearing strength of 20mm diameter bolt of grade 4.6 for joining main plates of 10mm thick using cover plate of 8mm thick using single cover butt joint.
a) 70.26 kN
b) 109.82 kN
c) 50.18 kN
d) 29.56 kN

View Answer

Answer: c [Reason:] diameter of bolt = 16mm, diameter of hole =16+2 =18mm e=1.5×18=27mm, p=2.5×16=40mm e/3d0 = 27/(3×18) = 0.5, p/3d0 -0.25 = 40/(3×18) -0.25=0.49, fub /fb = 400/410=0.975 kb = minimum of (e/3d0, p/3d0 -0.25, fub /fb,1) = 0.49 bearing strength = 2.5kbdtfu/1.25 = 2.5×0.49x16x8x400x10-3/1.25 = 50.18 kN.

7. Find the number of HSFG bolts of diameter 20mm, grade 88 for connection of member carrying factored tensile load of 200kN when no slip is permitted.
a) 5
b) 4
c) 3
d) 2

View Answer

Answer: b [Reason:] Fo=0.7fubAnb=0.7x800x0.78x(π/4)x202x10-3=137.22 kN Assume μf=0.5, ne=1, kh=1 Slip resistance of bolt = μf ne kh Fo/1.25 = 0.5x1x1x137.22/1.25 =54.88 kN Number of bolts required = 200/54.88 = 3.64 = 4(approximately).

8. What is the efficiency of joint when strength of bolt per pitch length is 60kN and strength of plate per pitch length is 150kN?
a) 25%
b) 30%
c) 35%
d) 40%

View Answer

Answer: d [Reason:] Efficiency = (strength of bolt per pitch length/ strength of plate per pitch length)x100 = 60×100/150 = 40%.

9. Strength of bolt is
a) minimum of shear strength and bearing capacity of bolt
b) maximum of shear strength and bearing capacity of bolt
c) shear strength of bolt
d) bearing capacity of bolt

View Answer

Answer: a [Reason:] Strength of bolt is minimum of shear strength and bearing capacity of bolt. Design shear strength = nominal shear capacity/1.25, Design bearing strength = nominal bearing capacity/1.25.

10. Prying forces are
a) friction forces
b) shear forced
c) tensile forces
d) bending forces

View Answer

Answer: c [Reason:] In connections subjected to tensile stresses, the flexibility of connected parts can lead to deformations that increases tension applied to bolts. This additional tension is called prying force.

Set 3

1. What is static 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 load

View Answer

Answer: b [Reason:] For a given frame and loading if there exists any distribution of bending moments throughout the frame which is both safe and statically admissible with set of load P, then value of load P must be less than or equal to collapse load.

2. 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 conditions

View Answer

Answer: d [Reason:] 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.

3. 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 load

View Answer

Answer: a [Reason:] 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.

4. 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 conditions

View Answer

Answer: c [Reason:] 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.

5. 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 load

View Answer

Answer: c [Reason:] For given frame and loading, if at least one safe and statically admissible bending moment distribution can be found, and if in this distribution bending 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.

6. 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 satisfied

View Answer

Answer: d [Reason:] Load that satisfies both static and kinematic theorem at the same time is called correct collapse load.

7. 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 assumed

View Answer

Answer: b [Reason:] In kinematic method, a mechanism is assumed and virtual work equations are used to determine the collapse load.

8. 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 – r

View Answer

Answer: d [Reason:] 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.

9. In static method of analysis, moment at any section is _______ plastic moment capacity.
a) greater than
b) two times
c) less than
d) three times

View Answer

Answer: c [Reason:] 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.

10. 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 + W

View Answer

Answer: a [Reason:] Load factor is ratio of collapse load to working load. It is represented by, F = Wc / W.

11. 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)v

View Answer

Answer: c [Reason:] 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.

12. 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) 3.98

View Answer

Answer: a [Reason:] Load factor, F = (fy/fbc)v. F = (1/0.66) x 0.14 = 1.27.

Set 4

1. Which of the following is the attribute of ideal column according to Euler?
a) material is non homogenous
b) material is isotropic
c) load does not act along centroidal axis
d) column ends are fixed

View Answer

Answer: b [Reason:] According to Euler, the following are the attributes of ideal column: (i) Material is isotropic, homogenous and is assumed to be perfectly elastic, (ii) column is initially straight and load acts along centroidal axis (no eccetricity of loads), (iii) column has no imperfections, (iv) column ends are hinged.

2. Euler critical load for column with both the ends hinged is given by
a) Pcr = 2π2EI/L2
b) Pcr= π2EIL2
c) Pcr = 2π2EIL2
d) Pcr = π2EI/L2

View Answer

Answer: d [Reason:] The Euler critical load for column with both the ends hinged is given by Pcr= π2EI/L2, where E is elastic modulus of material, I is moment of inertia and L is length of column.

3. Which of the following is true?
a) increasing member length causes reduction in stiffness
b) decreasing member length causes reduction in stiffness
c) member with high stiffness will buckle early than that with low stiffness
d) stiffness of member is not influenced by amount and distribution of material in cross section of column

View Answer

Answer: a [Reason:] Member with low stiffness will buckle early than that the one with high stiffness. Increasing member length causes reduction in stiffness. Stiffness of member is influenced by amount and distribution of material in cross section of column, the value of radius of gyration reflects the way in which material is distributed.

4. Euler critical load for column with both the ends fixed is given by
a) Pcr = 2π2EI/L2
b) Pcr = π2EIL2
c) Pcr = 4π2EI/L2
d) Pcr= π2EI/L2

View Answer

Answer: c [Reason:] The Euler critical load for column with both the ends hinged is given by Pcr = 4π2EI/L2, where E is elastic modulus of material, I is moment of inertia and L is length of column, effective length of column in this case = 2L.

5. What is the relation between critical stress and slenderness ratio?
a) critical stress is directly proportional to slenderness ratio
b) critical stress is inversely proportional to slenderness ratio
c) critical stress is square of slenderness ratio
d) critical stress is cube of slenderness ratio

View Answer

Answer: b [Reason:] fcr = Pcr/Ag = π2E/λ2, critical stress is inversely proportional to slenderness ratio of column and very large values can be obtained by using L/r → 0 .

6. Why is built up section used?
a) to sustain seismic loads only
b) for aesthetic appearance
c) used when rolled section do not furnish required sectional area
d) for resisting bending moment

View Answer

Answer: c [Reason:] Size and shape of rolled sections are limited because of limitation of rolling mills. When rolled section do not furnish required sectional area or when special shape or large radius of gyration is required in two different direction, a built up section is used.

7. The shear in column may not be due to
a) material used in column
b) lateral loads from wind
c) lateral loads from earthquake
d) end eccentricity of load

View Answer

Answer: a [Reason:] The shear in column may be due to (i) lateral loads from wind, earthquake, gravity or other loads, (ii)slope of column with respect to line of thrust due to both unintentional and initial curvature and increased curvature during bending, (iii)end of eccentricity of load due to either end connection or fabrication imperfections.

8. Which of the following is not true?
a) function of lacing and battens is to hold main component members
b) tie plates are provided at top and bottom of column
c) lacing bars and batten plates are not designed as load carrying elements
d) lacing bars and batten plates are designed as load carrying elements

View Answer

Answer: d [Reason:] Lacing bars and batten plates are not designed as load carrying elements. Their function is to hold main component members of built up column in relative position and equalize stress distribution in them. At ends and at intermediate points where it is necessary to interrupt the lacings, open sides are connected with tie plates. Tie plates are also provided at top and bottom of column.

9. Which of the following is true?
a) built up column lacings or battens are uneconomical if load carrying members permit greater reduction in weight than what is added by lacing or batten
b) built up column lacings or battens are economical if load carrying members permit greater reduction in weight than what is added by lacing or batten
c) no related shear stress force in plane of cross section
d) built up column designed as axially loaded column can never be eccentrically loaded

View Answer

Answer: b [Reason:] Built up column lacings or battens are economical if load carrying members permit greater reduction in weight than what is added by lacing or batten. Built up column designed as axially loaded column may be accidentally loaded eccentrically or may have initial crookedness. Variable bending moment will be induced in such column because of eccentricity between centroidal axis of column and line of action of applied load due to this, there will be related shear stress force in plane of cross section and in connecting system.

10. The buckling strength of latticed column is ___________ that of solid column having ame area and same slenderness ratio
a) smaller than
b) greater than
c) equal to
d) cannot be compared

View Answer

Answer: a [Reason:] The buckling strength of latticed column is smaller than that of solid column having same area and same slenderness ratio provided that solid column does not buckle locally because of thinness of material. This is because shear component of axial load produces deformation in lattice which tends to reduce overall stiffness of column and therefore reduce buckling strength of column.

11. Which of the following condition is not considered in design of built up column?
a) buckling of column as whole
b) failure of lattice member
c) material to be used for fabrication
d) buckling of component column

View Answer

Answer: c [Reason:] In design of built up column, following conditions are considered : (i) buckling of column as whole under axial load, (ii) buckling of component column, (iii) failure of lattice member, (iv) distortion of cross section.

Set 5

1. The slenderness ratio of each member when placed back-to-back or separated by small distance shall be
a) greater than 40
b) not greater than 40
c) 0.8 times the slenderness ratio of column as a whole
d) greater than 50

View Answer

Answer: b [Reason:] Two rolled sections placed back-to-back or separated by small distance should be connected together by rivets/bolts/welds so that slenderness ratio of each member when placed back-to-back or separated by small distance is not greater than 40 or 0.6 times the slenderness ratio of column as a whole.

2. Minimum number of bolts for connecting end of strut is
a) 0
b) 3
c) 1
d) 2

View Answer

Answer: d [Reason:] Ends of strut should be connected with minimum of 2 bolts/rivets or equivalent length of weld length (length must not be less than maximum width of member).

3. Which of the following is true?
a) when there is small spacing between the two sections placed back-to-back, washers and packing should be provided
b) when there is small spacing between the two sections placed back-to-back, washers and packing should not be provided
c) there should be additional connection in between along the length of member
d) when leg of angles greater than 125mm wide or web of channel is mm wide, minimum bolt is sufficient for connection

View Answer

Answer: a [Reason:] There should be minimum of two additional connection in between, spaced equidistant along the length of member. When there is small spacing between the two sections placed back-to-back, washers(in case of bolts) and packing(in case of welding) should be provided to make connection. When leg of angles greater than 125mm wide or web of channel is mm wide, minimum two bolts/rivets should be for connection.

4. Minimum diameter of bolt when member is less than 16mm thick is
a) 8
b) 10
c) 22
d) 20

View Answer

Answer: c [Reason:] Rivets/bolts should not be less than 16mm in diameter for member less than 10mm thick, 20mm in diameter for member less than 16mm thick and 22mm in diameter for member more than 16mm thick.

5. Which of the following is not true?
a) spacing of tack bolt should be less than 600mm
b) spacing of tack bolt should be greater than 600mm
c) if bolts are used, they should be spaced longitudinally at less than 4 times the bolt diameter
d) connection should extend at least 1.5 times the width of the member

View Answer

Answer: b [Reason:] Spacing of tack bolt should be less than 600mm. If bolts are used, they should be spaced longitudinally at less than 4 times the bolt diameter. Connection should extend at least 1.5 times the width of the member.

6. Members connected back-to-back connected by bolts should be
a) not be used
b) subjected to transverse loading in plane perpendicular to bolted surface
c) subjected to twice the transverse loading in plane perpendicular to bolted surface
d) not subjected to transverse loading in plane perpendicular to bolted surface

View Answer

Answer: d [Reason:] Members connected back-to-back connected by bolts should not be subjected to transverse loading in plane perpendicular to riveted/bolted/welded surface.

7. For members placed back-to-back, the spacing of bolt should not exceed
a) 12t
b) 16t
c) 18t
d) 20t

View Answer

Answer: a [Reason:] For members placed back-to-back, the spacing of bolt should not exceed 12t or 200mm, where t is thickness of member.

8. Longitudinal spacing between intermittent welds used for connection should be
a) greater than 18t
b) greater than 16t
c) not greater than 16t
d) equal to 18t

View Answer

Answer: c [Reason:] Longitudinal spacing between intermittent welds used for connection should not be greater than 16t, where t is thickness of thinner connection.