Multiple choice question for engineering
Set 1
1. A flange of radius 200mm is fastened to the machine screw by means of four cap screws of pitch circle radius 150mm. The external force P is 20kN which is loaded at 160mm from the machine screw. There are two dowel pins to take shear load. Determine the maximum effective tensile force acting on the bolt. The four bolts are equally spaced radially.
a) 4777.6N
b) 5777.6N
c) 6777.6N
d) 7777.7N
Answer
Answer: a [Reason:] Maximum force F=2Pl[a+bCos(90/2)] / 4(2a²+b²).
2. A flange of radius 200mm is fastened to the machine screw by means of four cap screws of pitch circle radius 150mm. The external force P is 20kN which is loaded at 160mm from the machine screw. There are two dowel pins to take shear load. Determine the core diameter of cap screw if permissible tensile stress in cap screw is 40N/mm².
a) None of the listed
b) 11.21mm
c) 12.33mm
d) 14.12mm
Answer
Answer: c [Reason:] Maximum tensile force=4777.6N or πd² x σ/4=4777.6N.
3. Answer the below questions i-iii with respect to this figure
A round flange bearing is fastened to the machine frame by means of 4 cap screws. The angle made by central axis of bolt 2 and horizontal is θ. Also the radii of flange and pitch circle are a & b respectively. Given: Sum of squares of distance of bolts from axis of tilt is 2a² +b²
3. i. Are all the four bolts under equal effective tensile load?
a) Yes
b) No
c) Depends on magnitude of load
d) Cannot be stated
Answer
Answer: b [Reason:] All the bolts are at different distance from the axis of tilt and hence unequally loaded.
4. ii. Determine the angle θ so that the force in the bolt 1 has maximum value.
a) 180’
b) 90’
c) 0’
d) 45’
Answer
Answer:a [Reason:] F₁=2Plx(a-b Cos θ)/4(2a²+b²), hence for force to be max Cosθ=-1 or θ=180’.
5. iii. Which bolt is under maximum stress if dowel pins are used to relieve cap screws from shear stress?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: c [Reason:] The screw farthest from the axis of tilt is under maximum stress.
6. Which of the following statements are true about eccentrically loaded circular base?
a) All bolts are symmetrically spaced with equal angle on either side of vertical
b) Bolts are not preloaded
c) Bearing is rigid
d) Dowel pins are used to relieve stresses
Answer
Answer: a [Reason:] Bolts are symmetrically placed only in a special case.
7. Certain types of radial bearings can also take thrust load.
a) True
b) False
Answer
Answer: a [Reason:] Sometimes radial bearings can take thrust load and vice versa.
8. Balls and the races of the deep groove ball may roll freely without any sliding.
a) True
b) False
Answer
Answer: a [Reason:] There is a point contact between the balls and the races.
Set 2
1. If core diameter of bolt is 13.8cm the it’s nominal diameter is given by?
a) 17.27mm
b) 15.34mm
c) 14.67mm
d) 16.34mm
Answer
Answer: a [Reason:] D=d/0.8.
Figure 1
2. Refer to fig 1.Two plates are fastened by means of two bolts. The yield strength of bolt is 400N/mm² and factor of safety is 4. Determine the permissible shear stress in the bolts.
a) 100N/mm²
b) 50N/mm²
c) 25N/mm²
d) 75N/mm²
Answer
Answer: b [Reason:] Permissible hear stress=0.5 x 400 /4 =50N/mm².
3. Refer to fig 1. Two plates are fastened by means of two bolts. The yield strength of bolt is 400N/mm² and factor of safety is 4. Determine the size of the bolts.
a) 8mm
b) 9mm
c) 10mm
d) 11mm
Answer
Answer: a [Reason:] Permissible hear stress=0.5 x 400 /4 =50N/mm². P=2 (π xd²/4) x τ or 5000=2 x π x d² x 50/4 or d=7.97mm.
Figure 2
The structure shown is subjected to an eccentric force P=5kN and eccentricity=500mm. The horizontal distance between two bolts is 200mm and vertical distance between bolts is 150mm. The yield strength of bolts is 400N/mm² and factor of safety is 3.
4. Referring to fig 2, Determine the primary shear force.
a) 625N
b) 1250N
c) 2500N
d) 1000N
Answer
Answer: b [Reason:] Primary shear force P₁=P₂=P₃=P₄=P/4 =1250N.
5. Referring to fig 3, Determine the secondary shear force.
a) 2000N
b) 2500N
c) 1500N
d) 5000N
Answer
Answer: d [Reason:] Secondary shear force=Moment about CG x distance from CG/sum of squares of distance of bolts from CG. F= (Pe)xr₁/(r₁²+r₂²+r₃²+r₄²). Here r₁=r₂=r₃=r₄=125mm hence F= 5000 x 500/(4×125) or F=5000N.
6. Referring to figure 2, determine the resultant shear force on the bolt lying left and above the CG.
a) 4068.58N
b) 4168.58N
c) 5068.58N
d) 5168.65N
Answer
Answer: a [Reason:] After breaking shear force into primary and secondary shear force, primary acts along the vertical positively and secondary acts at an angle of 180-36.87’ ACW from the vertical. Hence net shear force= √ [5000cos36.8-1250]²+ [5000sin36.8]² =4068.58N.
7. In figure 2, determine the resultant shear force on the bolt lying right and above the CG.
a) 654334N
b) 6047.44N
c) 5047.44N
d) 5989.32N
Answer
Answer: b [Reason:] After breaking net shear force into primary and secondary shear force, primary acts along the vertical positively and secondary acts at an angle of 36.8’CW from the vertical. Hence net shear force = √ [5000cos36.8+1250]²+ [5000sin36.8]².
8. In figure 2, determine the size of the bolts.
a) 10.74mm
b) 9.23mm
c) 11.54mm
d) 8.68mm
Answer
Answer: a [Reason:] The maximum shear force to which any bolt is subjected is 6047.44N. Hence 0.5 x 400/3= 4 x 6047.44/πd² or d=10.74mm.
9. Figure 3
Answer the questions i-v while considering figure 3
9. (i) Which bolt is under maximum shear stress?
a) 1
b) 2
c) 3
d) All are under equivalent shear stress
Answer
Answer: c [Reason:] Primary shear force acts equally on the three bolts in the vertically upward direction while the moment is CW along CG so its effect on bolts will be ACW. Hence secondary shear force acts vertically upward on bolt 3 and vertically downward on bolt 1.
10. (ii) Arrange the bolts in order of decreasing shear stresses.
a) 1>2>3
b) 2>1>3
c) 3>1>2
d) 3>2>1
Answer
Answer: d [Reason:] On bolt 3,primary and secondary shear stress act in same direction, on bolt 2 there is no secondary shear stress and on bolt 1 the two act in opposite direction.
11. (iii) Determine the primary shear stress to which the bolts are subjected if P=3kN.
a) 3000N
b) 1000N
c) 2000N
d) None of the listed
Answer
Answer: b [Reason:] Primary shear force=P/3.
12. (iv) Determine the secondary shear stress acting on the bolt 3 and its direction. The bolts are equidistant having separated by 60mm and the margin to the left and right is 25mm.Also P=5kN acts at a distance of 200mm from the channel.
a) 6100N vertically up
b) 4500N vertically down
c) 6100N vertically up
d) 4500N vertically down
Answer
Answer: c [Reason:] Moment about CG=3000 x (75+30+200). On bolt 3 secondary shear force will be M x r₁/ (r₁²+r₂²) and will act in a direction perpendicular the line joining CG and bolt 3. As moment is CW about CG, o bolts its effect will be ACW.
13. (v) Determine the size of the bolts if yield strength of bolt is 400N/mm² and factor of safety is 4. The bolts are equidistant having separated by 60mm and the margin to the left and right is 25mm. Also P=5kN acts at a distance of 200mm from the channel.
a) 14.34mm
b) 13.44mm
c) 15.44mm
d) 12.66mm
Answer
Answer: b [Reason:] Clearly bolt 3 is under maximum shear stress. Net shear stress= Primary shear stress + Secondary shear stress or τ= (1000+6100) N or 0.5 x 400//4=7100 x 4/πd².
Set 3
Figure 1
Answer the following questions in context to figure 1.
1. A bracket is attached to a vertical wall by means of four rivets. Find the rivet which is under maximum stress.
a) 1 and 4
b) 2 and 3
c) 3 and 1
d) 4 and 2
Answer
Answer: a [Reason:] Primary Shear act vertically upward on all the rivets. Secondary shear is proportional to the distance from CG. Hence rivets 1 and 4 are under maximum stress.
2. Calculate the primary shear stress on each rivet if P=30kN and diameter of rivets is 15mm.
a) None of the listed
b) 21.2 N/mm²
c) 42.4N/mm²
d) 169.6 N/mm²
Answer
Answer: c [Reason:] τ₁=P/4A or 7500/A.
3. If secondary shear stress acting on any bolts is given by Cxr₁ where r₁ is the distance of bolt from CG, then find the value of C. Bolts are equidistant with spacing of 100mm. Force P=30kN.
a) 46
b) 50
c) 72
d) 64
Answer
Answer: c [Reason:] C=Pe/[r₁²+r₂²+r₃²+r₄²].
4. Calculate the effective force in vector form to which rivet 2 is subjected. Bolts are separated by 100mm and force P=30kN.
a) 7500j+3600i
b) 7500j-3600i
c) -7500j-3600i
d) None of the listed
Answer
Answer: b [Reason:] Primary force=7500j, Sec Force=Cx50(-i) where C=72.
5. Calculate the effective force in vector form to which rivet 4 is subjected. Bolts are separated by 100mm and force P=30kN.
a) 7500j+7200i
b) None of the listed
c) 7500j+7200i
d) 7500j+7200i
Answer
Answer: d [Reason:] Primary force=7500j, Sec Force=Cx100 i where C=72.
6. Calculate the diameter of the rivets if permissible shear stress is 60N/mm². Bolts are separated by 100mm and force P=30kN.
a) 10mm
b) 12mm
c) 13mm
d) 15mm
Answer
Answer: d [Reason:] Magnitude of force acting on Bolt 4 or Bolt 1=√7500²+7200² or 10396.6=πd²τ/4.
7. Are the bolts 2 and 3 under same force?
a) Yes
b) No
c) Depends on nature of force whether it is of shear or bending in nature
d) Can’t be determined
Answer
Answer: b [Reason:] Primary shear force is same but secondary shear forces direction are different in the two bolts.
8. Are the bolts 2 and 3 subjected to same magnitude of force?
a) True
b) False
Answer
Answer: a [Reason:] Although the vectors are different for the forces on two bolts but their magnitudes are same.
9. Can we use the rivets of diameter 18mm in the following case if P=30kN and bolts are separated by 100mm each. Maximum permissible shear stress is 60N/mm².
a) Always
b) Never
c) In some cases
d) Cannot be determined
Answer
Answer: a [Reason:] The minimum diameter required for the rivets is obtained 15mm. Any rivet greater than this diameter is safe to use.
Set 4
1. A mechanical component may fail as a result of which of the following
a) elastic deflection
b) general yielding
c) fracture
d) each of the mentioned
Answer
Answer: a [Reason:] Failing simply means unable to perform its function satisfactorily.
2. Type of load affects factor of safety.
a) True
b) False
Answer
Answer: a [Reason:] Dynamic load has higher factor of safety as compared to static loading.
3. For cast iron components, which of the following strength are considered to be the failure criterion?
a) Ultimate tensile strength
b) Yield Strength
c) Endurance limit
d) None of the mentioned
Answer
Answer: a [Reason:] Ultimate tensile strength is the highest stress a component can undergo before failingand hence is used as a criterion.
4. For components made of ductile materials like steel, subjected to static loading which of the following strength is used as a failure of criterion?
a) Yield strength
b) Ultimate strength
c) Endurance limit
d) None of the mentioned
Answer
Answer: a [Reason:] In elastic material there is considerable plastic deformation at yielding point.
5. Pitting occurs on _____ of the component.
a) Surface
b) Inner body
c) Inside or on surface
Answer
Answer: a [Reason:] Pitting is a process in which small holes occur on a surface of component.
6. Buckling is elastic instability which leads to sudden large lateral deflection.
a) True
b) False
Answer
Answer: a [Reason:] Definition of buckling.
7. The critical buckling load depends upon which of the following parameters?
a) Yield strength
b) Modulus of elasticity
c) Radius of gyration
d) Each of the mentioned
Answer
Answer: d [Reason:] It depends on moment of inertia(which further depends on radius of gyration), elasticity and yield strength.
8. If there are residual stresses in the material, than lower factor of safety is used.
a) True
b) False
Answer
Answer: b [Reason:] Residual stress increases the chance of failure.
9. Which of the following relationship is true? (p=Poisson’s ratio)
a) E=2G (1+p)
b) E=G (2+p)
c) E= 2(G+ p)
d) No relation exist between E, G and p
Answer
Answer: a [Reason:] Formula.
10. Modulus of rigidity for carbon steels is greater than that of grey cast iron.
a) True
b) False
Answer
Answer: a [Reason:] Modulus of rigidity is double for carbon steels as compared to grey cast iron.
11. According to principal stress theory, which option represents the correct relation between yield strength in shear (YSS) and the yield strength in tension (YST)?
a) YSS=0.5YST
b) YSS=0.577YST
c) YST=0.5YSS
d) YST=0.577YSS
Answer
Answer: a [Reason:] Shear diagonal is at 45’ and by equation of shear stress theory, the required relation is obtained.
12 .A beam subjected to bending moment undergoes which of the following stresses?
a) Compressive
b) Tensile
c) Both compressive & tensile
d) None of the mentioned
Answer
Answer: c [Reason:] The portion above the neutral axis is under compression and the portion below it is under tensile stress.
13. The bending stress varies _______ with the distance from the neutral axis.
a) Linearly
b) Inversely
c) Squarely
d) Bending stress is independent of distance from the neutral axis
Answer
Answer: a [Reason:] Bending Stress= (Bending Moment x distance from neutral axis/ Moment of inertia).
Set 5
Figure 1
Two plates are joined by double strap butt joints with rivets arrangement as shown in the figure. Width of the plate is 300mm and diameter of rivets 30mm. The thickness of the plates can be taken as 20mm.Allowable stress in tension, shear and compression are 80 N/mm²,60 N/mm² and 160N/mm².
Answer the following questions with respect to figure 1
1. Calculate the shear resistance of 1 rivet along A-A.
a) 21.2kN
b) 42.4kN
c) 84.8kN
d) None of the listed
Answer
Answer: c [Reason:] P=2x[τ x πd²/4].
2. Calculate the tensile strength of rivet along A-A.
a) 2592kN
b) 864kN
c) 216kN
d) 432kN
Answer
Answer: d [Reason:] Strength=(w-d)xtxσ.
3. Calculate the tensile strength of rivets along B-B.
a) 84.8kN
b) 468.8kN
c) 384kN
d) None of the listed
Answer
Answer: b [Reason:] To fail along B-B, rivet in outermost row must undergo shear failure. Strength = (w-2d)x σ x t + 1xP(shear) = (300-60)x80x20/1000 + 84.8.
4. Calculate the tensile strength of rivets along C-C.
a) 772kN
b) None of the listed
c) 336kN
d) 590.4kN
Answer
Answer: d [Reason:] To fail along C-C, rivets in outer rows must undergo shear failure. Strength= (w-3d)x σ x t + 3xP(shear) = (300-90)x80x20/1000 + 3×84.8.
5. Calculate efficiency of the joint.
a) 45%
b) 85%
c) 90%
d) 100%
Answer
Answer: c
Explanatio: Strength along A-A, B-B and C-C are 432kN, 468.8kN and 590.4kN and strength of solid plate=wσt=480kN. Lowes is along AA. Hence efficiency=432/480.
6. Calculate the crushing resistance of 1 rivet.
a) 1153kN
b) 108kN
c) 96kN
d) 576kN
Answer
Answer: c [Reason:] P=dtσ= 30x20x160=96kN.
7. Calculate the pitch of the rivets.
a) None of the listed
b) 210mm
c) 105mm
d) 52.5mm
Answer
Answer: c [Reason:] m+p+p+m=w where m=1.5d.
8. The type of arrangement shown is
a) Lozenge Joint
b) Lap Joint
c) Star Joint
d) Darwin’s Joint
Answer
Answer: a [Reason:] Lozenge joint is other name of diamond shape joint.
9. If circumferential stress is P, then longitudinal stress in the pressure vessel is?
a) 2P
b) P/2
c) P
d) None of the listed
Answer
Answer: b [Reason:] Circumferential stress=2x longitudinal stress.
10. If width of plate=80mm ad diameter of rivets 10mm, then pitch will be?
a) 65mm
b) 60mm
c) 50mm
d) 70mm
Answer
Answer: c [Reason:] p+2m=w or p+2×1.5×10=80.