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Multiple choice question for engineering

Set 1

1. Piston rod is an example of column.
a) True
b) False

View Answer

Answer: a [Reason:] An slender machine component having its length considerable in proportion to its width qualifies for a column criterion.

2. Bucking of column means
a) Lateral deflection
b) Axial deflection
c) Torsional deflection
d) None of the listed

View Answer

Answer: d [Reason:] Buckling is characterised by lateral deflection but it is different from lateral deflection as there is sudden lateral deflection in buckling unlike lateral deflection where there is gradual deflection.

3. Slenderness ratio is [l= length of column and k= least radius of gyration of cross section about its axis].
a) l/k
b) k/l
c) l/2k
d) k/2l

View Answer

Answer: a [Reason:] It is a ratio of length to least radius of gyration.

4. Columns with what slenderness ratio are not designed with respect to buckling but are designed for compressive stresses.
a) >1
b) <1
c) >30
d) <30

View Answer

Answer: d [Reason:] When slenderness ratio is <30, there is no effect of buckling.

5. If slenderness ratio=45, which mode of failure will dominate?
a) Buckling
b) Compressive Stresses
c) Both buckling and compressive stress
d) Can’t be stated

View Answer

Answer: a [Reason:] If slenderness ratio>30, column shall be more prone to buckling.

6. Short column and long column are classified on the basis of
a) Slenderness ratio
b) Diameter
c) Length
d) None of the listed

View Answer

Answer: a [Reason:] Slenderness ratio takes into consideration length and radius of gyration and thus is preferred.

7. Cast iron column with a slenderness ratio of 75 are
a) Short Columns
b) Long Columns
c) Very short columns
d) None of the listed

View Answer

Answer: a [Reason:] Cast iron column with a slenderness ratio <80are short columns.

8. Steel columns with a slenderness ratio of 95 are
a) Short Columns
b) Long columns
c) Very long columns
d) None of the listed

View Answer

Answer: a [Reason:] Steel columns with slenderness ratio <100 are classified in short columns.

9. Which of the following are true for End fixity coefficient
a) Dimensionless number
b) Used in Euler’s equation
c) Provides condition of restraint at two ends
d) All of the listed

View Answer

Answer: d [Reason:] It is a dimensionless number used in Euler’s equation to take into account the restraints at the two ends.

10. Value of end fixity coefficient for both hands fixed is
a) 1
b) 4
c) 2
d) 0.25

View Answer

Answer: b [Reason:] Lower the mobility higher is the end fixity component.

Set 2

1. Calculate the coefficient of fluctuation of speed if maximum speed is 2500rpm and minimum speed is 1800rpm.
a) 1.44
b) 1.33
c) 0.33
d) 0.44

View Answer

Answer: c [Reason:] C=2[(Max-Min)/(Max+Min)].

2. Calculate the coefficient of steadiness if maximum speed of flywheel is 2500rpm and minimum speed is 1800rpm.
a) 3.03
b) 0.33
c) 0.44
d) 3.99

View Answer

Answer: a [Reason:] C=1.Coeffecient of fluctuation of speed.

3. Figure 1

Answer the questions i-v with respect to figure 1.
machine-design-questions-answers-coeffecient-flywheel-q3

3. i. Calculate the mean torque T supplied by motor (Torque vs. theta graph of a flywheel is shown above).
a) 3000N-m
b) 1500N-m
c) 2000N-m
d) 2500N-m

View Answer

Answer: b [Reason:] T[1000+2000/2]x2π/2π.

4. ii. The point where angular velocity is maximum is?
a) A
b) B
c) C
d) D

View Answer

Answer: c [Reason:] The flywheel is decelerated in the portion BC and hence rpm is least at C.

5. iii. The point where velocity of flywheel is maximum is
a) A
b) B
c) C
d) D

View Answer

Answer: b [Reason:] Flywheel is accelerated in the portion AB and hence velocity is maximum at B.

6. iv. Calculate the energy output from flywheel.
a) 500π
b) 150π
c) 200π
d) 250π

View Answer

Answer: d [Reason:] U=Area of ΔBCD as max and min velocities are at B and C.

7. iv. Torque vs. theta graph of a flywheel is shown. Coefficient of fluctuation of speed is 0.2 and the mean angular velocity of the machine is 25 rad/s. Thickness of flywheel is 20mm and density of steel used is 7700kg/mᵌ.

Calculate the outer radius of the flywheel.
a) 402mm
b) 202mm
c) 540mm
d) 468mm

View Answer

Answer: a [Reason:] I=U/ω²C and I=πptR⁴/2.

8. Figure 2
The turning moment diagram of a multi cylinder engine is shown. The intercepted areas between torque developed by the engine and the mean resisting torque of the machine taken in order from one end are -300,+850 -500, + 900, -550, + 430 and -700 mm².
machine-design-questions-answers-coeffecient-flywheel-q8
Answer questions i-v with respect to figure 2

8. i. Calculate the energy at point B if energy assumed at point A is U.
a) U
b) U-300
c) U+300
d) U-150

View Answer

Answer: b [Reason:] Simple addition of energy.

9. ii. Calculate the energy at point C if energy at point A is U.
a) U-550
b) None of the listed
c) U+550
d) U+550

View Answer

Answer: c [Reason:] E=U-300+ 850.

10. iii. Calculate the energy at point F if energy at A is u.
a) U+340
b) U+260
c) U+400
d) Can’t be determined

View Answer

Answer: c [Reason:] E=U-300+850-500+900-550.

11. iv. Calculate the maximum energy.
a) U+1000
b) U+950
c) U+630
d) U+600

View Answer

Answer: b [Reason:] Maximum energy is at E. E=U-300+850-500+900.

12. iv. Calculate the point of minimum energy.
a) U+480
b) U+650
c) U-300
d) U-450

View Answer

Answer: c [Reason:] Minimum energy is at B. E=U-300.

13. v. Calculate the net energy output from the flywheel.
a) 1300mm²
b) 1400mm²
c) 1250mm²
d) 1500mm²

View Answer

Answer: c [Reason:] Maximum energy at E[U+950] and minimum at B[U-300]. Thus net energy=[U+950]-[U-300].

14. Feather key can be used to prevent axial motion between two elements.
a) True
b) False

View Answer

Answer: b [Reason:] All types of key but feather key can be used to prevent axial motion between two machine elements.

15. Keyed joints never lead to stress concentration on shafts.
a) True
b) False

View Answer

Answer: b [Reason:] Keyway results in stress concentration in the shaft and makes the part weak.

Set 3

1. Autofrettage is beneficial for the high pressure cylinder.
a) True
b) False

View Answer

Answer: a [Reason:] It increases pressure capacity of the cylinder and reduces compressive stresses.

2. Autogreggate is a process of ___ stressing the cylinder.
a) Pre
b) Post
c) Over
d) None of the listed

View Answer

Answer: a [Reason:] It is a pre stressing phenomenon to improve pressure capacity.

3. Can we pre-stress the cylinder by subjecting cylindrical portion near inner diameter in plastic range and outer diameter is still in the elastic range.
a) True
b) False

View Answer

Answer: a [Reason:] On releasing the pressure, outer portion contracts exerting pressure on the inner portion which has undergone permanent deformation. This induces residual compressive stresses at the inner surface.

4. A compound cylinder consists of
a) 2 cylinders
b) Cylinder and a jacket
c) 2 jackets
d) At least two cylinders

View Answer

Answer: b [Reason:] Inner diameter of jacket increase and outer diameter of cylinder decreases when the jacket is heated.

5. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the shrinkage pressure.
a) 5.88N/mm²
b) 2.28N/mm²
c) 4.56N/mm²
d) 3.66N/mm²

View Answer

Answer: b [Reason:] σ=P[D₃²+D₂²]/[D₃²-D₂²] where D₂=40mm and D₃=60mm.

6. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the interference [E=210kN/mm²].
a) 2.8mm
b) 4.6mm
c) 5.4mm
d) 4.8mm

View Answer

Answer: a [Reason:] Δ=PD₂[2D₂²(D₃²-D₁²)] / Ex[(D₃²-D₂²)(D₂²-D₁²)].

7. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the radial stresses due to shrink shift in jacket.
a) +2.56[(45/r) ² – 1]
b) 1.824[(45/r) ² – 1]
c) -1.824[(45/r) ² – 1]
d) None of the listed

View Answer

Answer: c [Reason:] σ(r)=-PD₂²[D₃²/4r² – 1]/ [D₃²-D₂²].

8. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the tangential stresses due to shrink shift in jacket.
a) +2.56[(45/r) ² – 1]
b) 1.824[(45/r) ² – 1]
c) -1.824[(45/r) ² – 1]
d) None of the listed

View Answer

Answer: b [Reason:] σ(t)=+PD₂²[D₃²/4r² – 1]/ [D₃²-D₂²].

9. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the radial stress due to shrink shift in inner tube.
a) +3.04[1-(15/r) ²]
b) -3.04[1-(15/r) ²]
c) -3.04[1-(10/r) ²]
d) +3.04[1-(10/r) ²]

View Answer

Answer: b [Reason:] σ(r)= σ(r)=-PD₂²[1-D₁²/4r² ]/ [D₂²-D₁²].

10. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the tangential stress due to shrink shift in inner tube.
a) +3.04[1+ (15/r) ²]
b) -3.04[1+ (15/r) ²]
c) -3.04[1-(10/r) ²]
d) +3.04[1-(10/r) ²]

View Answer

Answer: b [Reason:] σ(r)= σ(r)=-PD₂²[1+D₁²/4r² ]/ [D₂²-D₁²].

11. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². In service the cylinder is further subjected to an internal pressure of 25MPa. Calculate the radial stress in compound cylinder.
a) -3.75[(45/r) ² – 1]
b) +3.75[(45/r) ² – 1]
c) -3.75[(45/r) ² -+1]
d) +3.75[(45/r) ² +1]

View Answer

Answer: a [Reason:] σ(r)=-PD₁²[D₃²/4r² – 1]/ [D₃²-D₁²]. Here P=30.

12. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². In service the cylinder is further subjected to an internal pressure of 25MPa. Calculate the tangential stress in compound cylinder.
a) -6.75[(45/r) ² + 1]
b) +3.75[(45/r) ² +1]
c) -3.75[(45/r) ² +1]
d) -3.75[(45/r) ² -1]

View Answer

Answer: b [Reason:] σ(t)=+PD₁²[1+ D₃²/4r² ]/ [D₃²-D₁²]. Here P=30.

13. A gasket is a device to allow easy diffusion of fluids across mating surfaces of a mechanical assembly.
a) True
b) False

View Answer

Answer: b [Reason:] A gasket acts as a barrier between two mating surfaces to prevent fluid flow.

14. Asbestos gaskets like other non-metallic gaskets can be used only up to a temperature of 70⁰C.
a) True
b) False

View Answer

Answer: b [Reason:] Asbestos can be used upto a temperature of 250⁰C while all other non-metallic gaskets can be used only upto a temperature of 70⁰C.

Set 4

1. A cylinder is considered thin when the ratio of inner diameter to wall thickness is more than 5.
a) True
b) False

View Answer

Answer: b [Reason:] A cylinder is considered thin when the ratio of inner diameter to wall thickness is more than 15.

2. Tangential stress in a cylinder is given by [symbols have their usual meanings].
a) PD/2t
b) 2PD/t
c) PD/4t
d) 4PD/t

View Answer

Answer: a [Reason:] Considering equilibrium in half portion of cylinder of unit length, DP=2σt.

3. Longitudinal stress in a cylinder is given by [symbols have their usual meanings].
a) PD/2t
b) 2PD/t
c) PD/4t
d) 4PD/t

View Answer

Answer: c [Reason:] Considering equilibrium PxπD²/4=σxπDt.

4. A seamless cylinder of storage capacity of 0.03mᵌis subjected to an internal pressure of 21MPa. The ultimate strength of material of cylinder is 350N/mm².Determine the length of the cylinder if it is twice the diameter of the cylinder.
a) 540mm
b) 270mm
c) 400mm
d) 350mm

View Answer

Answer: a [Reason:] 0.03=πd²L/4 and L=2d.

5. A seamless cylinder of storage capacity of 0.03mᵌis subjected to an internal pressure of 21MPa. The ultimate strength of material of cylinder is 350N/mm².Determine the thickness of the cylinder if it is twice the diameter of the cylinder.
a) 12mm
b) 4mm
c) 8mm
d) 16mm

View Answer

Answer: c [Reason:] t=PD/2σ.

6. Cylinder having inner diameter to wall thickness ratio less than 15 are
a) Thin cylinders
b) Thick Cylinders
c) Moderate cylinders
d) None of the listed

View Answer

Answer: b [Reason:] Smaller dia to thickness ratio implies more thickness and hence these are classified under thick cylinder.

7. Lame’s equation used to find the thickness of the cylinder is based on maximum strain failure.
a) True
b) False

View Answer

Answer: b [Reason:] It is based on maximum principal stress theory.

8. Lame’s equation is generally used for ductile materials.
a) True
b) False

View Answer

Answer: b [Reason:] Lame’s equation is used to determine thickness of the brittle as it used principal stress theory.

9. The piston rod of a hydraulic cylinder exerts an operating force of 10kN. The allowable stress in the cylinder is 45N/mm². Calculate the thickness of the cylinder using Lame’s equation. Diameter of the cylinder is 40mm and pressure in cylinder is 10MPa.
a) 2.05mm
b) 4.2mm
c) 5.07mm
d) None of the listed

View Answer

Answer: c [Reason:] t=D/2[√[σ+ P /σ-P] -1 ].

10. The piston rod of a hydraulic cylinder exerts an operating force of 10kN. The allowable stress in the cylinder is 70N/mm². Calculate the thickness of the cylinder using Clavarinoe’s equation. Diameter of the cylinder is 240mm.μ=0.3 and pressure in cylinder is 15MPa.
a) 35mm
b) 30mm
c) 27mm
d) None of the listed

View Answer

Answer: c [Reason:] t=D/2[√[σ+(1-2μ) P /σ-(1+μ)P] -1 ].

Set 5

1. Compression Ratio in diesel engine is lesser than that in Spark Ignition engine.
a) True
b) No
c) They are equal
d) Doesn’t matter

View Answer

Answer: b [Reason:] In diesel engine, there is no spark to ignite, charge is compressed to greater extent to support self-ignition.

2. Petrol engine are more economical than diesel engine.
a) True
b) False

View Answer

Answer: b [Reason:] Diesel engine have better thermal efficiency and hence are more economical.

3. There are ____ types of liners.
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] Dry liners and wet liners.

4. If no re boring allowance is to be given, then thickness of cylinder wall is max pressure is 3.5N/mm², bore diameter=200mm and permissible tensile stress is 40N/mm².
a) 7mm
b) 8mm
c) 9mm
d) 10mm

View Answer

Answer: c [Reason:] t=pD/2σ=8.75mm.

5. If diameter of cylinder of bore is 120mm, then thickness of the cylinder will be
a) Information not sufficient
b) 7mm
c) 12mm
d) 6mm

View Answer

Answer: b [Reason:] t=0.045D + 1.6.

6. If diameter of cylinder bore is 120mm, then thickness of dry liner will be
a) 2.2mm
b) 3.6mm
c) 4.8mm
d) 6mm

View Answer

Answer: b [Reason:] t= 0.03D to 0.035D.

7. If diameter of cylinder bore is 120mm, then thickness of water jacket wall will be
a) 4.26mm
b) 5.25mm
c) 2.56mm
d) All of the listed

View Answer

Answer: d [Reason:] p=t/3 to 3t/4, where t= thickness of cylinder wall=7mm.

8. If diameter of cylinder bore is 120mm, then thickness of water cylinder flange will be
a) 8.8mm
b) 10.2mm
c) 7.8mm
d) 12mm

View Answer

Answer: a [Reason:] q=1.2t to 1.4t.

9. If nominal diameter of bolt used is 20mm, then find the radial distance between outer diameter of flange and pitch circle diameter of studs.
a) 24mm
b) 28mm
c) 32mm
d) 36mm

View Answer

Answer: b [Reason:] r=d+6 to 1.5d i.e. 26mm to 30mm.

10. The circumferential hoop stress and longitudinal stress are both shear stress.
a) True
b) False

View Answer

Answer: b [Reason:] They both are tensile stress.

11. The circumferential hoop stress and longitudinal stress act in same direction and hence are straight away added.
a) True
b) False

View Answer

Answer: b [Reason:] They both act perpendicularly to each other and thus net stress in these directions is reduced.