1. Stress intensity factor is the critical value of stress at which crack extension occurs.
a) True
b) False
Answer: b
Explanation: Stress intensity specifies the stress intensity at the tip of the crack.

2. The critical value at which crack extension occurs is called
a) Stress Intensity Factor
b) Toughness
c) Fracture Toughness
d) None of the mentioned
Answer: c
Explanation: Fracture toughness is the critical value of stress intensity at which crack extension occurs.

3. Fracture toughness does not depend upon geometry of the part containing crack
a) True
b) False
Answer: b
Explanation: Fracture toughness is directly proportional to a factor Y that depends upon geometry of the part having crack.

4. How many modes are there for crack propagation?
a) 2
b) 3
c) 4
d) 5
Answer: b
Explanation: Opening, sliding and tearing are the 3 modes.

5. A curved beam has neutral axis is curved while loaded and straight when unloaded.
a) True
b) False
Answer: b
Explanation: Curved beam’s neutral axis is always curved irrespective of the loading.

6. The bending stress in a straight beam varies linearly with the distance from neural axis like that in a curved beam.
a) True
b) False
Answer: b
Explanation: Bending stress in a curved beam varies hyperbolically with the distance from neutral axis.

7. If for a curved beam of trapezoidal cross section, radius of neutral axis is 89.1816mm and radius of centroidal axis is 100mm, then find the bending stress at inner fibre whose radius is 50mm. Area of cross section of beam is 7200mm² and the beam is loaded with 100kN of load.
a) 97.3
b) 95.8
c) 100.6
d) None of the mentioned
Answer: c
Explanation: e=100-89.816=10.8184mm, h=89.1816-50=39.1816mm, M=100 x 100 N-m
Therefore σ=Mh/AeR or σ=10000 x 39.1816/ [7200 x 10.8184 x 50] or σ=100.6N/mm².

8. A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.
a) 14.80mm
b) 13.95mm
c) 16.5mm
d) 17.2mm
Answer: b
Explanation: σ(b)=Mh/ AeR or σ(b)=1000x4Dx(4D-0.2D-3.5D)/ 0.8D²x0.02Dx3.5D , σ(b)=21428.6/D²
21428.6/D² = 110 or D=13.95mm.

9. A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm² and considering only direct tensile stress.
a) 4.7mm
b) 6.8mm
c) 13.95mm
d) 3.4mm
Answer: d
Explanation: Direct Tensile Stress=1000/0.8D² or σ (t) =1250/D²
1250/D²=110 or D=3.4mm.

10. A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm² and considering combined effect of direct stress and bending stress.
a) 15.8mm
b) 14.35mm
c) 17.9mm
d) 18.1mm
Answer: b
Explanation: σ(b)=Mh/ AeR or σ(b)=1000x4Dx(4D-0.2D-3.5D)/ 0.8D²x0.02Dx3.5D , σ(b)=21428.6/D²
DirectTensile Stress=1000/0.8D² or σ (t) =1250/D²
Total stress=22678.6/D² N/mm²= 110 or D=14.35mm.

11. _______ is as the maximum energy that can be absorbed within the proportionality limit.
a) Proof resilience
b) Modulus of resilience
c) Impact resilience
d) Resilience
Answer: a
Explanation: Proof resilience is defined as the maximum that can be absorbed with in the proportionality limit without creating a permanent distortion.

12. The compressive strength of brittle materials is _________ its tensile strength.
a) Equal to
b) Less than
c) Greater than
d) As same as
Answer: c
Explanation: The compressive strength of brittle materials is greater than its tensile strength. The tensile strength of ductile material is greater than its compressive strength.

13. The tensile test is carried on ________ material.
a) Ductile
b) Brittle
c) Malleable
d) Plastic
Answer: a
Explanation: The tensile stress is carried on the tensile materials. In the same way, the compression test is carried on brittle materials.

14. The breaking stress is ____________ the ultimate stress.
a) Equal to
b) Less than
c) Greater than
d) As same as
Answer: b
Explanation: The stress developed in a material without any permanent stress is called elastic limit and the breaking stress is always less than the ultimate stress.

15. The ductility of a material is __________ to the increase in percentage reduction in an area.
a) inversely proportional
b) directly proportional
c) equal
d) uniform
Answer: a
Explanation: The ductility of material increases with the increase in percentage reduction in area of a specimen under tensile stress.

16. The odour of water can be determined by _________
a) Jackson turbidometer
b) Osmoscope
c) Thermometer
d) Sonoscope
Answer: b
Explanation: The main causes of odour in water are algae, sewage and dissolved gases. Taste and odour can also be expressed in terms of odour density. Odour can be estimated by osmoscope.

17. The colour of water is expressed in terms of ________
a) pH value
b) Silica scale
c) Platinum cobalt scale
d) Ppm
Answer: c
Explanation: Colour is caused by the presence of colloidal substance is aquatic growth etc. in water should be distinguished from turbidity which is termed as apparent colour. The colour is expressed in Platinum Cobalt scale. Colour may be removed by coagulation and adsorption method.

18. High turbidity of water can be determined by __________
a) Hellipe turbidometer
b) Baylis turbidometer
c) Jackson’s turbidometer
d) Turbidity rod
Answer: b
Explanation: The turbidity of potable water should be within 10 PPM or with in 10 units on the silica scale. High turbidity of water can be determined by Jackson turbidity metre and low turbidity of water can be determined by baylis turbidity metre.

19. The maximum permissible total solid content in water for domestic purposes should not exceed.
a) 350 ppm
b) 600 ppm
c) 500 ppm
d) 1000 ppm
Answer: c
Explanation: Analytically, the total solids content of water is defined as all the matter that remains as residue upon evaporation. The standards for drinking water is acceptable limit is 500 ppm.

20. Membrane filter technique is used for testing?
a) Copper
b) E -coli
c) Bacteria
d) Boron
Answer: b
Explanation: Membrane filter technique is considered as superior method. In this procedure, unknown volume of water sample is filtered through a membrane with opening less than 0.5 microns.

21. If a hollow steel tube is heated from a temperature of 25’C to 250’C then fid the expansion of tube if area of the cross section is 300mm²,length of tube=200mm and coefficient of thermal expansion is 10.8 x 10⁻⁶ per ⁰C.
a) 1.22mm
b) 0.486mm
c) 0.878mm
d) 1.52mm
Answer: b
Explanation: Expansion=ἀxlx∆T= 10.8 x 10⁻⁶ x 200 x 225=0.486mm.

22. All type of stresses vanishes after as soon as the applied load is removed.
a) True
b) False
Answer: b
Explanation: Residual stresses are independent of the load.

23. Residual stresses are always added in the load stresses and hence are always harmful.
a) True
b) False
Answer: b
Explanation: Residual stresses may be beneficial when they are opposite to load stresses and hence are subtracted from load stresses.

24. E – coli was formerly known as _________
a) F. Coli
b) B. Coli
c) G. Coli
d) R. Coli
Answer: b
Explanation: The pathogenic bacteria are generally inherent in the qualifying group of Bacteria of which the bacillus coli (B. Coli) now called as Escherichia coli (E.Coli ) is prominent.

25. ______ sample collected at an instant particularly.
a) Composite
b) Grab
c) Integrated
d) Differential
Answer: b
Explanation: To determine the character of the sample, at that particular instant is known as grab sample. The frequency of grab sampling depends upon the magnitude of fluctuation in the quality of source.

26. Which of the following samples is also known as catch sample?
a) Integrated
b) Composite
c) Grab
d) Scratch
Answer: c
Explanation: In sampling, catch sample collected from the sampling spot at any instant. It is also known as grabbing sample. It is influenced by the nature of tests are to be conducted.

27. If fluoride concentration in drinking water increases to more than ______ ppm, it causes fluorosis.
a) 2.5
b) 2
c) 1.5
d) 3
Answer: c
Explanation: When the concentration of fluoride increases to more than 1.5 ppm, a disfigurement involving staining of teeth known as mottled tooth enamel is caused. This disease is also termed as fluorosis.

28. What is the desirable limit for sulphates in drinking water?
a) 180 ng/L
b) 230 mg/L
c) 150 mg/L
d) 340 mg/L
Answer: c
Explanation: Sulphates ion is one of the major ions occurring in natural waters. In drinking water, sulphate causes a laxative effect and leads to scale formation in boilers. The desirable limit in drinking water is 150 mg /L.

1. Cotter joint is used when the members are subjected to which type of stresses?
a) Axial tensile
b) Axial compressive
c) Axial tensile or compressive
d) None of the mentioned
Answer: c
Explanation: Cotter joint is used when axial forces are applied.

2. The principle of wedge action is used in cotter joint.
a) True
b) False
Answer: a
Explanation: Wedge action imparts tightening to the cotter joint.

3. Can the cotter joint be used to connect slide spindle and fork of valve mechanism?
a) True
b) False
Answer: a
Explanation: As long as axial forces act, cotter joint can be employed.

4. Which of the following is not a part of cotter joint?
a) Socket
b) Spigot
c) Cotter
d) Collar
Answer: d
Explanation: There is no point of mentioning collar alone in a cotter joint. It has to be a spigot collar or socket collar.

5. Cold riveting holds the connected parts better than hot riveting.
a) True
b) False
Answer: b
Explanation: The compression of connected parts in hot riveting causes friction, which resist sliding of one part with respect to other. This force is greater in hot riveting.

6 .In cold riveting like hot riveting shank is subjected to majorly tensile stress.
a) True
b) False
Answer: b
Explanation: In cold riveting there is no reduction in length and hence no tensile stress. Shear stress dominates.

7. Determine the width of the cotter used in cotter joint connecting two rods subjected to axial load of 50kN and permissible shear stress in cotter is 50N/(mm² ). Given thickness of cotter=10mm
a) 5omm
b) 100mm
c) 150mm
d) 25mm
Answer: a
Explanation: Cotter is subjected to double shear hence width=P/(2*τ*t).

8. If joint is to fail by crushing of socket collar then estimate the diameter of socket collar. Given Permissible compressive stress= 126.67 N/mm².; Spigot dia=65mm; thickness 0f collar=15mm
a) 131mm
b) 139mm
c) 141mm
d) 149mm
Answer: a
Explanation: Compressive stress= P/ [(socket dia-spigot dia)*thickness].

9. Cotter joint can be used to connect two rods for torque transmission purpose.
a) True
b) False
Answer: b
Explanation: Cotter Joint is never used to connect two rods for torque transmission purpose.

10. Thickness of plate is required more in welding than in riveting.
a) True
b) False
Answer: b
Explanation: In riveting, cross section is weakened due to the holes and to compensate this, thicker plates are required in riveting.

11. Knuckle Joint can’t be used to connect two intersecting rods.
a) Yes
b) No, it can’t be used
c) It can be used with some modificatios
d) It is expensive and hence isn’t used
Answer: b
Explanation: Knuckle Joint is used to connect two rods whose axes coincide or intersect and lie in a same plane.

12. A knuckle joint is unsuitable for two rotating shafts, which transmit torque
a) True
b) False
Answer: a
Explanation: Knuckle joint can’t be used for torque transmission.

13. A maximum of how many roads may be connected using a knuckle joint?
a) 2
b) 3
c) 4
d) 5
Answer: b
Explanation: In rare explanation, two rods with forks and one rod with eye is connected.

14. A knuckle joint is also called socket pin joint.
a) True
b) False
Answer: b
Explanation: A knuckle joint is also called a Forked Pin Joint

15. Which of the following are important parts of knuckle joint?
a) Eye
b) Pin
c) Fork
d) Each of the mentioned
Answer: d
Explanation: All the mentioned parts are important components of knuckle joint.

16. Calculate the diameter of pin from shear consideration with maximum shear stress allowed is 40NN/mm² and an axial tensile force of 50kN is acting on the rod.
a) 39mm
b) 44mm
c) 49mm
d) 52mm
Answer: a
Explanation: As the pin is subjected to double shear diameter (D) = √(2P/π x τ) = 38.80mm.

17. If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.
a) 40mm
b) 50mm
c) 60mm
d) 70mm
Answer: a
Explanation: d=P/2aσ = 40mm.

18. If any cross section is subjected to direct tensile stress and bending stress, then find the dimension of cross section. Given length & breadth are t and 2t respectively. F=25kN acts on the top fibre of the cross section, M=F x t . Also maximum allowable tensile stress =100N/mm².
a) 25.5mm
b) 30.2mm
c) 27.55mm
d) None of the mentioned
Answer: a
Explanation: σ= [P/A] + [My/I], where y=t & I=t(2t)ᴲ/12.

19. A knuckle joint can be used in valve mechanism of a reciprocating engine.
a) Yes
b) No
c) Yes but there are stress probles
d) No as it is very dangerous to use
Answer: a
Explanation: Knuckle joint can be used till the rods coincide or intersect in a plane.

20. Calculate the direct shear stress in the welds by ignoring torsional shear stress if P=8kN and thickness of weld is 5mm.
a) None of the listed
b) 12.33N/mm
c) 13.33N/mm
d) 14.33N/mm
Answer: c
Explanation: Direct Shear=P/2xtx60.

 

Answer the question 21-27 with respect to figure 1
21. The two welds shown in the figure experience equal resisting force?
a) True
b) False
Answer: b
Explanation: As the welded joints are unsymetrically loaded hence resisting forces are different in the two welds.

22. The moment of forces about the CG is ?
a) Zero
b) Infinite
c) P₁y₁+P₂y₂
d) None of the listed
Answer: a
Explanation: External force passes through CG hence moment is zero.

23. Which equations can be used to find the resisting force in the two welds?
a) P=P₁+P₂
b) P₁+P₂=0
c) Both P=P₁+P₂ and P₁y₁=P₂y₂
d) P₁y₁=P₂y₂
Answer: c
Explanation: One equation is of equilibrium and the other conservation of moment about CG.

24. For the following welded joint l₁y₁=l₂y₂.
a) True
b) False
Answer: a
Explanation: P₁y₁=P₂y₂ and P₁=0.707hl₁τ and P₂=0.707hl₂τ.

25. Find the total length of weld required to with stand the load of 100kN. Permissible shear stress in weld is 70kN/mm².
a) None of the listed
b) 101.03mm
c) 202.06mm
d) 30309mm
Answer: c
Explanation: 100000=0.707x10xlx70.

26. Find the length of weld 2 required to with stand the load of 100kN. Permissible shear stress in weld is 70kN/mm².
a) None of the listed
b) 102.22mm
c) 132.4mm
d) 70.16mm
Answer: c
Explanation: By finding CG y₂=103.9mm, therefore l₁y₁=l₂y₂ and l₁+l₂=202.06mm.

7. Find the length of weld 1 required to with stand the load of 100kN. Permissible shear stress in weld is 70kN/mm².
a) None of the listed
b) 102.22mm
c) 132.4mm
d) 70.16mm
Answer: d
Explanation: By finding CG y₂=103.9mm, therefore l₁y₁=l₂y₂ and l₁+l₂=202.06mm.

Figure 2

Answer the following questions with respect to figure 2

8. The following welded joint is an example of
a) Eccentric Load in plane of welds
b) Axial load in plane of welds
c) Axial load in plane perpendicular to plane of welds
d) None of the mentioned
Answer: a
Explanation: The figure clearly depicts.

29. What kind of stresses does the welded joint undergo?
a) Torsional shear stress
b) Direct shear stress
c) Direct and torsional shear stress
d) None of the listed
Answer: c
Explanation: After shifting force to CG, we have a force and a moment about CG.

30. While considering moment of inertia for calculating torsional shear stress, J=I(xx) + I(yy), which of the following can be neglected in context with the following figure?
a) I(xx)
b) I(yy)
c) Both I(xx) and I(yy)
d) None of the listed
Answer: a
Explanation: I(xx) is negligible as compared to I(yy) as I(xx)=ltᵌ/12 and t is very less as compared to l so I(xx) is neglected in comparison to I(yy).

31. A power screw is only used to convert rotary motion into linear motion and not for transmitting power.
a) True
b) False
Answer: b
Explanation: Power screw converts the motion from rotary to linear and is used for power transmission.

32. Depending upon the holding arrangement, power screws operate in how many different arrangements.
a) 2
b) 3
c) 4
d) 5
Answer: a
Explanation: There are two types of arrangements. In one screw rotates while nut remains stationary and vice versa.

33. A power screw has no problem of wear as there is very less amount of friction associated.
a) True
b) False
Answer: b
Explanation: Wear is a serious problem in power screws as there is high friction in threads.

34. V threads are highly recommended for fastening as well as power transmission purpose.
a) Yes
b) Never
c) In some cases
d) Can’t be stated
Answer: b
Explanation: In fastening, high frictional force is required and hence V threads are used whereas in power transmission, reduction in frictional forces is required.

35. Trapezoidal threads are better than square threads as there is radial pressure or side thrust on the nut.
a) True
b) False
Answer: b
Explanation: Trapezoidal and not square threads suffer from the problem of bursting.

36. Trapezoidal threads screws have less load carrying capacity as compared to square thread screws.
a) True
b) False
Answer: b
Explanation: Square threads have less thickness at the core diameter and hence lower load carrying capacity.

37. Which of the following are true for buttress threads?
a) Combination of square and trapezoidal threads
b) Transmit motion in one direction only
c) They are used in vices
d) All of the mentioned
Answer: d
Explanation: As force is applied only in one direction in a vice so buttress threads are used.

38. Tr 40 x 14(P 7), here 14 indicates
a) Pitch
b) Lead
c) Diameter
d) None of the mentioned
Answer: b
Explanation: Tr=Trapezoidal threads,14=Lead(mm),7=Pitch(mm).

39. Nominal diameter of the screw thread is the same as core diameter.
a) True
b) False
Answer: b
Explanation: Nominal diameter is the largest diameter while core diameter is the smallest diameter of the screw thread.

40. If nominal diameter of screw thread=50mm and pitch=10mm then the mean diameter of the screw thread will be?
a) 40mm
b) 45mm
c) 60mm
d) 55mm
Answer: b
Explanation: Diameter(mean)=Diameter(nominal) – 0.5P .

41. If the load itself begin to the screw and descend down, unless a restraining torque is applied then the condition is termed as
a) Halting
b) Overhaulting
c) Front driving
d) None of the mentioned
Answer: b
Explanation: Overhaulting is the condition when load itsel begin to turn the screw.

42. Self-locking takes place when
a) Coefficient of friction is equal to or greater than the tangent of the helix angle
b) Coefficient of friction is lesser than or equal to the tangent of the helix angle
c) Coefficient of friction is equal to or greater than the tangent of the helix angle
d) None of the mentioned
Answer: c
Explanation: Self locking takes place if load does not descend on its own and that is possible only in c condition.

1. Stress concentration is defined as the localization of high stresses due to irregularities present in the component and no changes of the cross section.
a) True
b) False
Answer: b
Explanation: The given statement is true apart from the fact that there is no change in the cross section.

2. Stress Concentration Factor is the ratio of nominal stress obtained by elementary equations for minimum cross-section and highest value of actual stress near discontinuity.
a) True
b) False
Answer: b
Explanation: The stress concentration factor is just the reciprocal of that cited in the question.

3. If a flat plate with a circular hole is subjected to tensile force, then its theoretical stress concentration factor is?
a) 2
b) 3
c) 4
d) 1
Answer: b
Explanation: For any ellipse, K=1+2 x (semi major axis/semi minor axis).

4. For an elliptical hole on a flat plate, if width of the hole in direction of the load decrease, Stress Concentration Factor will______
a) Increase
b) Decrease
c) Remains constant
d) Can’t be determined. Varies from material to material
Answer: a
Explanation: K=1+2 x (semi major axis/semi minor axis. Hence K is inversely proportional to the semi minor axis.

5. In which of the following case stress concentration factor is ignored?
a) Ductile material under static load
b) Ductile material under fluctuating load
c) Brittle material under static load
d) Brittle material under fluctuating load
Answer: a
Explanation: In ductile materials under static load, there is plastic deformation near yielding point and hence redistribution of stresses take place. The plastic deformation is restricted to a smaller area and hence no perceptible damage take place.

6. Is it logical to use fluid analogy to understand the phenomenon of stress concentration?
a) True
b) False
Answer: a
Explanation: There is a similarity between velocity distribution in fluid flow in a channel and the stress distribution in an axially loaded plate. The equations for flow potential and in fluid mechanics and stress potential in solid mechanics are same.

7. Use of multiple notches in a V shaped flat plate will
a) Reduce the stress concentration
b) Increase the stress concentration
c) No effect
d) Cannot be determined
Answer: a
Explanation: The sharp bending of a force flow line is reduced due to multiple notches.

8. Which of the following reduces the stress concentration?
a) Use of multiple notches
b) Drilling additional holes
c) Removal of undesired material
d) Each of the mentioned
Answer: d
Explanation: All the mentioned options reduce the sharp bending of a force flow line.

9. A flat plate 30mm wide and “t”mm wide is subjected to a tensile force of 5kN. The plate has a circular hole of diameter 15mm with the centre coinciding with the diagonal intersection point of the rectangle. If stress concentration factor=2.16, find the thickness of the plate if maximum allowable tensile stress is 80N/mm².
a) 8mm
b) 9mm
c) 10mm
d) 12mm
Answer: b
Explanation: σ=P/ (w-d) x t or σ=5000/(30-15)xt ; σ(max)=K x σ or σ(max)=2.16 x σ or σ(max)=720/t; Also σ(max)=80.

10. The stress represented by sin (t) + 1 belongs to which category?
a) Fluctuating Stresses
b) Alternating stresses
c) Repeated Stresses
d) Reversed Stresses
Answer: c
Explanation: The minimum stress value is zero and hence is belongs to Repeated Stress category.

11. The stress represented by sin (t) + 2 belongs to which category?
a) Fluctuating Stresses
b) None of the mentioned
c) Repeated Stresses
d) Reversed Stresses
Answer:a
Explanation: The mean as well as the amplitude value is non zero hence it belongs to Fluctuation Stress Category.

12. The stress represented by sin (t) + 4 belongs to which category?
a) Alternating Stresses
b) None of the mentioned
c) Repeated Stresses
d) Reversed Stresses
Answer: a
Explanation: The mean as well as the amplitude value is non zero hence it belongs to Alternation Stresses category which is the other name of Fluctuation Stresses.

13. The stress represented by cos (t) belongs to which category?
a) Fluctuating Stresses
b) Alternating Stresses
c) Repeated Stresses
d) Reversed Stresses
Answer: d
Explanation: Half cycle is in tensile stress and other half in compressive stress, hence it belongs to Reversed Stresses Category.

14. If the mean stress value for a sinusoidal stress function is zero, then this type of stress falls in which category?
a) Fluctuating Stresses
b) Alternating Stresses
c) Repeated Stresses
d) Reversed Stresses
Answer: d
Explanation: If mean is to be zero, then there must be compressive as well as tensile stresses and hence belongs to reversed stresses category.

15. The phenomenon of decreased resistance of the materials to fluctuating stresses is the main characteristic of _____ failure.
a) Fracture
b) Fatigue
c) Yielding
d) None of the mentioned
Answer: b
Explanation: Fatigue failure of the material is the failure at low stress levels under fluctuating syresses.

16. Fatigue failure is time dependent failure.
a) True
b) False
Answer: a
Explanation: Fatigue failure is defined as time delayed fracture under cyclic loading.

17. There is sufficient plastic deformation prior to fatigue failure, which gives a warning well in advance.
a) True
b) False
Answer: b
Explanation: Fatigue cracks are not visible until they reach the surface and by that time the failure has already taken place. Material nevers enters in the plastic range.

1. The shaft is always stepped with ________ diameter at the middle portion and __________ diameter at the shaft ends.
a) Minimum, maximum
b) Maximum, minimum
c) Minimum, minimum
d) Zero, infinity
Answer: b
Explanation: Maximum diameter is in the middle portion while it is minimum at the ends.

2. ______ is used for a shaft that supports rotating elements like wheels, drums or rope sleaves.
a) Spindle
b) Axle
c) Shaf
d) None of the listed
Answer: a
Explanation: That is axle and not a spindle.

3. Axle is frequently used in torque transmission.
a) True
b) False
Answer: b
Explanation: Axle is a shaft supporting rotating elements.

4. Is it necessary for an axle to be ______ with respect to rotating element?
a) Stationary
b) Moving
c) Moving or stationary
d) None of the listed
Answer: c
Explanation: The axle may be stationary or rotate with the element.

5. Counter shaft is a secondary shaft.
a) True
b) False
Answer: a
Explanation: It is a secondary shaft used to counter the direction of main shaft.

6. Hot rolling produces a stronger shaft then cold rolling.
a) True
b) False
Answer: b
Explanation: Cold rolling produces stronger shaft as grain structure isn’t deformed in cold working.

7. Shafts are subjected to ______ forces.
a) Compressive
b) Tensile
c) Shear
d) None of the listed
Answer: b
Explanation: Shafts are subjected to tensile forces.

8. Which of the following act on shafts?
a) Torsional moment
b) Bending Moment
c) Both torsional and bending
d) None of the mentioned
Answer: c
Explanation: Shaft is subjected to torsional moment as well as bending moment.

9. When the shaft is subjected to pure bending moment, the bending stress is given by?
a) None of the listed
b) 32M/πdᵌ
c) 16M/πdᵌ
d) 8M/πdᵌ
Answer: b
Explanation: Stress =My/I where y=d/2 and I=πd⁴/64.

10. When the shaft is subjected to pure torsional moment, the torsional stress is given by?
a) None of the listed
b) 32M/πdᵌ
c) 16M/πdᵌ
d) 8M/πdᵌ
Answer: c
Explanation: Stress=Mr/J where r=d/2 and J=πd⁴/64.

11. While designing shaft on the basis of torsional rigidity, angle of twist is given by?
a) Ml/Gd⁴
b) 584Ml/Gd⁴
c) 292 Ml/Gd⁴
d) None of the mentioned
Answer: b
Explanation: θ=(180/π)xMl/GJ where J=πd⁴/32.

12. According to ASME code, maximum allowable shear stress is taken as X% of yield strength or Y% of ultimate strength.
a) X=30 Y=18
b) X=30 Y=30
c) X=18 Y=18
d) X=18 Y=30
Answer: a
Explanation: ASME Standard. The lesser value is taken among the two.

3. Does ASME Standard take into consideration shock and fatigue factors?
a) Yes
b) No
Answer: a
Explanation: Moment is multiplied by a number to consider these factors while designing the shaft.

Figure 1

The layout of a shaft supported on bearings at A & B is shown. Power is supplied by means of a vertical belt on pulley B which is then transmitted to pulley C carrying a horizontal belt. The angle of wrap is 180’ and coefficient of friction is 0.3. Maximum permissible tension in the rope is 3kN. The radius of pulley at B & C is 300mm and 150mm.
For questions 4-7, refer to figure 1.

14. Calculate the torque supplied to the shaft.
a) 453.5N-m
b) 549.3N-m
c) 657.3N-m
d) None of the listed
Answer: b
Explanation: M=(P₁-P₂)xR where P₁=3000N, P₂=P₁/e^(Coeff of friction x Angle of wrap).

15. Calculate the tension in the rope of pulley C.
a) 6778.3N and 7765.3N
b) 5948.15N and 2288.75N
c) 5468.4N ad 8678.3N
d) None of the listed
Answer: b
Explanation: (P₃-P₄)x150=549.3×1000 and P₃/P₄=2.6. Hence P₃=5948.15N and P₄=2288.75N.

16. If allowable shear stress in the shaft is 70N/mm² and torsional and bending moments are M=1185000N-mm and m=330000N-mm, find the diameter of the shaft.
a) 36.8mm
b) 39.7mm
c) 44.7mm
d) 40.3mm
Answer: c
Explanation: 70=(16/πdᵌ)x√M²+m².

17. If bending moment on point B in horizontal plate is M and in vertical plane is m, then the net bending moment at point B is?
a) M
b) m
c) M+m
d) √M²+m²
Answer: d
Explanation: The two moments act perpendicularly to each other.

8. Calculate the shaft diameter on rigidity basis if torsional moment is 196000N-mm, length of shaft is 1000mm. Permissible angle of twist per meter is 0.5’ and take G=79300N/mm².
a) None of the listed
b) 41.2mm
c) 35.8mm
d) 38.8mm
Answer: b
Explanation: d⁴=584Ml/Gθ

19. If yielding strength=400N/mm², the find the permissible shear stress according to ASME standards.
a) 72 N/mm²
b) 76 N/mm²
c) 268 N/mm²
d) 422 N/mm²
Answer: a
Explanation: 0.18×400.

20. The stiffness of solid shaft is more than the stiffness of hollow shaft with same weight.
a) True
b) False
Answer: b
Explanation: Hollow shaft is more stiff.

21. A force 2P is acting on the double transverse fillet weld. Leg of weld is h and length l. Determine the shear stress in a plane inclined at θ with horizontal.
a) PSinθ(Sinθ+Cosθ)/hl
b) P(Sinθ+Cosθ)/hl
c) Pcosθ(Sinθ+Cosθ)/hl
d) None of the listed
Answer: a
Explanation: F=PSinθ and width=h/(Sinθ+Cosθ).

22. Maximum shear stress in transverse fillet weld of leg h and length l is
a) P/hl
b) 1.21P/hl
c) P/1.21hl
d) None of the listed
Answer: b
Explanation: τ= PSinθ(Sinθ+Cosθ)/hl, by maximising it θ=67.5’ and hence find corresponding τ.

23. A sunk key fits in the keyway of the _____ only.
a) Hub
b) Sleeve
c) Both hub and sleeve
d) Neither hub nor sleeve
Answer: a
Explanation: Sunk key fits halfway in the hub and halfway in the shaft.

24. Hollow saddle key is superior to flat saddle key as far as power transmitting capability is concerned.
a) True
b) False
Answer: b
Explanation: The resistance to slip in case of flat key is more.

25. Saddle key is more suitable than sunk key for heavy duty applications.
a) True
b) False
Answer: b
Explanation: In sunk key, relative motion is also prevented by shear resistance of sunk key and hence sunk key is recommended.

26. The main advantage of sunk key is that it is a _____ drive.
a) Positive
b) Negative
c) Neutral
d) None of the listed
Answer: a
Explanation: Sunk key is a positive drive and no slip occurs.

27. Woodruff key permits _____ movement b/w shaft and the hub.
a) Axial
b) Radial
c) Eccentric
d) None of the listed
Answer: b
Explanation: Woodruff key is a sunk key and doesn’t permit axial moment.

28. Determine the length of kennedy key required to transmit 1200N-m and allowable shear in the key is 40N/mm². The diameter of shaft and width of key can be taken as 40mm and 10mm respectively.
a) 49mm
b) 36mm
c) 46mm
d) 53mm
Answer: d
Explanation: l=M/[dbτ√2].

29. Splines are keys.
a) True
b) False
Answer: a
Explanation: Splines are keys made with shafts.

30. Involute splines have stub teeth with a pressure angle of ___
a) 30
b) 45
c) 60
d) Can’t be determined
Answer: b
Explannation: Pressure angle is 30’ and not 60’.

31. A coupling is a mechanical device that temporarily joins two rotating shafts to each other.
a) True
b) False
Answer: b
Explanation: A coupling permanently joins two rotating shafts.

32. In distortion energy theorem, if a unit cube is subjected to biaxial stress, then S(yt) is given by which of the following?
a) √ (σ₁²- σ₁σ₂ +σ₂²)
b) σ₁²- σ₁σ₂ +σ₂²
c) √ (σ₁²+ σ₁σ₂ +σ₂²)
d) σ₁²+ σ₁σ₂ +σ₂²
Answer: a
Explanation: S(yt)=√[(σ₁ -σ₂)²+(σ₂ -σ₃)²+(σ₃ -σ₁)²]/2, for σ₃=0; expression a is obtained.

33. Oldham coupling i used to connect two shafts having intersecting axes.
a) True
b) False
Answer: b
Explanation: Oldham coupling is used to connect two parallel shafts separated by a small distance.

34. A machine started malfunctioning due to some issues with the coupling. The coupling emplaced in the machine was Oldham. The only coupling available in the workshop is Hooke’s coupling. So Oldham coupling can be replaced by Hooke’s coupling.
a) True
b) False
Answer: b
Explanation: It is not possible to replace Oldham by Hooke’s coupling as Hooke’s is used to connect intersecting axes and not parallel axes shafts.

35. A muff coupling is connecting two shafts. The torque involved is 650N-m. The shaft diameter is 45mm with length and breadth of the key being 14mm and 80mm respectively. Find the shear stress induced in the key.
a) 30.2N/mm²
b) 25.8N/mm²
c) 34.4N/mm²
d) None of the listed
Answer: b
Explanation: τ =2M/dbl.

36. The region of safety for biaxial stresses is of which shape in the case of maximum distortion energy theorem.
a) Ellipse
b) Circle
c) Rectangle
d) Square
Answer: a
Explanation: S(yt)= √ (σ₁²- σ₁σ₂ +σ₂²) or S²(yt)= (σ₁²- σ₁σ₂ +σ₂²). This is an equation of ellipse.

37. Distortion energy theorem is not recommended for ductile materials.
a) True
b) False
Answer: b
Explanation: Experimental investigations have shown that distortion energy theorem gives good results for failure of ductile component.

38. Among maximum shear stress theory and distortion energy theory, which gives the higher value shear yield strength?
a) Maximum shear stress theory
b) Distortion energy theory
c) Both give equal values
d) Vary from material to material
Answer: b
Explanation: Maximum shear stress theory gives S(sy)=0.5S(yt) while Distortion energy theorem gives S(sy)=0.577(Syt).

39. A muff coupling is connecting two shafts. The torque involved is 650N-m. The shaft diameter is 45mm with length and height of the key being 14mm and 80mm respectively. Find the compressive stress induced in the key.
a) 70.1 N/mm²
b) 51.6N/mm²
c) 45.5N/mm²
d) None of the listed
Answer: b
Explanation: σ=4M/dhl.

40. The strength of hollow shaft is more than the strength of solid shaft of same weight.
a) True
b) False
Answer: a
Explanation: Outer fibers are more effective in resisting the applied moments. In hollow shafts material is removed and spread on a larger radius.

41. Solid shaft is costlier than hollow shaft of same weight.
a) True
b) False
Answer: b
Explanation: Hollow shaft cost is more as material is to be selectively emplaced.

42. Solid shafts are used in epicyclic gearboxes.
a) True
b) False
Answer: b
Explanation: In epicyclic gears, one shaft rotates inside other and hence hollow shafts are used.

43. Flexible shafts have ___ rigidity in torsion making them flexible.
a) Low
b) High
c) Very high
d) Infinitely small
Answer: b
Explanation: Flexible shafts have high rigidity in torsion making then capable to transmit torque.

44. Flexible shafts have ______ rigidity in bending moment.
a) High
b) Low
c) Very high
d) Extremely low
Answer: b
Explanation: Flexible shafts have low rigidity in bending moments making them flexible.

1. Which of the following function can the spring perform?
a) Store energy
b) Absorb shock
c) Measure force
d) All of the mentioned
Answer: d
Explanation: Spring can easily perform all the listed functions.

2. The helix angle is very small about 2⁰. The spring is open coiled spring.
a) Yes
b) It is closed coiled spring
c) That small angle isn’t possible
d) None of the listed
Answer: b
Explanation: When the helix angle is small, the plane containing each coil is almost at right angles and hence it is called closed coiled spring.

3. The helical spring ad wire of helical torsion spring, both are subjected to torsional shear stresses.
a) True
b) False
Answer: b
Explanation: The wire of helical torsion sprig is subjected to bending stresses.

4. The longest leaf in a leaf spring is called centre leaf.
a) It is called middle leaf
b) It is called master leaf
c) Yes
d) None of the listed
Answer: b
Explanation: It is called master leaf.

5. Multi leaf springs are not recommended for automobile and rail road suspensions.
a) True
b) False
Answer: b
Explanation: They are highly used in automobile and rail road suspensions.

6. The spring index is the ratio of wire diameter to mean coil diameter.
a) True
b) False
Answer: a
Explanation: It is the ratio of mean coil diameter to wire diameter.

7. If spring index=2.5, what can be concluded about stresses in the wire?
a) They are high
b) They are negligible
c) They are moderate
d) Cannot be determined
Answer: a
Explanation: If indexis <3 then stresses are high due to curvature effect.

8. A spring with index=15 is prone to buckling.
a) True
b) False
Answer: a
Explanation: Due to large variation, such a spring is prone to buckling.

9. If the spring is compressed completely and the adjacent coils touch each other,the the length of spring is called as?
a) Solid length
b) Compressed length
c) Free length
d) None of the mentioned
Answer: a
Explanation: Terminology.

10. If number of coils are 8 and wire diameter of spring 3mm, then solid length is given by?
a) None of the listed
b) 27mm
c) 24mm
d) 21mm
Answer: c
Explanation: Solid length=8×3.

11. Compressed length is smaller than the solid length.
a) True
b) False
Answer: b
Explanation: Compressed is length of spring under maximum compressive force. There is some gap between the coils under maximum load.

12. Pitch of coil is defined as axial distance in compressed state of the coil.
a) Yes
b) It is measured in uncompressed state
c) It is same in uncompressed or compressed state
d) None of the listed
Answer: b
Explanation: Pitch is measured in uncompressed state.

13. If uncompressed length of spring is 40mm and number of coils 10mm, then pitch of coil is?
a) 4
b) 40/9
c) 40/11
d) None of the mentioned
Answer: b
Explanation: Pitch=Uncompressed length/N-1.

14. Active and inactive, both types of coils support the load although both don’t participate in spring action.
a) Active coils don’t support the load
b) Inactive coils don’t support the load
c) Both active and inactive don’t support the load
d) Both active and inactive support the load
Answer: b
Explanation: Inactive coils don’t support the load.

15. If a spring has plain ends then number of inactive coils is?
a) 1
b) 2
c) 3
d) 0
Answer: d
Explanation: There are no inactive coils in plain ends.

16. Spring having square ends has 1 inactive coil.
a) True
b) False
Answer: b
Explanation: There are 2 inactive coils.

17. Martin’s factor compensates for curvature effect in springs.
a) True
b) False
Answer: b
Explanation: Wahl’s factor accommodates curvature effect while designing spring.

18. The angle of twist for the equivalent bar to a spring is given by? (Symbols have their usual meaning)
a) 8PD²N/Gd⁴
b) 16PD²N/Gd⁴
c) 16PDN/Gdᵌ
d) 8PDN/Gdᵌ
Answer: b
Explanation: θ=Ml/GJ where M=PD/2, l=πDN and J=πd⁴/32.

19. The axial deflection of spring for the small angle of θ is given by?
a) 328PDᵌN/Gd⁴
b) 8PDᵌN/Gd⁴
c) 16PDᵌN/Gd⁴
d) 8PD²N/Gdᵌ
Answer: b
Explanation: Deflection=θxD/2.

20. A spring of stiffness constant k is cut in two equal parts. The stiffness constant of new spring will be k/2.
a) True
b) False
Answer: b
Explanation: k=Gd⁴/8DᵌN, hence k is inversely proportional to number of coils. Thus result will be 2k.

21. For two spring connected in series, the force acting on each spring is same and equal to half of the external force.
a) True
b) False
Answer: b
Explanation: Te force on each spring is equal to the external force.

22. For two springs connected in series, the net deflection is equal to the sum of deflection in two springs.
a) True
b) False
Answer: a
Explanation: The net deflection is sum of the deflection of sprigs connected in series.

23. For two springs connected in parallel, net force is equal to the sum of force in each spring.
a) True
b) False
Answer: a
Explanation: Net force applied is distributed in the two springs.

24. Patenting is defined as the cooling below the freezing point of water.
a) True
b) False
Answer: b
Explanation: Patenting is heating steel above critical range followed by rapid cooling.

25. Find the Wahl’s factor if spring index is 6.
a) 1.2020
b) 1.2424
c) 1.2525
d) 1.5252
Answer: c
Explanation: K=[4C-1/4C-4]+0.615/C.

26. Find the shear stress in the spring wire used to design a helical compression sprig if a load of 1200N is applied on the spring. Spring index is 6, and wire diameter 7mm.
a) 452.2N/mm²
b) 468.6N/mm²
c) 512.2N/mm²
d) None of the listed
Answer: b
Explanation: τ=K x 8PC/πd² where K=[4C-1/4C-4]+0.615/C.

27. Find the mean coil diameter of a helical compression sprig if a load of 1200N is applied on the spring. Spring index is 6, and wire diameter 7mm.
a) 7/6mm
b) 42mm
c) 1200×6/7 mm
d) None of the listed
Answer: b
Explanation: D=Cd.

28. Find total number coils in a spring having square and ground ends. Deflection in the spring is 6mm when load of 1100N is applied. Modulus of rigidity is 81370N/mm². Wire diameter and pitch circle diameter are 10mm and 50mm respectively.
a) 7
b) 6
c) 5
d) 4
Answer: a
Explanation: Deflection=8PDᵌN/Gd⁴ or N=4.4 or 5. Total coils=5+2(square grounded ends).

29. A railway wagon moving with a speed of 1.5m/s is brought to rest by bumper consisting of two springs. Mass of wagon is 100kg. The springs are compressed by 125mm. Calculate the maximum force acting on each spring.
a) 1200N
b) 1500N
c) 1800N
d) 2000N
Answer: c
Explanation: mv²/2=Pxdeflection/2.

30. For a helical torsion sprig, the stress concentration factor at outer fibre is? Give spring index=5.
a) 0.78
b ) 0.87
c) 1.87
d) 0.69
Answer: b
Explanation: K=4C²+C-1/4C(C+1).

31. In concentric springs, vibrations called surge are amplified.
a) True
b) False
Answer: b
Explantion: Surge are eliminated.

32. Can concentric springs be used to obtain a force which is not proportional to its deflection?
a) True
b) False
Answer: a
Explanation: Two sprigs having different free length can be meshed to obtain such a result.

33. The load shared by each spring is inversely proportional to the cross section of wire.
a) Yes
b) No, it is directly proportional
c) It is proportional to its square
d) It is proportional to its square root
Answer: b
Explanation: It is directly proportional to the cross section of wire.

34. A concentric spring consists of 2 sprigs of diameter 10mm and 4mm. The net force acting on the composite spring is 5000N. Find the force acting on each of the two springs.
a) 1232.2N and 3767.8N
b) 786.4N and 4213.6N
c) 689.7N and 4310.3N
d) 645.3N and 4354.7N
Answer: c
Explanation: P₁/P₂=d₁²/d₂² and P₁+P₂=5000.

35. If the spring have same solid length and number of coils in the two springs are 8 and 10, then find the diameter of the spring with 8 coils. It is given diameter of spring with 10 coils is 12mm.
a) 9.6mm
b) 9mm
c) 12mm
d) 15mm
Answer: b
Explanation: N₁d₁=N₂d₂.

36. Two spring having stiffness constants of 22N/mm and 25N/mm are connected in parallel. They are to be replaced by a single spring to have same effect. The stiffness of that spring will be?
a) None of the mentioned.
b) 3N/mm
c) 47N/mm
d) 11.7N/mm
Answer: c
Explanation: k=22+25.

37. What will happen if stresses induced due to surge in the spring exceeds the endurance limit stress of the spring.
a) Fatigue Failure
b) Fracture
c) None of the listed
d) Nipping
Answer: a
Explanation: If endurance limit is passed, fatigue failure will follow.

38. Surge is caused by resonance effect in the spring.
a) True
b) False
Answer: a
Explanation: When the natural frequency of vibrations of the spring coincides with frequency of external periodic force, resonance occurs.

39. Surge is a desirable effect in the sprigs.
a) True
b) False
Answer: b
Explanation: Surge means vibratory motion which isn’t welcomed anywhere in the machinery.

40. For a helical torsion sprig, the stress concentration factor at inner fibre is? Give spring index=5.
a) 1.005
b ) 1.175
c) 1.223
d) 1.545
Answer: b
Explanation: K=4C²-C-1/4C(C-1).

41. Bending stress in graduated length leaves are more than that in full length leaves.
a) Yes
b) No
c) In some cases
d) Can’t be stated
Answer: b
Explanation: Bending stress in full length leaves is around 50% higher than graduated length leaves.

42. Nip is the initial gap between extra full length leaf and the graduated length leaf before the assembly.
a) True
b) False
Answer: a
Explanation: Nipping is done to balance the bending stress in the full length leaf and graduated length leaf.

43. Nipping is defined as leaving the gap between full length leaf and graduated length leaf.
a) True
b) False
Answer: b
Explanation: Nipping is pre stressing achieved by a difference in radii of curvature.

44. A leaf spring consists of 3 extra full length leaves and 14 graduated length leaves. The maximum force that can act on the spring is 70kN and the distance between eyes of the spring is 1.2m. Width and thickness of the leaves are 100mm and 12mm respectively. If modulus of elasticity is 207000N/mm², calculate the initial nip.
a) 26.8mm
b) 24.9mm
c) 22.5mm
d) 23.1mm
Answer: b
Explanation: C=2PLᵌ/Enbtᵌ where L=1.2/2, n=3+14,P=70/2.

45. A leaf spring consists of 3 extra full length leaves and 14 graduated length leaves. The maximum force that can act on the spring is 70kN and the distance between eyes of the spring is 1.2m. Width and thickness of the leaves are 100mm and 12mm respectively. Calculate the initial pre load required to close the nip.
a) 4332.2N
b) 4674.1N
c) 4985.4N
d) Can’t be determined
Answer: b
Explanation: P=2x3x14x35000/17(3×3+2×14).

46. Belleville spring can be used in clutch application.
a) True
b) False
Answer: a
Explanation: When h/t=2.1, the Belleville spring can be used in clutches.

47. Belleville spring can only produce linear load deflection characteristics.
a) Only linear
b) Linear as well as non linear
c) Non-linear
d) None of the mentioned
Answer: b
Explanation: Beliville spring can provide any linear or non-linear load deflection characteristics.

48. When two Belleville sprigs are arranged in series, half deflection is obtained for same force.
a) One fourth deflection
b) Double deflection
c) Four time deflection
d) None of the listed
Answer: b
Explanation: Double deflection is obtained.

49. When two Belleville springs are in parallel, half force is obtained for a given deflection.
a) Half force
b) Double force
c) Same force
d) Can’t be determined
Answer: b
Explanation: Double force is obtained.

50. Propagation of fatigue failure is always due to compressive stresses.
a) Due to bending
b) Due to tensile
c) Due to fatigue
d) None of the listed
Answer: b
Explanation: Propagation is always due to tensile stresses.

51. The strain energy stored in a spiral spring is given by?
a) 12M²L/Ebtᵌ
b) 6M²L/Ebtᵌ
c) 8M²L/Ebtᵌ
d) None of the listed
Answer: b
Explanation: U=Mθ/2 where θ=12ML/Ebtᵌ.