## Machine Design – 1 [MCQ’s]

#### Module 01

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
Explanation: Failing simply means unable to perform its function satisfactorily.

2. Type of load affects factor of safety.
a) True
b) False

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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: Definition of buckling.

7. The critical buckling load depends upon which of the following parameters?
a) Yield strength
b) Modulus of elasticity
d) Each of the mentioned
Explanation: 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
Explanation: 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
Explanation: Formula.

10. Modulus of rigidity for carbon steels is greater than that of grey cast iron.
a) True
b) False
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: Bending Stress= (Bending Moment x distance from neutral axis/ Moment of inertia).

14. The normal stress is perpendicular to the area under considerations, while the shear stress acts over the area.
a) True
b) False
Explanation: This is the convention used.

15. If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 20N/mm²Find the maximum amount of shear stress to which the body is subjected.
a) 22.4mm
b) 25mm
c) 26.3mm
d) 27.2mm
Explanation: τ(max)=√( [σ(x)-σ(y) ]²/2² + τ²).

16. If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 10N/mm². Find the inclination of the plane in which shear stress is maximal.
a) 45’
b) 30’
c) 60’
d) 15’
Explanation: tan (2Ǿ)=2τ/[σ(x) – σ(y)].

17. If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 20N/mm². Find the maximum normal stress.
a) 90
b) 92.4
c) 94.2
d) 96
Explanation: σ=[σ(x) +σ(y)]/2 + √( [σ(x)-σ(y) ]²/2² + τ²).

18. If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 20N/mm². Find the minimum normal stress.
a) 45.4
b) 47.6
c) 48.2
d) 50.6
Explanation: σ=[σ(x) +σ(y)]/2 – √( [σ(x)-σ(y) ]²/2² + τ²).

19. If compressive yield stress and tensile yield stress are equivalent, then region of safety from maximum principal stress theory is of which shape?
a) Rectangle
b) Square
c) Circle
d) Ellipse

20. Maximum Principal Stress Theory is not good for brittle materials.
a) True
b) False
Explanation: Experimental investigations have shown that maximum principle stress theory gives good results for brittle materials.

21. The region of safety in maximum shear stress theory contains which of the given shape
a) Hexagon
b) Rectangle
c) Square
d) None of the mentioned
Explanation: In maximum shear stress theory we have the following equations: σ1= ±S(yt)
σ2= ±S (yt), σ1 – σ2 =±S (yt) assuming S(yt)=S(yc).

22. The total strain energy for a unit cube subjected to three principal stresses is given by?
a) U= [(σέ) ₁ + (σέ) ₂+ (σέ) ₃]/3
b) U= [(σ₁²+σ₂²+σ₃²)/2E] – (σ₁σ₂+σ₂σ₃+σ₃σ₁)2μ
c) U= [(σέ) ₁ + (σέ) ₂+ (σέ) ₃]/4
d) None of the mentioned
Explanation: U= [(σέ) ₁ + (σέ) ₂+ (σέ) ₃]/2 is the basic formula. After substituting values of έ₁, έ₂ and έ₃, we get the expression b.

23. Distortion energy theory is slightly liberal as compared to maximum shear stress theory.
a) True
b) False
Explanation: The hexagon of maximum shear theory falls completely inside the ellipse of distortion energy theorem.

#### Module 02

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

#### Module 03

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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Explanation: Trapezoidal and not square threads suffer from the problem of bursting.

a) True
b) False
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
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
c) Diameter
d) None of the mentioned

39. Nominal diameter of the screw thread is the same as core diameter.
a) True
b) False
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
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
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
Explanation: Self locking takes place if load does not descend on its own and that is possible only in c condition.

#### Module 04

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
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
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
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
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
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
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
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
c) Removal of undesired material
d) Each of the mentioned
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
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
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
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
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
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
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
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
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
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.

#### Module 05

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
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
Explanation: That is axle and not a spindle.

3. Axle is frequently used in torque transmission.
a) True
b) False
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
Explanation: The axle may be stationary or rotate with the element.

5. Counter shaft is a secondary shaft.
a) True
b) False
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
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
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
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
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
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
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
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
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
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
d) None of the listed
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
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
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
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
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
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
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
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
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
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
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
Explanation: Sunk key is a positive drive and no slip occurs.

27. Woodruff key permits _____ movement b/w shaft and the hub.
a) Axial
c) Eccentric
d) None of the listed
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
Explanation: l=M/[dbτ√2].

29. Splines are keys.
a) True
b) False
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Explanation: Flexible shafts have low rigidity in bending moments making them flexible.

#### Module 06

1. Which of the following function can the spring perform?
a) Store energy
b) Absorb shock
c) Measure force
d) All of the mentioned
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
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
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
Explanation: It is called master leaf.

5. Multi leaf springs are not recommended for automobile and rail road suspensions.
a) True
b) False
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
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
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
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
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
Explanation: Solid length=8×3.

11. Compressed length is smaller than the solid length.
a) True
b) False
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
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
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
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
Explanation: There are no inactive coils in plain ends.

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

17. Martin’s factor compensates for curvature effect in springs.
a) True
b) False
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
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
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
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
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
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
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
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
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
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
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
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
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
Explanation: K=4C²+C-1/4C(C+1).

31. In concentric springs, vibrations called surge are amplified.
a) True
b) False
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
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
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
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
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
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
Explanation: If endurance limit is passed, fatigue failure will follow.

38. Surge is caused by resonance effect in the spring.
a) True
b) False
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
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
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
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
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
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
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
Explanation: P=2x3x14x35000/17(3×3+2×14).

46. Belleville spring can be used in clutch application.
a) True
b) False
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
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
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
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
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