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v. Impact vi. Reliability

7. What are the types of variable stresses?a. Completely reversed or cyclic stressesb.Fluctuating stresses, c.Repeated stresses

8. Differentiate between repeated stress andreversed stress.

Repeated stress refers to a stress varying from zero to amaximum value of same nature. Reversed stress of cyclicstress varies from one value of tension to the same value of compression.

9. What are the types of fracture?a. Ductile fracture b.Brittle fracture

10. Distinguish between brittle fracture and ductilefracture.

In brittle fracture, crack growth is up to a small depth of the material. In ductile fracture large amount of plasticdeformation is present to a higher depth.11. Define stress concentration and stress

concentration factor. Apl 09Stress concentration is the increase in local stresses atpoints of rapid change in cross section or discontinuities.Stress concentration factor is the ratio of maximum stress atcritical section to the nominal stress12. Explain size factor in endurance strength.

Size factor is used to consider the effect of the size onendurance strength. A large size object will have more

defects compared to a small one. So, endurance strength isreduced.If K is the size factor, then

Actual endurance strength = Theoretical endurance limit x K 13. What is Griffith theory. (Or) State the condition

for crack growth .

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A crack can propagate if the energy release rate of crackis greater than crack resistance.14. What are the modes of fracture?

a. Mode I (Opening mode) Displacement is normal tocrack surface.

b.Mode II (Sliding mode) Displacement is in the plane of the plate.

C.Mode III (Tearing mode) Out of plane shear.15. What are the factors to be considered in the

selection of materials for a machine element?I.Required material properties, ii.Manufacturing easeIii.Material availability, iv.Cost.

16. What are various theories of failure?i.Maximum principal stress theory. ii.Maximum shear

stress theory. iii.Maximum principal strain theory.iv.Distortionenergy theory. v.Maximum strain energy theory.17. List out the factors involved in arriving at factor of safety

i.material properties, ii.nature of loads

iii.presence of localized stresses, iv.mode of failures18. Give some methods of reducing stressconcentration.

(Nov 08)i.Avoiding sharp corners, ii.Providing fillets.iii.Use of multiple holes instead of single holeiv.Undercutting the shoulder parts.

19. Explain notch sensitivity.

State the relation between stress concentration factor,fatigue stress concentration factor and notch sensitivity.Notch sensitivity (q) is the degree to which the theoreticaleffect of stress concentration is actually reached. Therelation is, K f = 1 + q(Kt-1)20. What are the factors that affect notch sensitivity?

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i. Material, ii. Notch radius, iii.Size of componentiv.Type of loading, v.Grain Structure

21. What is the use of Goodman & Soderbergdiagrams?

They are used to solve the problems of variable stresses.22. Define machinability

It is the property of the material, which refers to arelative case with which a material can be cut. It is measuredin a number of ways such as comparing the tool life forcutting different material23. What is an S-N Curve?

An S- N curve has fatigue stress on Y axis and number of loading cycles in X axis. It is used to find the fatigue stressvalue corresponding to a given number of cycles.24. Define Ductility

It is the property of the material enabling it to be drawninto wire, with the application of tensile force. It must be bothstrong and plastic. It is usually measured in terms of percentage elongation and reduction in area. (eg) Ni, Al, Cu

25. Define fatigueWhen a material is subjected to repeated stress, it failsat stresses below the yield point stress; such type of failureof the material is called fatigue.26. What is curved beam?

In curved beam the neutral axis does not coincide withthe centroidal axis.27. Give some example for curved beam.

C frame, crane hook 28. What is principle stress and principle plane?A plane which has no shear stress is called principle

plane the corresponding stress is called principle stress.29. Write the bending equation.

M/I = E/R = Fs/Y. M Bending moment

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I - Moment of intertia E - Youngs modulus R - Radiusof the shaft Fs Shear stress Y - Distance from neutral axis

30. Write the torsion equation . T/J = C/L = Fs/R, T Torque , J - Polar moment of

inertia C- Rigidity modulus Angle of twist L Length of the shaft Fs Shear stress R - Radius of the shaft31.Differentiate the stress distribution in abarsubjected to axial force and bending force

Principal stress is used to know about design stressadopted for a machine componentsubjected to two types of stresses combinely

32. For ductile material , which of the strength isconsidered for designing APL 08

(a)Component subjected to static loading(b) Component subjected to fatigue loading

Tensile stress is considered for static loadingCompressive stress is considered for fatigue loading.

33.Mention some standard codes of specification of steels

Steels are specified by x.C.ywhere x- number indicating 100 times the average

percent of carbon and y- number indicating 100 times theaverage percent of manganese.34.How will you account for stress concentration indesign of the machine elements

Mathematical method based on theory of plasticity

Experimental method based on theory of photoelasticity.35. What you mean by optimum design?Optimum design is the process of maximizing the desired

quantity or minimizing the undesired one.It is the processused to increase the productivity.

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36. State Rankines theory(Nov 07)Rankines theory states that when a copmponent is subjectedto biaxial or uniaxial stresses the failure occurs when themaximum normal stress is equal to the tensile yield strengthof the material in a simple tension test.37. What is gerber theory?Nov 09Gerber parabola joins the endurance stress and ultimatestress like goodman line. According to Gerber method, therelationship between m. a, u and -1 is given by

a = -1[1-( m/ u)]

38.What are the factors that govern selection of materials while designing a machine

component? Strength & stiffness

Surface finish and tolerances

Manufacturability

Ergonomics and aesthetics

Working atmosphere39.Write soderberg equation for machine componentsubjected to (a) Combination of mean and variabletorques

(b) Combination of mean and variable bendingmoments.

eq = y/n =m+kf*a*y/-1

eq = y/n = m+kf* a* y/ -1

40. State three conditions where tap bolts areused.

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One of the parts being joined has enough thicknessto accommodate a threaded hole.

Insufficient apace for a nut.

Material is strong enough so that the threads havelong life.

41.What are the reasons of replacing riveted jointby welded joint in modern equipment ?

Material is saved in welding joints and hence the

machine element will be light if welded joints are usedinstead id riveted joints. Lead proof joints can be easilyobtained by welded joints compared riveted joints.

42.What is the effect of increase in wire diameteron the allowable stress value?

The direct shear stress (1/d) and torsional shearstress (1/d). Hence, increase in wire diameter reducesthat allowable stress value.

43.What type of stresses is produced in a discflywheel?

Tensile stress due to centrifugal force

Tensile bending stress caused by the restraint of the arms and

The shrinkage stresses due to unequal rate of cooling of casting.

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44.For a journal bearing, the maximum operatingtemperature must be less than 80c.why?

Temperature rise will result in the reduction of the

viscosity as the oil used in the bearing. This would leadto metal to metal contact, thereby affecting the bearingperformance & life.

45.Why is piston end of a connecting rod kept smallerthan the crank pin end?

The piston end of the connecting rod experiences lessbending moment than the crack end. Hence, on the basisof beam of uniform strength, the piston end of theconnecting rod is smaller.

16 MARKS

1. A medium force fit on a 50mm shaft requires ahole tolerance of 0.025mm and a shaft toleranceof 0.016mm. The maximum interference is to be0.042mm. How will you dimension the hole andthe shaft, if hole deviation is H? Dec 2010

Given data:

Basic diameter of shaft, d=50mm

Hole tolerance = 0.025mm

Shaft tolerance = 0.016mm

Maximum interference = 0.042mm

Hole deviation = H

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Solution:

Hole:

Since this is a H system, the hole has the basicsize. In other words, lower limit of hole = basic size

Dmin = 50mm

Hole tolerance = Dmax- Dmin = 0.025

Dmax = 0.025+ Dmin = 0.025+50=50.025

The hole is dimensioned as 50-0.000+0.025

Shaft:Maximum interference = max. dia of shaft (d max)-Min.dia of hole (Dmin)

0.042 = d max-50

Shaft tolerance = d max- d min = 0.016mm

(or) 50.042-d min = 0.016

(or) d min = 50.042-0.016 = 50.026mm

The shaft is dimensioned as 50-0.000+0.025

2. Explain the following with mathematicalexpressions. Maximum principal stress theory; Von-mises stress theory. MAY 08Maximum principal stress theory:

This theory states that the failure of themechanical component occurs when subjected to biaxial oruniaxial stresses, when the maximum principal stress reachesthe ultimate or yield strength of the material.according to thistheory,

The failure occurs when, Su or Sy. If F.O.S isconsidered then,

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(Su or Sy)/N f .Where, -max principal stress,

Su- Ultimate stress, Sy- yield strength of the material.

Von-mises stress theory : This theory is also known asDistortion energy theory. This theory states that the failure of the mechanical component occurs when subjected to biaxialor uniaxial stresses, when the distortion energy per unitvolume reaches the limiting distortion energy per volume atthe yield point.acc to this theory,

[ 1 - 1 2 + 2 ] S yt /N f Where S yt Yield tensile strength of the material.

3. A cost iron pulley transmits 12 kW at 330 rpm. Thediameter of the pulley is 1.3 m and it has four straightarms of elliptical cross section in which the major axisis twice the minor axis.Determine the dimensions of the arm if the allowable bending stress is 18 MPa.MAY 08Given data :

R=D/2=1300/2=650 mm

Power=12kW,n=330 rpm ,No of arms = 4,a= 2b,b = 18 n/mm

Solution: Torque actin on pulley (T):

P =2nT/(60*1000) 12 =2 * 330*T/60000 ,

T=347.247Nm.

Force on each arm(F) :F= T/NR= 347247/(4*650) F = 133.56 N

Maximum bending moment M:M= F(R-r h) ,

where r h= radius of hub,as r h

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M = 86811.75 NmmDimensions of arm cross section :

Z= /32 ab for elliptical cross section,Z= b/8,

also b =

M/Z

18= 86811.75/( b/8)),b = 12281.35,b= 23.07 mma = 2b= 23.07*2a =46.2mm.

4. A shaft of 760 mm length is simply supported at itsends. It is subjected to a central concentrated cyclicload that varies from 12 to 36 kN. Determine thediameter of the shaft assuming a factor of safety of 2,size correction factor of 0.8, and surface correctionfactor of 0.85. The material properties are ultimatestrength = 500 MPa; yield strength = 280 MPa; andendurance limit = 250 MPa. Fatigue stressconcentration factor = 1.5. May 08Given data:

L= 760 mm,Fmax= 36 kN,

Fmin =12 kN,Nf= 2Kb= 0.8,Ka= 0.85,Kf=1.5,Sut= 500 N/mmSyt=280 N/mm ,Se= 250 N/mm

Solution:Ke = 1/Kf Ke =1/1.5= 0.6667Se= Ka.Kb.Kf.Se = 0.85*0.8*0.6667*250

= 113.33 N/mm.Mean stress and stress amplitude:

M= F/2 * L/2Mmax= 36000/2*760/2 =68.4 10 Nmm

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UNIT II

Design of Shafts and couplings

TWO MARKS

1. Define the term critical speed.

The speed, at which the shaft runs so that the additionaldeflection of the shaft from the axis of rotation becomesinfinite, is known as critical or whirling speed. 2. What are the factors considered to design a shaft?

i.strength, ii.stiffness3. What is key?

A key is device, which is used for connecting twomachine parts for preventing relative motion of rotation withrespect to each other.4. What are the types of keys?

i.Saddle key, ii.Tangent key, iii.Sunk keyiv.Round key and taper pins

5. What is the main use of woodruff keys?A woodruff key is used to transmit small value of torque

in automotive and machine tool industries. The keyway in theshaft is milled in a curved shape whereas the key way in thehub is usually straight.6. List the various failures occurred in sunk keys.

1.Shear failure 2.Crushing failure

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7. What is the function of a coupling between twoshafts?

Couplings are used to connect sections of longtransmission shafts and to connect the shaft of a drivingmachine to the shaft of a driven machine.

8. Under what circumstances flexible couplings areused?Nov 08, Nov 09

They are used to join the abutting ends of shafts when

they are not in exact alignment. They are used to permit anaxial misalignment of the shafts without under absorption of the power, which the shafts are transmitting.

9.What are the purposes in machinery for whichcouplings are used?

i.To provide the connection of shafts of units those aremanufactured separately such as motor and generator and to

provide for disconnection for repairs or alterations.ii.To provide misalignment of the shafts or to introducemechanical flexibility.

iii.To reduce the transmission of shock from one shaft toanother.

iv.To introduce protection against over load.

10.What are the main functions of the knuckle joints?

It is used to transmit axial load from one machineelement to other.11. Why a hollow shaft has greater stiffness andstrength than solid shaft of equal weight?April 08

For same weight, the moment of inertia as well as polarmoment of inertia of hollow shaft is higher than the solid

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shaft. Therefore the strength,stiffness and rigidity are higherfor hollow shaft.12.Why is maximum shear stress used for shaft

The shaft is made of ductile material so, the maximumshear stress is used for finding thefailure.13.What types of stresses are developed in the key?Direct shear stress and Crushing or compressive stresses are

developed in the key.

14 .Differentiate between keys and splinesS.No

KEYS SPLINES

1 It is used on the shafts tosecure the rotatingelements like gears pulleysor sprockets to prevent therelative motion betweenthem

The Splines are the multiplekeys which are made integralwith the shafts

2. Keys prevent the relativerotary as well as the axialmotion between the shaftand hub.

Splines prevent the relativerotary motion but permits theaxial motion between theshaft and hub

3. Torque transmissioncapacity of key is moderate

Torque transmission capacityof splines is high.

4. It is used in couplings tosecure the gears pulleysand couplings or sprocketsto the shaft.

The splines are used inautomobile gear boxes andmachine tool gear boxes.

15. What is the significance of slenderness ratio in

shaft design?Nov 08 The higher slenderness ratio will have the followingeffects on shafts:

(i) If the shaft is subjected to compressive force, thehigher slenderness ratio will increase the possibility of buckling failure

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(ii) The higher slenderness ratio will increase the torsionaland lateral deflection on the shaft .

16.What do you mean by stiffness and rigidity withreference to shafts?

Stiffness is the resistance offered by the shaft fortwisting and rigidity is the resistance offered by the shaftfor lateral bending.

17.Suggest suitable couplings for

Shafts with parallel misalignment

Shafts with angular misalignment of 10

Shafts in perfect alignment.

Shafts with parallel misalignment

Flexible coupling such as spring coupling can beused for shafts with parallel misalignment

Shafts with angular misalignment of 10

Universal coupling is suitable for shafts with angularmisalignment of 10

Shafts in perfect alignment

Rigid coupling can be used for shafts in perfectalignment.

16 MARKS

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1 .Design a bushed-pin type of flexible coupling toconnect a pump shaft to a motor shafttransmitting 30 kW at 900 rpm. The overall torque is15% more than mean torque. The

material allowable properties are as follows : (in crushing for shaft and key material) = 80 MPa (in shear for shaft and key material) = 40 MPa (in shear for cast iron) = 15 MPa.Bearing pressure for rubber bush = 0.8 MPaMaterial of the pin as same as that of shaft and keyMay 08

Given data: p =30 kW; n = 900 rpm; Tmax = 1.15 TmeanSolution:

P =2 nT/(60*10 Tmax=1.15*318.31 = 366.06 Nm max = 16 Tmax /d40 = 16* 366056/ d

d = 40 mm;

Design of key :From PSGDB P.NO :5.21, for d= 40 mm;W= 12 mm and h= 8 mm,

l= 1.5 d=1.5*h o =60 mm,crushing of key : c = (Tmax/d/2)/(h/2*l)subsitituting the values we get l =91.51 =92 mmShearing of key : = (Tmax/d/2)/wlSolving this equation we get l = 57.19 =58 mm

Taking larger value ,l=92 mmDesign of hub :Length of hub: l =92 mmouter dia of hub, D=2dD=2*40=80 mmTorsional sheat stress in hub: h= 16 Tmax/(D(1-K 4))

h= 3.88

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Direct shear stress induced : f = Tmax/D/2/Dt f = 2*366056/ Dt f = 1.82 < 14N/mm,Hence safe.

Design of pin:

For d= 40 mm, no of pins, N = 4Nominal dia of pin: d b = 0.5d/N =10mm

2.A solid shaft of diameter d is used in powertransmission.Due to modification of existingtransmission system, it is required to replace the solidshaft by a hollow shaft of the same material and equalstrength in torsion. If the recommended weight of thehollow shaft per meter length is half of the solid shaft,determine the diameter of the hollow shaft in terms of d. Dec 2009Solution

For solid shaft: max = 16T/d

T = * max* d/16 . eqn(1)For hollow shaft :

max = 16T/ d o[1-k 4]= 16T/ d o[1- (d i/d o)4]

= 16Td o/[d o4 - d i4]

T = max [d o4

- d i4

]/ 16d o.eqn(2) Equal strength solid and hollow shaft:

* max* d/16 = max [d o4 - d i4]/ 16d o

d i4 = d o4-d od .eqn (3)Weight of solid and hollow shafts:Weight of solid shaft per metre length :

Ws = dg/4 ..eqn(4)Weight of solid shaft per metre length :Wh = [d o - d i]g/4..eqn(5)

Wh =1/2 WsFrom eqn (4) and eqn (5) , we get

[do - di]g/4 = dg/4

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d o - d i = d/2d i = d o - d/2d i4 = d o4 - d o d 2 + d 4/4 eqn(6)

Outer diameter of hollow shaftFrom eqn (3) and eqn (6)d o4-d od = d o4 - d o d 2 + d 4/4d o d 2 - d od - d 2/4 = 0,

solving the above eqn we get ,d o = d 2( d 2 + d 2)/2,

= [d 2 + 2 d]/2d o = [ (1 + 2)/2]d

3.Design a rigid type of flange coupling to connect two

shafts. The input shaft transmits 37.5kW power at 180rpm to the output shaft through the coupling. Theservice factor for the application is 1.5. Select suitablematerial for various parts of the coupling. NOV 2010Given data:Power, p = 37.5kW = 37.5*10 WSpeed, N =180 rpmService factor = 1.5

Solution: Diameter of shaft, d:

Torque transmitted by the shaft, Mt

Mt = 1.5*p*60/2 N = 1.5*37.5*10*60/2 *180

= 2984.15 N-m = 2.984*10 6 N-mm ..(1)

Assume that the shaft key and bolts are made of mild steel

which is havingAllowable shear strength for steel = 60/mm

Allowable crushing strength for steel = 120 N/mm

Torque transmitted by the shaft, Mt

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Mt = /16*s*d (2)

From eqn (1) and (2)

2.984*10 6 = /16*s*d

D= 63.27mm

Say, diameter of the shaft, d = 64mm

Outside diameter of hub, D

D = 2d =2*64 =128mm

Pitch circle diameter of bolts, D1

D1 = 3d= 3*64 = 192mm

Outer diameter of flange, D 2

D2 = 4d = 4* 64 256mm

Length of hub, L

L = 1.5d = 1.5*64 = 96mm

Thickness of flange, tf

Tf = 0.5d =0.5*64 = 32mm

Design of hub:

Assume that the hub is made of cast iron the allowableshear strength of cast iron is 15N/mm. We know that

Since h is less than the allowable shear stress 15N/mm.Hence the design is safe.

Design of key:

From PSGMB 5.16, corresponding to d = 64mm

Width of the key, b = 18mm

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Height of the key, h = 11mm

Length of the key, l = 96mm (same as hub length)

Check for shear

This is less than the allowable value 60N/mm. Hencethe design is safe.

Check for crushing

Since, ck is more than the allowable value of 90N/mm.

The design is not safe. Hence increase the height of the key.

2.984*10 6 = 96*h/2*120*64/2

H = 16.18mm, say the revised height of key, h 17mm

Design of flange

Since h is less than the allowable shear stress15N/mm. Hence the design is safe.

Diameter of bolt, db

Torque transmitted by the bolts due to shear stress

Assume number of bolt, n = 4

Say, diameter of bolt, db = 13mm

Check for crushing

Since cb is less than 120N/mm. Hence, the design is safe.

4.Design a bushed-pin type of flexible coupling toconnect a pump shaft to a motor shaft

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transmitting 30 kW at 900 rpm. The overall torque is15% more than mean torque. Thematerial allowable properties are as follows : (in crushing for shaft and key material) = 80 MPa

(in shear for shaft and key material) = 40 MPa (in shear for cast iron) = 15 MPa.Bearing pressure for rubber bush = 0.8 MPaMaterial of the pin as same as that of shaft and keyMAY 2008Given data: p =30 kW; n = 900 rpm; Tmax = 1.15 TmeanSolution:

P =2 nT/(60*1000) = 2**900*T/60000 T = 318.31 Nm ; Tmax=1.15*318.31 = 366.06 Nm max = 16 Tmax /d ; 40 = 16* 366056/ dd = 40 mm;Design of key :From PSGDB P.NO :5.21, for d= 40 mm;W= 12 mm and h= 8 mm,l= 1.5 d=1.5*h o =60 mm,crushing of key : c = (Tmax/d/2)/(h/2*l)subsitituting the values we get l =91.51 =92 mmShearing of key : = (Tmax/d/2)/wlSolving this equation we get l = 57.19 =58 mm

Taking larger value ,l=92 mmDesign of hub :Length of hub: l =92 mmouter dia of hub, D=2dD=2*40=80 mmTorsional sheat stress in hub: h= 16 Tmax/(D(1-K 4))

h= 3.88

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5. A shaft is supported by two bearings placed 1 mapart. A 600 mm diameter pulley is mounted at adistance of 300 mm to the right of left hand bearingand this drivesa pulley directly below it with the help of belt havingmaximum tension of 2.25 kN. Another pulley 400 mmdiameter is placed 200 mm to the left of right handbearing and is driven with the help of electric motorand belt, which is placed horizontally to the right. Theangle of contact for both the pulleys is 180 and=0.24. Determine the suitable diameter for a solid

shaft, allowing working stress of 60 MPa in tension and40 MPa in shear for the material of shaft. .Assume thatthe torque on one pulley is equal to that on the otherpulley. Dec 2009Given data: d A = 600 mm,d B = 400 mm,FA1 = 2250 N, = rad ,

= 0.24, t = 60 N/mm , =40 N/mmSolution:

FA1 / F A2 = e = e 0.24 *

2250/ F A2 = 2.125

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FA2 = 1058.6 NVertical force on pulley A ,

FA = F A1 + F A2 = 3308.6 NTorque on shaft T

T = (FA1

- FA2

) dA/

2 = (2250 1058.6)300 T = 357.42 * 10 Nmm.Horizontal force on pulley B

FB1/FB2 = e = 2.125

FB1 = 2.125 F B2 T = [F B1 - FB2]* d B /2357.42* 10 = (2.125 F B2 - FB2)200FB2 = 1588.53 NFB1 = 2.125*1588.53 = 3375.63 N

Horizontal force on pulley B F B = F B1+F B2 = 4964.16 N

Bending moment on shaft:Considering vertical loading diagram

300 FA 1000C A BDRCV RDV

Taking moment about point A,- FA * 300 + R DV * 1000 = 0

- 3308.6*300 + R DV * 1000 = 0

RDV = 992.6NRCV + R DV - 3308.6 = 0 RCV = 2316 NVertical bending moments at point A and B are

MAV = R CV * 300 = 2316 * 300 = 694800 NMBV = R DV * 200 = 992.6 * 200 = 198520 N

Horizontal loading diagram:

c A B D

Taking moment about point C- FB * 800 + R DH * 1000 = 0

- 4964.16* 800 + R DH * 1000 = 0

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RDH = 3971.33 NRCH + R DH -4964.16 = 0RCH = 992.83 N

Horizontal bending moments at points A and B are:

MAH

= RCH

* 300 = 992.83 * 300 = 297850 NMBH = R DH * 200 = 3971.83 * 200 = 794270 NResultant bending moments at A and B are:MA = [ M AH + M AV] = [ 694800 + 297850]

MA = 756000 N mmMB = [ M BV + M BH ] = [ 198520 + 794270]

MB = 818700 N mmMax Bending moment M = M B = 818700 N mm

Torque on shaft T = (F A1 FA2)*d A/2

= (2250-1058.6)*300= 357420 Nmm

Equivalent torque T e Te = [(K bM)+ (K t T)

Assume K b =1 , K t = 1= [ (1 * 818700) + (1*

357420)]= 893320 Nmm

max = 16T e /d

40 = 16*893320/ dd= 48.5 mmd =50 mm.

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UNIT-IIIDesign of temporary and permanent joints

TWO MARKS1. How is a bolt designated ?

A bolt is designated by a letter M followed by nominaldiameter and pitch in mm. It is given as Mdp

2. What factors influence the amount of initialtension?

i.External load ,ii.Material used, iii.Bolt diameter3. What is bolt of uniform strength?

A bolt of uniform strength has equal strength at thethread and shank portion.4. What are the ways to produce bolts of uniform

strength?

i.Reducing shank diameter equal to root diameter.ii.Drilling axial hole5 . What stresses act on screw fastenings?

i.Initial stresses due to screwing upii.Stresses due to external forcesiii.Combined stresses.

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12. What are the types of welded joints?i.Butt joint ii.Lap joint iii.T joint iv.Corner joint v.Edge

joint. 13.What are the two types of stresses are induced ineccentric loading of loaded joint?

1. Direct shear stress. 2. Bending or torsional shearstress. 14 Define butt and lap joint

Butt joint: The joint is made by welding the ends oredges of two plates.

Lap joint: The two plates are overlapping each other for acertain distance. Then welded. Such welding is called fillet

weld.15 When will the edge preparation need?

If the two plates to be welded have more than 6mmthickness, the edge preparation should be carried out.

16. What are the two types of fillet weld?i.Longitudinal or parallel fillet weld ii.Transverse fillet

weld

17. State the two types of eccentric weldedconnections.i.Welded connections subjected to moment in a plane of

the weld.ii.Welded connections subjected to moment in a plane

normal to the plane of the weld.18. What are the practical applications of welded

joints?

It has employed in manufacturing of machine frames,automobile bodies, aircraft, and structural works.19.State any two advantages of welded joints overrivet joints April 08,09

1. Light in weight,2. Low number of labour.

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3.Leak proof 20. What is the meaning of bolt M242M242 represents nominal diameter of bolt is 24 mm and thepitch is 2mm.

21.Name the possible modes of failure of riveted joints Tearing at weakened section and shearing and crushing

of the rivets.What is threaded joint?Threaded joint is defined as the seperable joint of two or

more machine parts held together by means of threadedfastenings such as bolt, nut etc..22. What is a stud ?

Stud is a bolt in which the head is replaced by athreaded end.It passes through one of the parts to be connected and isscrewed to the other part.

23.What is the minimum size for fillet weld? If the

required weld size from strength consideration is toosmall how will you fulfil the condition of minimum weldsize. Nov 08

It is defined as the minimum size of the weld for a giventhicknees of the thinner part joined or plate to avoid coldcracking by escaping the rapid cooling.

Size of the weld, h = 2 throat thickness.

16 MARKS1.A plate 75 mm wide and 12.5 mm thick is joined withanother plate by a single transverse weld and adouble parallel fillet weld as shown in Fig. 1. Themaximum tensile and shear stresses are 70 MPa and56 MPa respectively. Find the length of each parallel

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fillet weld, if the joint is subjected to both static andfatigue loading. Stress concentration factor fortransverse weld is 1.5 and for parallel fillet weld is2.7 May 08

Given data:b = 75 mm; h = 12.5 mm; t = 70 N/mm = 56 N/mm, kft = 1.5; Kfb = 2.7

Solution:Load carrying capacity:

P = t bh =70*75*12.5 = 65625 N Joint subjected to static loading P 1 = bt=bh/2

= 56*75*12.5/2= 37123.1 N

The load carried by double parallel welds is P 2P2 = (l+l)t = 2lh/2 = 989.95 l N

The total load P= P 1+ P 2; 65624 = 37123.1 + 989.95 l; l= 28.8 mm\

Joint subjected to subjective loading : The load carried by asingle transverse load isP1=bt/Ka = bt/kft 2 = 24748.74 N

The load carried by double parallel weldsP2 = /Kfb (l+l)t = 366.65 NTotal load P = P 1+ P 2 ;

l = 111.5 mm2. A welded joint as shown in Fig. 2 is subjected to aneccentric load of 2 kN. Find the size of weld, if themaximum shear stress in the weld is 25 MPa. May08

Given data :P = 2000 N; = 25 N/mm; e= 120 mm, l = 40mmSolution:

I.Moment of inertia for weld group abouthorizontal axis through C.G:

Ixx = 2[1/12 lt] = 1/6*40 t = 10666.67 t mm 4II.direct shear stress d:A = A1+A2 lt+ lt= 2lt= 2*40*t = 80t d =P/A = 2000/80t 25/t,Moment induced shear stress b

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M = Pe = 2000*120=240000 Nmm, y= l/2= 20 mm The max moment induced shear stress

d = M.Y/Ixx = 450/t N/mm Resultant shear stress:

= ( d +

b

) = 450.69/t N/mmWeld size (h)=25 = 450.69/t,t = 18.028 mmh= 2 t= 25.49 =26 mmt = h/2= 18.38 mm

3.Design a knuckle joint to with stand a tensile load of 70 KN using steelwith the following permissible stresses (in tension) = 60 MPa(in crushing) = 72 MPa(in shear) = 48 MPa. May 08

Given dataLoad p = 70000 N, (in tension) = 60 MPa

(in crushing) = 72 MPa ,(in shear) = 48 MPaSolution :

`Diameter of rod,dt = P/(2(.d/4)=38.54 =40 mm

Diametr of knuckle pin (d p): = P/(2( d p )/4), d p = 30.46 = 31 mm. Thickness of single eye(t): The crushing stress , c = P/( d p*t)72 = 70000/31tt 1 = 31.36 =32 mm

also, t 1 = 1.25 d = 1.25 40 = 50mm.Thickness of fork (t f ):The crushing stress induced in fork, c = P/(231 t f )

t f = 13.68 =16 mm.the minimum thickness of the fork = 0.75 d

= 0.75(40)=30 mmOuter diameter of eye (D):

Tensile stress induced in asingle eye, t= P/(D- d p)t60 = 70000/(D-31)50:D =54.33; D =55 mmshear stress induced in a single eye = P/(D- d p)t

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48 = 70000/(D-31)50;D =60.26 =61 mm.

failure of fork in tension: , t = P/(D- d p) t 1

t = 70000//2(61-31)30= 38.89 < 60 N/mmhence fork end is safe in tensionfailure of fork end in tension = P/2(D- d p) t 1 = 70000/2(61-31)30

= 38.89 < 48 N/mmhence fork end is safe in shear

UNIT-IVDesign of energy storing elements

TWO MARKS1. What is a spring?

A spring is an elastic member, which deflects, or distortsunder the action of load and regains its original shape afterthe load is removed.

2. State any two functions of springs.i. To measure forces in spring balance, meters and

engine indicators. ii.To store energy.3. What are the various types of springs?

i.Helical springsii.Spiral springs

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iii.Leaf springs iv.Disc spring or Belleville springs4. Classify the helical springs.

a.Close coiled or tension helical spring.b.Open coiled or compression helical spring.

5. Define Leaf springsA leaf spring consists of flat bars of varying lengths

clamped together and supported at both ends, thus acting asa simply supported beam.

6. Define Belleville SpringsThey are made in the form of a cone disc to carry a high

compressive force. In order to improve their load carryingcapacity, they may be stacked up together. The major

stresses are tensile and compressive.7. What is spring index (C)?

The ratio of mean or pitch diameter to the diameter of wire for the spring is called the spring index.8. What is pitch?

The axial distance between adjacent coils inuncompressed state.

9. What is solid length? The length of a spring under the maximum compression

is called its solid length. It is the product of total number of coils and the diameter of wire. Ls = n x d

Where, nt = total number of coils.

10. What are the requirements of spring whiledesigning?

a.Spring must carry the service load without the stressexceeding the safe value.

b.The spring rate must be satisfactory for the givenapplication.11. What are the end conditions of spring?

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a.Plain end.b.Plain and ground end

c.Squared endd.Squared and ground end.

12. What is buckling of springs? The helical compression spring behaves like a column

and buckles at a comparative small load when the length of the spring is more than 4 times the mean coil diameter.

13. What is surge in springs? The material is subjected to higher stresses, which may

cause early fatigue failure. This effect is called as springsurge.

14. What is a laminated leaf spring?In order to increase, the load carrying capacity, number

of flat plates are placed and below the other.15. What semi elliptical leaf springs?

The spring consists of number of leaves, which are heldtogether by U- clips. The long leaf fastened to the supportedis called master leaf. Remaining leaves are called asgraduated leaves.

16. What is nipping of laminated leaf spring?Prestressing of leaf springs is obtained by a difference of

radii of curvature known as nipping.17. What are the various application of springs?

The springs are used in various applications, theyare

a.Used to absorb energy or shocks (e.g. shockabsorbers, buffers, e.t.c.)b.To apply forces as in brakes clutches, spring-loaded

valves, e.t.c.c.To measure forces as in spring balances and

engine indicators

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d.To store energy as in watches18 Define free length.Apl 2011

Free length of the spring is the length of the springwhen it is free or unloaded condition. It is equal to the solidlength plus the maximum deflection or compression plus clashallowance.L f= solid length + Ymax + 0.15 Ymax

19. Define spring index .Spring index (C) is defined as the ratio of the mean

diameter of the coil to the diameter of the wire.C =D/d20. Define spring rate (stiffness).

The spring stiffness or spring constant is defined as theload required per unit deflection of the spring.

K= W/y , Where W-load ; y-deflection21. Define pitch .

Pitch of the spring is defined as the axial distancebetween the adjacent coils in uncompressed state.Mathematically Pitch=free length n-1

22. What are the points to be taken into considerationwhile selecting the pitch of the spring?

The points taken into consideration of selecting thepitch of the springs are

a.The pitch of the coil should be such that if the spring isaccidentally compressed the stress does not increase theyield point stress in torsion.

b.The spring should not be close up before themaximum service load is reached.

23. Define active turns.Active turns of the spring are defined as the number of

turns, which impart spring action while loaded. As loadincreases the no of active coils decreases.24. Define inactive turns.

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Inactive turns of the spring is defined as thenumber of turns which does not contribute to the springaction while loaded. As load increases number of inactive coilsincreases from 0.5 to 1 turn.

25. What are the different kinds of end connectionsfor compression helical springs? Nov 09

The different kinds of end connection forcompression helical springs area.Plain ends

b.Ground endsc.Squared ends

d.Ground & square ends26. Write about the eccentric loading of springs?

If the load acting on the spring does not coincide withthe axis of the spring, then spring is said to be have eccentricload. In eccentric loading the safe load of the springdecreases and the stiffness of the spring is also affected.

27. Explain about surge in springs?When one end of the spring is resting on a rigid supportand the other end is loaded suddenly, all the coils of springdoes not deflect equally, because some time is required forthe propagation of stress along the wire. Thus a wave of compression propagates to the fixed end from where it isreflected back to the deflected end this wave passes throughthe spring indefinitely. If the time interval between the load

application and that of the wave to propagate are equal, thenresonance will occur. This will result in very high stresses andcause failure. This phenomenon is called surge28. What are the methods used for eliminating surgein springs?

The methods used for eliminating surge are

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a.By using dampers on the center coil so that the wavepropagation dies out

b.By using springs having high natural frequency.

29. What are the disadvantages of using helicalspring of non-circular wires?a.The quality of the spring is not goodb.The shape of the wire does not remain constant while

forming the helix. It reduces the energy absorbing capacity of the spring.

c.The stress distribution is not favorable as in circularwires. But this effect is negligible where loading is of staticnature.30. Why concentric springs are used?

a.To get greater spring force within a given spaceb.To insure the operation of a mechanism in the event of

failure of one of the spring31. What is the advantage of leaf spring over helicalspring?

The advantage of leaf spring over helical spring is thatthe end of the spring may be guided along a definite path as itdeflects to act a structural member in addition to energyabsorbing device. 32. Write notes on the master leaf & graduated leaf?

The longest leaf of the spring is known as main leaf ormaster leaf has its ends in the form of an eye through whichbolts are passed to secure the spring. The leaf of the springother than master leaf is called the graduated leaves.33. What is meant by nip in leaf springs?

By giving greater radius of curvature to the full lengthleaves than the graduated leaves, before the leaves areassembled to form a spring thus a gap or c

34. What is the application of leaf spring?

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41. What is the function of flywheel? Apl 08A flywheel used in machine serves as a reservoir which

stores energy during the period when the supply of energy ismore than the requirement and releases it dulling the periodwhen the requirement of energy is more than the supply.

42. Define the term fluctuation of speed andfluctuation of energy .Nov 08,Nov 09

The ratio of maximum fluctuation of speed to the meanspeed is called co efficient of fluctuation of speed. The ratioof fluctuation of energy to the mean energy is calledcoefficient of fluctuation of energy. 43. State the type of stresses induced in a rim

flywheel?1.Tensile stress due to centrifugal force2.Tensile bending stress caused by the restraint of the

arms and3.The shrinkage stresses due to unequal rate of cooling

of casting.44.What are the stresses induced in flywheel arms?1.Tensile stress due to centrifugal force.

2.Bending stress due to torque3.Stress due to belt tension.45. How does the function of flywheel differ from thatof governor?

A governor regulates the mean speed of an engine whenthere are variations in the mean loads. It automaticallycontrols the supply of working fluid to engine with the varyingload condition & keeps the mean speed with contain limits. It

does not control the speed variation caused by the varyingload. A flywheel does not maintain const speed.46.State any two applications of leaf springsNov 08

1.structural members,2Energy absorbing devices

47.What are the applications of concentric springs

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May 2010Valve springs in heavy duty diesel engines, rail road

suspensions etc..

48.Two springs of stiffness K 1 , K 2 are connected inseries what is the stiffness of connection?May 2010

Two springs of stiffness K 1 , K 2 are connected in series

Then combined stiffness, 1/K =1/ K 1 + 1/ K 2

16 MARKS

1.Design a close-coiled helical spring of silicon-

manganese steel for the valve of and IC engine capableof exerting a net force of 65N when the valve is openand 54N when the valve is closed. The internal andexternal diameters are governed by space limitations,as it has to fit over bushing of 19mm outside diameterand go inside a space of 38mm diameter. The valve liftis 6mm.

Given data :

P min = 54N

P max = 65N

D min = 19mm

D max = 38mm

Y o = 6mm

Solution :Since material is not specified, let us take 20Mn2

alloy steel which is commonly used for spring.

From PSGDB page 1.13,

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y = 44kg/mm = 44*9.81 = 432N/mm

Assuming, yield shear stress, y = 0.5 y

We have y = max = 0.5*432 = 216N/mm

We can also assume that

D = D min+ D max/2 = 19+38/2 = 28.5mm

Let us assume that d= 5mm

C = D/d = 28.5/5 = 5.7

K s = 4c-1/4c-4+0.615/c

= 4(5.7)-1/4(5.7)-4 + 0.615/5.7

= 1.26

Let us find the shear stress developed in the spring for theassumed value of d =5mm

We know that

= k s*8*p max C/ d =1.26*8*65*5.7/ *5 = 47.83N/mm

since < max, design is safe.

We know that Y = 8pcn/Gd

Assuming, G = 84*10N/mm,

n= 26.1 say 27.

2. Design and draw suitable flywheel for a four strokefour cylinder 133kW engine running at 375rpm. Due tospace restriction the flywheel diameter should notexceed 1.2m.

Given data:

Power = 133kW = 133 *10w

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Speed, N = 375 rpm

Maximum diameter of the rim, D = 1.2m

Solution:

Since it is given that the flywheel diameter shouldnot exceed 1.2m, let us assume that the mean diameter of the flywheel rim as 1.1 m. R = 0.55m.

Also, assume

Maximum fluctuation of energy, E =20% of energydeveloped per revolution (E) and

Speed variation is 1% either way from mean.

Therefore, the total fluctuation of speed,

N1 N2 = 2% of mean speed = 0.02N

Mass of the flywheel rim:

Let m = mass of the flywheel rim in kg

Work done (or) energy developed per revolution.E = p*60/N = 133*10*60/375 = 21280N-m

Maximum fluctuation of energy,

E =0.2E = 0.2*21280 = 4256N-m

Since the speed variation is 1% either way from mean,therefore the total fluctuation of speed, N1- N2 = 2% of

mean speed = 0.02N and co-efficient of fluctuation of speed,

K s = n1-n2/n1 = 0.02

Velocity of the flywheel,

V = DN/60 = *1.1*375/60 = 21.598m/sec

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We know that the maximum fluctuation of energy (E)

4256 = mV Ks =m (21.598) 0.02

m = 21.598*0.02/4256 = 456.17kg ans.

Cross-sectional dimensions of the flywheel rim :

Let us assume, h = depth (or) thickness of the rim inmeters and b = width of the rim in meters, 2h

Cross-sectional are of rim,

A = b *h = 2h*h = 2h

We know that mass of the flywheel rim (m),M = A D*p (assume, = 7200kg/m)

456.17 = A* D* = 2h* *1.1*7200

h = 0.0957m = 95.7mm ans.

b = 2h = 2*95.7 =191.4mm ans.

Diameter and length of hub:

Let us assume, d = diameter of the hub

d 1= diameter of the shaft and

l = length of hub

We know that mean torque transmitted [(Mt) max] by theshaft is twice the mean torque.

(mt)max =2*(mt)mean = 2*3386.8(mt)max = 6773.6N-m = 6.773*10 6 N-mm

We also know that maximum torque transmitted by theshaft

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6.773*10 6 = /16**d1 (assume , = 40N/mm)

= /16*40*d1

d 1= 95.2mm ans.

The diameter of the hub (d) is made equal to twice thediameter of the shaft (d1) and length of hub (l) is equal tothe width of rim (b).

D =2d1 =2*95.2 = 190.4mm ans.

And l=b = 190.4mm ans.

Cross-sectional dimensions of the elliptical arms:

Let, a = major axis

C = minor axis

Assume minor axis = 0.5 * major axis

C = 0.5a

N = number of arms (assume, n=6)

b = bending stresses for the material of thearms (assume b = 14Mpa) b =14N/mm

We know that the maximum bending moment in the armat the hub end, which is assumed as cantilever, is givenby

M = (Mt)/D.n (D-d) = 3386.8*10/1.1*6(1.1-0.1904)

= 466.76*10N-mmAnd section modulus for the cross-section of arm,

Z = /32 ca = /32 (0.5a)a = 0.05a

We know that the bending stress ( b),

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14 =M/Z

14 = 466.76*10/0.05a

a = 87.36mm

Say, major axis, a = 88mm ans.

c= 0.5a

c= 0.5*88 = 44mm ans.

A punching press pierces 30 holes per minute in aplate using 12 kN-m of energy per hole during eachrevolution. Each piercing takes 35% of the time needed

to make one revolution. The punch receives powerthrough a gear reduction unit which in turn is fed by amotor driven belt pulley 750 mm diameter and turningat 240 rpm. Find the power of the electric motor if overall efficiency of the transmission unit is 80%.Design a cast iron flywheel to be used with thepunching machine for a coefficient of fluctuation of speed is 0.05, if the space considerations limits themaximum diameter to 1.3 m.Allowable shear stress in the shaft mateial = 48 MPaAllowable tensile stress for cast iron = 5 MPaDensity of cast iron = 7200 Kg/mGiven data: n p = 30 rpm,E = 12 10 J,n = 240 rpm,= 0.8 ,Cs = 0.05 ,

Dmax = 1.3 m,all = 5 N/mm, = 7200 kg/m , = 48 N/mmSolution:The punching operation takes place during 35% of rotation of

the punching machine crank shaft ( = 126). Hence during65% of the rotation of the crank shaft ( =234), the energy isstored in the flywheel wheel. Therefore, the max fluctuation of energy is E = 0.65E = 0.65 12000=7800 J.Assumptions :

1. the flywheel used is a rim type flywheel with C= 0.9

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2. the width of flywheel rim is taken as twice its thickness3. the no of flywheel arms is N= 6.4. For flywheel arms , the elliptical cross section is

a 1 =2b 1

5. The max torque is taken as twice the mean torque.Dimensions of flywheel rim :Mean angular speed = 2n/ 60 = 2*240/60

= 25.13 rad/s. The max diameter of the flywheel is 1.3 m. let take meandiameter as D m = 1.1 mRm =0.55 m ,E = ICs,I = 247.02 kg m,

For a flywheel rim, I r = CI =0.9*247.02=222.32 kgmIr = m r Rm = (2 R m bt)

222.32 = 2 * 0.55 * 2t *t *7200t = 121.5 mm =122 mm

Stresses in flywheel rim:V =R m = 0.55* 25.13 = 13.82 m/s.

The resultant stress induced in flywheel rim is rim = V = [ 0.75 +(0.5R m /Nt)]

= 7200*13.82*[0.75 + 0.5*0.55/(6*0.122)]= 1.88 *10 6 N/mm < 5 N/mm

rim < all

Hence the flywheel rim is safe.Diameter of flywheel shaft: The average torque on punching machine shaft

Tavp = E/2 = 12000/2 Tavp = 1909.86 Nm

The reduction ratio from flywheel shaft to punching machineshaft,

G = 240/30 = 8 The avg torque on flywheel shaft

Tavg = T avp / G = 1909.86/8 = 238.73 Nm The maximum torque on flywheel shaft,

Tmax = 2 T avg = 2*238.73=477.465 Nm =16 T max / d ,48 = 16*477465/ d

d = 37 mm = 40 mmradius of hub, r b = d= 40 mm

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From PSGDB P.No 5.21, for d = 40 mmWidth of key w = 12 mmHeight of key,h= 8 mmConsidering shearing of key:

d

= 2Tmax

/dwl , 48 = 2*477465/(40*12*l)l = 41.44 =42 mmConsidering crushing of key

c = 4T max / dhl , 96 = 4*477465/(40*8*l)l = 63 mmKey dimensions

W = 12 mm, h = 8 mm, l = 63 mmDimensions of flywheel arms

The resultant stress induced in the flywheel arm isarm = V + Ta (R m - rb)/(NR m Za )

= * 7200*13.82 + (477465*(0.55-0.04))/(6*0.55*Z a )

= 1.031 *10 6 + 73.8 / Z aFor the safety of flywheel arms

arm = all1.031 *10 6 + 73.8 / Z a = 5 * 10 6

73.8/ Z a = 3.988 * 10 6Za = 1.86 *10 -5 m .eqn 3

For elliptical cross section of the arm at the hub end,

Za = /32*a 12

*b1= /32(2 b 1)b 1 = /8 b 1 ..eqn 4From eqn 3 and 4,

/8 b 1 =1.86 *10 -5b 1= 0.03617 m = 36.17 mm = 37mm

Power requirement of electric motorPower input to the punching machine P P

= energy input to the punching machine perstroke*

No of punching strokes per second=E*n P/60= 12000*30/60= 6000 W

Power requirement of electric motor,PP = P P/ m = 6/0.8 = 7.5 kW

Flywheel dimensions:Flywheel rim Rm = 550 mm,

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t = 122 mm,b =244 mm

flywheel arms : N =6a 1 =74 mm, b 1 =37 mm

flywheel hub: db = 2 rb =80mmlb =63 mmShaft : d = 125 mm,Key : w = 12 mm,

h = 8 mm,l= 63 mm .

UNIT - V

Design of Bearings and miscellaneous elements

TWO MARKS

1. What is bearing?Bearing is a stationery machine element which supports

a rotating shafts or axles and confines its motion.2. Classify the types of bearings.

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(i). Depending upon the type of load coming upon the shaft:a. Radial bearing, b.Thrust bearings.

(ii).Depending upon the nature of contact:a.Sliding contact, b.Rolling contact bearings or

Antifriction bearings.3.What are the required properties of bearingmaterials?

Bearing material should have the following properties,i.High compressive strength, ii.Low coefficient of

friction,iii.High thermal conductivity, iv.High resistance to

corrosion,

v.Sufficient fatigue strength, vi.It should be soft with alow modulus of elasticity and vii. Bearing materials shouldnot get weld easily to the journal material.4. What is a journal bearing?

A journal bearing is a sliding contact bearing whichgives lateral support to the rotating shaft.5. What are the types of journal bearings depending

upon the nature of contact?

i.Full journal bearing, ii. Partial bearing and iii. Fittedbearing.6. What are the types of journal bearing dependingupon the nature of lubrication?i.Thick film type, ii.Thin film typeiii.Hydrostatic bearings, iv.Hydrodynamic bearing.7. What is known as self acting bearing? Nov 07

The pressure is created within the system due to rotation

of the shaft, this type of bearing is known as self actingbearing.

8.What is the advantage of Teflon which is used forbearing?May 2008

1.It has low coefficient of friction,

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2.It can be used at higher temperature3.Chemically inert.

9.What is meant by life of antifriction bearing?Nov 08It is defined as the life of 90% of group of identical

bearing will complete or exceed before fatigue failure.

10.What is a Quill bearing? Apl 2010Quill bearings are characterized by cylindrical rollers of

very small diameter and relatively long.they are also called asneedle bearings.11.For ajournal bearing, the maximum operatingtemperature must be less than 80C why? Nov 2010

Temperature rise will result in the reduction of theviscosity of the oil used in the bearing.This would lead tometal to metal contact, thereby affecting the bearingperformance and life.12.Why is a piston end of a connecting rod keptsmaller than the crank pin end Nov 2010

The piston end of the connecting rod experiences theless bending moment than the crank end. Hence on the basis

of uniform strength, the piston end of the connecting rod issmaller.13.List any foyr advantages of the rolling bearing oversliding bearings Apl 09

1.Starting friction is low,2.Lubrication is simple,3.It reqires less axial spaceand more diametral force4.Heavier loads , higher speeds are permissible.

14.State the disadvantages of the thrust ball bearingNov 09High initial cost,less capacity to withstand shock,noisy

operation at high speed,life is infinite, Design of bearinghousing is complicated.

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15. What are the stresses setup in an IC engineconnecting rod?

1. Tensile stresses, 2.compressive stresses, 3. Bendingstresses due to inertia forces

16. What is the method of manufacturing connectingrod?

Forging is the method of manufacturing the connectingrod.

16 MARKS1.A single row deep groove ball bearing no.6002 is

subjected to an axial thrust load of 1000N and aradial load of 2200N. Find the expected life that 50%of the bearings will complete under thiscondition .Nov 2010

Solution:

From PSGDB 4.14, for bearing no. 6002

Co = 2850kgf = 28500N

C = 5590kgf = 55900N

Fa/co = 1000/28500 = 0.035

From table 5.1.2 or PSGDB 4.4, corresponding to

Fa /co = 0.035, the value of e = 0.235 (byinterpolation).

Since fa/fr =1000/2200 = 0.454>e, from table 5.1.2 orPSGDB 4.4., the radial load factor x = 0.56 and thrustload factor y = 1.83 (by interpolation). The service factoris selected from table 5.1.3 or PSGDB 4.2 consideringrotary machine with no impact condition.

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S = 1.1 to 1.5 say 1.3

Equivalent load, p = (xfr + y fa) s =(0.56*1.83*1000)1.3

P = 39806N

We know that the rating life of the bearing

L = (C/P) a

= (55900/39806) = 2.77 million revolutions.

Expected life at 50% reliability :

L50 /L 90 = [ln (1/R 50 )/ln(1/R 90 )] 1/3

L50 = ln (1/0.5)/ (ln1/R 90 )1/1.17 * 2.77*5

L50 = 13.85 million revolutions

2.A bearing for an axial flow compressor is to carry aradial load of 2500 N and thrust of 1500 N. The serviceimposes light shock and the bearing will be in use for40 hours/week for 5 years. The speed of the shaft is1000 rpm. Select suitable ball bearing for the purposeand give the required tolerances on the shaft and thehousing. Diameter of the shaft is 50 mm. May 08Given data:Radial load, Fr =2500 N; axial load, Fa =1500 N

Lh10 = 40*52*5 =10400 hrs, n= 1000 rpmd=50 mm

Solution:Rating life and service factor:

L10 =L h10 *60*n/10 6=624 million revolutions,

For light shock, Ka = 1.5 from PSGDB P.No 4.2From PSGDB P.no 4.12, For d = 50 mm

Select bearing no 6010 withC= 1700 kgf =16677 N,Co = 1370kgf =13439 NAnd Fa/Co = 0.11From PsgDB p.no 4.4 ,

For Fa/Co = 0.11, e = 0.297 (by linear interpolation)

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Fa/Fr = 1500/2500 =0.6On comparing Fa/Fr > e

Take X= 0.56 and Y = 1.47 (by linear interpolation)Equivalent dynamic load Pe = (X v Fr + YFa)Ka

= (0.56*1*2500 +1.47*1500)1.5= 5407.5 NRequired dynamic capacity Cr = (L 10 )1 /3 Pe

Cr = 46208.8N >16677 N

So. it is unsafe,Select

higher bearing from PSGDB P.no 4.15for d = 50 mmC = 7000 kgf = 68670 N

Co = 5300kgf =51993NFa/Co = 0.029 From PSGDB P.No 4.4 for Fa/Co=0.029

e=0.225 , Fa/Fr = 1500/2500 =0.6,Fa/Fr >e

Take X= 0.56, Y = 1.946 by linear interpolationEquivalent dynamic load Pe = (X v Fr + YFa)Ka

= (0.56*1*2500+1.946*1500)1.5 = 6478.5 NRequired dynamic capacity Cr = (L 10 )1 /3 Pe

Cr = ss360.83 < 68670 N Hence it is safe.Tolerance on shaft :

From PSGDB P.No for d =50 mm, service conditionof light shock with rotating inner race , the tolerance on theshaft is

50 j6 ie 50 ( +0.011/-0.005 ) mm. from PSGDB P.NO 3.8Tolerance on housing:

From PSGDB P.NO 4.15 for bearing no 6410, theouter diameter of bearing or inner diameter on housing is 130mm.

From PSGDB P.NO 4.9 for service condition of lightshock and stationary outer race, the tolerance on housing is130J7 ie 130 (+0.026/-0.014) from PSGDB P.NO 3.9

3..The journal bearing supports a load of 150 kN due toturbine shaft of 300 mm diameter running ar 1800rpm. The bearing clearance is 0.25 mm, while the

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allowable bearing pressure is 1.6 N/mm.if theviscosity of lubriacating oil at bearing temperature of 60C is 0.02 kg/ms.determine (i)the length of bearingand (ii) the amount of heat to be removed by thelubricant per minute. May 2009Given data: Load W = 150kN,d =300 mm, speed , n =1800 rpm ,

bearing clearance 2C = 0.25 mm,p = 1.6 N/mm , Z = 0.02 kg/m-s

Ta = 60C Solution:1.length of bearing,(l):

p= W/(ld), l= W/pd = 150000/(1.6*300)=312.5 mm2.Dynamic viscosity ,Z= 0.02 kg/m-s = 20*10 -9 N-s/m3. Radial clearance(C) and journal radius (r),

2c = 0.25 mmc = 0.125 mm,

r =d/2 = 300/2 =150 mm.4.Coefficient of friction ,

= 19.5 (r/c)(Zn/p)+ K , where K = 0.002 for

0.75

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