designing of center crankshaft

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    DESIGN OF

    CENTRE CRANKSHAFT

    Presented by:Ashvath sharma (1109122011)

    Jatin mukheja (1109100023)

    Vaibhav kumar banka (1109120112)

    ME-B(6th

    SEM)

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    OBJECTIVE

    To give a brief idea about the crankshaft, its types,materials of which it is made up of, its importance in anIC engine.

    Basic designing procedure of a centre crankshaft.

    To give a practical example of designing of crankshaftfor a 4 stroke diesel engine of INDICA VISTA.

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    CONTENTS

    Introduction

    Types of crankshaft. Materials used for crankshaft.

    Designing procedure.

    Practical example. References.

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    INTRODUCTION

    The crankshaft is an important part of the IC engine that converts thereciprocating motion of the piston into rotary motion through theconnecting rod.

    The crankshaft consists of three portionscrankpin, crank web and shaft.

    The big end of the connecting rod is attached to the crank pin.

    The crank web connects the crank pin to the shaft portion. The shaft portion rotates in the main bearings and transmits power to the

    outside source.

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    TYPES OF CRANKSHAFT

    The crankshaft, depending upon the position of crank,may be divided into the following two types:

    1. Side crankshaft or overhung crankshaft

    2. Centre crankshaft

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    MATERIAL SELECTION

    The crankshafts are subjected to shock and fatigue loads. Thusmaterial of the crankshaft should be tough and fatigue resistant.

    The alloying elements typically used in these carbon steels are

    manganese, chromium, molybdenum, nickel, silicon, cobalt,vanadium, and sometimes aluminium and titanium. Each of thoseelements adds specific properties in a given material.

    The carbon content is the main determinant of the ultimate strengthand hardness to which such an alloy can be heat treated.

    The crankshafts are generally made of carbon steel, special steel orspecial cast iron.

    The crankshafts are made by drop forging or casting process but theformer method is more common.

    The surface of the crankpin is hardened by case carburizing,

    nitriding or induction hardening.

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    DESIGN PROCEDURE

    The following procedure may be adopted for designing a crankshaft:

    1. The crankshaft must be designed or checked for at least two crank

    positions.a) When the crank-shaft is subjected to maximum bending moment

    and

    b) When the crankshaft is subjected to maximum twisting momentor torque.

    2. The additional moment due to weight of flywheel, belt tension andother forces must be considered.

    3. It is assumed that the effect of bending moment does not exceed twobearings between which a force is considered.

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    WHEN THE CRANK-SHAFT IS SUBJECTED TO

    MAXIMUM BENDING MOMENT

    Piston gas load

    Fp= /4*D2*p

    Assume that the distance (b)between the bearings 1 and 2 is equal to twicethe piston diameter.

    b=2D

    and b1=b2=b/2

    We know that due to the piston gas load, there will be two horizontalreactions H1 and H2 at bearings 1 and 2 respectively, such thatH1=(Fp*b1)/b

    H2=(Fp*b2)/b

    Assume that the length of the main bearings to be equal, i.e., c1 = c2 = c / 2.

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    We know that due to the weight of the flywheel acting downwards, therewill be two vertical reactions V2 and V3 at bearings 2 and 3 respectively,

    such thatV2=(W*c1)/c={(W*c/2)}/c=W/2

    V3=(W*c2)/c={(W*c/2)}/c=W/2

    Due to the resultant belt tension (T1 + T2) acting horizontally, there will betwo horizontal reactions H2 and H3 respectively, such that

    H2=(T1+T2)c1/c={(T1+T2)c/2}/c=(T1+T2)/2

    H3=(T1+T2)c2/c={(T1+T2)c/2}/c=(T1+T2)/2

    Now the various parts of the crankshaft are designed as discussed below:(a) Design of crankpin

    Let dc= Diameter of the crankpin in mm ;

    lc= Length of the crankpin in mm ; and

    b= Allowable bending stress for the crankpin.

    We know that the bending moment at the centre of the crankpin:

    Mc=H1*b2

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    We also know that

    Mc=/32*dc3*b

    Equating equations, value of dc is obtained.

    We know that length of the crankpin,

    lc=Fp/(dc*pb) (by using pb=Fp/ lc*dc)

    (b) Design of left hand crank web

    We know that thickness of the crank web

    t=0.65dc+ 6.35 mm

    and width of the crank web

    w=1.125dc+12.7 mm

    We know that maximum bending moment on the crank web,

    M=H1(b2-lc/2-t/2)

    Section modulus,

    Z=(1/6*w*t2)

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    Bending stress

    b=M/Z

    We know that direct compressive stress on the crank web,

    c=H1/w*t

    Total stress on the crank web

    b+c

    If the total stress on the crank web is less than the allowable bendingstress then design is safe.

    (c) Design of right hand crank web

    From the balancing point of view, the dimensions of the right hand crankweb (i.e. thickness and width) are made equal to the dimensions of the lefthand crank web.

    (d) Design of shaft under the flywheel

    Let ds = Diameter of the shaft in mm

    Since the lengths of the main bearings are equal, therefore

    l1=l2=l3=2*(b/2-lc/2-t)

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    Assuming width of the flywheel,

    Now we have ,value of c.c1=c2=c/2

    We know that bending moment due to the weight of flywheel,

    MW=V3*c1

    And bending moment due to the belt pull,

    MT=H3*c1

    Resultant bending moment on the shaft,

    Ms={(M

    W)2+(M

    T)2}1/2

    We also know that bending moment on the shaft (Ms)

    Ms=(/32*ds3*b)

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    PRACTICAL EXAMPLE:

    TATA INDICA VISTA4 STROKE DIESEL

    ENGINE

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    PART SPECIFICATIONNumber of Cylinders

    Type of Engine ( Inline / Vee engine )

    4

    1248 cc, inline diesel engineBore/stroke(D/L)

    Cylinder spacing

    Power @speed

    Torque @speed

    Reciprocating mass

    Rotating mass

    Connecting rod length

    Compression ratio

    Engine type

    Mass of piston

    Mass of connecting rod

    Crankpin mass

    Mass of webCenter of gravity radius

    Crank radius

    Reciprocating mass

    Rotating mass

    Ratio of r/l

    Cylinder pitch

    Weight of flywheel

    9.6/82

    Assume

    55KW @ 4000 rpm

    190 Nm @ 1750 rpm

    Assume

    Assume

    Assume

    17.6:1

    Compression ignition engine

    1.36 kg

    0.60 kg

    0.25 kg

    0.25 kg37.96 mm

    39.5 mm

    1.56 kg

    1.12 kg

    0.31

    84 mm

    1 kg

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    pmax = (power*no of cylinders)/(volume of cylinders*rpm)

    = (55*4)/(1248*10-6*4000)

    = 44.07 kN/mm2

    Piston gas load

    Fp= /4*D2* pmax= /4*(69.6)2*44.07=167668.5 N=167.67 kN,

    Assume that the distance (b)between the bearings 1 and 2 is equal to twicethe piston diameter.

    b=2D=2*69.6=139.2 N,

    and b1=b2=b/2=139.2/2=69.6 mm,

    We know that due to the piston gas load, there will be two horizontalreactions H1 and H2 at bearings 1 and 2 respectively, such that

    H1=(Fp*b1)/b=(167.67*69.6)/139.2 =83.83 kN

    H2=(Fp*b2)/b=(167.67*69.6)/139.2 =83.83 kN

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    Assume that the length of the main bearings to be equal, i.e.c1 = c2 = c / 2

    We know that due to the weight of the flywheel acting downwards, therewill be two vertical reactions V2 and V3 at bearings 2 and 3 respectively,such that

    V2=(W*c1)/c={(W*c/2)}/c=W/2=9.8/2=4.9N

    V3=(W*c2)/c={(W*c/2)}/c=W/2=9.8/2=4.9N

    Due to the resultant belt tension (T1 + T2) acting horizontally, there will betwo horizontal reactions H2 and H3 respectively, such that

    H2=(T1+T2)c1/c={(T1+T2)c/2}/c=(T1+T2)/2

    H3=(T1+T2)c2/c={(T1+T2)c/2}/c=(T1+T2)/2

    But in case of TATA INDICA VISTA , belt is absent so

    T1+T2 = 0

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    Now the various parts of the crankshaft are designed as discussed below:

    (a) Design of crankpin

    Let dc= Diameter of the crankpin in mm ;

    lc= Length of the crankpin in mm ; and

    b= Allowable bearing stress for the crankpin. It may be

    assumed as 83 kg/mm2.

    We know that the bending moment at the centre of the crankpin

    Mc = H1*b2= 83.83*69.6 = 5834.56 kN-mm

    We also know that

    Mc= /32*dc3*b = /32*dc3*83 = 8.148 dc3N-mm

    = 8.148*10-3 * dc3 kN- mm

    Equating equations ,we have

    dc3=62840/8.148 *10-3

    dc= 89.46 mm or say 90mm

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    We know that length of the crankpin,

    lc = Fp/(dc*pb) = 167.67*103/(90*10) = 186.28

    (by using pb=Fp/ lc*dc) (say pb = 10)

    (b) Design of left hand crank web

    We know that thickness of the crank web

    t = 0.65dc+ 6.35 mm = (0.65*90)+6.35 = 64.85 say 65 mm

    and width of the crank web

    w = 1.125dc+12.7 mm = (1.125*90)+12.7 = 113.95 say 115 mm

    We know that maximum bending moment on the crank web,

    M = H1(b

    2-l

    c/2-t/2) = 83.83*(69.6-186.28/2-65/2) = -4697.83 N-mm

    The bending moment is negative, therefore the design is not safe thus thedimensions are on higher side.

    Now lets assume, dc = 45 mm

    lc = 373.57 mm

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    This is very high, which require huge length crank shaft .To have optimumdimension of crankshaft lets assume length of crank web as lc=24mm andcheck whether these dimension are suitable for load exerted by the piston ,,

    and other forces.

    Therefore , dimensions of crank web are :

    t =35.6 mm

    and l = 63.32 say 68 mm

    But the thickness of crankshaft is also on higher side so lets assume thethickness as t = 13.2mm

    Bending moment,

    M=4275.33kN mm

    Section modulus,

    Z=(1/6*w*t2)=(1/6*68*13.22)=1974.72mm3

    Bending stress

    b=M/Z=4275.33/(1974.72)=2.165kN/mm2=49.6N/mm2

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    We know that direct compressive stress on the crank web,

    c=H1/w*t=83.83/(68*13.2)=.09339kN/mm2

    Total stress on the crank web

    b+c=2.165+.09339=2.2583N/mm2

    Since the total stress on the crank web is less than the allowable bendingstress of 83 MPa,therefore, the design of the left hand crank web is safe.

    (c) Design of right hand crank web

    From the balancing point of view, the dimensions of the right hand crankweb (i.e. thickness and width) are made equal to the dimensions of the lefthand crank web.

    (d) Design of shaft under the flywheel

    Let ds = Diameter of the shaft in mm

    Since the lengths of the main bearings are equal, therefore

    l1=l2=l3=2*(b/2-lc/2-t)=2*(139.2/2-24/2-13.2)=88.8 mm

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    Assuming width of the flywheel as 200 mm, we havec=88.88+200 =288.88 mm

    Allowing space for gearing and clearance, let us take c = 300 mm.

    c1=c2=c/2=300/2=150 mm

    We know that bending moment due to the weight of flywheel,

    MW=V3*c1=4900*300=1470000 kN-mm=1.47*106N-mm

    We also know that bending moment on the shaft

    1.47*106=(/32*ds

    3*b

    )

    =(/32*ds3*83)

    =8.144*ds3

    ds=56.51 say 60 mm.

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    REFERENCES

    1) Design Data Hand Book, K. Mahadevan and K. Balaveera

    Reddy,CBS publication, 19892) A text Book of Machine Design, P.C.Sharma and D.K.Aggarwal, S

    K Kataria and Sons, 1993

    3) A text Book of Machine Design, R S Khurmi and J1. Design DataHand Book, K. Mahadevan and K. Balaveera Reddy, CBS

    publication, 19894) Automobile Mechanics, N K giri, Khanna Publishers, 2005

    5) Automotive Mechanics, Crouse/Anglin, Tata McGraw-Hill, 2003

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