designing of center crankshaft
TRANSCRIPT
-
8/11/2019 designing of center crankshaft
1/24
DESIGN OF
CENTRE CRANKSHAFT
Presented by:Ashvath sharma (1109122011)
Jatin mukheja (1109100023)
Vaibhav kumar banka (1109120112)
ME-B(6th
SEM)
-
8/11/2019 designing of center crankshaft
2/24
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.
-
8/11/2019 designing of center crankshaft
3/24
CONTENTS
Introduction
Types of crankshaft. Materials used for crankshaft.
Designing procedure.
Practical example. References.
-
8/11/2019 designing of center crankshaft
4/24
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.
-
8/11/2019 designing of center crankshaft
5/24
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
-
8/11/2019 designing of center crankshaft
6/24
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.
-
8/11/2019 designing of center crankshaft
7/24
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.
-
8/11/2019 designing of center crankshaft
8/24
-
8/11/2019 designing of center crankshaft
9/24
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.
-
8/11/2019 designing of center crankshaft
10/24
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
-
8/11/2019 designing of center crankshaft
11/24
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)
-
8/11/2019 designing of center crankshaft
12/24
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)
-
8/11/2019 designing of center crankshaft
13/24
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)
-
8/11/2019 designing of center crankshaft
14/24
PRACTICAL EXAMPLE:
TATA INDICA VISTA4 STROKE DIESEL
ENGINE
-
8/11/2019 designing of center crankshaft
15/24
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
-
8/11/2019 designing of center crankshaft
16/24
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
-
8/11/2019 designing of center crankshaft
17/24
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
-
8/11/2019 designing of center crankshaft
18/24
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
-
8/11/2019 designing of center crankshaft
19/24
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
-
8/11/2019 designing of center crankshaft
20/24
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
-
8/11/2019 designing of center crankshaft
21/24
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
-
8/11/2019 designing of center crankshaft
22/24
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.
-
8/11/2019 designing of center crankshaft
23/24
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
-
8/11/2019 designing of center crankshaft
24/24