connecting rod design

7/30/2019 connecting rod design

1/16

DESIGN AND ANALYSIS OF CONNECTING ROD

Submitted by

Ashok Mangina

Roll No.: 11011D8906

(ENGINEERING DESIGN)


2/16

The connecting rod forms an integral part of an internal combustionengine.

The connecting rod is acted upon by different types of loads while

undergoing its operation and are as follows,

Force on the piston due to gas pressure and inertia of the reciprocating parts

Force due to inertia of the connecting rod or inertia bending forces.

Force due to friction of the piston rings and of the piston, and

Force due to friction of the piston pin bearing and the crank pin bearing.

Heavy duty applications connecting rod was selected for the study.

The analytically calculated loads acting on the small end of connecting rod

were used to carry out the static analysis using ANSYS.

ABSTRACT


3/16

The main function of connecting rod is to transmit the translational motion ofpiston to rotational motion of crank shaft.

The function of the connecting rod also involves transmitting the thrust of the

piston to the connecting rod.

Connecting rod has three main zones. The piston pin end, the center shank

and the big end.

The piston pin end is the small end, the crank end is the big end and thecenter shank is of I-cross section.

Connecting rod is a pin jointed strut in which more weight is concentrated

towards the big end.

A Connecting rod subjected to an axial load W may buckle with X-axis as

neutral axis(i.e. in the plane of motion of the connecting rod) or Y-axis as

neutral axis (i.e. in the plane perpendicular to the plane of motion). The Connecting rod is considered like both ends hinged for buckling about X-

axis and both ends fixed for buckling about Y-axis.

INTRODUCTION


4/16

Compressive Yield stressc

xxkxxI

Radius of the gyration of the section about X-axis and Y-axisrespectivelyyykyyI Moment of Inertia of the section about X-axis and Y-axis

respectively

A Cross sectional area of the connecting rod

l length of the connecting rodWB Buckling load

L Effective length of the connecting rod

a Material Constant = 1/7500 for steel

NOTATIONS


5/16

WB =

According to Rankines formula, Buckling Load about X-axis is given by

= (For both ends hinged, L=l)WB =

2)(1xx

c

k

La

A

WB =

Buckling Load about Y-axis is given by

WB =2

)2(1 yy

c

k

l

a

A

=

(For both ends fixed, L=l/2)

WB =2

)(1yy

c

k

L

a

A

WB =

A Connecting rod should be equally strong in buckling about both the axes. i.e.,

WB about X-axis = WB about Y-axis

2)(1xx

c

k

la

A

WB =

X X

Y

Y

DESIGN


6/16

2][1xx

c

kla

A

2]2[1

yy

c

kla

A

=

2][xxkl 2]

2[

yykl

=

2

xxk =2

yyk4 xxI = yyI4

( A )I=2

k

DESIGN(cntd)


7/16

Let us consider an I-section of the connecting rod as shown

with the following proportions

Flange and Web thickness of the section t

Width of the section = 4t

Height of the section = 5t

4t

t

5tt

Area of the section A = 2x(4txt) + 3txt = 11t2

Ixx = 1/12[4tx(5t)3 3tx(3t)3] = 419/12 t4Ixx = 1/12[4tx(5t)3 3tx(3t)3] = 419/12 t4

Iyy = 1/12[2xtx(4t)3 3tx(t)3] = 131/12 t4

DESIGN(cntd)


8/16

DESIGN (input data)


9/16

Maximum force on Connecting rod = Fc =

or Gas pressure

pD

]4

[2

13]497[

2

=

= 96067.77 N

Buckling Load = Fc FoSWB =

= 96067.77 5

Buckling Load = Fc FoSWB = Buckling Load = Fc FoSWB =

= 480338.85 N

xxk = AIXX

2

4

11

1

12

419

t

t= = 1.78t

DESIGN(cntd)


10/16

2

)(1xx

c

k

la

A

WB = 480338.85 = =

2

2

)78.1

233(7500

11

11320

t

t

320

85.480338

= 2846.2

112

4

t

t

11t4-136.45t2-311.75 = 0

t2 =12

75.31114)45.136(45.1362

t = 12 mm

DESIGN(cntd)


11/16

Let us now check the design for the induced bending stress due to inertia

bending forces on the connecting rod,

We know that mass of the connecting rod per meter length,

m1 = Volume x Density = Area x Length x Density

= A x l x = 11t2 x l x

= 11(0.012)2 x 0.233 x 7860

= 2.9 Kg

We know that mass of the connecting rod per meter length,

m1 = Volume x Density = Area x Length x Density

= A x l x = 11t2 x l x

= 11(0.012)2 x 0.233 x 7860

= 2.9 Kg

Maximum Bending moment,

Mmax = m2r x = m12r x

= 2.9 x x 0.064

= 58.011 Nm = 58011 Nmm

39

l

39

2l

60

28752

39

233.02

DESIGN(cntd)


12/16

max

max

ZM

16.24140

58011 2.4 N/mm2= =(max)b =

Maximum bending stress due to inertia bending forces,

= 13.97 t3

= 13.97 x 123

= 24140.16 mm4

25t

Ixx

t

t

5

2

12

4194

=xxZ =

And Section modulus,

DESIGN(cntd)


13/16

GEOMETRIC MODEL

Geometric model created using Pro-Engineer


14/16

Stress plot

Maximum Stress = 207 N/mm2

ANALYSIS


15/16

displacement plot

Maximum Displacement = 0.16mm

ANALYSIS


16/16

OBSERVATIONS AND CONCLUSION

1. Stresses are high at the small end, of the order of 207 N/mm2

2. Design needs to be refined for reducing the stresses and

displacement also.

Documents

connecting rod design