A project Report
On
Design of Connecting rod
For an Automobile industry.
Under the guidance of
Prof K.V.S.S Narayan Rao
Prepared
by:
Pramod Kumar Moharana (Roll No 64)
Pramod Kumar Moharana (Roll No 64)
Samay Singh Meena (Roll No 86)
PGDIE-42
Sec-B
Content
1. Introduction
2. Functioning of connecting rod
3. Design process flow of connecting rod.
4. Manufacturing process flow of connecting rod.
5. Conclusion
6. References
Introduction
In a reciprocating piston engine, the connecting rod or conrod connects the piston to the crank or crankshaft. Together with the crank, they form a simple mechanism
that converts linear
motion into
rotating motion. Connecting rods may also convert rotating motion into linear
motion.
Historically,
before the development of engines, they were first used in this way.
As a connecting rod is rigid, it may transmit
either a push or a pull and so the rod may rotate the crank through both halves
of a revolution, i.e. piston pushing and piston pulling. Earlier mechanisms,
such as chains, could only pull. In a few two-stroke engines, the connecting
rod is only required to push.
Today,
connecting rods are best known through their use in internal combustion piston
engines, such as car engines. These are of a distinctly different design from earlier
forms of connecting rods, used in steam engines and steam locomotives.
Functioning
of connecting rod
Design
steps of Connecting rod.
Need or Aim:
The Primary purpose of connecting rod is to
transmit gas forces applied on the piston due to ignition to crankpin of crankshaft assembly.
Synthesis(Mechanism)
The above mechanism will give the required
motion to the connecting rod.
Analysis of
Forces in connecting rod.
The various forces acting on the connecting
rod are as follows:
1. Force on the piston due to gas pressure
and inertia of the reciprocating parts,
2. Force due to inertia of the connecting
rod or inertia bending forces,
3. Force due to friction of the piston
rings and of the piston,
4. Force due to friction of the piston pin
bearing and the crankpin bearing.
1. Force on the
piston due to gas pressure and inertia of reciprocating parts
Consider a connecting
rod PC as shown in Fig
Forces on the connecting rod
Let p =
Maximum pressure of gas,
D = Diameter
of piston,
A =
Cross-section area of piston = (π /4)D2
mR = Mass of
reciprocating parts, = Mass of piston, gudgeon pin etc. + 1/3 rd mass of
connecting rod,
ω = Angular
speed of crank,
φ = Angle of
inclination of the connecting rod with the line of stroke,
θ = Angle of
inclination of the crank from top dead centre,
r = Radius
of crank,
l = Length
of connecting rod, and
n = Ratio of
length of connecting rod to radius of crank = l / r.
We know that
the force on the piston due to pressure of gas,
FL =
Pressure × Area = p . A = p × (π /4)D2
FI = Mass ×
*Acceleration = mR .ω2. r (cos n θ + (cos
2 θ)/n)
It may be
noted that the inertia force of reciprocating parts opposes the force on the
piston when
it moves
during its downward stroke (i. e. when the piston moves from the top dead
centre to bottom
dead
centre). On the other hand, the inertia force of the reciprocating parts helps
the force on the piston when
it moves from the bottom dead centre to top dead centre.
Net force
acting on the piston or piston pin (or gudgeon pin or wrist pin),
The –ve sign
is used when piston moves from TDC to BDC and +ve sign is used when piston
moves from
BDC to TDC.
When weight
of the reciprocating parts (WR = mR . g) is to be taken into consideration,
then
F1 ±WR
2. Force due to inertia of the connecting rod or inertia bending forces
Consider a connecting rod PC and a crank OC rotating with uniform angular velocity ω rad / s. In order to find the acceleration of various points on the connecting rod, draw the Klien’s acceleration diagram CQNO as shown in Fig. 32.11 (a). CO represents the acceleration of C towards O and NO represents the acceleration of P towards O. The acceleration of other points such as D, E, F and G etc.,on the connecting rod PC may be found by drawing horizontal lines from these points to intresect CN at d, e, f, and g respectively. Now dO, eO, fO and gO respresents the acceleration of D, E, F and G all towards O. The inertia force acting on each point will be as follows:
Consider a connecting rod PC and a crank OC rotating with uniform angular velocity ω rad / s. In order to find the acceleration of various points on the connecting rod, draw the Klien’s acceleration diagram CQNO as shown in Fig. 32.11 (a). CO represents the acceleration of C towards O and NO represents the acceleration of P towards O. The acceleration of other points such as D, E, F and G etc.,on the connecting rod PC may be found by drawing horizontal lines from these points to intresect CN at d, e, f, and g respectively. Now dO, eO, fO and gO respresents the acceleration of D, E, F and G all towards O. The inertia force acting on each point will be as follows:
Inertia
force at C = m × ω2 × CO
Inertia
force at D = m × ω2 × dO
Inertia
force per unit length at the crankpin = m1 × ω2 r
and inertia
force per unit length at the piston pin = 0
Inertia
force due to small element of length dx at a distance x from the piston pin P,
Resultant inertia force,=m /2 × ω2 r
3. Force due to friction of piston rings and of the piston.
3. Force due to friction of piston rings and of the piston.
F = π D · tR · nR · pR · μ
where D =
Cylinder bore,
tR = Axial
width of rings,
nR = Number
of rings,
pR =
Pressure of rings (0.025 to 0.04 N/mm2), and
μ =
Coefficient of friction (about 0.1).
Since the
frictional force of the piston rings is usually very small, therefore, it may
be neglected.
The friction
of the piston is produced by the normal component of the piston pressure which varies
from
3 to 10
percent of the piston pressure. If the coefficient of friction is about 0.05 to
0.06, then the frictional force due to piston will be about 0.5 to 0.6 of the
piston pressure, which is very low. Thus, the frictional
force due to
piston is also neglected.
4. Force due to friction of the
piston pin bearing and crankpin bearing
The force
due to friction of the piston pin bearing and crankpin bearing, is to bend the
connecting rod and to increase the compressive stress on the connecting rod due
to the direct load. Thus, the maximum compressive
stress in the connecting rod will be
σc (max) = Direct compressive stress + Maximum bending or whipping stress due to inertia bending stress.
σc (max) = Direct compressive stress + Maximum bending or whipping stress due to inertia bending stress.
Material
Selection.
The connecting rods are usually manufactured by drop forging process and it should have adequate strength, stiffness and minimum weight. The material mostly used for connecting rods varies from mild carbon steels (having 0.35 to 0.45 percent carbon) to alloy steels (chrome-nickel or chrome molybdenum steels). The carbon steel having 0.35 percent carbon has an ultimate tensile strength of about 650 MPa when properly heat treated and a carbon steel with 0.45 percent carbon has a ultimate tensile strength of 750 MPa. These steels are used for connecting rods of industrial engines. The alloy steels have an ultimate tensile strength of about 1050 MPa and are used for connecting rods of aeroengines and automobile engines.
Design of Connecting Rod
In designing a connecting rod, the following dimensions are required to be
determined :
1. Dimensions of cross-section of the connecting rod,
2. Dimensions of the crankpin at the big end and the piston pin at the
small end,
3. Size of bolts for securing the big end cap, and
4. Thickness of the big end cap.
The procedure adopted in determining the above mentioned dimensions is
discussed as below :
# Design a connecting rod for an I.C. engine running at 1800 r.p.m. and
developing a maximum pressure of 3.15 N/mm2. The diameter of the piston is 100
mm ; mass of the reciprocating parts per cylinder 2.25 kg; length of connecting
rod 380 mm; stroke of piston 190 mm and compression ratio 6 : 1. Take a factor
of safety of 6 for the design. Take length to diameter ratio for big end bearing
as 1.3 and small end bearing
as 2 and the corresponding bearing pressures as 10 N/mm2 and 15 N/mm2. The
density of material of the rod may be taken as 8000 kg/m3 and the allowable stress
in the bolts as 60 N/mm2 and in cap as 80 N/mm2. The rod is to be of I-section
for which you can choose your own proportions. Draw a neat dimensioned sketch showing
provision for lubrication. Use Rankine formula for which the numerator
constant may be taken as 320 N/mm2 and the denominator constant 1 / 7500.
Given : N = 1800 r.p.m. ; p = 3.15 N/mm2 ; D = 100 mm ; mR = 2.25 kg ; l =
380 mm
= 0.38 m ; Stroke = 190 mm ; *Compression ratio = 6 : 1 ; F. S. = 6.
The connecting rod is designed as discussed below :
1. Dimension of I- section of
the connecting rod
Let us consider an I-section of the connecting rod, as shown in Fig. 32.14
(a), with the following
proportions :
Flange and web thickness of the section = t
Width of the section, B = 4t
and depth or height of the section, H = 5t
First of all, let us find whether the section chosen is satisfactory or
not.
The connecting rod is considered like both ends hinged for buckling about
X-axis and both ends fixed for buckling about Y-axis. The connecting rod should
be equally strong in buckling about both the axes. We know that in order to
have a connecting rod equally strong about both the axes, Ixx = 4 Iyy
where Ixx = Moment of inertia of the section about X-axis, and Iyy = Moment of inertia of the
section about Y-axis.
In actual practice, Ixx is kept
slightly less than 4 Iyy. It is usually taken between 3 and 3.5 and the connecting
rod is designed for buckling about X-axis.
Now, for the section as shown in Fig. area of the section,
Now let us find the dimensions of this I-section. Since the connecting rod
is designed by taking the force on the connecting rod (FC) equal to the maximum force on the
piston (FL) due to gas pressure, therefore,
We know that the connecting rod is designed for buckling about X-axis
(i.e. in the plane of
motion of the connecting rod) assuming both ends hinged. Since a factor of
safety is given as 6, therefore the buckling load,
WB = FC × F. S. = 24 740 × 6 = 148 440 N
We know that radius of gyration of the section about X-axis,
Length of crank, r = Stroke of piston/2 = 190/2= 95 mm
Length of the connecting rod, l = 380 mm ...(Given)
Equivalent length of the connecting rod for both ends hinged,
L = l = 380 mm
Now according to Rankine’s formula, we know that buckling load (WB),
Thus, the dimensions of I-section of the connecting rod are :
Thickness of flange and web of the section
Thickness of flange and web of the section
= t = 7 mm Width of the section, B = 4 t = 4 × 7 =
28 mm
and depth or height of the section,
and depth or height of the section,
H = 5 t = 5 × 7 = 35 mm
These dimensions are at the middle of the connecting rod. The width (B) is
kept constant throughout the length of the rod, but the depth (H) varies. The depth near the big
end or crank end is kept as 1.1H to 1.25H and the depth near the small end or piston end is kept as 0.75H
to 0.9H. Let us take
Depth near the big end, H1 = 1.2H = 1.2 × 35 = 42 mm and depth near the
small end,
H2 = 0.85H = 0.85 × 35 = 29.75 say 30 mm
Dimensions of the section near the big end = 42 mm × 28 mm .
Dimensions of the section near the big end = 42 mm × 28 mm .
and dimensions of the section near the small end = 30 mm × 28 mm
Since the connecting rod is manufactured by forging, therefore the sharp
corners of I-section
are rounded off, as shown in Fig., for easy removal of the section from the
dies.
2. Dimensions of the crankpin
or the big end bearing and piston pin or small end bearing
Let dc = Diameter of the crankpin or big end bearing,
lc = length of the crankpin or big end bearing = 1.3 dc ...(Given)
pbc = Bearing pressure = 10 N/mm2 ...(Given)
We know that load on the crankpin or big end bearing = Projected area ×
Bearing pressure
= dc .lc . pbc = dc × 1.3 dc × 10 = 13 (dc)2
Since the crankpin or the big end bearing is designed for the maximum gas
force (FL), therefore,
equating the load on the crankpin or big end bearing to the maximum gas
force,
i.e. 13 (dc)2 = FL = 24 740 N
∴ (dc )2 = 24 740 / 13 = 1903 or dc = 43.6 say 44 mm Ans.
and lc = 1.3 dc = 1.3 × 44 = 57.2 say 58 mm Ans.
The big end has removable precision bearing shells of brass or bronze or
steel with a thin lining
(1mm or less) of bearing metal such as babbit.
Again, let dp = Diameter of the piston pin or small end bearing,
lp = Length of the piston pin or small end bearing = 2dp ...(Given)
pbp = Bearing pressure = 15 N/mm2 ..(Given)
We know that the load on the piston pin or small end bearing = Project
area × Bearing pressure
= dp . lp . pbp = dp × 2 dp × 15 = 30 (dp)2
Since the piston pin or the small end bearing is designed for the maximum
gas force (FL),
therefore, equating the load on the piston pin or the small end bearing to
the maximum gas force,
∴ (dp)2 = 24 740 / 30 = 825 or dp = 28.7 say 29 mm and lp = 2
dp = 2 × 29 = 58 mm
dp = 2 × 29 = 58 mm
The small end bearing is usually a phosphor bronze bush of about 3 mm
thickness.
3. Size of bolts for securing
the big end cap
Let dcb = Core diameter of the bolts,
σt = Allowable tensile stress for the material of the bolts
= 60 N/mm2 ...(Given)
and nb = Number of bolts. Generally two bolts are used.
We know that force on the bolts
The bolts and the big end cap are subjected to tensile force which
corresponds to the inertia
force of the reciprocating parts at the top dead centre on the exhaust
stroke. We know that inertia force of the reciprocating parts,
We also know that at top dead centre on the exhaust stroke, θ = 0.
Equating the inertia force to the force on the bolts, we have
and nominal diameter of the bolt,
4. Thickness of the big end
cap
Let tc = Thickness of the big end cap, bc = Width of the big end cap. It is taken equal to the length of the crankpin
or big end bearing (lc) = 58 mm (calculated above)
σb = Allowable bending stress for the material of the cap = 80 N/mm2
...(Given)
The big end cap is designed as a beam freely supported at the cap bolt
centres and loaded by the inertia force at the top dead centre on the exhaust stroke (i.e. FI when θ = 0).
Since the load is assumed to act in between the uniformly distributed load and
the centrally concentrated load, therefore, maximum bending moment is taken as
= Dia. of crank pin or big end bearing + 2 × Thickness of bearing
liner + Nominal dia. of bolt + Clearance = (dc + 2 × 3 + db + 3) mm = 44 +
6 + 12 + 3 = 65 mm
Detailed
drawing.
Manufacturing Process of Connecting rod.
Conclusion
The project on the design of connecting rod
with respect to the Industrial engineering gave us platform to understand the design
process of connecting rod in detailed , also scope for further studies on advance
research on connecting rod and usage of IE tools for the manufacturing process like
JIT.
References
1. Machine Design book By R.S.Khurmi.
3. www.youtube.com
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