Connecting_Rod_Report_Kramer_Suls_Moon (1)

M E C H A N I C A L D E S I G N O F A
C O N N E C T I N G R O D
erik kramer1
, deaho moon1
, jordan suls1
abstract
The goal of this project was to design a connecting rod for a Ferrari F136 F engine which
holds up to all safety standards as well as remains both cost and weight effective. A
MATLAB study of engine loads and beam analysis was performed to check analytically
for buckling or yield failure in a preliminary simplified beam design. These studies
resulted in a Yield Safety Factor of 2.09 as well as buckling safety factors of 16.25 in the y
direction and 64.98 in the z direction. Following the simplified analytical approach, a 2D
numerical plane stress assumption finite element analysis (FEA) was done in COMSOL
to further evaluate the design. A more realistic connecting rod but simplified profile
was used with a constant thickness of 0.5 inches. This gave a yield safety factor of 1.19.
Lifetime analysis was performed on this design and resulted in an infinite lifetime using
S-N methods but only 14032 cycles using fatigue fracture methods. Two design iterations
were then performed to increase the factor of safety against yield to 2.62 and to increase
the fatigue fracture cycles to failure to 19300. Finally, using our final iteration 2D profile
as a starting point, two detailed 3D models were produced and simulated under working
conditions in Abaqus. From this we found a successful final design with a yield safety
factor of 1.97.
1 Department of Mechanical and Aerospace Engineering, University of California Los Angeles
1

Contents 2
contents
1 Introduction 2
2 Engine Assumptions & Operating Loads 3
3 Load Analysis 4
4 Beam Theory Analysis 5
5 Initial Design Solid Modeling 7
6 Two-Dimensional Elastic Finite Element Analysis 8
7 Lifetime Analysis 11
7.1 S-N Fatigue Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
7.2 Fracture Fatigue Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
8 Design Optimization 15
9 Three-Dimensional Solid Model 18
10 Three-Dimensional Elastic Finite Element Analysis 18
11 Discussion & Conclusions 21
12 Appendix A - Derivations 21
13 Appendix B - MATLAB Code 24
13.1 Load Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
13.2 Beam Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
13.3 S-N Fatigue Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
13.4 Fatigue Fracture Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
13.5 3D FEA Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
14 Appendix C - Final Design Documentation 46
1 introduction
Connecting rods are widely used as couplers in various engineering applications based
on slider-crank mechanisms. For example, depending on the kinematic inversion of the
mechanism, connecting rods are used in compressors, rotary engines found in early air-
craft, driving wheels of early steam locomotives, and garden water pumps [1].
Today connecting rods are mainly used in internal combustion engines. The purpose
of connecting rods is to convert linear motion of the pistons to rotational motion of the
crankshaft. The design of connecting rods for high strength with light weight is crucial
because they have to rapidly respond to the motion of the pistons under extreme levels of
stress exerted by pistons, chemical reactions, etc. Hence, they are one of the most critical
components of the engine because their performance and lifecycle directly affect those of
the engine.
Depending on the types of engines, connecting rod designs are varied based on the
beam shapes and materials. There are two basic types of connecting rods; I-Beam and
H-Beam. I-Beam rods are the most common type of connecting rods. Since I-Beams offer
high strength with light weight, I-Beams are generally used in high rpm engines. H-

engine assumptions & operating loads 3
Beam rods are used in low rpm engines with higher torque due to its greater compressive
strength than I-Beam rods [2]. The material selection plays a significant role in connecting
rod design. Light and strong materials are desired but material price is a limiting factor
in the connecting rod design. For example, using titanium for a connecting rod instead
of steel can reduce the weight by approximately 22% but will cost over twice as much
to manufacture [2]. Therefore, along with economical analysis, thorough material and
mechanical analysis is required for the optimal connecting rod design.
The purpose of this study is to optimize the design of a connecting rod for a selected
engine. The variable parameters for the connecting rod design are the dimensions, the
weight, the material used, and the cost. However, the cost will be the minimal factor to
be considered in this study because the connecting rod is designed for one of the most
expensive luxury car engines. The objective is to maximize the durability and resistance
to fatigue failure.
In this report we will present loading and bending analytical analysis which were de-
rived by hand and then computed via MATLAB. Following that we will present a simpli-
fied 2D model for a plane stress FEA as well as lifetime analysis. Several design iterations
will also be given. Lastly, a detailed 3D model will be shown which was used for a 3D
Abaqus FEA to prove its viability against static yield.
2 engine assumptions & operating loads
The connecting rod was designed for the Ferrari F136 F engine, which is primarily used
in the Ferrari 458 Italia, Spider and Speciale [3]. The specifications for this engine are as
follows:
Engine Displacement 4499 cm3
Engine Torque 540 N m
Engine Speed 6000 RPM
Stroke Length 81 mm
Connecting Rod Length 144 mm
Piston Mass 400 g
Along with these constant engine operation parameters, some preliminary design as-
sumptions and choices for the connecting rod’s properties were made such that a numer-
ical load analysis could be implemented with the use of MATLAB. The following design
values are based on a simple initial design which will be expanded upon in section 5.
Note that the values here were used for the initial analytical analysis but constantly re-
vised throughout the design optimization process to reflect the design being tested. Our
initial design assumed the use of AISI 4340 normalized steel.

load analysis 4
Connecting Rod Mass 557.64 g
Relative Center of Mass β 0.3250
Moment of Inertia 1.89x10−3 kg m2
Young’s Modulus 2.05x1011 Pa
Yield Strength 7.1x108 Pa
Ultimate Tensile Strength 1.11x109 Pa
3 load analysis
The connecting rod is subject to three time dependent loads: the inertial force due to
motion, the contact force from the piston, and the contact force crankshaft. Based on the
assumption that the crankshaft has a constant angular speed, the loads on the connecting
rod are known to be periodic.
Figure 1 shows the basic geometry of the study. The connecting rod is pinned to the
piston at P, to the crankshaft at R, and has its center of mass at G. The crankshaft rotates
about the origin at constant speed ω, the time derivative of its angle φ. The piston is
constrained to the y-axis and forms the angle γ with the connecting rod. Inspection
yields a geometric description of the positions of these three points. A basic kinematics
analysis, found in Appendix A, yields the position, speed, and acceleration of the three
key points.
• Angular Velocity of Connecting Rod: ˙γ = − ωλcosφ
(1−λ2 sin2 φ)1/2
• Angular Acceleration of Connecting Rod: ¨γ = ω2 λ(1−λ2 )sinφ
(1−λ2 sinφ)3/2
• Acceleration of Crankshaft Pin: ¨R = −ω2 R
• Acceleration of Piston Pin: ¨P = lλ[( ˙γ − ω)ωcosφ + ¨γsinφ]
• Acceleration of Center of Gravity: ¨G = (1 − β) ¨R + β ¨P
Figure 1: The simple assumed geometry of the piston crankshaft system

beam theory analysis 5
φ [rad]
-8 -6 -4 -2 0 2 4 6 8
AxialForces[N]
×105
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
ΦPRa
ΦSRa
-mg
ddGa
(a) Axial Forces vs Crankshaft Angle
φ [rad]
-8 -6 -4 -2 0 2 4 6 8
TransverseForces[N]
-8000
-6000
-4000
-2000
0
2000
4000
6000
8000
ΦPRt
ΦSRt
-mg
ddGt
(b) Transverse Forces vs Crankshaft Angle
Figure 2: Results of load analysis in MATLAB.
Combining these equations with the known combustion pressure on the piston, the
equations of motion for the system have been derived by balancing the linear momentum
for the piston and the angular momentum of the connecting rod. All derivations can be
found in Appendix A.
• Force of Piston on Rod: ΦPRy = −mP
¨P − p(φ)A
• Force of Crankshaft on Rod: ΦPRx = βmG
¨Gx −
JG ¨γ−[(β−1)ΦPRy +βΦSRy ]Rx
Py −Px
To allow for a focused ﬁnite element study of the connecting rod, the inertial reference
frame is changed to a non-inertial one. The origin of this frame is at the center of mass,
G. The x-axis lies along the line connecting points of the piston and crankshaft, and
the y-axis describes the transverse direction. To accurately describe the dynamics in this
reference frame, the inertial forces must be considered. Analysis of a generic material
particle at any location relative to the new reference frame yields a body force per unit
volume, which is related to the linear acceleration of the center of gravity, its angular
acceleration, and its centripetal acceleration.
Solving these equations with the use of the MATLAB Load Analysis code found in
Appendix B yields the axial and transverse components of force on the connecting rod
shown in Figure 2. We also calculate our Yield Safety Factor to be 2.09.
4 beam theory analysis
The results of the Load Analysis yielded values for the contact force at the crankshaft pin,
contact force at the piston pin, inertial force due to motion and inertial moment due to
motion. Using beam theory, the axial stresses on the connecting rod were found with the
aid of MATLAB. The geometry of our beam was set up as a rectangular cross section with
a thickness of 12.7mm (in the z-direction) and a height of 25.4mm (in the y-direction).

beam theory analysis 6
z [m]
-0.015 -0.01 -0.005 0 0.005 0.01 0.015
y[m]
-0.01
-0.005
0
0.005
0.01
cross section
Figure 3: The cross section of our beam used for preliminary analysis
The resulting data of the axial force, transverse force, axial load and bending moment
along the beam is shown in Figure 4. These graphs were constructed with the MATLAB
Beam Analysis code found in Appendix B.
The relevant values found from this Beam Analysis are as follows:
• Maximum stress in the beam: σmax = 339MPa
• Buckling Safety Z Factor: nz−axis = 64.98
• Buckling Safety Y Factor: ny−axis = 16.25

initial design solid modeling 7
φ [rad]
-10 -5 0 5 10
AxialForces[N]
×105
-1
-0.5
0
0.5
1
ΦPRa
ΦSRa
-mg
ddGa
φ [rad]
-10 -5 0 5 10
TransverseForces[N]
-8000
-6000
-4000
-2000
0
2000
4000
6000
8000
ΦPRt
ΦSRt
-mg
ddGt
x [m]
0 0.05 0.1
AxialLoad[N]
×104
-8
-6
-4
-2
0
x [m]
0 0.05 0.1
BendingMoment[Nm]
-200
-100
0
100
200
Figure 4: Results of a MATLAB beam analysis
5 initial design solid modeling
For our preliminary design we focused on a simple solution to quickly provide two dimen-
sional finite element stress results using a plane stress assumption. Some initial design
parameters were pulled directly from physical engine constraints such as the pin sizes
and rod length from center to center of the two pin holes. For the free parameters we
first chose a standard thickness of 12.7mm and width of 25.4mm with ease of manufac-
turing in mind. The two remaining free parameters, pin hole wall thicknesses and fillet
radius were chosen mostly aesthetically. Pin hole wall thicknesses were given a dimen-
sion of half of their respective pin hole radii and fillets were chosen to be 10mm. While
it is understood that these values are not necessarily optimal, they provide a somewhat
similar design to existing connecting rods and thus will serve as a good basis for initial
design analysis. Before being put into production, these values will most likely change to
be more optimal in both strength and weight conservation.
Production of a geometric model was done via two programs. The first was done in
SolidWorks using a simple sketch and extrude method. Fillets were added on once the
rod was extruded to the desired overall shape. A 3D model of our initial design can be
seen in Figure 5. Using SolidWorks material and mass properties function, we were able
to obtain accurate moments of inertia for the design as well as the location of the center
of mass which would be used in a 2D COMSOL analysis. Initial material for the part

two-dimensional elastic finite element analysis 8
Figure 5: A model of our initial design in 3D using SolidWorks
was chosen to be AISI 4340 normalized steel because it is a proven material for existing
connecting rods. Once the physical properties of the initial design were known, this
design was also created in COMSOL using their native geometric building tools. The
COMSOL model was done to be a 2D top down view of the full 3D part as shown in
Figure 7. More information on the COMSOL design and the resulting stress simulation
can be found in the following section.
6 two-dimensional elastic finite element analysis
To compute the stress and strain distributions on the connecting rod, a 2-D Finite Element
Analysis (FEA) in COMSOL was performed. A parametric simulation of the stress as a
function of the crank angle, φ, was produced with the use of body forces and certain fixed
constraints. The pressure distribution as a function of the crank angle is demonstrated in
Figure 6, where the maximum pressure placed on the connecting rod occurs at an angle
of approximately π/10.
In this case, the crankshaft pin was fixed in the x- and y-direction to restrict any motion.
For the piston pin, rigid body rotation was avoided by setting a prescribed displacement
of zero for the pin in the y-direction. Also, an axial force of FPRa, which represents the
combustion pressure, was transmitted from the piston pin onto the connecting rod, acting
in the x- direction. Finally, a body load, defined as a force per unit volume, was placed
on the connecting rod.

crank angle [rad]
-6 -4 -2 0 2 4 6
combustionpressure[Pa]
×106
0
5
10
15
Figure 6: Combustion pressure as a function of the crank angle
To ensure an accurate solution for the FEA, a ﬁne mesh was used such that the stress
concentration in the crucial areas could be visualized. The areas of concern were the ﬁl-
leted corners and the contact edges between the connecting rods and the crankshaft/pis-
ton pin. The resulting mesh is shown in Figure 7.
The results of this study is best shown graphically, so the areas of high stress can be
visualized. Figure 8 shows the stress distribution at the end of the cycle, with the default
color scale to see where the maximum stresses occur. It is clear that regions close to the
crankshaft and piston pin are subjected to the greatest compressive stresses.
For this analysis, it is useful to compare the stress to that of the yield strength to see
if the forces on the connecting rod are going to create plastic deformation. To do this,
the color scale was set from 0 to 7.1x108 Pa, which is the yield strength for AISI 4340
normalized steel. The crank angles of concern were 0 and π/10 radians, where the
pressure on the connecting rod is at a maximum. Figure 9 shows the stresses on the
connecting rod at those crank angles. The maximum stress values at 0 and π/10 are
3.34x108 Pa and 5.98x108 Pa, respectively. This equates to a safety factor against yield
of 1.19.

Figure 7: The 2D mesh used for COMSOL FEA.
Figure 8: Results of a COMSOL load analysis normalized to max stress calculated.
(a) A phi state where stress is at its min. (b) A phi state where stress is at its max.
Figure 9: FEA results normalized to yield strength.

lifetime analysis 11
7 lifetime analysis
While it is very important to design against static yielding and fracture in structural com-
ponents, many components are made to last numerous years before failure is acceptable.
Because of this, lifetime analysis is an important step in the design process. Depending
on the environment and loading conditions (both mechanically and thermally) compo-
nents can fail even if they have high static factors of safety. Lifetime failure comes in
many forms including fatigue, corrosion, creep, fouling, thermal shock, and wear. Most
lifetime failure mechanisms result from physical degradation of components to non ideal
operational conditions or outright yield or fracture. Some failure mechanisms can result
in technical faults without being destructive, like unacceptable increase of friction for a
sliding sleeve valve.[4] Due to the cyclic loading nature of a connecting rod, fatigue is one
of the main concerns of the rod s lifetime. The time for one engine revolution depends
on the rpm of the engine, therefore it is easier to measure lifetime of a connecting rod in
engine cycles instead of absolute time. For the reasons stated above we focus our lifetime
analysis on fatigue in this report. Fatigue has two main approaches, S-N (damage accu-
mulation) methods or crack propagation. In the following sections we present analysis
using these two theories.
7.1 S-N Fatigue Analysis
For the S-N fatigue we adopt a numerical rainflow method for the connecting rod s load-
ing cycle. The Palmgren-Miner rule can be written as a linear damage rule:
D = Σ
ni
Ni
where D is the accumulated damage, ni is the number of cycles at stress level σi, and Ni
is the number of cycles to failure at stress level σi.[1] To calculate the fatigue life, ni, Ni,
and Si should be determined, however, the fatigue data provided in S-N curves are not
directly interchangable for real-life mechanical components. Using the rainflow method,
Si is calculated by using Soderberg criteria and its corresponding Ni can be obtained
from S-N curves.
Due to lack of data availability at higher temperatures and for the consistency among
the values, material property constants are used at room temperatures. For our numerical
fatigue analysis, the input values are based on the SN curve for 4340 steel. The strengths
of the connecting rod material are Sut = 125ksi = 861MPa, Sy = 103ksi = 710MPa.
By considering that the material should be an un-notched steel bar with 0 stress ratio,
the SN curve of 4340 steel which satisfies the conditions is shown.[5] Reliability is the
probability that machine systems and components will perform their intended function
satisfactorily without failure.[1] For our fatigue analysis, we used the 95 percent reliability
factor, which is a high reliability. In other words, based on our design, 95 percent of
connecting rods are capable of enduring the cyclic loads.

Figure 10: The S-N curve for 4340 steel at room temperature.[5]
From the S-N curve shown in Figure 10, the yield and endurance cycles are found
as Ny = 4x104, Ne = 2x107. These values and the Sut and Sy of the material are
used as inputs in a numerical MATLAB rainflow fatigue analysis (code for which can be
found in Appendix B). Figure 11 shows the damage accumulation of our design using
this method. The results of this analysis found our initial connecting rod design to have
infinite lifetime.
7.2 Fracture Fatigue Analysis
In a fracture approach to fatigue analysis, it is assumed that every component has cracks
which form inherently during the manufacturing process. Though the size of these cracks
may be too small to detect with modern technology, they are nonetheless present. Fatigue
fracture analysis builds off the idea that under a load cycle these cracks propagate in size
until they hit a critical crack size when fast fracture occurs. A simplified version of crack
propagation in one dimension can be done via Paris law which gives a relation between
cycles N and the crack length based on constants C and m as well as shape factor b and
alternating stress ∆σ.
dA
dN
= C∆Km ∆K = ∆σβ
√
πa
Using this simplified version of crack propagation, the crack always grows each cycle.
This means that there exists a finite number of cycles before the crack reaches a critical
length and fast fracture occurs (KI > KIC for mode one cracks). Since the finest and the

Figure 11: The damage accumulation over the lifetime of our connecting rod.
most precise machining and manufacturing technologies will be used for the luxury car
engine, it is assumed that the initial crack size is the smallest detectable with current tech-
nology. According to Larsen, the minimum detectable crack depth of AISI 4340 steel is
50µm by using an ultrasonic method along with split spectrum processing.[6] By investi-
gating the highest stress concentration location on the connecting rod through COMSOL,
we located a likely mode-I crack on the ﬁllet surface between the bar and small pin sec-
tions. The Paris constants for AISI 4340 steel are C = 4.5x10−12 [m/(cycle MPa
√
m]
and m = 2.36.[7] For our numerical analysis, C should be converted into SI units. By mul-
tiplying the conversion factor 10−6m, C becomes 4.5x10−12−6m [m/(cycle Pa
√
m].
The critical stress intensity factor, KIC, for 4340 Steel is 99MPa
√
m.[1]
According to City Speed for Los Angeles [8], the average driving speed in LA is 26.8
mph with standard deviation of 8.8 mph. Since the engine we chose is for the 2012 Ferrari
458 Italia, one of the fastest cars, we assumed that the average driving speed of the car is
44.4 mph, which is two standard deviations above the average so that the car s average
driving speed is within the top 2.1 percent range in LA. The relationship between engine
rpm and car speed is approximately given as
V =
DF
336R
where V is the velocity in mph, D the tire diameter in inches, F the engine rpm, and
R the gear ratio.[9] Based on the relationship, we assume that the Ferrari drives at 44.4
mph in 5th gear at 3000 rpm. The Ferrari has a 5th gear ratio of 1.03 with a ﬁnal drive
axle ratio of 5.14. The front tire size is 235/35R20 which corresponds to 26.5 inches for

Figure 12: Possible initial crack locations allowed in our fatigue fracture study.
the tire diameter.[10] We use the front tire size because the engine rotates front wheels to
drive the car. The maximum number of miles driven by the car is assumed to be 300,000
miles and the maximum number of load cycles N can be derived from the relationship
between engine and wheel rpm along with transmission ratio. Note that a single load
cycle is equal to two engine cycles. Wheel revolution can be derived by dividing engine
crankshaft revolution by transmission ratio, which is the product of the gear ratio and
the final drive axle ratio. Therefore we conclude that 6.04x108 load cycles would be
considered infinite lifetime.
Ninf =
Distance Car Driven
T ire Circumference
x
T ransmission Ratio
2
Ninf = 300, 000miles x
63360 in
mile
x
1 wheel revolution
26.5π in
x
1.03x5.14
2
Ninf = 6.04x108
A MATLAB script (included in the Appendix B) was implemented to numerically solve
for the cycles to failure using fatigue fracture in mode one. It was designed to assume
the worst case scenario. Due to the greatest tensile stresses occurring in the middle bar
region of the connecting rod, the script only focused on that region and assumed that
cracks propagated only normal to the top and bottom surface of the connecting rod as
shown in Figure 12. After finding the point of highest edge stress from our 2D COMSOL
analysis and checking for fast fracture in the first cycle, the script uses Paris law to
propagate the crack for each subsequent cycle. During each cycle the script checks for fast

design optimization 15
Figure 13: Crack propagation as a function of loading cycles N.
fracture before propagating the crack. In order to obtain the most realistic results possible,
the maximum stress (σmax) used for ∆K in Paris law as well as in the fast fracture
check (KI > KIC) was the 2D COMSOL results for σx averaged over the crack length at
the crack surface. Because only tensile forces cause crack opening, the minimum stress
(σmin) was assumed to be zero. Beta was also resolved each cycle based on the current
crack length. Using this script a lifetime of 14032 cycles was found. Figure 13 shows
a plot of the crack size vs cycles for our initial design. While this number is far below
what we calculated to be infinite lifetime, we can attribute this to the many simplified
assumptions our method introduced. In reality the crack heals itself under compression
each cycle. Also, crack propagation is not a 2D phenomenon. Real cracks have a depth
dimension along side the length and can propagate in any direction. Additionally, they
are subject to multiple crack opening modes at any one time. Paris law, while a good
starting tool is not great for accurate results in complicated geometries. Due to the scope
of this project we will not be focusing on refining our code to reflect more realistic crack
propagation but instead look for increased cycles to failure with each design optimization
iteration.
8 design optimization
In order to find an optimal design we iterated through two more 2D designs before mov-
ing onto a 3D one. Each time the design was changed, a new run of 2D FEA and lifetime
analysis was performed. The first design iteration changed the shape profile of the con-

(a) The profile of our first design iteration. Thick-
ness into the page for this design is 0.5 inches.
(b) The profile of our second design iteration.
Thickness into the page for this design is 1
inches.
Figure 14: The profiles of our two design iterations.
necting rod while maintaining its original thickness of 0.5 inches. We increased the width
of the middle beam section while decreasing the thickness of the material surrounding
the crank shaft and piston pin. This was done in an attempt to increase its fatigue fracture
lifetime while not drastically increasing material use as shown in Figure 14a and Figure
15. While this produced more cycles to failure, COMSOL results still showed contact
stress still being very close to the yield strength as well as a drastically increased amount
of elastic deformation in each cycle. Another design iteration (shown in Figure Figure
14b and Figure 16) was performed keeping the shape profile the same but increasing the
beam thickness as well as the thickness of the material around crankshaft. This change
produced a more favorable safety factor against yield as well as increased cycles to failure.
The third design’s increased strength results and 1 inch thick contact areas were satisfac-
tory enough to allow us to move onto 3D analysis while feeling confident about removing
material to turn the connecting rod into an I-beam. The results of our design iterations
are produced in the table below.
Design One Design Two Design Three
Maximum Von Mises Stress (σ ) 5.98x108 Pa 5.44x108 Pa 2.71x108 Pa
Yield Safety Factor (ny) 1.19 1.31 2.62
S-N Lifetime Infinite Infinite Infinite
Fatigue Fracture Lifetime ny 14032 cycles 14578 cycles 19300 cycles

Figure 15: 2D FEA results of our ﬁrst design iteration. Thickness into the page for this design is
0.5 inches.
Figure 16: 2D FEA results of our second design iteration. Thickness into the page for this design
is 1 inch.

three-dimensional solid model 18
9 three-dimensional solid model
With the satisfactory results from our third redesign we began our move to 3D using the
2D profile of Design Three shown in Figure 14b. Modeling was done in SolidWorks and
continued to use AISI 4340 normalized steel as the material of choice due to its properties
and availability. Because 2D FEA simulations showed low stress in the beam portion
of the connecting rod even at the thinner thickness of 0.5 inches, we chose to use that
as the nominal thickness in our bar. However, in order to keep our advantage against
yield due to contact stresses at the piston pin, we kept the thickness of that area at 1
inch as simulated in Design Three of our 2D analysis. With this overall shape we added
Figure 17: Our first detailed 3D solid model design.
wings for the later addition of crankshaft bolts as well as numerous fillets to reduce stress
concentrations at design corners and edges. Material was also cut out on both sides of
the beam portion to reduce weight. This cutout also changed the rod’s cross-section to
that of an I-beam. The completed 3D model of our connecting rod can be seen in Figure
17.
10 three-dimensional elastic finite element analysis
The 3-D model created in SolidWorks was imported into ABAQUS to perform a Finite
Element Analysis on our fully designed connecting rod. Simulation was done for the
most critical crankshaft angle π/10 radians, which was found determined from the re-
sults of our 2D COMSOL FEA. By applying the axial and transverse body forces on the
connecting rod and the corresponding compressive force from the Piston pin, the stress
concentrations at every node of our 3-D model was found. This was done to determine

three-dimensional elastic finite element analysis 19
Figure 18: Abaqus results for our first 3D design.
which areas were developing the greatest stresses and whether the connecting rod would
yield under the stress. During the first run the nominal thickness of the bar was 0.5
inches with the pin contact areas having increased thickness of 1 inch as per our design
presented in the previous section. The maximum stress was found to be on the fillet of
the I-beam cutout closest to the Piston Pin, shown in Figure 18. The maximum Von Mises
stress was 4.76 MPa. When compared to the yield strength of Forged 4340 Steel, this gave
a safety factor of 1.49. This value was too low to safely implement into a car, so one itera-
tion of the 3D design was performed to increase nominal thickness of the rod increasing
it to 1 inch and the thickness of the connecting pin areas to 1.5 inch. Some alterations
of the fillets were also performed and the differences between the two 3D designs can be
seen in figure 19. Although this increases the amount of material required to manufacture
the connecting rod, which corresponds to an increase in the cost of the part, this design
alteration was necessary to ensure that yielding does not occur during a single loading
cycle. With a thicker connecting rod, the maximum stress was calculated to be 3.61 MPa,
giving a safety factor of 1.97. The safety factor for the second design is significantly more
conservative so it was selected as our final design. The stress state of the second and final
connecting rod is shown in Figure 20. Engineering drawings and a render of our final
design can be found in Appendix C.

three-dimensional elastic finite element analysis 20
Figure 19: A comparison of our two 3D designs.
Figure 20: Abaqus results for our second and ﬁnal 3D design.

discussion & conclusions 21
11 discussion & conclusions
Though it took several iterations at different points, we believe the final design of our
connecting rod is acceptable for further deeper analysis and/or prototyping. Overall,
we encountered favorable results in beam bending simulations and 2D FEA plane stress
analysis. In the analytical analysis, our design produced large safety factors for buckling
(ny−axis = 16.25 and nz−axis = 64.98). With some design changes we were also able to
reach a safety factor of 2.62 for yield in a 2D plane stress simulation.
In the S-N lifetime analysis, we achieved favorable infinite life results but had to use
fatigue data and all other material property values at room temperature due to the lack
of availability of data at engine operation temperatures. The normal car engine operating
temperature is between 195 F and 220 F, which is around 95 C[11]. As temperature affects
the strength of material, there must be an error in the analysis. However, around this tem-
perature, the ratio of tensile strength at operating temperature and at room temperature
for steel is 1.020.[1] The actual material strength at the engine operating temperature is
higher than at room temperature. By using the room temperature values, our analysis
corresponds to a worst case scenario. This means that we can safely say infinite life for
S-N fatigue analysis will hold at true engine temperatures.
Although overall our fatigue fracture lifetime was small in comparison to the expected
level for infinite lifetime, this was mainly due to the use of simplified crack propagation
techniques. We were however able to increase our cycles to failure significantly which
attests to the success of our design iterations. Finally, in the 3D analysis we similarly
produced satisfactory results to the 2D case after some small design tweaking.
The promising results from our analysis for a Ferrari F136 F engine connecting rod
supports moving forward with more refined 3D analysis and/or prototyping. While there
were initial issues with high contact forces and yielding, our final design was calculated
to not fail under the operative axial or radial forces. Using these results we might wish
to continue on with a rigorous cost and weight analysis to provide a satisfactory final
product for both the consumer and the manufacturer.
12 appendix a - derivations

appendix a - derivations 22
Figure 21: Kinematic equation derivations

appendix a - derivations 23
Figure 22: Dynamic equation derivations

appendix b - matlab code 24
13 appendix b - matlab code
13.1 Load Analysis
1 % MAE 296A Project, Fall 2015
2 % Load_Analysis
3 clc
4 clear
5 close all
6 disp('Load Analysis')
7
8
9 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
10 % Engine Parameters (choose you own values) - Ferrari F136F Engine Used in
11 % Ferrari 458 italia GT3 2011-2013
12 omega=6000; % Engine angular speed [rpm]
13 T=540; % Engine Torque at omega=6000 [Nm]
14 V=0.004499; % Engine total dispacement [m^3]
15 n=8; % Number of cylinders
16 r=0.0405; % half stroke length [m]
17 l=0.144; % connectin rod length [m]
18
19 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
20 % Mass Properties (input your own values. Values will be refined when you ...
have your solid model)
21 mp=0.4; % piston mass [kg]
22 mg=0.55764; % connecting rod mass [kg]
23 beta=0.3250; % ralative position of G [-] beta= where ...
beta=((l/2)-(Solidworks CoM))/l is center to center legnth of holes ...
(l=144m)
24 J=0.0018907464; % moment of inertia [Kg*m^2] Lyy from solidworks
25
26 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
27 % Automatically generated values (do not edit)
28 rpm2rads=2*pi/60; % conversion factor rpm to rad/s [rad/s/rpm]
29 omega=omega*rpm2rads; % Engine angular speed [rad/s]
30 Vc=V/n; % cylinder volume [m^3]
31 A=Vc/(2*r); % piston surface area [m^2]
32 lambda=r/l; % lambda ratio [-]
33 mep=4*pi*T/V; % mean effective pressure [Pa]
34 eta=0.8; % engine mechanical efficiency [-]
35 mip=mep/eta; % mean indicated pressure [Pa]
36
37 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
38 % load the combustion data (do not edit)
39 %load combustion_pressure
40 load cp_downsampled_v6
41
42 pressure=pressure*mip; % rescale presure based on engine parameters
43 figure(1)

44 plot(phi,pressure,'Linewidth',2)
45 xlim([-2*pi 2*pi])
46 grid on
47 xlabel('crank angle [rad]')
48 ylabel('combustion pressure [Pa]')
49 set(gcf , ...
50 'Color' , 'w' );
51 set(gca, ...
52 'FontName' , 'Helvetica' , ...
53 'Box' , 'off' , ...
54 'TickDir' , 'out' , ...
55 'TickLength' , [.02 .02] , ...
56 'XMinorTick' , 'on' , ...
57 'YMinorTick' , 'on' , ...
58 'XGrid' , 'on' , ...
59 'YGrid' , 'on' , ...
60 'XColor' , [.15 .15 .15] , ...
61 'YColor' , [.15 .15 .15] , ...
62 'LineWidth' , 1 );
63
64 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
65 % Calculate Inertial Components (complete here - done 10/9)
66 dgamma=-lambda*cos(phi)*omega./ ...
67 sqrt((1-lambda^2*sin(phi).^2));% first time derivative of gamma
68 ddgamma=(1-lambda^2)*lambda*sin(phi)*omega^2./ ...
69 (1-lambda^2*sin(phi).^2).^(3/2);% first time derivative of gamma
70 R=r*[-sin(phi) cos(phi)]; % position of R
71 P=l*[zeros(length(phi),1) lambda*cos(phi)+sqrt(1-lambda^2*sin(phi).^2)]; ...
% position of P
72 G=R+beta*(P-R); % position of G
73 ddR=-(omega^2)*(R); % acceleration of R
74 ddP=l*lambda*[zeros(length(phi),1) ...
cos(phi).*(dgamma-omega)*omega+sin(phi).*ddgamma]; % acceleration of P
75 ddG=(1-beta)*ddR+beta*ddP; %(My edit)
76 % acceleration of G
77
78 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
79 % Calculate Contact Forces (complete here - done 10/9)
80 FPRy=-mp*ddP(:,2)-pressure*A;
81 FSRy=mg*ddG(:,2)-FPRy; %my Edit
82 FPRx=beta*mg*ddG(:,1)-(J*ddgamma-((beta-1)*FPRy+beta*FSRy) ...
83 .*R(:,1))./(P(:,2)-R(:,2));
84 FSRx=mg*ddG(:,1)-FPRx; %my Edit
85
86 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
87 % Store forces in vector form for easier manipulation (do not edit)
88 FPR=[FPRx FPRy];
89 FSR=[FSRx FSRy];
90
91 %cross2Dv(FSR,G-R)+cross2Dv(FPR,G-P)-J*ddgamma
92 %return
93
94 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

95 % Find the axial components of forces (complete here - not edited 10/9)
96 a=1/l*(P-R); % axial unit vector
97 FPRa=dot2Dv(a,FPR);
98 FSRa=dot2Dv(a,FSR);
99 ddGa=dot2Dv(a,ddG);
100
101 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
102 % Find the transverse components of forces (complete here - not edited 10/9)
103 t=[-a(:,2) a(:,1)]; % unit vector along the trasverse direction
104 FPRt=dot2Dv(t,FPR);
105 FSRt=dot2Dv(t,FSR);
106 ddGt=dot2Dv(t,ddG);
107
108 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
109 % Store forces in vector form in the axial-transverse
110 % reference system (do not edit)
111 FPRat=[FPRa FPRt];
112 FSRat=[FSRa FSRt];
113 ddGat=[ddGa ddGt];
114
115 Gat=[ones(length(phi),1) zeros(length(phi),1)]*beta*l;
116 Rat=[zeros(length(phi),1) zeros(length(phi),1)];
117 Pat=[ones(length(phi),1) zeros(length(phi),1)]*l;
118 %cross2Dv(FSRat,Gat-Rat)+cross2Dv(FPRat,Gat-Pat)-J*ddgamma
119 %FPRat+FSRat-mg*ddGat
120 %return
121
122 figure(2)
123 plot(phi,FPRa,'r','Linewidth',2)
124 hold on
125 plot(phi,FSRa,'b','Linewidth',2)
126 plot(phi,-mg*ddGa,'k','Linewidth',2)
127 legend(texlabel('Phi_{PRa}'),texlabel('Phi_{SRa}'),texlabel('-m_g ddGa'))
128 grid on
129 xlabel([texlabel('phi ') '[rad]'])
130 ylabel([texlabel('Axial Forces ') '[N]'])
131 print(gcf, '-depsc', 'axial')
132 set(gcf , ...
133 'Color' , 'w' );
134 set(gca, ...
136 'Box' , 'off' , ...
137 'TickDir' , 'out' , ...
138 'TickLength' , [.02 .02] , ...
141 'XGrid' , 'on' , ...
142 'YGrid' , 'on' , ...
143 'XColor' , [.15 .15 .15] , ...
144 'YColor' , [.15 .15 .15] , ...
146
147 figure(3)

148 plot(phi,FPRt,'r','Linewidth',2)
149 hold on
150 plot(phi,FSRt,'b','Linewidth',2)
151 plot(phi,-mg*ddGt,'k','Linewidth',2)
152 legend(texlabel('Phi_{PRt}'),texlabel('Phi_{SRt}'),texlabel('-m_g ddGt'))
153 grid on
155 ylabel([texlabel('Transverse Forces ') '[N]'])
156 print(gcf, '-depsc', 'transverse')
157 set(gcf , ...
158 'Color' , 'w' );
159 set(gca, ...
161 'Box' , 'off' , ...
162 'TickDir' , 'out' , ...
163 'TickLength' , [.02 .02] , ...
166 'XGrid' , 'on' , ...
167 'YGrid' , 'on' , ...
168 'XColor' , [.15 .15 .15] , ...
169 'YColor' , [.15 .15 .15] , ...
171
172 error_axial=max(abs(FPRa+FSRa-mg*ddGa))
173 error_transverse=max(abs(FPRt+FSRt-mg*ddGt))
174
175 disp('Done')

13.2 Beam Analysis
1 % MAE 296A Project
2 % Beam_Analysis 1
3 disp('Beam Analysis 1')
4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
6 % Bending moment diagram
7 Mx=[zeros(length(phi),1) -FSRt*beta*l -FSRt*beta*l-J*ddgamma ...
zeros(length(phi),1)];
8 Mmax=max(max(Mx)); % maximum of bending moment over x and phi
9 Mmin=min(min(Mx)); % minimum of bending moment over x and phi
10 % Axial load diagram
11 Px=[-FSRa -FSRa FPRa FPRa];
12 Pmax=max(max(Px)); % maximum of axial load over x and phi
13 Pmin=min(min(Px)); % minimum of axial load over x and phi
14 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
15 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
16 % Movie
17 figure(4)
18 clf
19 subplot(2,2,1)
20 plot(phi,FPRa,'r','Linewidth',2)
21 hold on
22 plot(phi,FSRa,'b','Linewidth',2)
23 plot(phi,-mg*ddGa,'k','Linewidth',2)
24 legend(texlabel('Phi_{PRa}'),texlabel('Phi_{SRa}'),texlabel('-m_g ddGa'))
25 grid on
27 ylabel([texlabel('Axial Forces ') '[N]'])
28 v1=axis;
29
30 subplot(2,2,3)
31 plot(phi,FPRt,'r','Linewidth',2)
32 hold on
33 plot(phi,FSRt,'b','Linewidth',2)
34 plot(phi,-mg*ddGt,'k','Linewidth',2)
35 legend(texlabel('Phi_{PRt}'),texlabel('Phi_{SRt}'),texlabel('-m_g ddGt'))
36 grid on
38 ylabel([texlabel('Transverse Forces ') '[N]'])
39 v3=axis;
40
41 subplot(2,2,2)
42 xlabel('x [m]')
43 ylabel('Axial Load [N]')
44
45 subplot(2,2,4)
46 xlabel('x [m]')
47 ylabel('Bending Moment [Nm]')
48 set(gcf , ...

49 'Color' , 'w' );
50
51 for kk=1:length(phi)
52 subplot(2,2,1)
53 p1=plot([phi(kk) phi(kk)],[v1(3) v1(4)],'m','Linewidth',2);
54
55 set(gca, ...
57 'Box' , 'off' , ...
58 'TickDir' , 'out' , ...
59 'TickLength' , [.02 .02] , ...
62 'XGrid' , 'on' , ...
63 'YGrid' , 'on' , ...
64 'XColor' , [.15 .15 .15] , ...
65 'YColor' , [.15 .15 .15] , ...
67
68 subplot(2,2,3)
69 p2=plot([phi(kk) phi(kk)],[v3(3) v3(4)],'m','Linewidth',2);
70
71 set(gca, ...
73 'Box' , 'off' , ...
74 'TickDir' , 'out' , ...
75 'TickLength' , [.02 .02] , ...
78 'XGrid' , 'on' , ...
79 'YGrid' , 'on' , ...
80 'XColor' , [.15 .15 .15] , ...
81 'YColor' , [.15 .15 .15] , ...
83
84 subplot(2,2,4)
85 hold on
86 p3=plot([0 beta*l beta*l l]', Mx(kk,:),'Linewidth',2);
87 grid on
88 axis([0 l Mmin Mmax])
89
90 set(gca, ...
92 'Box' , 'off' , ...
93 'TickDir' , 'out' , ...
94 'TickLength' , [.02 .02] , ...
97 'XGrid' , 'on' , ...
98 'YGrid' , 'on' , ...
99 'XColor' , [.15 .15 .15] , ...
100 'YColor' , [.15 .15 .15] , ...

102
103 subplot(2,2,2)
104 hold on
105 p4=plot([0 beta*l beta*l l]', Px(kk,:),'Linewidth',2);
106 grid on
107 axis([0 l Pmin Pmax])
108
109 set(gca, ...
111 'Box' , 'off' , ...
112 'TickDir' , 'out' , ...
113 'TickLength' , [.02 .02] , ...
116 'XGrid' , 'on' , ...
117 'YGrid' , 'on' , ...
118 'XColor' , [.15 .15 .15] , ...
119 'YColor' , [.15 .15 .15] , ...
121
122 drawnow
123 if kk=length(phi)
124 delete(p1)
125 delete(p2)
126 delete(p3)
127 delete(p4)
128 end
129 end
130 disp('Done')
131 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1 % MAE 296A Project
2 % Beam_Analysis 2
3 disp('Beam Analysis 2')
4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
6 % Material properties (complete here - Edited 10/9)
7 E=2.05e11; % Young modulus (in Pa)
8 sy=710e6; % yield strength (in Pa)
9
10 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
11 % Geometric properties (complete here - Edited 10/9)
12 h=0.0254; % heigth of the beam (in m)
13 b=0.0127; % tickness of the beam (in m)
14 Ax=b*h; % cross section area
15 Iz=b*h^3/12; % moment of inertia with respect to the z axis
16 Iy=h*b^3/12; % moment of inertia with respect to the y axis
17 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
18 % The maximum axial stress over x and y is
19 Smax=abs(Px)/Ax+abs(Mx)*h/(2*Iz); % axial stress as a function ...
of x and phi

20 smax=max(max(Smax)); % maximum axial stress in the ...
beam
21 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
22 % Check for elastic yield (do not edit)
23 yield_safety_facor=sy/smax
24 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
25 % Check for buckling using Iz (do not edit)
26 Pcr_z=pi*2*E*Iz/l^2
27 buckling_z_safety_facor=Pcr_z/max(max(Px))
28 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
29 % Check for buckling using Iy (do not edit)
30 Pcr_y=pi*2*E*Iy/l^2
31 buckling_y_safety_facor=Pcr_y/max(max(Px))
32 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
33 % plot the cross section (do not edit)
34 figure(5)
35 clf
36 plot([-b/2 b/2 b/2 -b/2 -b/2],[-h/2 -h/2 h/2 h/2 -h/2],'Linewidth',2)
37 hold on
38 plot(1.1*[-b b]/2,[0 0],'-.k')
39 plot([0 0],1.1*[-h h]/2,'-.k')
40 axis equal
41 xlabel(['z [m]'])
42 ylabel(['y [m]'])
43 title('cross section')
44 disp('Done')
45 set(gcf , ...
46 'Color' , 'w' );
47 set(gca, ...
49 'Box' , 'off' , ...
50 'TickDir' , 'out' , ...
51 'TickLength' , [.02 .02] , ...
54 'XGrid' , 'on' , ...
55 'YGrid' , 'on' , ...
56 'XColor' , [.15 .15 .15] , ...
57 'YColor' , [.15 .15 .15] , ...

13.3 S-N Fatigue Analysis
1 %Fatigue Analysis using Rainflow counting method
2 clc
3 clear
4 sigmax = importdata('sigmaxSimple3.txt');
5 sigmay = importdata('sigmaySimple3.txt');
6 sigmaxy = importdata('sigmaxySimple3.txt');
7
8
9 %*********Input values************************************************
10 Sy = 71e7; %yield strength
11 Ny = 4e4; %yield cycles
12 Sut = 86.1e7; %ultimate strength
13 Se = modified_endurance_limit(Sut); % endurance stress limit
14 Ne = 2e7; % endurance cycles
15
16 % Sy ---^--------
17 % |
18 % |
19 % |
20 % Se ---+---+-----
21 % | |
22 % +---------> N
23 % Ny Ne
24 %*********************************************************************
25
26 %**** Inputing Stresses (x, y, xy)************************************
27 sx = sigmax; %[node,cycle]
28 sy = sigmay; %[node,cycle]
29 sxy = sigmaxy; %[node,cycle]
30
31 allstress = zeros(3,size(sx,2)-2,size(sx,1));
32 % store the stress components in a 3-dimensional array [component, ...
cycle,node]
33
34 for f=1:size(sx,1) %loop through nodes
35 allstress(1,:,f)=sx(f,3:end);
36 end
37 for f2=1:size(sy,1) %loop through nodes
38 allstress(2,:,f2)=sy(f2,3:end);
39 end
40 for f3=1:size(sxy,1) %loop through nodes
41 allstress(3,:,f3)=sxy(f3,3:end);
42 end
43 %*****************************************************************
44
45 %*****************Principle Stresses (sigma 1 and 2)**********************
46 %Transform the stress state at each state into principle stresses
47 %Because the stress state is two-dimensional, we can use Mohr's circle ...
equations

48
49 principlestress = zeros(2,size(allstress,2),size(allstress,3));
50 % store the stress components in a 3-dimensional array [principle ...
component, cycle, node]
51
52 for u1=1:size(principlestress,3) %loop through nodes
53 for u2=1:size(principlestress,2) %loop through cycle
54 sigma_x = allstress(1,u2,u1);
55 sigma_y = allstress(2,u2,u1);
56 sigma_xy = allstress(3,u2,u1);
57
58 sigma_1 = 0.5*(sigma_x + sigma_y) + sqrt( (0.5*(sigma_x - ...
sigma_y))^2 + sigma_xy^2 );
59 sigma_2 = 0.5*(sigma_x + sigma_y) - sqrt( (0.5*(sigma_x - ...
sigma_y))^2 + sigma_xy^2 );
60
61 principlestress(1,u2,u1) = sigma_1;
62 principlestress(2,u2,u1) = sigma_2;
63 end
64 end
65 %*****************************************************************
66
67 %*****************Damage Analysis*****************************************
68 oneNodeStresses = zeros(size(principlestress,2), 1); % [cycle]
69 Dc_damage = zeros(size(principlestress,3), 1); % [node]
70
71 for ii=1:1:size(principlestress,3) %loop through nodes
72
73 Both_Dc_damage = 0;
74 for kk=1:1:size(principlestress,1) %loop through components
75
76 for gg=1:1:size(principlestress,2) %loop through cycles
77 oneNodeStresses(gg,1) = principlestress(kk,gg,ii);
78 end
79 [maxstress, minstress, meanstress, alternatingstress] = ...
rainflow_counter(oneNodeStresses);
80
81 Dc = 0;
82 for pp=1:1:size(meanstress,1) %loop through peaks
83
84 sigma_m = meanstress(pp,1);
85 sigma_a = alternatingstress(pp,1);
86
87 %**** Soderberg criteria ******
88 Si = sigma_a/(1-(sigma_m/Sy));
89 %*******************************
90
91 %**** From S-N curve ***********
92 m =(Sy-Si)/(Sy-Se);
93 Ni = Ny*(Ne/Ny)^m;
94
95 if (Si Se)
96 Ni = inf;

97 end
98 if (Si Sy)
99 Ni = Ny;
100 end
101
102 %*******************************
103
104 %**** Palgren-Miner ***********
105 n_i = 0.5;
106 Di = n_i/Ni;
107 Dc = Dc + Di;
108
109 %*******************************
110 end %end of loop through peaks
111
112 Both_Dc_damage = Both_Dc_damage + Dc;
113
114 end %end of loop through components
115
116 Dc_damage(ii) = Both_Dc_damage;
117
118 end %end of loop through nodes
119
120 Nt = 1/max(max(Dc_damage))
121
122 %*************************************************************************
123
124 %*****************Plotting Damage*****************************************
125 figure(1)
126 scatter3(sx(:,1),sx(:,2),zeros(size(sx(:,1))),20,Dc_damage(:,1),'filled')
127 axis equal
128 view(0,90)
129 grid off
130 colorbar
131 title('Single Cycle Damage Parameter: Dc')
132 xlabel('X Coordinate [m]')
133 ylabel('Y Coordinate [m]')
134 %*************************************************************************
1 function [Se_mod] = modified_endurance_limit( Sut )
2
3 %****** Calculate modified endurance limit************
4
5 %********** Average Endurance Limit **********************
6 if (Sut 1400e6) %1400 MPa
7 Se = 0.5*Sut;
8 end
9 if (Sut > 1400e6) %1400 MPa
10 Se = 700e6; %700 MPa
11 end
12

13 %****************** Surface Factor, ka ************************
14
15 % Surface Finish | a | b |
16 % --------------------------------------------------
17 % Ground | 1.58 | -0.085 |
18 % Machined or cold-drawn | 4.51 | -0.265 |
19 % Hot-rolled | 57.7 | -0.718 |
20 % As-forged | 272.0 | -0.995 |
21
22 %Assuming machined surface finish, but you can modify the code for
23 %another choose
24
25 a = 4.51;
26 b = -0.265;
27
28 ka = a*(Sut/1e6)^b; %Sut is in Pa and ka = a*Sut^b uses Sut in MPa
29 %so 1e6 modification used.
30
31 %****************** Size Factor, kb ************************
32
33 %Axial loading causes the most fatigue damage.
34 %Therefore, the size factor will be based on axial loading.
35 %Axial loading has no size effect, so kb = 1;
36
37 kb = 1;
38
39 %****************** Loading Factor, kc ************************
40
41 % { 1 bending
42 % kc = { 0.85 axial
43 % { 0.59 pure torsion
44
45 %Axial loading is assumed to dominate.
46
47 kc = 0.85;
48
49 %****************** Temperature Factor, kd ************************
50
51 %This value can be modified for your engine temperature
52 Tf = 90; % degrees F where 70 Tf 1000
53
54 kd = ...
0.975+((0.432e-3)*Tf)-((0.115e-5)*Tf^2)+((0.104e-8)*Tf^3)-((0.595e-12)*Tf^4);
55
56 %****************** Reliability Factor, ke ************************
57
58 %This is statistical data for endurance strength
59
60 %Column 1: Reliability (%)
61 %Column 2: Reliability Factor
62 reliability_table = [50 1.0;
63 90 0.897;
64 95 0.868;

65 99 0.814;
66 99.9 0.753;
67 99.99 0.702;
68 99.999 0.659;
69 99.9999 0.620 ];
70
71
72 ke = reliability_table(3,2); %Using 95% reliability.
73 %You can choose your reliability percent
74
75 %********** Modified Endurance Limit **********************
76
77 Se_mod = ka*kb*kc*kd*ke*Se;
78
79 end
1 function [maxstress, minstress, meanstress, alternatingstress] = ...
rainflow_counter( stress )
2 % stress, maxstress, and minstress are column vectors.
3 %*******************************************************************
4 %Finding all of the min and max values (peaks) for rainflow counting
5 slope = 0;
6 old_slope = 0;
7 actual_start_stress = 0;
8 i = 1;
9 count = 1;
10 actual_stress = [0];
11 temp = [0];
12 first_peak = 0; %if = 1 first peak is a max
13 %if = -1 first peak is a min
14
15 %****For the first value to establish the first old_slope*****
16 p = 0;
17 while p < size(stress, 1)
18 start1_stress = stress(size(stress, 1)-p,1);
19 next1_stress = stress(1,1);
20 if (start1_stress = next1_stress)
21
22 if (start1_stress > next1_stress)
23 old_slope = -1;
24 p = size(stress, 1) + 1;
25 end
26 if (start1_stress < next1_stress)
27 old_slope = 1;
28 p = size(stress, 1) + 1;
29 end
30 end
31 p = p + 1;
32 %*******************************************************
33 end
34

35 while i < size(stress, 1) %size: 1 = row, 2 = column
36 start_stress = stress(i,1);
37 next_stress = stress(i+1,1);
38 if (start_stress = next_stress)
39 if (start_stress > next_stress)
40 slope = -1;
41 end
42 if (start_stress < next_stress)
43 slope = 1;
44 end
45 if (old_slope = 0)
46 if (old_slope = slope)
47 %*******Found a peak or valley*************
48 for x=1:1:size(actual_stress, 1)
49 temp(x,1)= actual_stress(x,1);
50 end
51 actual_stress = zeros(count, 1);
52 for y=1:1:size(temp, 1)
53 actual_stress(y,1)= temp(y,1);
54 end
55 actual_stress(count,1) = start_stress;
56 count = count + 1;
57 temp = zeros(count, 1);
58
59 if (first_peak == 0)
60 if (slope == -1)%peak/maximum
61 first_peak = 1;
62 end
63 if (slope == 1) %valley/minimum
64 first_peak = -1;
65 end
66 end
67 end
68 %****************************************
69 end
70 end
71 old_slope = slope;
72 i = i+1;
73 end
74 %*******************************************************************
75 %Adds the first peak value to 'actual_stress' as the last peak value
76 for t=1:1:size(actual_stress, 1)
77 temp(t,1)= actual_stress(t,1);
78 end
79 actual_stress = zeros(count, 1);
80 for g=1:1:size(temp, 1)
81 actual_stress(g,1)= temp(g,1);
82 end
83 actual_stress(count,1) = actual_stress(1,1);
84 %*******************************************************************
85
86 %**************************Rule 1, 2, and 3***************************
87 minstress = zeros((size(actual_stress, 1)-1),1);

88 maxstress = zeros((size(actual_stress, 1)-1),1);
89 meanstress = zeros((size(actual_stress, 1)-1),1);
90 alternatingstress = zeros((size(actual_stress, 1)-1),1);
91
92 ender = -1;
93 stressmarkers = zeros(size(actual_stress, 1),5);
94 %1: beginning value; values of -1 mean the marker is empty
95 %2: stop value
96 %3: stress value for top rainflow marker
97 %4: stress value for bottom rainflow marker
98 %5: 1 = top marker or 2 = bottom marker
99 for a=1:1:size(stressmarkers, 1)
100 stressmarkers(a,1) = -1;
101 end % setting all stress markers to empty
102 stressmarkercount = 1;
103 stoppedbymarker = 0;
104 for k=1:1:(size(actual_stress, 1)-1) %loop through all peaks and valleys
105 maxvalue = actual_stress(k,1);
106 nextmarker = 0;
107 %***loop through to find next highest peak or lowest valley*****
108 j = k;
109 while (j size(actual_stress, 1) )
111 if (maxvalue < actual_stress(j,1))
112 ender = j;
113 j = size(actual_stress, 1)+1;
114 end
115 end
116 if (first_peak == -1)
117 if (maxvalue > actual_stress(j,1))
118 ender = j;
119 j = size(actual_stress, 1)+1;
120 end
121 end
122 j = j + 2;
123 end
124 if (ender == -1)
125 ender = size(actual_stress, 1);
126 end
127 %*************************************************************
128
129 %***loop through to find next highest peak or lowest valley*****
130 %For the opposition (peak or valley)
131 maxopposite = actual_stress(k+1,1);
132 L = k+1;
133 while (L ender)
135 %*********Looping through markers**************
136 b = 1;
137 while (b size(stressmarkers, 1))
138 if (stressmarkers(b,1) = -1)
139 if (stressmarkers(b, 5) == 2)
140 endmarker = stressmarkers(b,2);

141 if (endmarker == L)
142 %******Rainflow from above has stopped it
144 maxopposite = stressmarkers(b,4); %bottom ...
value.
145 stressmarkers(b, 1) = -1;
146 b = size(stressmarkers, 1)+1;
147 %***********************************
148 end
149 end
150 end
151 b = b + 1;
152 end
153 %*********************************************
154 if (maxopposite > actual_stress(L,1))
155 stop = L;
156 if (oncethroughL == 1)
157 %*********Leaving a stop marker**************
158 stressmarkers(stressmarkercount, 1) = k+1;
159 stressmarkers(stressmarkercount, 2) = stop;
160 stressmarkers(stressmarkercount, 4) = ...
actual_stress(k+1+nextmarker,1);
161 stressmarkers(stressmarkercount, 5) = 2;
162 stressmarkercount = stressmarkercount +1;
163 nextmarker = L - (k+1);
164 %*********************************************
165 end
166 if (stoppedbymarker == 0)
167 maxopposite = actual_stress(L,1);
168 end
169 %*********************************************
170 end
171 end
172 if (first_peak == -1)
173 %*********Looping through markers**************
174 b = 1;
175 while (b size(stressmarkers, 1))
176 if (stressmarkers(b,1) = -1)
177 if (stressmarkers(b, 5) == 1)
178 endmarker = stressmarkers(b,2);
179 if (endmarker == L)
180 %******Rainflow from above has stopped it
182 maxopposite = stressmarkers(b,3); %top value.
183 stressmarkers(b, 1) = -1;
184 b = size(stressmarkers, 1)+1;
185 %***********************************
186 end
187 end
188 end
189 b = b + 1;
190 end
191 if (maxopposite < actual_stress(L,1))

192 %*********************************************
193 stop = L;
194 if (oncethroughL == 1)
195 %*********Leaving a stop marker**************
196 stressmarkers(stressmarkercount, 1) = k+1;
197 stressmarkers(stressmarkercount, 2) = stop;
198 stressmarkers(stressmarkercount, 3) = ...
actual_stress(k+1+nextmarker,1);
199 stressmarkers(stressmarkercount, 5) = 1;
200 stressmarkercount = stressmarkercount +1;
201 nextmarker = L - (k+1);
202 %*********************************************
203 end
205 maxopposite = actual_stress(L,1);
206 end
207 %*********************************************
208 end
209 end
210 oncethroughL = 1;
211 L = L+2;
213 L = ender+1;
214 end
215 end
216 oncethroughL = 0;
218
219 %*************************************************************
220 %assigns minimum and max stress values
221 if (maxopposite < actual_stress(k,1))
222 minstress(k,1) = maxopposite;
223 maxstress(k,1) = actual_stress(k,1);
224 end
225 if (maxopposite > actual_stress(k,1))
226 maxstress(k,1) = maxopposite;
227 minstress(k,1) = actual_stress(k,1);
228 end
229
230 %*************************************************************
232 first_peak = -1;
233 elseif (first_peak == -1)
234 first_peak = 1;
235 end
236 ender = -1;
237 end
238
239 for b1=1:1:(size(maxstress,1))
240 meanstress(b1,1) = 0.5*(maxstress(b1,1) + minstress(b1,1));
241 alternatingstress(b1,1) = 0.5*(maxstress(b1,1) - minstress(b1,1));
242 end
243

244 end %end of function
1 function N=S2N(Sy,Se,Ne,S)
2 % S ^
3 % |
4 % |
5 % |
6 % Se +---+-----
7 % | |
8 % +---------> N
9 % Ne
10
11
12
13 N=Ne*(S-Sy)/(Se-Sy); % equation of the slope
14 N(find(S<Se))=Inf; % Overwrite values for stresses lower than the ...
endurance limit S<Se
15 N(find(S>Sy))=0; % Overwrite values for stresses larger than the ...
yield strength
1 function S=soderberg(Sy,Sa,Sm)
2 S=Sa./(1-Sm/Sy);

13.4 Fatigue Fracture Analysis
1 % MAE 296A Project, Fall 2015
2 % Fatigue Fracture
3 % Erik Kramer
4 clc
5 clear
6
7 %% Import Data
8 sigmax = importdata('sigmaxSimple.txt');
9
10 %% Declare Initial Values
11 a=5e-5; %inital crack length in meters (decided from minimum inspection size)
12 asave(1)=a; %crack propogation vector
13 rodWidth=0.0245; %From design geometry
14 N=0; %start with no intital cycles
15 m=2.36; %Paris Constant for material type
16 C=4.5*10^(-12-6*m); %Paris Constant for material type
17 KIc=99e6; %Fracture property for material type
18 cycle=true; %inital set of fast frac conditions
19 Nmax=1.2e9; %value for infinite lifetime Back out N for 300,000 miles ...
(200k with safety factor of 1.5) 2 engine cycles per loading cycle N
20 startX=0; %minimum x dim where cracks are allowed to start
21 lastX=0.06; %max x dim where cracks are allowed to start
22 topEdge=0.0127; %Nominal Top surface y dim
23 topEdgeFlux=0.0001; %Top mesh nodes y dim deviation must be postive
24 botEdge=-0.0127; %Nominal bot surface max y dim
25 botEdgeFlux=0.0001; %Bot mesh nodes y dim deviation must be postive
26 meshResolution=0.003; %Given x variation between assumed close by left ...
right nodes
27
28 %% Read 2D COMSOL Results for Edge Smax and Smin and Find highest Edge Stress
29 %sigmax needs to be preloaded
30
31 %Finding Stresses at intersted edges
32 TopEdgeStress=sigmax(sigmax(:,2)<(topEdge+topEdgeFlux) & ...
sigmax(:,2)>(topEdge-topEdgeFlux) & ...
33 sigmax(:,1)<(lastX) & sigmax(:,1)>(startX),:);
34 BotEdgeStress=sigmax(sigmax(:,2)<(botEdge+botEdgeFlux) & ...
sigmax(:,2)>(botEdge-botEdgeFlux) & ...
35 sigmax(:,1)<(lastX) & sigmax(:,1)>(startX),:);
36 EdgeStress=[TopEdgeStress; BotEdgeStress];
37
38 %Finding Max Stress at the interested edges
39 stressOnlyEdge=EdgeStress;
40 stressOnlyEdge(:,[1 2])=[];
41 [EdgeMax, Imax]=max(stressOnlyEdge(:));
42 [EdgeMin, Imin]=min(stressOnlyEdge(:));
43 [rowMax, colMax]=ind2sub(size(stressOnlyEdge),Imax);
44 [rowMin, colMin]=ind2sub(size(stressOnlyEdge),Imin);
45 colMax=colMax+2;

46 colMin=colMin+2;
47
48
49 %Inital Crack location where max edge stress is
50 crackStartX=EdgeStress(rowMax,1);
51 crackStartY=EdgeStress(rowMax,2);
52
53 %Finding inital remote stresses
54 Smax=EdgeStress(rowMax,colMax);
55 if min(stressOnlyEdge(rowMax,:))<0
56 Smin=0; %Min Remote stress tensile from COMSOL DATA assumed zero ...
becasue tension-compression
57 else
58 Smin=min(stressOnlyEdge(rowMax,:));
59 end
60
61
62 %% Stresses Along the Crack Propagation Direction
63 CrackStress=sigmax(sigmax(:,1)>(crackStartX-meshResolution) & ...
sigmax(:,1)<(crackStartX+meshResolution),:);
64
65
66 %% Check for Immideate Fracture
67 KImax=BetaFinder(a/rodWidth)*sqrt(pi*a)*Smax;
68 if KImax KIc
69 cycle=false;
70 FastFrac=true;
71 end
72
73
74 %% Run Cycles
75 while cycle==true
76
77 %Progressing Cycles
78 N=N+1;
79
80 %Solves for beta using current a
81 beta=BetaFinder(a/rodWidth);
82
83 %Solves for new a via Paris law
84 a=ParisCrackProp(a,C,m,N,beta,Smax-Smin);
85 asave(N+1)=a;
86
87 %Smax FINISH TO AVERAGE STRESS
88 if crackStartY>0 %top edge
89 currentCrackStress=CrackStress(CrackStress(:,2) crackStartY-a,:);
90 else %bottom edge
91 currentCrackStress=CrackStress(CrackStress(:,2) crackStartY+a,:);
92 end
93 currentCrackStressOnly=currentCrackStress;
94 currentCrackStressOnly(:,[1 2])=[];
95
96 %average max stress over the length of the crack

97 maxCurrent=max(currentCrackStressOnly,[],2);
98 Smax=sum(maxCurrent)/length(maxCurrent);
99 minCurrent=min(currentCrackStressOnly,[],2);
100 minCurrent(minCurrent<0)=0;
101 Smin=sum(minCurrent)/length(minCurrent);
102
103 KImax=BetaFinder(a/rodWidth)*sqrt(pi*a)*Smax; %check for fast fracture
104 if KImax KIc
105 cycle=false;
106 FastFrac=true;
107 end
108
109 if cycle==true && N==Nmax
110 cycle=false;
111 FastFrac=false;
112 end
113 end
114
115 %% Give Results
116
117 if FastFrac==false
118 N=Inf;
119 end
120
121 sprintf('The cycles to failure are %d.',N)
122
123 figure(1)
124 plot(0:N,asave/0.0254,'b');
125 hold on
126 plot(N,asave(N+1)/0.0254,'rx');
127 legend('Crack Growth','Fast Fracture','Location', 'Best');
128 grid on
129 xlabel('Cycles');
130 ylabel('Crack Length (Inches)');
131 print(gcf, '-depsc', 'transverse')
132 set(gcf , ...
133 'Color' , 'w' );
134 set(gca, ...
136 'Box' , 'off' , ...
137 'TickDir' , 'out' , ...
138 'TickLength' , [.02 .02] , ...
141 'XGrid' , 'on' , ...
142 'YGrid' , 'on' , ...
143 'XColor' , [.15 .15 .15] , ...
144 'YColor' , [.15 .15 .15] , ...
146 hold off;

1 function [anew] = ParisCrackProp(a,C,m,N,beta,dS)
2 %solves for new crack length via Paris' Law
3 Q=C*(beta*sqrt(pi)*dS)^m;
4 z=(1-(m/2));
5 y=m-2;
6 K=((Q*N-((2*a^z)/(y)))*(y/-2));
7 anew=K^(1/z);
8
9 end
13.5 3D FEA Analysis
1 clc
2 clear
3 pressure4COMSOL = importdata('pressure4COMSOL.txt');
4
5 %Inputs [Put in crank angle]
6 %phi = -0.5585; % [rad] crank angle (Angle that causes bending) CRANK ...
ANGLE ONE IN ABAQUS
7 phi = 0.3142; % [rad] crank angle (Angle of MAX CONTACT STRESS from ...
COMSOL)CRANK ANGLE TWO IN ABAQUS
8
9 %Additional Inputs [Connecting rod parameters]
10 R = 0.0405; %[m] half stroke length
11 L = 0.144; %[m] connecting rod length
12 mp = 0.4; %[kg] piston mass
13 omega = 6000*2*pi/60; %[1/s] engine rotational velocity in rad/sec
14 beta = 0.2874; %[1] relative position of center of mass
15 A = 0.0069; %[m^2] piston area
16 rho = 10000 %[kg/m^3] density of connecting rod
17 Jg = 3.54150667e-3; %[kg m^2] moment of inertia Pz solidworks
18 vol = 133128.69e-9; %[m^3] volume of connecting rod
19 D_pin = 0.022; %[m] Diameter of piston pin
20 t_pin = 0.0354; %[m] Thickness of piston pin
21
22 %Equations ******* DO NOT CHANGE ****************************
23 volume_pin = pi()*t_pin*(D_pin/2)^2; %[m^3] Volume of piston pin
24 pressure = interp1(pressure4COMSOL(:,1),pressure4COMSOL(:,2),phi);% [Pa] ...
pressure
25 lam = R/L; % ratio R/L
26 mg =rho*vol; % mass of connecting rod
27 dgamma = -omega*lam*cos(phi)/(1-lam^2*sin(phi)^2)^0.5 % angular velocity ...
of the rod
28 ddgamma = omega^2*lam*(1-lam^2)*sin(phi)/(1-lam^2*sin(phi)^2)^1.5 % ...
angular acceleration of the rod
29
30 Rx = -R*sin(phi); %x position of Crankshaft pin
31 Ry = R*cos(phi); % y position of Crankshaft pin
32 Px =0; % Piston pin position x

appendix c - final design documentation 46
33 Py =L*(lam*cos(phi)+(1-lam^2*sin(phi)^2)^0.5); % Piston pin position y
34
35 ddRx = -omega^2*Rx; % x acceleration of Crankshaft pin
36 ddRy = -omega^2*Ry; % y acceleration of Crankshaft pin
37 ddPx = 0; % x acceleration of Piston pin
38 ddPy = L*lam*((dgamma-omega)*omega*cos(phi)+ddgamma*sin(phi)); % y ...
acceleration of Piston pin
39 ddGx = (1-beta)*ddRx+beta*ddPx; % x acceleration of Centroid
40 ddGy =(1-beta)*ddRy+beta*ddPy; % y acceleration of Centroid
41 ax =(Px-Rx)/((Px-Rx)^2+(Py-Ry)^2)^0.5; %x component of axial unit vector
42 ay =(Py-Ry)/((Px-Rx)^2+(Py-Ry)^2)^0.5; %y component of axial unit vector
43 tx =-ay; %x component of transverse unit vector
44 ty =ax; %y component of transverse unit vector
45
46 FPRy= -mp*ddPy-pressure*A; % y Piston Force on pin
47 FSRy =mg*ddGy-FPRy; % y Crankshaft Force on pin
48 FPRx= beta*mg*ddGx-(Jg*ddgamma-((beta-1)*FPRy+beta*FSRy)*Rx)/(Py-Ry); % x ...
Piston Force on pin
49 FPRa= FPRx*ax+FPRy*ay; % Axial Piston Force on pin
50 FPRaPerVol =FPRa/volume_pin
51 ddGa =ddGx*ax+ddGy*ay % axial acceleration of Centroid
52 ddGt= ddGx*tx+ddGy*ty % transverse acceleration of Centroid
53
54 %These are body load equations that you will put into Abaqus
55 %ba =-rho*(ddGa-ddgamma*Z-dgamma*dgamma*(X-0.0338)); % Body Load in x
56 %bt =-rho*(ddGt+ddgamma*(X-0.0338)-dgamma*dgamma*Z); % Body Load in y
14 appendix c - final design documentation

Figure 23: A 3D render of our ﬁnal design

1.00
1.50
7.70
5.67
1.25
1.99
3.47
0.79
R1.15
AA
2.11
0.25
0.87
R0.12
R0.12
R0.12
2.81
0.25
SECTION A-A
C
D
E
B
F
A
23 14
C
F
E
A
B
D
2 14 3
DRAWN
CHK'D
APPV'D
MFG
Q.A
UNLESS OTHERWISE SPECIFIED:
DIMENSIONS ARE IN MILLIMETERS
SURFACE FINISH:
TOLERANCES:
LINEAR:
ANGULAR:
FINISH: DEBURR AND
BREAK SHARP
EDGES
NAME SIGNATURE DATE
MATERIAL:
DO NOT SCALE DRAWING REVISION
TITLE:
DWG NO.
SCALE:1:5 SHEET 1 OF 1
A4
WEIGHT:
FINAL_ROD
Figure 24: An engineering drawing of our ﬁnal design.

References 49
references
[1] Shigley, Joseph Edward, and John Joseph Uicker. Theory of Machines and Mecha-
nisms. International ed. New York: McGraw-Hill, 1980: 15-16.
[2] Carley, Larry. "Connecting Rods: So Many Choices - Engine Builder Maga-
zine." Engine Builder Magazine. Babcox Media, Inc., 1 Sept. 2008. Web. 10
Oct. 2015. <http://www.enginebuildermag.com /2008/09/connecting-rods-so-many-
choices/>.
[3] "The 458 Italia’s V8 Voted Best Performance Engine and Best Engine." Ferrari GT. N.p.,
18 May 2015. Web. 19 Oct. 2015.
[4] http://www.exprobase.com/Default.aspx?page=11#
[5] Military Handbook - MIL-HDBK-5H Metallic Materials and Elements for Aerospace
Vehicle Structures (Knovel Interactive Edition). Place of Publication Not Identiﬁed:
U.S. Dept. of Defense. Print.
[6] Larsen, James M. Small-crack Test Methods. Philadelphia, PA: ASTM, 1992. Print.
[7] Putatunda, S.k., and J.m. Rigsbee. "Effect of Specimen Size on Fatigue Crack Growth
Rate in AISI 4340 Steel." Engineering Fracture Mechanics: 335-45. Print.
[8] http://infinitemonkeycorps.net/projects/cityspeed/city.html?city=
losangelesca
[9] http://www.dragtimes.com/2012-Ferrari-458-gear-ratios-top-speed-31.html?
tire_width=235&tire_aspect=35&wheel_diam=20&tire_diam=28.1299212598&axle_
ratio=5.140&redline=9000&gear1=3.080&gear2=2.190&gear3=1.630&gear4=1.290&
gear5=1.030&gear6=0.840&gear7=0.690&gear8=0.000&calc_button=Calculate
[10] http://www.driverside.com/specs/ferrari-458_italia-2012-30960-54122-0
[11] http://www.cars.com

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