5/10/20132013

Suspension DesignMini Baja – Spring Final Report

Southeastern Louisiana University Mechanical Engineering Technology Senior Design 2, ET-494 By: Brock Kerry Instructor: Dr. Cris Koutsougeras Advisor: Dr. Ho-Hoon Lee

Table of Contents :

1. Abstract……………………………….………………………………Page 2

2. Design Project…………….………………………………………….Pages 2

3. Newton’s Second Law…………..……………………………………Page 2

4. Simple Harmonic Oscillations………………………………………..Page 3

5. Natural Frequency….……….……………….……………………….Page 3

6. Damping…………….…………………………..……………………Page 4

7. Vibrations and Viscous Damping…….……………………………Page 4-6

8. Geometry…………………………………………………….……Page 6-13

9. Vehicle Weight.…..…………………………………………………Page 14

10. Forming the Equation.……………………………………………Page 14-16

References………………………………………………………………..Page 17

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1. Abstract

In this paper I will briefly introduce the project I had selected for my senior design course. I will then give a description of all of the topics I learned about this semester and explain how they come together in the analysis of suspensions. I hope to portray a general understanding of how to solve for the variables needed for the design process.

2. Design Project

I selected to design the suspension system for the mini Baja car for my senior project. The suspension of a vehicle is a system of springs, shock absorbers and linkages that connects a vehicle to the wheels. The suspension’s purpose is to contribute to the vehicle’s road contact, handling and braking. This goal contains many variables that must come together for a specific result. The idea is to design a suspension that, given specific conditions, allows the wheels of the mini baja car to maintain as much contact to the ground as possible throughout rough terrain and sharp turns while also keeping the driver safe and comfortable.

In order to design the suspension I had to identify, learn, and understand the variables that determine what type of suspension is needed for the vehicle.

3. Newton’s Second Law

To start understanding suspensions I had to become familiar with the 2nd order differential equation

m x+c x+kx=0

However, before we can begin working with this equation it is important to understand Newton’s second law of motion. His second law states that if the resultant force on a particle is not zero, the particle will undergo an acceleration having a magnitude directly proportional to the magnitude of the resultant force and having the same line of action and sense as the resultant force [1].

F=m a

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4. Simple Harmonic Oscillations

Simple harmonic motion occurs when the force on an object is proportional and in the opposite direction to the displacement of the object. Mathematically, this can be written:

F=−kx

F is force, x is displacement, and k is a positive constant. This is exactly the same as Hooke's Law, which states that the force F on an object at the end of a spring equals -kx, where k is the spring constant.

5. Natural Frequency

The time for one complete cycle of oscillation is called the period, τ , and is expressed mathematically as

τ=2πω

The period of oscillation is given in seconds and the circular frequency, ω, is usually specified in radians per second. The natural frequency ƒ, is the number of cycles of oscillation per unit time and is measured in hertz

ƒ=1τ= ω

2πHz

Note: one hertz is equal to one cycle per second.

One of the constraints in the design of the suspension is the natural frequency produced by the vehicle. The desired natural frequency of the system is 1 hertz. Why is this? Our entire lives, we as humans have been conditioned to be comfortable with a frequency of 1 hertz. When people walk our bodies naturally move at this frequency, anything other than 1Hz frequency could cause the driver discomfort. The equation that I will use for frequency is

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ω=√ km

The mass of the vehicle will be one of the first values that I am given so the application of this equation will be the starting point of my calculations for the suspension.

6. Damping

The process of energy dissipation is referred to in the study of vibrations as damping. Once we have the spring specifications identified, we want to be able to reduce the oscillations so the vehicle doesn’t continue to bounce up and down. Therefore, it is important to find a proper damper for our system. The type of damping we will be working with is viscous damping. A viscous damper is characterized by the resistive force exerted on a body moving in a viscous fluid. The common automobile shock absorber is a viscous damper [1]. Mathematically the damping force is

F=c x

Where c is known as the damping constant, x is the velocity of the mass acting in the direction opposite to the velocity of the mass.

7. Vibrations with Viscous Damping

When a viscous damper is connected to a simple spring-mass system and the system is at its equilibrium position, the mathematical equation is

m x=−kx−c x

Or

m x+c x+kx=0

Note that at equilibrium the damper exerts no force. This can be solved as a homogeneous differential equation. The equation should then, for simplicity, be written as

x+ cm

x+ km

x=0

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We then change the equation to its auxiliary equation

s2+ cm

s+ km

=0

The next step is to use the quadratic formula to solve for the roots s1∧s2

s1 , s2=−b ±√b2−4 ac

2a

There are three different ways to solve this equation and it depends on whether the value of √b2−4 ac is zero, positive, or negative. If the value is zero, the roots will be equal to each other and our system will be critically damped. If the value is negative, the system will be under damped and if the value is positive, the system will be over damped. Under damping will cause the vehicle to oscillate too much which will decrease the handling and cause motion sickness to the driver. Over damping will increase the time in which the vehicle takes to return back to its equilibrium state and could result in a rough ride. If the system is critically damped it should eliminate oscillations and return to equilibrium quicker. Therefore critical damping would be the most desirable target.

When the value is zero, the quadratic equation gives us

s=−c2 m

=−√ km

=−ω

Orc = 2mω

This value of c is known as the critical damping constant. This value is then used to find the damping ratio ζ. So

cm

=2ζ √ km

ζ = c2mω (Damping ratio)

The damping ratio should be equal to 1 for critical damping. I should be able to calculate the mass, spring constant, and damping constant for the suspension of the mini baja.

8. Geometry 5 | P a g e

The first thing I did this semester was collect material from Alfred Showers, who designed the steering system, and Robert Golmon, who designed the frame. Before I could make any calculations I had to find the dimensions of the frame and the locations where the shock and the bottom a-arm are going to be connected along with the length of the bottom a-arm.

(Figure 1: Inventor Model of frame)

Using the model from Inventor, I was able to identify some set parameters and determine what the variables were. The distance between where the shock and the bottom a-arm are connected to the frame is 12.685 inches and the length of the bottom a-arm is 19.052 inches. The shock will be connected to the bottom a-arm at distance of 9.3 inches from the frame. The variables will be the motion of the bottom a-arm and the angle between the a-arm and the frame. In order to find the relationship between the motion of the a-arm and the length of the shock I had to form three equations. Figures 2 and 3 show how I chose to identify the geometry.

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(Figure 2: Geometry of shock and bottom a-arm)

(Figure 3: Geometry of shock length and static ride height)

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The first step was to find the angle Beta,

h – x = l sinβ

β = sin−1[(h−x )

l]

(where, x, will be the displacement of the frame)

Next, we identify the angle Alpha,

α = 90o – θ + β

Now that we have Alpha, we can use the law of cosines to find the length of the spring, y. To do this we take,

c2=a2 + b2 -2ab cosθ

And form this,

y = √ l22+l3

2−2¿¿

The displacement of the frame and the change in length of the spring will not form a linear relationship so we have to take several measurements from several positions and form a general (linear) relationship between them. To make this easier I wrote a program that would help calculate the angles Alpha and Beta.

/*This program will calculate the relationship between the displacement of the* *frame and the change in length of the spring using the formula * * * Beta = asin*((h-x)/l)*(180/3.14159) */ #include <stdio.h> #include <math.h> int main ()

{ double h, x, B, A, y; //variables double l, l2, l3, theta;//parameters double asin ( double real ); double cos (double real ); double sqrt (double real ); //Calculation of all angles

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printf( "\n"); printf( " Please enter the vertical distance from the a-arm connection\n"); printf( " at the wheel, inches, to the bottom of the frame(static ride):"); scanf( "%lf", &h); printf( " \n Please enter the displacement of the frame, in inches: "); scanf( "%lf", &x ); printf( "\n Please enter the length, in inches, of the bottom a-arm: "); scanf( "%lf", &l); printf( "\n Please enter the angle of theta: "); scanf( "%lf", &theta); while ( h > x) { B = asin((h - x)/l)*(180/3.14159); A = B +(1.5708*(180/3.14159)-theta); printf( " \n\n You entered... \n"); printf( " A-arm length: %lf inches\n", l); printf( " Frame displacement: %lf inches\n", x); printf( " Static ride height: %lf inches\n", h); printf( " \n\nThe angle Beta = %lf Degrees\n", B); printf( " \nTherefore, \n"); printf("\n"); printf( " If Theta is equal to: %lf degrees \n", theta); printf( " Beta is equal to: %lf degrees \n", B); printf( " Alpha is equal to: %lf degrees \n\n", A); printf( " ***Rest of Program In Progress*** \n"); getch(); printf( "\n\n Now we will have to find the length of the shock...\n"); printf( " Please enter the length between where the bottom a-arm\n"); printf( " and the shock are connected to the frame(length 3):"); scanf( "%lf", &l3); printf("\n Now enter the distance on the a-arm (from the frame) where\n"); printf(" the shock is connected(length 2):"); scanf( " %lf", &l2); while ( B < A ) { y = sqrt(((l2)*(l2)) + ((l3)*(l3)) - (2)*(l2)*(l3)*cos(A*(3.14159/180))); printf( " the corresponding length of the shock is = %lf ", y); getch(); return 0; }}}

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I also wrote a second program that displays several different displacements of “x” and the corresponding length of “y”. Using these values I could enter them in Excel and find a linear relationship between them.

/*This program will use the parameters found in program 1 to calculate the relationship between the displacement of the car and the corresponding length of the shock. It will then display a range of values with "x" being the displacement of the car and "y" representing the length of the shock. */ #include <stdio.h> #include <math.h> int main () { double h, x, B, A, y, y2;//variables double l, l2, l3, theta;//parameters double asin ( double real ); double cos (double real ); double sqrt (double real ); printf( "\n"); printf(" We will now use our measurements to calculate several values \n"); printf(" of 'y' so we can find a linear relationship between the \n"); printf(" displacement of the car and the length of the shock. \n"); printf("\n"); printf(" Please enter the static ride height of the vehicle: "); scanf( " %lf", &h); printf("\n Please enter the length of the bottom a-arm: "); scanf( " %lf", &l); printf("\n Please enter the angle theta (in degrees): "); scanf( " %lf", &theta); printf("\n Please enter the value of length 2 (l2): "); scanf( " %lf", &l2); printf("\n Please enter the value of length 3 (l3): "); scanf( " %lf", &l3); printf("\n"); printf(" As the displacement of the frame increases \n"); printf(" your values for (x) and (y) will be: \n"); printf("\n");

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printf(" x-displacement(inches) y-length(inches)\n"); for ( x = 0.0; x <= h+.000000001; x += 0.10 ) { B = asin((h - x)/l); A = B +(1.5708-theta/180*3.14159); y = sqrt(((l2)*(l2)) + ((l3)*(l3)) - (2)*(l2)*(l3)*cos(A)); printf( " %.1f %.2f\n", x,y); } getch(); return ; }The results, given my measurements, were:

To make sure my calculations were accurate I created a few sketches on Inventor using my dimensions to measure the length of the shock at different positions.

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x y compression0 16.43 0

0.1 16.39 0.040.2 16.35 0.080.3 16.32 0.110.4 16.28 0.150.5 16.24 0.190.6 16.2 0.230.7 16.16 0.270.8 16.12 0.310.9 16.08 0.351 16.05 0.38

1.1 16.01 0.421.2 15.97 0.461.3 15.93 0.51.4 15.89 0.541.5 15.85 0.581.6 15.81 0.621.7 15.77 0.661.8 15.73 0.71.9 15.69 0.742 15.65 0.78

2.1 15.61 0.822.2 15.57 0.862.3 15.53 0.92.4 15.49 0.942.5 15.45 0.982.6 15.41 1.022.7 15.37 1.062.8 15.33 1.12.9 15.29 1.143 15.25 1.18

3.1 15.21 1.223.2 15.17 1.263.3 15.13 1.33.4 15.09 1.343.5 15.05 1.38

(Figure 4: Inventor sketch of geometry w/ specified measurements at static ride)

(Figure 5: Inventor sketch of geometry when parallel w/ ground)

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(Figure 6: Inventor sketch of geometry when shock is fully compressed)

My calculations matched up with the CAD measurements and using Excel and I was able to obtain a linear relationship between the displacement of the frame and the change in length of the shock. This relationship, which I found to be, Δy = .394x, will serve as a parameter when solving for the spring rate. Also, with the dimensions I calculated I was able to find a possible option for the shock.

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(Figure 7: Possible option for a shock, given our dimensions)

9. Vehicle Weight

In order to find the mass of the system we need to first find the weight of the vehicle. I estimate the weight of the car to be approximately 550 lbs. which should be slightly heavier than the actual weight. That weight will then be divided by 4 (for each wheel). I am also assuming that the weight distribution is roughly 40/60 for the front and rear suspension. Therefore, the mass acting on the system should be:

10. Forming the Equation

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This is where the fun begins. I now have specified values for the weight of the vehicle, the natural frequency, the damping ratio, and the relationship between the displacement of the car and the change in length of the spring. Therefore, I can use Newton’s second law to form the differential equation that will help me find the spring constant, k, and the damping constant. After some trial and error, I decided to use 386.4 in/s2 when calculating the mass so units for k would be lb/in.

m = 110 lbs

386.4∈¿ s2 = .285 lb−s2

¿

Δy = .394x

ζ = 1

ωn = 6.28 rad/s

Final Calculations:

F=m a

m d xdt

=−ky−c y

m d xdt

dx=(−ky−c y ) dy

m x d x=¿ −kydy−c ydy

m∫ x d x=¿ −k∫ ydy−c∫ y dy

At this point I substituted y with its x-value then integrated

12

m x2=¿ −12

k .155 x2−c∫ .155 x2dt

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ddt ( 12

m x2+12

k .155x2=¿ −c∫ .155 x2 dt )

m x x+k .155 x x=−c .155 x2

(m x+k .155 x) x=−c .155 x2

m x+c .155 x+k .155 x = 0

From this equation, k .155 can be used in the equation for natural frequency, ω=√(k/m), to find the spring constant for this system.

ωn=√ km

6.28 rad/s = √ k∗.155.285

39.4 = k (.155).285

k = (39.4 )(.285)

.155

= 72.45 lb¿

Next, c .155 can be plugged into the critical damping equation, ζ = c2mω , to find the

damping constant for this specific system.

ζ = c2mω 1 =

c∗.1552(.285)(6.28)

2 ( .285 ) (6.28 )=c (.155) c=2(.285)(6.28).155

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¿23.09 lb∗s¿

Deliverables:

1) Equation m x+c .155 x+k .155 x = 0

2) Spring Rate k = 72.45 lb¿

3) Damping Constant c=23.09 lb∗s¿

We now have all the information we need to select materials for the suspension. However, an important note is that these calculations are specific to the system that was designed but parameters may have to be changed when it comes to actually purchasing a spring. Getting custom parts built can be time consuming and also expensive. If the spring rate that was calculated is not a commercially available, then adjustments will have to be made in order to get a spring that exist. This can be done by changing the position on the a-arm where the shock is connected.

References

[1] Hutton, David. Applied Mechanical Vibrations. New York: McGraw-Hill, 1981. Print.

[2] Hibbeler, R.C. Engineering Mechanics Statics. 10th ed. Upper Saddle River: Pearson Prentice Hill, 2004. Print.

[3] Shelly, Joseph F. Engineering Mechanics, Dynamics. N.p.: McGraw-Hill Book, 1980. Print.

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