Quarter_Car_Suspension

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College of Engineering and Computer Science Department of Mechanical Engineering ME 384 Dr. C.T. Lin Quarter Car Suspension Joseph Sarhadian Due: May 02, 2016 Submitted: May 02, 2016

Transcript of Quarter_Car_Suspension

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College of Engineering and Computer Science

Department of Mechanical Engineering

ME 384

Dr. C.T. Lin

Quarter Car Suspension

Joseph Sarhadian

Due: May 02, 2016

Submitted: May 02, 2016

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Introduction:

A “Quarter-Car” Suspension as it appears in figure 1 was analyzed. The car with a mass

of m moves along a road. The system was initially at rest. Equation of motion was driven from

the free body diagram, and since the value of spring constant (k), damping constant (b) was

given, the natural frequency and damping coefficient was calculated. Lastly, the system was

further more analyzed by creating a block diagram and using Simulink to see the behavior of the

system.

Figure 1: Quarter-Car Suspension

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Procedure:

Obtain Equation of motion from Free Body Diagram:

The free body diagram was drawn as it shown in figure 2. The equation of motion was

driven from free body diagram by summing all forces.

∑ 𝐹 = 𝑚�̈�𝑚

−𝑘(𝑋𝑚 − 𝑋𝑟) − 𝑏(�̇�𝑚 − �̇�𝑟) = 𝑚�̈�𝑚

Figure 2: Free Body Diagram

Obtaining values of natural frequency and damping coefficient:

The equation of motion was further more simplified to transfer function by converting T-

domain S-domain.

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𝐾

𝑚𝑋𝑟(𝑆) +

𝑏

𝑚 𝑆𝑋𝑟 (𝑆) = 𝑆2𝑋𝑚 (𝑆) +

𝐾

𝑚𝑋𝑚 (𝑆) +

𝑏

𝑚 𝑆𝑋𝑚 (𝑆)

𝑋𝑟(𝑆) (𝐾

𝑚+

𝑏

𝑚𝑆) = 𝑋𝑚(𝑆)(𝑆2 +

𝐾

𝑚+

𝑏

𝑚𝑆)

𝑋𝑚(𝑆)

𝑋𝑟(𝑆)=

𝑏

𝑚𝑆+

𝐾

𝑚

𝑆2+𝑏

𝑚𝑆+

𝐾

𝑚

The roots of characteristic equation was found by setting characteristic equation to zero, and

using initial values that was given in statement problem which are

𝑘 = 39.475 𝑚 = 1.5 𝑏 = 7.5

𝑆2 +𝑏

𝑚𝑆 +

𝑘

𝑚= 0

𝑚𝑆2 + 𝑏𝑆 + 𝑘 = 0

𝑆1,2 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑆1,2 = −𝑏

2𝑚±

√𝑏2 − 4𝑚𝑘

2𝑚

𝑆1,2 = −7.5

2𝑚±

√7.52 − 4(1.5)(39.475)

2(1.5)

𝑆1,2 = −2.5 ± 𝑖 4.48

Characteristic

Equation

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The value of natural frequency and damping coefficient also was calculated

𝜔𝑛 = √𝑘

𝑚= √

39.478

1.5

𝜔𝑛 = 5.1 𝑟𝑎𝑑

𝑠𝑒𝑐

𝜁 =𝑏

2√𝑘𝑚=

7.5

2√(39.478)(1.5)

𝜁 = 0.487

Block Diagram and Simulink:

In order to draw block diagram the equation of motion needed to be rearrange

�̈�𝑚 =𝑏

𝑚�̇�𝑟 +

𝑘

𝑚𝑋𝑟 −

𝑏

𝑚�̇�𝑚 −

𝑘

𝑚𝑋𝑚

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Figure 3: Block Diagram

The block diagram in figure 3 was used in order to build the Simulink model as it shown in

figure 4.

Figure 4: Block Diagram

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Displacement and Velocity of the mass as the function of time:

Simulink was able to plot Velocity and displacement of the mass as the function of time

by having bus creator in its model. The Simulink model was call out in MATLAB script in order

to observe the graph. Figure 5 demonstrate the graph.

Figure 5: Velocity and Displacement

Inertia, Damping, and Spring Forces:

Equation of motion was rearranged in order to understand the behavior of the system.

The value of damping force was found by subtracting velocity input subtracted from velocity of

mass, and then multiplied it by the gain of damper. Also, the spring force value was found by

subtracting displacement input from displacement of mass. The �̈�𝑚 was multiplied by 1

𝑚 in order

to eliminate the extra mass unit in force of inertia because force has already the component of mass

and acceleration in it.

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−𝑘(𝑋𝑚 − 𝑋𝑟) − 𝑏(�̇�𝑚 − �̇�𝑟) = 𝑚�̈�𝑚

𝑘(𝑋𝑟 − 𝑋𝑚) + 𝑏(�̇�𝑟 − �̇�𝑚) = 𝑚�̈�𝑚

𝐹𝐼 = 𝑚�̈�𝑚

𝐹𝐼 =1

𝑚∗ (𝑚�̈�𝑚)

𝐹𝐼 = �̈�𝑚

𝐹𝑑 = 𝑏(�̇�𝑚 − �̇�𝑟)

𝐹𝑠 = 𝑘(𝑋𝑚 − 𝑋𝑟)

As it shown in figure 4 “To work space” component from Simulink library was added up

to the model. Then the component was connected to inertia, damping, and spring force branches

in order to plot the graph of forces vs time in MATLAB. Figure 6 represent force function of

inertia, spring, and damper.

Figure 6: Inertia, spring, and damper forces Vs Time

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The equation of motion can be rearranged to understand the relation between forces. The

equation demonstrate that summation of all forces has to be equal to zero which means the

summation of inertia, spring, and damper forces at 𝑡 = 4𝑠 is equal to zero.

−𝑘(𝑋𝑚 − 𝑋𝑟) − 𝑏(�̇�𝑚 − �̇�𝑟) = 𝑚�̈�𝑚

𝑚�̈�𝑚 + 𝑘(𝑋𝑚 − 𝑋𝑟) + 𝑏(�̇�𝑚 − �̇�𝑟) = 0

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‘Velocity and Displacement Plot’

%Calling Simulink Model

Sim_out=sim('Quarter_Car_Suspension');

%Plot Displacement and Velocity

plot(Xm,'b');

hold on

plot(Vm,'r');

legend('Velocity','Displacement')

grid

xlabel('Time')

ylabel('Velocity, Displacement')

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‘Time History Plot’

%Calling Simulink Model

Sim_out=sim('Quarter_Car_Suspension');

%Plot Inerti , Damping , and Spring Forces

plot(SF,'r');

hold on

plot(DF,'g');

hold on

plot(IF,'b');

legend('Mass','Damper','Spring')

grid

title('Force Vs Time')

xlabel('Time (S)')

ylabel('Force (N)')

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‘Block Diagram’