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School

Gwinnett Technical College *

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Course

203

Subject

Mechanical Engineering

Date

Oct 30, 2023

Type

pdf

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Uploaded by gimarkm

MATLAB/Simulink You can use the link below to learn Simulink in detail. https://www.mathworks.com/help/simulink/getting-started-with-simulink.html Example 1: For initial condition: ? = 3 ?? , ? = 1 𝑁𝑠 𝑚 , ? = 2 𝑁 𝑚 𝑎?? 𝑥 0 = 1? If we have an input; 𝐹 = ???? ???? . ( 𝐹𝑖?𝑎? 𝑉𝑎??? 𝑖? 1) , ? = 3 ?? , ? = 1 𝑁? ? , ? = 2 𝑁 ? 𝑎?? 𝑖?𝑖?𝑖𝑎? ????𝑖?𝑖??? 𝑎?? 𝑧??? . If we want to control this system:

Example-2 Designing an automotive suspension system is an interesting and challenging control problem. When the suspension system is designed, a 1/4 model (one of the four wheels) is used to simplify the problem to a 1-D multiple spring-damper system. A diagram of this system is shown below. This model is for an active suspension system where an actuator is included that is able to generate the control force U to control the motion of the bus body. Design requirements A good automotive suspension system should have satisfactory road holding ability, while still providing comfort when riding over bumps and holes in the road. When the vehicle is experiencing any road disturbance (i.e. pot holes, cracks, and uneven pavement),the vehicle body should not have large oscillations, and the oscillations should dissipate quickly. Since the distance X1-W is very difficult to measure, and the deformation of the tire (X2-W) is negligible, we will use the distance X1-X2 instead of X1-W as the output in our problem. Keep in mind that this is an estimation. The road disturbance (W) in this problem will be simulated by a step input. This step could represent the vehicle coming out of a pothole. We want to design a feedback controller so that the output (X1-X2) has an overshoot less than 5% and a settling time shorter than 5 seconds. For example, when the vehicle runs onto a 10 cm high step, the vehicle body will oscillate within a range of +/- 5 mm and return to a smooth ride within 5 seconds.

(M1) 1/4 bus body mass 2500 kg (M2) suspension mass 320 kg (K1) spring constant of suspension system 80,000 N/m (K2) spring constant of wheel and tire 500,000 N/m (b1) damping constant of suspension system 350 N.s/m (b2) damping constant of wheel and tire 15,020 N.s/m (U) control force From the picture above and Newton's law, we can obtain the dynamic equations as the following: For W;  Insert a Step block in the lower left area of your model window. Label it "W".  Edit it's Step Time to "0" and it's Final Value to "0". (We will assume a flat road surface for now). The last force is the input U acting between the two masses.  Insert a Step block (from the Sources library) in the upper left of the model window.  Edit this Step block's Step Time to "0" and leave its Final Value "1".  Label this Step block "U". m1 = 2500; m2 = 320; k1 = 80000; k2 = 500000; b1 = 350; b2 = 15020;

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