An engineering company is designing and testing a car suspension system. The system has a conventional suspension design, consisting of a shock-absorber and spring at each wheel. The shock-absorber provides a damping effect that is proportional to the vertical speed (i.e. up and down motion) of the wheel, and the spring's resistance is proportional to the vertical displacement of the wheel.
The design team analyses the suspension system in two separate parts: Part 1 - dynamics of the spring-damper system; Part 2 - stress analysis of the coil spring.
PART 1
The company would like to investigate how the actual performance of the suspension system compares to that of theory. Dynamic theory states that the equation of motion of the wheel in the vertical direction can be expressed as,
or
where m is the mass of the wheel, k is the spring stiffness and c is the damping coefficient.
This is a second order differential equation, and if for example, the car hits a hole at t = 0, such that it is displaced from its equilibrium position with y = y0, and dy/dt = 0, it will have a solution of the form,
where and provided
To analyse the actual performance of the design, the suspension system was built and tested with the following displacements of the wheel recorded during the time period of 2.6 and 3.1 seconds (this dataset is known to contain experimental error):
Time (s) |
2.6 |
2.65 |
2.7 |
2.75 |
2.8 |
2.85 |
2.9 |
2.95 |
3 |
3.05 |
3.1 |
Displacement (mm) |
0.074 |
0.094 |
0.106 |
0.113 |
0.127 |
0.145 |
0.145 |
0.148 |
0.154 |
0.162 |
0.158 |
The design team are interested if the dataset can be characterised by expressions which are less complex that the above theory.
PART 2
The design team performed stress analysis on the coil spring, and it was determined that the stress at a specific point could be characterised by:
Similar calculations were performed to calculate the maximum principal stress at 250 locations along a non-linear path through the coil spring. The following frequency distribution of the calculated maximum principal stresses was produced:
Max. Principal Stress (MPa) |
0.5 |
1 |
1.5 |
2 |
2.5 |
3 |
3.5 |
4 |
4.5 |
5 |
Frequency |
10 |
15 |
19 |
24 |
36 |
50 |
42 |
29 |
20 |
5 |
The design team is interested in the significance of these results and therefore need to calculate the area under the graph which contains this dataset.
The design team believe any further calculations of principal stresses and determining the area under the graph will be time consuming (and potentially error prone), thus would like an automated method for performing this.
Submission
Create two separate MATLAB scripts for Part 1 and Part 2:
PART 1 - Create a MATLAB script which is capable of performing the following:
PART 2 – Create a MATLAB script that is capable for performing the following:
Make sure all sections of the MATLAB script are annotated to explain the role of each line within the script.
PART --- 1
Code :=
%%%%%%%%%% Theroritical Ploting and Roots %%%%%%%%%%%%% m = 3.6*10^3; c = 1.5*10^3; k = 1.5*10^4; y0 = 0.3; p = sqrt((k/m)-(c^2/(4*m^2))); n = c/(2*m); y = @(t) exp(-n*t).*(y0*cos(p*t) + y0*(n/p)*sin(p*t)); t = linspace(0,15); plot(t,y(t)) % ploting theritical displacement yline(0); % Ploting zero line to identify root locations xlabel("Time [s]"); ylabel("Displacement [m]"); title("Theoritical Plot") % Form figure it's evident that first 3 roots are near 0.5, 2 and 4 % To calculate exact values we will use fzero function root1 = fzero(y,0.5); % 1st root using initial guess of 0.5 root2 = fzero(y,2); % 2nd root using initial guess of 2 root3 = fzero(y,4); % 3rd root using initial guess of 4 fprintf("\n First 3 roots : [%.4f %.4f %.4f] sec",[root1,root2,root3]) %%%%%%%%% Experimental Calculations %%%%%%%%%%%%%%% warning off Time = [2.6 2.65 2.7 2.75 2.8 2.85 2.9 2.95 3 3.05 3.1]; Disp = [0.074 0.094 0.106 0.113 0.127 0.145 0.145 0.148 0.154 0.162 0.158]; figure(2) plot(Time,Disp,'*-') [lin_Fit,gof_LinFit] = fit(Time',Disp','poly1'); % Linear Fit % defininig fit type for saturation growth myfittype = fittype('a*x/(b+x)','coefficients',{'a','b'}); [sat_growth_fit,gof_SatFit] = fit(Time',Disp',myfittype); hold on time_plot = linspace(2.6,3.1); plot(time_plot,lin_Fit(time_plot),'LineWidth',1.2); % Ploting Linear Fit plot(time_plot,sat_growth_fit(time_plot),'LineWidth',1.2); % ploting saturation growth fit xlabel("Time"); ylabel("Displacement"); legend(["Experimental Data","Linear Fit","Saturation Growth fit"],'Location','NorthWest') title("Experimental Plot"); % R2 and standard error for both R2_LinFit = gof_LinFit.rsquare; % Extracting Rsquare from goodness of fit (gof) Std_err_LinFit = gof_LinFit.rmse; % Extracting Standard error from gof R2_SatFit = gof_SatFit.rsquare; Std_err_SatFit = gof_SatFit.rmse; fprintf("\n\n For Linear Fit : R2 = %.4f , Standard Error = %.4f", R2_LinFit,Std_err_LinFit); fprintf("\n For Saturation Growth Fit : R2 = %.4f , Standard Error = %.4f\n", R2_SatFit,Std_err_SatFit); warning on
Results:-
PART----- 2
Max Principal Stress for given condition : 5.3652 MPa
Area under Curve = 121.9053
Text :-
sigma_y = 3.1; sigma_z = 4.5; tau_yz = -1.4; % Formula for maximum principal Stress sigma_max_P = (sigma_y + sigma_z)/2 + sqrt(((sigma_y - sigma_z)/2)^2 + tau_yz^2); sigma = 0.5:0.5:5; freq = [10 15 19 24 36 50 42 29 20 5]; plot(sigma,freq) % Ploting Frequncy Distribution spline_fit = fit(sigma',freq','SmoothingSpline'); % Fitting spline to data hold on p = plot(spline_fit,sigma,freq); % Ploting Fitted Spline p(1).MarkerSize = 15; p(2).LineWidth = 1.6; legend(["Distribution","Data Points","Fitted Spline"],'Location','Best') xlabel("\sigma [MPa]"); ylabel("Frequency"); grid on % For area under curve we will make more fine data to get accurate value sigma_finer = linspace(0.5,5,100); % generating 100 points for sigma between 0.5 & 5 freq_finer = spline_fit(sigma_finer); % Finding Corresponding frequencies by fitted spline area = trapz(sigma_finer,freq_finer); % Finding area under curve using trapz fprintf("\n Area under Curve = %.4f",area) % Priniting area
An engineering company is designing and testing a car suspension system. The system has a convent...
An engineering company is designing and testing a car suspension system. The system has a conventional suspension design, consisting of a shock-absorber and spring at each wheel. The shock-absorber provides a damping effect that is proportional to the vertical speed (i.e. up and down motion) of the wheel, and the spring's resistance is proportional to the vertical displacement of the wheel.The design team analyses the suspension system in two separate parts: Part 1 - dynamics of the spring-damper system; Part 2...
Create a MATLAB script An engineering company is designing and testing a car suspension system. The system has a conventional suspension design, consisting of a shock-absorber and spring at vertical speed (ie. up and down motion) of the wheel, and the spring's resistance is proportional to the vertical displacement of the wheel. The company would like to investigate how the actual performance of the suspension system compares to that of theory. Dynamic theory states that the equation of motion of...
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