Post on 14-Feb-2022
Wright State University Fall 2015
Department of Mechanical and Materials Engineering
ME 1020-02: ENGINEERING PROGRAMMING WITH MATLAB
MID-TERM EXAM 3 Open Book, Closed Notes
Create a Separate MATLAB Script File for Each Problem
Submit all MATLAB files (.m) to Dropbox on Pilot
Each MATLAB File Must Have the Following Header:
NAME:
CIRCLE ONE: GRADE MY EXAM DO NOT GRADE MY EXAM
Problem 1 (5 Points): The useful life of a machine bearing depends on its operating temperature, as the
following data show. Using Function Discovery principles, determine whether the function is linear, power-law,
or exponential, and obtain a functional description of these data. Plot the function and the original data on the
same plot. The plot must have axis labels and a title. Estimate a bearing’s life if it operates at 190°F. Show this
point on the plot. Do not use any inputs from the user, such as the Basic Curve Fitting Interface or ginput.
Temperature (°F) 100 120 140 160 180 200 220
Bearing Life (Hours) 28,000 21,000 15,000 11,000 8,000 6,000 4,000
Problem 2 (5 Points): Data analysis of the breaking strength of a certain fabric shows that it is normally
distributed with a mean of 300 lb and a variance of 9. Create a MATLAB Script File to estimate the percentage
of fabric samples that will have a breaking strength:
a. Less than or equal to 297 lb,
b. Within 3 lb of the mean,
c. Greater than 303 lb.
Problem 3 and 4 are on the back of this page.
Problem 3 (5 Points): Create a MATLAB Script File to solve the following problem:
𝑥 − 3𝑦 = 2
𝑥 + 5𝑦 = 18
4𝑥 − 6𝑦 = 10
Plot the equations and the solution over the range 0 ≤ 𝑥 ≤ 10. The plot must have axis labels and a title.
Problem 4 (5 Points): Create a MATLAB Script File to calculate and plot the value of 𝑦 for −5 ≤ 𝑥 ≤ 5.5
using a WHILE loop. Make sure to use a sufficient number of points to create a smooth plot.
𝑦 = 𝑒𝑥+1 for 𝑥 < −1
𝑦 = 2 + cos(𝜋𝑥) for − 1 ≤ 𝑥 < 5
𝑦 = 1 + 10(𝑥 − 5) for 𝑥 ≥ 5
Make sure to give the plot a title and axis labels.
% Problem 1, Mid-Term Exam 3, ME 1020, Fall 2015
clear
clc
disp('Problem 1, Mid-Term Exam 3, ME 1020, Fall 2015: Scott Thomas')
format short
Temperature = [100 120 140 160 180 200 220]
Life = [28000 21000 15000 11000 8000 6000 4000]
figure
plot(Temperature, Life,'o')
xlabel('Temperature (F)')
ylabel('Bearing Life (Hr)')
title('Problem 1: Rectilinear Plot')
figure
loglog(Temperature, Life,'o')
xlabel('Temperature (F)')
ylabel('Bearing Life (Hr)')
title('Problem 1: Loglog Plot')
figure
semilogy(Temperature, Life,'o')
xlabel('Temperature (F)')
ylabel('Bearing Life (Hr)')
title('Problem 1: Semilog y Plot')
disp('Semi-log y plot is linear: Exponential Function')
format shortEng
p = polyfit(Temperature, log10(Life),1);
m = p(1)
b = 10^p(2)
xplot = linspace(90,230,100);
yplot = b*10.^(m*xplot);
x190 = 190
y190 = b*10.^(m*x190)
figure
plot(Temperature, Life,'o',xplot,yplot,x190,y190,'r*')
xlabel('Temperature (F)')
ylabel('Bearing Life (Hr)')
title('Problem 1: Bearing Life versus Temperature')
text(110,32000,'Bearing Life = (141.8603e+003)*10^{-6.9579e-003*Temperature}')
text(110,3000,'Bearing Life(T = 190F) = 6.7586e+003 Hours')
Problem 1, Mid-Term Exam 3, ME 1020, Fall 2015: Scott Thomas
Temperature =
100 120 140 160 180 200 220
Life =
Columns 1 through 6
28000 21000 15000 11000 8000 6000
Column 7
4000
Semi-log y plot is linear: Exponential Function
m =
-6.9579e-003
b =
141.8603e+003
x190 =
190.0000e+000
y190 =
6.7586e+003
% Problem 2, Mid-Term Exam 3, ME 1020, Fall 2015
clear
clc
disp('Problem 2, Mid-Term Exam 3, ME 1020, Fall 2015: Scott Thomas')
format short
mean = 300
sigma_squared = 9
a = 297
b = 303
disp('Part a:')
P_a = 100*0.5*(1 + erf((a - mean)/(sqrt(sigma_squared)*sqrt(2))))
disp('Part b:')
P_b = 100*0.5*(erf((b - mean)/(sqrt(sigma_squared)*sqrt(2))) - erf((a - mean)/(sqrt(sigma_squared)*sqrt(2))))
disp('Part c:')
P_c = 100 - P_a - P_b
Problem 2, Mid-Term Exam 3, ME 1020, Fall 2015: Scott Thomas
mean =
300
sigma_squared =
9
a =
297
b =
303
Part a:
P_a =
15.8655
Part b:
P_b =
68.2689
Part c:
P_c =
15.8655
Published with MATLAB® R2012b
5 4 3 2 1 0 11 10 11 10 12 4
% Problem 3, Mid-Term Exam 3, ME 1020, Fall 2015
clear
clc
disp('Problem 3, Mid-Term Exam 3, ME 1020, Fall 2015: Scott Thomas')
format short
A = [1 -3; 1 5; 4 -6]
b = [2; 18; 10]
% The system is overdetermined since the # of equations > # of unknowns
RankA = rank(A)
RankAb = rank([A b])
% Since rank(A) = 2 and rank([A b]) = 3, no solution exists. The left
% division method can be used to compute the least-squares solution,
% which is not an exact solution.
x = A\b
xplot = linspace(0,10,2);
y1 = (xplot - 2)/3;
y2 = (-xplot + 18)/5;
y3 = (4*xplot - 10)/6;
plot(xplot,y1,xplot,y2,xplot,y3,x(1),x(2),'r*'), xlabel('x'), ylabel('y')
title('Problem 3, Mid-Term Exam 3, ME 1020, Fall 2015: Scott Thomas')
legend('x - 3y = 2','x + 5y = 18','4x - 6y = 10','x = A\b','Location','Best')
grid on
Problem 3, Mid-Term Exam 3, ME 1020, Fall 2015: Scott Thomas
A =
1 -3
1 5
4 -6
b =
2
18
10
RankA =
2
RankAb =
3
x =
6.0928
2.2577
Published with MATLAB® R2012b
5 4 3 2 1 0
12 10 30 6 0 0
% Problem 4: Mid-Term Exam 3, ME 1020, Fall 2015
clear
clc
disp('Problem 4: Scott Thomas')
x = -5;
dx = 0.001;
k=1;
y=0;
while x <= 5.5
if x <= -1
y(k)=exp(x+1);
else if x < 5 & x >= -1
y(k) = 2 + cos(pi*x);
else
y(k) = 1 + 10*(x - 5);
end
end
x = x + dx;
xplot(k) = x;
k = k + 1;
end
xplot;
y;
plot(xplot,y), title('Problem 4: Scott Thomas')
xlabel('x'), ylabel('y')
Problem 4: Scott Thomas
Published with MATLAB® R2012b