matlab sheet 5 - solution functions - ahmed...

Faculty of Engineering Spring 2017 Mechanical Engineering Department

Dr./ Ahmed Nagib Elmekawy 1 of 5 Matlab Sheet 5 - Solution

Matlab Sheet 5 - Solution

Functions 1. The output of the MATLAB atan2 function is in radians. Write a function called

atan2d that produces an output in degrees.

M-File: function angle_deg = atan2d(y,x)

% Computes atan2(y,x) in degrees.

angle_rad = atan2(y,x);

angle_deg = angle_rad*180/pi;

In command window >>atan2d(1,0))

ans =

90

>>atan2d(1,-1))

ans =

135

2. Write a function that accepts temperature in degrees Fahrenheit (°F) and

computes the corresponding value in degrees Celsius (°C). The relation between

the two is

𝑇 ℃ =5

9(𝑇 ℉ − 32)

Test your function at 32, 50 and 100.

M-File: function y = temp(x)

% Converts from Fahrenheit to Celsius.

y = (5/9)*(x-32);

In command window >> temp([32,212])

ans =

0 100



3. An object thrown vertically with a speed 𝑣0 reaches a height ℎ at time 𝑡, where

ℎ = 𝑣0𝑡 −1

2𝑔𝑡2

Write and test a function that computes the time 𝑡 required to reach a specified

height ℎ, for a given value of 𝑣0. The function’s inputs should be h, 𝑣0, and 𝑔.

Test your function for the case where ℎ = 100 m, 𝑣0 = 50 m/s, and 𝑔 = 9.81 m/s2.

Interpret both answers.

M-File: function t = time(h,v0,g)

% Computes time t to reach a specified height h, with

initial speed v0.

roots([0.5*g,-v0,h])

In command window >> time(100,50,9.81)

ans =

7.4612

2.7324

The smaller value is the time to reach the height while ascending; the larger

value is the time to reach the height while descending.



Writing and reading from command window and files

4. When a belt is wrapped around a cylinder, the relation between the belt forces

on each side of the cylinder is

𝐹1 = 𝐹2𝑒𝜇𝛽

where 𝛽 is the angle of wrap of the belt in radian and 𝜇 is the friction coefficient.

Write a script file that first prompts a user to specify 𝛽 , 𝜇 , and 𝐹2 and then

computes the force 𝐹1. Test your program with the values 𝛽 = 130°,

𝜇 = 0.3, and 𝐹2 = 100 N. (Hint: Be careful with 𝛽!)

M-File: beta_deg = input(‘Enter a value for beta in degrees:’);

% Convert to radians.

beta = (pi/180)*beta_deg;

mu = input(‘Enter a value for mu:’);

F2 = input(‘Enter a value for force F2:’);

F1 = F2*exp(mu*beta);

disp(‘The force F1 is:’)

disp(F1)

In command window Enter a value for beta in degrees:130

Enter a value for mu:0.3

Enter a value for force F2:100

The force F1 is:

197.5217



5. The intrinsic electrical conductivity 𝜎 of a semiconductor can be approximated

by:

𝜎 = 𝑒(𝐶−

𝐸𝑔

2𝑘𝑇)

Where 𝜎 is measured in (Ω − 𝑚)−1, 𝐸𝑔 is the band gap energy, 𝑘 is Boltzmann’s

constant (8.26 × 10-5 ev/K), and 𝑇 is temperature in kelvins. For Germanium, 𝐶

= 13.83 and 𝐸𝑔 = 0.67 ev. Write a program in a script file that calculates the

intrinsic electrical conductivity for Germanium for various temperatures. The

values of the temperature should be read from an xls spreadsheet using the

xlsread command. The output should be presented as a table where the first

column is the temperature and the second column is the intrinsic electrical

conductivity. Use the following values for temperature: 400, 435, 475, 500, 520,

and 545 K.

M-File: clear, clc

C=13.83; Eg=0.67; k=8.62e-5;

T=xlsread('Germanium_data.xlsx');

sigma=exp(C-Eg./(2*k*T));

tbl=[T sigma];

disp(' Intrinsic')

disp(' Temperature Conductivity')

disp(' deg K (ohm-m)^-1')

fprintf(' %4.0f %5.1f\n',tbl')

Excel File



In Command Window

6. The graph of the function 𝑓(𝑥) = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑 passes through the points

(–2.6, –68), (0.5, 5.7), (1.5, 4.9), and (3.5, 88). Determine the constants a, b, c,

and d. (Write a system of four equations with four unknowns, and use MATLAB

to solve the equations.)

M-File: clear, clc

x=[-2.6 0.5 1.5 3.5]; y=[-68; 5.7; 4.9; 88]; power=3:-1:0;

X=[x(1).^power; x(2).^power; x(3).^power; x(4).^power];

coefs=X\y;

fprintf('\nThe equation is f(x)=%.3fx^3 + %.3fx^2 + %.3fx +

%.3f\n',coefs)

In Command Window



7. The pressure drop in Pa for a fluid flowing in a pipe with a sudden increase in

diameter is given by:

∆𝑝 =1

2[1 − (

𝑑

𝐷)

2

]

2

𝜌𝑣2

where 𝜌 is the density of the fluid, 𝑣, the velocity of the flow, and 𝑑 and D are

defined in the figure. Write a program in a script file that calculates the

pressure drop ∆𝑝. When the script file is executed it request the user to input

the density in kg/m3, the velocity in m/s, and values of the non-dimensional

ratio as a vector. The program displays the inputted values of and 𝑣 followed

by a table with the values of in the first column and the corresponding values

of ∆𝑝 in the second column.

Execute the program assuming flow of gasoline (𝜌 = 737 kg/m3) at 𝑣 = 5 m/s

and the following ratios of diameters 𝑑/𝐷 = 0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.2.

M-File: clear, clc

rho=input('Please input the fluid density in kg/m^3: ');

v=input('Please input the fluid velocity in m/s: ');

d_ratio=input('Please input the pipe diameter ratio as a vector [x

x x]: ');

delP=0.5*(1-d_ratio.^2).^2*rho*v^2;

fprintf('\nFor gasoline with a density of %.0f kg/m^3 and a flow

',rho)

fprintf('velocity of %.1f m/s\n\n',v)

tbl=[d_ratio;delP];

disp(' delta P')

disp(' d/D (Pa)')

fprintf(' %3.1f %6.1f\n',tbl)



In Command Window



8. A truss is a structure made of members joined at their ends. For the truss shown

in the figure

the forces in the nine members are determined by solving the following system

of nine equations:

𝐹2 + cos(48.81°)𝐹1 = 0

𝐹6 + cos(48.81°)𝐹5 − 𝐹2 = 0

sin(48.81°) 𝐹5 + 𝐹3 = 0

−cos(48.81°)𝐹1 + 𝐹4 = 0

−sin(48.81°) 𝐹1 + 𝐹3 = 1800

−𝐹4 − cos(48.81°)𝐹5 = 1200

−𝐹7 − sin(48.81°)𝐹5 − sin(45°)𝐹9 = 0

sin(45°)𝐹9 = 1500

−cos(45°)𝐹9 + 𝐹8 = 0

Write the equations in matrix form and use MATLAB to determine the forces in

the members. A positive force means tensile force and a negative force means

compressive force. Display the results in a table where the first column displays

the member number and the second column displays the corresponding force.

M-File: clear, clc

T=[cosd(48.81) 1 0 0 0 0 0 0 0

0 -1 0 0 cosd(48.81) 1 0 0 0

0 0 1 0 sind(48.81) 0 0 0 0

-cosd(48.81) 0 0 1 0 0 0 0 0



-sind(48.84) 0 -1 0 0 0 0 0 0

0 0 0 -1 -cosd(48.81) 0 0 0 0

0 0 0 0 -sind(48.81) 0 -1 0 -sind(45)

0 0 0 0 0 0 0 0 sind(45)

0 0 0 0 0 0 0 -1 -cosd(45)];

A=[0; 0; 0; 0; 1800; 1200; 0; 1500; 0];

N=1:9;

F=T\A;

tbl=[N;F'];

disp(' ')

disp(' Member Force')

disp(' No. lbf')

fprintf(' %1i %7.1f\n',tbl)

In Command Window



Interpolation

9. It Interpolation is useful when one or more data points are missing. This

situation often occurs with environmental measurements, such as temperature,

because of the difficulty of making measurements around the clock. The

following table of temperature versus time data is missing readings at 5 and 9

hours. Use linear interpolation with MATLAB to estimate the temperature at

those times.

M-File: Temp = [1:4,6:8,10:12];

Time = [10,9,18,24,21,20,18,15,13,11];

Temp_int = [5,9];

interp1(Temp,Time, Temp_int)

In Command Window

ans =

22.5000 16.5000

Thus the estimated temperatures at 5 P.M. and 9 P.M. are 22.5° and 16.5°.



10. Computer-controlled machines are used to cut and to form metal and other

materials when manufacturing products. These machines often use cubic

splines to specify the path to be cut or the contour of the part to be shaped. The

following coordinates specify the shape of a certain car’s front fender. Fit a

series of cubic splines to the coordinates, and plot the splines along with the

coordinate points.

M-File: x = [0,.25,.75,1.25,1.5,1.75,1.875,2,2.125,2.25];

y = [1.2,1.18,1.1,1,.92,.8,.7,.55,.35,0];

x_int = [0:.01:2.25];

y_int = spline(x,y,x_int);

plot(x,y,’o’,x_int,y_int),xlabel(‘x (ft)’),ylabel(‘y (ft)’)

axis([0 2.25 0 1.5]);

The Plot

0 0.5 1 1.5 20

0.5

1

1.5

x (ft)

y (

ft)



11. A The following data are the measured temperature T of water owing from a hot

water faucet after it is turned on at time t = 0.

a. Plot the data, connecting them first with straight lines and then with a

cubic spline.

b. Estimate the temperature values at the following times, using linear

interpolation and then cubic spline interpolation: t = 0.6, 2.5, 4.7, 8.9.

c. Use both the linear and cubic spline interpolations to estimate the time

it will take for the temperature to equal the following values: T = 75, 85,

90, 105.

Part a

M-File: t = [0:10];

T = [72.5, 78.1, 86.4, 92.3, 110.6, 111.5, 109.3, 110.2,

110.5, 109.9, 110.2];

t_int = [0:.05:10];

T_int = spline(t,T,t_int);

plot(t,T,t,T,’*’,t_int,T_int)

xlabel(‘Time (sec)’),ylabel(‘Temperature (deg F)’)

The Plot

0 1 2 3 4 5 6 7 8 9 10

70

75

80

85

90

95

100

105

110

115

Time (sec)

Tem

pera

ture

(deg F

)



Part b Insert the following lines at the end of the script file from Part (a). t_est = [0.6,2.5,4.7,9.8];

T_est_spline = interp1(t,T,t_est,0spline0)

T_est_linear = interp1(t,T,t_est)

The results are T_est_spline = [74.7235, 88.2044, 112.6234, 109.9696] and

T_est_linear

= [75.8600, 89.3500, 111.2300, 110.1400].

Part c

Insert the following code after the ylabel command in part (a).

grid,[tlinear,Tlinear] =

ginput(4),[tspline,Tspline] = ginput(4)

and use the mouse to select the points corresponding to T = 75, 85, 90, and

105.

The results are

tlinear = [0.4493, 1.8088, 2.5922, 3.6982] and

tspline = [0.6336, 1.7396,

2.7995, 3.6751].

matlab sheet 5 - solution functions - ahmed...

Documents