Give comments about each line (most lines)(% etc) Do in a very simple way of cod

Question

Give comments about each line (most lines)(% etc) Do in a very simple way of coding (MATLAB) no complex codes (I am 4 weeks in my course) Copy and paste code here 22. 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 solv- ing the following system of nine equa- 1800 lb 1200 lb all 1500 lb . 420 42R 1 48D F2+ cos(48.81°)F 0 F6+ cos(48.81°)Fs-F2 0, sin(48.81°)F,+F3-0 -cos(48.81°)F, +F, = 0 -sin(48.81*)FitF,-1800, -F4-cos(48.81*)F, = 1200, -F,-sin(48.81*)F,-sin(45°)F, = 0, sin(45°)F, = 1500, -cos(45°F,-F. = 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 correspond-

Explanation / Answer

%F1*cos(48.81)+F2+F3*0 +F4*0 +F5*0+F6*0+F7*0+F8*0+F9*0=0
%F1*0 -F2+F3*0 +F4*0 +F5*cos(48.81)+F6+F7*0+F8*0+F9*0=0
%F1*0+F2*0+F3 +F4*0 +F5*sin(48.81)+F6*0+F7*0+F8*0+F9*0=0
%-F1*cos(48.81)+F2*0+F3*0 +F4 +F5*0+F6*0+F7*0+F8*0+F9*0=0
%-F1*sin(48.81)+F2*0+F3 +F4*0 +F5*0+F6*0+F7*0+F8*0+F9*0=1800
%F1*0+F2*0+F3*0 -F4 -F5*cos(48.81)+F6*0-F7*0+F8*0-F9*0=1200
%F1*0+F2*0+F3*0 -F4 -F5*sin(48.81)+F6*0-F7+F8*0-F9*sin(45)=0
%F1*0+F28+F3*0 +F4*0 +F5*0+F6*0+F7*0+F8*0+F9*sin(45)=1500
%F1*0+F280+F3*0 +F4*0 +F5*0+F6*0+F7*0-F8-F9*cos(45)=0
% The left hand side of the matrix form can be written as
theta= 48.81*pi/180;
phi= 45*pi/180;

A= [ cos(theta), 1,0,0,0,0,0,0,0;
0,-1,0,0,cos(theta),1,0,0,0;
0,0,1,0,sin(theta),0,0,0,0;
-cos(theta),0,0,1,0,0,0,0,0;
-sin(theta),0,1,0,0,0,0,0,0;
0,0,0,-1,-cos(theta),0,0,0,0;
0,0,0,0,-sin(theta),0,-1,0,sin(phi);
0,0,0,0,0,0,0,0,sin(phi);
0,0,0,0,0,0,0,1,-cos(phi);]
%The right hand side of the matrix form can be written as
Y= [ 0,0,0,0,1800,1200,0,1500,0;]
% SOLUTION
z=A/Y;
x=1:9
M=z(1:9)
r={'memebr' 'force'}
f=figure('position',[100 100 300 500])
t=uitable(f,'data',z,'columnname',r,'position',[0 0 250 500])

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