Matlab Torsion Problem A solid shaft with a cylindrical cross section has a leng
ID: 2086101 • Letter: M
Question
Matlab Torsion Problem A solid shaft with a cylindrical cross section has a length L, diameter D, shear modulus G and yield stress Ty A torque T is applied about the z axis and increased until the angle of twist has reached a certain value (say ????), then the torque is removed, write matlab code that determines for a given ????- The maximum torque applied The maximum shear stress generated The permanent angle of twist The residual shear stress at the elastic-plastic interface Choose 16 values for ???? that range from zero to 3 times the angle of twist at the onset of yield. Apply the code to the special case: L 35 ft, D-2.4in, G-11.2x106 psi and -22 ksi. (Hint, for T=75 kip in, the residual shear stress at rp-5.02 ksi).Explanation / Answer
clear all;close all;clc;commandwindow;
%getting values from the user...
G=input('enter the value of shear modulus G : ');
D=input('enter the diameter in inches : ');
J=(pi/32)*(D^4);
L=input('enter the length of the shaft in feet : ');
L=L*12; %feet to inches conversion...
phi=input('give the angle of twist at the onset of the yield : ');
%computing for 16 values of phi_max...
i=1;
for phi_max=0:(1/16):(3*phi)
%computing maximum torque T_Max...
T_Max(:,i)=(G*J*phi_max)/L;
%computing maximum shear stress Tou_max...
Tou_max(:,i)=(G*D*phi_max)/(2*L);
i=i+1;
end
%computing permanent angle of twist phi_y...
tou_y=input('enter the yield stress tou_y : ');
phi_y=(tou_y*2*L)/(G*D);
display('Maximum troque for 16 values of phi_max is : ');
display(T_Max);
display('maximum shear stress for 16 values of phi_max is : ');
display(Tou_max);
display('computing permanent angle of twist phi_y is :');
display(phi_y);
OUTPUT:
T_Max =
1.0e+05 *
Columns 1 through 8
0 0.0543 0.1086 0.1629 0.2171 0.2714 0.3257 0.3800
Columns 9 through 16
0.4343 0.4886 0.5429 0.5972 0.6514 0.7057 0.7600 0.8143
Columns 17 through 24
0.8686 0.9229 0.9772 1.0314 1.0857 1.1400 1.1943 1.2486
Columns 25 through 32
1.3029 1.3572 1.4115 1.4657 1.5200 1.5743 1.6286 1.6829
Columns 33 through 40
1.7372 1.7915 1.8457 1.9000 1.9543 2.0086 2.0629 2.1172
Columns 41 through 48
2.1715 2.2258 2.2800 2.3343 2.3886 2.4429 2.4972 2.5515
Column 49
2.6058
maximum shear stress for 16 values of phi_max is :
Tou_max =
Columns 1 through 6
0 2000 4000 6000 8000 10000
Columns 7 through 12
12000 14000 16000 18000 20000 22000
Columns 13 through 18
24000 26000 28000 30000 32000 34000
Columns 19 through 24
36000 38000 40000 42000 44000 46000
Columns 25 through 30
48000 50000 52000 54000 56000 58000
Columns 31 through 36
60000 62000 64000 66000 68000 70000
Columns 37 through 42
72000 74000 76000 78000 80000 82000
Columns 43 through 48
84000 86000 88000 90000 92000 94000
Column 49
96000
computing permanent angle of twist phi_y is :
phi_y =
0.0031