1. Using MATLAB, plot the three elements of the strain vector when the above material is oriented at (1) 0° and (2) at 30°. This is accomplished by incrementally applying stress (i.e., 5 psi, 10 psi, and so on) and then determining the strain vector at each step. find the following load conditions:

1. A load applied in the one direction
2. A biaxial load is applied

Start with a small number and increase until you can make a reasonable plot. Note: Two points do not make a reasonable plot. take note that at 0° [Q] = [Qbar]. Since this is the case, it would be a good thought experiment to work through what Qbar enables. Using the below script fill in the missing part and the Matlab output should generate the Reduced Stiffness Matrix (Qbar in psi) for 0 degrees and Reduced Stiffness Matrix (Qbar in psi) for 30 degrees.  Thanks

 

 

Image transcription text

in which Q1 1 = Q1 1 Cose + 2(Q12 + 2086) sine cose + Q22 sinte Q12 = (Q11 + @22 - 4068) sin e cos e
+ Q,2( since + cos e) Q22 = Q41 sin*e + 2(Q12+ 2065) Sin e coste + Q22 cose (2.85) Q16= (Q11
-12-2065) Sin e cos 8 + (Q12 - Q22 + 2065) Sine cos e Q26 = (Q41 - Q12 -2066) sin e cos ... 

Image transcription text

S, 1 = S,, cos e + (2$12 + Seg) sin e cose + Smaz since S12 = S,2( sine + cos"0) + (S,1 + S22 - S:)
since cos e Sa2 = S, 1 sine + (2S12 + $56) since cose + Sma cos*e (2.88) S16 =(2S,1 -2512 - S66) sin e
cos"e -(2522 - 2S12 - Sep) sine cos 0 S26 = (2S, 1 -2512 - Seg) sin"e cos e -(2S... 

 

%Material Properties

E1 = 30*10^6;

E2 = 1*10^6;

G12 = 0.5*10^6;

v12 = .3;

 

%derived

v21 = v12*(E2/E1);

 

%%%%///\\\%%%%%% Primary Calculations - Review Carefully %%%%%///\\\%%%%%%

 

% The system of equations needed for this calculation are:

% {strain} = [Sbar]*{applied stress}

% where: {strain} and {applied stress} are {1x3} vectors as shown in the

% notes. As shown in the notes and text, Sbar is the inverse of Qbar and vice versa.

% We are going to use the transformed, reduced stiffness matrix (Qbar) in future

% course assignments. So as opposed to computing the compliance matrix (Sbar) directly, we will compute

% the reduced stiffness matrix (Qbar) and then invert it. That it is

% available for future assignments. Matrix inversion is a built in function

% in MATLAB, so it makes this very easy to do as shown in the next few

% steps. If you are rusty as to what is happening go back and review some

% of the material on linear algebra in MATH 345 or even some earlier work

% using Crammer's rule to solve a system of equations.

 

 

%set the orientation and get the reduced stiffness matrix using the function

% at the end of this file

 

%%%%% need to set orientation here

theta = 0;

%%%%%%

 

[Qbar] = ReducedStiffnessMatrix(E1, E2, G12, v12, v21, theta);

Sbar = inv(Qbar);

 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%% first plot is for the unidirectional load %%%%%%%%%%%%%

 

figure

subplot (2,1,1);

hold on

 

%% Uniaxial Loading Condition

multiplier = 10^3; % think about this as applying stress in ksi. by definint

% it here it can be changed at just one location incase

% we want to change the load to say Msi (or 10^6)

 

for stress = 0:10:100 %%performs 11 increments of stress {0, 10, 20,..., 90, 100}

appliedStress = [stress;0;0]*multiplier; %unidirectional loading

strain = Sbar*appliedStress; % solving equation for strain

 

% note that this reports stress to the plot in ksi

plot(strain(1), appliedStress(1)/multiplier, '-.b*',strain(2), appliedStress(1)/multiplier, 'go', strain(3), appliedStress(1)/multiplier, 'r+' );

end

title(['Stress Strain Plot for \theta = ', num2str(theta), ' Degrees - Uniaxial Load Condition']);

xlabel('strain (in/in)');

ylabel('stress (ksi)');

legend('\epsilon_1', '\epsilon_2', '\tau_12');

 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%% biaxial Loading Condition %%%%%%%%%%%%%%%%%%%%%%%%%%%

 

subplot (2,1,2)

hold on

 

for stress = 0:10:100

 

appliedStress = [stress;stress;0]*multiplier; %bidirectional loading

strain = Sbar*appliedStress; % solving for strain

 

% note that this reports stress to the plot in ksi

plot(strain(1), appliedStress(1)/multiplier, '-.b*',strain(2), appliedStress(1)/multiplier, 'go', strain(3), appliedStress(1)/multiplier, 'r+' );

end

 

%% Plot notation

title(['Stress Strain Plot for \theta = ', num2str(theta), ' Degrees - Biaxial Load Condition']);

xlabel('strain (in/in)');

ylabel('stress (Ksi)');

legend('\epsilon_1', '\epsilon_2', '\epsilon_3');

 

hold off

 

 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Functions %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

 

% Determine the elements of the reduced stiffness matrix

function [Qbar] = ReducedStiffnessMatrix(E1, E2, G12, v12, v21, theta)

 

 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Below is a Reduced Stiffness Matrix - Qbar. It is used just to show how to

% create the plots above. For your homework, you need to comment that

% out and fill in the elements of Qbar below

 

% Note that the units in PSI

Qbar = [30, .5, 0; .5, 2, 0; 0,0,1]*10^6;

 

%note this is a [3x3] so Q66 % as shown in the text will correspond to

% the element in the [3,3] slot

 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

 

% you will need to add some code here for your assignment

 

%To help organize your code, first define defining the elements of

% of Q (the reduced stiffness matrix) first

%%%%%insert your code here using eqn for Q such as

 

Q(1,1) = E1/(1-v12*v21);

Q(1,2) = ;

Q(2,1) = Q(1,2);

Q(2,2) = ;

Q(3,3) = ;

 

%%%%%%%%%%% Once you have Q now find for Qbar%%%%%%%%%%

%%%%% use the equations for Qbar as shown in the notes and text

%%%%% DO NOT USE A TRANSFORMATION MATRIX T. That was shown in the notes

%%%%% just to illustrate how Qbar was derived

 

%insert your code here for Qbar such as:

Qbar(1,1) = Q(1,1)*cosd(theta)^4 + 2*(Q(1,2) + 2*Q(3,3))*(sind(theta)^2)*(cosd(theta)^2) + Q(2,2)*sind(theta)^4;

Qbar(1,2) = ;

Qbar(1,3) = ;

Qbar(2,1) = Qbar(1,2);

Qbar(1,2);

Qbar(2,2) = ;

Qbar(2,3) = ;

Qbar(3,1) = Qbar(1,3);

Qbar(3,2) = Qbar(2,3);

Qbar(3,3) = ;

 

end