User:Eml4507.s13.team2.cc/R6

Honor Pledge
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. MATLAB code from our own Problem 3.1 was used.

Given: 3-Bar Truss
A three-bar plane truss with identical members is presented, and is shown in the image below.

Element 2 is vertical, and all members make equal angles with each other. A force is applied at Node 1, while Nodes 2 and 3 can have non-zero or zero prescribed displacement degrees of freedom.



The following member properties are given:

$$ E = 206 \ GPa $$

$$ A = 1*10^{-4} \ m^2 $$

$$ L = 1 \ m $$

The following force and displacement DOFs are given:

$$ F = 20,000 \ N $$

$$ d_3= 2 \ cm $$

$$ d_4= -1 \ cm $$

$$ d_5= -3 \ cm $$

$$ d_6= 5 \ cm $$

$$ d_7= 0 \ cm $$

$$ d_8= 0 \ cm $$

Compute the reactions for non-zero displacement dofs
Compute the reactions for the case of non-zero prescribed displacement degrees of freedom.

Compute the reactions for zero displacement dofs
Compute the reactions for the case of zero prescribed displacement degrees of freedom.

Compare the reactions
Compare the reactions found for the two different cases.

Solving for Non-Zero Displacement DOFs
In order to solve for the desired values, the following MATLAB code from Problem 3.1 was used:

Output of MATLAB Code
The above code returns the following output:

k = 1.545e+007 -8.9201e+006 -1.545e+007  8.9201e+006 -8.9201e+006   5.15e+006  8.9201e+006   -5.15e+006 -1.545e+007 8.9201e+006   1.545e+007 -8.9201e+006 8.9201e+006  -5.15e+006 -8.9201e+006    5.15e+006 k = 7.7238e-026 1.2614e-009 -7.7238e-026 -1.2614e-009 1.2614e-009   2.06e+007 -1.2614e-009   -2.06e+007 -7.7238e-026 -1.2614e-009 7.7238e-026  1.2614e-009 -1.2614e-009  -2.06e+007  1.2614e-009    2.06e+007 k = 1.545e+007 8.9201e+006  -1.545e+007 -8.9201e+006 8.9201e+006   5.15e+006 -8.9201e+006   -5.15e+006 -1.545e+007 -8.9201e+006  1.545e+007  8.9201e+006 -8.9201e+006  -5.15e+006  8.9201e+006    5.15e+006 K = Columns 1 through 5 3.09e+007 1.8626e-009 -7.7238e-026 -1.2614e-009  -1.545e+007 1.8626e-009   3.09e+007 -1.2614e-009   -2.06e+007  8.9201e+006 -7.7238e-026 -1.2614e-009 7.7238e-026  1.2614e-009            0 -1.2614e-009  -2.06e+007  1.2614e-009    2.06e+007            0 -1.545e+007 8.9201e+006            0            0   1.545e+007 8.9201e+006  -5.15e+006            0            0 -8.9201e+006 -1.545e+007 -8.9201e+006           0            0            0 -8.9201e+006  -5.15e+006            0            0            0 Columns 6 through 8 8.9201e+006 -1.545e+007 -8.9201e+006 -5.15e+006 -8.9201e+006  -5.15e+006 0           0            0            0            0            0 -8.9201e+006            0            0 5.15e+006           0            0 0  1.545e+007  8.9201e+006 0 8.9201e+006    5.15e+006 R = 14142       14142            0            0            0            0            0            0 d = -0.028976    0.010785         0.02        -0.01        -0.03         0.05            0            0 rf = -2.6217e-011 -4.2816e+005 -3.6562e+005 2.1109e+005 3.5148e+005 2.0293e+005 forces = -4.2219e+005 -4.2816e+005 -4.0586e+005

The first, second, and third outputs k give the element stiffness matrices for element 1, 2, and 3, respectively.

K gives the global stiffness matrix.

R gives the RHS external force matrix.

d gives the global displacement DOFs.

rf gives the reaction force components.

forces gives the member forces.

Plot of Deformed and Undeformed Shape
To plot the figure, the following MATLAB code was used.

The resulting figure for non-zero displacement DOFs is shown below.



Output of MATLAB Code
By changing the prescribed displacement degrees of freedom in the MATLAB code ("ebcVals") to zero, the following results are obtained.

k = 1.545e+007 -8.9201e+006 -1.545e+007  8.9201e+006 -8.9201e+006   5.15e+006  8.9201e+006   -5.15e+006 -1.545e+007 8.9201e+006   1.545e+007 -8.9201e+006 8.9201e+006  -5.15e+006 -8.9201e+006    5.15e+006 k = 7.7238e-026 1.2614e-009 -7.7238e-026 -1.2614e-009 1.2614e-009   2.06e+007 -1.2614e-009   -2.06e+007 -7.7238e-026 -1.2614e-009 7.7238e-026  1.2614e-009 -1.2614e-009  -2.06e+007  1.2614e-009    2.06e+007 k = 1.545e+007 8.9201e+006  -1.545e+007 -8.9201e+006 8.9201e+006   5.15e+006 -8.9201e+006   -5.15e+006 -1.545e+007 -8.9201e+006  1.545e+007  8.9201e+006 -8.9201e+006  -5.15e+006  8.9201e+006    5.15e+006 K = Columns 1 through 4 3.09e+007 1.8626e-009 -7.7238e-026 -1.2614e-009 1.8626e-009   3.09e+007 -1.2614e-009   -2.06e+007 -7.7238e-026 -1.2614e-009 7.7238e-026  1.2614e-009 -1.2614e-009  -2.06e+007  1.2614e-009    2.06e+007 -1.545e+007 8.9201e+006            0            0 8.9201e+006  -5.15e+006            0            0 -1.545e+007 -8.9201e+006           0            0 -8.9201e+006  -5.15e+006            0            0 Columns 5 through 8 -1.545e+007 8.9201e+006  -1.545e+007 -8.9201e+006 8.9201e+006  -5.15e+006 -8.9201e+006   -5.15e+006 0           0            0            0            0            0            0            0   1.545e+007 -8.9201e+006            0            0 -8.9201e+006   5.15e+006            0            0 0           0   1.545e+007  8.9201e+006 0           0  8.9201e+006    5.15e+006 R = 14142       14142            0            0            0            0            0            0 d = 0.00045767  0.00045767            0            0            0            0            0            0 rf = -5.773e-013 -9428.1     -2988.6       1725.5       -11154      -6439.5 forces = -3450.9     -9428.1        12879

Plot of Deformed and Undeformed Shape
The resulting figure for non-zero displacement DOFs is shown below. Displacements were increased by a factor of 100 to clearly show the difference.



Comparing the reactions
For the non-zero case, the reactions are:

rf = -2.6217e-011 -4.2816e+005 -3.6562e+005 2.1109e+005 3.5148e+005 2.0293e+005

For the zero case, the reactions are:

rf = -5.773e-013 -9428.1    -2988.6      1725.5      -11154     -6439.5

As can be seen, the reactions for the zero displacement DOF case are several orders of magnitude smaller.