User:Eml4500.f08.wiki1.schaet/hw5

Problem Statement
3-D Three Bar Truss Problem p230 in Bhatti Consider the three bar truss problem shown below. The cross sectional are of elements 1 and 2 is 200mm2 and that of element 3 is 600mm2. All elements are made of the same material with E=200Gpa. The applied load is P=20kN. We were also asked to determine if this truss system was statically determinate. If it was, we were to solve it using statics and compare the solution to the FEA solution generated with matlab. Below is the picture, similar to the one given in the book. The nodal coordinates are as follows: The solution given in the book is as follows:

Matlab Code
Below is the code for the 3-D Three Bar Truss System:  Three Bar Space Truss

Needed m-files for Space Truss Problem
This code uses three m-files. The first mfile, SpaceTrussElement.m, defines the element stiffness matrix. The second m-file, SpaceTrussResults.m, computes the element strain, stress and axial forces. The third m-file, NodalSoln.m, was given in an earlier homework assignment. It computes the nodal solutions.  SpaceTrussElement.m  SpaceTrussResults.m  NodalSoln.m

conn and lmm Arrays of Space Truss Problem
Below are the arrays lmm and conn used to solve the 3-D three-bar truss system: conn= $$ \begin{bmatrix} 1 & 4 \\ 2 & 4 \\ 3 & 4\end{bmatrix} $$ lmm= $$ \begin{bmatrix} 1 & 2 & 3 & 10 & 11 & 12 \\ 4 & 5 & 6 & 10 & 11 & 12 \\ 7 & 8 & 9 & 10 & 11 & 12 \end{bmatrix}$$

Images of Space Truss Problem
Below are the four images of the undeformed shape vs. the deformed shape. The undeformed shape is the blue dotted line and the deformed shape is the red solid line. Their perspective plot is from the point (-2, -2, 3) as suggested. From point (-2,-2,3) YZ plane XZ plane XY plane

Matlab Code for the Images of Space Truss Problem
Below is the matlab code used to obtain the previous images:  Image Code

Statics Solution of the Space Truss Problem
This problem, unfortunately for us, is statically determinate. In order to solve this problem, we first note that there will be three reaction forces at node 1, node 2, and node 3. Each one of these reaction forces can be divided into the x, y, and z components. Doing so gives us the following equation: $$ R_1 = R_{1x}\hat i + R_{1y}\hat j + R_{1z}\hat k $$ $$ R_2 = R_{2x}\hat i + R_{2y}\hat j + R_{2z}\hat k $$ $$ R_3 = R_{3x}\hat i + R_{3y}\hat j + R_{3z}\hat k $$ In order to solve for these reactions, we need to find what each reaction components are. Noting that R3x and R3y are both zero, we have the following equations to work off of: $$\mathbf{R_{1x}} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1}$$ $$\mathbf{R_{1y}} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1}$$ $$\mathbf{R_{1z}} = \mathbf{R_1}\sin{\phi_1}$$ $$\mathbf{R_{2x}} = \mathbf{R_2}\cos{\phi_2}\cos{\theta_2}$$ $$\mathbf{R_{2y}} = \mathbf{R_2}\cos{\phi_2}\sin{\theta_2}$$ $$\mathbf{R_{2z}} = \mathbf{R_2}\sin{\phi_2}$$ $$\mathbf{R_{3z}} = \mathbf{R_3}$$ Using the pythagorean theorem and trigonometry the following calculations were made in order to find the values for θ and Φ. $$\theta_1 = 63.435^\circ$$ $$\theta_2 = -45^\circ$$ $$\phi_1 = 42.975^\circ$$ $$\phi_2 = 44.482^\circ$$ Using these equations, we set up the equations for the forces in the following steps: $$ \sum F = 0 $$ $$\sum \mathbf{F_x} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1} + \mathbf{R_2}\cos{\phi_2}\cos{\theta_2} = 0$$ $$\sum \mathbf{F_y} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1} + \mathbf{R_2}\cos{\phi_2}\sin{\theta_2} - \mathbf{P} = 0$$ $$\sum \mathbf{F_z} = \mathbf{R_1\sin{\phi_1}} + \mathbf{R_2\sin{\phi_2}} + \mathbf{R_3} = 0$$ Plugging in all of the known values, given and calculated, the following results are obtained: R1 = 20374.6588 R2 = 13214.4037 R3 = -23148.1172 These values are just like the ones received using Matlab and the FEA method.