User:EML4500.f08.RAMROD.E/HW2

Step 4) Elimination of known degrees of freedom in order to reduce the global Force-Displacement relationship.

For this problem, the reduction is due to the fact that global nodes 1 and 3 are fixed and therefore have zero displacement, or:

$$d_{1} = d_{2} = d_{5} = d_{6} = 0$$

Setting these displacements to zero is the equivalent of eliminating the corresponding columns (1,2,5 & 6) of the global stiffness matrix, K. This elimination is shown below.

$$K = \begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix}$$

In addition to removing these columns because of the fixed boundary conditions, rows 1, 2, 5, & 6 can also be eliminated due to the principle of virtual work (PVW). After deletion of these rows, the resulting K becomes:

$$K = \begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44}\\ \end{bmatrix}$$

And, the resulting global Force-Displacement relationship is:

$$\begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44}\\ \end{bmatrix} * \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \begin{Bmatrix} F_{3}\\ F_{4} \end{Bmatrix}$$

We have already solved for each element in K previously such that

$$K = \begin{bmatrix} 3.0625 & -2.175\\ -2.175 & 2.6875\\ \end{bmatrix}$$

Looking at the problem drawing, we can see that F3 is zero and that F4 is equal to -7. Plugging these values and the values of K into the above global F-D relationship gives the relationship:

$$ \begin{bmatrix} 3.0625 & -2.175\\ -2.175 & 2.6875\\ \end{bmatrix} * \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \begin{Bmatrix} 0\\ -7 \end{Bmatrix}$$

Now, to solve for the unknown displacements, we can multiply both sides by K-1, where:

$$K^{-1} = \frac{1}{det(K)}\begin{bmatrix} K_{33} & -K_{34}\\ -K_{43} & K_{44} \end{bmatrix} = \begin{bmatrix} 0.875 & 0.621\\ 0.621 & 0.768 \end{bmatrix}$$

It is important to not that K-1 is not the same as the transpose of K(KT).

Therefore, the equation to solve for the unknown displacements are:

$$\begin{bmatrix} 0.875 & 0.621\\ 0.621 & 0.768 \end{bmatrix}*\begin{Bmatrix} 0\\ -7 \end{Bmatrix} = \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix}$$

Solving for these displacements gives

$$d_{3} = -4.347 $$ $$d_{4} = -5.376$$

 Step 5: Finding Reactions  The final step for this problem is to solve for the unknown reactions at global nodes 1 and 3. For this we use the element F-D relationships:

$$K^{(1)}* d^{(1)} = F^{(1)}$$ $$K^{(2)}* d^{(2)} = F^{(2)}$$

In the above equation, K(e), d(e), and  half of f(e) are known for each element, e. For this situation, the F-D relationships with all known values substituted is:

$$\begin{Bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16}\ & \frac{3}{16} \end{Bmatrix} \begin{Bmatrix} -4.347\\ -5.376 \end{Bmatrix}= \begin{Bmatrix} f^{(1)}_1\\ f^{(1)}_2\end{Bmatrix}$$

$$\begin{Bmatrix} \frac{5}{2} & \frac{-5}{2}\\ \frac{-5}{2} & \frac{5}{2} \end{Bmatrix} \begin{Bmatrix} -4.347\\ -5.376 \end{Bmatrix}= \begin{Bmatrix} f^{(2)}_3\\ f^{(2)}_4 \end{Bmatrix}$$

Using this equation, we can now solve for the f(1) matrix and thus we find the values for the reactions: $$f^{(1)}_{1} = F_{1} = -4.191$$ $$ f^{(1)}_{2} = F_{2} = -2.420$$

Using this same procedure applied to element 2, we can solve for the values of the other two unknown reactions giving us: $$f^{(2)}_{3} = F_{5} = 2.564$$ $$f^{(2)}_{4} = F_{6} = -2.564$$