User:EML4507.s13.team4ever.Tobin/R.2.6

Problem Statement Find the displacements of the nodes as well as the forces in the springs(tension/compression). Also determine the reaction forces at the walls.

Nodal Equilibrium Equations $$f_1^{(1)}+ f_1^{(4)}=F_1 = R_1$$ $$f_2^{(1)}+ f_2^{(2)}+ f_2^{(3)}=F_2 = 0 $$ $$f_3^{(3)}+ f_3^{(4)}+ f_3^{(5)}=F_3 = 1000$$ $$f_4^{(2)}+ f_4^{(5)}+ f_4^{(6)}=F_4 = 0$$ $$f_5^{(6)}=F_5=R_5$$

$$\begin{bmatrix}K_s\end{bmatrix}\begin{Bmatrix}Q_s\end{Bmatrix}=\begin{Bmatrix}F_s \end{Bmatrix}$$

$$ 100\begin{bmatrix}7 &-5 &-2  &0  &0 \\  -5&15  &-6  &-4  &0 \\  -2&-6  &12  &-4  &0 \\  0&-4&-4  &11  &-3 \\ 0&0  &0  &-3  &3 \end{bmatrix}\begin{Bmatrix}u_1\\ u_2\\ u_3\\ u_4\\ u_5\end{Bmatrix}=\begin{Bmatrix}R_1\\ 0\\ 1000\\ 0\\ R_5\end{Bmatrix}$$ $$100\begin{bmatrix}15 &-6  &-4\\ -6  &12  &-4\\-4&-4  &11\\ \end{bmatrix}\begin{Bmatrix} u_2\\ u_3\\ u_4\end{Bmatrix}=\begin{Bmatrix}0\\ 1000\\ 0\end{Bmatrix}$$

Using Matlab Solutions Displacements These are the displacements at $$u_2$$,  $$u_3$$, and    $$u_4$$ respectively. The negative sign shows that the nodes are moving in the direction of the applied force which is against the convention of positive displacement left to right.

Now it is possible to find the forces in the spring. Negative forces denote compression and positive forces denote tension. Forces in Springs $$P=k(u_j-u_i)$$ $$ P^{(1)}=500(-0.8542-0)=-427.1N$$ $$P^{(2)}=400(-0.8750+0.8542)=-8.32N$$ $$P^{(3)}=600(-1.5521+0.8542)=-419N$$ $$P^{(4)}=200(-0.8750-0)=-310N$$ $$P^{(5)}=400(-0.8750+1.5521)=270.8N$$ $$P^{(6)}=300(0+0.8750)=262.5N$$ Now, Reaction Forces $$ f_1^{(1)}+ f_1^{(4)}=F_1 = R_1$$

$$R_1=-427N-310N=-737N$$

$$ f_5^{(6)}=F_5=R_5$$

$$R_5=262.5N$$

Verification with Calfem

a values are the nodal displacements 1 through 5. r values are the reaction forces at the walls. es1-es6 are the forces in the springs. Negative values denote compression and positive values denote tension. These answers match those previously calculated.