Displacements of a spring assembly

Element stiffness matrices

For a spring element, $[k^{(e)}] = k\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$.

 wall |--k1--o--k2--o--k3--o--> 600 N
      1      2      3      4
                    --> 400 N (axial, at node 3)

(a) Global stiffness matrix (N/mm), degrees of freedom $u_1, \dots, u_4$

$$[K] = \begin{bmatrix} 200 & -200 & 0 & 0 \\ -200 & 500 & -300 & 0 \\ 0 & -300 & 400 & -100 \\ 0 & 0 & -100 & 100 \end{bmatrix}$$

(b) Apply boundary condition $u_1 = 0$

Delete row and column 1:

$$\begin{bmatrix} 500 & -300 & 0 \\ -300 & 400 & -100 \\ 0 & -100 & 100 \end{bmatrix}\begin{Bmatrix} u_2 \\ u_3 \\ u_4 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 400 \\ 600 \end{Bmatrix}$$

From row 3: $-100u_3 + 100u_4 = 600$, so $u_4 = u_3 + 6$.

Adding rows 2 and 3 gives $-300u_2 + 300u_3 = 1000$, so $u_3 = u_2 + 3.333$.

Row 1: $500u_2 - 300(u_2 + 3.333) = 0$ gives $200u_2 = 1000$, so $u_2 = 5.000$ mm.

Then $u_3 = 8.333$ mm and $u_4 = 14.333$ mm.

(c) Reaction

From row 1 of the full system: $R_1 = 200u_1 - 200u_2 = -200(5) = -1000$ N. The wall pulls back with 1000 N, which equals the total applied load $400 + 600$.

(d) Spring forces

Check at node 3: $1000 - 600 = 400$ N, equal to the applied load.

Answer: $u_2 = 5.00$ mm, $u_3 = 8.33$ mm, $u_4 = 14.33$ mm; reaction at node 1 = 1000 N (acting towards the wall, opposite to the loads); spring forces 1000 N, 1000 N and 600 N.