Two-bar plane truss displacement and member forces

Geometry

Both bars have length $L = \sqrt{2^2 + 3^2} = 3.606$ m $= 3606$ mm. Direction cosines (measured from the support towards joint 3):

                   --> 10 kN
           3 (2,3) o
                  / \     20 kN acts
                 /   \    downward at 3
                /     \
     (0,0)  1  ^       ^  2  (4,0)
              pin     pin

Element stiffness contribution to joint 3

For a truss element, the part belonging to joint 3 is $\dfrac{AE}{L}\begin{bmatrix} c^2 & cs \\ cs & s^2 \end{bmatrix}$.

$\dfrac{AE}{L} = \dfrac{500(200\times10^3)}{3606} = 27\,732$ N/mm.

$$\begin{aligned} K_{xx} &= 27\,732(0.5547^2 + 0.5547^2) = 27\,732(0.6154) = 17\,068\ \text{N/mm} \\ K_{yy} &= 27\,732(0.8321^2 + 0.8321^2) = 27\,732(1.3846) = 38\,402\ \text{N/mm} \\ K_{xy} &= 27\,732[(0.5547)(0.8321) + (-0.5547)(0.8321)] = 0 \end{aligned}$$

Solve

$$\begin{bmatrix} 17\,068 & 0 \\ 0 & 38\,402 \end{bmatrix}\begin{Bmatrix} u_3 \\ v_3 \end{Bmatrix} = \begin{Bmatrix} 10\,000 \\ -20\,000 \end{Bmatrix}$$ $$u_3 = 0.5859\ \text{mm},\quad v_3 = -0.5208\ \text{mm}$$

Member forces

Axial elongation of a bar: $\delta = c\,u_3 + s\,v_3$ (since the other end is fixed). Force $= \dfrac{AE}{L}\delta$.

Check equilibrium at joint 3 (bars in compression push the joint away from their supports): horizontal components $3.005(0.5547) - 21.033(0.5547) = -10.0$ kN, balancing the applied $+10$ kN; vertical components $3.005(0.8321) + 21.033(0.8321) = 20.0$ kN, balancing the applied 20 kN downward.

Answer: $u_3 = 0.586$ mm, $v_3 = -0.521$ mm; bar 1-3 force $= 3.0$ kN (compression); bar 2-3 force $= 21.0$ kN (compression).