Truss: force method with EA = 10^5 kN

Reading of the figure: A (0, 0) hinged, B (4, 0) roller, D (4, 3) and C (0, 3) at the top; members AB, BD, DC, CA and the two diagonals CB and AD (a single diagonal would leave the truss determinate, so both are taken). The 50 kN load acts horizontally at D towards the right. $D_s = 6 + 3 - 8 = 1$. Take the force in AD as the redundant $X$. $EA = 10^5$ kN is the same for all members, so it cancels from the forces.

Reactions

$\Sigma M_A = 0$: $V_B(4) = 50(3) \Rightarrow V_B = 37.5$ kN ($\uparrow$); $V_A = 37.5$ kN ($\downarrow$); $H_A = 50$ kN ($\leftarrow$).

Tables (tension +)

$$f_{11} = \sum\frac{n^2L}{AE} = \frac{17.280}{AE} = 1.728\times10^{-4}\ \text{m/kN}, \qquad \Delta_{10} = \sum\frac{S_0nL}{AE} = \frac{-700.000}{AE}$$

Compatibility

$$\Delta_{10} + f_{11}X = 0 \ \Rightarrow\ X = -\frac{-700.000}{17.280} = 40.51\ \text{kN}$$

Final forces $S = S_0 + nX$

Answer: AD $= 40.51$ kN (T); AB $= +17.59$ kN, BD $= -24.31$ kN, DC $= +17.59$ kN, CA $= +13.19$ kN, CB $= -21.99$ kN (positive = tension).