Truss: force method, 4 m x 3 m panel

Reading of the figure: A (0, 0) hinged and D (4, 0) roller, B (0, 3) and C (4, 3) at the top; members AB, CD (3 m), BC, AD (4 m) and diagonals AC, BD (5 m). The 50 kN load acts horizontally at C (towards the right). $AE$ is constant. $D_s = 6 + 3 - 8 = 1$. Take the force in AC as the redundant $X$.

Reactions

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

Force tables (tension +; $S_0$ with AC removed; $n$ for $X=1$)

$$f_{11} = \frac{\sum n^2L}{AE} = \frac{17.280}{AE}, \qquad \Delta_{10} = \frac{\sum S_0nL}{AE} = \frac{-700.000}{AE}$$

Compatibility

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

Final forces $S = S_0 + nX$

Answer: AC $= +40.51$ kN, BD $= -21.99$ kN, AB $= +13.19$ kN, BC $= +17.59$ kN, CD $= -24.31$ kN, AD $= +17.59$ kN (positive = tension).