Indeterminate truss: bar forces and reactions

Reading of the figure: A (0, 0) hinged and B (6.5, 0) roller at the base, C (6.5, 5) and D (0, 5) at the top; members AB, CD (6.5 m), BC, DA (5 m), diagonals AC and BD (8.20 m). The 400 kN load acts horizontally at C (towards the right). $E = 240\times10^6$ kN/m$^2$. Areas: BD $= 20$ cm$^2$, all others $15$ cm$^2$. So $EA_{BD} = 480000$ kN and $EA = 360000$ kN for the other members.

$D_s = 6 + 3 - 8 = 1$. Take the force in BD as the redundant $X$ (tension +).

Reactions

$\Sigma M_A = 0$: $V_B(6.5) = 400(5) \Rightarrow V_B = 307.69$ kN ($\uparrow$); $V_A = 307.69$ kN ($\downarrow$); $H_A = 400$ kN ($\leftarrow$).

Force table (tension +; $S_0$ with BD removed)

$$\Delta_{10} = \sum\frac{S_0nL}{EA} = 0.014101\ \text{m},\qquad f_{11} = \sum\frac{n^2L}{EA} = 7.2877\times10^{-5}\ \text{m/kN}$$

Compatibility

$$\Delta_{10} + f_{11}X = 0 \ \Rightarrow\ X = -\frac{0.014101}{7.2877\times10^{-5}} = -193.49\ \text{kN}$$

Final forces $S = S_0 + nX$

Reactions

$H_A = 400$ kN ($\leftarrow$), $V_A = 307.69$ kN ($\downarrow$), $V_B = 307.69$ kN ($\uparrow$).

Answer: AB $= +153.37$, BC $= -189.72$, CD $= +153.37$, DA $= +117.98$, AC $= +311.16$, BD $= -193.49$ kN (positive = tension).