Truss: force method, 60 kN at D

Reading of the figure: A (0, 0) hinged, E (8, 0) roller; D (4, 0) on the bottom chord; top joints B (2, 3) and C (6, 3); members AB, BC, CE (top and sides), AD, DE (bottom chord), BD, CD and the two crossing diagonals AC and BE. The 60 kN load acts downward at D. $EA$ is constant. (The member arrangement is read from the description; the method is the same for any arrangement.)

$m = 9$, $j = 5$, $r = 3$, so $D_s = m + r - 2j = 2$. Choose $X_1$ = force in AC and $X_2$ = force in BE (tension +).

Reactions

By symmetry $V_A = V_E = 30$ kN and $H_A = 0$.

Force tables (tension +)

Flexibility coefficients (in $1/AE$ units)

$$\begin{aligned} f_{11} &= f_{22} = 12.677, \qquad f_{12} = f_{21} = -0.661 \\ \Delta_{10} &= \sum S_0n_1L = 117.576, \qquad \Delta_{20} = \sum S_0n_2L = 117.576 \end{aligned}$$

Compatibility

$$\begin{aligned} 12.677\,X_1 + (-0.661)\,X_2 &= -117.576 \\ -0.661\,X_1 + 12.677\,X_2 &= -117.576 \end{aligned}$$ $$X_1 = S_{AC} = -9.78\ \text{kN},\qquad X_2 = S_{BE} = -9.78\ \text{kN}$$

Final member forces $S = S_0 + n_1X_1 + n_2X_2$

Answer: AB $= CE = -30.80$ kN, BC $= -28.33$ kN, AD $= DE = +25.83$ kN, BD $= CD = +36.06$ kN, AC $= BE = -9.78$ kN (positive = tension).