Truss: force method, double-diagonal truss

Reading of the figure: bottom joints A (0, 0), E (3, 0), F (6, 0); top joints B (0, 4), C (3, 4), D (6, 4). Members: chords AE, EF, BC, CD; verticals AB, CE, DF; and two crossing diagonals in each panel (BE, AC and CF, ED). A is hinged, F is a roller. Loads: 50 kN downward at C and 20 kN horizontal at D (towards the right). $AE$ is constant.

$m = 11$, $j = 6$, $r = 3$, so $D_s = 11 + 3 - 12 = 2$. Take $X_1$ = force in AC and $X_2$ = force in CF (tension +).

Reactions

$\Sigma M_A = 0$: $V_F(6) = 50(3) + 20(4) \Rightarrow V_F = 38.333$ kN ($\uparrow$); $V_A = 50 - 38.333 = 11.667$ kN ($\uparrow$); $H_A = 20$ kN ($\leftarrow$).

Force tables (tension +)

Flexibility coefficients (in $1/AE$ units)

$$\begin{aligned} f_{11} &= \sum n_1^2L = 17.280, \quad f_{22} = \sum n_2^2L = 17.280, \quad f_{12} = \sum n_1n_2L = 2.560 \\ \Delta_{10} &= \sum S_0n_1L = 250.000, \quad \Delta_{20} = \sum S_0n_2L = 538.000 \end{aligned}$$

Compatibility

$$\begin{aligned} 17.280\,X_1 + 2.560\,X_2 &= -(250.000) \\ 2.560\,X_1 + 17.280\,X_2 &= -(538.000) \end{aligned}$$ $$X_1 = S_{AC} = -10.08\ \text{kN},\qquad X_2 = S_{CF} = -29.64\ \text{kN}$$

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

Answer: AE $= +26.05$, EF $= +17.78$, BC $= -2.70$, CD $= +9.03$, AB $= -3.61$, CE $= -18.23$, DF $= -14.62$, BE $= +4.51$, AC $= -10.08$, CF $= -29.64$, ED $= +18.28$ kN (positive = tension).