Truss bar forces BG, HC, CF

Reading of the figure: bottom joints A (0, 0), H (6, 0), G (12, 0), F (18, 0), E (24, 0); top joints B (6, 6), C (12, 6), D (18, 6); end posts AB and DE; verticals BH, CG, DF; chords AH, HG, GF, FE, BC, CD; X-bracing in the two middle panels (diagonals BG, CH and CF, DG). A is hinged, E is a roller, 8 kN acts downward at H. $AE$ is constant.

$m = 15$, $j = 8$, $r = 3$, so $D_s = m + r - 2j = 2$ (internal). Choose the redundants $X_1$ = force in BG and $X_2$ = force in CF (tension +). The primary structure keeps CH and DG as the single diagonals of the two panels.

Reactions

$V_E = 8\times6/24 = 2$ kN, $V_A = 6$ kN, $H_A = 0$.

Force tables (tension +; $S_0$ = loads on the primary, $n_1$, $n_2$ = unit redundants)

Flexibility coefficients (in $1/AE$)

$$\begin{aligned} f_{11} &= \sum n_1^2 L = 28.971, \quad f_{22} = \sum n_2^2L = 28.971, \quad f_{12} = \sum n_1n_2L = 3.000 \\ \Delta_{10} &= \sum n_1S_0L = 15.515, \quad \Delta_{20} = \sum n_2S_0L = 40.971 \end{aligned}$$

Compatibility equations

$$\begin{aligned} 28.971\,X_1 + 3.000\,X_2 &= -15.515 \\ 3.000\,X_1 + 28.971\,X_2 &= -40.971 \end{aligned}$$ $$X_1 = S_{BG} = -0.393\ \text{kN},\qquad X_2 = S_{CF} = -1.373\ \text{kN}$$

Member forces required

• BG: $S = X_1 = -0.393$ kN (compression)
• CF: $S = X_2 = -1.373$ kN (compression)
• HC (CH): $S = S_0 + n_1X_1 + n_2X_2 = 2.828 + (1)(-0.393) + 0 = 2.435$ kN (tension)

Answer: BG $= -0.393$ kN, HC $= 2.435$ kN, CF $= -1.373$ kN (positive = tension).