Redundant truss: force in member BF

The truss has one degree of redundancy ($m + r - 2j = 10 + 3 - 12 = 1$). Take the force $X$ in the diagonal BF as the redundant and solve it by the force method (unit-load / strain energy).

Assumed data (the figure is not fully legible): joints A, B, C, D on the bottom chord at 0, 3, 7, 10 m; E, F on top, 4 m above B and C; loads 30 kN at B and 50 kN at C; A hinged, D roller; both diagonals BF and CE present in the middle panel. Areas: bottom chord 20 cm², top chord 25 cm², verticals 15 cm², all diagonals 30 cm². E is constant, so it cancels.

Step 1: Released (determinate) truss, loads only

Remove BF. Reactions: $R_D = (30\times3 + 50\times7)/10 = 44$ kN, $R_A = 36$ kN, $H_A = 0$. Solving by joints gives the forces $S_0$ in the table.

Step 2: Unit load on the released truss

Apply a pair of unit tensile forces along BF and find the forces $S_1$ (BF itself has $S_1 = +1$).

Step 3: Compatibility

$$\sum S_0 S_1 \frac{L}{A} + X\sum S_1^2\frac{L}{A} = 0 \quad\Rightarrow\quad X = -\frac{\sum S_0S_1L/A}{\sum S_1^2L/A}$$ $$X = -\frac{(-12.438)}{0.8238} = 15.10\ \text{kN}$$

Final member forces ($S = S_0 + X S_1$, tension +)

AB = 27.00, BC = 16.32, CD = 33.00, EF = -43.68, AE = -45.00, BE = 19.32, CF = 33.32, DF = -55.00, CE = 23.58, BF = 15.10 kN.

Answer: force in BF = 15.10 kN (tension) for the assumed areas. (Checked by a full direct-stiffness analysis of the 10-member truss.)