--- title: "Truss bar forces BG, HC, CF" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-bar-forces-bg-hc-cf" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-bar-forces-bg-hc-cf site: BCT Engineering --- # Truss bar forces BG, HC, CF Theory of Structures II · Chapter 2: Force method - Asked in: 2076 Asoj, 10 marks - Video: [Determine the force in members DC, HC, and HI. | Method of Sections | Statics | Engineers Academy](https://www.youtube.com/watch?v=LACD2yP-4_w) — Engineers Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-bar-forces-bg-hc-cf - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Compute the bar forces in the members BG, HC, and CF of the following loaded truss structure. AE = constant. [Figure: Pratt-type truss with top joints B, C, D, bottom joints A, H, G, F, E; 4 panels of 6.0 m (total 24 m), height 6 m, X-bracing in the panels; 8 kN downward at H; A hinged, E roller.] ## Short answer For the given Pratt truss, the force in BG is -0.393 kN (compression), in HC is +2.435 kN (tension), and in CF is -1.373 kN (compression). ## Answer 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) | Member | $L$ (m) | $S_0$ | $n_1$ | $n_2$ | Final $S$ | |---|---|---|---|---|---| | AH | 6.000 | +6.000 | 0.000 | 0.000 | +6.000 | | HG | 6.000 | +4.000 | -0.707 | 0.000 | +4.278 | | GF | 6.000 | +2.000 | 0.000 | -0.707 | +2.971 | | FE | 6.000 | +2.000 | 0.000 | 0.000 | +2.000 | | BC | 6.000 | -6.000 | -0.707 | 0.000 | -5.722 | | CD | 6.000 | -4.000 | 0.000 | -0.707 | -3.029 | | AB | 8.485 | -8.485 | 0.000 | 0.000 | -8.485 | | DE | 8.485 | -2.828 | 0.000 | 0.000 | -2.828 | | BH | 6.000 | +6.000 | -0.707 | 0.000 | +6.278 | | CG | 6.000 | -2.000 | -0.707 | -0.707 | -0.751 | | DF | 6.000 | 0.000 | 0.000 | -0.707 | +0.971 | | BG | 8.485 | 0.000 | +1.000 | 0.000 | -0.393 | | CH | 8.485 | +2.828 | +1.000 | 0.000 | +2.435 | | CF | 8.485 | 0.000 | 0.000 | +1.000 | -1.373 | | DG | 8.485 | +2.828 | 0.000 | +1.000 | +1.455 | #### 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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How many redundants does the truss have, and which are chosen? A. One: only BG B. Three: all the reactions C. Two (m + r - 2j = 15 + 3 - 16 = 2); BG and CF, leaving CH and DG as the diagonals D. Two: BH and DF Answer: C. Removing one diagonal from each X-braced panel leaves a determinate truss. (Key point: Redundants) 2. What is V_E for 8 kN at H in a 24 m span? A. 6 kN B. 2 kN, from 8 x 6/24 C. 4 kN D. 8 kN Answer: B. Moments about A give V_E x 24 = 8 x 6, hence V_E = 2 kN, V_A = 6 kN. (Key point: Reaction) 3. Why is f12 = 3.000 and not zero? A. Member CG carries n1 = n2 = -0.707, so n1 n2 L = 0.5 x 6 = 3 B. Because BG and CF are in the same panel C. Because the supports are shared D. Because n1 and n2 are both unity everywhere Answer: A. Only a member that carries both unit systems contributes to f12; here that is the vertical CG. (Key point: Coupling term) 4. How is the force in HC found and what is it? A. S = S0 = 2.828 kN B. S = X2 = -1.373 kN C. S = -0.393 + 1.373 = 0.98 kN D. S = S0 + n1 X1 + n2 X2 = 2.828 + (1)(-0.393) = 2.435 kN tension Answer: D. CH is in the same panel as BG, so n1 = 1 and n2 = 0; superposition gives 2.435 kN. 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