--- title: "Truss: force method, double-diagonal truss" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-force-method-50kn-20kn-6m" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-force-method-50kn-20kn-6m site: BCT Engineering --- # Truss: force method, double-diagonal truss Theory of Structures II · Chapter 2: Force method - Asked in: 2073 Shrawan, 12 marks - 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-force-method-50kn-20kn-6m - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine the forces in all members of the truss shown in figure below using the force method. AE is constant for all members. [Figure: truss with joints A, B (left), C, D (top) and E, F (bottom); panels 3 m + 3 m wide and 4 m high, double diagonals; 50 kN downward at C and 20 kN horizontal at D; A hinged, F roller.] ## Short answer Force method with redundants AC and CF forces. Flexibility coefficients computed from virtual work, compatibility equations solved. Final forces: AE=+26.05 kN, EF=+17.78 kN, BC=-2.70 kN, CD=+9.03 kN, AB=-3.61 kN, CE=-18.23 kN, DF=-14.62 kN, BE=+4.51 kN, AC=-10.08 kN, CF=-29.64 kN, ED=+18.28 kN (positive tension). ## Answer 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 +) | Member | $L$ (m) | $S_0$ | $n_1$ | $n_2$ | Final $S$ | |---|---|---|---|---|---| | AE | 3.000 | +20.000 | -0.600 | 0.000 | +26.046 | | EF | 3.000 | 0.000 | 0.000 | -0.600 | +17.785 | | BC | 3.000 | -8.750 | -0.600 | 0.000 | -2.704 | | CD | 3.000 | -8.750 | 0.000 | -0.600 | +9.035 | | AB | 4.000 | -11.667 | -0.800 | 0.000 | -3.606 | | CE | 4.000 | -50.000 | -0.800 | -0.800 | -18.226 | | DF | 4.000 | -38.333 | 0.000 | -0.800 | -14.620 | | BE | 5.000 | +14.583 | +1.000 | 0.000 | +4.507 | | AC | 5.000 | 0.000 | +1.000 | 0.000 | -10.076 | | CF | 5.000 | 0.000 | 0.000 | +1.000 | -29.641 | | ED | 5.000 | +47.917 | 0.000 | +1.000 | +18.275 | #### 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 $S = S_0 + n_1X_1 + n_2X_2$ | Member | Force (kN) | Nature | |---|---|---| | AE | +26.05 | tension | | EF | +17.78 | tension | | BC | -2.70 | compression | | CD | +9.03 | tension | | AB | -3.61 | compression | | CE | -18.23 | compression | | DF | -14.62 | compression | | BE | +4.51 | tension | | AC | -10.08 | compression | | CF | -29.64 | compression | | ED | +18.28 | tension | **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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The truss has m = 11 members, j = 6 joints and r = 3 reactions. What is the degree of static indeterminacy? A. 1 B. 2 C. 3 D. 0 Answer: B. D_s = m + r - 2j = 11 + 3 - 12 = 2, so two redundants (the diagonals AC and CF) are taken. (Key point: Degree of indeterminacy) 2. What do the two compatibility equations f11 X1 + f12 X2 = -D10 and f21 X1 + f22 X2 = -D20 enforce? A. Joint equilibrium at C and D B. Zero total reaction at the roller F C. The redundant diagonals AC and CF must fit, with no net gap or overlap in the real truss D. Equal force in both diagonals Answer: C. The redundants are chosen so the displacements of the cut diagonals in the primary structure are cancelled, which restores compatibility. (Key point: Compatibility condition) 3. How is the final force in each member obtained? A. S = S0 + n1 X1 + n2 X2 B. S = S0 - n1 X1 - n2 X2 without using n C. S = n1 X1 only D. S = S0 divided by (n1 + n2) Answer: A. S0 is the force due to the loads in the primary structure; n1, n2 are forces due to unit loads at the redundants, scaled by X1, X2. (Key point: Superposition of member forces) 4. Which member carries the largest force and what is its nature? A. AE, 26.05 kN compression B. ED, 18.28 kN compression C. CE, 18.23 kN tension D. CF, 29.64 kN compression Answer: D. Solving the equations gives X2 = S_CF = -29.64 kN, the biggest magnitude, and negative means compression. 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