--- title: "Truss: force method, EA and 2EA members" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-force-method-ea-2ea-50kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-force-method-ea-2ea-50kn site: BCT Engineering --- # Truss: force method, EA and 2EA members Theory of Structures II · Chapter 2: Force method - Asked in: 2069 Chaitra, 18 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-ea-2ea-50kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Find out the member forces in the truss shown in figure below using the force method. The axial rigidity of all vertical and horizontal members is EA and that for all inclined members is 2EA. [Figure: two-panel tall truss ABCDEF, 3 m wide, two panels of 4 m height each, with X-bracing; 50 kN horizontal at C and 50 kN horizontal at B; A hinged, D roller.] ## Short answer Using the force method with diagonal CE as redundant, the member forces (kN, + tension) are: AB 66.67, BC 45.06, DE -200, EF -21.61, BE -66.21, CF -16.21, AE 166.67, BD 0, BF 27.01, CE -56.32 (compression). ## Answer **Assumptions (figure not fully clear):** joints A(0,0), B(0,4), C(0,8) on the left and D(3,0), E(3,4), F(3,8) on the right (3 m wide, two panels of 4 m). Members: verticals AB, BC, DE, EF and horizontals BE, CF (axial rigidity EA); diagonals AE, BD, BF, CE (axial rigidity 2EA, lengths 5 m). Both 50 kN loads act horizontally to the right at B and C. A is a hinge and D a roller (vertical reaction only). There is no member joining A and D (the supports hold them). #### Degree of indeterminacy $m=10$, $r=3$, $j=6$: $D_s=m+r-2j=10+3-12=1$. Take the force $X$ in the diagonal CE as the redundant. #### Reactions $\sum M_A=0$: $R_D\times3=50\times8+50\times4\Rightarrow R_D=200$ kN (up), $R_{Ay}=200$ kN (down), $H_A=100$ kN (to the left). #### Forces $S_0$ (CE removed, loads only) and $S_1$ (unit tension in CE) Solved joint by joint. Compatibility: $\sum S_0S_1\dfrac{L}{AE}+X\sum S_1^2\dfrac{L}{AE}=0$ (the $L/AE$ values are in units of $1/EA$, with the $2EA$ of the diagonals already taken into account). | Member | L (m) | AE | $S_0$ (kN) | $S_1$ | $S_0S_1L/AE$ | $S_1^2L/AE$ | Final $S$ (kN) | |---|---|---|---|---|---|---|---| | AB | 4.000 | EA | 66.67 | 0.000 | 0.000 | 0.0000 | 66.67 | | BC | 4.000 | EA | 0.00 | -0.800 | 0.000 | 2.5600 | 45.06 | | DE | 4.000 | EA | -200.00 | 0.000 | 0.000 | 0.0000 | -200.00 | | EF | 4.000 | EA | -66.67 | -0.800 | 213.333 | 2.5600 | -21.61 | | BE | 3.000 | EA | -100.00 | -0.600 | 180.000 | 1.0800 | -66.21 | | CF | 3.000 | EA | -50.00 | -0.600 | 90.000 | 1.0800 | -16.21 | | AE | 5.000 | 2EA | 166.67 | 0.000 | 0.000 | 0.0000 | 166.67 | | BD | 5.000 | 2EA | 0.00 | 0.000 | 0.000 | 0.0000 | 0.00 | | BF | 5.000 | 2EA | 83.33 | 1.000 | 208.333 | 2.5000 | 27.01 | | CE | 5.000 | 2EA | 0.00 | 1.000 | 0.000 | 2.5000 | -56.32 | | **Sum** | | | | | **691.667** | **12.2800** | | $$ X=-\frac{691.667}{12.2800}=-56.32\ \text{kN} $$ #### Final member forces ($S=S_0+XS_1$; + tension, - compression) Verticals: AB = 66.67, BC = 45.06, DE = -200.00, EF = -21.61 kN. Horizontals: BE = -66.21, CF = -16.21 kN. Diagonals: AE = 166.67, BD = 0.00, BF = 27.01, CE = -56.32 kN. As a check, the whole truss was also analysed by the direct stiffness method and gives identical forces. **Answer: CE = -56.32 kN (compression), AE = 166.67 kN, BF = 27.01 kN, BD = 0.00 kN, verticals and horizontals as tabulated.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Using D_s = m + r - 2j, what is the degree of indeterminacy of this truss? A. 0 B. 1 C. 2 D. 3 Answer: B. m = 10, r = 3, j = 6 gives 10 + 3 - 12 = 1, so one redundant such as the force in CE. (Key point: Degree of indeterminacy) 2. What equation is used to find the redundant X? A. Sum of S0 S1 L/AE + X sum of S1^2 L/AE = 0 B. Sum of S0 L/AE = X C. Sum of S1 L/AE = 0 D. X = sum of S0 S1 / sum of S1 Answer: A. The relative displacement at the cut member must vanish, giving X = -sum(S0 S1 L/AE) / sum(S1^2 L/AE). (Key point: Compatibility equation) 3. How do the different rigidities (EA for verticals and horizontals, 2EA for diagonals) enter the calculation? A. They are ignored B. Each member's L/AE uses its own rigidity, so diagonals' terms are halved C. Only the diagonals are included D. Rigidity is taken as EA for all members Answer: B. The sums use L/(AE) of each member, so a member with 2EA contributes half the L/EA of one with EA. (Key point: Different axial rigidities) 4. What are S0 and S1? A. S0: forces with the redundant removed under loads; S1: forces due to a unit tension in the redundant member B. S0: final forces; S1: forces from the temperature change C. S0: forces due to unit load; S1: forces due to the real loads D. S0: reactions; S1: member lengths Answer: A. Final forces are S = S0 + X S1, superposing the released truss under loads and the unit redundant. (Key point: Role of S0 and S1) 5. What is the force in the redundant diagonal CE? A. +56.32 kN (tension) B. -56.32 kN (compression) C. 0 kN D. -166.67 kN Answer: B. X = -691.667/12.28 = -56.32 kN, i.e. compression. 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