--- title: "Truss: rectangular, diagonal 2 cm too short" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-rectangular-lack-of-fit-100kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-rectangular-lack-of-fit-100kn site: BCT Engineering --- # Truss: rectangular, diagonal 2 cm too short Theory of Structures II · Chapter 2: Force method - Asked in: 2066 Bhadra, 10 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-rectangular-lack-of-fit-100kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question A rectangular horizontal truss of span 12 m and height 9 m is with two diagonals and is supported by two hinges fixed at the base. A horizontal force of magnitude 100 kN is acting toward the truss at the left top joint. The diagonal connecting the loaded joint was manufactured 2 cm shorter than the assigned length. Calculate the forces induced in every member assuming Young's modulus and cross-sectional areas of every member to be $2\times10^{5}$ MPa and 1000 mm² respectively. ## Short answer Take H_B as redundant; compatibility includes BD lack-of-fit 20 mm. Final forces (kN): BC=-85.33, CD=-113.77, DA=-10.33, AC=142.21, BD=17.21 (+ tension, – compression); AB zero. ## Answer **Setup:** joints A(0,0) and B(12,0) at the base (both hinges), C(12,9) and D(0,9) at the top. Members: verticals AD, BC; top chord CD; diagonals AC and BD (bottom chord AB is not a member between the two hinges, or if present it carries no force because A and B cannot move). 100 kN acts horizontally to the right at D (the loaded joint). The diagonal BD joining the loaded joint is 20 mm short. $E=2\times10^5$ MPa, $A=1000$ mm², so $EA=2\times10^5$ kN and lengths: AD = BC = 9 m, CD = 12 m, AC = BD = 15 m. #### Degree of indeterminacy $m=5$, $r=4$, $j=4$: $D_s=5+4-8=1$. Take the horizontal reaction at B, $X=H_B$ (to the right +), as the redundant. In the primary truss B is a roller on the base line. #### Forces in the primary truss $S_0$ = bar forces due to the 100 kN load (the lack of fit is treated separately). $S_1$ = bar forces due to a unit force at B acting to the right. | Bar | L (m) | $S_0$ (kN) | $S_1$ | $S_0S_1L$ | $S_1^2L$ | Final $S$ (kN) | |---|---|---|---|---|---|---| | BC | 9.000 | -75.00 | -0.7500 | 506.250 | 5.0625 | -85.33 | | CD | 12.000 | -100.00 | -1.0000 | 1200.000 | 12.0000 | -113.77 | | DA | 9.000 | 0.00 | -0.7500 | 0.000 | 5.0625 | -10.33 | | AC | 15.000 | 125.00 | 1.2500 | 2343.750 | 23.4375 | 142.21 | | BD | 15.000 | 0.00 | 1.2500 | 0.000 | 23.4375 | 17.21 | | **Sum** | | | | **4050.00** | **69.000** | | #### Lack of fit term The bar BD is shorter by 20 mm, i.e. its free elongation is $e_0=-0.02$ m. It contributes $S_1e_0=1.2500\times(-0.02)=-0.02500$ m to the displacement at B. #### Compatibility ($H_B$ gives no horizontal movement at B) $$ \frac{\sum S_0S_1L}{EA}+\sum S_1e_0+X\frac{\sum S_1^2L}{EA}=0 $$ $$ \frac{4050.000}{2\times10^5}+(-0.02500)+X\frac{69.0000}{2\times10^5}=0\;\Rightarrow\; X=H_B=13.77\ \text{kN} $$ (positive: $H_B$ acts to the right, in the same direction as the 100 kN load.) #### Bar forces ($S=S_0+XS_1$, + tension, - compression) BC = -85.33, CD = -113.77, DA = -10.33, AC = 142.21, BD = 17.21 kN. Reactions: $H_B=13.77$ kN (right), $H_A=-113.77$ kN, i.e. 113.77 kN to the left; vertical reactions $V_A=-75.00$ kN and $V_B=75.00$ kN (couple of the 100 kN load). The result was checked with a full stiffness analysis. **Answer: BC = -85.33, CD = -113.77, DA = -10.33, AC = 142.21, BD = 17.21 kN (+ tension, - compression); AB carries zero.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For this hinged-base rectangular truss with two diagonals, what is the degree of indeterminacy and what is chosen as redundant? A. Ds = 1; the horizontal reaction H_B B. Ds = 2; both diagonals C. Ds = 0; the truss is determinate D. Ds = 1; the vertical reaction V_A Answer: A. m + r - 2j = 5 + 4 - 8 = 1. H_B is released so B becomes a roller in the primary truss. (Key point: Degree of indeterminacy and redundant) 2. BD is 20 mm short and S1 for BD is 1.25. What does the lack of fit contribute to the horizontal displacement at B? A. +0.025 m B. -0.020 m C. -0.025 m D. -0.0125 m Answer: C. The term is S1 x e0 = 1.25 x (-0.02) = -0.025 m, because the free elongation is negative for a short bar. (Key point: Lack of fit term) 3. Which equation gives the redundant X = H_B? A. Sum(S0 S1 L)/EA + Sum(S1 e0) + X Sum(S1^2 L)/EA = 0 B. Sum(S0 L)/EA = X C. X = 100 kN divided by the number of bars D. Sum(S1 e0) = 0 only Answer: A. B cannot move horizontally, so the displacement from the load, the lack of fit and the redundant must add to zero. This gives X = 13.77 kN. (Key point: Compatibility equation) 4. How is the final force in each bar obtained, and what is the force in BD? A. S = S0 + X S1; BD = 17.21 kN tension B. S = S0 - X S1; BD = 17.21 kN compression C. S = S0 only; BD = 0 D. S = X S1 only; BD = 142.21 kN Answer: A. Superposition S = S0 + X S1. For BD, S0 = 0 and S1 = 1.25, so S = 13.77 x 1.25 = 17.21 kN tension. 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