--- title: "Indeterminate truss: bar forces and reactions" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:indeterminate-truss-bar-forces-400kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/indeterminate-truss-bar-forces-400kn site: BCT Engineering --- # Indeterminate truss: bar forces and reactions Theory of Structures II · Chapter 2: Force method - Asked in: 2072 Chaitra, 12 marks - Video: [SA26: Force Method (Truss Analysis)](https://www.youtube.com/watch?v=AuBcV1oehD8) — Dr. Structure - 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/indeterminate-truss-bar-forces-400kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine the bar forces and reactions that develop in the statically indeterminate truss shown in figure below. [Figure: rectangular truss ABCD, 6.5 m wide and 5 m high, with both diagonals; A hinged, B roller; 400 kN horizontal at C. Cross-sectional area: member BD = 20 cm², other members = 15 cm²; Young's modulus = $240\times10^{6}\ \text{kN/m}^2$.] ## Short answer Force method with redundant BD force. Compatibility gives X = -193.49 kN. Final forces: AB=+153.37 kN, BC=-189.72 kN, CD=+153.37 kN, DA=+117.98 kN, AC=+311.16 kN, BD=-193.49 kN (positive tension). Reactions: H_A=400 kN left, V_A=307.69 kN down, V_B=307.69 kN up. ## Answer Reading of the figure: A (0, 0) hinged and B (6.5, 0) roller at the base, C (6.5, 5) and D (0, 5) at the top; members AB, CD (6.5 m), BC, DA (5 m), diagonals AC and BD (8.20 m). The 400 kN load acts horizontally at C (towards the right). $E = 240\times10^6$ kN/m$^2$. Areas: BD $= 20$ cm$^2$, all others $15$ cm$^2$. So $EA_{BD} = 480000$ kN and $EA = 360000$ kN for the other members. $D_s = 6 + 3 - 8 = 1$. Take the force in BD as the redundant $X$ (tension +). #### Reactions $\Sigma M_A = 0$: $V_B(6.5) = 400(5) \Rightarrow V_B = 307.69$ kN ($\uparrow$); $V_A = 307.69$ kN ($\downarrow$); $H_A = 400$ kN ($\leftarrow$). #### Force table (tension +; $S_0$ with BD removed) | Member | $L$ (m) | $EA$ (kN) | $S_0$ (kN) | $n$ | $S_0nL/EA\ (\times10^{-3})$ | $n^2L/EA\ (\times10^{-3})$ | |---|---|---|---|---|---|---| | AB | 6.500 | 360000 | 0.00 | -0.7926 | +0.0000 | 0.0113 | | BC | 5.000 | 360000 | -307.69 | -0.6097 | +2.6056 | 0.0052 | | CD | 6.500 | 360000 | 0.00 | -0.7926 | +0.0000 | 0.0113 | | DA | 5.000 | 360000 | 0.00 | -0.6097 | +0.0000 | 0.0052 | | AC | 8.201 | 360000 | +504.65 | +1.0000 | +11.4957 | 0.0228 | | BD | 8.201 | 480000 | +0.00 | +1.0000 | +0.0000 | 0.0171 | $$ \Delta_{10} = \sum\frac{S_0nL}{EA} = 0.014101\ \text{m},\qquad f_{11} = \sum\frac{n^2L}{EA} = 7.2877\times10^{-5}\ \text{m/kN} $$ #### Compatibility $$ \Delta_{10} + f_{11}X = 0 \ \Rightarrow\ X = -\frac{0.014101}{7.2877\times10^{-5}} = -193.49\ \text{kN} $$ #### Final forces $S = S_0 + nX$ | Member | Force (kN) | Nature | |---|---|---| | AB | +153.37 | tension | | BC | -189.72 | compression | | CD | +153.37 | tension | | DA | +117.98 | tension | | AC | +311.16 | tension | | BD | -193.49 | compression | #### Reactions $H_A = 400$ kN ($\leftarrow$), $V_A = 307.69$ kN ($\downarrow$), $V_B = 307.69$ kN ($\uparrow$). **Answer: AB $= +153.37$, BC $= -189.72$, CD $= +153.37$, DA $= +117.98$, AC $= +311.16$, BD $= -193.49$ kN (positive = tension).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 6 members, 4 joints and 3 reaction components, what is D_s and what is taken as the redundant? A. D_s = 2, reactions at A B. D_s = 0, none C. D_s = 3, the three reactions D. D_s = 1, the force in BD Answer: D. D_s = 6 + 3 - 8 = 1, so one redundant, here the diagonal BD. (Key point: Degree of indeterminacy) 2. How do the different areas (BD = 20 cm^2, others 15 cm^2) enter the solution? A. They change the reactions B. They cancel out of the equations C. They change the loads D. Each term S0 n L / EA and n^2 L / EA uses its own EA, with EA of BD = 480000 kN Answer: D. Flexibility sums are member by member with L/EA, so BD uses a larger EA than the others. (Key point: Unequal member areas) 3. From D10 + f11 X = 0 the redundant is X = -193.49 kN. What does the sign mean? A. BD is in tension B. BD is in compression C. BD carries no force D. BD has failed Answer: B. Tension is positive, so the negative value means BD is in compression. (Key point: Compatibility and sign of X) 4. How is V_B found for the 400 kN horizontal load at C (5 m high, 6.5 m wide)? A. V_B = 400 x 6.5 / 5 B. V_B = 400 / 2 C. V_B = 400 x 5 / 6.5 = 307.69 kN upward, from moments about A D. V_B = 0 Answer: C. Taking moments about A: V_B(6.5) = 400(5), so V_B = 307.69 kN, and V_A is equal and downward. 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