--- title: "Two-bar plane truss displacement and member forces" subject: "Finite Element Method" chapter: 3 key: "semester_8/finite_element_method:two-bar-truss-member-forces" source: https://www.bctengineer.com/q/semester_8/finite_element_method/two-bar-truss-member-forces site: BCT Engineering --- # Two-bar plane truss displacement and member forces Finite Element Method · Chapter 3: Direct Stiffness Method: Discrete Finite Elements - Asked in: year not given, 10 marks - Video: [FEA Truss Problems | FEM truss Problems and Solutions | FEm Truss Problems](https://www.youtube.com/watch?v=n9CFPvU-bno) — Shruti Mahesh - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/two-bar-truss-member-forces - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-3 ## Question Two pin-jointed bars form a plane truss. Joints 1 and 2 are fixed supports at (0, 0) m and (4, 0) m. Joint 3 is at (2, 3) m and is joined to joints 1 and 2 by bars 1-3 and 2-3. Both bars have $A = 500\ \text{mm}^2$ and $E = 200$ GPa. A load of 10 kN in the +x direction and 20 kN in the -y direction acts at joint 3. Using the truss element stiffness matrix, find the displacement of joint 3 and the axial force in each bar. ## Short answer With direction cosines c1=0.5547, s1=0.8321; c2=-0.5547, s2=0.8321 and AE/L=27732 N/mm, joint 3 stiffness is assembled. Solve: u3=0.586 mm, v3=-0.521 mm. Bar forces: 3.0 kN (compression) in bar 1-3, 21.0 kN (compression) in bar 2-3. ## Answer #### Geometry Both bars have length $L = \sqrt{2^2 + 3^2} = 3.606$ m $= 3606$ mm. Direction cosines (measured from the support towards joint 3): | Bar | $c = \cos\theta$ | $s = \sin\theta$ | |---|---|---| | 1-3 | $2/3.606 = 0.5547$ | $3/3.606 = 0.8321$ | | 2-3 | $-2/3.606 = -0.5547$ | $0.8321$ | ```text --> 10 kN 3 (2,3) o / \ 20 kN acts / \ downward at 3 / \ (0,0) 1 ^ ^ 2 (4,0) pin pin ``` #### Element stiffness contribution to joint 3 For a truss element, the part belonging to joint 3 is $\dfrac{AE}{L}\begin{bmatrix} c^2 & cs \\ cs & s^2 \end{bmatrix}$. $\dfrac{AE}{L} = \dfrac{500(200\times10^3)}{3606} = 27\,732$ N/mm. $$ \begin{aligned} K_{xx} &= 27\,732(0.5547^2 + 0.5547^2) = 27\,732(0.6154) = 17\,068\ \text{N/mm} \\ K_{yy} &= 27\,732(0.8321^2 + 0.8321^2) = 27\,732(1.3846) = 38\,402\ \text{N/mm} \\ K_{xy} &= 27\,732[(0.5547)(0.8321) + (-0.5547)(0.8321)] = 0 \end{aligned} $$ #### Solve $$ \begin{bmatrix} 17\,068 & 0 \\ 0 & 38\,402 \end{bmatrix}\begin{Bmatrix} u_3 \\ v_3 \end{Bmatrix} = \begin{Bmatrix} 10\,000 \\ -20\,000 \end{Bmatrix} $$ $$ u_3 = 0.5859\ \text{mm},\quad v_3 = -0.5208\ \text{mm} $$ #### Member forces Axial elongation of a bar: $\delta = c\,u_3 + s\,v_3$ (since the other end is fixed). Force $= \dfrac{AE}{L}\delta$. | Bar | $\delta$ (mm) | Force (N) | |---|---|---| | 1-3 | $0.5547(0.5859) + 0.8321(-0.5208) = -0.1083$ | $27\,732(-0.1083) = -3005$ | | 2-3 | $-0.5547(0.5859) + 0.8321(-0.5208) = -0.7585$ | $27\,732(-0.7585) = -21\,033$ | Check equilibrium at joint 3 (bars in compression push the joint away from their supports): horizontal components $3.005(0.5547) - 21.033(0.5547) = -10.0$ kN, balancing the applied $+10$ kN; vertical components $3.005(0.8321) + 21.033(0.8321) = 20.0$ kN, balancing the applied 20 kN downward. **Answer:** $u_3 = 0.586$ mm, $v_3 = -0.521$ mm; bar 1-3 force $= 3.0$ kN (compression); bar 2-3 force $= 21.0$ kN (compression). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For bar 2-3, measured from the support towards joint 3, what are c and s? A. c = -0.5547, s = 0.8321 B. c = 0.5547, s = 0.8321 C. c = -0.8321, s = 0.5547 D. c = 0.8321, s = -0.5547 Answer: A. Joint 2 is at (4,0) and joint 3 at (2,3), so dx = -2 and dy = 3; c = -2/3.606 and s = 3/3.606. (Key point: Direction cosines) 2. Why is Kxy at joint 3 zero? A. The bars are vertical B. The cs terms of the two symmetric bars are equal and opposite and cancel C. The load has no y component D. AE is zero Answer: B. Bar 1-3 gives +cs and bar 2-3 gives -cs, so they sum to zero and the x and y equations decouple. (Key point: Stiffness terms) 3. What is the displacement of joint 3? A. u = -0.586 mm, v = 0.521 mm B. u = 0.521 mm, v = -0.586 mm C. u = 0.586 mm, v = -0.521 mm D. u = 0.586 mm, v = 0.521 mm Answer: C. u = 10000/17068 = 0.586 mm and v = -20000/38402 = -0.521 mm. (Key point: Displacement) 4. How is the axial elongation of a bar computed from the joint displacement when the other end is fixed? A. delta = u + v B. delta = s u + c v C. delta = u - v D. delta = c u + s v Answer: D. The joint displacement is projected on the bar axis: delta = c u3 + s v3. Force = (AE/L) delta. 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