--- title: "Truss: nodal displacements, reactions, member forces" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:truss-fem-displacements-forces" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/truss-fem-displacements-forces site: BCT Engineering --- # Truss: nodal displacements, reactions, member forces Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2073 Magh, 10 marks - Video: [Find member forces, Displacements, and ILD for truss using Truss 101 software | EXAMPLE -1](https://www.youtube.com/watch?v=7CDHJ0r9gIc) — CIVIL MANIA - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/truss-fem-displacements-forces - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Determine the nodal displacements, reaction forces and member forces of the given truss structure, loaded as shown in the figure. Given that for each member, sectional area $A = 2\times10^{-3}$ m$^2$ and modulus of elasticity $E = 2\times10^5$ MPa. [Figure: three-member truss; node 1 at the origin (pinned), node 2 at 4 m to the right of node 1, node 3 at 3 m above node 1 (pinned); members 1-2, 2-3 and 1-3; at node 2 a 50 kN upward load and a 100 kN horizontal load (to the right)] ## Short answer Direct stiffness solution gives u2=1.667 mm right and v2=3.958 mm up. Member forces are 166.67 kN tension in 1-2, 83.33 kN compression in 2-3, and zero in 1-3. Reactions are (-166.67, 0) kN at node 1 and (66.67, -50) kN at node 3. ## Answer **Data:** node 1 $(0,0)$ pinned, node 2 $(4,0)$ loaded, node 3 $(0,3)$ pinned. $A=2\times10^{-3}$ m$^2$, $E=2\times10^5$ MPa $=2\times10^8$ kN/m$^2$, so $EA=4\times10^5$ kN. Loads at node 2: $F_x=100$ kN, $F_y=50$ kN. Units: kN, m. ```text 3 (0,3) |\ | \ 5 m 3 m \ | \ 50 kN 1----2 -> 100 kN 4 m ``` #### Element data Direction cosines $c=\dfrac{x_j-x_i}{L}$, $s=\dfrac{y_j-y_i}{L}$; axial stiffness $EA/L$. | Member | $L$ (m) | $c$ | $s$ | $EA/L$ (kN/m) | |---|---|---|---|---| | 1-2 | 4 | 1.0 | 0.0 | 100000 | | 2-3 | 5 | -0.8 | 0.6 | 80000 | | 1-3 | 3 | 0.0 | 1.0 | 133333 | Element stiffness in global axes: $$ [k]=\frac{EA}{L}\begin{bmatrix}c^2&cs&-c^2&-cs\\cs&s^2&-cs&-s^2\\-c^2&-cs&c^2&cs\\-cs&-s^2&cs&s^2\end{bmatrix} $$ - Member 1-2: $10^5\begin{bmatrix}1&0&-1&0\\0&0&0&0\\-1&0&1&0\\0&0&0&0\end{bmatrix}$ (DOFs $u_1,v_1,u_2,v_2$) - Member 2-3: $10^4\begin{bmatrix}5.12&-3.84&-5.12&3.84\\-3.84&2.88&3.84&-2.88\\-5.12&3.84&5.12&-3.84\\3.84&-2.88&-3.84&2.88\end{bmatrix}$ (DOFs $u_2,v_2,u_3,v_3$) - Member 1-3: $1.3333\times10^5$ in the $v_1,v_3$ terms only. #### Assembly and solution Nodes 1 and 3 are pinned ($u_1=v_1=u_3=v_3=0$); free DOFs are $u_2,v_2$. Adding the node-2 terms of members 1-2 and 2-3: $$ \begin{bmatrix}151\,200&-38\,400\\-38\,400&28\,800\end{bmatrix}\begin{Bmatrix}u_2\\v_2\end{Bmatrix}=\begin{Bmatrix}100\\50\end{Bmatrix} $$ $$ u_2=1.6667\ \text{mm}\ (\rightarrow),\qquad v_2=3.9583\ \text{mm}\ (\uparrow) $$ #### Member forces $F=\dfrac{EA}{L}\,[-c\ \ -s\ \ c\ \ s]\{d\}$: - Member 1-2: $166.67$ kN (tension) - Member 2-3: $-83.33$ kN (compression) - Member 1-3: $0.00$ kN (zero force, both ends fixed in the vertical direction) #### Reactions $\{R\}=[K]\{d\}-\{F\}$ - Node 1: $R_x=-166.67$ kN, $R_y=0.00$ kN - Node 3: $R_x=66.67$ kN, $R_y=-50.00$ kN Check: $\Sigma F_x=-166.67+66.67+100=0$; $\Sigma F_y=0-50+50=0$; $\Sigma M_1=50\times4-66.67\times3=0$. **Answer:** $u_2=1.667$ mm, $v_2=3.958$ mm; member forces 166.67 kN (T), 83.33 kN (C), 0; reactions $(-166.67,\,0)$ kN at node 1 and $(66.67,\,-50)$ kN at node 3. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For member 2-3 (node 2 at (4,0), node 3 at (0,3)), what are c and s? A. c = 0.8, s = 0.6 B. c = -0.8, s = 0.6 C. c = -0.6, s = 0.8 D. c = 0.8, s = -0.6 Answer: B. c = (0-4)/5 = -0.8 and s = (3-0)/5 = 0.6. (Key point: Direction cosines) 2. Why are only u2 and v2 unknown? A. Node 2 is the only node with load and it is free; nodes 1 and 3 are pinned B. Node 2 is pinned C. Member 1-3 removes them D. Only horizontal motion is possible Answer: A. Pins at 1 and 3 fix u1, v1, u3, v3, leaving the two DOFs of node 2. (Key point: Free DOFs) 3. Why is member 1-3 a zero-force member? A. Its area is zero B. Both ends have v = 0, so it has no change in length (it is vertical and only u moves, which does not stretch it) C. It is in compression D. It is the longest member Answer: B. Node 1 and node 3 are both pinned, so the member does not elongate and the force is 0. (Key point: Zero-force member) 4. Which check confirms the reactions (R1x = -166.67, R3x = 66.67, R3y = -50)? A. Sum Fx = -166.67 + 66.67 + 100 = 0 and Sum Fy = 0 - 50 + 50 = 0 B. Sum Fx = 100 only C. The member forces sum to zero D. R3y must equal 100 Answer: A. Reactions plus applied loads (100 kN horizontal, 50 kN vertical) must give zero in both directions. 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