--- title: "Simply supported beam with two elements" subject: "Finite Element Method" chapter: 3 key: "semester_8/finite_element_method:simply-supported-beam-two-elements" source: https://www.bctengineer.com/q/semester_8/finite_element_method/simply-supported-beam-two-elements site: BCT Engineering --- # Simply supported beam with two elements Finite Element Method · Chapter 3: Direct Stiffness Method: Discrete Finite Elements - Asked in: year not given, 8 marks - Video: [Nodal Force Vector for Beam with UDL and Point Load | FEA | Combined UDL & Point Load using FEM](https://www.youtube.com/watch?v=WQMCEG8g7BA) — Mahesh Gadwantikar - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/simply-supported-beam-two-elements - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-3 ## Question A simply supported beam of span 4 m carries a point load of 10 kN at mid-span. $E = 200$ GPa and $I = 8\times10^6\ \text{mm}^4$. Model the beam with two equal beam elements. Find the deflection at mid-span, the rotations at the supports and the support reactions, and compare the deflection with the exact value. ## Short answer With two 2 m beam elements (EI=1.6×10⁶ N·m²), symmetry gives θ₂=0 and θ₃=−θ₁. Solving yields mid-span deflection 8.33 mm downward, support rotations ±6.25×10⁻³ rad. Reactions are 5 kN each. Results match the exact solution PL³/(48EI) exactly because the cubic element is exact for nodal loads. ## Answer #### Data $EI = 200\times10^9 \times 8\times10^{-6} = 1.6\times10^6\ \text{N m}^2$, $L_e = 2$ m for each element. $\dfrac{EI}{L_e^3} = \dfrac{1.6\times10^6}{8} = 2\times10^5$ N/m. Nodes: 1 (left support), 2 (mid-span), 3 (right support). Degrees of freedom: $(v_1, \theta_1, v_2, \theta_2, v_3, \theta_3)$. Boundary conditions: $v_1 = v_3 = 0$. ```text ^ 10 kN | | o======= (1) ===o======= (2) ===o 1 2 3 |<------ 2 m ---->|<------ 2 m -->| ``` #### Element matrix (each element) With $L_e = 2$: $$ [k] = 2\times10^5\begin{bmatrix} 12 & 12 & -12 & 12 \\ 12 & 16 & -12 & 8 \\ -12 & -12 & 12 & -12 \\ 12 & 8 & -12 & 16 \end{bmatrix} $$ #### Reduced system Free DOFs: $\theta_1, v_2, \theta_2, \theta_3$. By symmetry about mid-span, $\theta_2 = 0$ and $\theta_3 = -\theta_1$. Row for $\theta_1$ (element 1, second row), with $v_1 = 0$, $\theta_2 = 0$: $$ 2\times10^5\,(16\theta_1 - 12v_2) = 0 $$ Row for $v_2$ (sum of element 1 third row and element 2 first row), with $v_1 = v_3 = 0$, $\theta_2 = 0$, $\theta_3 = -\theta_1$: $$ 2\times10^5\,(-12\theta_1 + 24v_2 + 12\theta_3) = 2\times10^5\,(-24\theta_1 + 24v_2) = -10\,000 $$ So $\theta_1 = 0.75v_2$, and $-18v_2 + 24v_2 = -0.05$ gives $$ v_2 = -0.008333\ \text{m},\qquad \theta_1 = -0.00625\ \text{rad} $$ #### Reactions $R_1 = 2\times10^5(12v_1 + 12\theta_1 - 12v_2 + 12\theta_2) = 2.4\times10^6(-0.00625 + 0.008333) = 5000$ N, and by symmetry $R_3 = 5000$ N. #### Comparison Exact: $v_{max} = \dfrac{PL^3}{48EI} = \dfrac{10\,000(4)^3}{48(1.6\times10^6)} = 8.333$ mm; exact support rotation $\dfrac{PL^2}{16EI} = 6.25\times10^{-3}$ rad. The FE results match exactly because the cubic element is exact for a load applied at a node. **Answer:** mid-span deflection = 8.33 mm downward; support rotations = 6.25 $\times10^{-3}$ rad (left clockwise, right anticlockwise); reactions = 5 kN each; the FE result equals the exact value. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is EI/Le^3 for each element (EI = 1.6 x 10^6 N m^2, Le = 2 m)? A. 2 x 10^5 N/m B. 8 x 10^5 N/m C. 1.6 x 10^6 N/m D. 4 x 10^5 N/m Answer: A. EI/Le^3 = 1.6 x 10^6 / 8 = 2 x 10^5 N/m. (Key point: Element data) 2. Which relation follows from symmetry about mid-span? A. v2 = 0 B. theta2 = 0 and theta3 = -theta1 C. theta1 = theta3 D. theta2 = theta1 Answer: B. At mid-span the slope is zero by symmetry, and the end rotations are equal in size but opposite in sense. (Key point: Symmetry) 3. What is the mid-span deflection? A. 0.833 mm B. 83.3 mm C. 8.33 mm downward D. 20 mm Answer: C. v2 = -0.008333 m, so the mid-span deflection is 8.33 mm downward. (Key point: Deflection) 4. What are the support reactions? A. 10 kN each B. 2.5 kN each C. 10 kN and 0 D. 5 kN each Answer: D. By symmetry each support carries half of the 10 kN load. 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