--- title: "Tapered bar analysis using two elements" subject: "Finite Element Method" chapter: 5 key: "semester_8/finite_element_method:tapered-bar-two-elements" source: https://www.bctengineer.com/q/semester_8/finite_element_method/tapered-bar-two-elements site: BCT Engineering --- # Tapered bar analysis using two elements Finite Element Method · Chapter 5: Interpolation Functions - Asked in: year not given, 8 marks - Video: [Lecture Video 4_18ME61_Module 2_Problem on Taper bar using Elimination method_Vinod Biradar](https://www.youtube.com/watch?v=DnfJGwa9wqA) — Sairam MECH Video - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/tapered-bar-two-elements - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-5 ## Question A tapered steel bar of length 400 mm is fixed at its wide end, where the area is 600 mm$^2$. The area decreases linearly to 200 mm$^2$ at the free end. $E = 200$ GPa. An axial load of 15 kN acts at the free end. Model the bar with two elements of equal length, each with area equal to the average of the areas at its ends. Find the nodal displacements and the stress in each element. Compare the tip displacement with the exact value. ## Short answer Use element areas 500 and 300 mm^2, stiffnesses 5e5 and 3e5 N/mm. Nodal displacements: u2=0.030 mm, u3=0.080 mm; element stresses 30 MPa and 50 MPa. Exact tip displacement is 0.0824 mm, about 2.9% higher. ## Answer #### Mesh and element areas Area varies as $A(x) = 600 - x$ (mm$^2$, $x$ in mm from the fixed end). | Element | Length (mm) | End areas (mm$^2$) | Average area (mm$^2$) | |---|---|---|---| | 1 | 200 | 600, 400 | 500 | | 2 | 200 | 400, 200 | 300 | ```text |#### 600 -> 400 -> 200 |###################>---> 15 kN 1 2 3 ``` #### Element stiffness $$ \begin{aligned} k_1 &= \frac{A_1E}{L} = \frac{500(200\times10^3)}{200} = 5\times10^5\ \text{N/mm} \\ k_2 &= \frac{300(200\times10^3)}{200} = 3\times10^5\ \text{N/mm} \end{aligned} $$ #### Global equations with $u_1 = 0$ $$ \begin{bmatrix} 8\times10^5 & -3\times10^5 \\ -3\times10^5 & 3\times10^5 \end{bmatrix}\begin{Bmatrix} u_2 \\ u_3 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 15\,000 \end{Bmatrix} $$ Both elements carry the same force, 15 kN, so: $$ u_2 = \frac{15\,000}{5\times10^5} = 0.03\ \text{mm},\qquad u_3 = u_2 + \frac{15\,000}{3\times10^5} = 0.03 + 0.05 = 0.08\ \text{mm} $$ #### Stresses $$ \sigma_1 = \frac{15\,000}{500} = 30\ \text{MPa},\qquad \sigma_2 = \frac{15\,000}{300} = 50\ \text{MPa} $$ #### Exact solution $$ u_{tip} = \int_0^{400}\frac{P\,dx}{E(600 - x)} = \frac{15\,000}{200\times10^3}\ln\frac{600}{200} = 0.075(1.0986) = 0.08240\ \text{mm} $$ The FE tip value is 0.0800 mm, which is 2.9 % below the exact value. The stress at the fixed end, $15\,000/600 = 25$ MPa, and at the tip, $75$ MPa, differ from the element values (30 and 50 MPa); the error reduces with more elements. **Answer:** $u_2 = 0.030$ mm, $u_3 = 0.080$ mm; $\sigma_1 = 30$ MPa, $\sigma_2 = 50$ MPa; exact tip displacement 0.0824 mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the area of each element chosen when the tapered bar is modelled with two bar elements? A. The area at the free end of that element B. The area at the fixed end of that element C. The average of the areas at its two ends D. The area at the bar's mid-length for both elements Answer: C. Each element is treated as a uniform bar with the average of its end areas: 500 mm^2 for element 1 and 300 mm^2 for element 2. (Key point: Element area) 2. Which expression and value give the stiffness of element 1 (A = 500 mm^2, L = 200 mm, E = 200 GPa)? A. AE/L = 3 x 10^5 N/mm B. AL/E = 0.5 N/mm C. E/(AL) = 2 x 10^3 N/mm D. AE/L = 5 x 10^5 N/mm Answer: D. k = AE/L = 500(200x10^3)/200 = 5 x 10^5 N/mm. (Key point: Element stiffness) 3. Why does each element carry the full 15 kN, and what are u2 and u3? A. Only element 2 carries load; u2 = 0 and u3 = 0.05 mm B. Both carry 15 kN by equilibrium; u2 = 0.05 mm, u3 = 0.08 mm C. Both carry 15 kN by equilibrium; u2 = 0.03 mm, u3 = 0.08 mm D. Load is shared equally, so u2 = 0.015 mm and u3 = 0.04 mm Answer: C. Only the tip is loaded, so the internal force is 15 kN everywhere. u2 = 15000/5x10^5 = 0.03 mm and u3 = 0.03 + 15000/3x10^5 = 0.08 mm. (Key point: Nodal displacements) 4. How do the FE results compare with the exact solution? A. FE tip displacement is above the exact value because the average area is too large B. FE tip displacement 0.08 mm is about 2.9% below the exact 0.0824 mm; element stresses 30 and 50 MPa are average values C. FE stresses are 25 and 75 MPa, equal to the exact end values D. FE tip displacement is exact; stresses are exact Answer: B. Exact tip displacement is 0.075 ln3 = 0.0824 mm, so the 0.08 mm FE value is 2.9% low. Element stresses 30 and 50 MPa are averages, unlike the exact end values 25 and 75 MPa. 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