--- title: "Axisymmetric triangle strains and stresses at centroid" subject: "Finite Element Method" chapter: 6 key: "semester_8/finite_element_method:axisymmetric-triangle-centroid-strains" source: https://www.bctengineer.com/q/semester_8/finite_element_method/axisymmetric-triangle-centroid-strains site: BCT Engineering --- # Axisymmetric triangle strains and stresses at centroid Finite Element Method · Chapter 6: Applications in Solid Mechanics - Asked in: year not given, 8 marks - Video: [Axisymmetric (2D) element problem | Finite element Analysis | FEA in Tamil](https://www.youtube.com/watch?v=Bf1ZiBwAFQw) — Focus Academy Lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/axisymmetric-triangle-centroid-strains - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-6 ## Question An axisymmetric triangular element has nodes 1 ($r = 20$, $z = 0$), 2 ($r = 40$, $z = 0$) and 3 ($r = 20$, $z = 30$), coordinates in mm. $E = 200$ GPa and $\nu = 0.3$. The nodal displacements (mm) are $u_1 = 0.01$, $w_1 = 0$, $u_2 = 0.02$, $w_2 = 0.01$, $u_3 = 0$, $w_3 = 0.015$. Find the area, the centroid, the matrix $[B]$ at the centroid, the strains and the stresses at the centroid. ## Short answer Area A=300 mm^2, centroid (26.67,10) mm; the centroidal [B] has hoop-row coefficients 7.5. Strains: epsilon_r=5.0e-4, epsilon_theta=3.75e-4, epsilon_z=5.0e-4, gamma_rz=1.67e-4. Stresses: sigma_r=235.6, sigma_theta=216.3, sigma_z=235.6, tau_rz=12.8 MPa. ## Answer #### Area and centroid $$ 2A = r_1(z_2 - z_3) + r_2(z_3 - z_1) + r_3(z_1 - z_2) = 20(-30) + 40(30) + 0 = 600 \Rightarrow A = 300\ \text{mm}^2 $$ $$ \bar r = \frac{20 + 40 + 20}{3} = 26.67\ \text{mm},\qquad \bar z = \frac{0 + 0 + 30}{3} = 10\ \text{mm} $$ #### Coefficients $\alpha_1 = r_2z_3 - r_3z_2 = 1200$, $\alpha_2 = r_3z_1 - r_1z_3 = -600$, $\alpha_3 = r_1z_2 - r_2z_1 = 0$. | $i$ | $\beta_i = z_j - z_k$ | $\gamma_i = r_k - r_j$ | |---|---|---| | 1 | $0 - 30 = -30$ | $20 - 40 = -20$ | | 2 | $30 - 0 = 30$ | $20 - 20 = 0$ | | 3 | $0 - 0 = 0$ | $40 - 20 = 20$ | Hoop-strain coefficient at the centroid, $f_i = \dfrac{\alpha_i}{\bar r} + \beta_i + \dfrac{\gamma_i\bar z}{\bar r}$: $$ f_1 = \frac{1200}{26.67} - 30 - \frac{200}{26.67} = 45 - 30 - 7.5 = 7.5,\quad f_2 = -22.5 + 30 + 0 = 7.5,\quad f_3 = 0 + 0 + 7.5 = 7.5 $$ (Each is $7.5$ because $N_i = 1/3$ at the centroid, so $f_i = 2A/(3\bar r) = 7.5$.) #### $[B]$ at the centroid $$ [\bar B] = \frac{1}{600}\begin{bmatrix} -30 & 0 & 30 & 0 & 0 & 0 \\ 7.5 & 0 & 7.5 & 0 & 7.5 & 0 \\ 0 & -20 & 0 & 0 & 0 & 20 \\ -20 & -30 & 0 & 30 & 20 & 0 \end{bmatrix} $$ #### Strains $$ \begin{aligned} \varepsilon_r &= \frac{-30(0.01) + 30(0.02)}{600} = 5.0\times10^{-4} \\ \varepsilon_\theta &= \frac{7.5(0.01 + 0.02 + 0)}{600} = 3.75\times10^{-4} \\ \varepsilon_z &= \frac{-20(0) + 0(0.01) + 20(0.015)}{600} = 5.0\times10^{-4} \\ \gamma_{rz} &= \frac{-20(0.01) + (-30)(0) + 0 + 30(0.01) + 0 + 0}{600} = 1.67\times10^{-4} \end{aligned} $$ #### Stresses $\dfrac{E}{(1 + \nu)(1 - 2\nu)} = 384\,615$ MPa. $$ \begin{aligned} \sigma_r &= 384\,615\,[0.7(5.0) + 0.3(3.75 + 5.0)]\times10^{-4} = 235.6\ \text{MPa} \\ \sigma_\theta &= 384\,615\,[0.7(3.75) + 0.3(5.0 + 5.0)]\times10^{-4} = 216.3\ \text{MPa} \\ \sigma_z &= 384\,615\,[0.7(5.0) + 0.3(5.0 + 3.75)]\times10^{-4} = 235.6\ \text{MPa} \\ \tau_{rz} &= 384\,615(0.2)(1.67\times10^{-4}) = 12.8\ \text{MPa} \end{aligned} $$ **Answer:** $A = 300\ \text{mm}^2$, centroid $(26.67,\ 10)$ mm; $\varepsilon_r = 5.0\times10^{-4}$, $\varepsilon_\theta = 3.75\times10^{-4}$, $\varepsilon_z = 5.0\times10^{-4}$, $\gamma_{rz} = 1.67\times10^{-4}$; $\sigma_r = 235.6$, $\sigma_\theta = 216.3$, $\sigma_z = 235.6$, $\tau_{rz} = 12.8$ MPa. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For nodes (20,0), (40,0), (20,30), what are the area and centroid? A. A = 300 mm^2; centroid (26.67, 10) B. A = 300 mm^2; centroid (30, 15) C. A = 300 mm^2; centroid (20, 10) D. A = 600 mm^2; centroid (26.67, 10) Answer: A. 2A = 600 gives A = 300; the centroid is the average of the nodal coordinates. (Key point: Area and centroid) 2. What value does each hoop coefficient f_i take at the centroid, and why? A. 7.5, because N_i = 1/3 so f_i = 2A/(3 r_bar) B. 30, because beta_1 = -30 C. 26.67, because it equals r_bar D. 0, because the hoop strain is zero at the centroid Answer: A. At the centroid each Ni = 1/3, so f_i = 600/(3 x 26.67) = 7.5. (Key point: Hoop strain row) 3. What are eps_r and eps_theta at the centroid? A. 5.0 x 10^-4 and 3.75 x 10^-4 B. 0 and 5.0 x 10^-4 C. 3.75 x 10^-4 and 5.0 x 10^-4 D. 1.67 x 10^-4 and 5.0 x 10^-4 Answer: A. eps_r = [-30(0.01) + 30(0.02)]/600 = 5x10^-4 and eps_theta = 7.5(0.03)/600 = 3.75x10^-4. (Key point: Strains) 4. Which [D] and sigma_r result is used for E = 200 GPa and nu = 0.3? A. Factor 384,615 MPa with 0.7, 0.3 pattern; sigma_r = 235.6 MPa B. Factor 200,000 MPa; sigma_r = 50 MPa C. Factor 219,780 MPa; sigma_r = 100 MPa D. Factor 384,615 MPa; sigma_r = 12.8 MPa Answer: A. sigma_r = 384,615[0.7(5.0) + 0.3(3.75 + 5.0)]x10^-4 = 235.6 MPa; 12.8 MPa is the shear stress. 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