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Axisymmetric triangle strains and stresses at 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.

An axisymmetric triangular element has nodes 1 (r=20r = 20, z=0z = 0), 2 (r=40r = 40, z=0z = 0) and 3 (r=20r = 20, z=30z = 30), coordinates in mm. E=200E = 200 GPa and ν=0.3\nu = 0.3. The nodal displacements (mm) are u1=0.01u_1 = 0.01, w1=0w_1 = 0, u2=0.02u_2 = 0.02, w2=0.01w_2 = 0.01, u3=0u_3 = 0, w3=0.015w_3 = 0.015. Find the area, the centroid, the matrix [B][B] at the centroid, the strains and the stresses at the centroid.

Area and centroid

2A=r1(z2−z3)+r2(z3−z1)+r3(z1−z2)=20(−30)+40(30)+0=600⇒A=300 mm22A = 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 rˉ=20+40+203=26.67 mm,zˉ=0+0+303=10 mm\bar r = \frac{20 + 40 + 20}{3} = 26.67\ \text{mm},\qquad \bar z = \frac{0 + 0 + 30}{3} = 10\ \text{mm}

Coefficients

α1=r2z3−r3z2=1200\alpha_1 = r_2z_3 - r_3z_2 = 1200, α2=r3z1−r1z3=−600\alpha_2 = r_3z_1 - r_1z_3 = -600, α3=r1z2−r2z1=0\alpha_3 = r_1z_2 - r_2z_1 = 0.

iiβi=zj−zk\beta_i = z_j - z_kγi=rk−rj\gamma_i = r_k - r_j
10−30=−300 - 30 = -3020−40=−2020 - 40 = -20
230−0=3030 - 0 = 3020−20=020 - 20 = 0
30−0=00 - 0 = 040−20=2040 - 20 = 20

Hoop-strain coefficient at the centroid, fi=αirˉ+βi+γizˉrˉf_i = \dfrac{\alpha_i}{\bar r} + \beta_i + \dfrac{\gamma_i\bar z}{\bar r}:

f1=120026.67−30−20026.67=45−30−7.5=7.5,f2=−22.5+30+0=7.5,f3=0+0+7.5=7.5f_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 7.57.5 because Ni=1/3N_i = 1/3 at the centroid, so fi=2A/(3rˉ)=7.5f_i = 2A/(3\bar r) = 7.5.)

[B][B] at the centroid

[Bˉ]=1600[−300300007.507.507.500−2000020−20−30030200][\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

εr=−30(0.01)+30(0.02)600=5.0×10−4εθ=7.5(0.01+0.02+0)600=3.75×10−4εz=−20(0)+0(0.01)+20(0.015)600=5.0×10−4γrz=−20(0.01)+(−30)(0)+0+30(0.01)+0+0600=1.67×10−4\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

E(1+ν)(1−2ν)=384 615\dfrac{E}{(1 + \nu)(1 - 2\nu)} = 384\,615 MPa.

σr=384 615 [0.7(5.0)+0.3(3.75+5.0)]×10−4=235.6 MPaσθ=384 615 [0.7(3.75)+0.3(5.0+5.0)]×10−4=216.3 MPaσz=384 615 [0.7(5.0)+0.3(5.0+3.75)]×10−4=235.6 MPaτrz=384 615(0.2)(1.67×10−4)=12.8 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 mm2A = 300\ \text{mm}^2, centroid (26.67, 10)(26.67,\ 10) mm; εr=5.0×10−4\varepsilon_r = 5.0\times10^{-4}, εθ=3.75×10−4\varepsilon_\theta = 3.75\times10^{-4}, εz=5.0×10−4\varepsilon_z = 5.0\times10^{-4}, γrz=1.67×10−4\gamma_{rz} = 1.67\times10^{-4}; σr=235.6\sigma_r = 235.6, σθ=216.3\sigma_\theta = 216.3, σz=235.6\sigma_z = 235.6, τrz=12.8\tau_{rz} = 12.8 MPa.

✓ Answer checked by Claude Opus 5.5 (high effort)

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