# Solution for the Plane Strain Element Let's solve the given problem step by step. --- ## **Given Data** - **Node Coordinates** (mm): - \( i = 1: (5, 15) \) - \( j = 2: (15, 5) \) - \( m = 3: (25, 15) \) - **Nodal Displacements** (mm): - \( u_1 = 0.005, \ v_1 = 0.002 \) - \( u_2 = 0, \ v_2 = 0 \) - \( u_3 = 0.005, \ v_3 = 0 \) - **Material Properties:** - \( E = 105 \ \text{GPa} = 105 \times 10^3 \ \text{MPa} \) - \( \nu = 0.3 \) - \( t = 1 \) - **All coordinates in mm.** --- # **Step 1: Calculate Strains** For a linear triangular element, the strains are **constant** throughout the element. ## **a) Calculate the Area of the Triangle** \[ A = \frac{1}{2} | (x_2y_3 + x_3y_1 + x_1y_2) - (x_3y_2 + x_1y_3 + x_2y_1) | \] Plug in the coordinates: - \( (x_1, y_1) = (5, 15) \) - \( (x_2, y_2) = (15, 5) \) - \( (x_3, y_3) = (25, 15) \) \[ A = \frac{1}{2} | 15 \times 15 + 25 \times 15 + 5 \times 5 - (25 \times 5 + 5 \times 15 + 15 \times 15) | \] \[ = \frac{1}{2} | 225 + 375 + 25 - (125 + 75 + 225) | \] \[ = \frac{1}{2} | 625 - 425 | = \frac{1}{2} \times 200 = 100 \ \text{mm}^2 \] --- ## **b) Strain-Displacement Matrix \([B]\)** For 2D linear triangle: \[ [B] = \frac{1}{2A} \begin{bmatrix} y_2-y_3 & 0 & y_3-y_1 & 0 & y_1-y_2 & 0 \\ 0 & x_3-x_2 & 0 & x_1-x_3 & 0 & x_2-x_1 \\ x_3-x_2 & y_2-y_3 & x_1-x_3 & y_3-y_1 & x_2-x_1 & y_1-y_2 \end{bmatrix} \] Calculate differences: - \( y_2 - y_3 = 5 - 15 = -10 \) - \( y_3 - y_1 = 15 - 15 = 0 \) - \( y_1 - y_2 = 15 - 5 = 10 \) - \( x_3 - x_2 = 25 - 15 = 10 \) - \( x_1 - x_3 = 5 - 25 = -20 \) - \( x_2 - x_1 = 15 - 5 = 10 \) So, \[ [B] = \frac{1}{2 \times 100} \begin{bmatrix} -10 & 0 & 0 & 0 & 10 & 0 \\ 0 & 10 & 0 & -20 & 0 & 10 \\ 10 & -10 & -20 & 0 & 10 & 10 \end{bmatrix} = \frac{1}{200} \begin{bmatrix} -10 & 0 & 0 & 10 & 10 & 0 \\ 0 & 10 & 0 & -20 & 0 & 10 \\ 10 & -10 & -20 & 0 & 10 & 10 \end{bmatrix} \] --- ## **c) Nodal Displacement Vector** \[ \mathbf{d} = [u_1, v_1, u_2, v_2, u_3, v_3]^T = [0.005, 0.002, 0, 0, 0.005, 0]^T \ \text{(mm)} \] --- ## **d) Compute Strains** \[ \varepsilon = [B]\mathbf{d} \] Expand: ### Row 1: \( \varepsilon_x \) \[ \varepsilon_x = \frac{1}{200}\left(-10 \times 0.005 + 0 \times 0.002 + 0 \times 0 + 10 \times 0 + 10 \times 0.005 + 0 \times 0 \right) \] \[ = \frac{1}{200}( -0.05 + 0 + 0 + 0 + 0.05 + 0 ) = \frac{0}{200} = 0 \] ### Row 2: \( \varepsilon_y \) \[ \varepsilon_y = \frac{1}{200}(0 \times 0.005 + 10 \times 0.002 + 0 \times 0 + (-20) \times 0 + 0 \times 0.005 + 10 \times 0) \] \[ = \frac{1}{200}(0 + 0.02 + 0 + 0 + 0 + 0) = \frac{0.02}{200} = 0.0001 \] ### Row 3: \( \gamma_{xy} \) \[ \gamma_{xy} = \frac{1}{200}(10 \times 0.005 + (-10) \times 0.002 + (-20) \times 0 + 0 \times 0 + 10 \times 0.005 + 10 \times 0) \] \[ = \frac{1}{200}(0.05 - 0.02 + 0 + 0 + 0.05 + 0) = \frac{1}{200}(0.08) = 0.0004 \] **So, the strains are:** \[ \boxed{ \begin{aligned} \varepsilon_x &= 0 \\ \varepsilon_y &= 0.0001 \\ \gamma_{xy} &= 0.0004 \end{aligned} } \] --- # **Step 2: Plane Strain Stresses** For plane strain: \[ \begin{aligned} \sigma_x &= \frac{E}{(1+\nu)(1-2\nu)}\Big[(1-\nu)\varepsilon_x + \nu \varepsilon_y\Big] \\ \sigma_y &= \frac{E}{(1+\nu)(1-2\nu)}\Big[\nu \varepsilon_x + (1-\nu)\varepsilon_y\Big] \\ \tau_{xy} &= \frac{E}{2(1+\nu)}\gamma_{xy} \end{aligned} \] Given \( E = 105 \times 10^3 \) MPa, \( \nu = 0.3 \): \[ (1+\nu)(1-2\nu) = (1.3)(0.4) = 0.52 \] \[ \frac{E}{(1+\nu)(1-2\nu)} = \frac{105000}{0.52} = 201923 \text{ MPa} \] \[ \frac{E}{2(1+\nu)} = \frac{105000}{2.6} = 40385 \text{ MPa} \] Now: ### \( \sigma_x \) \[ \sigma_x = 201923[(1-0.3) \times 0 + 0.3 \times 0.0001] = 201923[0.00003] = 6.06 \ \text{MPa} \] ### \( \sigma_y \) \[ \sigma_y = 201923[0.3 \times 0 + (1-0.3) \times 0.0001] = 201923[0.00007] = 14.13 \ \text{MPa} \] ### \( \tau_{xy} \) \[ \tau_{xy} = 40385 \times 0.0004 = 16.15 \ \text{MPa} \] --- # **Step 3: Principal Stresses** \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] \[ \frac{\sigma_x + \sigma_y}{2} = \frac{6.06 + 14.13}{2} = 10.095 \ \text{MPa} \] \[ \frac{\sigma_x - \sigma_y}{2} = \frac{6.06 - 14.13}{2} = -4.035 \ \text{MPa} \] \[ \sqrt{(-4.035)^2 + (16.15)^2} = \sqrt{16.28 + 260.82} = \sqrt{277.1} = 16.65 \ \text{MPa} \] So, \[ \sigma_1 = 10.095 + 16.65 = 26.75 \ \text{MPa} \] \[ \sigma_2 = 10.095 - 16.65 = -6.56 \ \text{MPa} \] --- # **Step 4: Principal Angle** \[ \tan(2\theta_p) = \frac{2\tau_{xy}}{(\sigma_x - \sigma_y)} = \frac{2 \times 16.15}{6.06 - 14.13} = \frac{32.3}{-8.07} = -4.004 \] \[ 2\theta_p = \arctan(-4.004) = -75.98^\circ \] \[ \theta_p = -37.99^\circ \] --- # **Summary Table** | Quantity | Value | |----------------------|--------------------| | \( \sigma_x \) | 6.06 MPa | | \( \sigma_y \) | 14.13 MPa | | \( \tau_{xy} \) | 16.15 MPa | | \( \sigma_1 \) | 26.75 MPa | | \( \sigma_2 \) | -6.56 MPa | | \( \theta_p \) | -37.99° | --- ## **Final Answers** **a)** Element stresses: \[ \boxed{ \sigma_x = 6.06\ \text{MPa}, \quad \sigma_y = 14.13\ \text{MPa}, \quad \tau_{xy} = 16.15\ \text{MPa} } \] **b)** Principal stress \( \sigma_1 \): \[ \boxed{26.75\ \text{MPa}} \] **c)** Principal stress \( \sigma_2 \): \[ \boxed{-6.56\ \text{MPa}} \] **d)** Principal angle \( \theta_p \): \[ \boxed{-37.99^\circ} \] --- Let me know if you need more details or a step-by-step derivation for any part!