# Step-by-Step Solution Let's solve the problem step by step as per the requirements. --- ## **Section 1: Strains (\(\epsilon_x\), \(\epsilon_y\), \(\gamma_{xy}\))** ### **Step 1.1: Identify Nodes of an Element** Let's work with the triangle formed by nodes 1, 2, and 3: - Node 1: \((0, 0)\) - Node 2: \((2, 0)\) - Node 3: \((0, 6)\) Nodal displacements: - \(u_1 = 0,\ v_1 = 0\) - \(u_2 = 2.5 \times 10^{-3}\ \text{cm},\ v_2 = 5 \times 10^{-3}\ \text{cm}\) - \(u_3 = 1.25 \times 10^{-3}\ \text{cm},\ v_3 = 5 \times 10^{-4}\ \text{cm}\) ### **Step 1.2: Linear Strain-Displacement Relations** For a linear triangular (CST) element, the strain-displacement matrix (\(B\)) is constant over the element. The strain vector is: \[ \left\{ \begin{matrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{matrix} \right\} = B \left\{ \begin{matrix} u_1 \\ v_1 \\ u_2 \\ v_2 \\ u_3 \\ v_3 \end{matrix} \right\} \] #### **Element Geometry** Let the nodes be: | Node | \(x\) (cm) | \(y\) (cm) | |------|------------|------------| | 1 | 0 | 0 | | 2 | 2 | 0 | | 3 | 0 | 6 | #### **Shape Function Derivatives** For a 2D linear triangle: \[ \frac{\partial N_1}{\partial x} = \frac{y_2 - y_3}{2A},\quad \frac{\partial N_2}{\partial x} = \frac{y_3 - y_1}{2A},\quad \frac{\partial N_3}{\partial x} = \frac{y_1 - y_2}{2A} \] \[ \frac{\partial N_1}{\partial y} = \frac{x_3 - x_2}{2A},\quad \frac{\partial N_2}{\partial y} = \frac{x_1 - x_3}{2A},\quad \frac{\partial N_3}{\partial y} = \frac{x_2 - x_1}{2A} \] #### **Calculate Area (\(A\))** \[ A = \frac{1}{2}|(x_2y_3 + x_3y_1 + x_1y_2) - (x_3y_2 + x_1y_3 + x_2y_1)| \] \[ A = \frac{1}{2}|(2 \times 6 + 0 \times 0 + 0 \times 0) - (0 \times 0 + 0 \times 6 + 2 \times 0)| = \frac{1}{2}(12 - 0) = 6\ \text{cm}^2 \] #### **Derivatives** - \(y_2 = 0, y_3 = 6\): \(y_2 - y_3 = -6\) - \(y_3 - y_1 = 6 - 0 = 6\) - \(y_1 - y_2 = 0 - 0 = 0\) - \(x_3 = 0, x_2 = 2\): \(x_3 - x_2 = -2\) - \(x_1 - x_3 = 0 - 0 = 0\) - \(x_2 - x_1 = 2 - 0 = 2\) So, \[ \frac{\partial N_1}{\partial x} = \frac{-6}{12} = -0.5,\quad \frac{\partial N_2}{\partial x} = \frac{6}{12} = 0.5,\quad \frac{\partial N_3}{\partial x} = 0 \] \[ \frac{\partial N_1}{\partial y} = \frac{-2}{12} = -0.167,\quad \frac{\partial N_2}{\partial y} = 0,\quad \frac{\partial N_3}{\partial y} = \frac{2}{12} = 0.167 \] ### **Step 1.3: Construct B Matrix** \[ B = \begin{bmatrix} \frac{\partial N_1}{\partial x} & 0 & \frac{\partial N_2}{\partial x} & 0 & \frac{\partial N_3}{\partial x} & 0 \\ 0 & \frac{\partial N_1}{\partial y} & 0 & \frac{\partial N_2}{\partial y} & 0 & \frac{\partial N_3}{\partial y} \\ \frac{\partial N_1}{\partial y} & \frac{\partial N_1}{\partial x} & \frac{\partial N_2}{\partial y} & \frac{\partial N_2}{\partial x} & \frac{\partial N_3}{\partial y} & \frac{\partial N_3}{\partial x} \end{bmatrix} \] Insert values: \[ B = \begin{bmatrix} -0.5 & 0 & 0.5 & 0 & 0 & 0 \\ 0 & -0.167 & 0 & 0 & 0 & 0.167 \\ -0.167 & -0.5 & 0 & 0.5 & 0.167 & 0 \end{bmatrix} \] ### **Step 1.4: Nodal Displacements Vector** \[ d = \begin{bmatrix} u_1 \\ v_1 \\ u_2 \\ v_2 \\ u_3 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 2.5 \times 10^{-3} \\ 5 \times 10^{-3} \\ 1.25 \times 10^{-3} \\ 5 \times 10^{-4} \end{bmatrix} \] ### **Step 1.5: Compute Strain** \[ \begin{bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{bmatrix} = B \cdot d \] - \(\varepsilon_x = -0.5 \times 0 + 0.5 \times 2.5 \times 10^{-3} + 0 \times 1.25 \times 10^{-3}\) - \(= 0 + 1.25 \times 10^{-3} + 0 = 1.25 \times 10^{-3}\) - \(\varepsilon_y = -0.167 \times 0 + 0 \times 5 \times 10^{-3} + 0.167 \times 5 \times 10^{-4}\) - \(= 0 + 0 + 8.35 \times 10^{-5} = 8.35 \times 10^{-5}\) - \(\gamma_{xy} = -0.167 \times 0 - 0.5 \times 0 + 0 \times 2.5 \times 10^{-3} + 0.5 \times 5 \times 10^{-3} + 0.167 \times 5 \times 10^{-4} + 0 \times 1.25 \times 10^{-3}\) - \(= 0 + 0 + 0 + 2.5 \times 10^{-3} + 8.35 \times 10^{-5} + 0\) - \(= 2.5 \times 10^{-3} + 8.35 \times 10^{-5} = 2.5835 \times 10^{-3}\) **Convert to standard notation (\(\gamma_{xy}\) is engineering shear strain):** | Strain | Value | |-------------|------------------------------| | \(\epsilon_x\) | \(1.25 \times 10^{-3}\) | | \(\epsilon_y\) | \(8.35 \times 10^{-5}\) | | \(\gamma_{xy}\) | \(2.5835 \times 10^{-3}\) | --- ## **Section 2: Stresses (\(\sigma_x, \sigma_y, \tau_{xy}\))** ### **Step 2.1: Constitutive Matrix for Plane Stress** Given: - \(E = 210\, \text{GPa} = 210 \times 10^9\, \text{Pa}\) - \(\nu = 0.25\) Plane stress matrix: \[ D = \frac{E}{1-\nu^2} \begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1-\nu}{2} \end{bmatrix} \] Plug in values: \[ \frac{E}{1-\nu^2} = \frac{210 \times 10^9}{1-0.25^2} = \frac{210 \times 10^9}{0.9375} = 224 \times 10^9\, \text{Pa} \] So, \[ D = 224 \times 10^9 \begin{bmatrix} 1 & 0.25 & 0 \\ 0.25 & 1 & 0 \\ 0 & 0 & 0.375 \end{bmatrix} \] ### **Step 2.2: Calculate Stresses** \[ \begin{bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{bmatrix} = D \cdot \begin{bmatrix} \epsilon_x \\ \epsilon_y \\ \gamma_{xy} \end{bmatrix} \] Plug in values: \[ \left( \begin{matrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{matrix} \right) = 224 \times 10^9 \left[ \begin{matrix} 1 & 0.25 & 0 \\ 0.25 & 1 & 0 \\ 0 & 0 & 0.375 \end{matrix} \right] \left( \begin{matrix} 1.25 \times 10^{-3} \\ 8.35 \times 10^{-5} \\ 2.5835 \times 10^{-3} \end{matrix} \right) \] Calculate each: - \(\sigma_x = 224 \times 10^9 [1 \times 1.25 \times 10^{-3} + 0.25 \times 8.35 \times 10^{-5}]\) - \(= 224 \times 10^9 [1.25 \times 10^{-3} + 2.0875 \times 10^{-5}]\) - \(= 224 \times 10^9 \times 1.270875 \times 10^{-3}\) - \(= 284,659,600 \text{ Pa} = 284.66 \text{ MPa}\) - \(\sigma_y = 224 \times 10^9 [0.25 \times 1.25 \times 10^{-3} + 1 \times 8.35 \times 10^{-5}]\) - \(= 224 \times 10^9 [3.125 \times 10^{-4} + 8.35 \times 10^{-5}]\) - \(= 224 \times 10^9 \times 3.96 \times 10^{-4}\) - \(= 88,704,000 \text{ Pa} = 88.70 \text{ MPa}\) - \(\tau_{xy} = 224 \times 10^9 \times 0.375 \times 2.5835 \times 10^{-3}\) - \(= 224 \times 10^9 \times 0.9688 \times 10^{-3}\) - \(= 224 \times 10^9 \times 9.688 \times 10^{-4}\) - \(= 216,000,000 \text{ Pa} = 216.00 \text{ MPa}\) --- ## **Section 3: Final Answers** ### **a) Strains at centroid** \[ \boxed{ \begin{aligned} \varepsilon_x & = 1.25 \times 10^{-3} \\ \varepsilon_y & = 8.35 \times 10^{-5} \\ \gamma_{xy} & = 2.5835 \times 10^{-3} \end{aligned} } \] ### **b) Stresses at centroid** \[ \boxed{ \begin{aligned} \sigma_x & = 284.66\ \text{MPa} \\ \sigma_y & =