Let's solve the problem step-by-step: --- ### (a) Develop the relation \( \varepsilon = Bq \) by developing the **B** matrix #### **Step 1: Displacement interpolation** The displacement field is: \[ u(\xi) = N_1 q_1 + N_2 q_2 \] where \[ N_1 = \frac{\cos\left(\frac{\pi}{4}(1+\xi)\right)}{4}, \quad N_2 = \frac{\cos\left(\frac{\pi}{4}(1-\xi)\right)}{4} \] #### **Step 2: Strain-displacement relation** Strain in 1D: \[ \epsilon = \frac{du}{dx} \] But \( u \) is a function of \( \xi \), so use the chain rule: \[ \epsilon = \frac{du}{dx} = \frac{du}{d\xi} \cdot \frac{d\xi}{dx} \] #### **Step 3: Compute \( \frac{du}{d\xi} \)** \[ \frac{du}{d\xi} = \frac{dN_1}{d\xi}q_1 + \frac{dN_2}{d\xi}q_2 \] #### **Step 4: Compute \( \frac{d\xi}{dx} \)** Given: \[ \xi = \frac{2}{x_2-x_1}(x-x_1) - 1 \] \[ \frac{d\xi}{dx} = \frac{2}{x_2-x_1} \] So, \[ \frac{dx}{d\xi} = \frac{x_2-x_1}{2} \] #### **Step 5: Assemble the B matrix** \[ \epsilon = \frac{du}{dx} = \frac{du}{d\xi} \cdot \frac{d\xi}{dx} \] So, \[ \epsilon = \left( \frac{dN_1}{d\xi}q_1 + \frac{dN_2}{d\xi}q_2 \right) \frac{d\xi}{dx} \] \[ \epsilon = \left[\frac{dN_1}{d\xi} \ \frac{dN_2}{d\xi}\right] \frac{d\xi}{dx} \begin{Bmatrix} q_1 \\ q_2 \end{Bmatrix} \] But since \( \frac{d\xi}{dx} = \frac{2}{x_2-x_1} \): \[ \boxed{ \epsilon = \begin{bmatrix} \frac{dN_1}{d\xi} & \frac{dN_2}{d\xi} \end{bmatrix} \frac{2}{x_2-x_1} \begin{Bmatrix} q_1 \\ q_2 \end{Bmatrix} } \] Therefore, the **B matrix** is: \[ \boxed{ B = \begin{bmatrix} \frac{dN_1}{d\xi} & \frac{dN_2}{d\xi} \end{bmatrix} \frac{2}{x_2-x_1} } \] --- ### (b) Develop the stiffness matrix \( k^e \) For 1D, the element stiffness matrix is: \[ k^e = \int_{x_1}^{x_2} B^T E B \, dx \] Transform to intrinsic (\( \xi \)) coordinates (\( dx = \frac{x_2-x_1}{2} d\xi \)), and limits \(\xi = -1\) to \(+1\): \[ k^e = \int_{-1}^{+1} B^T E B \, \frac{x_2-x_1}{2} d\xi \] #### Substitute B: \[ B = \left[\frac{dN_1}{d\xi} \ \frac{dN_2}{d\xi}\right] \frac{2}{x_2-x_1} \] So: \[ k^e = \int_{-1}^{+1} \left( \frac{2}{x_2-x_1} \begin{bmatrix} \frac{dN_1}{d\xi} \\ \frac{dN_2}{d\xi} \end{bmatrix} \right) E \left( \begin{bmatrix} \frac{dN_1}{d\xi} & \frac{dN_2}{d\xi} \end{bmatrix} \frac{2}{x_2-x_1} \right) \frac{x_2-x_1}{2} d\xi \] \[ = \int_{-1}^{+1} E \begin{bmatrix} \frac{dN_1}{d\xi} \\ \frac{dN_2}{d\xi} \end{bmatrix} \begin{bmatrix} \frac{dN_1}{d\xi} & \frac{dN_2}{d\xi} \end{bmatrix} \frac{4}{x_2-x_1} d\xi \] \[ = \int_{-1}^{+1} E \frac{4}{x_2-x_1} \begin{bmatrix} \frac{dN_1}{d\xi}\frac{dN_1}{d\xi} & \frac{dN_1}{d\xi}\frac{dN_2}{d\xi} \\ \frac{dN_2}{d\xi}\frac{dN_1}{d\xi} & \frac{dN_2}{d\xi}\frac{dN_2}{d\xi} \end{bmatrix} d\xi \] --- ### **Final Answers** #### **Part (a):** \[ \boxed{ \epsilon = Bq, \quad \text{where} \quad B = \begin{bmatrix} \frac{dN_1}{d\xi} & \frac{dN_2}{d\xi} \end{bmatrix} \frac{2}{x_2-x_1} } \] #### **Part (b):** \[ \boxed{ k^e = \int_{-1}^{+1} E \frac{4}{x_2-x_1} \begin{bmatrix} \left(\frac{dN_1}{d\xi}\right)^2 & \frac{dN_1}{d\xi}\frac{dN_2}{d\xi} \\ \frac{dN_2}{d\xi}\frac{dN_1}{d\xi} & \left(\frac{dN_2}{d\xi}\right)^2 \end{bmatrix} d\xi } \] The integral is written in terms of the natural coordinate \(\xi\), as required.