Let’s solve the problem step by step: **Given:** - The original rectangle \(ABCD\) is distorted as shown. - We are to find: 1. Shear strain \(\gamma_{xy}\) at point **C** 2. Shear strain \(\gamma_{xy}\) at point **D** 3. Normal strain \(\epsilon_x\) of side **CD** --- ### **Step 1: Coordinates Before and After Deformation** Let’s assign the original coordinates: | Point | Original (x, y) | Deformed (x', y') | |-------|-------------------|----------------------| | A | (500, 0) | (500+6, 0+3) = (506, 3) | | B | (500, 350) | (500+14, 350+9) = (514, 359) | | C | (0, 350) | (0+4, 350+5) = (4, 355) | | D | (0, 0) | (0, 0) | --- ### **Step 2: Strain Calculations** #### **A. Shear Strain \(\gamma_{xy}\) at Point C** Shear strain at a point is given by: \[ \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \] where \(u\) and \(v\) are the displacements in \(x\) and \(y\). - For small strain and for a rectangle, approximate: - \(\frac{\partial u}{\partial y}\) is the change in \(x\)-displacement per unit \(y\) (vertical edge) - \(\frac{\partial v}{\partial x}\) is the change in \(y\)-displacement per unit \(x\) (horizontal edge) ##### **Find Displacements:** - At C: \(u_C = 4\), \(v_C = 5\) - At D: \(u_D = 0\), \(v_D = 0\) - At B: \(u_B = 14\), \(v_B = 9\) - At A: \(u_A = 6\), \(v_A = 3\) ##### **Partial Derivatives at C:** \[ \frac{\partial u}{\partial y} \approx \frac{u_B - u_A - (u_C - u_D)}{(y_B - y_A) + (y_C - y_D)} \] But since C and D are on the left, use: \[ \frac{\partial u}{\partial y} \approx \frac{u_C - u_D}{y_C - y_D} = \frac{4 - 0}{350 - 0} = \frac{4}{350} = 0.01143 \] \[ \frac{\partial v}{\partial x} \approx \frac{v_A - v_D}{x_A - x_D} = \frac{3 - 0}{500 - 0} = \frac{3}{500} = 0.006 \] \[ \gamma_{xy, C} = 0.01143 + 0.006 = \boxed{0.01743} \] --- #### **B. Shear Strain \(\gamma_{xy}\) at Point D** \[ \frac{\partial u}{\partial y} \approx \frac{u_A - u_D}{y_A - y_D} = \frac{6 - 0}{0 - 0} = \text{undefined} \ (\text{so use C-D instead}) \] But along vertical edge, use C-D as above: \[ \frac{\partial u}{\partial y} \approx \frac{u_C - u_D}{y_C - y_D} = \frac{4}{350} = 0.01143 \] \[ \frac{\partial v}{\partial x} \approx \frac{v_D - v_C}{x_D - x_C} = \frac{0 - 5}{0 - 4} = \frac{-5}{-4} = 1.25 \] But this seems odd; check along D-A: \[ \frac{\partial v}{\partial x} \approx \frac{v_A - v_D}{x_A - x_D} = \frac{3}{500} = 0.006 \] So at D: \[ \gamma_{xy, D} = 0.01143 + 0.006 = \boxed{0.01743} \] (Same as at C, since both are on the left edge.) --- #### **C. Normal (Axial) Strain \(\epsilon_x\) of Side CD** \[ \epsilon_x = \frac{\Delta L}{L_0} \] Where \(L_0 = 500 \text{ mm}\) (original length CD), and \(\Delta L\) is increase in length after deformation. - CD original: from \((0,0)\) to \((500,0)\) - After deformation: from \((0,0)\) to \((4,355)\) Find new length: \[ L = \sqrt{(4-0)^2 + (355-0)^2} = \sqrt{4^2 + 355^2} = \sqrt{16 + 126025} = \sqrt{126041} \approx 355.16 \text{ mm} \] Wait, this is vertical side. For CD (horizontal): Original: \(D(0,0)\) to \(C(0,350)\) (vertical edge, length 350 mm) After deformation: \(D(0,0)\) to \(C(4,355)\) \[ L = \sqrt{(4-0)^2 + (355-0)^2} = \sqrt{16 + 126025} = \sqrt{126041} \approx 355.17 \text{ mm} \] So, \[ \Delta L = 355.17 - 350 = 5.17 \text{ mm} \] \[ \epsilon_x = \frac{5.17}{350} = 0.0148 \] --- ## **Final Answers:** 1. **Shear strain \(\gamma_{xy}\) at C:** \(\boxed{0.01743}\) 2. **Shear strain \(\gamma_{xy}\) at D:** \(\boxed{0.01743}\) 3. **Normal strain \(\epsilon_x\) of side CD:** \(\boxed{0.0148}\) If you need more details or clarification, let me know!