Let's break down the problem step by step: --- ## **Problem Restatement** - The rectangular plate is deformed into a parallelogram as shown - You are asked **determine the average shear strain corner A**. --- ## **Shear Strain (\(\gamma\)) Concept** Shear strain at a corner is defined as the change in angle (in radians) between two originally perpendicular sides. \[ \text{Shear strain} = \gamma = \tan \theta \approx \theta \quad \text{(for small angles, in radians)} \] Where \(\theta\) is the change in angle from \(90^\circ\) due to deformation. --- ## **Step 1: Identify Deformation** - Initially, \(\angle DAB = 90^\circ\). - After deformation, point D moves horizontally by **5 mm**. --- ## **Step 2: Calculate Shear Strain** Shear strain at point A is the change in angle between sides **AD** and **AB**: - **Original:** AD is vertical, AB is horizontal (\(90^\circ\)). - **Deformed:** D moves right by 5 mm, so the new line AD' is slanted. The small angle \(\theta\) at A is: \[ \tan\theta = \frac{\text{horizontal displacement}}{\text{vertical length}} \] \[ \tan\theta = \frac{5}{300} \] Since \(\theta\) is small, \(\tan\theta \approx \theta\) (in radians): \[ \gamma = \theta = \frac{5}{300} = .01667 \text{ radians} \] --- ## **Step 3: Final Answer** **So, the average shear strain at corner A is:** \[ \boxed{.0167 \ \text{rad}} \] --- ### **Matching to Options** - **A) 1.671 rad** - **B) .0292 rad** - **C) .0167 rad** ⟵ **(Correct)** - **D) .0125 rad** --- ## **Summary Table** | Step | Description | Value | |------|-----------------------------------|--------| | 1 | Horizontal displacement (mm) | 5 | | 2 | Vertical length (mm) | 300 | | 3 | Shear strain (\(\gamma\)) | .0167 rad | --- **Let me know if you need a diagram or further explanation!**