The normal strain (\( \varepsilon \)) on side DC of the deformed plate is approximately \(0.00175\). The shear strain (\( \gamma_B \)) at corner B is approximately \(-1.75 \times 10^{-6}\), considering small deformations and given displacement values.
The normal strain (\( \varepsilon \)) is defined as the change in length (\( \Delta L \)) divided by the original length (\( L \)):
\[ \varepsilon = \frac{\Delta L}{L} \]
For side DC, the normal strain (\( \varepsilon_{DC} \)) is given by:
\[ \varepsilon_{DC} = \frac{\Delta D_x}{D} \]
Given \( \Delta D_x = 1 \) mm and \( D = 571 \) mm:
\[ \varepsilon_{DC} = \frac{1 \, \text{mm}}{571 \, \text{mm}} \]
Now, calculate \( \varepsilon_{DC} \):
\[ \varepsilon_{DC} \approx 0.00175 \]
So, the normal strain of side DC is approximately \(0.00175\).
The shear strain (\( \gamma \)) at the corner B is given by the formula:
\[ \gamma = \frac{\text{change in angle}}{\text{original angle}} \]
For small deformations, the shear strain at B (\( \gamma_B \)) can be expressed in terms of the displacements \( \Delta B_x \), \( \Delta C_x \), \( \Delta C_y \), \( \Delta D_x \), and \( \Delta D_y \):
\[ \gamma_B = \frac{1}{2} \left( \frac{\Delta C_x - \Delta B_x}{BC} + \frac{\Delta D_y - \Delta C_y}{CD} \right) \]
Given \( \Delta B_x = 2 \) mm, \( \Delta C_x = 4 \) mm, \( \Delta C_y = 2 \) mm, \( \Delta D_x = 1 \) mm, \( \Delta D_y = 1 \) mm, \( BC = 675 \) mm, and \( CD = 571 \) mm:
\[ \gamma_B = \frac{1}{2} \left( \frac{4 - 2}{675} + \frac{1 - 2}{571} \right) \]
Now, calculate \( \gamma_B \):
\[ \gamma_B \approx -1.75 \times 10^{-6} \]
So, the shear strain at the corner B is approximately \(-1.75 \times 10^{-6}\).