Question 2.16: For the square plate and the deformation described in Exampl......
For the square plate and the deformation described in Example 2.15, determine an expression for γ_{xy}(x, y).
Plan the Solution By comparing the “before deformation” and “after deformation” figures, we can see that there is a definite change in the right angle between x lines and y lines. We need to determine this change in angle so that we can evaluate the shear strain from Eq. 2.36.
\gamma_{n t}(P)=\lim _{\substack{Q \rightarrow P \text { along } n \\ R \rightarrow P \text { along } t}}\left(\frac{\pi}{2}-\theta^*\right) Shear Strain (2.36)
Equation 2.36 can be written as
\gamma_{xy}(P)=γ_{xy}(x, y)=\lim _{\substack{Δx \rightarrow 0 \\ Δy \rightarrow 0 }}\left(\frac{\pi}{2}-\theta^*\right)where Δx, Δy, and θ* are indicated in Fig. 1.
θ^* = \frac{\pi}{2} – \tan^{-1}\left(\frac{x_{N^*} – x_{M^*}}{ a}\right)x_{N^*} = x + \left(1 – \frac{x} {a}\right)(0.2a)
x_{N^*}- x_{M^*}= \left(1 – \frac{x} {a}\right)(0.2a)
Therefore,
θ^* = \frac{\pi}{2} – \tan^{-1}\left[0.2\left(1 – \frac{x}{ a}\right)\right]Since the x lines and y lines remain straight, θ* doesn’t depend on the lengths of Δx and Δy, and we don’t need the limit operation. So,
γ_{xy}(x, y) =\frac{\pi}{2} – θ^* =\tan^{-1}\left[0.2\left(1 – \frac{x}{ a}\right)\right]Review the Solution From Fig. lb, we see that γ_{xy} should be independent of y, as our final result indicates. Also, from Fig. lb we see that γ_{xy} should be greatest when x = 0 and should be zero at x = a. These observations are consistent with our answer. Finally, the largest shear strain is at x = 0,where γ_{xy}(x, y) = tan^{-1}(0.2) = 0.1974 ≈ 0.2.Therefore,
γ_{xy} could be approximated by γ_{xy}(x, y) = 0.2\left(1 – \frac{x}{ a}\right).
