Question 2.15: The thin, square plate ABCD in Fig. 1a undergoes deformation......
The thin, square plate ABCD in Fig. 1a undergoes deformation in which no point in the plate displaces in the y direction. Every horizontal line in the plate, except line AD, is uniformly shortened as the edge AB remains straight and rotates clockwise. Determine an expression for ε_x(x, y).
Plan the Solution We can use the definition of extensional strain ε_n(P), which, in this case, is ε_x(x, y).We will have to use the geometry of deformation to determine an expression for Δs* in Eq. 2.35.
{{\epsilon}}_{n}(P)=\operatorname*{lim}_{Q\to P a l o n{\mathrm{g}},n}\left(\frac{\Delta s^{*}-\Delta s}{\Delta s}\right) Extensional Strain (2.35)
To determine ε_x(x, y), let Eq. 2.35 be written as
\epsilon_x(x, y)=\lim _{\Delta x \rightarrow 0}\left(\frac{\Delta x^*-\Delta x}{\Delta x}\right)where Δx and Δx* are defined in Fig. 2.
To get an expression for ε_x(x, y) we need an expression for Δx*.We are told that every horizontal line is uniformly shortened, and we see that AD remains its original length while B*C* is shorter than BC by 20%. Furthermore, the shortening of horizontal lines is linearly related to y. Therefore,
Δx* = Δx – \left(\frac{y} {a}\right) (0.2 Δx)
So,
\epsilon_x(x, y)=\lim _{\Delta x \rightarrow 0}\left[\frac{-(y/a)(0.2 Δx)}{Δx}\right]or
ε_x(x, y) = -\left(\frac{y} {a}\right) (0.2)Review the Solution The answer is negative, which indicates a shortening, and the answer is dimensionless, as it should be for strain. Also, according to the answer, there is no strain at y = 0 (along AD), and there is a 20% shortening at y = a. These results agree with the stated geometry of deformation.
