Question 7.4: In a material in a state of plane strain, it is known that t...
In a material in a state of plane strain, it is known that the horizontal side of a 10 × 10-mm square elongates by 4 μm, while its vertical side remains unchanged, and that the angle at the lower left corner increases by 0.4 × 10^{-3} rad (Fig. 7.67). Determine (a) the principal axes and principal strains, (b) the maximum shearing strain and the corresponding normal strain.
(a) Principal Axes and Principal Strains. We first determine the coordinates of points X and Y on Mohr’s circle for strain. We have
\epsilon_{x}=\frac{+4 \times 10^{-6} m }{10 \times 10^{3} m }=+400 m \epsilon_{y}=0 \left|\frac{ g _{x y}}{2}\right|=200 m
Since the side of the square associated with rotates clockwise, point X of coordinates \epsilon_{x} and \left| g _{x y} / 2\right| is plotted above the horizontal axis. Since \epsilon_{y}=0 and the corresponding side rotates counterclockwise, point Y is plotted directly below the origin (Fig. 7.68). Drawing the diameter XY, we determine the center C of Mohr’s circle and its radius R. We have
O C=\frac{\epsilon_{x}+\epsilon_{y}}{2}=200 \mu OY = 200 μ
R=\sqrt{(O C)^{2}+(O Y)^{2}}=\sqrt{(200 m )^{2}+(200 m )^{2}}=283 m
The principal strains are defined by the abscissas of points A and B. We write
\epsilon_{a}=O A=O C+R=200 m +283 m =483 m\epsilon _{b}=O B=O C-R=200 m -283 m =-83 m
The principal axes Oa and Ob are shown in Fig. 7.69. Since OC = OY, the angle at C in triangle OCY is 45°. Thus, the angle 2 u _{p} that brings XY into AB is 45°\circlearrowright and the angle u _{p} bringing Ox into Oa is 22.5°\circlearrowright.
(b) Maximum Shearing Strain. Points D and E define the maximum in-plane shearing strain which, since the principal strains have opposite signs, is also the actual maximum shearing strain (see Sec. 7.12). We have
\frac{ g _{\max }}{2}=R=283 m g_{\max }=566 m
The corresponding normal strains are both equal to
\epsilon ^{\prime}=O C=200 mThe axes of maximum shearing strain are shown in Fig. 7.70.
