Question 1.5: Strains in a Plate Given A thin, triangular plate ABC is uni......
Strains in a Plate
Given
A thin, triangular plate ABC is uniformly deformed into a shape ABC, as depicted by the dashed lines in Figure 1.11.
Find
a. The normal strain along the centerline OC.
b. The normal strain along the edge AC.
c. The shear strain between the edges AC and BC.
Assumptions
The edge AB is built into a rigid frame. The deformed edges AC=BC’ are straight lines.
Solution
We have L_{O C}=a \text { and } L_{A C}=L_{B C}=a \sqrt{2}=1.41421 a (Figure 1.11).
a. Normal strain along OC. Since the contraction in length OC is \Delta a=-0.0015 a , Equation (1.20) gives
\varepsilon=\frac{\delta}{L} (1.20)
\varepsilon_{O C}=-\frac{0.0015 a}{a}=-0.0015=-1500 \mu
b. Normal strain along AC and BC. The lengths of the deformed edges are equal to L_{A C}=L_{B C}=\left[a^2+(a-0.0015)^2\right]^{1 / 2}=1.41315 a . It follows that
\varepsilon_{A C}=\varepsilon_{A C}=-\frac{1.41315 a-1.41421 a}{1.41421 a}=-750 \mu
c. Shear strain between AC and BC. After deformation, angle ACB is therefore
A C^{\prime} B=2 \tan ^{-1}\left[\frac{a}{a-0.0015 a}\right]=90.086^{\circ}
So, the change in the right angle is 90 − 90.086=− 0.086°. The associated shear strain (in radians) equals \gamma=-0.086\left\lgroup\frac{\pi}{180}\right\rgroup=-1501 \mu
Comments: Inasmuch as the angle ACB is increased, the shear strain is negative. The MATLAB solution of this sample problem and many others are on the website (see Appendix E).