Question 9.9: For the given state of plane strain in Figure 9.33(a), use M...

We again show the undeformed and deformed geometries considering a small rectangular element in Figure 9.33(a) itself.

We now show the corresponding Mohr’s circle of strain in Figure 9.34:

From the figure,

\begin{aligned} BC & = CA = GC \\ & =\sqrt{(240-150)^2+25^2} \text { unit } \\ & =93.41 \text { unit } \end{aligned}

where C is the centre of the circle. Clearly, the principal strains are:

∈_1= OF = OC + CF =(150+93.41) \times 10^{-6}=243.41 \mu

and          ∈_2= OE = OC – CE =(150-93.41) \times 10^{-6}=56.59 \mu

\tan 2 \phi=\frac{25}{(240-150)} \Rightarrow \phi=7.76^{\circ} ⦪

We show the principal strains and their orientations in Figure 9.33(b). In the figure, the oriented unstrained differential element is A _1 B _1 C _1 D _1 and on deformation due to the principal strain, the element becomes A _1 B _1^{\prime} C _1^{\prime} D _1^{\prime} . Corresponding changes in the dimensions of the undeformed element are also shown in the figure. The maximum in-plane shear strain is γ/2

CG = 93.41μ

or          \left(\gamma_{x y}\right)_{\max }=186.82 \mu  ⦨ 52.76°

This orientation is also shown in Figure 9.33(b). The figure shows the oriented undeformed element A _2 B _2 C _2 D _2 and its deformed configuration is shown as A _2^{\prime} B _2^{\prime} C _2^{\prime} D _2^{\prime} . Finally, considering the planestrain condition we note that the principal strain components are

∈_1=243.41 \mu, ∈_2=56.59 \mu, ∈_3=0\left(∈_1>∈_2>∈_3\right)

These strain components are plotted in three-dimensional Mohr’s circle for strain (note that it is a similar concept of three-dimensional Mohr’s circle for stresses) in Figure 9.35:

Thus, the maximum shear strain is

\frac{1}{2} \gamma_{\max }=121.71 \mu \Rightarrow \gamma_{\max }=243.41 \mu