Question 9.12: The strains determined by the use of the rosette as shown in...

For plane-strain condition \left(\in_{x x}, \in_{y y}, \gamma_{x y}\right) , strain at any arbitrary direction defined by the angle θ with the x-axis is

∈_{\theta \theta}=\frac{1}{2}\left(∈_{x x}+∈_{y y}\right)+\frac{1}{2}\left(∈_{x x}-∈_{y y}\right) \cos 2 \theta+\frac{\gamma_{x y}}{2} \sin 2 \theta        (9.109)  [refer Eq.(9.109)]

Now putting θ= 30°, –30° and 90°, respectively, in the above equation, we get

\begin{aligned} & ∈_1=\frac{1}{2}\left(∈_{x x}+∈_{y y}\right)+\frac{1}{2}\left(∈_{x x}-∈_{y y}\right) \cos 60^{\circ}+\frac{\gamma_{x y}}{2} \sin 60^{\circ} \\ & ∈_2=\frac{1}{2}\left(∈_{x x}+∈_{y y}\right)+\frac{1}{2}\left(∈_{x x}-∈_{y y}\right) \cos 60^{\circ}+\frac{\gamma_{x y}}{2} \sin 60^{\circ} \\ & ∈_3=∈_{y y} \end{aligned}               (1)

Subtracting the first two relations:

\gamma_{x y} \sin 60^{\circ}=\left(∈_1+∈_2\right) \Rightarrow \gamma_{x y}=\frac{2}{\sqrt{3}}\left(∈_1+∈_2\right)             (2)

Putting ∈_{y y}=∈_3 \text { and } \gamma_{x y}=(2 / \sqrt{3})\left(∈_1+∈_2\right) into the first relation of Eq. (1) we obtain:

\frac{1}{2}\left(∈_{x x}+∈_3\right)+\frac{1}{2}\left(∈_{x x}+∈_3\right) \cos 60^{\circ}+\frac{1}{2}\left(∈_1+∈_2\right)=∈_1

or

\begin{gathered} ∈_{x x} \cos ^2 30^{\circ}+∈_3 \sin ^2 30^{\circ}=\frac{1}{2}\left(∈_1+∈_2\right) \\ \frac{3}{4} ∈_{x x}=\frac{1}{2}\left(∈_1+∈_2\right)-\frac{∈_3}{4} \\ =\frac{1}{4}\left[2\left(∈_1+∈_2\right)-∈_3\right] \end{gathered}

Therefore,

∈_{x x}=\frac{1}{3}\left[2\left(∈_1+∈_2\right)-∈_3\right]          (3)

Thus, assimilating the relations in Eqs. (1)–(3) we get

∈_{x x}=\frac{1}{3}\left[2\left(∈_1+∈_2\right)-∈_3\right], \quad ∈_{y y}=∈_3, \quad \gamma_{x y}=\frac{2}{\sqrt{3}}\left(∈_1+∈_2\right)

Now putting the numerical values of ∈_1, ∈_2 \text { and } ∈_3 into the above equation, we get

\begin{aligned} & ∈_{x x}=\frac{1}{3}[2(600+450)-(-75)] \mu=725 \mu \\ & ∈_{y y}=-75 \mu \text { and } \quad \gamma_{x y}=173.21 \mu \end{aligned}

Thus, the in-plane principal strains are

  \begin{aligned} ∈_1 & =\frac{1}{2}\left(∈_{x x}-∈_{y y}\right)+\frac{1}{2} \sqrt{\left(∈_{x x}+∈_{y y}\right)^2+\gamma_{x y}^2} \\ & =\frac{1}{2}(725-75)+\frac{1}{2} \sqrt{(725+75)^2+(100 \sqrt{3})^2} \\ & =734.27 \mu \end{aligned}

and          ∈_2=\frac{1}{2}\left(∈_{x x}+∈_{y y}\right)-\frac{1}{2} \sqrt{\left(∈_{x x}+∈_{y y}\right)^2+\gamma_{x y}^2}=-84.27 \mu

(a) The in-plane principal strains are

∈_1=734.27 \mu \text { and } ∈_2=-84.27 \mu

(b) The in-plane maximum shearing strain are

\frac{\left(\gamma_{x y}\right)_{\max }}{2}=\frac{1}{2}\left(∈_1+∈_2\right) \Rightarrow\left(\gamma_{x y}\right)_{\max }=818.54 \mu