Question 9.14: Calculate the maximum in-plane normal strain if the followin...
Calculate the maximum in-plane normal strain if the following strains have been obtained by the use of the rosette shown in the Figure 9.42:
∈_1=-50 \mu, \quad ∈_2=+360 \mu \text { and } ∈_3=+315 \mu
We know that
∈_{\theta \theta}=\bar{∈}+∈_{ V } \cos 2 \theta+\frac{\gamma_{x y}}{2} \sin 2 \theta
where \bar{∈}=\left(∈_{x x}+∈_{y y}\right) / 2 ; ∈_{ V }=\left(∈_{x x}-∈_{y y}\right) / 2 . Now putting θ = 45°, 180° and 135°, we get
∈_1=\left\lgroup\bar{∈}+\frac{\gamma_{x y}}{2} \right\rgroup (1)
∈_2=\bar{∈}+∈_{ V }=∈_{x x} (2)
and ∈_3=\left\lgroup\bar{∈}-\frac{\gamma_{x y}}{2} \right\rgroup (3)
So from Eqs. (1) and (3),
2 \bar{∈}=∈_{x x}+∈_{y y}=\left(∈_1+∈_3\right)
and \gamma_{x y}=\left(∈_1+∈_3\right)
Again from Eq. (2),
∈_{y y}=\left(∈_1+∈_3-∈_2\right)
Thus, we get
\begin{aligned} & ∈_{x x}=∈_2=+360 \mu \\ & ∈_{y y}=\left(∈_1+∈_3-∈_2\right)=-95 \mu \end{aligned}
and \gamma_{x y}=-365 \mu
Thus, the maximum in-plane normal strain is given by
\begin{aligned} ∈_{\max } & =\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}(360-95)+\frac{1}{2} \sqrt{(360+95)^2+(-365)^2} \\ & =424.15 \mu \end{aligned}