Question 13.34: The strain components εx, εy, and γxy are given for a point ...

The strain components \varepsilon_{x}, \varepsilon_{y}, \text { and } \gamma_{x y} are given for a point in a body subjected to plane strain. Using Mohr’s circle, determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle \theta_{p}, the principal strain deformations, and the maximum in-plane shear strain distortion in a sketch.
\varepsilon_{x}=515  \mu \varepsilon \quad \varepsilon_{y}=-265  \mu \varepsilon \quad \gamma_{x y}=-1,030  \mu rad

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The basic Mohr’s circle is shown.

\begin{aligned}&C=\frac{(515  \mu)+(-265  \mu)}{2}=125  \mu \varepsilon \\&R=\sqrt{(390  \mu)^{2}+(515  \mu)^{2}}=646.0070  \mu \varepsilon\end{aligned}

 

\begin{aligned}\varepsilon_{p 1} &=C+R=125  \mu \varepsilon+646.0070  \mu \varepsilon=771  \mu \varepsilon \\\varepsilon_{p 2} &=C-R=125  \mu \varepsilon-646.0070  \mu \varepsilon=-521  \mu \varepsilon \\\gamma_{\max } &=2 R=1,292  \mu rad\end{aligned}

 

The magnitude of the angle 2 \theta_{p} between point x and point 1 (i.e., the principal plane associated with \varepsilon_{p 1}) is found from:

\tan 2 \theta_{p}=\frac{1,030  \mu}{|(515  \mu)-(-265  \mu)|}=\frac{1,030  \mu}{780  \mu}=1.32051 \therefore 2 \theta_{p}=52.8640^{\circ} \text { thus, } \theta_{p}=26.4^{\circ}

By inspection, the angle \theta_{p} from point x to point 1 is turned clockwise.

Since \varepsilon_{p 1} is positive and \varepsilon_{p 2} is negative, the absolute maximum shear strain is the maximum in-plane shear strain:

\gamma_{ abs \max }=\gamma_{\max }=1,292  \mu rad

A sketch of the principal strain deformations and the maximum in-plane shear strain distortions is shown below.

 

 

 

 

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