The state of strain at the point on the pin leaf has | StudySoup

Textbook Solutions for Mechanics of Materials

Chapter 10 Problem 10-3

Question

The state of strain at the point on the pin leaf has components of \(\epsilon_{x}=200\left(10^{-6}\right), \quad \epsilon_{y}=180\left(10^{-6}\right)\), and \(\gamma_{x y}=-300\left(10^{-6}\right)\). Use the strain transformation equations and determine the equivalent in-plane strains on an element oriented at an angle of \(\theta=60^{\circ}\) counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.

Solution

Step 1 of 4

Write the relation to find the in plane strain along \(x^1\) by using the following relation.

\(\varepsilon_{X 1}=\frac{\varepsilon_{x}+\varepsilon_{y}}{2}+\frac{\varepsilon_{x}-\varepsilon_{y}}{2} \cos 2 \theta+\frac{\gamma_{x y}}{2} \sin 2 \theta\)

Here, \(\varepsilon_{\mathrm{X}}\), \(\varepsilon_{\mathrm{y}}\) are the plane strains along x and y direction \(\gamma_{x y}\) is the shear strain, and \(\theta\) is the plane angle.

Substitute \(200 \times 10^{-6}\) for \(\varepsilon_{\mathrm{X}}\), \(180 \times 10^{-6}\) for \(\varepsilon_{\mathrm{y}}\), \(-300 \times 10^{-6}\) for \(\gamma_{x y}\) is the shear strain, and \(\theta\) is the plane angle.

\(\begin{aligned} \varepsilon_{\times 1} & =\left[\left(\frac{200+180}{2}\right)+\left(\frac{200180}{2}\right) \cos 2\left(60^{\circ}\right)+\frac{(-300)}{2} \sin 2\left(60^{\circ}\right)\right] \times 10^{-6} \\ & =\left[190+10 \cos 120^{\circ}-150 \sin 120^{\circ}\right) \times 10^{-6} \\ & =55.1 \times 10^{-6} \end{aligned}\)

Therefore, plane strain along \(x^1\) is \(55.1 \times 10^{-6}\).

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full solution

Title Mechanics of Materials 10 
Author Russell C. Hibbeler
ISBN 9780134319650

The state of strain at the point on the pin leaf has

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