The strain rosette shown in the figure was used to obtain normal strain data at a point on the free surface of a machine part. (a) Determine the strain components εx, εy, and γxy at the point. (b) Determine the principal strains and the maximum in-plane shear strain at the point. (c) Draw a sketch

Question

The strain rosette shown in the figure was used to obtain normal strain data at a point on the free surface of a machine part.
(a) Determine the strain components [latex]\varepsilon_{x}, \varepsilon_{y}, ext { and } \gamma_{x y}[/latex] at the point.
(b) Determine the principal strains and the maximum in-plane shear strain at the point.
(c) Draw a sketch showing the angle [latex] heta_{p}[/latex], the principal strain deformations, and the maximum in-plane shear strain distortions.
(d) Determine the magnitude of the absolute maximum shear strain.
[latex]\varepsilon_{a}=410 \mu \varepsilon, \quad \varepsilon_{b}=-540 \mu \varepsilon, \quad \varepsilon_{c}=-330 \mu \varepsilon, \quad v=0.30[/latex]

Accepted Answer

(a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where [latex]\varepsilon_{n}[/latex] is the measured normal strain. In each equation, the angle θ associated with each strain gage will be referenced from the positive x axis.

[latex]\begin{aligned}&410 \mu \varepsilon=\varepsilon_{x} \cos ^{2}\left(0^{\circ} ight)+\varepsilon_{y} \sin ^{2}\left(0^{\circ} ight)+\gamma_{x y} \sin \left(0^{\circ} ight) \cos \left(0^{\circ} ight) (a)\&-540 \mu \varepsilon=\varepsilon_{x} \cos ^{2}\left(45^{\circ} ight)+\varepsilon_{y} \sin ^{2}\left(45^{\circ} ight)+\gamma_{x y} \sin \left(45^{\circ} ight) \cos \left(45^{\circ} ight) (b)\&-330 \mu \varepsilon=\varepsilon_{x} \cos ^{2}\left(90^{\circ} ight)+\varepsilon_{y} \sin ^{2}\left(90^{\circ} ight)+\gamma_{x y} \sin \left(90^{\circ} ight) \cos \left(90^{\circ} ight) (c)\end{aligned}[/latex]

From Eq. (a):

[latex]\varepsilon_{x}=410 \mu \varepsilon[/latex]

and from Eq. (c):

[latex]\varepsilon_{y}=-330 \mu \varepsilon[/latex]

Using these two results, solve Eq. (b) to find [latex]\gamma_{x y}:[/latex]

[latex]\begin{aligned}-540 \mu \varepsilon &=(410 \mu \varepsilon) \cos ^{2}\left(45^{\circ} ight)+(-330 \mu \varepsilon) \sin ^{2}\left(45^{\circ} ight)+\gamma_{x y} \sin \left(45^{\circ} ight) \cos \left(45^{\circ} ight) \& herefore \gamma_{x y}=-1,160 \mu rad\end{aligned}[/latex]

(b) Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

[latex]\begin{aligned}\varepsilon_{p 1, p 2} &=\frac{\varepsilon_{x}+\varepsilon_{y}}{2} \pm \sqrt{\left(\frac{\varepsilon_{x}-\varepsilon_{y}}{2} ight)^{2}+\left(\frac{\gamma_{x y}}{2} ight)^{2}} \&=\frac{(410 \mu)+(-330 \mu)}{2} \pm \sqrt{\left(\frac{(410 \mu)-(-330 \mu)}{2} ight)^{2}+\left(\frac{-1,160 \mu}{2} ight)^{2}} \&=40 \mu \pm 687.9680 \mu\end{aligned}[/latex]

[latex]\begin{aligned}&\varepsilon_{p 1}=728 \mu \varepsilon ext { and } \varepsilon_{p 2}=-648 \mu \varepsilon \&\gamma_{\max }=1,376 \mu rad \quad ext { (maximum in-plane shear strain) }\end{aligned}[/latex]

[latex]\begin{aligned} an 2 heta_{p} &=\frac{\gamma_{x y}}{\left(\varepsilon_{x}-\varepsilon_{y} ight)}=\frac{-1,160 \mu}{[(410 \mu)-(-330 \mu)]}=\frac{-1,160 \mu}{740 \mu}=-1.5676 \&\left. herefore heta_{p}=-28.7^{\circ} \quad ext { (clockwise from the } x ext { axis to the direction of } \varepsilon_{p 1} ight)\end{aligned}[/latex]

(c) The principal strain deformations and the maximum in-plane shear strain distortions are shown on the sketch below.

(d) The problem states that the strain readings were obtained from the free surface of a machine part.
From this statement, we can conclude that a state of plane stress exists. For plane stress, the third principal strain [latex]\varepsilon_{2}=\varepsilon_{p 3}[/latex] is not equal to zero. The normal strain in the z direction can be computed from Eq. 13.15:

[latex]\varepsilon_{z}=-\frac{v}{1-v}\left(\varepsilon_{x}+\varepsilon_{y} ight)=-\frac{0.30}{1-0.30}[(410 \mu)+(-330 \mu)]=-34.2857 \mu \varepsilon[/latex]

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

[latex]\gamma_{ abs \max }=\gamma_{\max }=1,376 \mu rad[/latex]

Question 13.39: The strain rosette shown in the figure was used to obtain no...

Related Answered Questions