The strain rosette shown in the Figures P13.73–P13.76 was used to obtain normal strain data at a point on the free surface of a machine component. (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

Question

The strain rosette shown in the Figures P13.73–P13.76 was used to obtain normal strain data at a point on the free surface of a machine component.
(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) Using the results from part (b), determine the principal stresses and the maximum in-plane shear stress. Show these stresses on an appropriate sketch that indicates the orientation of the principal planes and the planes of maximum in-plane shear stress.
(d) Determine the magnitude of the absolute maximum shear stress at the point.

[latex]\begin{array}{|c|c|c|c|c|c|}\hline ext { Problem } & \varepsilon_{a} & \varepsilon_{b} & \varepsilon_{c} & E & v \\hline ext { P13.76 } & 55 \mu \varepsilon & -110 \mu \varepsilon & -35 \mu \varepsilon & 212 GPa & 0.30 \\hline\end{array}[/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}&55 \mu \varepsilon=\varepsilon_{x} \cos ^{2}\left(315^{\circ} ight)+\varepsilon_{y} \sin ^{2}\left(315^{\circ} ight)+\gamma_{x y} \sin \left(315^{\circ} ight) \cos \left(315^{\circ} ight) (a)\&-110 \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) (b) \&-35 \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) (c)\end{aligned}[/latex]

From Eq. (b):

[latex]\varepsilon_{x}=-110 \mu \varepsilon[/latex]

Using this result, solve Eqs. (a) and (c) simultaneously to obtain:

[latex]\varepsilon_{y}=130 \mu \varepsilon ext { and } \quad \gamma_{x y}=-90 \mu rad[/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{(-110 \mu)+(130 \mu)}{2} \pm \sqrt{\left(\frac{(-110 \mu)-(130 \mu)}{2} ight)^{2}+\left(\frac{-90 \mu}{2} ight)^{2}} \&=10 \mu \pm 128.1601 \mu\end{aligned}[/latex]

[latex]\begin{aligned}&\varepsilon_{p 1}=138.1601 \mu \varepsilon=138.2 \mu \varepsilon ext { and } \varepsilon_{p 2}=-118.1601 \mu \varepsilon=-118.2 \mu \varepsilon \&\gamma_{\max }=256.3201 \mu rad =256 \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{-90 \mu}{[(-110 \mu)-(130 \mu)]}=\frac{-90 \mu}{-240 \mu}=0.3750 \&\left. herefore heta_{p}=10.28^{\circ} \quad ext { (counterclockwise from the } x ext { axis to the direction of } \varepsilon_{p 2} ight)\end{aligned}[/latex]

(c) Use [latex]\varepsilon_{p 1} ext { and } \varepsilon_{p 2}[/latex] in Eqs. (13.23) to compute [latex]\sigma_{p 1}[/latex]:

[latex]\begin{aligned}\sigma_{p 1} &=\frac{E}{1-v^{2}}\left(\varepsilon_{p 1}+v \varepsilon_{p 2} ight)=\frac{212,000 MPa }{1-(0.3)^{2}}\left[\left(138.1601 imes 10^{-6} ight)+(0.3)\left(-118.1601 imes 10^{-6} ight) ight] \&=23.9285 MPa =23.9 MPa ( T )\end{aligned}[/latex]

and [latex]\sigma_{p 2}[/latex]:

[latex]\begin{aligned}\sigma_{p 2} &=\frac{E}{1-v^{2}}\left(\varepsilon_{p 2}+v \varepsilon_{p 1} ight)=\frac{212,000 MPa }{1-(0.3)^{2}}\left[\left(-118.1601 imes 10^{-6} ight)+(0.3)\left(138.1601 imes 10^{-6} ight) ight] \&=-17.8714 MPa =17.87 MPa ( C )\end{aligned}[/latex]

The maximum in-plane shear stress can be computed from the two principal stresses:

[latex] au_{\max }=\frac{\left|\sigma_{p 1}-\sigma_{p 2} ight|}{2}=\frac{|(23.9285 MPa )-(-17.8714 MPa )|}{2}=20.8999 MPa =20.9 MPa[/latex]

and the normal stress on the plane of maximum shear stress is

[latex]\sigma_{ avg }=\frac{\sigma_{p 1}+\sigma_{p 2}}{2}=\frac{(23.9285 MPa )+(-17.8714 MPa )}{2}=3.0286 MPa =3.03 MPa ( T )[/latex]

(d) For plane stress, [latex]\sigma_{z}=\sigma_{p 3}=0[/latex]. Since [latex]\sigma_{p 1} ext { and } \sigma_{p 2}[/latex] are of opposite signs,

[latex] au_{ abs \max }= au_{\max }=20.9 MPa[/latex]

Question 13.76: The strain rosette shown in the Figures P13.73–P13.76 was us...

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