A solid 18-mm-diameter shaft is subjected to an axial load P. The shaft is made of aluminum [E = 70 GPa; ν = 0.33]. A strain gage is mounted on the shaft at the orientation shown in Figure P13.77.
(a) If P = 14.7 kN, determine the strain reading that would be expected from the gage.
(b) If the gage indicates a strain value of ε = 810 με, determine the axial force P applied to the shaft.

Question

Accepted Answer

(a) From the given diameter, the cross-sectional area of the shaft is

[latex]A=\frac{\pi}{4}(18 mm )^{2}=254.46900 mm ^{2}[/latex]

and thus, the normal stress in the shaft is

[latex]\sigma_{x}=\frac{P}{A}=\frac{(14.7 kN )(1,000 N / kN )}{254.46900 mm ^{2}}=57.767 MPa[/latex]

At the location of the strain gage, the stresses in the shaft can be summarized as

[latex]\sigma_{x}=57.767 MPa , \quad \sigma_{y}=0 MPa , \quad au_{x y}=0 MPa[/latex]

From Eqs. (13.21), the normal strains in the x and y directions are

[latex]\begin{aligned}&\varepsilon_{x}=\frac{1}{E}\left(\sigma_{x}-v \sigma_{y} ight)=\frac{1}{70,000 MPa }[57.767 MPa -(0.33)(0 MPa )]=825.248 imes 10^{-6} mm / mm \&\varepsilon_{y}=\frac{1}{E}\left(\sigma_{y}-v \sigma_{x} ight)=\frac{1}{70,000 MPa }[0 MPa -(0.33)(57.767 MPa )]=-272.332 imes 10^{-6} mm / mm\end{aligned}[/latex]

and since the shear stress is zero, the shear strain is also zero: [latex]\gamma_{x y}[/latex] = 0.

Write a normal strain transformation equation for the gage oriented at θ = 145°:

[latex]\begin{aligned}\varepsilon_{n} &=\varepsilon_{x} \cos ^{2} heta+\varepsilon_{y} \sin ^{2} heta+\gamma_{x y} \sin heta \cos heta \&=(825.248 \mu \varepsilon) \cos ^{2}\left(145^{\circ} ight)+(-272.332 \mu \varepsilon) \sin ^{2}\left(145^{\circ} ight)+(0 \mu rad ) \sin \left(145^{\circ} ight) \cos \left(145^{\circ} ight) \&=464.155 \mu \varepsilon\end{aligned}[/latex]

Therefore, the strain gage should be expected to read a normal strain of

[latex]\varepsilon_{n}=464 \mu \varepsilon[/latex]

(b) A normal strain transformation equation can be written for the gage:

[latex]\begin{gathered}\varepsilon_{n}=\varepsilon_{x} \cos ^{2} heta+\varepsilon_{y} \sin ^{2} heta+\gamma_{x y} \sin heta \cos heta \ herefore 810 \mu \varepsilon=\varepsilon_{x} \cos ^{2}\left(145^{\circ} ight)+\varepsilon_{y} \sin ^{2}\left(145^{\circ} ight)\end{gathered}[/latex]

Recognize that there is no shear stress [latex] au_{x y}=0[/latex], and hence, [latex]\gamma_{x y}=0[/latex].

From Eqs. (13.21), substitute for [latex]\varepsilon_{x} ext { and } \varepsilon_{y}:[/latex]

[latex]810 \mu \varepsilon=\frac{1}{E}\left(\sigma_{x}-v \sigma_{y} ight) \cos ^{2}\left(145^{\circ} ight)+\frac{1}{E}\left(\sigma_{y}-v \sigma_{x} ight) \sin ^{2}\left(145^{\circ} ight)[/latex]

and eliminate terms of [latex]\sigma_{y}[/latex] since [latex]\sigma_{y}[/latex] = 0 for the shaft:

[latex]810 \mu \varepsilon=\frac{\sigma_{x}}{E} \cos ^{2}\left(145^{\circ} ight)-\frac{v \sigma_{x}}{E} \sin ^{2}\left(145^{\circ} ight)[/latex]

Solve for [latex]\sigma_{x}[/latex]:

[latex]\begin{aligned}\sigma_{x} &=\frac{(810 \mu \varepsilon) E}{\left[\cos ^{2}\left(145^{\circ} ight)-v \sin ^{2}\left(145^{\circ} ight) ight]} \&=\frac{\left(810 imes 10^{-6} ight)(70,000 MPa )}{\left[\cos ^{2}\left(145^{\circ} ight)-(0.33) \sin ^{2}\left(145^{\circ} ight) ight]} \&=100.810 MPa\end{aligned}[/latex]

The axial load P that causes this normal stress is

[latex]P=\sigma_{x} A=\left(100.810 N / mm ^{2} ight)\left(254.46900 mm ^{2} ight)=25,653.06 N =25.7 kN[/latex]

Question 13.77: A solid 18-mm-diameter shaft is subjected to an axial load P...

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