A hollow shaft with an outside diameter of 57 mm and an inside diameter of 47 mm is subjected to torque T. 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 P13.78.
(a) If T = 900 N-m, determine the strain reading that would be expected from the gage.
(b) If the gage indicates a strain value of ε = −1,400 με, determine the torque T applied to the shaft.

Question

Accepted Answer

(a) The cross-sectional area of the hollow shaft is

[latex]A=\frac{\pi}{4}\left(D^{2}-d^{2} ight)=\frac{\pi}{4}\left[(57 mm )^{2}-(47 mm )^{2} ight]=816.814 mm ^{2}[/latex]

and the polar moment of inertia for the shaft is

[latex]J=\frac{\pi}{32}\left[D^{4}-d^{4} ight]=\frac{\pi}{32}\left[(57 mm )^{4}-(47 mm )^{4} ight]=557,271.413 mm ^{4}[/latex]

The maximum shear stress in the hollow aluminum shaft (i.e., the shear stress on the outer surface) is found from the elastic torsion formula:

[latex] au_{\max }=\frac{T c}{J}=\frac{(900 N - m )(57 mm / 2)(1,000 mm / m )}{557,271.413 mm ^{4}}=46.028 MPa[/latex]

The normal stresses in the x and y directions are zero; therefore, the stresses in the shaft at the location of the strain gage can be summarized as

[latex]\sigma_{x}=0 MPa , \quad \sigma_{y}=0 MPa , \quad au_{x y}=-46.028 MPa[/latex]

Note: The negative sign on [latex] au_{x y}[/latex] is determined by inspection. The stress element at the location of the strain gage looks like this:

From Eq. (13.18), determine the shear modulus G:

[latex]G=\frac{E}{2(1+v)}=\frac{70,000 MPa }{2(1+0.33)}=26,315.8 MPa[/latex]

and compute the shear strain [latex]\gamma_{x y}[/latex] from Eq. (13.22):

[latex]\gamma_{x y}=\frac{ au_{x y}}{G}=\frac{-46.028 MPa }{26,315.8 MPa }=-1,749.058 imes 10^{-6} rad[/latex]

Write a normal strain transformation equation for the gage oriented at θ = 55°:

[latex]\begin{aligned}\varepsilon_{n} &=\varepsilon_{x} \cos ^{2} heta+\varepsilon_{y} \sin ^{2} heta+\gamma_{x y} \sin heta \cos heta \&=(0 \mu \varepsilon) \cos ^{2}\left(55^{\circ} ight)+(0 \mu \varepsilon) \sin ^{2}\left(55^{\circ} ight)+(-1,749.058 \mu rad ) \sin \left(55^{\circ} ight) \cos \left(55^{\circ} ight) \&=-821.788 \mu \varepsilon\end{aligned}[/latex]

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

[latex]\varepsilon_{n}=-822 \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 -1,400 \mu \varepsilon=\gamma_{x y} \sin \left(55^{\circ} ight) \cos \left(55^{\circ} ight)\end{gathered}[/latex]

recognizing that the normal stresses in the x and y directions are zero.

From Eq. (13.22), substitute for [latex]\gamma_{x y}[/latex]:

[latex]-1,400 \mu \varepsilon=\frac{ au_{x y}}{G} \sin \left(55^{\circ} ight) \cos \left(55^{\circ} ight)[/latex]

Solve for [latex] au_{x y}[/latex]:

[latex] au_{x y}=\frac{\left(-1,400 imes 10^{-6} rad ight)(26,315.8 MPa )}{\sin \left(55^{\circ} ight) \cos \left(55^{\circ} ight)}=-78.413 MPa[/latex]

The torque T that causes this shear stress is

[latex]T=\frac{ au_{x y} J}{c}=\frac{\left(78.413 N / mm ^{2} ight)\left(557,271.413 mm ^{4} ight)}{(57 mm / 2)}=1,533,241 N - mm =1.533 kN - m[/latex]

Question 13.78: A hollow shaft with an outside diameter of 57 mm and an insi...

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