A steel shaft with an outside diameter of 20 mm is supported in flexible bearings at its ends. Two pulleys are keyed to the shaft, and the pulleys carry belt tensions as shown in Figure P15.77. The yield strength of the steel is σY = 350 MPa. (a) Determine the factors of safety predicted at points

Question

A steel shaft with an outside diameter of 20 mm is supported in flexible bearings at its ends. Two pulleys are keyed to the shaft, and the pulleys carry belt tensions as shown in Figure P15.77. The yield strength of the steel is [latex]\sigma_{Y}[/latex] = 350 MPa.
(a) Determine the factors of safety predicted at points H and K by the maximum-shearstress theory of failure.
(b) Determine the Mises equivalent stresses at points H and K.
(c) Determine the factors of safety at points H and K predicted by the maximumdistortion-energy theory.

Accepted Answer

Section properties:

[latex]\begin{array}{ll}A=\frac{\pi}{4}(20 mm )^{2}=314.159 mm ^{2} & J=\frac{\pi}{32}(20 mm )^{4}=15,707.963 mm ^{4} \Q=\frac{(20 mm )^{3}}{12}=666.667 mm ^{3} & I_{y}=I_{z}=\frac{\pi}{64}(20 mm )^{4}=7,853.982 mm ^{4}\end{array}[/latex]

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Equilibrium of entire shaft:

[latex]\begin{aligned}&\Sigma F_{z}=-A_{z}-D_{z}+300 N +2,100 N +1,100 N +200 N =0 \&\Sigma M_{A, y ext { axis }}=-(300 N )(150 mm )-(2,100 N )(150 mm ) \&\quad-(1,100 N )(450 mm )-(200 N )(450 mm )+(600 mm ) D_{z}=0\end{aligned}[/latex]

therefore

[latex]D_{z}=1,575 N ext { and } A_{z}=2,125 N[/latex]

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Detail of equivalent forces at H and K: Detail of equivalent moments at H and K:

[latex]\begin{aligned}F_{x} &=0 N \F_{y} &=0 N \F_{z} &=1,100 N +200 N -1,575 N \&=-275 N\end{aligned}[/latex] [latex]\begin{aligned}M_{x}=&(1,100 N )(120 mm )-(200 N )(120 mm ) \=& 108,000 N - mm \M_{y}=&(1,575 N )(300 mm )-(1,100 N )(150 mm ) \&-(200 N )(150 mm )=277,500 N - mm \M_{z}=& 0 N - mm\end{aligned}[/latex]

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Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest.

Consider point H.
Force [latex]F _{z}[/latex] creates a transverse shear stress in the xz plane at H. The magnitude of this shear stress is:

[latex] au_{x z}=\frac{(275 N )\left(666.667 mm ^{3} ight)}{\left(7,853.982 mm ^{4} ight)(20 mm )}=1.167 MPa[/latex]

Moment [latex]M _{ x }[/latex], which is a torque, creates a torsion shear stress in the xz plane at H. The magnitude of this shear stress is:

[latex] au_{x z}=\frac{M_{x} c}{J}=\frac{(108,000 N - mm )(20 mm / 2)}{15,707.963 mm ^{4}}=68.755 MPa[/latex]

Moment [latex]M _{ y }[/latex] does not create bending stress at H because H is located on the neutral axis for bending about the y axis.

Summary of stresses at H:

[latex]\begin{aligned}\sigma_{x} &=0 MPa \\sigma_{z} &=0 MPa \ au_{x z} &=-1.167 MPa +68.755 MPa =67.588 MPa\end{aligned}[/latex]

Principal stress calculations for point H:

[latex]\begin{aligned}\sigma_{p 1, p 2} &=\frac{(0 MPa )+(0 MPa )}{2} \pm \sqrt{\left(\frac{(0 MPa )-(0 MPa )}{2} ight)^{2}+(-67.588 MPa )^{2}} \&=0 MPa \pm 67.588 MPa\end{aligned}[/latex]

therefore, [latex]\sigma_{p 1}=67.588 MPa[/latex] and [latex]\sigma_{p 2}=-67.588 MPa[/latex]

Consider point K.
Force [latex]F _{z}[/latex] does not cause either a normal stress or a shear stress at K.

Moment [latex]M _{x}[/latex], which is a torque, creates a torsion shear stress in the xy plane at K. The magnitude of this shear stress is:

[latex] au_{x y}=\frac{M_{x} c}{J}=\frac{(108,000 N - mm )(20 mm / 2)}{15,707.963 mm ^{4}}=68.755 MPa[/latex]

Moment [latex]M _{y}[/latex] creates bending stress at K. The magnitude of this stress is:

[latex]\sigma_{x}=\frac{M_{y} z}{I_{y}}=\frac{(277,500 N - mm )(20 mm / 2)}{7,853.982 mm ^{4}}=353.324 MPa[/latex]

Moment [latex]M _{z}[/latex] does not create bending stress at K because K is located on the neutral axis for bending about the z axis.

Summary of stresses at K:

[latex]\begin{aligned}\sigma_{x} &=353.324 MPa \\sigma_{y} &=0 MPa \ au_{x y} &=-68.755 MPa\end{aligned}[/latex]

Principal stress calculations for point K:

[latex]\begin{aligned}\sigma_{p 1, p 2} &=\frac{(353.324 MPa )+(0 MPa )}{2} \pm \sqrt{\left(\frac{(353.324 MPa )-(0 MPa )}{2} ight)^{2}+(-68.755 MPa )^{2}} \&=176.662 MPa \pm 189.570 MPa\end{aligned}[/latex]

therefore, [latex]\sigma_{p 1}=366.232 MPa[/latex] and [latex]\sigma_{p 2}=-12.908 MPa[/latex]

(a) Maximum-Shear-Stress Theory
Element H:

[latex]\left|\sigma_{p 1}-\sigma_{p 2} ight|=|67.588 MPa -(-67.588 MPa )|=135.176 MPa[/latex]

The factor of safety associated with this state of stress is:

[latex]FS _{H}=\frac{350 MPa }{135.176 MPa }=2.59[/latex]

Element K:

[latex]\left|\sigma_{p 1}-\sigma_{p 2} ight|=|366.232 MPa -(-12.908 MPa )|=379.140 MPa[/latex]

The factor of safety associated with this state of stress is:

[latex]FS _{K}=\frac{350 MPa }{379.140 MPa }=0.923[/latex]

(b) Mises equivalent stresses at points H and K:
Element H:

[latex]\begin{aligned}\sigma_{M, H} &=\left[\sigma_{p 1}^{2}-\sigma_{p 1} \sigma_{p 2}+\sigma_{p 2}^{2} ight]^{1 / 2} \&=\left[(67.588 MPa )^{2}-(67.588 MPa )(-67.588 MPa )+(-67.588 MPa )^{2} ight]^{1 / 2} \&=117.066 MPa =117.1 MPa\end{aligned}[/latex]

Element K:

[latex]\begin{aligned}\sigma_{M, K} &=\left[\sigma_{p 1}^{2}-\sigma_{p 1} \sigma_{p 2}+\sigma_{p 2}^{2} ight]^{1 / 2} \&=\left[(366.232 MPa )^{2}-(366.232 MPa )(-12.908 MPa )+(-12.908 MPa )^{2} ight]^{1 / 2} \&=372.853 MPa =373 MPa\end{aligned}[/latex]

(c) Maximum-Distortion-Energy Theory:
Element H:

[latex]FS _{H}=\frac{350 MPa }{117.066 MPa }=2.99[/latex]

Element K:

[latex]FS _{K}=\frac{350 MPa }{372.853 MPa }=0.939[/latex]

Question 15.77: A steel shaft with an outside diameter of 20 mm is supported...

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