A 1.25-in.-diameter solid shaft is subjected to an axial force of P = 7,000 lb, a horizontal force of V = 1,400 lb, and a concentrated torque of T = 220 lb-ft, acting in the directions shown in Figure P15.85. Assume L = 6.0 in. The ultimate failure strengths for this material are 36 ksi in tension

Question

A 1.25-in.-diameter solid shaft is subjected to an axial force of P = 7,000 lb, a horizontal force of V = 1,400 lb, and a concentrated torque of T = 220 lb-ft, acting in the directions shown in Figure P15.85. Assume L = 6.0 in. The ultimate failure strengths for this material are 36 ksi in tension and 50 ksi in compression. Use the Mohr failure criterion to evaluate the safety of this component at points H and K. Support your answers with appropriate documentation.

Accepted Answer

Section properties:

[latex]\begin{array}{ll}A=\frac{\pi}{4}(1.25 in. )^{2}=1.227185 in .^{2} \quad J=\frac{\pi}{32}(1.25 in .)^{4}=0.239684 in .^{4} \Q=\frac{(1.25 in .)^{3}}{12}=0.162760 in .^{3} \quad I_{y}=I_{z}=\frac{\pi}{64}(1.25 in .)^{4}=0.119842 in. { }^{4}\end{array}[/latex]

Equivalent forces at H and K:

[latex]\begin{aligned}&F_{x}=-7,000 lb \&F_{y}=0 lb \&F_{z}=1,400 lb\end{aligned}[/latex]

Equivalent moments at H and K:

[latex]\begin{aligned}&M_{x}=220 lb - ft =2,640 lb - in . \&M_{y}=-(1,400 lb )(6 in .)=-8,400 lb ext {-in. } \&M_{z}=0 lb - in.\end{aligned}[/latex]

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 _{x}[/latex] creates an axial stress at H. The magnitude of this normal stress is:

[latex]\sigma_{x}=\frac{7,000 lb }{1.227185 in .^{2}}=5,704.113 psi[/latex]

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{(1,400 lb )\left(0.162760 in .^{3} ight)}{\left(0.119842 in. ^{4} ight)(1.25 in .)}=1,521.097 psi[/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{(2,640 lb - in. )(1.25 in . / 2)}{0.239684 in .^{4}}=6,884.050 psi[/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}=-5,704.113 psi \&\sigma_{z}=0 psi \& au_{x z}=1,521.097 psi +6,884.050 psi =8,405.147 psi\end{aligned}[/latex]

Principal stress calculations for point H:

[latex]\begin{aligned}\sigma_{p 1, p 2} &=\frac{(-5,704.113 psi )+(0 psi )}{2} \pm \sqrt{\left(\frac{(-5,704.113 psi )-(0 psi )}{2} ight)^{2}+(-8,405.147 psi )^{2}} \&=-2,852.057 psi \pm 8,875.850 psi\end{aligned}[/latex]

therefore, [latex]\sigma_{p 1}=6,023.794 psi[/latex] and [latex]\sigma_{p 2}=-11,727.907 psi[/latex]

Mohr failure criterion at point H:

[latex]\begin{aligned}\frac{\sigma_{p 1}}{\sigma_{U T}}-\frac{\sigma_{p 2}}{\sigma_{U C}} &=\frac{6,023.794 psi }{36,000 psi }-\frac{-11,727.907 psi }{50,000 psi } \&=0.167-(-0.235) \&=0.402 \quad herefore ext { acceptable }\end{aligned}[/latex]

Consider point K.
Force [latex]F _{x}[/latex] creates an axial stress at K. The magnitude of this normal stress is:

[latex]\sigma_{x}=\frac{7,000 lb }{1.227185 in .^{2}}=5,704.113 psi[/latex]

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{(2,640 lb - in .)(1.25 in . / 2)}{0.239684 in. ^{4}}=6,884.050 psi[/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{(8,400 lb - in .)(1.25 in . / 2)}{0.119842 in .^{4}}=43,807.591 psi[/latex]

Summary of stresses at K:

[latex]\begin{aligned}\sigma_{x} &=-5,704.113 psi -43,807.591 psi =-49,511.704 psi \\sigma_{y} &=0 psi \ au_{x y} &=-6,884.050 psi\end{aligned}[/latex]

Principal stress calculations for point K:

[latex]\begin{aligned}\sigma_{p 1, p 2} &=\frac{(-49,511.704 psi )+(0 psi )}{2} \pm \sqrt{\left(\frac{(-49,511.704)-(0 psi )}{2} ight)^{2}+(-6,884.050 psi )^{2}} \&=-24,755.852 psi \pm 25,695.182 psi\end{aligned}[/latex]

therefore, [latex]\sigma_{p 1}=939.330 psi[/latex] and [latex]\sigma_{p 2}=-50,451.034 psi[/latex]

Mohr failure criterion at point K:

[latex]\begin{aligned}\frac{\sigma_{p 1}}{\sigma_{U T}}-\frac{\sigma_{p 2}}{\sigma_{U C}} &=\frac{939.330 psi }{36,000 psi }-\frac{-50,451.034 psi }{50,000 psi } \&=0.026-(-1.009) \&=1.035 \quad herefore ext { not acceptable }\end{aligned}[/latex]

Question 15.85: A 1.25-in.-diameter solid shaft is subjected to an axial for...

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