A 2.5-in.-diameter solid aluminum post is subjected to a horizontal force of V = 9 kips, a vertical force of P = 20 kips, and a concentrated torque of T = 4 kip-ft, acting in the directions shown in Figure P15.76. Assume L = 3.5 in. The yield strength of the aluminum is σY = 50 ksi, and a minimum

Question

A 2.5-in.-diameter solid aluminum post is subjected to a horizontal force of V = 9 kips, a vertical force of P = 20 kips, and a concentrated torque of T = 4 kip-ft, acting in the directions shown in Figure P15.76. Assume L = 3.5 in. The yield strength of the aluminum is [latex]\sigma_{Y}[/latex] = 50 ksi, and a minimum factor of safety of [latex]FS _{\min }[/latex] = 1.67 is required by specification. Consider points H and K, and determine whether the aluminum post satisfies the specifications according to:
(a) the maximum-shear-stress theory.
(b) the maximum-distortion-energy theory.

Accepted Answer

Section properties:

[latex]\begin{array}{ll}A=\frac{\pi}{4}(2.5 ext { in. })^{2}=4.908739 ext { in } .^{2} \quad J=\frac{\pi}{32}(2.5 ext { in. })^{4}=3.834952 ext { in. }{ }^{4} \Q=\frac{(2.5 in. )^{3}}{12}=1.302083 in .^{3} \quad I_{x}=I_{z}=\frac{\pi}{64}(2.5 in .)^{4}=1.917476 in .{ }^{4}\end{array}[/latex]

Equivalent forces at H and K:

[latex]\begin{array}{lll}F_{x}=-9 ext { kips } & F_{y}=-20 ext { kips } & F_{z}=0 ext { kips } \M_{x}=0 ext { kip-in. } & M_{y}=48 ext { kip-in. } & M_{z}=(9 ext { kips })(3.5 in .)=31.5 ext { kip-in. }\end{array}[/latex]

Axial stress magnitude at H due to [latex]F _{y}[/latex]:

[latex]\sigma_{y}=\frac{20,000 lb }{4.908739 in. ^{2}}=4,074.367 psi[/latex]

Shear stress magnitude at H due to [latex]F _{x}[/latex]:

[latex] au_{x y}=\frac{(9,000 lb )\left(1.302083 in.^{3} ight)}{\left(1.917476 in .^{4} ight)(2.5 in .)}=2,444.620 psi[/latex]

Torsion shear stress magnitude at H due to [latex]M _{y}[/latex]:

[latex] au_{x y}=\frac{M_{y} c}{J}=\frac{(48,000 lb - in. )(2.5 in . / 2)}{3.834952 in .^{4}}=15,645.568 psi[/latex]

Summary of stresses at H:

[latex]\begin{aligned}\sigma_{x} &=0 psi \\sigma_{y} &=-4,074.367 psi \ au_{x y} &=-2,444.620 psi +15,645.568 psi \&=13,200.948 psi\end{aligned}[/latex]

Principal stress calculations for point H:

[latex]\begin{aligned}\sigma_{p 1, p 2} &=\frac{(0 psi )+(-4,074.367 psi )}{2} \pm \sqrt{\left(\frac{(0 psi )-(-4,074.367 psi )}{2} ight)^{2}+(13,200.948 psi )^{2}} \&=-2,037.183 psi \pm 13,357.213 psi\end{aligned}[/latex]

therefore, [latex]\sigma_{p 1}=11.320 ksi[/latex] and [latex]\sigma_{p 2}=-15.394 ksi[/latex]

Bending stress magnitude at K due to [latex]M _{ z }[/latex]:

[latex]\sigma_{y}=\frac{M_{z} x}{I_{z}}=\frac{(31,500 lb - in. )(2.50 in . / 2)}{1.917476 in .^{4}}=20,534.808 psi[/latex]

Shear stress magnitude at K due to [latex]M _{ y }[/latex]:

[latex] au_{y z}=\frac{M_{y} c}{J}=\frac{(48,000 lb - in .)(2.5 in . / 2)}{3.834952 in .{ }^{4}}=15,645.568 psi[/latex]

Summary of stresses at K:

[latex]\begin{aligned}&\sigma_{z}=0 psi \&\sigma_{y}=-4,074.367 psi +20,534.808 psi =16,460.442 psi \& au_{y z}=-15,645.568 psi\end{aligned}[/latex]

Principal stress calculations for point K:

[latex]\begin{aligned}\sigma_{p 1, p 2} &=\frac{(0 psi )+(16,460.442 psi )}{2} \pm \sqrt{\left(\frac{(0 psi )-(16,460.442 psi )}{2} ight)^{2}+(15,645.568 psi )^{2}} \&=8,230.221 psi \pm 17,678.245 psi\end{aligned}[/latex]

therefore, [latex]\sigma_{p 1}=25.908 ksi[/latex] and [latex]\sigma_{p 2}=-9.448 ksi[/latex]

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

[latex]\left|\sigma_{p 1}-\sigma_{p 2} ight|=|11.320 ksi -(-15.394 ksi )|=26.714 ksi[/latex]

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

[latex]FS _{H}=\frac{50 ksi }{26.714 ksi }=1.872[/latex]

Element K:

[latex]\left|\sigma_{p 1}-\sigma_{p 2} ight|=|25.908 ksi -(-9.448 ksi )|=35.356 ksi[/latex]

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

[latex]FS _{K}=\frac{50 ksi }{35.356 ksi }=1.414[/latex]

Since the factor of safety at K is less than the minimum required factor of safety, the aluminum post does not satisfy the specifications.

(b) Maximum-Distortion-Energy Theory:
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[(11.320 ksi )^{2}-(11.320 ksi )(-15.394 ksi )+(-15.394 ksi )^{2} ight]^{1 / 2} \&=23.225 ksi\end{aligned}[/latex]

[latex]FS _{H}=\frac{50 ksi }{23.225 ksi }=2.15[/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[(25.908 ksi )^{2}-(25.908 ksi )(-9.448 ksi )+(-9.448 ksi )^{2} ight]^{1 / 2} \&=31.706 ksi\end{aligned}[/latex]

[latex]FS _{K}=\frac{50 ksi }{31.706 ksi }=1.577[/latex]

Since the factor of safety at K is less than the minimum required factor of safety, the aluminum post does not satisfy the specifications.

Question 15.76: A 2.5-in.-diameter solid aluminum post is subjected to a hor...

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