A 1.25-in.-diameter solid shaft is subjected to an axial force of P = 520 lb, a horizontal shear force of V = 275 lb, and a concentrated torque of T = 880 lb-in., acting in the directions shown in Figure P12.14. Assume L = 7.0 in. Determine the normal and shear stresses at (a) point H and (b) point

Question

A 1.25-in.-diameter solid shaft is subjected to an axial force of P = 520 lb, a horizontal shear force of V = 275 lb, and a concentrated torque of T = 880 lb-in., acting in the directions shown in Figure P12.14. Assume L = 7.0 in. Determine the normal and shear stresses at (a) point H and (b) point K. For each point, show these stresses on a stress element.

Accepted Answer

Section properties:

[latex]\begin{array}{ll}A=\frac{\pi}{4}(1.25 in .)^{2}=1.227185 in. ^{2} & J=\frac{\pi}{32}(1.25 in .)^{4}=0.239684 in. ^{4} \Q=\frac{(1.25 in .)^{3}}{12}=0.162760 in .{ }^{3} & 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}=-520 lb \&F_{y}=0 lb \&F_{z}=275 lb\end{aligned}[/latex]

Equivalent moments at H and K:

[latex]\begin{aligned}M_{x} &=880 lb - in .\M_{y} &=-(275 lb )(7 in .)=-1,925 lb - in . \M_{z} &=0 lb ext {-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.

(a) 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{520 lb }{1.227185 in .{ }^{2}}=423.734 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{(275 lb )\left(0.162760 in. ^{3} ight)}{\left(0.119842 in. ^{4} ight)(1.25 in .)}=298.787 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{(880 lb - in .)(1.25 in. / 2)}{0.239684 in. ^{4}}=2,294.683 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} &=-424 psi \\sigma_{z} &=0 psi \ au_{x z} &=298.787 psi +2,294.683 psi =2,590 psi\end{aligned}[/latex]

(b) 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{520 lb }{1.227185 in .^{2}}=423.734 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{(880 lb - in .)(1.25 in . / 2)}{0.239684 in. ^{4}}=2,294.683 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{(1,925 lb - in .)(1.25 in . / 2)}{0.119842 in .^{4}}=10,039.240 psi[/latex]

Summary of stresses at K:

[latex]\begin{aligned}\sigma_{x} &=-423.734 psi -10,039.240 psi \&=-10,460 psi \\sigma_{y} &=0 psi \ au_{x y} &=-2,290 psi\end{aligned}[/latex]

Question 12.14: A 1.25-in.-diameter solid shaft is subjected to an axial for...

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