A 0.875-in.-diameter by 15-ft-long steel rod (1) is stress free after being attached to rigid supports. A clevis-and-bolt connection, as shown in Figure P5.50/51, connects the rod with the support at A. The normal stress in the steel rod must be limited to 18 ksi, and the shear stress in the bolt

Question

A 0.875-in.-diameter by 15-ft-long steel rod (1) is stress free after being attached to rigid supports. A clevis-and-bolt connection, as shown in Figure P5.50/51, connects the rod with the support at A. The normal stress in the steel rod must be limited to 18 ksi, and the shear stress in the bolt must be limited to 42 ksi. Assume E = 29,000 ksi and [latex]\alpha=6.6 imes 10^{-6} /{ }^{\circ} F[/latex] and determine:
(a) the temperature decrease that can be safely accommodated by rod (1) based on the allowable normal stress.
(b) the minimum required diameter for the bolt at A using the temperature decrease found in part (a).

Accepted Answer

Section properties:
For the 0.875-in.-diameter rod, the cross-sectional area is:

[latex]A_{1}=\frac{\pi}{4}(0.875 in .)^{2}=0.601320 in .^{2}[/latex]

Force-Temperature-Deformation Relationship
The relationship between internal force, temperature change, and deformation of an axial member is:

[latex]\delta_{1}=\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T L_{1}[/latex]

Since the rod is attached to rigid supports, [latex]\delta_{1}=0[/latex].

[latex]\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T L_{1}=0[/latex]

which can also be expressed in terms of the rod normal stress:

[latex]\sigma_{1} \frac{L_{1}}{E_{1}}+\alpha_{1} \Delta T L_{1}=0[/latex]

Solve for ΔT corresponding to a 24 ksi normal stress in the steel rod:

[latex]\begin{aligned}\Delta T &=-\sigma_{1} \frac{L_{1}}{E_{1}} \frac{1}{\alpha_{1} L_{1}}=-\frac{\sigma_{1}}{\alpha_{1} E_{1}} \&=-\frac{18 ksi }{\left(6.6 imes 10^{-6} /{ }^{\circ} F ight)(29,000 ksi )}=-94.0^{\circ} F\end{aligned}[/latex]

The normal force in the steel rod is:

[latex]F_{1}=(18 ksi )\left(0.601320 in. ^{2} ight)=10.823768 ext { kips }[/latex]

If the allowable shear stress in the bolt is 42 ksi, the minimum diameter required for the double shear bolt is

[latex]\begin{aligned}2\left[\frac{\pi}{4} d_{ bolt }^{2} ight] & \geq \frac{10.823768 kips }{42 ksi }=0.257709 in.^{2} \& herefore d_{ bolt } \geq 0.405 in.\end{aligned}[/latex]

Question 5.51 : A 0.875-in.-diameter by 15-ft-long steel rod (1) is stress f...

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