Solve Example Problem 19–2 again, but assume that the joint has a flexible elastomeric gasket separating the clamping members and that the stiffness of the bolt is then 10 times that of the joint.

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Accepted Answer

The procedure will be the same as that used previously, but now [latex]k_{b}=10 k_{c}[/latex]. Thus,
[latex]\begin{aligned}&F_{b}=P+\frac{k_{b}}{k_{b}+k_{c}} F_{e}=P+\frac{10 k_{c}}{10 k_{c}+k_{c}}F_{e}=P+\frac{10 k_{c}}{11 k_{c}} F_{e} \&F_{b}=P+10 F_{e}/ 11=4000+10(3000) / 11=6727\mathrm{Ib} \&F_{c}=P-\frac{k_{c}}{k_{b}+k_{c}} F_{e}=P-\frac{k_{c}}{10 k_{c}+k_{c}} F_{e}=P-\frac{k_{c}}{11 k_{c}} F_{e} \&F_{c}=P-F_{e} / 11=4000-3000 / 11=3727\mathrm{lb}\end{aligned}[/latex]

The stress in the bolt would be

[latex]\sigma=\frac{6727 lb}{0.0775 in^2}=86800[/latex] psi

This exceeds the proof strength of the Grade 5 material and is dangerously close to the yield strength.

Question 19.3: Solve Example Problem 19–2 again, but assume that the joint ...

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