Tubes are expanded into a rectangular header that is 1.25 in. thick. The holes are 0.875-in. diameter with two 0.3125-in.-deep grooves in the hole for holding power. The grooves are 0.125 in. high with 0.25-in. spacing between them. The top groove is 0.25 in. from the top edge. What is the membrane

Question

Tubes are expanded into a rectangular header that is 1.25 in. thick. The holes are 0.875-in. diameter with two 0.3125-in.-deep grooves in the hole for holding power. The grooves are 0.125 in. high with 0.25-in. spacing between them. The top groove is 0.25 in. from the top edge. What is the membrane ligament efficiency if the openings are on 3 -in. centers?

Accepted Answer

Calculate the equivalent diameter [latex]D_{E}[/latex] using Eq. 17.4:

[latex]D_{E}=\frac{1}{t}\left(d_{0} T_{0}+d_{1} T_{1}+d_{2} T_{2}+\ldots+d_{n} T_{n} ight)[/latex] (17.4)

p = 3 in.

[latex]d_{0}=0.875 ext { in. } \quad T_{0}=0.25 ext { in. }[/latex]

[latex]d_{1}=0.9375 ext { in. } \quad T_{1}=0.125 ext { in. }[/latex]

[latex]d_{2}=0.875 ext { in. } \quad T_{2}=0.25[/latex]

[latex]d_{3}=0.9375 ext { in. } \quad T_{3}=0.125 ext { in. }[/latex]

[latex]d_{4}=0.875 ext { in. } \quad T_{4}=0.50 ext { in. }[/latex]

[latex]\begin{aligned}D_{E}=& \frac{1}{1.25}(0.875 imes 0.25+0.9375 imes 0.125+0.875 imes 0.25+\&0.9375 imes 0.125+0.875 imes 0.5)=0.888\end{aligned}[/latex]

The membrane ligament efficiency is

[latex]e_{m}=\frac{3-0.888}{3}=0.704[/latex]

Question 17.4: Tubes are expanded into a rectangular header that is 1.25 in...

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