386 BELL SYSTEM TECHNICAL JOURNAL 



300, .T2 = I, we find P/db = 120, x./db = .35. This falls slightly above the 

 ko — 30(0) curve as required. The whole experimental curve is then 

 plotted to the coordinates P/2.5 vs .V2/2.5 and is found to fit as closely as 

 necessary. Hence the parameters are adopted as ^0 = 305, db = 2.5. 



We can now calculate the maximum acceleration that the tube will 

 receive in, say, a three-foot drop test. First calculate, from equations (1) 

 and (2), 



' 2hW2 ^ /2 X 36 X 22.5 ^ ^ 31 



2hko _ , / - ^N ^^^^^^^^^^ _ .. 

 22.5 



Then db/do = 1.08. Entering Fig. 4 with this value we find G,n/Go = 1.82. 

 Hence the maximum acceleration is: 



Grn = 31.3 X 1.82 = 57g. 



Finally, entering Fig. 5 with db/do = 1.08 we find d,n/db = 0.8. Hence the 

 maximum displacement is dm = 0.8 X 2.5 = 2.0 inches. This indicates 

 that the load-displacement test was carried far enough to cover the range 

 up to a three-foot drop. 



It may be observed that the results obtained, by treating the same data 

 as Class B or Class C cushioning, agree within a few per cent. This is 

 because, in the example chosen, both B and C curves can be made to fit the 

 experimental load-displacement curve. 



1.12 Consequences of x\brupt Bottoming (Class D) 

 It is useful to examine cushioning systems that can bottom more abruptly 

 than Class C cushioning. Abrupt bottoming is possible, for example, in a 

 tension spring package lacking a snubbing device. An estimate of the 

 increase in acceleration can be made by studying the case of bilinear elasticity 

 (Fig. 1.4.4). Here we have a spring rate ^0 up to a displacement ds, follow- 

 ing which the cushioning has a different spring rate kb . ^0 represents the 

 average spring rate before bottoming and kb can represent the much greater 

 stiffness of the wall of the container. 

 If do > ds, that is, if 



1/ 



2^^^^i>^., (1.12.1) 



the suspended article will bottom and the maximum displacement and 

 acceleration are obtained by using both of the equations (1.4.4) in evaluating 

 the integral in (1.2.15). Thus, 



[ ' koX2 dx2 + I "' [kbX; - {kb - ko)d,\ dx, = W,h. (1.12.2) 



Jo Jrl. 



