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Design and Stress Analysis of Extraterrestrial ... - The Black Vault

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We assume w.= A0 +Alr + A 2 r2 + A 3r3; substituting this value<br />

into equation (5.46), we obtain<br />

w,= A, + 2A 2 r 4.3A 3 0; w, 2A 2 -6A 3 r; wV i= 6eA; 5=0.<br />

Substituting the values <strong>of</strong> wL; w'; w"; w, 1 wq into equation (5.46)<br />

1i 4 14<br />

<strong>and</strong> equating the terms with identical functions relative to r, we<br />

obtain<br />

A. = 0; A, = a tg OBo;" A 2 = a tg.e OB; A 3 = a tg 9D 2 .<br />

Finally,<br />

u,,, = 523,5 a tg O.r + 151 a tg , 2 -<br />

6,6la tg Or.<br />

3. Let us find the general solution to equation (5.46):<br />

lp = !o[e-0 (C, sin + C 2 cos Li) + eQ (C 3 s;InQ+ C 4 CO- Q)0 + w,.<br />

Here<br />

'-ro)<br />

23 F 12 (1 tL2)<br />

2-, t9 11<br />

Let us establish the length ,;<br />

jj<br />

the shell from formula<br />

3 VfoI.<br />

Usually, in the problems which we are examining the sheils are<br />

long. In our example, the shell is short since<br />

1, 3V0,412,8 5, 7R c..e.<br />

569

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