(Here, the symbols q and E, represent the same real quantities as in the 

 section on the undamped beam; and, if e = 1 and g = 0, Equation [30] 

 becomes the same as Equation [3a] in that section.) 



Basic solutions for Equation [30] must now be sought. Assume that 

 M varies as e''" "*" . Then Equation [30] divided through by M gives for 

 T: 



X + 2^q (e - ig) A - q (e - ig) = 



. -2 . 

 Solving as a quadratic in X gives: 



—O o * o li OH 1/2 



X = - Cq (e - ig) + [^q (e - ig) + q (e - ig) ] 



Since the square root indicated here has tvo possible values, each 



the negative of the other, plus or minus alternatives are indicated twice 



-2 

 over, whereas x has really only two alternative values. It is conven- 

 ient, therefore, to select a particular value of the square root, which 

 will be denoted by a subscript + and will be defined presently. This 

 square root is chosen so that when e = 1 and g = it becomes the quantity 



q Vl + E, occurring in the theory of the undamped beam. In this latter 



2 2 



theory, it wsis found convenient to write X = q„ when the positive sign 



2 2 



in the plus or minus alternative was chosen but X = - q, when the 



— 2 

 negative sign was chosen. Analoguously , the alternative values of X 



— 2 — 2 — 2 



obtained from the last equation are taken to be x = - q or q where: 



[31a] 



—2 2/ 2 2 '^'^\ 



q, = q I ^(e - ig) + [e - ig + C (e - ig) ]^ \ 



q2 = q^ H(e - ig) + [e - ig + C^ (e - igf]^ ^} [31b] 



31 



