dd 

 so that for G — = 1 

 dx 



ro.i9o^ rK,^ 



-.^-0.024 > = the solution -JK^^, obtained by matrix inversion and multiplication as 



I-O.273J tK3j 



indicated in Sheet 3. Columns (2) and (3) of Sheet 3 come from Columns (l5) and nA) at 



the bottom of Sheet 1. Column (Z) = (?) • Kj + (s) • K2 + (6) • K3 , and is the shear flow 



H f) 

 per unit G — ; see statement following Equation [37]. The net torque is (2) • (?) , and 

 dx 



when Column (j) is divided by twice (for symmetry) this sum, then Column (8) is obtained.* 



The torsional flexibility is computed at the right of Sheet 3; see Equation [51] of Appendix 



A. 2 and pages 84 and 85. 



Comparison of these computed results shown on Sheets 1 through 3 can be made with 



Figure 2 and Table 3. The principal difference between these calculations and those used 



by the program are the units (in sample problem, scaling was done after computing, in program 



before computing) and the method of solving simultaneous equations. The weight calculation 



for the sample problem (discussed in section Inertial Parameters in Appendix A. 2, which 



gives results agreeing with Table 3a) is presented in Table 4. 



CONCLUSIONS 



A procedure has been developed for computing the inertia-elastic parameters of a ship 

 hull in a mechanized manner by use of a digital computer. This procedure requires the routine 

 tabulation of basic data systematically obtained in a prescribed fashion directly from ship 

 plans for use as input to the digital computer. The computer then calculates the ship param- 

 eters as output to be used in the finite difference form of the beam vibration equations 

 developed in Reference 1. Such mechanization fits the trend toward routinizing complex 

 calculations leading to eventual design utility. 



^y ^z "~ ''out "" ''' '^i' i^ found from Equation [s?] of Appendix A. 2 for G = 1. Then from Equations 



dx 



[11], [13], and [38] iq.| ^ _ = kj^^pi d^ = [l ] JK-! dO = |q. ! which is represented 



dx dx dx 



by Column (j). From Equation [26] the net torque is Z(^ • (7) where (7)and, therefore, the torque corresponds 



dfe* ^ d^ 

 to a value of G -— = 1. The shear flow for V = V = 0, M^ = 1 (but G ^ 1) is obtained from Equation 



[40]. Thusiqi ^ =0= .v'/^ /^= =0.888 • 10-3 • (2) or q . 10^ = 0.888 • 10^ which is 



' y "z "^ 2L{2)-{^ 2-563 ^^ J 



Column (¥). The factor 2 in the denominator has been added for symmetry. 



42 



