Three solutions for the q's are needed, one for M^ = i, V = V^ = 0; one for M^= 0, 



V = 1, V =0; etc. M is the twisting moment about the x-axis. There is no dependence of 

 (a^x)- on twisting moment. For that reason the solution for plate shear flows corresponding 

 toM =1,V =V =0 has no component due to q ^^^ , which is based on (^qout)^ and, there- 

 fore, on (a ). ; this solution arises entirely from C[^ for the various loops (see Item 4 given 

 previously). The reason for three solutions for q., the shear flows in the plates, being needed 

 is covered extensively elsewhere in Appendix A. 2. In particular, refer to the section Shear 

 and Torsion, the section following Equation [26] through page 75. 



The solution per unit torque M^^ is the solution for the shear flows in the plates q. 

 existing when V = V = 0, M =1. It is found as follows: First find the most general set 



D y Z ' X " 



of shear flows which satisfy the condition that the sum of the q's out of every node vanishes 

 (see pages 54-55 and 75-76). This set can be found in terms of circulating flows. ("Circulating 

 flows" refer to the plate shears in the loops, q^^^ . The section Shear and Torsion in 

 Appendix A. 2 elaborates on the point.) In the computer program this was done by first finding 

 a tree, that is, a set of plates so that one and only one path exists between every two nodes. 

 For each plate not on the tree, there exists a closed loop through that plate and others in the 

 tree. The most general set of shear flows that have zero net flow out of each node consists 

 of a linear combination of flows in these loops. The new unknowns are the flows in the loops. 

 By integrating around those loops (see Item 4, page 56), there will be sufficient equations to 

 solve for these unknowns, but a new unknown (6d^/d\) is introduced (Equation [37] of 

 Appendix A. 2). The shear flows can now be written in terms of this one unknown, which can 

 be found since the resultant torque is to be unity (Equations [38]— [41] of Appendix A. 2). 



The details of the method of solution for the plate shears q. existing for unit y-shear 



V (or per unit z-shear V ) are presented in the section Shear and Torsion in Appendix A. 2. 

 The location of the center of shear y, z" is also determined from V^ shear and V shear, 

 respectively. In particular, see the paragraph following Equation [31]. 



The inertial parameters come from structural and nonstructural items. Structural items 

 include ship hull, deck, longitudinal members, etc. Nonstructural items include machinery, 

 cargo, superstructure, transverse bulkheads, the virtual mass of the water, etc. The inertial 

 parameters which must be calculated are mass of a section LM, the position of the center of 

 gravity Y-C.G., Z-C.G.; the rotary inertias I-YY, I-ZZ, I-MYZ; and the polar moment of interia 

 I-MX. For symmetric motions of a symmetric ship, only SM and I-ZZ are needed. For anti- 

 symmetric motion of a symmetric ship only SM, Z-C.C, I-YY, and I-MX are needed. For the 

 general case, however, all parameters are needed. Y-C.G. and I-MYZ vanish in the case of 

 symmetry. SM is obtained by adding all mass items in a Ax section. The Y-C.G. and Z-C.G. 

 come from dividing the mass m.oments by the total mass. The rotary inertia will not be calcu- 

 lated by the equation on page 35 of Reference 1, since this can be properly evaluated from the 

 input data (see section Inertial Parameters in Appendix A. 2). 



58 



