4. By appealing to the equations of elasticity, and making assumptions about no change 



of shape of the cross section, it can be shown that the integral around any closed path per 



ds 

 unit length through the section ^ q equals twice the enclosed area times the rate of 



^^x 1 ^qds 



twist, or = . (Rate of twist = dd /dx); see Chapters 16 and 17 of Reference 



dx 2A Gt ^ " ' 



8, or Chapter VII of Reference 9; also, see pages 69-72 of Appendix A. 2. By assum.ing that 



the plate segments are straight lines, it is possible to compute these areas by knowing about 



the connections made and the locations of the nodes. Since q is a constant along each path, 



Y^ As 



the integral may be replaced by a sum z_, (-q) , where +q. is used if the positive 



^ Gt 

 direction assigned to the unknown q. is in the same direction as the positive direction of the 



qds 

 closed loop; see Equation [23] of Appendix A. 2. The integral $ — — is evaluated around 



Gt 



each of the loops. In the sample problem, remember that the tree is formed by omitting plates 

 5 and 9. For example, the loop formed by the tree and plate 9 is shown heavy in Figure 8, 

 and the area associated with this loop is shaded. This area can be written (see page 72 of 

 Appendix A.2): ^ ^ ^ §. (RAs). = ^ ^S. (y^^ - y,z,)j 



where the summation is over Plates 2, 3, 8, and 9 which make up the loop. Here R is the 

 perpendicular distance from the origin to the plate and As is the length of the plate. See 

 Figure 9, in which the contribution of Plate 2 to this summation is shaded. The. proper sign 

 of (RAs). is assured by the head-tail polarity of the plate. The entry into the summation is 

 multiplied by (5= +1) if the direction around the loop agrees with the sense of plate j and by 

 (S. = -1) if the direction around the loop is contrary to the polarity of plate j. In this example, 

 §,. 5c Sg, and §„ are -1, -1, -1, and 1, respectively. (See Column (§) , Table 4, Sheet 3, 

 and the discussion of the [L.J matrix in Appendix A.2, pages 68-75.) q is constant on 

 any plate. Positive values of q in the plates are given by the arrows in Figure 8. 



5. The resultant torque must be equal to the moment of the shear flows (see Chapter 6 

 and Figure 6.15 of Reference 8 or page 219 of Reference 9.) The resultant torque or twisting 

 moment about the x-axis is M^ = Sq. (yjZ^- ~ yh'^f^ (^®® Equation [26], Appendix A.2). 



It represents moment about the x-axis. Its polarity is given by the 9^ arrow in Figure 3. 

 Since the shear flow on a plate is a constant, the total force is just q times the length. The 

 moment about the origin is the net force times the moment arm. Therefore, the net torque due 

 to one plate about the origin is given simply by the shear flow times twice the area of the 

 triangle which would be formed if the ends of the plate were joined to the origin by straight 

 lines (see Equation [26] of Appendix A.2). 



The above assumptions give exactly the correct amount of equations to yield a unique 

 solution for the shear flows in the panels in terms of M^, V and V . Only a general discus- 

 sion of the method of solution is given here. See pages 71-78 for a detailed discussion. 



56 



