92 Mr. P. F. Ward on the Transverse 



Therefore the point of inflexion is at 



•V = 9-91236, z/l = -651829. 

 The form of the rod is shown in fig. 2. 



Fig-. 2. 



In the third mode x 1 = 30'020, and therefore 



u = -C{7003-06^ (^) + -178460^0 W}. 

 To find the nodes we have 



V ' x. 



X. 



o (jf). 



U*)- 



u/C. 



J3-82 

 13-84 



/7-00 

 17-02 



3-6481 

 3-6864 



12-25 

 12-3201 



4-39630 



44-5826 

 45-2971 



- -00173794 



- -00133795 

 + ■000377234 



-18-0871 

 + 11-3863 



+ 1-41354 

 -10-72551 



Hence the nodes are given by 



x = 3-67160, 12-25816, 

 zjl= -122305, -408334. 



To find the points of maximum excursion we have 



2*Jx. 



x. 



dx ' 



dx 



4:/°- 



rs-12 



15-14 



J8-46 

 18-48 



65536 

 66049 



17-8929 

 17-9776 



300776 

 304395 



28-6109 

 290341 



- -000810898 

 + -000225034 



- -000646653 

 - -000943240 



+5-14201 

 -211915 



- -57735 

 + 1-42413 



The required points together with the actual displacement 

 are then as given below : 



2s/x. 



x. 



(p Q (x). 



10-72545 

 146-746 



til*). 



u/C. 



z/h 



5-13417 

 8-46577 



6-58993 

 1791733 



-132279 

 + -0644071 



+924445 

 -477-235 



•219523 

 •596815 



