Skating on Tltin Ice. 11 



as before in (1) § 10, with a solution 



7] = B x ch qx cos qx -f B 2 sh ## sin gw, . . (4) 



to satisfy the condition of symmetry on each side of 0. 



The bending moment at the free end of the beam is zero, 

 and so at x= + Z, 



J? = ~~ 2Biq2 sh ql sin ql+ 2B ^ 2 ch ql cos ql = °' ( 5 ) 

 and we put 



~ /ch q,x cos ff.r , sh qx sin oa'\ /0 \ 



V = C( t 7 . 7 4— xS 7 )• • • ( 6 ) 



ysng'/sin^Z ch ql cos ql/ 



The load W is supported by the extra buoyancy due to 

 the downward displacement tj ; so that 



W = ijvdx, (7) 



in which 



j; 



q ch qx cos qxdx = %{ sh #Z cos ql + ch ^Z sin <^), (8) 



i 



1 # sh qx sin g^d^ = J( — sh <?Z cos^Z+ch t^Z sin ^Z), (9) 



ttt_ C / sh gZ cos gZ + ch ql sin gZ — sh ql cos ^Z-f-ch #Z sin ql\ 

 g\ sh ql sin ql ch ^Z cos #Z / 



0/ 1 i 1 \ = 2pC/ 1 1 \. (1Q) 



<? \sin <jZ cos <?Z shqlchql) q \sin 2^Z sh 2ql) ' 



so that if a was the extra draft of the plank due to the 

 load W distributed uniformly over the upper surface, 

 W = 2pla, 



n , sh2qlsin2ql . 



C = a * l ^2qlT^2qV (11) 



V_ _ a j cn 9l cos ql °h <l x cos qx + sh ql sin <^Z sh ^ sin qx ( . 

 a~ q ~ sh 2ql + sin 2ql ~' [U) 



12. With a series of parallel line-loads, W, at equal 

 interval 21, the surface of the ice will be bent into equal 

 waves joined together, given between x= ±Z by the general 

 solution of (4) § 11, with the origin midway between two 

 consecutive loads ; and with symmetry in the wave on each 



