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Unformatted text preview: =12640.6N (2) lets suppose the vertical distance of the centre of pressure is h, then h/ sinF=ydF that is to sayg hA(x)= gsinh/ sinI(x) so h/ sin= I(x)/ A(x)= I(x)/(yA) since I(x)=I+y*y*A so h/ sin=y+I/(yA) I=(1/4)* *(D/2) *(D/2) *(D/2) *(D/2)=(1/4) **(1.25/2) *(1.25/2) *(1.25/2) *(1.25/2)=0.1198 sin=60/x=150/(125+x), then x=250/3, so sin=18/25 h= [y+I/(yA)] sin={[1.05/(18/25)]+0.1198/[1.05/(18/25)**(1.25/2) *(1.25/2)]}* (18/25) =1.098m 3.15 A pair of lock gates, each 3 m wide, have their lower hinges at the bottom of the gates and their upper hinges 5 m from the bottom. The width of the lock is 5.5m. Find the reaction between the gates when the water level is 4.5 m above the bottom on one side and 1.5 m on the other. Assuming that this force acts at the same height as the resultant force due to the water pressure find the reaction forces on the hinges....
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This note was uploaded on 03/22/2010 for the course GEOMATICS LSGI taught by Professor Kady during the Spring '07 term at Hong Kong Shue Yan.
 Spring '07
 KADY

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