This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View Full
Document
 Spring '07
 KADY

Click to edit the document details