30 



HYDRAULICS AND ITS APPLICATIONS 



Where the plane of the rectangle is vertical = and the above 



i ues 



more 



From the form of this result it is clear that as h increases, X becomes 

 nearly equal to (h + ^) *'.., the Centre of Pressure approaches more 



nearly to the centroid of the area. 



The position of the Centre of Pressure 

 may, in some instances, be deduced by 

 elementary methods. 



E.g., Parallelogram with base in sur- 

 face (Fig. 13 b). Divide the surface into 

 a series of elementary horizontal strips 

 of equal width. The pressure on each is 

 proportional to its distance from the 

 surface, and' will be represented by the 

 ordinate of the triangle E F H erected 

 on the strip and perpendicular to the 

 area as shown. This triangle may be 

 taken to represent the load diagram. 

 The single resultant of this load will 

 pass through the C. G. of the load area ; 

 will be perpendicular to the surface ; and 

 will therefore cut the median line E F at 



a distance from E equal to - E F, i.e., at P. 



o 



If the upper edge of the parallelogram be at a depth h below the 

 surface, the load diagram will now be a quadrilateral E K L F, such that 

 L K when produced meets F E produced in the surface. 



If E K = pi and FL = /7 2 , considering the pressure per unit width, we 

 may divide the load area into two, E K F and F K L (Fig. 14). 



The resultant of first 



second is 



= .EF = b. 



a 2i 



= ^ . E F = & b. 



This varying pressure may then be replaced by two single forces RI 

 and .R a , acting through the C. G* . G\ and <jr 2 of the two load areas, i.e., at 

 points distance $ E F from E and F respectively. 



