EQUATION OF ELASTIC CURVE *59 



Fig. I. Assume that each differential load, y dx = fxdx, acts through 

 its center of gravity. Now construct a force polygon as in (c) and 

 an equilibrium polygon as in (b) Fig. I. 



Now, in (b) the tangent of the angle between any side of the 



dy 



equilibrium polygon and the X-axis is tan a . If the string b o in 



d x 



(b) is produced until it cuts the vertical line through 3, it will cut off 

 the intercept 3-3 1 , which is the difference between two consecutive 

 values of d y and therefore equals d 2 y. 



Now, it has been proved that the moment of the force acting 

 through point 2 in (b) about point 3, is equal to the intercept 3-3 1 

 multiplied by the pole distance //, is equal to 3-3 1 X H = d 2 y H, 

 But the moment of the differential load fxdx, which acts through 

 point 2, about point 3, is fx dx 2 , and 



fx dx 2 = d-y H 

 and 



- CD 



dx* H 



It is evident that (i) is the differential equation of the equi- 

 librium polygon in (b). 



Now, if the loading is taken so that 3; = fx = M, where M 

 represents the bending moment at any given point x, due to a given 

 loading, the equation for the equilibrium polygon becomes 



d-y M 



= (2) 



dx 2 PI 



From mechanics we have the relation that 



which differs from (2) only in having El substituted for H, E being 



the modulus of elasticity and / the moment of inertia of the given beam. 



This relation may be deduced as follows : In (d) Fig. I, let 



equilibrium polygon 123 represent the neutral axis of a beam as in 



