On the Best Form for a Balance-Beam. 21 



given strength is a minimum, and this can be accomplished 

 by the use of the methods of the differential calculus as 

 follows : — 



Let us suppose the material of the beam to possess equal 

 strength in compression and tension, then the beam will be 

 symmetrical about the line BC, the lower half being the 

 exact counterpart of the upper; let the length BC = 21, 

 DE = 2x, then under a given load W acting at B, the 

 tension on BD will be 



vv DE ~ vv 2x 



The amount of material in each part of the frame will be 

 proportional to the product of the stress into the length, 

 therefore, the amount of material in BD will be 



By symmetry the stresses on the four bars BD, DC, CE, EB 

 will be equal, and the material required for them will be 



I 2 + x 2 



4cW 



2x 



The compression on DA and the tension on AE will each 

 equal W, and the amount of material in them will 

 be 2c Wx 



The total material in the frame is 



m (l 2 + a? x 



And we wish to find the value of x, for which the quantity 

 in the bracket is a minimum. 

 Let 



I 2 

 V = 2x~ +X 



dx ~ 2x 2 ^ L W 



When y is a minimum or maximum -^- = 

 9 dx 



