June, 1920] Springs of Minimum Inertia 323 



have in hand. Introducing the values of K, N, I, and Z in 

 equations 1 and 2 we find after reduction the following: 



BEAM OF CONSTANT STRENGTH, DEPTH CONSTANT, FIXED AT ONE END. 



(a) bdL = 6(E/f2)PA 



(b) bd- = C)PL/f 



(c) A = 6PLV(bd'^E) Check 



RECTANGULAR BEAMS FIXED AT ONE END. 



(a') bdL = 9(E/f2)PA 



(b') bd2 =(5PL/(f) 



(c') A = 4PLV(bd3E) Check 



SOLID ROUND ROD, FIXED AT ONE END. 



(a") 7rr2L=12(E/f2)PA 



(b'O r^= 4PL/(7rf) 



(c") A= 4PLV(37rErO Check 



At first sight it might seem that the three equations in each 

 of the above cases were independent of one another, but that is 

 not the case as the expression for the volume was obtained from 

 a combination of the other two. The equation for the flexture 

 is introduced to serve to check the computation. In the first 

 two cases there are three unknowns, b, d, and L; so that any 

 one of them may be assumed and the other two computed 

 which gives great flexibility in design; in the third case, how- 

 ever, there are but two unknowns and hence but a single solution 

 of the problem. As an illustration, let us assume — 



P = 100 lbs. 

 A = 0.1" 



f = 40,000 lbs. per square inch. 

 E = 29,000,000 lbs. per sq. in. 



First Case. From fa) bdL= 1.0872, from (b) bd-' = 0.015L. 

 Let us assume d = |" and we will then have L = 4.26", b = 1.02". 

 Second Case. From (a') bdL= 1.631, from (b') bd2 = 0.015L 

 and as above assume 6. = \" and we will have L = 5.21" and 

 b = 1.25". Third Case. From (a") r-L = 0.6922, from (b'O 

 t3 = 0.0031S1L, whence r= 0.2942 and L = 8.002. 



SPRINGS DEPENDING UPON TORSION (COILED SPRINGS). 



Under this class we will consider those springs which are 

 wound in the form of a helix and in whichi the action is such as to 

 :stretch or compress the spring along the axis of the helix. In 



