570 hersey: stiffness of elastic systems 



applications to the aneroid barometer, which will be treated 

 elsewhere, but their applicability is not confined to that par- 

 ticular elastic system. 



The stiffness of coupled systems. 3 Having given the individual 

 stiffness s a and s b of two systems A and B (ordinarily but not 

 necessarily single bodies) which are coupled together, let it be 

 required to find the stiffness, S, of the coupled system. It is 

 understood that s a and s b refer to forces and displacements at 

 the coupling, while S refers to a displacement at the coupling, 

 but to a force applied anywhere. The desired general relation 

 is found to be 4 



S = \(s a + s b ) . (1) 



where X is a dimensionless characteristic of the component to 

 which the external force' is applied, and denotes the ratio of its 

 stiffness with respect to that force, to its stiffness (s a or s b ) 

 with respect to a force at the coupling. If the external force is 

 itself applied at the coupling, X = 1, and the stiffness of the 

 coupled system equals the sum of the stiffnesses of the compo- 

 nents. This can be shown by a simple experiment. Make a 

 rigid frame about a foot square and cut off two five-inch lengths 

 of the same helical spring. Fasten the two springs, respec- 

 tively, each by one end, to two opposite sides of the frame, in 



3 In treating coupled systems, the displacements are assumed small. This 

 restriction does not apply to the subsequent discussion of a single body. 



4 Let x be the displacement of a reference mark at the fraction r of the dis- 

 tance along the coupling piece from A toward B, while x a and x b are the respec- 

 tive displacements necessary for coupling the two members, t being the tension 

 in the coupling, and F the particular external force, applied to A, with respect 

 to which we seek the stiffness of the coupled system, S. Then, by definition, 



F . . 



S = — , while the conditions of equilibrium and constraint are expressed by the 

 x 



four equations 



x =(l — r)x a — rxb x a + x b = constant 



F t t 



x a = -| h const. xb = h const. 



Xs a s a s b 



in which the forces and displacements are interpreted vectorially. Eliminating 

 the three quantities x a , x b , and t gives 



F = \(s a + s b )x + K 

 in which K denotes a term involving r, s a , and s b , but obviously vanishing if 

 the system is so adjusted that x = when F = 0. Dividing through by x now 

 gives the result (1). 



