1909.] Elasticity of Rubber Balloons and Hollow Viscera. 497 



6. In the bladder of a large dog, giving sufficient range between the 

 assumption of globular form and the elastic limit to allow analysis of the 

 graph of pressure against radius, it was found that the equation 



(r — a)(p — b) = c 



was followed. In this case a was practically zero ; but like the rubber 

 balloon the behaviour was not that of a perfectly elastic and isotropic 

 substance. 



Note on the foregoing Paper by W. Sutherland. 



From the purely physical point of view the simplest way to prepare for a 

 theoretical interpretation of experiments such as these is to fix attention in the 

 first instance on tension per unit area. 



Let the tension per cm. 2 be t in the balloon or bladder which has radius r 

 and thickness z. Let initial values of these, when t — 0, be r and z - 

 Consider the equilibrium of a hemisphere. It experiences a pull 2irrzt 

 from the other hemisphere. But on account of the excess p of the pressure 

 inside the sphere over that outside the hemisphere is subject to a thrust 

 irr 2 p ; thus 



irr 2 p = 2 irrzt or pr 2 = 2 rzt. (1) 



If, as in studying the surface tension of bubbles, we fix attention on zt, the 

 total tension across unit width of cross-section of the bounding wall, and call 

 it T, we have 



p = 2T/r. (2) 

 According to Hooke's law, we write 



t = E(r-ro)/r , (3) 



where E is a modulus of elasticity appropriate to the conditions of the 

 experiment, which in the present case are equal tensions in two dimensions 

 and no external stress in the third dimension. For substances such as 

 rubber and most organic tissues which have a compressibility, small in 

 comparison with their deforniability, E for small strains is twice the ordinary 

 Young's modulus for small strains. But when large strains are used, as in 

 these experiments, E can no longer be treated as a constant. It is a function 

 of the strain. This appears when we compare Prof. Osborne's formula 

 (p — b)(r — a) = c with (1) and (3), after elimination of z by the relation 

 r 2 z = r % expressing incompressibility, for we get 



= b(r-a) + c _ 2 Esq (r-r )r _ ^ 



This makes E a complicated function of r. 



