328 Hertz's Theory of the Contact of Elastic Bodies. 



As 6 increases from to 90°, a|t also runs through the same 

 range but the variation of yjr is marked by a sudden transition 

 from a slow and steady increase over most of the range to a 

 very rapid increase near the end. For example, with 

 ^=1 we find ^ = 10*28' for = 80°, 



t/t = 90° for = 90°. 



This means that p l keeps near the direction of r as the 

 normal is departed from, being inclined to r on the side of 

 the normal ; but just before the radius vector reaches the 

 boundary there is a rapid tilting of the direction of p l9 

 bringing it perpendicular to r, i. e. to the boundary. Along 

 with this, of course, the magnitude of p x becomes zero. 



For the purpose of obtaining the forms of the lines of 

 principal stress, it is convenient to tabulate (6—-^), the 

 inclination of pi to the fixed direction of the normal. This 

 is found to have a maximum value at about 6 = 7 9°. The 

 lines of principal stress are thus two families of similar 

 curves, intersecting at right angles and having points of 

 inflexion along the radii 6— ±79°. Since the direction of the 

 tangent is known for each point of the plane, an approxi- 

 mation to the forms of the curves can be found*. The 

 result is shown on fig. 6. The forms are similar to those 

 given by Fuchs f for the outer curves in the more general 

 case. The broken line drawn to the intersection of the two 

 curves is the locus of inflexions. 



A remark may be added about the displacements in this 

 case. In the account of Boussin esq's solution given by 

 Love % it is mentioned that the particles move towards or 

 from the line of action of the applied force according as 

 they lie outside or inside a cone whose angle is given by 



cos 2 + cos0 = l-2cr. 



It may not have been noticed that the differential equation 

 of the "lines of displacement"" can be integrated in finite 

 form. Using polars, the differential equation is 



dr/r= -d0(l 4- cos 6) U(l— a) cos 0-(l- 2a) }/ 



sin 6{ (3 -4o-) cos + 2(1 -<r)}, 

 with the integral 



— log r = A log (1 — cos 6) 4- B log (cos 0+C)+ const., 



# d'Ocagne, Calcul Graphique et Nomographic, p. 155. 

 t See Love, loc. cit. p. 196. 

 t Loc. cit. p. 190. 



