G. F. Becker — Finite Elastic Stress-Strain Function. 339 



For the axes of the shear the tangential stress must vanish, so 

 that X or ft must become zero, and therefore the axes of x and 

 y are the shear axes. If SR X and 9? y are the normal axial 

 stresses, one then has 



-W,a = %/a= 6/3. . 



A physical interpretation must now be given to the quantity 

 a. In a finite shearing strain of ratio a, it is easy to see that 

 the normal to the planes of no distortion makes an angle with 

 the contractile axis of shear the cotangent of which is a. If 

 the tensile axis of the shear is the axis of y, and the contractile 

 axis coincides with x, this cotangent is A/ ft. Hence in the 

 preceding formulas a is simply the ratio of shear. 



In a shear of ratio a with a tensile axis in the direction of 

 oy, minus %? x a is the negative stress acting in the direction of 

 the x axis into the area a on which it acts. It is therefore the 

 load or initial stress acting as a pressure in this direction. 

 Similarly ^l y /a is the total load or initial stress acting as a ten- 

 sion or positively in the direction oy. Hence a simple finite 

 shearing strain must result from the action of two equal loads 

 or initial stresses of opposite signs at right angles to one another 

 the common value of the loads being in the terms employed 



It is now easy to pass to a simple traction in the direction of 

 oy since the principle of superposition is applicable to this 

 case. Imagine two equal shears in planes at right angles to 

 one another combined by their tensile axes in the direction oy, 

 and let the component forces each have the value Q/S. To 

 this system add a system of dilational forces acting positively 

 and equally in all directions with an intensity Q/S. Then the 

 sum of the forces acting in the direction of oy is Q and the 

 sum of forces acting at right angles to oy is zero. 



Inversely a simple finite load or initial stress of value Q is 

 resoluble into two shears and a dilation, each axial component 

 of each elementary initial stress being exactly one-third of the 

 total load. Thus the partition of force in a finite traction is 

 exactly the same as it is well known to be in an infinitesimal 

 traction, provided that the stress is regarded as initial and not 

 final.f 



* This proposition I have also deduced directly from the conditions of equilib- 

 rium in Bull. Geol. Soc. Amer., vol. iv, 1893, page 36. It may not be amiss here 

 to mention one or two properties of the stresses in a shear which are not essential 

 to the demonstration in view. The equation of the shear ellipse may be written 

 in polar coordinates l/r'^ay + a-^A Hence the resultant load on any plane 

 whatever is 9ftr= ±Q/3. The final tangential stress is well known to be maxi- 

 mum for planes making angles of 45° with the axes ; but it is easy to prove that 

 the tangential load, Xr, is maximum for the planes of no distortion. These are 

 also the planes of maximum tangential strain. Rupture by shearing is deter- 

 mined by maximum tangential load, not stress. 



f Thomson and Tait, Nat. Phil., section 682. 



