FORCES IN A SHEAR. 37 



The position of the plane remaining" constant, it is permissible to com- 

 bine J 57 and G like simple forces to a. tangential component, T, acting in 

 the direction of a c, and a normal component. iV, acting- perpendicularly 

 across ac. Evidently, if P and Q are considered as in general positive 



quantities — 



T= — Fs!n 9 — G cos 9 = (P-f- Q) sin 9 cos ■>. 

 N= —Fcos 9 — G sin 9 = P cos' 9 -| Q sin' 9, 



and T will be a maximum with reference to 9 when — 



cos 2 9 = sm 2 -V or 9 = ± 45°. 



Although the tangential stress is greatest for this angle, one has no 

 right to infer that the maximum tangential strain is at 45°, because there 

 is a normal stress on the plane at this angle amounting to (P-f Q) '-■ 

 On the contrary, it was shown above (page 34) that the maximum tan- 

 gential strain in a shear occurs for planes which make an angle with ox 

 the tangent of which is 1 H. or the normal to which is given by tan 9 = a. 

 The conditions of this plane are also such that there can be no normal 

 stress acting upon it, and hence N= 0, so that one of the stresses must 

 have a negative value and — ■ 



tan? 9 = 



Q 



This relation enables one to determine the forces which produce a 

 finite shear. The area on which the stress Q acts is a, and the force 

 acting on the distorted cube in this direction is minus Q a. The area on 

 which P acts is 1/a, and the lateral force is therefore P/« ; hut by the last 

 equation — Qa = Pja, so that a finite shear, as well as an infinitesimal 

 one. results from the action of two equal forces acting at right angles to 

 one another in opposite senses.* 



Simple Pressure. — -Knowing the composition of a shear enables one to 

 pass synthetically to the case of simple pressure or traction. If two 

 equal shears at right angles to one another are combined, the contractile 

 axes coinciding, each mu3t produce the same effect as the other if the 

 mass is isotropic. Each must also produce the same effect as if it acted 

 alone. This statement does not imply a relation between stress and 

 strain, for the shear in the xy plane leaves the mass unstrained in the 

 y : plane. Hence two equal shears, each of ratio «, reduce the unit cube 



* I have met with no demonstration of this relation between finite shearing stress and strain, bul 

 I am not prepared to state that none has been published. 



