SHEAEING. 17 



or cube whose end is distorted into the form of a rhombus, 

 as has been shown in the case already mentioned (Fig. 5). 



Under the shearing stress upon it the prism becomes 

 simply distorted from A B C D to A b c D, (Fig. 7), 

 the sides remaining of the same length and parallel, and the 

 diagonal D B being shortened and A C lengthened. In 

 order that this distortion may occur four equal stresses S 

 must act tangentially to the sides, as shown. The result 

 of these stresses will be a distortion without a perceptible 

 change of volume. The shearing stresses Si and S 2 result 

 in forces 



S 2 X A B acting along B A = F 2 

 and Si x D A acting along D A = FI. 



Similarly, the other pair of stresses S and S 4 result in 

 forces F 3 and F 4 . Combining these, the resultants of FI 

 and F 2 will be a force along the diagonal C A ; and there 

 will be an equal and opposite force acting along the 

 diagonal in the opposite direction A C. 



Thus the effect of the shearing stresses is the same as 

 would be produced by a tensile stress acting along the 

 direction of the diagonal A C. 



The total force thus acting along A C, the side of square 

 being taken as unity is 



-^ + 4 = V2 S, 

 A/2 A/2 



Si and S 2 being equal. 



Similarly, the pairs S 2 , S, S 4 , Si may again be resolved so 

 that they result m a compressive force along the other 

 diagonal' B D. This will, by the same course of reasoning, 



also be 



A/2S. 



These are the total forces along the diagonals. The tensile 

 stress parallel to A C will be at right angles to D B, and 

 be of intensity 



A/2 



and the compressive stress at right angles to the diagonal 

 A C will also be 



/ V2S s 

 * ~vT = 



So that it is thus proved that the four equal shearing forces 

 acting along the sides of the square are equivalent to two 



o 



