﻿of Ductile Materials under Combined Stress. 545 



When determining the size of a shaft under both loadings 

 it should be calculated from the working tensile strength, hj 

 using the equivalent bending moment if this is less than twice 

 the working torsional shear strength, or from the shear 

 strength if this is less than one half the tensile strength. In 

 other words, the tensile strength should be taken as twice 

 the shear strength, and the smaller value taken when obtaining 

 these from the values given by tests, or used as working 

 stresses. A simple case will make this quite clear. Unwin 

 gives 13200 lbs/in.' 2 and 5400 lbs/in. 2 as the tensile and 

 torsional strengths of mild steel under the same conditions 

 of loading. These should be taken as 10800 and 5400 lbs/in. 2 

 when working with the formula given, thus taking the 

 smaller values. If the twisting moment was small, there 

 would be no danger in using 13200 and 6G00 lbs/in. 2 , thus 

 approximating to the ellipse given in tig. 5. The formula 

 based on the ellipse is slightly more complicated and may be 

 given as 



T 2 M 2 T e 2 



or 



"*/ torsion J tension "i/ tors 



T 2 + / ytor,ion \ 2 M2 ^ Tg2 . 

 \ /tension / 



the letters having the same meanings as before. 



13. An Explanation of the Valuation of the Maximum 

 Shear Stress. 



The complete theory of elastic strength has yet to be 

 determined. The table, section 10, gives the principal stresses, 

 the maximum in one case being almost twice that in another. 

 The maximum shear stress is nearly constant and may be 

 taken as the determinant to a first approximation. After 

 allowing for experimental errors, it is difficult to explain th« 

 differences in the values. Mr. Guest suo-o-esteda force corre- 

 spending to friction due to the force normal to the plane or. 

 greatest shear, the same having been mentioned by other 

 writers. It is now necessary to examine the results more closely 

 to determine whether they justify this explanation. When 

 the principal stresses areT/2 + \Z& + T*J£, the distribution 

 is that indicated in fig. 6. Assuming a force similar to 

 friction this would assist the shearing force, so that the yield 

 in tension would take place with a lower stress. With a force 

 the reverse of friction, opposing the shear stress if there is 

 tension across the plane, the tensile strength should be more 

 than twice the shear strength. In the case of bending, there 



