Materials under Combined Stress. 119 



calculated from the formula 



in which 



T is the torque ; 



D is the diameter of the bar. 



For the tubes employed in the later tests the formula 

 becomps 



in which 



D is the external diameter of a tube ; 

 d is its internal diameter. 



Having found p and S, the maximum and minimum prin- 

 cipal stresses are represented by the expressions 



i±\/? + ss ' 



one of which is positive, and the other is negative. The 

 third principal stress is zero, because in the case of solid 

 bars the stress due to the shearing force on the section is zero 

 where the bending stress is a maximum. Since the tubes 

 are bent by couples, there is no direct shearing force on a 

 cross-section. 



The stress difference is the difference between the maxi- 

 mum and minimum principal stresses, and therefore equals 



V! 



S 2 . 



It is twice the maximum shearing >tress. 

 The maximum strain is given by 



E 

 in which 



1\ is the maximum principal stress ; 



P 2 and P 3 are the other principal stress - 



i] is Poisson's Ratio : 



E is Young's Modulus of Elasticity. 





