32 STRENGTH OF MATERIALS 



Let p e denote the equivalent unit stress. Then, from the al> 

 definition of p e , it must be so -chosen as to produce either s l or * f , 

 wliichever is the greater. Consequently 



(13) p = Es\ or p e = Es y 

 Comparing equations (12) and (13), 



(14) Pe=Pi P* or Pe = P* Pi- 



\ / 'KYI? ' "' 



The value of the equivalent stress can thus be calculated diivrtly 

 from the two principal stresses. In order that the strain be safe, 

 the greater of the two values of p e found from equation (14) mu>t 

 not exceed the maximum allowable unit stress. 



Problem 33. Find the value of the equivalent stress in Problem 24, and compare 



it with the principal stresses. 



36. Combined bending and torsion. One of the most important 



applications of the preceding paragraph is to the calculation of tin- 

 equivalent stress in a beam subjected simultaneously to bending and 

 torsion. 



Let the axis of X be drawn in the direction of the axis of tlu* 

 beam. Then on any cross section of the beam there will be a normal 

 stress p x due to bending, and a shearing stress q due tx> t: 

 while the stress between adjacent longitudinal fibers is zero, that is, 

 p y = 0. Therefore, from equation (7), the principal stresses are 



Consequently, from equation (14), the equivalent stress is 



Problem 34. A round steel shaft used for transmitting power bears a trans- 



verse load. At the most dangerous section the normal stress due to bt'iiilini: is 

 5000 Ib./in.a and the shear due to torsion is 8000 Ib./inA Calculate the intr. 



of the equivalent stress. 



