62 



STRENGTH OF MATERIALS 



Neglecting the rounded corners, 

 f 



Jk 



Consequently, from formula (28), the unit shear at P is 

 Q 



7(1466.6) 

 At the point P, therefore, p x = 15,300 lb./in.*, p, = 0, and q = 64 lb./in.. 



Hence, from formula (7), Article 26, 



>* = 16,804 



To calculate the equivalent stress it is necessary to find the principal stresses, 

 which are, from the above, 



Pl = 15,304 lb./in. 2 and p, = 



Hence, from formula (14), Article 36, for m = 3J the equivalent stress at P is 

 p e = 16,306 Ib./in- 2 . 



58. Oblique loading. If, for any cross section, the plane of the 

 external bending moment does not pass through a principal axis of 



the siM-timi. tin* loading is said 

 to be oblique. In this case t In- 

 lending moment M can be re- 

 -nlvr.l into components parallel 

 to the principal axes, naim-ly. 

 Mcoea and M sin , where a 

 is the angle \vhirh the plane 

 containing M makes with one 

 of the principal axes. 



For materials which conform to Hooke's law it has been found 

 that the stress due to several sets of external forces can be calculated 

 for each set separately and then combined into a single resultant. 

 This is called the law of superposition. Applying this law to th- 

 present case, 



Mcosa Jfsina Jlfcosa 



FIG. 47 



/A\ 

 (30) 



where e y , e z are the distances of the extreme fibers of the beam from 

 the axes of Y and Z respectively, and S 9 , S f are the corresponding 



section moduli. 



