56 



STRENGTH OF MATERIALS 



stress, and it is assumed to be uniformly distributed over the area 6x, p a bx = P; 

 whence p a = (see Fig. 41, a). 



Similarly, let p b denote the maximum intensity of the stress forming the stress 

 couple. Then, taking moments about the center C of the portion AB, since the 

 stress forming the couple is also distributed over the area fez, we have 



bx* 



and M = 

 Me 



Therefore, substituting in the formula p = , we have 



P 



Pb = 



12 



6P 



Consequently, 

 whence 



and 



Pmax = Pb Pa = 

 min 



2P 



As a numerical example of the above, leU = 6ft. = 60 in., P = 200 lb., 6 = 4 in., 

 and p = 600 lb./in. 2 (for ordinary brick work). Solving the above equation by the 



formula for quadratics, 



^1 2P V4P' + GbpPl 



4 -*- 



whence, by substituting the above values, 



z= 6.<5 in. 



|55. Distribution of shear over 

 rectangular cross section. 

 sider a cross section of a rectan- 

 gular beam at a distance x i 

 the left support, as J/TV.tfS in Ki.n.4L', 

 and let P be a point in this cross 

 section at a distance y from the neutral axis. Then, by equation (17), 



Article 43, the unit normal stress at P is p = ^ If the cross 



. 42 



* Bach, Elasticitat u. Festigkeitslehre, p. 430. 



^ Strength Of Materials 



remainder of this chapter may 



