ANALYSIS OF STKESS IN BEAMS 



61 



p pi 



shear on any section is > and the maximum bending moment is 



Hence the maximum unit normal stress is 



M 3 PI 



Also, since Q = and I ydF = -- > the maximum unit shear is 



o 



Now let K denote the ratio between the tensile strength in the direc- 

 tion of the fiber and the shearing strength parallel to the fiber. 

 Then, in nnU-r that tin- beam shall be equally safe against normal 

 and shearing stress, p = icq, or 



3 PI 3 P . 



whence 



2/ 



In general; * is not greater than 10. If * = 10, 1 = 5h. Consequently, 

 if the length of a beam is greater than ."> times its 



depth, the shear is not likely to cause rupture. ~ ' "^~T 



Problem 67. Th- bending moment and shear at a certain 

 negie I-beam, Efo, i:-J. "f th- dimensions 

 given in Fig. 46, are M = 200,000 ft. Ib. and Q = 15,000 Ib. 

 respectively. Calculate the maximum normal stress and the 

 equivalent stress for a point directly undi -r the flange, and 

 compare these values with the normal stress in the extreme 20' 



fiber, 



in the Carnegie handbook, the moment of | 



inertia <>f tins MM-tion about a neutral axis perpendicular 

 t'. the web is 1= 1466.5 in. 4 . Consequently, the normal 

 stress in the extreme fibei 



Me 



I 



1466.5 



= 16,365 Ib./in.a, 



and the normal stress at a point P under the flange is 



2.400.000 (0.35) = 

 1466.5 



FIG. 46 



