Lit. 75. SHEARING STRESSES IN A SOLID BEAM. 



109 



beam, three planes should be passed through it. In the usual 

 case, which is the only one here considered, the cross section is 

 symmetrical about the plane of the bending moment, and there 

 is no variation of stress parallel to the neutral axis, or perpen- 

 dicular to the plane of the bending moment. It remains to find 

 how the stress varies vertically, and horizontally lengthwise of 

 beam. That there are horizontal shearing stresses will now be 

 shown. If we find how the horizontal shearing stresses are dis- 

 tributed, the distribution of the vertical shearing stresses will 

 be known because, according to Art. 17, their intensities at any 

 point must be equal. 



A numerical case will be first worked out and a general 

 formula derived afterward. 



Fig. 82. 



Considering the simple case of a rectangular beam, 10"X 

 16", carrying a single load as shown in Fig. 82, the shearing 

 stresses at all sections between the left reaction and the load will 

 be alike, because the shear at each section is 3000 Ibs. It follows 

 that the horizontal shear does not vary lengthwise of the beam, 

 in this part of it. 



Considering two cross sections AB and A'#', one inch apart 

 the difference in the bending moments is 3000 in. Ibs. If the 

 equilibrium of a block cut out by these two sections and the 

 neutral plane be considered, the forces acting upon it come 

 from the stresses cut in these three planes as illustrated in Fig. 

 82. 



The difference in the extreme fiber unit stresses is A'C' AC 



