Art. 75. SHEARING STRESSES IN A SOLID BEAM. 



Ill 



through the point and is opposite to 8. This makes the ver- 

 tical components of the unit stresses along P'P' equal, and since 

 the horizontal components increase toward the points P' t the in- 

 tensity is greatest at these points. 



To derive a general formula for s s at any point of a cross 

 section, the procedure is similar to that for the numerical case 

 above. In place of taking the two sections one inch apart, as in 

 Fig. 82, they will be taken a distance dx apart, so that the stress 

 in this distance may be considered of constant intensity, and the 

 horizontal section will be taken a distance v above the ncntr.-il 

 axis. The longitudinal stress on an element of area 2w wide and 

 dv deep is 2wdvs and the total stress, above the horizontal sec- 

 tion is / 2ivsdv. Substituting for s its value from equation 



/ - /**/i) 



(12), the total stress on one end of the block is-j- / 2wvdv; the 



is4 A 



J v 



M' /? 



is -j I 2u 



total stress on the other end of the block is -j- / 2wvdv. The 



J v 



difference between these stresses is equal to the horizontal shear 

 on the area 2wdx. M'M = dM = Sdx according to equation 



total hori- 



(21); therefore, W 1 M I 2wvdv = ^p I 2wvdv = 

 J v J v 



zontal shear. 



2wvdv = unit shear in a strip whose 



sides are vertical = s s cos<j> in a strip whose sides are not vertical 

 (Fig. 84). Therefore, 



s 2w cos<t> I 



rv\ 

 I 2wvdv =2^^ 



/ v 



WCOS<j>I 



(22) 



in which M s is the statical moment of that part of the section 

 lying above a horizontal line through the point where s s is to be 

 calculated. 



For a rectangular cross section < becomes zero and cos$ = 1. 



Equation (22) reduces to s = 

 is, for .stress at the neutral axis, 

 s. 



12Z P\ 



=-w / vdv== 



Jo 



bh s 



/ 



J v 



3 S 

 2 A 



bvdv Whenv => that 



(23) 



