SHEARS AND MOMENTS. 



ture which is under consideration. The same principle holds at 

 any section of a beam (74), so that if the shear and moment at 

 any section are known, the shear and moment at any other sec- 

 tion may be found. See Figs. 110 and 113. 



If a timber beam acts as a cantilever 20 ft. long with a load 

 of 3000 Ibs. at its end, 



M maK = 3000X20 = 60000 ft. Ibs. = 720000 in. Ibs., neglect- 

 ing the beam's own weight for the sake of simplicity. 

 From equation (12), if the working stress is 1500 Ibs. per sq. in.. 



720000 _ , ftn _ 7 

 ~I500~ = : ~V = 



If h is assumed as 16 in., b 



W 



6X480 



16X16 



A commercial size would be 12"xl6 

 fiber stress 



Hfvi 720000 X 8 



= 11.25 in. 



making the extreme 



= 1405 Ibs. per sq. in. 

 The shear is 3000 Ibs. at each section. The mean unit shear is 



3000 



P er sc l' * n ' 



max i mum shearin stress 



12X16 ~ "' 



(at the neutral axis) is 15.6X1% = 23.4 Ibs. per sq. in. (75), 

 which is much smaller than what is usually allowed. (Cambria 

 p. 3l)l). 



Fig. 100. The origin is taken at A ; the same results can, of 

 course, be gotten if it is taken at B. 



In this case the shear diagram is a straight line since 

 8 = Iv'i wx is the equation of a right line. The equation for M 

 is that of a parabola. 



It will be noted that the maximum deflection is much less 

 than in the above case, for the same total load. 



Fig. 104. The equations for a triangular load are easily 

 derived, if it is remembered that its center of gravity is at JL 

 from B. 



93. Shears and Moments for a Beam Supported at 

 Both Ends. Fig. 96. The reactions are gotten by the law of 

 the lever, and the shears are equal to them. M ma ^ =R l l l =R 2 l 2 

 at the load, where S =0. i/ max does not occur at the load but where 



=0, which is evidently in the longer segment. 

 dx 



Fig. 97. The reactions are apparent. The moment between 

 the loads must be constant, hecause the shear is zero (74), or 



