288 STRENGTH OF MATERIALS 



The reaction Ri can be found in a similar manner by taking 

 moments about the support B. Their sum R\-\-Ri must be 

 equal to the sum of loads Wi + Wz+Wa. 



EXAMPLE. Find the reactions of a cantilever bridge loaded 

 as shown in Fig. 2. 



SOLUTION. Substituting given values in the formula and 

 noting that the moment of Pt about B is of opposite sign to 

 the moments of the other loads, 



10.000X120+8.000X90+15.000X40-20,000X30 



R\ ! ' ' ~~ 



150 



= 12,800 Ib. 

 and 



10.000X30+8.000X60+15.000X110+20,000X180 



1?2 == 



150 



= 40.200 Ib. 



The sum of the loads is 10.000+8.000 + 15,000+20,000 

 = 53,000. The sum of the reactions is 40.200+ 12,800 = 53,000. 



External Shear and Bending Moment. The forces acting 

 on a beam tend, on the one hand, to shear its fibers vertically 

 and, on the other hand, to bend it, producing compressional 

 stresses in the fibers on one side of the neutral axis and tensional 

 on the other side. The tendency to shear the fibers vertically 

 is determined by the external shear, and that of bending by 

 the bending moment. 



The external shear at any section of a beam is the algebraic 

 sum of all the external forces (loads and reactions) on one 

 side of the section. Forces acting upwards are considered 

 positive, and those acting downwards, negative. For brev- 

 ity, external shear is often called simply shear, but it must 

 not be confused with shearing stress at the section. The 

 external shear is equal to either reaction minus the sum of 

 the loads between that reaction and the section considered. 

 The maximum shear is always equal to the greater reaction. 

 For a simple beam with a uniformly distributed load, the 

 maximum shear is at the supports, and is equal to one-half the 

 load, or to the reaction; the shear changes at every point of 

 the loaded length, the minimum shear being zero at the center 

 of the span. The maximum shear in a simple beam having a 



