106 MECHANICS 



It can readily be seen that in a cantilever beam the con- 

 ditions are somewhat reversed; that is, in a beam having 

 one support at the middle, the shear in the part to the left 

 of the section taken to the left ot the support is negative, 

 while if the section is taken to the right of the support, it is 

 positive. Like in a simple beam, however, there is an 

 intermediate point where the shear changes sign. 



FIG. 11 



EXAMPLE. At what point in the beam loaded as shown 

 in Fig. 11 does the shear change sign? 



SOLUTION. Compute the reaction RI as follows: With 

 the center of moments at /?2, the moments of the loads, in 

 foot-pounds, are 



9,000X10 = 90000 



4,000X26 =104000 



3,000 X 10 X 17 = 510000 



Total, 704000 



The reaction at /?, is therefore 704,000 + 30 = 23,4662 Ib. 

 Proceeding from R } , the first load that occurs is c of 4,000 Ib. 

 Then, 23,4662-4,000 = 19,4662 Ib. The next load that 

 occurs on the beam is the uniform load of 3,000 Ib. per 

 running ft. There being altogether 30,000 Ib. in this load, 

 it is evident that the load will more than counteract the 

 remaining amount of the reaction R^ \ the point where the 

 change of sign occurs must consequently be somewhere in 

 that part of the beam covered by the uniform load. The 

 load being 3,000 Ib. per running ft., if the remaining part 



