110 MECHANICS 



at p. The shear at this point, just to the left of the con- 

 centrated load, is +5,848 -500 -(1,000X4)= +1,348 Ib. 

 This value laid off above ab gives the point g. Draw fg. 

 From g lay off vertically downwards 200 Ib. to the point h. 

 From a, on the line ab, lay off 11 ft., as at q. The shear at q 

 is +5,848 -500 -(1,000X6) -200= -852 Ib. From q lay 

 off, vertically, 852 Ib. to i, downwards in this case because 

 the shear is negative. Join h and i. If correctly drawn, 

 hi should be parallel to fg. From i draw ij horizontal and 

 equal to 4 ft.; from j draw jk vertically downwards, equal 

 to 1,000 Ib.; from k draw kl horizontally, equal to 6 ft.; 

 from / draw Im vertically downwards, equal to 5,000 Ib. ; 

 and from m draw mn horizontally, equal to 4 ft. Then, if 

 the diagram is correctly constructed, bn should be vertical and 

 equal to R 2 , or 6,852 Ib. Then acde . . . mnb is the shear 

 diagram, and ordinates drawn from any point on ab across 

 the diagram will, when measured by the proper scale, indi- 

 cate the shear at that section, which is positive if above the 

 line ab and negative if below it. 



BENDING MOMENTS 



In order to illustrate the method of calculating the bend- 

 ing moment, or the moment of a force that tends to bend 

 a beam, the beam shown in Fig. 14 will be considered. At 

 the point a is a joint, or hinge. It is evident that when loads 



:; 



FIG. 14 



are applied, as shown by the arrows, the beam will bend 

 at the joint and take the position indicated by the dotted 

 lines; or, to be more exact, will bend still farther until it 

 falls off the supports entirely. 



The loads and reactions (and weight of the beam itself, if 

 this is considered) are what cause the beam to bend. Con- 



