MECHANICS 109 



the line ab, namely, 600 Ib. At this point the shear is 

 reduced by 1,200 Ib., shown on the diagram by the line /g, 

 and becomes negative. At the third load, counting from 

 the left, the shear is still further reduced by 400 Ib., and is 

 represented by the line hi in the diagram. From this load 

 on to the right-hand reaction the shear is negative in sign 

 and equal to 1,000 Ib. This right-hand reaction drawn 

 from j to b closes the diagram. 



It may now be seen that to find the shear at any point in a 

 beam it is necessary to draw an ordinate from the corre- 

 sponding point on the line ab and to measure the length of 

 the part included in the shaded diagram with the correct 

 scale. For instance, if it is desired to ascertain the shear 

 and its sign at the point k, draw a perpendicular line km 

 crossing the shear diagram. The length Int measured by 

 the scale selected for the shear will show that the shear at 

 this point is 600 Ib., and, as Im is located below the base 

 line ab, the shear is negative. 



As a general example, the shear diagram of the beam 

 shown in Fig. 13 may be plotted. As the load here is not 

 symmetrical, it is first necessary to calculate the reactions 

 RI and R- 2 . The positive moments about the left-hand end 

 of the beam, in foot-pounds, are: 



500X 3= 1500 



(1,000X6) X 8= 48000 



200X 9= 1800 



1,000X15= 15000 



5,000X21 = 105000 



Total, 171300 



The total negative moment is R 2 X25. Therefore, R 2 

 = 171,300-*- 25 =6,852 Ib. The sum of the loads is 500 

 + (1,OOOX6) + 200+1,000 + 5,000 = 12,700 Ib. RI is there- 

 fore equal to 12,700-6,852 = 5,848 Ib. 



The plotting of the shear diagram shown in Fig. 13 may 

 now be started. Draw the line ab equal to the length of the 

 beam; ac equal to RI, or 5,848 Ib. ; cd, horizontally, equal 

 to 3 ft. ; de, vertically downwards, equal to 500 Ib. ; and ef, 

 horizontally, equal to 2 ft. From a on ab lay off 9 ft., as 



