78 RESISTANCE OF MATERIALS 



SHEARS AT SUPPORTS FOR EQUAL SPANS AND UNIFORM LOAD 



2 2 



1 2 



0|3 

 8 



65 30 



8 8 



2 _ 3 



.0|4 6|5 5|6 4|0 



10 10 10 10 



1 2 3 4 



0|ll 17|15 13 1 13 15 1 17 11|Q 



28 28 28 28 28 



12345 



0|15 23 1 20 18J19 19J18 20J23 15|0 

 38 38 38 38 38 38 

 1 2 3 4 5 6 



Q|41 63 1 55 49 1 51 53 [53 51|49 55 [63 4j| 



104 104 104 104 104 104 104 



1234567 



0|56 86 1 75 6?|70 72 1 71 71J72 70J67 75J86 5e| 



142 142 142 142 142 142 142 142 



153. Calculate the reactions of the supports in problem 151. 



Solution. The reaction at any support may be found by finding the shears close 

 to the support on each side. The sum of these two shears is then equal to the re- 

 action. Thus, in the present case, to find any given reaction, say J? 2 , consider the 

 portion of the beam between R l and E 2 , as shown in Fig. 70, and form the moment 

 equation for this segment. Then 



and therefore, since M t = and M 2 = ^ g wl?, Sf = ^| wl. Similarly, the moment 

 equation for the segment of the beam between R 2 and E 8 is 



whence, by substituting"^ = ^ 5 wl 2 and M 2 = 5 3 g loZ 2 , we have 



Consequently, E z = Sf + 8f = w i = w i, 



28 28 



This method applies when it is required to find one reaction only, independently 

 of the others. If all the reactions are required, it is simpler to calculate them in 

 succession, starting at one end, without reference to the shears. For instance, to 

 find R lt take a section through R 2 and consider the loads on the left of the section. 

 Then the moment equation for this portion is 



whence 



