INFLUENCE DIAGRAMS 79 



Solving, we have 



p l Pt+p, 



- = r- (c) 



a L 



From (c) it follows that the maximum bending moment at 2 occurs 

 when the average load on the left of the section is the same as the 

 average load on the entire bridge. 



Uniform Loads. In Fig. 5oa, the bending moment at 2' due to a 

 uniform load p d x will be p y d x in (a) . But 3; d x is the area of the 

 influence diagram under the uniform load, and the bending moment 

 at 2' due to a uniform load will be equal to the area of the influence 

 diagram covered by the load, multiplied by the load per unit of length. 

 For a uniform load, p, covering the entire span the bending moment 

 at 2' will be p times the area of the influence diagram 1-2-3. For a 

 uniform load the bridge must be fully loaded to obtain maximum bend- 

 ing moment at any point. It will be seen that the general criterion for 

 maximum moment is satisfied when the bridge is fully loaded with a 

 uniform load. 



Maximum Shear in a Truss. Let P 19 P 2 , and P 3 in Fig. sob 

 represent the loads on the left of the panel, on the panel, and on the 

 right of the (n + i)st. panel, respectively. It is required to find the 

 position of the loads for a maximum shear in the panel. 



m 



With a load unity at 2' the shear in the panel is , and 1-2 is the 



n 



influence shear line for loads to the left of the panel. With a load unity 



n m i 



at 3' the shear in the panel is - - , and 3-4 is the m- 



n 



fluence shear line for loads to the right of the panel. For a load on 



m n m I 



the panel the shear will vary from at '2' to at 3 , 



n n 



and the line 2-3 is the influence shear line for loads on the panel. 



The influence diagram for the entire span is the polygon 1-2-3-4. 

 It will be seen that the lines 1-2 and 3-4 are parallel, and are at a 

 distance unity apart. 



The total shear in the panel will then be 



