498 Profs. John Perry and W. E. Ayrton. [Dec. 11, 



and the shear at any point P is equal to S minus the load between 

 and P, therefore we can draw the diagram of shearing force. Also 

 the deflection at any point P equals 



As a familiar example, we may consider a beam OQ, of which the 

 ends are fixed horizontally and at the same level, here 



oc = 0, 



and «i=0. 



We can prove that, when we know the angle of the beam at one end 

 of a span, and the bending moment at the other end, we have sufficient 

 data for calculation, having the above summations m, &c. This applies 

 to the familiar case of a beam fixed horizontally at one end and merely 

 supported at the other end, the supports being level, as before. 



The following well-known examples illustrate the use of the for- 

 mulae : — 



In all cases when the cross- section of the beam remains constant, 

 the summations d, f, n, and q may easily be shown to have the follow- 

 ing values : — 



d= 



X 



1 EI 



2 EI' 



2EI' 



f — h~ 

 J1 2EI' 



7 3 



/ 2 



2EI' 



7h= zk 



«,=-£, 



' 2E1 



a* 



6Ef 





l_3 



so that the theorem of three moments becomes — 



+ 2M 1 (Z 1 + Z 2 ) + M 2 L-2m l l 1 -mJ, + 6El(^ + ^-S\=0. 



In all cases where the load is spread uniformly over the span, being 

 w per unit length, it is easy to show that — 



wx 2 

 : "2"' 



wx ?i ivP the section being con- 



^~6EI' ^ 1_ 6E1' stant on the span. 



^ wx 4 * rn _ wP 



C = _ , -Pi-: 



24EI 1 24EI 



When the spans are equal in length, and all cross-sections are the 



