494 



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



Heppel found it convenient for his special case to make the summations 

 twice for the same span, working forwards and backwards from every 

 point of support, but in the following the summations are all effected 

 working in the same direction, only one summation being necessary 

 for each span. 



3. Suppose OQ to be one span of a continuous beam which has 

 loads and supporting forces beyond and Q. . To find the bending 

 moment at a point P, it is sufficient to know what is the bending 

 moment at 0, and the shearing force at a very short distance inside 

 0. If we suppose the bending moment to be called positive when it 

 tends to make the beam convex upwards, and the shearing force at a 

 section positive when it tends to elevate the part of the beam to the 

 right of the section, then M, the bending moment at a point P, is 

 equal to 



M Q -S xOP + m, 



where M and S are the bending moment and shear at 0, and where 

 m is the sum of all such expressions as — an element of load at any 

 place P', between and P, multiplied by P'P. 



Again, if Mis the bending moment at a section, E the modulus of 

 elasticity, I the moment of inertia of the section about its neutral 

 line, y the vertical displacement of a point in the neutral axis of the 

 beam from its horizontal position, x the horizontal distance of this 

 point from some fixed point in the neutral axis as 0, then for all 

 beams and girders used by engineers — 



dx* EI ' W ' 



The only restriction that we shall use is to suppose all points of sup- 

 port in the same straight line. But as is well known, if this case 

 is solved, the unrestricted case may also be solved, and the case in 

 which there is a yielding in the supports. 



Let H Q Mi be the bending moments at the points of support and 

 Q. Let S be the shearing force just inside 0. When we know the 

 nature of the loading — - 



Mo P _Mj 



A A 

 Q 

 < . . . . li > 



we can find m the moment at P due to the load between and P. 

 Thus if w be the load per unit length at any place, and if we know 

 how iv alters, we can find the sum — 



for for 



I w(OP — x)dx— \ wzdz—m ...... (2), 



Jo Jo 



