214 CONTIGUOUS GIRDER. [CHAP. XHI. 



r + 1 in the second, we obtain the formnlse of Art. 137. Foi 

 any other support left of r, or right of r + 1, we have only to 

 give the proper values to m, s and r for any given case, and 

 find the corresponding Clapeyronian numbers. 



14O. Practical Rule, and Table. The moments at the 

 supports of the loaded span having been found by the formulae 

 of Art. 137, or the triangle of moments of Art. 138, instead of 

 using the above formula, we may find the moments at the 

 other supports as follows : 



For all supports left of the loaded span: Commencing at 

 the left end support, place over each support the Clapeyronian 

 numbers 



. 1 4 15 56 209 780, etc. 



Take the last number thus obtained, before reaching the left 

 support of the loaded span, as a common denominator. Then 

 the moment at the left end is of course zero. At the second 

 support 1, at the third 4, at the fourth 15, at the fifth 56, etc., 

 all divided by this common denominator, will express the frac- 

 tional part of the moment at the left support of the loaded 

 span, which the moment at the support in question is. For the 

 moments at the supports right of the loaded span, proceed simi- 

 larly, only count from the right end. 



Thus, for a girder of ten spans, sixth span from left loaded : 

 The moments M and M 7 due to load being found, suppose we 

 wish the moments left of M 6 . Commencing at left end, num- 

 ber the supports 1, 4, 15, 56, 209. (Let the reader draw a figure 

 representing the case.) The number 209 is the last before 

 reaching the sixth support. We take this, therefore, for a com- 



mon denominator. Then we have M t = M 2 = 



M <= M anaM = M '- 



2 T9r9 



So for supports to the right of support 7, we have 



1 4- 1 



M u = M 10 =iM 7 M 9 =M 7 M 8 = 



Remembering that M 6 and M 7 are both positive, and that 

 the moments alternate either way from these supports, we find 

 easily the proper signs for the moments right and left. 



We can now, therefore, find the moments at D and E due to 

 the first and second cases of loading of Fig. 87 (Art. 126). 



