CHAP. XTH.] ANALYTICAL FORMULA. 233 



Here 8=5, n = %, j} = ~L, r = 2 ; therefore, from Art. 150, 



4,Ac 5 + A'ct frAfr-f A'q. 



I c* + % % I C 4 + I c 6 



and G! = 0, <% = 1, <%=:, c 4 = 13, c 5 = 48.5. 



w Z 8 

 Since, then, A = A' = for uniform load, we have 



M 1= 0, M^ = 



If the load is two tons per ft., wfi = 20,000, and 

 M! = O, M,j = 414.7, Mg = 1036.6, M 4 = - 279.5, 1^ = 79.7. 



Find the shears and reactions at each support. 



Ex. 5. A beam of four equal spans, has the second span from 

 left covered with full load. What is the moment and shear at 

 left of load f 



Ans. M 2 = -^y w I*, S 2 = $1$ w I. 



What at right of load ? 



Ans. M 3 = -/% w I, S' s = w I. 



What are the formulae for concentrated load ? 



A /Y\ & 1\/T . i v^ Z* n EC Z2 i "I Q M | TJ 7 



^J./vO JLTAo -' K / I " fkj ^^ j!t/ /v "T" JL7 tit I J^ v 



Tfc 



83 = = [56 - 58 It + 3 ^ - #*], 



Examples might be multiplied indefinitely. 



The above is sufficient to show the comprehensiveness of our 

 formulae, and the ease with which results may be obtained, 

 which, by the usual methods, would require long and intricate 

 mathematical discussions. The points of inflection and the 

 deflection may also in any case be easily determined, and gen- 

 eral equations similar to the above deduced, but, as we have 

 seen, the above are sufficient for full and complete calculation. 



154. Tables for moments. From the formulae of Art. 150 

 we can easily find the moments for both uniform and concen- 

 trated load in a single span for various numbers of spans. If 

 these results are tabulated we, shall obtain tables from which 



