STKKSSKS IN (OMINTors UKAMS. 



56. ComittUOUS BEAMS, UniFORrlLMDS, COfiSTAHTMOWNTOFlHfRJIA AflDflODULuSOFf LAST/CITY. 

 Shear, \V' 



'sfeksi HI 



unitlengtfy 



Span, ^ i^span 

 Length, ^ I, 



Support, I 2 



Reaction, R, RI 



/1oment,f1, A 



flLj 

 3 



A 



*----*! i 



.~fff3R9GL.. i*?^! s J? a . n . 



in I inn 



fin., /% 



ml 



Relation between momentsat supports for the n^ and (ntlj spans, 



Shear to right ofn&supportj 



5/jear to left offnu)^ support. 



r- 

 Shear to right of (ntlj- support, 



I/' S i . , 



' - - + w ml ( ml 



? 

 inn I 



Shear at any point i 



a j ynn 



/ 

 Reaction at (m/J^ support, 



(O 



fe) 



Moment at any point in n&span 

 -* 



Point of max. positive momentinn^ 1 span, Maximum positive moment in n^span, 



V f V' 2 



/* j (h) n=n n f-Q ; (i) 



EXPLAtiATiOfiOF FORMULAS; n* number of- 'first span considered or its left support. 



Given a continuous beam of several spans uniformly ' loaded '(for spans withno/oddw-0). 

 Apply formula (a) to /-and? spans at the left end making n=l. Three unknown moments 

 appear,M / ,r1 2 ,andMj. Ifbeamissimp/ysupportedatleftendfl^O. Next apply formu/a(a) to? 

 and 3 spans making n*2. Again there will be three unknowns fi z , M 3 andty. Continue unfit 

 last two spans have been considered (never consider last span alone). If beam is simp/y supported 

 at right end, thef1Forthatsupport=0. There are now as many equations as there are unknowns 

 sobysolving, the momentsat all of the supports maybe Found. IF the beam is symmetrical 

 as to loading and dimensions, the calculations may be shortened by eguating moments which 

 are known, by inspection,^ be equal. Knowing the moments at the supports; the shear atanypoint, 

 the reactions,and the moment at any point may be calculated. (R,= fond R For last support 

 eguaby" for last span). For Fixed ends imagine the beam toextend one span beyond the Fixed 

 end and apply the for mulas,as above, equating the length and load of the imaginary span to 

 zero and the moment at the extreme end of the imaginary span to zero.Care should be taken 

 that shears and moments are us edwith their proper sign. 



SPECIAL CASES; 



For a beam of equal spans with equal uniform loads, Formula (a) reduces to- 



M n / 4M m/ +M n + 2 = - wt*j (See also 57, of this chapter.) (j) 



Fora beam of two unequal spans with unequal uniform loads and simpfy supported 

 at the ends, M, = 0,Mj=0 and from formula (a) 



