CORRUGATED BAR COMPANY, INC, 



The condition of simply supported ends assumed in Problems I and II is that which 

 occurs when the beam frames into a brick wall. If, however, the ends frame into a 

 column or girder, the moment at the end support will not equal zero, but will have a 



value depending upon the degree of fixity. This may be as large as -r^ where small 



beams frame into heavy columns. For ordinary conditions the Joint Committee 



recommends a value of -r^ 

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We will now take Problem II, Case 1, and substitute for Mu a value -^ and for 



Mi, -r^ and note the efiFect on the various moments and shears. 

 Id 



M,=M. = -^'=- ^^^^^^ = -117.500 ft. lb. 



10 10 



From symmetry M2 = Mi 

 Inserting numerical values, 



(-117,500) (20)+120M2+40M2 



(4,700) (20)3 (4,700) (40) 



4 4 



I6OM2 = (-9.400,000-75,200,000)+2,350,000 



n^ n^ (-84,600,000+2,350,000) ri^nAQf* ik 



M2 = Mz = r^TT = — 514,063 ft. lb. 



y^ ^ (-514,063)-(-117,500) ^ (4,700) (20) ^ _^27,172 lb. 



Z\) z 



F'2= 27,172- (4,700) (20) = -66,828 lb. 

 ^^ ^ (-514,063)-(-514,063) ^ (4,700) (40) ^ ^^^^^ ^^ 



Ri = 27.172 lb. 



^2 =94,000+66,828 = 160,828 lb. 



27,172 

 "^^ = 1:700" = ^-^^ ^*- 



Moment at xi, M = (-117,500) + (27,172) (5.78) -iil??^i5^^ =-38,955 ft. lb. 

 94,000 „^.. 



^^ =1:700 =20ft. 



Moment at 3:2, M = (-514,063)+(94,000) (20) - -^^^^5^^^' = + 425,937 ft. lb. 



It will be noted that for Problem II, Case 1, with the ends partially restrained, 

 negative moment exists throughout span h. 



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