CORRUGATED BAR COMPANY, INC. 



seven spans and Diagram 11 gives shear coefficients of wl at supports for the same 



case. 



Example. For a beam of three spans the negative moment at the first interior 



5 

 support is . lOwl^. The shear at the end of the middle span is —wl and at the inner 



R 



support of the end span it is rrxwl. 



Equal Concentrated Loads on All Spans: Diagram 12, page 46, shows three 

 cases of loading. (1) Loads at middle points. (2) Loads at third points. (3) 

 Loads at middle and quarter points. The full irregular line is the moment curve and 

 the broken line represents the shear line for each case of loading. The ordinates to the 

 moment line are coefficients of PI and the ordinates to the shear line are coefficients of 

 P. The numerical coefficients are given at critical sections. 



Example : At the central span of a five-span girder loaded at the third points, the 

 negative moment at the adjacent support is 0.211 PI; the positive moment at either 

 of the loads is 0. 122P/ and the reaction at the adjacent support is 1.93P. 



The moments and shears of any uniform load should be combined with those of the 

 concentrated loads on the girder. 



Partial Uniform Load: Diagrams 13, 14, 15 and 16, pages 47 and 48, for two 

 and three span beams, give moment and shear coefficients that are maximum for the 

 indicated positions of the uniform load; the ends of the beams being either free or 

 fixed. 



Unequal Spans. Uniform Load: In the design of schools, hospitals, hotels 

 public buildings, garages and shop buildings, the layout usually involves continuous 

 beams of unequal span. The application of the three moment theorem in such 

 cases can best be illustrated by means of problems. 



^ a:j > i-r a-2----> 



%- H.^-- ::^fc__.,,.,o.-^ ......,,.»..-.. 4 



Live load per foot of span = 800 lbs. 



Dead =1000 " 



Fig. 6 



Problem L Assume a beam of three unequal spans as shown in Fig. 6, carrying a 

 live load of 800 pounds per linear foot and a dead load of 1,000 pounds per linear foot. 

 Find the critical moments, shears and reactions, the ends being assumed simply 

 supported. 



Ca^e 1 



liiiii iiiiiiiiiitiiii iiiiiaiiiiaiii 



Live Load on All Spans. The ends being simply supported Mi=Mi=0. 

 The moments Mz and M3 at the intermediate supports can now be found. 



40 



