22 



PRACTICAL STRUCTURAL DESIGN 



support. The length of the beam therefore to the right support 

 = 13 ft., and one-half = 6.5 ft. The weight = 13 x 50 = 650 Ibs. 



M -(13.5x751.56) -[(2x200) +(5x300) +(6.5x650) +(9x250)] = 



1771.06 ft. Ibs. 



In Fig. 14 the section xx is taken 5 ft. from the left sup- 

 port. This leaves the load of 200 Ibs. to the left of the section, 



Fig. 14 Another Example of Uniform Load and Several Concentrated 

 . Loads on Beam on Two Supports 



so it is omitted from the calculations. The load was used in ob- 

 taining the reactions, but in this present example it will be noticed 

 that the moment arm from the section is only 10.5 ft. to the 

 reaction. The beam length = 15 - 5 = 10 ft. and the weight 

 = 10 x 50 = 500 Ibs., with a moment arm to the center of gravity 

 = 5 ft. 



M=(10.5x751.56)-[(2x300)+(5x500)+(6x250)]=3291.38ft.lbs. 



Fig. 15 Still Another Example of Uniform Load and Several 

 Concentrated Loads on Beam on Two Supports 



In Fig. 15 the section xx, is taken 8 ft. from the left sup- 

 port. The beam length = 15 - 8 = 7ft. and the weight = 7 X 50 

 = 350 Ibs., with a moment arm of 3.75 ft. to the center of gravity. 



M = (7.5 x 751.56) - [(3 x 250) + (3.75 X 350)] = 



3574.2 ft. Ibs. 



