26 PRACTICAL STRUCTURAL DESIGN 



understood. The second quantity is negative. By clearing 

 away the parenthesis and multiplying each quantity within by 



PT PT 



the load we obtain + -=- and r, which of course cancel each 



Z i 



other, leaving Pa, for P = reaction when weight of beam is 

 neglected. 



Example (Fig. 19). A beam 15 ft. long, weighing 9 Ibs. per 

 lineal foot, carries two loads of 300 Ibs. each at points distant 5 ft. 



. _5'_ from the ends. Find the 



^k Aj bending moment. 



eg. h First. Under each 



IS 1 - 



load for the loads only. 



M = Pa = 300 x 5 = 



1500 ft. Ibs. 

 Fig. 19 -Diagram Illustrating the Example Thig moment is con . 



stant for each point on the beam between the loads. 

 Second. Under each load for the load and the weight of the 



beam. 



W Q v 1 ^ 



Ri = R z = P + --= 300 + ~=2 = 367.5 lbs . 



M = (367.5 x 5) - = 1725 ft. Ibs. 



This moment is not constant, for the weight of the beam between 

 the loads must be considered. The increase is slight and usually 

 is negligible, except in the case of concrete beams, in which the 

 dead load often equals or exceeds the live load. This example 

 will be worked out in detail, there being four distinct steps. 



First. Total moment at middle of beam = R 2 X -r- 



i 



Second. Moment due to weight of half the beam. The 

 weight = -- Multiply this by the distance from the middle of 



m 



the span to the center of gravity of half the beam = -r This 

 reduces to -~- x -r = - T~, to be subtracted from the total moment, 



LI 4 o 



which took into consideration the weight of the beam. This gives 



_ ZA 

 M = Rt X - - 



