178 HOMOGENEOUS BEAMS 



From the construction of the formulas, it will be noted 

 that the size of a beam required to sustain a certain falling 

 load cannot be found direct. The size of beam must be 

 assumed; then the formulas are used to ascertain whether 

 the beam will meet the requirements. 



EXAMPLE. A 12-in., 40-lb. I beam carries, besides its 

 own weight, a uniform load of 260 Ib. per ft. The span is 

 10 ft. If a load of 400 Ib. drops on the beam from a distance 

 of 18 in., will it develop a unit stress beyond the safe unit 

 stress of 12,500 Ib. ? 



SOLUTION. The total static load per foot on the beam 

 is 260 + weight of beam per foot = 260 + 40 = 300 Ib. per ft. 

 The total static load on the beam, therefore, is 300 X span 

 = 300 X 10 = 3,000 Ib. The deflection due to the falling load 

 of 400 Ib., according to column 13, of the table on page 000 

 is .00000505 X10 3 X. 4 = .00202 in. The constant a thus 

 equals 1 1 



- = .2142 

 W 2 3,000 



1 + .489-- 2 1 + -4S9X - 



Therefore, 



= 25,11 

 The maximum bending moment due to this load is 



25.117X10 = 62 7Q2 5 ft lb Qr 62i 792.5 X 12 = 753,510 in.-lb. 



4 



The maximum bending moment due to the static or dead 

 o nnr) y 1 



load is =3,750 ft.-lb., or 3,750 X 12 = 45,000 in.-lb. 



O 



The total bending moment of both the static and sudden 

 load is therefore 753,510 + 45,000 = 798,510 in.-lb. =Ss. 

 From the table on page 148, S = 41. Therefore, 798,510 

 = 41s, and s = 798,510 -H 41 = 19,476 lb. per sq. in. 



This is greater than 12,500, which was assumed as the 

 allowable unit stress. Even if 16,000 lb. were taken as the 

 allowable unit stress, the actual stress would still be too 

 large and a beam of larger size would have to be assumed. 



