82 PRACTICAL STRUCTURAL DESIGN 



maximum fiber stress of 1000 Ibs. per square inch to carry a 

 uniformly distributed load of 6000 Ibs. on a span of 14 ft. 



Answer. Assume the beam to weigh 20 Ibs. per linear foot 

 = 14 x 20 = 280 Ibs. The total load = 6280 Ibs. 



M = 628 X Q 14 X 12 = 1.5 x 6280 x 14 = 131,880 in. Ibs. 



o 



Assume a depth of 12 ins., which will give a beam 11.5 ins. the 

 usual depth of a commercial size 12-in. beam. 



M r = ^- = 167 bh 2 = 131,880 in. Ibs. 



M r 131,880 _ 



" 167 h* " 167 x 11.5 2 " 



This calls for a commercial size beam 7 ins. x 12 ins., the actual 

 size of which will probably be about 6.5 ins. X 11.5 ins. The weight 

 per linear foot, assuming wood to weigh 35 Ibs. per cubic foot, will 



be r-r-; = 18.2 Ibs. This is so close to the weight 



144 



assumed that we will let it stand. 

 Investigate for shear. 



4bhs 4 x 6.5 x 11.5 x 100 __ 



W = 5 = o = 9967 Ibs. 



o o 



The load of 6280 Ibs. is therefore safe. 

 The deflection is to be kept below -5$-$ of the span 



12x14 



360 



= 0.466 in. 



n U 14x14 L 



~ 441 ~ 44 x 11.5 " "443" 



Find the span on which the safe bending load is equal to the safe 



shearing load. 



, 8 M 143,500 



12 W 1.5 x 9967 



= 9.67 ft. 



This beam cannot be safely loaded with more than 9967 Ibs. 

 on any span of less than 9 ft. 8 ins. no matter what the safe load 

 hi bending may be. (The moment used here is the actual resisting 

 moment of the beam which had to be selected to carry the load, 

 that is, M = 167 x 6.5 x 11.5 2 = 143,500 in. Ibs., the actual bend- 



