104 PRACTICAL STRUCTURAL DESIGN 



In a mill-constructed building 7 ins. x 14 ins. white pine beams 

 spaced 5 ft. on centers are used with a span of 20 ft. The allow- 

 able maximum fiber stress is 800 Ibs. per square inch and the beams 

 are to be strengthened so the total floor load can be increased to 

 100 Ibs. per square foot, inclusive of floor, beams, and live load. 



Testing first the strength of the beams against failure by longi- 

 tudinal shear on the neutral axis, the unit shear being one-tenth 

 the allowable fiber stress, 



TT7 4 x 7 x 14 x 80 



W = - Q - = 10,450 Ibs. 



o 



The total panel load will be 5 x 20 X 100 = 10,000 Ibs., so the 

 beam will carry the additional load without failing in shear. 



M b = 1.5 X 10,000 x 20 = 300,000 in. Ibs. 



,, 800 x 7 x 14 2 



M r = ^ = 182,933 in. Ibs. 



b 



Then the difference between the bending moment and the 

 resisting moment is 300,000 - 182,933 = 117,067 in. Ibs., which 

 difference must be cared for by reinforcement. To secure equal 

 deflection the reinforcement should be the same wood, white pine, 

 but the difference will not be appreciable in this case, and to use 

 yellow pine will give a smaller piece for reinforcement because 

 of the higher allowable fiber stress. The beam is in an old build- 

 ing and quite likely the maximum deflection in the white pine has 

 been reached, and there is a decided permanent set. The reinforce- 

 ment should be added when the floor is unloaded in order to enable 

 the old beam and the new pieces to deflect together when the 

 live load is added, the difference in deflection between the two kinds 

 of wood being cared for by the deflection due to dead load in the 

 wood having the greatest deflection. 



Assuming, therefore, yellow pine with a fiber stress of 1300 Ibs. 

 per square inch and a depth of 14 ins. the thickness is to be com- 

 puted. Let 



R = 1300 -s- 6 = 217 

 M 117,067 

 6 - W = 2I7-X-IP - 2 ' 76 mS - 



Use two If-in. planks, one on each side. When surfaced the 

 thickness will be practically 2f ins. 



The load is uniformly distributed ; the original beam is large 

 enough to carry the required load without a shearing failure; 



