PROBLEMS IN DESIGN OF BEAMS 99 



Let L ** span in feet, 



b = breadth in inches, 

 h - depth in inches (required for strength), 

 x = depth in inches (required for deflection), 

 D = deflection found in inches, 

 d - allowable deflection in inches. 



Then 



|W.. ./1.585 X 3.625* _ 5 15 mg 



nr \ 0534- 



Use nominal 2-in. x 6-in. planks. 

 The formula is developed as follows : 



dU PL 3 



bh 3 ~ btf ' 



which becomes x 



bh 3 A DL 3 



CanceUing common factors, we get JJT, and x 3 = 3-, the formula 



used above. 



7. Assuming a floor with same load and fiber stress is to be 

 carried on joists find the size required for joists 12 ins. center to 

 center and 16 ins. center to center. Deflection to govern. 



The allowable deflection = ~^ = 0.534 in. 



U 16 2 



10.4 ins. 



4QD 46 x 0.534 

 The depth in this example needed to avoid undue deflection is 

 based on the fiber stress used in design, for the breadth is governed 

 by the depth. In the case of the laminated floor a constant breadth 

 of 12 ins. was used and the deflection was fixed by a lower fiber 

 stress than that used for strength only. 



The bending moment for a width of 1 ft. = 18,432 in. Ibs. (from 

 the last example). The thickness of the joist will be 



18,432 



b = 133 x 10.625' - L23 ms ' 



Use nominal 1.5-in x 11-in. joists, 12 ins. center to center. . 

 With joists spaced 16 ins. center to center the bending moment 

 is increased one-third, 18,432 x 1.333 - 24,600 in. Ibs. 



24,600 



6 - 133 X 10.625' - L64 "*' 

 Use nominal 2-in. x 11-in. joists, 16 ins. center to center. 



