88 PRACTICAL STRUCTURAL DESIGN 



books, it will be well to study the subjects thoroughly in one 

 and then become familiar with the similar matter presented in 

 the others. 



Only rolled shapes have been considered so far. Compound 

 shapes, i.e. plate girders and trusses, will be taken up later. 



Practical Problems in Design 



1. Find the resisting moment of flooring f in. thick; f in. 

 thick; 1| in. thick; If ins. thick. 



Answer. f in. = 0.625 in.; in. = 0.875 in.; 1| ins. = 1.125 

 ins. ; If ins. = 1.75 ins. The width will be taken as 12 ins., as floor 

 loads are generally given in pounds per square foot. The flooring 

 is white pine having a fiber stress of 800 Ibs. per square inch. 

 [In all problems it is understood that by fiber stress is meant 

 the maximum (skin) stress.] The unit moment of resistance 

 = 800 -=- 6 = 133.33. 



M r = 133.33 x 12 x 0.625 2 = 625 in. Ibs. 



Mr = 133.33 x 12 x 0.875 2 = 1225 in. Ibs. 



Mr = 133.33 x 12 x 1.125 2 = 2025 in. Ibs. 



Mr = 133.33 X 12 x 1.75 2 = 4900 in. Ibs. 



2. What is the greatest spacing permissible between joists if 

 the deflection is to be limited the usual amount? 



Flooring comes in long pieces and thus, extending over a number 

 of supports, to each of which it is nailed, the thickness can be equal 

 in inches to one-half the span in feet. This gives a maximum span 

 for the f-in. of 2 x 0.625 = 1.25 ft. (15 ins.) ; f-in., 2 x 0.875 = 1.75 

 ft. (21 ins.) ; lf-ins., 2 x 1.125 = 2.25 ft. (27 ins.) ; lf-ins., 2 x 1.75 

 = 3.5 ft. (42 ins.). 



Floors generally have greater stiffness than is here shown 

 because of the tongue and groove along the edges, but this is fre- 

 quently nullified by the fact that the loads brought on floors are 

 more often concentrated than uniformly distributed. The above 

 rule for deflection is arbitrary, and if the spans mentioned are 

 actually used it will be well to check the deflection by a proper 

 formula. Refer to the table of relative strength and stiffness of 

 beams. The deflection formula gives deflection for uniform loads 

 on beams resting freely on two end supports. First find the de- 

 flection by the formula and multiply it by the constant found in 

 the column of relative deflections, opposite the condition of loading 

 to which the case under consideration may apply. 



