Dec. 3* 1917 
Tests of a Large-Sized Reinforced-Concrete Slab 
5i9 
the effective width of the centrally loaded slab, but that when D is 
reduced to less than — the corresponding value of (be)=~+D. The 
approximate coincidence of this curve with that of the test values 
would seem to confirm the truth of the assumed relation. Unfortu¬ 
nately, however, the results can not be regarded as entirely conclusive, 
owing to the fact that the recent test involved only one span length 
and yielded only two points on the critical part of the curve between 
the total widths of 18 and 22 feet. It is hoped that this point will be 
cleared up by further tests. 
CONCLUSIONS AFFECTING THE DESIGN OF SLABS 
To provide against the frequently realized condition of a heavy con¬ 
centrated load applied near the parapet, the test conclusively shows 
that the resisting moment required in the portion of a bridge slab near 
the outer edges is greater than that which is necessary in the central 
portion. 
Further than this, if the relation indicated above be verified and 
shown to include other span lengths, it would seem that in designing a 
slab the necessary allowance for the concentrated load near the outer 
edge can be made very simple in the following manner: 
(1) Use the formulas for narrow rectangular beams, substituting 
for the breadth ( b ) the value obtained from the foregoing table for 
central concentrated loads; (2) determine the loss in effective width 
due to the assumed eccentricity of the load; and (3) supply the defi¬ 
ciency by designing the curb of the parapet to provide a resisting moment 
equal to that of a slab of width equal to the loss in effective width due 
to eccentricity, making allowance for the greater stiffness of the section 
under the parapet. Thus, suppose a slab of 16-foot span and 20-foot 
width is to be designed to carry a concentrated load of 20,000 pounds 
applied at a point 4 feet from one edge, then 
Total width _2o_ i 
Span 16 
from the table for central concentrated loading, the effective width = 0.69 X 
16 feet= 11.04 feet. Consider the load of 20,000 pounds to be carried by 
a width of 11.04 feet, use the ordinary formulae for rectangular-beam 
design and determine the effective depth of the slab and the area of 
steel required. Now, by the relation indicated above, determine the 
effective width with the load in the critical position 4 feet from one 
edge, then 
(6^=51^+4=5.52+4=9.52 feet 
2 
the difference between the values of be and be is 11.04 —9.52 = 1.52 feet. 
