USEFUL DATA 



i i 



J! 



4:1"--- 



-18-%"0bars, 9-3"each way 



FOOTING TABLES 



The purpose of a footing is to distribute the column load uniformly over the soil 

 so that unequal settlements may be avoided. To accomplish this one of three general 

 tj-pes of foundation may be used: (a) square spread footings; (b) combined spread 

 footings; (c) spread footings supported by piles. Tables covering these three types 



are given on pages 136 to 142. In special 

 cases there may be employed cantilever 

 footings or a raft foundation extending 

 over the entire lot area. To cover these 

 in tabular form, however, would be 

 difficult. 



In the table for square column foot- 

 ings the method of design outlined in 

 Bulletin 67, University of Illinois, has 

 been followed and its application is 

 illustrated in the following problem. 

 Problem — Given a column load of 

 500,000 pounds. Design a square spread 

 footing, reinforced in two directions, for 

 an allowable soil pressure of 6,000 

 pounds per square foot; diameter of 

 column 25 inches; punching shear not 

 to exceed 120 pounds per square inch; 

 bond stress limited to 100 pounds 

 per square inch ; fa = 16,000 and/c = 650. 

 Approximating the weight of the 

 footing at 400 pounds per square foot, 

 the net soil reaction will be 6,000 — 

 400 = 5,600 pounds per square foot. 

 The area of the footing will be, 500,- 

 0004-5,600 = 90.4 sq. ft. and we will 

 use a footing 9' 6" x 9' 6". 



Fig. 9 



The effective depth of footing is found by dividing the column load, less the net 

 soil reaction directly under the column area, by the perimeter of the column 



multiplied by the unit punching shear, or Eff. depth = 



500,000-19,000 



=51 in. 



(25) (3.14) (120) 

 To the effective depth should be added 3 inches of concrete for the protection 



of the metal, giving a total depth of footing of 4' 6*. 



The area of steel, in one direction, to resist the moment of the net soil reaction 



about the edge BC of the cap, is 



A = 



^ 4.08 + 9.50 ^ (2 . 71) (5,600) (1 . 53) (12) 



= 4.10 sq. in. 



^) (33) (16,000) 



This area is equivalent to 14 — Y^' round bars. 



As it is required to maintain the bond stress at 100 pounds per square inch 



133 



