174 RESISTAKCE OF MATERIALS 



106. Dimensions of spider. The size and number of rods in the 

 spider are determined from the condition that the total area shall 

 be sufficient to carry the shear, as will now be shown. 



It is customary to place a fillet at the top of each column, as 

 shown in Fig. Ill, the depth of the fillet being not less than the 

 thickness of the slab, and its diameter about twice the diameter of 

 the column. The area of concrete in shear at the face of the column 

 is then 7rd(2f), where d denotes the diameter of the column and t 

 the thickness of the slab, and at the outside of the fillet is 7r(2 d)t, 

 which is the same -amount. Consequently, a fillet of these dimen- 

 sions doubles the area of concrete in shear. The chief purpose of 

 the fillet, however, is to avoid the weakening effect of an angle 

 and thus enable the slab to develop its full strength at its junction 

 with the column. 



Since there are more slab rods to carry the shear at the outside 

 of the fillet than at the face of the column, the latter section is 

 weakest in shear. To illustrate the method of dimensioning for 

 shear, consider the same numerical problem as above ; namely, a 

 panel 20 x 20 ft. with a total live and dead load of 320 lb./ft. 2 

 The total load on each column, or the total shear, is then 

 320 x 400 = 128,000 Ib. Assuming that the columns supporting 

 the floor are 1 ft. in diameter, and taking a cylindrical section at 

 the face of the column, the total area of concrete in shear, includ- 

 ing the fillet, will be TT x 12 x 2 x 9.5 = 716 sq. in. For a working 

 stress in shear of 30 lb./in. 2 , the shearing strength of the concrete 

 alone is therefore 716 x 30 = 21,480 Ib. 



The amount of metal in the slab rods was determined previously 

 as 0.628 in. 2 /ft. Since the column is 1 ft. in diameter, and since 

 each set of rods is in double shear and there are four sets of 

 rods, the total area of metal in shear at the surface of the column 

 is 8 X 0.628 = 5 sq. in. Assuming the working stress in shear of 

 the metal as 10,000 lb./in. 2 , the shearing strength developed by 

 the slab rods alone is 5 x 10,000 = 50,000 Ib. Since the total load 

 is 126,000 Ib., there still remains to be taken care of 126,000- 

 (21,480 + 50,000)= 54,520 Ib. of shear. 



To design the spider to carry this shear, assume that it is made 

 up of 8 rods, as shown in Fig. 111. Then the amount of shear on 



