Double Reinforced of long span and nearly square in plan may be supported on all four sides and reinforced in both directions. It is not feasible to make an accurate analysis of the dis tribution of loadings in such a slab. When the length and width of slab are equal, it is assumed that the reinforcement in each direction carries one-half the load as uniformly distributed. The loads carried by the mid sections (aaaa, bbbb, Fig. 54) are, however, greater than those carried by the sections next the supports, and the reinforcement should be spaced closer in the middle than at the sides. It is sug gested that the mid-half area (aaaa, bbbb, Fig. SO) of the slab be con sidered as carrying 13 times the average load, and the side sections (acca, bceb) two-thirds of the average load per square foot of slab.

When the slabs are not square, the reinforcement parallel to its shorter dimension carries the greater part of the load. The Joint Contrnittce makes the following recommendation concerning the division of the loads in such slabs: Floor slabs having the supports extending along the four sides should be designed and reinforced as continuous over the supports. If the length of the slab exceeds 1.5 times its width the entire load should be carried by transverse reinforcement.

Vor uniformly distributed loads on square slabs, one-half the live and dead load may be used in the calculations of moment to be resisted in each direction. For oblong slabs, the length of which is not greater than one and one-half times their width, the moment to be resisted by the transverse reinforcement may be found by using a proportion of the live and dead load equal to that given by the formula r=b where 1=length and b =breadth of slab. The longitudinal reinforcement should then be proportioned to carry the remainder of the load. In placing reinforcement in such slabs account may well be taken of the fact that the bending moment is greater near the center of the slab than near the edges. For this purpose two-thirds of the previously calculated moments may be assumed as carried by the center half of the slab and one-third by the oustide quarters.

An interesting discussion of the distribution of stresses in double reinforced slabs may he found in a paper by Mr. A. C. Janni in Trans actions of the American Society of Civil Engineers, 1917.

118. Problems in Design.—The use of the formulas and tables which have been given, in designing slab and beams, will he illustrated by the solution of a few problems. In these examples, the working stresses recommended by the Joint Committee for 2000 pounds con crete will he used.

Example 19.—A concrete slab is to be supported by T-beams 6 feet apart c. to c., and to carry a live load of 250 pounds per square foot. The T-beams have a clear span of 17 feet and are built into brick walls at the ends. Design the slab and beams.

Solution.—Assume the weight of slab as 50 pounds per square foot, giving a total load of 300 pounds per linear foot for a section of slab 12 inches wide. Taking the slab as fully continuous.

From Table VII, for b= 16,000 and = 650, R= 108, p = .0078 and j= .874. Formula (9) gives 10,800/ 10S = 100, and d= 2.9 inches, use 3 inches. A = pbd = 3 X12 X .0078 = .277 From Table XV (p. 199), we select 1-inch round bars spaced 4.5 inches apart, A = .29 If concrete extends inch below steel, the thickness of slab is 3i inches, and the weight of slab is 150X3.75/12=47 pounds per square foot, which agrees with the assumed load.

Reinforcement for negative moment over the supports should be the same as for positive moment at mid-span, and will be provided by turning up every alternate bar at the quarter point on each side of the support and continuing them over the support to the one-third point.. Transverse reinforcement to prevent cracks will be provided by using -inch bars spaced 12 inches apart.

No diagonal tension reinforcement is necessary.

T-beam.—Assuming the weight. of the web of the T-beam as 150 pounds per linear foot, the total load on the T-beam is 6(250+-17) +150=1930 pounds per linear foot. Taking the bearing upon the wall as 6 inches the effective length of 'J'-beain between centers of bearings is 17.5+.5= IS feet.

and from Table X, six a-inch round bars in two rows, 2 inches c. to c., spaced 2.75 inches apart in the rows and 1.75 inches from side of web. A =3.61 As the ends of the beam are built into the walls, some negative moment may be developed at the supports, which might cause cracks to occur unless reinforced. The upper layer of reinforcement will therefore be turned up, two rods at the quarter point and the other midway between the quarter point and support, and extend to the end of the beam (see Fig. 55).

If the concrete extend 2 inches below the steel, the weight of web below the slab is 9(1S.5-l-3-3.75)X 150/144=166 pounds per linear foot. This is a little greater than the assumed value, but would add less than 1 per cent to the total load and need not be redesigned. For the three bars in bottom of beam at the support. Table X gives 1o=3X2.75=S.25, and the unit bond stress =131 This is too great for safety, and the bars should be bent into hooks at the ends.