The maximum unit shear v = 40793 18 X .S74 X40 = 65 lb., Diag onal tension reinforcement will be needed from support to first load (60 inches). This may be supplied by bending up horizontal steel. The bending moment at first load is This is about three-fourths of the moment at the middle and two bars may be bent up at this point. For two bars Ad= 1.12 and Formula (15) Turn up pairs of bars at 20, 40 and 60 inches from support. The bond stress on four horizontal bars at end of beam is This is rather large unless deformed bars are used, and bars should be bent into hooks at ends.
Example 21.—A reinforced concrete slab, divided into panels 12 ft. X 14 feet, by T-beam supports is to carry a live load of 150 pounds per square foot. The T-beams are supported by columns at the corners of the panels; their ends resting upon side walls. Design the slab and beams.
the weight of slab at 70 pounds per square foot. The proportion of load carried by the 12-foot span is 14, 12 —0.5=.67 (see Section 117). The load on the slab in the 12-foot length is (150+70)X.67=147 lb. per square foot and in the 14-foot length 220X.34=75 pounds per square foot. If 4,i"3 of the average load per square foot he borne by the midsection, the load to be carried by a 12-inch width will be 147 X4/3 = 196 pounds per linear foot.
25224/10S = 261 and d=4.75 inches. If the concrete extend inch below steel, the total depth of slab will be 5.5 inches and the weight of slab 150 X5.5/12= 69 pounds per square foot, as assumed.
A = .0078 X 12 X4.75= .44 in? From Table XV, i-inch square bars spaced 3.5 inch c. to c. may be used.
Alternate bars in each span will be turned up at the quarter points for negative shear at the supports.
The side-sections of the shorter span will carry one-half the moment of the mid-sections, and will need about one-half the rein forcement. We will space the t--inch bars 6 inches apart for the side sections.
For the longer span (14 feet) the load upon the mid-section will be 75 X4/3 =100 pounds per linear foot, and the bending moment M=-- 100 X in.-lb. If we place the reinforce ment in the 14-foot direction on top of that in the shorter span, the effective depth will be about i-inch less, or =4.25 inches, 19600 X4.25 = 86. From Table VII, we find that and Id = bclr if and R=86, p=.006 and X4.25=.306 in and Table XV gives :-inch round bars spaced 4 inches apart, A= .33 Use these for mid-half of slab and bars of the same size spaced 7 inches apart for the side-sections.
T-Beams.—Assuming the longer T-beam to weigh 250 pounds per linear foot, the total load will be 196 X 12X 14+250X 14 =36,428 pounds. The maximum shear will be V= 36,428 X.6=21,857 pounds
and section needed for shear Try b'=10, d=21 inches.
The load at the middle of the beam is greater than that at the ends; this somewhat increases the moment, but the error will not be more than about 2 per cent if the load be taken as uniformly distributed.
ml 35428X 14X 12 :1f 10 10 =611,990 in. lb.- Taking the width of flange as one-fourth the length of beam b=45 in.
11I_ 611990 =11S bt(d . ' ' From Diagram I, we find that the neutral axis is in the flange and the beam should be designed as a rectangular section.
alt = 611990 _ 45X21 X21-29. From 'fable VII for and R=29, vve see that f, will be less than 350 ib., and the steel needed is p = .0021 (p = 7R 100000 approximately) or A = .0021 X45 X20=1.59 Four 11-inch square bars may be used. Two of these bars to be turned up at the quarter point on each side of the support to provide for tension clue to negative moment.
We now have at the support a double reinforced beam in which b= 10 inches, d= 21 inches, A =-f' =1.S') in.', p = p' = .0090, d'= 2 inches, d',/d = .095 and G= 611990 10X21X21 =136. From Table XII, for f G=136 and d' 'd=.095 we find that and p'=.0042 are required. The reinforcement for tension is a little small, but as the beam will be strengthened by the slab rein forcement parallel to it, the 16-inch bars will probably be ample.
Diagonal tension reinforcement will be needed for 5 feet from the supports. If stirrups be spaced 9 inches apart, seven stirrups will be needed. The first stirrup will require -inch round bars bent to Li-shape will answer for the first three stirrups, the four next the middle of the beam may be -inch.
The loads upon the shorter beams, assuming the beans to weigh 150 pounds. per foot, are 100X 14X 12+150X 12 =18600 pounds. The maximum shear is 18600 X .6= 11160 pounds. b'd 105 =106. A section 7 inches X 16 inches might be used, but assuming that the depth must he the same as for the longer beams, we may use 7 inches X 21 inches. Then and b=1/4=36 inches. As before, the neutral axis is in the flange, R = 36X221 X 21 =17, X and Table VII, f will be small and 7)=.0012. A = .0012X36 X 21 =0.90 Three-inch round bars will be used. Part of these bars will be turned up, two on one side and one on the other of each support to provide for tension due to negative moment. Then tension reinforcement, is needed 72X47/87=40 inches from support. Spacing stirrups 10 inches apart, for end stirrups 2X 16000