v is 120 at the supports, and diagonal tension reinforce ment is needed where v is more than 40 lb., Stirrups will be needed for two-thirds of the distance from the support to the mid span, or 6 feet. If the stirrups be spaced s=d 2=9 inches apart, eight stirrups will be needed at each end of the beam. For the stirrups next the support (34) Two 1-inch round bars, bent as shown (Fig. 55) may be used for the first four stirrups, and -ineli bars for the four nearer the middle of the beam.
Example 20.—A reinforced concrete slab, to carry a live load of 200 pounds per square font, is to rest upon a series of T-beams 5 feet apart e. to c. The T-beams are to be continuous for three spans over girders 15 feet c. to e. The girders are supported by walls at the ends and have a span of 20 feet. Design the slab and beams.
Solution.—Asstune the weight of slab at 40 pounds per square foot ; t hen 3I = 240X iX 5 X 12 = 6000 in. lb. From Table v11, R=10S, 6000, 108=55.5 and Take in. A =225X12X.0078=.21 If concrete extend a-inch below steel, the total depth of slab is 3 inches, and weight of slab is 150 X pounds per square foot.
From Table XV —A-inch round bars spaced 4 inches apart give A=23 Negative moment at supports will be provided for by bending these up at the quarter points. For lateral reinforcement to prevent cracks, round bars spaced 12 inches c. to c. will be used.
T-beams.—Assuming weight of web of T-beam as 12.5 pounds per linear foot, load upon T-beam is 5(200+40)+125=1325 pounds per linear foot and total span load is 1325X 15=19,875 pounds.
Maximum shear in end span next girder is F=19,S75X.6=11, 925 pounds, and b'd=1' 105=113 in.'- 7X 16 or 8 14 might be used. Try- 7 X16, then .lf =117/10= 19,875 X15X 12 10 = 357,7.50 in.-lb. Taking overhang of flange as six times its depth, b= 2 X 6 X3+7 =43 inches and Formula (31) From diagram I, f=325 lb. 'in.= and p= .0024. A =1.65 Table X, six b-inch round bars, .1=1.S4 in." in two rows, 11 inches apart and spaced 2 inches c. to c. and 1.5 inches from side of web.
If concrete extends 2 inches below steel, the weight of web below slab is 7 X 16X 150, 144=117 pounds per linear foot, which is within the assumed load.
The negative moment at crossing of girder is equal to the posi tive moment already found. Turn up the upper row of bars on each
side to provide for tension at top of beam and run the lower ones through at bottom to provide compression reinforcement as shown in Fig. 56. We now have a beam with compression reinforcement, in which b=7, A=A'=1.84, 112=.0164, Formula (48) gives G=-- 1, 7XI6X1G _ bd 357750 = 200, and Table XII, for L. 16000, L= 650, G=200, and d/d = .18, we find that p = .0139 and p'=.0219 are required. The area of steel in compression (p .0164) is not sufficient and we must either increase the area of compression steel in the bottom of beam or increase the area of concrete section over the support. Try making d= 17 inches. Then p=p'=1.84/119=.0157,d' Id= 2.875/17 =.17 and r 357750 =177.
X17X17 Now from Table XII, we find that p=.0123 and p'=.0156 are required. The reinforcement is now sufficient and we will increase the depth to 17 inches at the girder, sloping the haunches as shown in Fig. 56.
Diagonal Te71sion.—Assuming J as .85, the maximum unit shear 11925 next the girder is r=,— =118 1h./ in? If stirrups be spaced 7XS5X17 8 inches apart, the area required for the end stirrups is (34) Two finch bars will answer, or a ;-inch bar bent to LT-shape around horizontal reinforcement. Stirrups will be needed to 6 feet from girder and 4 feet from end support in the end spans and 5 feet from girder on each end of the middle span.
girders are simple rectangular beams carrying three concentrated loads at the middle and quarter points. Each load is 1.1 times a span load of the T-beam, or 1.1 X 19S75=21862 pounds; assuming that the girder weighs 800 pounds per linear foot, the reaction or shear at the support is 1.5 X21862+800 X10=40793 pounds t',nd the maximum bending moment 1H=-IO,793X(120 20,SG2) X GO = :3,10:3,110 in.-lb. = 3,103,110, 108 = 28,735. For b=20, for b=1S, Try b=1S, then A 1S X40=5.61 Ten ;-inch square bars (A =5.02) placed in two rows 1 inches c. to c., six bars in lower and four in upper layer (Fig. 56). if concrete extend 2; inches below center of lower layer of steel, the beam is 43 inches deep and weighs 1SX43X 150, 144=806 pounds per linear foot, which agrees with the assumed weight.