This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1921 Excerpt: ...9. A T-beam has the following dimensions, 6=48 inches, =4 inches, d = 22 inches, b'= 10 inches. The steel reinforcement consists of six f-inch round rods. If the safe unit stresses of steel and concrete are 15,000 and 600 lb./in.2 respectively, and n=15, what is the safe resisting moment of the beam? 2 65 Solution- ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1921 Excerpt: ...9. A T-beam has the following dimensions, 6=48 inches, =4 inches, d = 22 inches, b'= 10 inches. The steel reinforcement consists of six f-inch round rods. If the safe unit stresses of steel and concrete are 15,000 and 600 lb./in.2 respectively, and n=15, what is the safe resisting moment of the beam? 2 65 Solution--From Table X, A=2.65 in.2, and p = 00s..Q =.0025; formula (27) gives 0025X15+KA)2=.0025X15+ Using (28) we find id-22 3X-247-24_ Jd22 2X.247-.3-20-39. From (22), /, _ 15(1-.247) U 47 45? If/ = 600 lb./in.2, /, =600X45.7 = 27,420 lb./in.2 This is greater than the safe unit stress on steel, and the safe moment will be that which causes a stress of 15,000 lbs./in.2 on the steel, or from (29), M = 2.65 X15000 X 20.39 = 810000 in.-lb. 10. The flange of the T-beam is 26 inches wide and 4 inches thick. The beam is to carry a bending moment of 520,000 in.-lb. The safe unit stresses for concrete and steel are 600 and 16,000 lb./in.2 respectively. What area of steel and depth of beam are needed. Solution.---By (23) k = Yqqqx15 X 600=-360" We must noW find d by assuming values and testing their suitability. Try d = 18; from (28) we have (9) gives C = M/jd = 520000/16.3 = 31900. C From (24) fc = jzrt-ot=Q lb./in.2 This is a safe value, but 2kd a less depth will answer. Trying 15 inches, we find C = 38,000 pounds, and fc--580 lb./in.2; 15 inches is, therefore, approximately the minimum value for d. For this value of d, Formula (25) gives, A = T/f, =38000/15000 = 2.375 in.2 Width of Flange.--T-beams without lateral reinforcement in the flanges should have a width of flange not more than three times the width of web, 6 = 36'. When the flange is reinforced at right angles to the length of beam, as in a slab floor with T-beams support, experience indicates ...
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