. The strength of materials; a text-book for engineers and architects. n be plotted in very convenient form as in Fig. 109. Thesecurves may be likened to the tables of section moduli for steelsections. Numerical Example. — A reinforced concrete beam isrequired to carry a bending moment of 240,000 in. lbs. Designthe section for stresses c = 600, t = 16,000, assuming that thebreadth of the beam is 10 inches. By equation {U) d = ^ ^^-^ 240,000M 95 X 10= 15*9 inches. .-. A- ^ -00675a 0 A, = -00675 X 15-9 x 10 = 1*07 sq. inches. Adopt 2 — J bars [giving an area 1-2]. Third Method—General No-terisio
. The strength of materials; a text-book for engineers and architects. n be plotted in very convenient form as in Fig. 109. Thesecurves may be likened to the tables of section moduli for steelsections. Numerical Example. — A reinforced concrete beam isrequired to carry a bending moment of 240,000 in. lbs. Designthe section for stresses c = 600, t = 16,000, assuming that thebreadth of the beam is 10 inches. By equation {U) d = ^ ^^-^ 240,000M 95 X 10= 15*9 inches. .-. A- ^ -00675a 0 A, = -00675 X 15-9 x 10 = 1*07 sq. inches. Adopt 2 — J bars [giving an area 1-2]. Third Method—General No-terision Method.—In thismethod we will, as before, assume that all the tensile stress istaken by the steel, but we will assume that the stress-straincurve for concrete is not straight but some other curve. 228 THE STRENGTH OF IVIATERIALS In this way we get the stress diagram, Fig. 110, from thestrain diagram. Suppose that its area = k . c . ii and that its centroid is atdistance y from the top. Then, as in equations (8), (9) we get {d — n) m .c n — . n d 1 + mc. /. »~ A n d , d- r. cL, ------- -•-- r Fig. 110.—Reinforced Concrete Beams. Methods. Now, since the total compressive stress must be equal to thetotal tensile stress, we have tAj = khnc (16) {d — n) m A^ k 0 n. •. kh 11^ ^ m A^{d — n).. k b 71^ -{- m Aj71 — m Aj d A^m ( 0 ^ ^ 4:dkb \ ^ 2X6 W ^ + m A, ~ ^/ or. d r TTi (2l\ r m J (H)(18) (19) Then moment of resistance = =tA,{d-y) for tension (20) = kb 71 c {d — y) for compression . (21) STRESSES IN BEAMS 229 Numerical Example with Stress-strain Curve Para-bola.—Take the section that we have worked for theprevious formulae. (See Fig. 106.) If the stress-strain curve is a parabola tangential at thecompression edge we have 2 3ri » ?frko mTTon cr^/^firkn w — ^ j / 1 1. 6 . tilt; glVCll OCOulUll fi — jj 1 A / X 1 1 ;? 1 ,AA ,3 = 5-12 .-. d -n = 10 - 5-12 = 4-88 Safe for concrete = 1 X 6 X 5-12 X 500 (4-88 -f 3-20) in. lbs. = 3 X j2 X
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