. The strength of materials; a text-book for engineers and architects. , but will calculate it soas to give c = 600 lbs. -per sq. = 6000 lbs. per sq. = 15 d m c15 Then we have n = 1 + 1 + 6000 = 5-4 ins. 15 + 600 -^ ^ »^ 1 / 3 — /o — Fig. 114. •. From equation (22) 16,000A, = ^00^^g_38xl:9:|2, I 5*4 J .. A^ = 438 sq. ins. .•. Adopt, say, 3 bars If^ diameter. Then working by the equivalent moment of Inertia I, = 15 X 4-49 X 9-62 + 48 x \^ - ^^ /<i9 = 8,641 .-. Safe = ^^^ ^,^^^^ = 960,000 in. lbs. 54 For other cases of reinforced concrete beams the reader is STRESSES IN BEA
. The strength of materials; a text-book for engineers and architects. , but will calculate it soas to give c = 600 lbs. -per sq. = 6000 lbs. per sq. = 15 d m c15 Then we have n = 1 + 1 + 6000 = 5-4 ins. 15 + 600 -^ ^ »^ 1 / 3 — /o — Fig. 114. •. From equation (22) 16,000A, = ^00^^g_38xl:9:|2, I 5*4 J .. A^ = 438 sq. ins. .•. Adopt, say, 3 bars If^ diameter. Then working by the equivalent moment of Inertia I, = 15 X 4-49 X 9-62 + 48 x \^ - ^^ /<i9 = 8,641 .-. Safe = ^^^ ^,^^^^ = 960,000 in. lbs. 54 For other cases of reinforced concrete beams the reader is STRESSES IN BEAMS 235 referred to the authors Elementary Principles of ReinforcedConcrete Construction (Scott Greenwood & Son, London). COMBINED BENDING AND DIRECT STRESSES If the loadmg on a beam is such as to cause a direct stressin addition to bending stresses, then the resultant stressesacross the section will be obtained by adding together theseparate stresses. Let b d, Fig. 115, represent the elevationof a section of a beam, c being the centroid of the section. Fig. 115. whose area is A and whose compression and tensile moduliare Z,. and Z^, d being the compression side and b the tensionside. Then, if the direct force is a thrust Q, there will be a uniform compression stress of -^ over the section. If the bending moment is equal to M, the maximum compression and tensile M Mstresses due to bending are equal respectively to ^y and ^. Therefore we have Resultant maximum compressive stress = /, = ^ + ^ .. (1) Resultant maximum tensile stress = /^ = ^ — ^ . (2) The distribution of the combined stresses across the section 236 THE STRENGTH OF MATERIALS is then as shown in Fig. 115, fh representing the maximumcompressive stress, and g e the maximum tensile stress. Theneutral axis then is at the point n, where the stress is the direct force is a pull T instead of a thrust Q, we have Resultant maximum tensile stress = maximum compressive stress T M M T Z,. A ^ Stres
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