Essentials in the theory of framed structures . as 06y^/ues of c Pig. 132. single load is adjacent to one of the supports, and jfe is on thepoint of becoming zero. Under this condition the point ofmaximum deflection cannot be at a distance greater than the center of the span. Any second load apphed to the 220 THEORY OF FRAMED STRUCTURES Chap. V beam must necessarily throw the point of maximum deflectionnearer the center. Hence the point of maximum deflection of asimple beam of uniform cross-section, loaded in any manner,will be near the center and not more than of its length
Essentials in the theory of framed structures . as 06y^/ues of c Pig. 132. single load is adjacent to one of the supports, and jfe is on thepoint of becoming zero. Under this condition the point ofmaximum deflection cannot be at a distance greater than the center of the span. Any second load apphed to the 220 THEORY OF FRAMED STRUCTURES Chap. V beam must necessarily throw the point of maximum deflectionnearer the center. Hence the point of maximum deflection of asimple beam of uniform cross-section, loaded in any manner,will be near the center and not more than of its lengthfrom the center. 143. A 20-in. 6s-lb. I-beam supports two loads of 30,000 (Fig. 133). Since the loads are symmetrically placed,the elastic curve and M-diagram are symmetrical about thecenter. The tangent to the elastic curve at the center, drawn. = in. through T, is horizontal and A = i. £ = 29,000,000 ^.,I = 1, in*.; hence EI = 33,915,500,000 in^.-lb. Area-moment of PQSU about P-area PQF 150,000 X X = 1,250,000area QSUV 150,000 X X = 9,843,750 ,093-750 ft^^ = i= .o93;75o X 1,72833>9i5)5oo,ooo When the length of a beam is expressed in feet, and the loadsare expressed in pounds, the area-moment will be expressed infoot-pounds; and the factor 1,728 is introduced if EI is ex-pressed in inch^-pounds. EI may be expressed in foot^-pounds by dividing by 144,whenceEI = 235,521 , then A = t= .093,750 _235,521 The deflection at the center may be found from Table I. ft. = in. Sec. II DEFLECTION OF BEAMS 221 k = and c = ; therefore F — for each load, then 2Pf . 60,000(25 X 12)^ _,_ . A = -—-^ X = ? ^ ^ - = in. 6£/ 6 X 33,915,500,000 144. Deflection Under Uniform Load.—The beam in Fig. 134 supports a uniform load and the M-
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Keywords: ., bookcentury1900, bookdecade1920, booksubjectstructu, bookyear1922
