. The strength of materials; a text-book for engineers and architects. can calculate the maximum deflections by reasoningbased on Mohrs Theorem, and we will deal with such casesnow (Fig. 123). (1) Simply Supported Beam with Central Load W.—Let AB represent a simply supported beam of span I with acentral load W. Then adb is the diagram, the maximum ordinate being equal to - . Let A^ c^ B^ be the elastic line of the beam; then, according to Mohrs Theorem, the shape of this elasticline is the same as that of an imaginary cable of the same spanloaded with the curve and subjected to a hor
. The strength of materials; a text-book for engineers and architects. can calculate the maximum deflections by reasoningbased on Mohrs Theorem, and we will deal with such casesnow (Fig. 123). (1) Simply Supported Beam with Central Load W.—Let AB represent a simply supported beam of span I with acentral load W. Then adb is the diagram, the maximum ordinate being equal to - . Let A^ c^ B^ be the elastic line of the beam; then, according to Mohrs Theorem, the shape of this elasticline is the same as that of an imaginary cable of the same spanloaded with the curve and subjected to a horizontal pullequal to the flexural rigidity. Now consider the stability of one half of this cable. It iskept in equilibrium by three forces : the horizontal pull H DEFLECTIONS OF BEAMS 253 at the point c^; the resultant load P on half the cable; andthe tension T at the point A^. Take moments about the point Ai, then we have H X 8 - P X ?/ H In this case P = area of one-half of diagram_l I Wl _ WJ^~ 2 2 ^ 4 ~ 16y = distance of centroid of shaded triangle from a. ) ^P ^^ 1 ^^ T^ — j[ ^ ITP 1 Fig. 123.—Deflections of simply supported Beams. I ~ 3H = EI WP I• ^ 16 ^ WP~ 48 E 1 (2) Simply Supported Beam with Uniform Load.—LetAB represent a simply supported beam of span I, with auniformly distributed load W. The diagram is a parabola, the height being equal to Wl^-. Then considering the stability of half the imaginary cable, we have as before 8 =- P X «/ H 254 THE STRENGTH OF ]\LVTERIALS In this case P = area of one-half of diagram 12 W Z _ WJ2 ~ 2 3 8 ~ 24 51 H = EI . WZ2 51 5WP • 24 16 E I 384 E 1 •(3) Cantilever with ax Isolated Load not at FreeEnd.—Let a cantilever of sj)an l (Fig. 124) carr3ing a load Wat a pomt at distance I from the fixed end a. Then the diagram is a triangle, a d being equal toW /, Ai B^ represents the elastic line of the beam and theimaginary cable. In this case we must imagine the load asacting upwards. The cable is
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