. The strength of materials; a text-book for engineers and architects. ordinate of diagram. Alternative Construction.—The following alternativeconstruction is usually more accurate in practice. On a hori-zontal base Ag B2, Fig. 60, set up Cg m equal to , ^. e. the Ij due to an isolated load W placed at c, the centre of theload. Join m Ag, M B2, cutting the verticals through d in nand D respectively, and join n d. On Eg Do draw a parabola W I Eg Q D2 of height equal to -^ (the due to a uniform 134 THE STRENGTH OF MATERIALS load on a span e d), then the diagram comes
. The strength of materials; a text-book for engineers and architects. ordinate of diagram. Alternative Construction.—The following alternativeconstruction is usually more accurate in practice. On a hori-zontal base Ag B2, Fig. 60, set up Cg m equal to , ^. e. the Ij due to an isolated load W placed at c, the centre of theload. Join m Ag, M B2, cutting the verticals through d in nand D respectively, and join n d. On Eg Do draw a parabola W I Eg Q D2 of height equal to -^ (the due to a uniform 134 THE STRENGTH OF MATERIALS load on a span e d), then the diagram comes as shownshaded. If it is desired to have the diagram on a straightbase, the parabola may be drawn on the inclined base n d asindicated in dotted lines. The proof of this construction isleft as an exercise to the student. Case 5. Irregular Load.—Graphical Construction.—Let a number of loads Wi, Wg, W3, and W^, be placed knj-where along a span ab. Number the spaces between theloads and set down, 0, 1; 1,2; 2, 3 ; 3, 4, as a vertical vector - I . ooocooorrxTrr). ^ 8 Fig. 60.—^Alternative Construction for Uniform Load overPortion of Span. line to represent the loads to some convenient scale, and inany position take a point p at convenient polar distance pfrom the vector line, and join p 0, p 1, P 2, etc.• Across space 0 then draw a h parallel to p 0; across space 1draw h c parallel to p 1 and so on until e / is reached, thisbeing parallel to P 4. Join a f, then the figure a, b, c, d, e, /, a, will give the for the given load system. Now draw p x parallel to a /,/the closing link of the linkpolygon then on the vector line, 4 :«: = R„ and xO = R^. To draw the shear diagram, draw a horizontal line through BENDING MOMENTS AND SHEARING FORCES 135 X right across the span : this gives the base line for project the point 0 horizontally across space 0; projectpoint 1 across space 1 and so on, the stepped diagram thusobtained being the shear diagram. Proof
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