Mechanics of engineeringComprising statics and dynamics of solids: and the mechanics of the materials of constructions, or strength and elasticity of beams, columns, arches, shafts, etc . e respective sections of the beam. Theywill be laid off upward or downward according as J isfound to be upward or downward, when the free body con-sidered extends from the section toward the right. In these diagrams the moment ordinates are set off onan arbitrary scale of so many inch-pounds, or foot-pounds,to the linear inch of paper; the shears being simplypounds, or some other unit of force, on a scale of
Mechanics of engineeringComprising statics and dynamics of solids: and the mechanics of the materials of constructions, or strength and elasticity of beams, columns, arches, shafts, etc . e respective sections of the beam. Theywill be laid off upward or downward according as J isfound to be upward or downward, when the free body con-sidered extends from the section toward the right. In these diagrams the moment ordinates are set off onan arbitrary scale of so many inch-pounds, or foot-pounds,to the linear inch of paper; the shears being simplypounds, or some other unit of force, on a scale of so manypounds to the inch of paper. The scale on which thebeam is drawn is so many feet, or inches, to the inch ofpaper. 241. Safe Load at the Middle of a Prismatic Beam Support-ed at the Ends.—Fig. 234. The reaction at each supportis y^P. Make a section n at any distance x<~L from B*Consider the portion nB free, putting in the proper elas-tic and external forces. The weight of beam is JT(mom. about n)=Q we have 266 MECHANICS OF ENGINEERING. e 2 Evidently ilt^is proportional to x, and the ordinates repre-senting it will therefore be limited by the straight line. ,Fig. 234. BR, forming a triangle BRA. From symmetry, anothertriangle ORA forms the other half of the moment dia-gram. From inspection, the maximum M is seen to be inthe middle where x= yil, and hence (Mm&TL.)=Mm=%Pl (1) Again by putting 2Yvert. compons.)=0, for the free bodynB we have *5 and must point downward since £ points upward. Hencethe shear is constant and = ^Pat any section in the righthand half. If n be taken in the left half we would have,nB being free, from Invert. com.)=0, j=p—y2p=y2p FLEXURE. SAFE LOADS. 267 the same numerical value as before ; but J must point up-ward, since t at B and J at n must balance the downwardP at A. At A, then, the shear changes sign suddenly,that is, passes through the value zero; also at A, M is amaximum, thus illustrating the statement in § 240. Notice
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Keywords: ., bookcentury1800, bookdecade1880, booksubjectenginee, bookyear1888
