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 . we have 0=%pv-yiP?-±pi?... pc=±p From symmetry PB also = -^P, while P0 must = fPrsince PB + P0 + Pa = 2 P (whole beam free). [Note.—If the supports were not on a level, but if, (for instance)the middle support 0 were a small distance = h^ belowthe level line joining the others, we should put x = I andy = —ho in eq. (6), and thus obtain PB = Pc= ^ P + 3EI~!, which depends on the material and form of thet prism
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 . we have 0=%pv-yiP?-±pi?... pc=±p From symmetry PB also = -^P, while P0 must = fPrsince PB + P0 + Pa = 2 P (whole beam free). [Note.—If the supports were not on a level, but if, (for instance)the middle support 0 were a small distance = h^ belowthe level line joining the others, we should put x = I andy = —ho in eq. (6), and thus obtain PB = Pc= ^ P + 3EI~!, which depends on the material and form of thet prismatic beam and upon the length of one span, (whereaswith supports aU on a level, PB = Pc = A P is independentof the material and form of the beam so long as it is ho-mogeneous and prismatic.) If P0, which would then =?! P — 6 EI (Jiq-^-W), is found to be negative, it shows that0 requires a support from above, instead of below, tocause it to occupy a position Jiq below the other supports, the beam must be latched down at 0.] The moment diagram of this case can now be easily con-structed ; Fig. 277. For any free body nG, n lying in DCrwe have 324 MECHANICS OF Fig. 277. figure) and at 0, where x= I, becomes ??- M= , varies directly as x, un-c til x passes D when, for anypoint on DO, M=%Px-P{x-1-) which is =0, (point of in-flection of elastic curve)for x=8/u I (note that x ismeasured from G in thisPI |PZ; MG=0; MB=3\Pl; and Jfo«0 Hence, since M max. =^Pl3 the equation for safe loadingis RI 6 tvi /y\ AP, while on DO e 32^ The shear at 0 and anywhere on CD=it =^P in the opposite direction . . • • (8) The moment and shear diagrams are easily constructed,as shown in , the former being svmmetrical abouta vertical line through 0, the latter about the point 0Both are bounded by right lines. 273. Two Equal Spans. Uniformly Distributed Load Over Whole Length. Prismatic ^ ^JOX* N c —Fig. 278. Supports B, 0, £ij I iJJ 11 Ujljll G> on
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Keywords: ., bookcentury1800, bookdecade1880, booksubjectenginee, bookyear1888
