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 . kea new pole on a horizontal through n, with the sameH1, and draw a corresponding equil. polygon; in the lat-ter vm would be horizontal. We might also shift thisnew trial polygon upward so as to make vm and vmcoincide. It would satisfy conditions (7) and (8), havingthe same ss as the first trial polygon ; but to satisfy con-dition (9) it must have its ss altered in a certain ratio,which we must now find. But
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 . kea new pole on a horizontal through n, with the sameH1, and draw a corresponding equil. polygon; in the lat-ter vm would be horizontal. We might also shift thisnew trial polygon upward so as to make vm and vmcoincide. It would satisfy conditions (7) and (8), havingthe same ss as the first trial polygon ; but to satisfy con-dition (9) it must have its ss altered in a certain ratio,which we must now find. But we can deal with the indi-vidual sTs just as well in their present positions in ] The points E and L in vm, vertically over Er andL in vm!, are now fixed; they are the intersections of thespecial polygon required, with vm. The ordinates between vmf and the trial equilibriumpolygon have been called z instead of z\ they are pro-portional to the respective 2s of the required specialpolygon. The next step is to find in what ratio the (z)s need tobe altered (or H altered in inverse ratio) in order to be-come the (z)s; , in order to fulfil condition (9), viz.: AKCH-KLBS. 409. Z\{yz) = l\(yz>) . (9) This may be done pre-cisely as for the rib withtwo hinges, but the nega-tive (s)s must be prop-erly considered (§ 375)See Fig. 431 for the de~tail. Negative 2s or sspoint upward. From Fig. 431a I\{yz)=HJc,\ from symmetry Z%Vz)=2H„ Fig. 4316 we have Fig. 431. and from Fig. 431c II(yzff)=HQkr[The same pole distance H0 is taken in all these construc-tions] .-. I\{yz)=H{){k^kT). If, then, H0 {k\-\-kv) = 2R0k condition (9) is satisfied by thezrs. If not, the true pole distance for the special of Fig. 430 will be kY-\-kc H- 2k H With this pole distance and a pole in the horizontal throughn (Fig. 430) the force diagram may be completed for therequired special polygon ; and this latter may be con-structed as follows : Beginning at the point E, in vm,through it
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Keywords: ., bookcentury1800, bookdecade1880, booksubjectenginee, bookyear1888
