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 . w a vertical through the middleof each. Let the loads borne by the respective Jss beFlt P2, etc., and with them form a vertical load-line A C tosome convenient scale. With any convenient pole 0draw a trial force diagram 0 AC, and a correspondingtrial equilibrium polygon F G, beginning at any point inthe vertical F. Its ordinates %, 22, etc., are propor-tional to those of the special equil. pol. sought (whosea
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 . w a vertical through the middleof each. Let the loads borne by the respective Jss beFlt P2, etc., and with them form a vertical load-line A C tosome convenient scale. With any convenient pole 0draw a trial force diagram 0 AC, and a correspondingtrial equilibrium polygon F G, beginning at any point inthe vertical F. Its ordinates %, 22, etc., are propor-tional to those of the special equil. pol. sought (whoseabutment line is OB) [§ 374a (2)]. We next use it to de-termine n [see § 374a]. We know that OB is the abut-ment-line of the required special polygon, and that . *.its pole must lie on a horizontal through n. It remainsto determine its H, or pole distance, by equation (1) justgiven, viz.: 2? yy = 2?yz. First by § 375 find the valueof the summation 2f(yy), which, from symmetry, we maywrite = 221%yy)=2[y1yl+y&i+y&i+y#J yx Hence, Fig. 426, we obtain 2\ (yy)=2[H0k] Next, also by § 375, see , using the same pole dis-tance H0 as in Fig. 426, wefind 2\(yz)=H<A\ ,. 2/i*i +2/2*2Hnh. + 2/3% +y& = 2\(yz) = H0(k+kr Again, since 2\ (yz)= y8z8 + 2/7*7 + 2/6*6 + 2/5*5 whichfrom symmetry (of rib) = 2/l*8 + 2/2*7+2/3*6 + 2/4*5, we obtain, Fig. 428, Il(2/*) = #o&r,(same H0); and .-. If now we find that ki+kt=2h 464 MECHANICS OE ENGINEERING. the condition 2f (yy) = 2\ (yz) is satisfied, and the poledistance of our trial polygon in Fig. 425, is also that ofthe special polygon sought; , the z identical invalue with the zs of Fig. 424. In general, of course, wedo not find that k-\-h = 2k. Hence the z s must allbe increased in the ratio 2k : (&/+&/) to become equal tofche 2s. That is, the pole distance H of the spec, equil-polygon must be jt_ k{-\-k jr,, (in which H = the pole distance of the2k trial polygon) since from §339 the ordi- nates of two equilibrium polygons (for
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Keywords: ., bookcentury1800, bookdecade1880, booksubjectenginee, bookyear1888
