. The strength of materials; a text-book for engineers and architects. Fig. 213.—Continuous Beams—Graphical Treatment. Now in our problem Ave do not know the position of thepoints 0 and 3, and we see that these would be known if themid links were found, so that our j^roblem now reduces tothat of finding these mid links. On both sides of the centroid vertical g g draw lines atdistance p, and set down lengths l^, 2^ equal to 1, 2 and jointhem across, intersecting on the centroid vertical. Theselines are called the cross lines. Now draw any vertical u u, then clearly the intercepts madeby the ver
. The strength of materials; a text-book for engineers and architects. Fig. 213.—Continuous Beams—Graphical Treatment. Now in our problem Ave do not know the position of thepoints 0 and 3, and we see that these would be known if themid links were found, so that our j^roblem now reduces tothat of finding these mid links. On both sides of the centroid vertical g g draw lines atdistance p, and set down lengths l^, 2^ equal to 1, 2 and jointhem across, intersecting on the centroid vertical. Theselines are called the cross lines. Now draw any vertical u u, then clearly the intercepts madeby the vertical on the mid links and cross lines are equal. FIXED AND CONTINUOUS BEAMS * 453 From this it follows that if a point on one mid link is known,a point vertically below it on the other mid link can be , let the right-hand mid link of this span meet theleft-hand mid link of the next span in a point j, Fig. 214, ona vertical line q Fig. 214.—Continuous Beams—^Fixed Points. They are similar Then consider the triangles g j k, f 2 3. j k _ x-^ p X j k = 2, 3 X X;^ = Xj^ X area 6 as 454 THE STRENGTH OF MATERIALS Similarly considering the triangle on the other side of Q Qwe should have Where Zg is the length of the next span. Further, x-^^ -\~ X2 = k (li + h) h• • ^1 ~ 3 x<, - 3 . •. Q Q is at a distance = -^ from y y, and is thus called an inverted third line. Determination of Fixed Points.—Let abc (Fig. 214)represent two consecutive spans of a continuous beam, andlet the third lines be drawn as shown. Suppose that we know that the right-hand mid link of thespan A B passes through a fixed point r. Let this mid linkcut the inverted third line q q in J and the third line y Y in l,then L b must be a support tangent. Produce l b to meetthe first third line of the span b c in l, then j l is the left-hand mid link; and then join fb and produce it to meetJ l in f, then f will be a fixed point on the mid link
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