. The strength of materials; a text-book for engineers and architects. x y square ft. tons. Since each portion of the vector line is — of the n ordinates, the portion 2, 3 of the vector line represents the area of its corresponding section of the diagram to a scale V = n X e X X X y square ft. tons. Now calculate the length of E I on this scale. This will be too large for EI practical use, so take a pole p at distance — , where r is some convenient whole number. With this pole p, draw the linkpolygon a c b\ then this is the elastic line of the beam for 262 THE STRENGTH OF MATERIALS the gi
. The strength of materials; a text-book for engineers and architects. x y square ft. tons. Since each portion of the vector line is — of the n ordinates, the portion 2, 3 of the vector line represents the area of its corresponding section of the diagram to a scale V = n X e X X X y square ft. tons. Now calculate the length of E I on this scale. This will be too large for EI practical use, so take a pole p at distance — , where r is some convenient whole number. With this pole p, draw the linkpolygon a c b\ then this is the elastic line of the beam for 262 THE STRENGTH OF MATERIALS the given loading, or, more strictly speaking, a c b, when reduced to a horizontal base, would give the elastic line. The scale to which the deflections are to be read is then obtained as follows— If the polar distance were taken equal to E I, the deflections would be to the space scale 1^ = x feet, but as the polar dis- . EI X tance is , the deflections will be to a scale V =— feet. The r r following numerical example should clear up the difficulty as to scale—. Fig. 127.—Graphical Construction for Deflections. Numerical Example.—^ 16 x 6 x 62 lb. rolled steeljoist of 24 ft. span carries a uniformly distributed load {in-cluding its own weight) of 8 tons, and also an isolated load of5 tons, at a point 6 ft. from the left-hand support. Find themaximum deflection (Fig. 128). In this case E = 12,500 tons per sq. =- 725-7 inch ,500 X 725-7 EI 144 = 62,980 sq. ft. tons. First draw the diagrams for each of the loads, takingas linear scale, say V = 4 ft., and for the scale, sayV = 20 ft. tons. Now divide the diagram into a con-venient number of equal parts, say 12, and draw the mid DEFLECTIONS OF BEAMS 263 ordinate of each part, treating these as force lines, then setthese ordinates down a vector line, 0, 1, 2, etc. .. 12 to areduced scale, say one-fourth for convenience. Then 1 in. down the vector line represents 4 X 4 X 20 160 sq. ft. tons, beca
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