Essentials in the theory of framed structures . \/l\ y:h: :x(l — x): from which the following observation may be made in connec-tion with Fig. 59. Any two ordinates // and KH are to eachother as the products of the two parts into which each ordinatedivides the base line UV. This equation gives a clue to a verysimple method of locating any desired number of pointsthrough which a parabola is to be drawn. Suppose it isdesired to locate seven points on the curve. Divide the lineUV into eight equal parts by seven points. The ordinate at 94 THEORY OF FRAMED STRUCTURES Chap. II each point is proporti
Essentials in the theory of framed structures . \/l\ y:h: :x(l — x): from which the following observation may be made in connec-tion with Fig. 59. Any two ordinates // and KH are to eachother as the products of the two parts into which each ordinatedivides the base line UV. This equation gives a clue to a verysimple method of locating any desired number of pointsthrough which a parabola is to be drawn. Suppose it isdesired to locate seven points on the curve. Divide the lineUV into eight equal parts by seven points. The ordinate at 94 THEORY OF FRAMED STRUCTURES Chap. II each point is proportional to the product of the number ofparts on either side of it, thus: 1X7=7 2 X 6 = 12 3 X 5 = IS4X4= 16 5X3= IS6X2 = 12 7X1=7 Now if we wish the middle ordinate to equal 6,400 instead of 16,we multiply that ordinate and all the others by 400, whence 7 X 400 = 2,80012 X 400 = 4,800 15 X 400 = 6,000 16 X 400 = 6,400, etc. (6) Graphic Method.—If the tangents at the extremities of aparaboHc curve are known, the parabola may be constructed A %. Pig. 62. as shown in Fig. 62. Suppose we have two tangents AC andBC and the points of tangency A and B. Divide AC into anynumber of equal parts, and BC into the same number of equalparts. Number the points from A to C and from C to i-i, 2-2, etc. The parabolic curve lies tangent tothese lines as shown. Sec. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM 95 Draw tangents through U and V (Fig. 59) intersecting at may be observed from the shear diagram that the positiveslope of the tangent UD is 1,600 vertical to i ft. hori-zontal. The tangent DV has a corresponding negative slope,hence the intersection D is directly above K and the ordinateHD = 12,800 Construct the paraboUc bending momentdiagram on these tangents, and check the lengths of severalordinates in the moment diagram by the algebraic method. 60. Illustrative Problem. The shear and bending moment diagrams for a beam (Fig. 63)supporting a uniform load o
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