Essentials in the theory of framed structures . at which the shear changes from positive tonegative is located. The ordinates in the shear diagramindicate that the moment diagram is composed of the straightline OA, the parabolic curve ABCD, and the straight lines £)£and EX. The maximum ordinate RC is at the section of zeroshear. The following ordinates are obtained from the area ofthe shear diagram: o at 0-|-4,ooo X S = -|-20,000 -f-20,000 = NAi(4,ooo -f 1,000)15 = + + 57,500 = JB-l-i,ooo X i X 5 = +2,500 -(-60,000 = RC— 2,000 X 5 X 10 = — + 50,000 = WD— 2,000 X 15 = —30,000 -[-20,
Essentials in the theory of framed structures . at which the shear changes from positive tonegative is located. The ordinates in the shear diagramindicate that the moment diagram is composed of the straightline OA, the parabolic curve ABCD, and the straight lines £)£and EX. The maximum ordinate RC is at the section of zeroshear. The following ordinates are obtained from the area ofthe shear diagram: o at 0-|-4,ooo X S = -|-20,000 -f-20,000 = NAi(4,ooo -f 1,000)15 = + + 57,500 = JB-l-i,ooo X i X 5 = +2,500 -(-60,000 = RC— 2,000 X 5 X 10 = — + 50,000 = WD— 2,000 X 15 = —30,000 -[-20,000 = ZE— 5,000 X 4 = —20,000 oat X Lay off to scale the ordinates NA, WD and ZE; and draw thelines OA, DE and EX. The line OA, having the same slopeat A as the parabola, is tangent to it; likewise the line DE istangent to the parabola at D. If the lines OA and ED are 96 THEORY OF FRAMED STRUCTURES Chap. II produced, they will intersect at F; a, point directly under thecenter of the uniform load. If the uniform load of 6,000 Fig. 63. were concentrated at its center, the bending moment diagramwould be OF EX. The parabola may now be drawn on thetangents FA and FD. Sec. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM 97 The parabolic segment ABDG has several interesting prop-erties. Let PS represent any ordinate in the bending momentdiagram between AN and DW, and let x represent the distanceof this ordinate from the left support; then the bendingmoment at x is M. = PS = 4,000^ - ^°°^^ --^ 2 = — looa;^ + s,oooa; — 2,500 (i) Let QS = y be the ordinate to the line AD, then from similartriangles QT:DH::TA:HA QT ?= y — AN = y — 2o,cooBE = DW - AN = 30,000 TA = X- s HA =30therefore y — 20,000:30,000:: a; — 5:30 or y = 1,000a; + 15,000 then PQ = M^ — y = m = — looa;^ + 4,oooie — 17,500Let AT = X — $ = Vor X = V -\- ^ then m = — loo(z) + 5)^ + 4,000(1 + 5) — 17,500or m = iooi(30 — v) (2) Equation (2) is also the expression for the bending moment
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Keywords: ., bookcentury1900, bookdecade1920, booksubjectstructu, bookyear1922
