Essentials in the theory of framed structures . R2 and R3 will now bedeveloped in connection with Fig. 159. Let the tangent to theelastic curve be drawn through C and let h and h represent thetangential deviations at A and B respectively; then Elh = Pk{i - k)h /i/iVfyfe/i +^(i - k)h] + Mik.)[\ = m(^.)(^ ti. — = —till Since k is less than unity, ib — ^^ is positive; hence M is anegative bending moment, and Rz is a negative reaction actingdownward. When the two spans are of equal length I, the reactions are ^1 = -{k -5^ + 4) 4 (9) i?2 = -{-2k^ + 6k)4 (10) Rz = -{k - k)4 (11) 163. I
Essentials in the theory of framed structures . R2 and R3 will now bedeveloped in connection with Fig. 159. Let the tangent to theelastic curve be drawn through C and let h and h represent thetangential deviations at A and B respectively; then Elh = Pk{i - k)h /i/iVfyfe/i +^(i - k)h] + Mik.)[\ = m(^.)(^ ti. — = —till Since k is less than unity, ib — ^^ is positive; hence M is anegative bending moment, and Rz is a negative reaction actingdownward. When the two spans are of equal length I, the reactions are ^1 = -{k -5^ + 4) 4 (9) i?2 = -{-2k^ + 6k)4 (10) Rz = -{k - k)4 (11) 163. In Fig. 160a the parabola PQS is the M-diagram, whenylC is considered as a simple beam; and the parabola STV is theM-diagram when CB is considered as a simple beam. Thetriangle PUV is added to represent the bending moment onaccount of the continuity. If the tangent to the elastic curve 2s8 THEORY OF FRAMED STRUCTURES Chap. VI be drawn through C, and h and fe represent the tangentialdeviations at A and B respectively; then BOOO Elk = (18,000 X 12 X - X 6) + M(6 X 8) = 864,000 + 48M Elh = (81,000 X 18 X - X 9) + M(9 X 12) = 8,748,000 + loSikf3/1 = — 2hM s5,8oo Sec. II RESTRAINED AND CONTINUOUS BEAMS 259 The M-diagram may now be drawn to scale as shown in statics —55,800 = 12R1 — (12,000 X 6) = i8i?3— (36,000 X 9) Ri = 1,350Ri = 31,750 Ri = 14,900164. Two Unequal Spans, Supporting Unequal UniformLoads.—The general expressions for Ri, R2 and R3 for a con-
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