? C? H?l 1?l 2 dC = │ Dl1 │ dH = dC + Dl3 = 0.4 mm = 0.6 mm (1) (2) ? C? H?l 1?l 2 Dl1 = – 0.2 mm Dl2 = – 0.4 mm Dl3 = 0.2 mm dH = dC + Dl3 = 0.267 mm = 0.467 mm ? C = │?l 1│+ │?l 2│3 13 2 位移分析! P64 35-3 A BC l = 4 m h = 3 m 5m q = 10 kN /m SMA = 0 F F BC BCF F A Ay yF F A Ax x F BC = kN 3 100 FN = (1) 斜杆用钢丝索 A = n4 ?d 2?= A F N≤ [s ] ∴取 n = 67 或(2) 斜杆用两根等边角钢 A = 2A1 ?= A F N≤ [s ] ∴取∟ 20 ×3 ( A1 = 1.13 mm2 ) 结论! n ≥ 66.3 66 A1 ≥ 1.044 mm2 P64 35-4 F F AC B ①② 45o60o F F N N2 2F F N N1 1 SFx = 0 SFy = 0 FN1 = 0.897 F FN2 = 0.732 F ? 1 = A 1F N1≤ [s ]1 F ≤ 107 kN ? 2 = A 2F N2≤ [s ]2 F ≤ 123 kN 则 [ F ]= 107 kN P65 36-1 AB F F F F FN = 5 kN ?= A F N = 25 MPa s = Ee1 E = ? 1?= 208 GPa ??= ? 1│? 2│= 15 4 = 0.267
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