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Berechnung von Lagerkräften

+. x. z. B. Berechnung von Lagerkräften. Gegeben:. Gesucht:. Freikörperbild. L 5 = 1m. F 2 = 10 kN. M 1 = 10 kNm. 0,5 m = L 4. A. ●. 45°. 0,5 m = L 3. F BZ. F AX. 0,5 m = L 1. 0,5 m = L 2. F AZ. I Σ F x = 0 II Σ F z = 0 III Σ M (A) = 0. F AX.

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Berechnung von Lagerkräften

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  1. + x z B Berechnung von Lagerkräften Gegeben: Gesucht: Freikörperbild L5 = 1m F2 = 10 kN M1 = 10 kNm 0,5 m = L4 A ● 45° 0,5 m = L3 FBZ FAX 0,5 m = L1 0,5 m = L2 FAZ

  2. I ΣFx = 0 II ΣFz = 0 III ΣM(A) = 0 FAX Gleichgewichtsbedingung Berechnung der Lagerkräfte L5 = 1m F2 = 10 kN B M1 = 10 kNm I∑FX = 0 =FAX + sin 45° * F2 = FAX +sin 45° * 10 kN = FAX + 7,07 kN FAX= -7,07 kN II ∑FZ = 0 = - FAZ + cos 45° * F2 – FBZ = - FAZ + cos 45° * 10 kN – FBZ = -FAZ + 7,07 kN –FBZ FAZ= 7,07 kN - FBZ 0,5 m = L4 A ● 45° 0,5 m = L3 FBZ 0,5 m = L1 0,5 m = L2 FAZ

  3. FAX L5 = 1m F2 = 10 kN 10 kNm 2,7m 7,07 kN * 1,35m 2,7m 7,07 kN * 0,35m 2,7m + + M1 = 10 kNm 0,5 m = L4 A ● 45° 0,5 m = L3 FBZ B 0,5 m = L1 0,5 m = L2 FAZ IIIΣM(A) = 0 = M1 – FAX * 1,35m – FAZ * 0,35m + FBZ * LGes = 10 kNm -7,07 kN * 1,35m – 7,07 kN * 0,35m + FBZ * 2,7m FB= = - 3,70 kN + 3,54 kN + 0,92 kN FB = -0,75 kN III → II: → FAZ = 7,07 kN - FBZ = 7,07 kN – (- 0,75 kN) FA= 7,82 kN

  4. + x z B Belastungsschema FBZ = 0,75 kN L5 = 1m 7,07 kN F2 = 10 kN 7,07 kN M1 = 10 kNm 0,5 m = L4 A ● 45° 0,5 m = L3 FAZ = 7,07 kN 0,5 m = L1 0,5 m = L2 FAZ = 7,82 kN

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