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Structural Analysis of Forces on Timber Beams

Learn about equilibrium conditions, force analysis, and moment calculations on timber beams. Clear explanations and examples provided for better understanding.

fulton-howard
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Structural Analysis of Forces on Timber Beams

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  1. Aufgabe 4 45° A B F = 10 kN 0,5 m 0,5 m H. S C H O L Z - O1 FV 1a - 10. 01. 06 M+ x z

  2. Freischneiden: FB FBZ A 45° FAX FAZ FBX B F = 10 kN H. S C H O L Z - O1 FV 1a - 10. 01. 06 0,5 m 0,5 m M+ x z

  3. FB FBZ FAX 45° FAZ FBX F = 10 kN 0,5 m 0,5 m M+ x A B z Gleichgewichtsbedingungen I. Σ FX = 0 = FAX - FBX II. Σ FZ = 0 = FAZ – F + FBZ H. S C H O L Z - O1 FV 1a - 10. 01. 06 III. Σ M(A) = 0 = 10kN * 0,5m – FBZ * 1m [FBX kann kein Moment erzeugen]

  4. FB FBZ FAX 45° FAZ FBX F = 10 kN 0,5 m 0,5 m A I. Σ FX = 0 = FAX - FBX II. Σ FZ = 0 = FAZ – F +FBZ III.Σ M(A) = 0 = 10kN * 0,5m – FBZ * 1m B NR für FB sin α = a / c ; cos α = b / c ; sin 45° = cos 45° = √2 / 2 = 0,7071 a = b = √2 / 2 * c H. S C H O L Z - O1 FV 1a - 10. 01. 06 →FBX = FBZ = √2 / 2 *FB

  5. FB FBZ FAX 45° FAZ FBX F = 10 kN 0,5 m 0,5 m A I. Σ FX = 0 = FAX - FBX II. Σ FZ = 0 = FAZ – F +FBZ III.Σ M(A) = 0 = 10kN * 0,5m – FBZ * 1m B I. FAX = FBX FAX = √2 / 2 * FB II. FAZ = 10 kN - √2 / 2 * FB III. 10kN * 0,5m = √2 / 2 * FB * 1m H. S C H O L Z - O1 FV 1a - 10. 01. 06 5kNm = √2 / 2 * m * FB 5kN = √2 / 2 * FB FB = 5 * 2 / √2 kN = 10 / √2 kN = 7,071 kN

  6. FB FBZ FAX 45° FAZ FBX F = 10 kN 0,5 m 0,5 m A I. Σ FX = 0 = FAX - FBX II. Σ FZ = 0 = FAZ – F +FBZ III.Σ M(A) = 0 = 10kN * 0,5m – FBZ * 1m B III. → II. FAZ = 10 kN - √2 / 2 * FB FAZ = 10 kN - 5 kN FAZ = 5 kN H. S C H O L Z - O1 FV 1a - 10. 01. 06 III. → I. FAX = (√2 / 2) * (10 / √2) kN FAX = 5 kN

  7. FB = 7,071 kN FBZ FAX = 5 kN A 45° FAZ = 5 kN FBX B F = 10 kN H. S C H O L Z - O1 FV 1a - 10. 01. 06 0,5 m 0,5 m M+ x z

  8. Zusammenstellung: FB = 7,071 kN FBZ FAX = 5 kN 45° FAZ = 5 kN FBX A B F = 10 kN 0,5 m 0,5 m I. Σ FX = 0 = FAX - FBX II. Σ FZ = 0 = FAZ – F +FBZ III.Σ M(A) = 0 = 10kN * 0,5m – FBZ * 1m NR für FB sin α = a / c cos α = b / c sin 45° = cos 45° = √2 / 2 = 0,7071 a = b = √2 / 2 * c I. FAX = FBX FAX = √2 / 2 * FB II. FAZ = 10 kN - √2 / 2 * FB →FBX = FBZ = √2 / 2 * FB III. 10kN * 0,5m = √2 / 2 FB * 1m 5kNm = √2 / 2 * m * FB H. S C H O L Z - O1 FV 1a - 10. 01. 06 5kN = √2 / 2 * FB FB = 5 * 2 / √2 kN = 10 / √2 kN = 7,071 kN M+ III. → II. FAZ = 10 kN – 5 kN x FAZ = 5 kN III. → I. FAX = (√2 / 2) * (10 / √2) kN z FAX = 5 kN

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