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Design of Bending Members in Timber

Design of Bending Members in Timber. An Example of Timber Beams. What can go wrong ?. TIMBER BEAMS: Bending failure Lateral torsional buckling Shear failure Notch failure Bearing failure Excessive deflections. Bending Strength . Linear elastic stresses. y. M. Design Equation:.

Samuel
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Design of Bending Members in Timber

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  1. Design of Bending Members in Timber

  2. An Example of Timber Beams

  3. What can go wrong ? TIMBER BEAMS: • Bending failure • Lateral torsional buckling • Shear failure • Notch failure • Bearing failure • Excessive deflections

  4. Bending Strength Linear elastic stresses y M Design Equation: Where Fb is the characteristic bending strength For timber it is Fb = fb (KDKHKSbKT)

  5. Bending failure in compression • Only likely for very high grade material • Benign failure mode

  6. Logging bridge near Pemberton, BC Glulam I-beam

  7. Bending failure in tension • Most likely failure mode • Brittle • Combination of tension and shear, although tension fracture is the initiating mode

  8. Lateral torsional buckling Bending capacity Mr = φ Fb S KZb KL where φ = 0.9 and Fb = fb (KD KH KSb KT )

  9. Glulam beams in a Gerber system

  10. Glued-laminated beams better laminations 20f-E and 24f-E grades

  11. better laminations Glued-laminated beams 20f-EX and 24f-EX grades

  12. Elastic buckling: Mcr = π / Le √(G J E Iy ) Lateral bending stiffness Torsional stiffness y y Δx x Δy θ x x y y Lateral torsional buckling of timber beams Le Note: The warping stiffness for rectangular shapes is small compared to the torsional and bending stiffness

  13. Lateral torsional buckling of deep I-joists

  14. Mu Mr = φ Fb S KZb Mr = φ Fb S KZb KL material failure combination of material failure and lateral torsional buckling elastic lateral torsional buckling Capacity of a timber beam subject to lateral torsional buckling Mr Le

  15. Lateral torsional buckling factor KL KL 1.0 KL = 1 KL = 1 – 1/3 (CB / CK)4 0.67 practical limit 0.5 KL = (0.65 E KSE KT) / (CB2 Fb KX) CK = ( 0.97 E KSE KT / Fb )0.5 0 CB 0 10 20 30 40 50 Slenderness ratio CB = ( Le d / b2 )0.5

  16. Deep glulam beam

  17. d b Prevention of lateral torsional buckling < 610 mm KL = 1.0 when lateral support is provided as shown < 610 mm < 8d

  18. Bridging for floor joists

  19. A τ y d τmax = V(0.5A)(d/4) (bd3/12)b =1.5 V/A N.A. b Shear stress in a beam

  20. Shear in a timber beam As σv(max) Vr = φ Fv 2/3 A KZv where φ = 0.9 and Fv = fv (KD KH KSv KT ) σv(avg) σv(max) = 1.5 σv(avg) = 1.5 V / A

  21. UNBC Prince George, BC

  22. This part of the load transferred in direct compression • Direct compression transfer of loads in the end zones reduces the total shear force to be carried. 45o critical section Shear failures • One of the very weak properties of wood • Shrinkage cracks often occur at the ends of beams in the zone of maximum shear stress

  23. Shear design of glulam beams A simple approach for beams where the volume < 2.0 m3: Vr = φ Fv 2/3 A KN where φ = 0.9 and Fv = fv (KD KH KSv KT ) KN = notch factor (see next section) For larger beams this is usually quite conservative and a more sophisticated approach is used (see clause 6.5.7.3)

  24. Notch factor for Glulam beams dn d dn e KN = ( 1 – dn/d )2 For e > d : KN = ( 1 – dn/d ) For e < d : KN = 1 – dne/[d(d – dn)]

  25. Notch effect in sawn lumber • For notches on the tension side of supports (sawn lumber) • In new code: Reaction calculation NEW !! • Fr =  Ft A KN •  = 0.9 • Ft = ft (KD KH KSt KT) • where ft = specified reaction force strength = 0.5 MPa for sawn lumber • KSt = 1.0 for dry and 0.7 for wet service conditions • A = gross cross-section area • KN = notch factor Area A

  26. d1 e d dn Notch factorKN Based on Fracture Mechanics theory

  27. Not only compression perpendicular to grain but also tension of the fibres along edges compression perpendicular to grain tension of fibres along the edges Bearing failure in a timber beam • The “soft” property of wood • Often governs

  28. < 150mm no high bending stress > 75mm Bearing factor Bearing resistance Ab Qr =  Fcp Ab KZcp KB = 0.8 Fcp = fcp (KScp KT)

  29. Critical bearing areas in woodframe construction

  30. Bearing resistance (double bearing) Ab2 Abavg= 0.5(Ab1 +Ab2) but ≤ 1.5 Ab1 Ab1 45 deg Qr = (2/3)  Fcp Abavg KZcp KB = 0.8 Fcp = fcp (KD KScp KT)

  31. Bearing factor KB

  32. Deflections • A serviceability criterion • Avoid damage to cladding etc. (Δ ≤ L/180) • Avoid vibrations (Δ ≤ L/360) • Aesthetics (Δ ≤ L/240) • Use unfactored loads • Typically not part of the code • Δ

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