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 M Design Equation: Where F b is the characteristic bending strength For timber it is F b = f b (K D K H K Sb K T ) y

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. Bending capacity M r = φ F b S K Zb K L where φ = 0.9 and F b = f b (K D K H K Sb K T ) Lateral torsional buckling

10. Glulam beams in a Gerber system

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

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

14. Lateral torsional buckling of timber beams x x x y y y y ΔxΔx ΔyΔy θ LeLe Elastic buckling: M cr = π / L e √(G J E I y ) Torsional stiffness Lateral bending stiffness Note: The warping stiffness for rectangular shapes is small compared to the torsional and bending stiffness

15. Lateral torsional buckling of deep I-joists

16. Capacity of a timber beam subject to lateral torsional buckling LeLe MrMr combination of material failure and lateral torsional buckling M r = φ F b S K Zb K L elastic lateral torsional buckling  Mu Mu material failure M r = φ F b S K Zb

17. Lateral torsional buckling factor K L KLKL 1.0 10304020500 0 0.5 practical limit Slenderness ratio C B = ( L e d / b 2 ) 0.5 K L = 1 C K = ( 0.97 E K SE K T / F b ) 0.5 0.67 CBCB K L = 1 – 1/3 (C B / C K ) 4 K L = (0.65 E K SE K T ) / (C B 2 F b K X )

18. Deep glulam beam

19. Prevention of lateral torsional buckling d/b Lateral support at spacing: < 4 no support < 5 purlins or tie rods < 6.5 compression edge held by decking or joists < 7.5 top edge plus bridging < 9 both edges K L = 1.0 when lateral support is provided as shown d b < 610 mm < 8d

20. Bridging for floor joists

21. Shear stress in a beam N.A. y A b τ d τ max = V(0.5A)(d/4) (bd 3 /12)b =1.5 V/A

22. Shear in a timber beam V r = φ F v 2/3 A K Zv where φ = 0.9 and F v = f v (K D K H K Sv K T ) σ v(max) = 1.5 σ v(avg) = 1.5 V / A σ v(max) σ v(avg) AsAs

23. UNBC Prince George, BC

24. 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 Direct compression transfer of loads in the end zones reduces the total shear force to be carried. 45 o critical section This part of the load transferred in direct compression

25. Shear design of glulam beams A simple approach for beams where the volume < 2.0 m 3 : V r = φ F v 2/3 A K N where φ = 0.9 and F v = f v (K D K H K Sv K T ) K N = 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)

26. Notch factor for Glulam beams dndn dndn d e K N = ( 1 – d n /d ) 2 For e > d : K N = ( 1 – d n /d ) For e < d : K N = 1 – d n e/[d(d – d n )]

28. Notch effect in sawn lumber For notches on the tension side of supports (sawn lumber) In new code: Reaction calculation F r =  F t A K N  = 0.9 F t = f t (K D K H K St K T ) where f t = specified reaction force strength = 0.5 MPa for sawn lumber K St = 1.0 for dry and 0.7 for wet service conditions A = gross cross-section area K N = notch factor Area A NEW !!

29. Notch factor K N e d1d1 dndn d Based on Fracture Mechanics theory

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

33. Bearing resistance AbAb Q r =  F cp A b K Zcp K B  = 0.8 F cp = f cp (K Scp K T ) > 75mm no high bending stress < 150mm Bearing factor

34. Critical bearing areas in woodframe construction

35. Bearing resistance (double bearing) A b2 Q r = (2/3)  F cp A bavg K Zcp K B  = 0.8 F cp = f cp (K D K Scp K T ) A b1 45 deg A bavg = 0.5(A b1 +A b2 ) but ≤ 1.5 A b1

36. Bearing factor K B Bearing length or diameter (mm) Bearing factor K B < 12.51.75 251.38 381.25 501.19 751.13 1001.10 > 1501.0

38. 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