1. Wood A. Types Hardwoods or Angiosperms (some are soft) Softwoods or Gymnosperms (some are hard)

2. Terminology Hardwood Angiosperm Deciduous Broad-leafed Porous Softwoods Gymnosperm Coniferous Needle-leafed Non-porous

3. Typical Hardwoods Oak Ash Walnut Yellow-Poplar Cherry Birch Hickory Maple

4. Typical Softwoods Pine Spruce Fir Hemlock Redwood

5. Many species world-wide

6. Variability of wood

7. Wood-based composites

8. FIGURE 6-53 (a) Plywood. (b)Marking used by American Plywood Association. (continued) James A. Jacobs & Thomas F. Kilduff Engineering Materials Technology, Fourth Edition Copyright ©2001 by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

9. Basic Terminology

10. Boards Surfaces Face Edge End or transverse

11. Dimensions Thickness Width Length

12. Axes Radial Tangential Longitudinal

13. Grain or Figure Quarter-sawn Flat-sawn

14. FIGURE 6-55 (a)Top segment illustrates end grain. (b) Middle segment is quarter sawed. (c) Bottom segment is plain sawed. These match the 3-D sketch above. James A. Jacobs & Thomas F. Kilduff Engineering Materials Technology, Fourth Edition Copyright ©2001 by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

15. Warp Bow Crook Cup Twist

16. Grades Hardwood Grades Softwood Grades

17. Defects

18. Wood Anatomy Pith Annual Growth Rings Heartwood Sapwood Earlywood or springwood Latewood or summerwood Cambium Juvenile wood

19. Wood Anatomy Xylem Phloem Cell wall Tracheid Ray Fiber Vessel element or pore Resin canal Pits

20. FIGURE 6-49 Section cut from tree shows patterns from Figure 6-48 James A. Jacobs & Thomas F. Kilduff Engineering Materials Technology, Fourth Edition Copyright ©2001 by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

21. Reaction wood Compression wood Tension wood

22. Tree growth Mechanisms Growth requirements New techniques

23. Board Measurement: Nominal size / Actual size Board foot: 1bf = 144 cu. in. (Thickness(in) x Width(in) x Length(in))/144 = bf orT(in) x W(ft) x L(ft) = bf or (T(in) x W(in) x L(ft))/12 = bf

24. Calculate the board feet in: A. 1" x 12" x 5’ 1 x 1 x 5 = 5 bf B. 2" x 12" x 10‘ 2 x 1 x 10 = 20 bf C. 2" x 4 x 10‘ 2 x 1/3 x 10 = 6 2/3 bf

25. Calculate the board feet in: D. 1" x 6" x 18" 1 x 1/2 x 1 1/2 = 3/4 bf E. 1 1/2" x 7" x 8 " 1 1/2 x 7/12 x 2/3 = 3/2 x 14/36 = 42/72 = 7/12 bf

26. Material properties of wood

27. Chemical

28. Chemical composition of wood Cellulose (40-50%) (Long-chain polymer with low solubility) Hemicellulose (20-35%) (Noncellulosic polysaccharides, readily soluble in dilute alkali) Lignin (15-35%) Ash (0.1-0.5%) Extractives (Extraneous materials: tannins, polyphenols, gums, fatty acids, waxes, etc.)

29. FIGURE 6-47 (a)Cellulose formula. (b) Lignin formula. James A. Jacobs & Thomas F. Kilduff Engineering Materials Technology, Fourth Edition Copyright ©2001 by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

30. Microscopic structure of wood

31. Physical properties of wood Grain or orientation Earlywood/Latewood Heartwood/Sapwood Growth ring size Fibril orientation

32. FIGURE 6-50 (a) Grain direction (U.S. Forest Products Laboratory, Madison, Wisconsin) James A. Jacobs & Thomas F. Kilduff Engineering Materials Technology, Fourth Edition Copyright ©2001 by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

33. Figure C-16 White oak and red oak showing: top is end-grained; middle is edgegrain, quarter-sawed; and bottom is flat-grained, plainsawed. [Forest Products Laboratory] James A. Jacobs & Thomas F. Kilduff Engineering Materials Technology, Fourth Edition Copyright ©2001 by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

34. Moisture content Wood is Hygroscopic MC = (M-M o )/M o in percent (to be discussed later) Fiber Saturation Point Equilibrium Moisture Content Volumetric changes (to be discussed later)

35. Color in wood

36. Figure C-15 Hardwoods and softwoods 1) Birch, 2) Hickory, 3) Eastern Red Cedar, 4) Southern Yellow Pine, 5) White Oak Group, 6) Western Red Cedar, 7) Douglas Fir, 8) White Pine, 9) Black Walnut, 10) Red Gum, 11) Redwood, 12) Hard Maple, 13) Red Oak Group, 14) White Ash [Heritage Workshop] James A. Jacobs & Thomas F. Kilduff Engineering Materials Technology, Fourth Edition Copyright ©2001 by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

37. Porosity

38. Density and specific gravity OD sp.gr. of cell wall = 1.53 Typical sp. gr. Values

39. Mechanical properties of wood Strength varies directly with sp.gr. and inversely with MC Creep Hardness Other

40. Wood: Resource Allocation*

41. Harvesting Wood Many strategies: clear cutting selective cutting cable logging, etc. Environmental impacts habitat loss erosion, etc. Transportation & Storage

42. Primary Processing Debarking Sawing strategies* Drying Kiln-drying Air-drying Grading

43. Wood Moisture Wood is wet at harvest Wood is dried to: prevent rot (fungal attack) ensure dimensional stability (lower weight) Final moisture content depends on use

44. Water in the air Wood is hygroscopic Readily absorbs moisture from the air Equilibrium Moisture Content (EMC) relative humidity & temperature climactic zones

45. Moisture Content mc = mass of water / mass of dry wood (expressed as a percentage) mc = (M - M o ) / M o where M = mass of wet wood M o = oven-dry mass

46. mc Problem A Initial weight = 10 pounds. OD weight = 8 pounds. (Oven-dry weight is at 0% mc.) Determine the initial moisture content.

47. Problem A’s Solution mc = mass of water / mass of dry wood mc = (M - M o ) / M o mc = (10 Lbs - 8 Lbs) / 8 Lbs mc = 2 Lbs / 8 Lbs mc =.25 mc = 25 %

48. mc Problem B Initial weight = 15 ton. Weight at 10% mc = 11 tons. Determine the weight at 5% mc.

49. Problem B’s Solution First, find M o (see derivation) M o = M / (mc + 1) M o = 11 tons / (10% + 1) M o = 11 tons / 1.10 M o = 10 tons

50. Problem B’s Solution Next, use 10 tons as M o Try to visualize it by adding 10 tons of dry wood 5% of 10 tons in water 10 tons + 1/2 ton = 10.5 tons M = M o (mc + 1) M = 10 tons (105%) M = 10.5 tons or 21,000 Lbs

51. Where is the water? 1. Free water Inside the cell cavity or lumen 2. Bound water In the cell wall itself 3. Chemically combined water.

52. Stages of drying 1. Free water is lost from the lumen. No shrinkage From 100%+ mc down to FSP 2. Fiber Saturation Point (usu. 30% mc.) Lumen is dry Wall is saturated 3. Below FSP, bound water is lost Wood shrinks

53. As wood dries, it shrinks it weighs less it can warp cup, crook, bow, twist it can develop other defects checking splitting case-hardening, etc

54. Wood Shrinkage Longitudinal shrinkage = 0.1 to 0.2% Radial shrinkage = 2.2 to 7.8% Tangential shrinkage = 4.2 to 12.8% (latewood bands build up considerable tension.) Volumetric shrinkage = 6.8 to 16.8%

55. Review Terminology & Classifications Board measurement Tree growth and removal Primary processing Moisture content mc = mass of water / mass of dry wood mc = (M - M o ) / M o EMC and FSP Wood shrinkage