Engineering with Wood Tension & Compression

1. Engineering with WoodTension & Compression Presenters: David W. Boehm, P.E. Gary Sweeny, P.E.

2. The information presented in this seminar is based on the knowledge and experience of the engineering staff at Engineering Ventures. Background and support information comes from various sources including but not limited to: • NDS • IBC • BOCA • Simplified Design for Wind and Earthquake Forces, Ambrose & Vergun • The project files at Engineering Ventures All designs of structures must be prepared under the direct supervision of a registered Professional Engineer.

3. Wood Compression & Tension Members(ALLOWABLE STRESS DESIGN)

4. Wood Compression and Tension Members • Definitions • Parameters of design • Design procedure – Axial compression • Bending and Axial compression • Bending and Axial Tension • Sample Problems • Questions

5. What are compression members? Structural members whose primary loads are axial compression Length is several times greater than its least dimension Columns and studs Some truss members

6. What Are Tension Members? • Structural members whose primary loads are axial tension • Some truss members • Rafter collar ties • Connections are critical

7. Types of wood columns • Simple solid column • square, rectangular, circular • Built-up column • mechanically laminated, nailed, bolted • Glued laminated column • Studs

8. Column Failure Modes • Crushing – short • Crushing and Buckling - intermediate • Buckling - long

9. Slenderness ratio: SlendernessRatio NDS=National Design Specification The larger the slenderness ratio, the greater the instability of the column

10. Effective Column Length, l • When end fixity conditions are known:

11. Simple Solid Column ll1& l2= distances between points of lateral support d1& d2= cross-sectional dimensions

12. Slenderness Ratio • Simple solid columns: < 50 • Except during construction < 75 • A large slenderness ratio indicates a greater instability and tendency to buckle under lower axial load

13. Design of Wood Columns • fc≤ F’c (Allowable Stress Design) • fc = P/A, Actual compressive stress = load divided by area • F’c = Allowable compressive stress

14. NDS table values Design of Wood ColumnsDetermination of Allowable Stress, F’c • Compressive stress parallel to grain adjustment factors: • Load duration • Wet service • Temperature • Size • Incising • Column stability

15. Adjustment Factors

16. Column Stability Factor, Cp • Where: KcE is defined by the Code (NDS) for the particular type of wood selected Modulus of Elasticity is adjusted by the following factors: CM, Ct, Ci, CT

17. Column Stability Factor, Cp • Compression members supported throughout the length: Cp = 1.0 • C is given in the Code (NDS) for the type of column selected c is given in the Code (NDS) for the type of column selected.F*c is Fc multiplied by all of the adjustment factors exceptCp

18. Stress Check

19. Example #1 – Column Design Example: Find the capacity of a 6x6 (Nominal) wood column. Given: Height of Column = 12’-0” End conditions are pinned top & bottom Wood species & Grade = Spruce-Pine-Fir No. 1 Visually graded by NLGA Interior, dry conditions, normal use for floor load support (DL &LL) Tabulated Properties: (from NDS Tables) E= 1,300,000 psi Fc= 700 psi Factors: Compression members (from NDS) CD=1.0 (duration) CM=1.0 (moisture) Ct=1.0 (Temp) Temperature CF=1.1 (Size)(Table) Ci=1.0 (Incising) Cp=TBD

22. Slenderness Ratio For columns, slenderness Ratio = le1/d1 or le2/d2 whichever is larger l = 12’-0” le= Kel Pin-pin: Ke = 1.0 le= 1.0 (12)= 12.0 Slenderness Ratio =

23. = 0.58 C = 0.8 For sawn lumber Fc* = Fc x CD CM Ct CF Ci = 700 (1)(1)(1)(1.1)(1) = 770 psi KcE= 0.3 For Visually Graded Lumber E’ = 1,300,000 psi l e/d = 26.2 Cp – Stability Factor

24. Column Capacity

25. Tension • Adjustment Factors

26. Bending and Axial Tension and Where: Fb* = tabulated bending design value multiplied by all applicable adjustment factors except CL Fb** = tabulated bending design value multiplied by all applicable adjustment factors except Cv

27. Bending and Axial Compression

28. fc Compression Zone- Axial Compression Compression Zone- Bending, X-X fb x-x fb y-y Compression Zone- Bending, Y-Y Combined Max. Stress Max. Compression Zone- Combined Bending and Axial Compression

29. Problem: Is a 2x6 @ 24” oc wall stud pattern adequate for a one story exterior wall? Given: Height of stud = 8’–6” Assume Pin-Pin End conditions and exterior face is braced by wall sheathing Wood species/grade = Spruce-Pine-Fir No.1/No.2 Visually graded by NLGA rules Interior, Dry conditions, normal use Subject to wind & roof loads (snow) Loads: 20 psf wind; 3.0k axial compression Example #2Exterior Wall Stud Design

30. Exterior Wall Stud Design Solution: Combined bending & axial compression Wood properties: (from NDS Tables) CD=1.15 Fc = 1150 psi Cm=1.0 E = 1,400,000 psi Ct=1.0 Cf=1.1 Ci=1.0 Cp=TBD Fc* = 1150 (1.15)(1)(1)(1.1)(1.0)= 1455 psi

31. For Pin-Pin, Ke = 1.0 (Sheathed/nailed) Therefore 18.5 governs Slenderness Ratio for Compression Calculation

32. c = 0.8 for sawn lumber (table) Cp = 0.63 F’c = Fc(CDCMCtCFCiCp) F’c = 1150(1.15)(1)(1)(1.1)(1)(.63) F’c = 1455(.63) = 917 psi Allowable Compressive Stress

33. Fb = 875 psi F’b1= FbCDCMCtCLCFCfuCi Cr CD = 1.6 (wind) CM = 1.0 = 875(1.6)(1.0)(1.0)(1.0)(1.3)(1.0)(1.0)(1.15) Ct = 1.0 CL = 1.0 = 2093 psi CF = 1.3 (Table 4A) Cfu = 1.0 Ci = 1.0 Cr = 1.15 (repetitive) F’b2 = ignore since no load in “b2” direction Allowable Bending Stress

34. Combined Stresses

35. Combined Stress Index (Check deflection and shear)

36. Questions???

37. (Assume blocked) Effective Length for Bending Calculation