1. Engineering with Wood Tension & 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 Built-up 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 Slenderness ratio: 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
l l1& 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. Design of Wood Columns Determination of Allowable Stress, F’c
Compressive stress parallel to grain adjustment factors:
Load duration
Wet service
Temperature
Size
Incising
Column stability
NDS table values

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 except Cp

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

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

24. Column Capacity

25. Tension
Adjustment Factors

26. Bending and Axial Tension
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. Bending and Axial Compressionfc
Compression Zone- Axial Compression
fb x-x
Compression Zone- Bending, X-X
fb y-y
Compression Zone- Bending, Y-Y
Combined Max. Stress
Max. Compression Zone- Combined

29. Example #2 Exterior Wall Stud Design
Problem: Is a 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

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

32. Allowable Compressive Stress
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

33. Allowable Bending Stress
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

34. Combined Stresses

35. Combined Stress Index
(Check deflection and shear)

36. Questions???

37. Effective Length for Bending Calculation
(Assume blocked)