Engineering with Wood Tension & Compression Presenters: David W. Boehm, P.E. Gary Sweeny, P.E.

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.

Wood Compression & Tension Members (ALLOWABLE STRESS DESIGN)

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

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

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

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

Column Failure Modes Crushing – short Crushing and Buckling - intermediate Buckling - long

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

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

Simple Solid Column l l1& l2 = distances between points of lateral support d1& d2 = cross-sectional dimensions

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

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

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

Adjustment Factors

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

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

Stress Check

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

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

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

Column Capacity

Tension Adjustment Factors

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

Bending and Axial Compression

Bending and Axial Compression fc 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

Example #2 Exterior Wall Stud Design 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

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

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

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

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

Combined Stresses

Combined Stress Index (Check deflection and shear)

Questions???

Effective Length for Bending Calculation (Assume blocked)