Architecture 544 Wood Columns • Failure Modes • Euler Equation • End Conditions and Lateral Bracing • Analysis of Wood Columns • Design of Wood Columns Solemar, Bad Dürrheim Klaus Linkwitz, 1987 University of Michigan, TCAUP Arch 544 Slide 1 of 38 Failure Modes FLEXURE AXIAL Strength Stability Serviceability Deflection Bearing (crushing limit) University of Michigan, TCAUP Arch 544 Slide 2 of 38 Leonhard Euler (1707 – 1783) Euler Buckling (elastic buckling) = – A = Cross sectional area (in2) – E = Modulus of elasticity of the material (lb/in2) – K = Stiffness (curvature mode) factor portrait by Emanuel Handmann,1753 – L = Column length between pinned ends (in.) –le = K L – r = radius of gyration (in.) 0.822 E′min / 12 University of Michigan, TCAUP Arch 544 Slide 3 of 38 Failure Mode - Strength Short Columns – fail by crushing –fc = Actual compressive stress – A = Cross-sectional area of column (in2) – P = Load on the column –Fc = Allowable compressive stress per codes University of Michigan, TCAUP Arch 544 Slide 4 of 38 Failure Modes – Stability Long Columns – fail by buckling Traditional Euler NDS Equation 0.822 – E = Modulus of elasticity of the column material (psi) – E’min = reduced E modulus (psi) – K = Stiffness (curvature mode) factor – L = Column length between ends –le = Ke l (inches) (inches) – d (inches) – r = radius of gyration = I/A (inches) –0.822 = /12 / 12 University of Michigan, TCAUP Arch 544 Slide 5 of 38 y-y Slenderness Ratio le/d x-x Slenderness Ratios: The larger ratio will govern. Try to balance for efficiency. d = 3.5 Slenderness Limited to < 50 b = 1.5 ratios for an 8 ft long 2x4: University of Michigan, TCAUP Arch 544 Slide 6 of 38 End Support Conditions NDS 3.7.1.2 Ke is a constant based on the end conditions l is the actual length le is the effective length (curved part) le = Kel use these University of Michigan, TCAUP Arch 544 Slide 7 of 38 Allowable Flexure Stress Fc’ Actual Flexure Stress fb Fc from tables determined by species and grade fc = P/A Fc’ = Fc (adjustment factors) Fc’ fc University of Michigan, TCAUP Arch 544 Slide 8 of 38 Adjustment Factors Sawn Lumber - 4 University of Michigan, TCAUP Arch 544 Slide 9 of 38 Allowable Flexure Stress Fc’ Fc from tables determined by species and grade Fc’ = Fc (CD CM Ct CF Ci CP ) Adjustment factors for compression: CD Load Duration Factor Ct Temperature Factor University of Michigan, TCAUP Arch 544 Slide 10 of 38 Allowable Flexure Stress Fc’ (For Dimensioned Lumber) Fc from tables determined by species and grade Fc’ = Fc (CD CM Ct CF Ci CP ) Adjustment factors for compression: CM Moisture Factor CF Size Factor University of Michigan, TCAUP Arch 544 Slide 11 of 38 Allowable Flexure Stress Fc’ (For Timbers) Fc from tables determined by species and grade Fc’ = Fc (CD CM Ct CF Ci CP ) Adjustment factors for compression: CM Moisture Factor CF Size Factor University of Michigan, TCAUP Arch 544 Slide 12 of 38 Allowable Flexure Stress Fc’ Fc from tables determined by species and grade Fc’ = Fc (CD CM Ct CF Ci CP ) Adjustment factors for compression : Ci Incising Factor University of Michigan, TCAUP Arch 544 Slide 13 of 38 Allowable Flexure Stress Fc’ Fc from tables determined by species and grade Fc’ = Fc (CD CM Ct CF Ci CP ) University of Michigan, TCAUP Arch 544 Slide 14 of 38 CP estimation University of Michigan, TCAUP Arch 544 Slide 15 of 38 Analysis of Wood Columns Data: • Column – size, length • Support conditions • Material properties – Fc , E • Load Required: • Pass/Fail or margin of safety 1. Calculate slenderness ratio le/d largest ratio governs. Must be < 50 2. Find adjustment factors (all except CP) CD CM Ct CF Ci 3. Calculate CP 4. Determine F’c by multiplying the tabulated Fc by all the above factors 5. Calculate the actual stress: fc = P/A 6. Compare Allowable and Actual stress. F’c > fc passes University of Michigan, TCAUP Arch 544 Slide 16 of 38 Analysis Example: Data: section 4x8 (nominal) Douglas Fir-Larch No1 M.C. 15% P = 7000 LBS (Snow Load) Find: Pass/Fail From NDS Supplement Table 4A Fc = 1500 psi Emin = 620000 psi CD = 1.15 (snow) CM = 1.0 Ct = 1.0 CF = 1.05 (4x8) Ci = 1.0 CP = ? University of Michigan, TCAUP Arch 544 Slide 17 of 38 Analysis Example: Calculate CP University of Michigan, TCAUP Arch 544 Slide 18 of 38 Analysis Example: Calculate CP University of Michigan, TCAUP Arch 544 Slide 19 of 38 Analysis Example: Calculate CP Compare Allowable and Actual stress F’c > fc passes University of Michigan, TCAUP Arch 544 Slide 20 of 38 Capacity Analysis of Columns Data: • Column – size, length • Support conditions • Material properties – Fc , E Required: • Maximum Load, Pmax 1. Calculate slenderness ratio le/d largest ratio governs. Must be < 50 2. Find adjustment factors (all except CP) CD CM Ct CF Ci 3. Calculate CP 4. Determine F’c by multiplying the tabulated Fc by all the above factors 5. Set actual stress = allowable, fc = F’c 6. Find the maximum allowable load Pmax = F’c A University of Michigan, TCAUP Arch 544 Slide 21 of 38 Capacity Example Data: • 4x10 • Hem – Fir, No 2 M.C. = 20% • Wind Load •L1 = 8’ L2 = 4’ Ke=1.0 Required: • Maximum Load, Pmax From NDS Supplement Table 4A Fc = 1300 psi Emin = 470000 psi CD = 1.6 CMc = 0.8 CME = 0.9 Ct = 1.0 CF = 1.0 Ci = 1.0 CP = ? University of Michigan, TCAUP Arch 544 Slide 22 of 38 Allowable Flexure Stress Fc’ 4 x 10 M.C 20% Fc = 1300psi Fc from tables determined by species and grade Fc’ = Fc (CD CM Ct CF Ci CP ) Adjustment factors for compression: CM Moisture Factor CF Size Factor University of Michigan, TCAUP Arch 544 Slide 23 of 24 Capacity Example Find CP Find the maximum load, Pmax University of Michigan, TCAUP Arch 544 Slide 24 of 24 Stud Wall Design Given: • Lumber species, grade and size • Conditions of use • Load Required: • Stud spacing 1. Calculate slenderness ratio le/d largest ratio governs. Must be < 50 2. Find adjustment factors (all except CP) CD CM Ct CF Ci 3. Calculate CP 4. Determine F’c by multiplying the tabulated Fc by all the above factors 5. Set actual stress = allowable: fc = F’c 6. Find the capacity of one stud: Pmax = F’c A 7. Find allowable spacing (12”, 16” or 24” o.c.) 8. Check bearing. University of Michigan, TCAUP Arch 544 Slide 25 of 38 Stud Wall Example Data: •2x6 • S-P-F, Stud M.C. = 12% • D+L Load = 2500 PLF • Braced as shown Ke=1.0 Required: • o.c. spacing From NDS Supplement Table 4A Fc = 725 psi Emin = 440000 psi CD = 1.0 (LL) CMc = 1.0 CME = 1.0 Ct = 1.0 CF = 1.0 (stud) Ci = 1.0 CP = ? University of Michigan, TCAUP Arch 544 Slide 26 of 38 Stud Wall Example Data: •2x6 • S-P-F, Stud M.C. = 12% • D+L Load = 2500 PLF • Braced as shown Ke=1.0 CF = 1.0 (stud) University of Michigan, TCAUP Arch 544 Slide 27 of 38 Stud Wall Example From NDS Supplement Table 4A Fc = 725 psi Emin = 440000 psi University of Michigan, TCAUP Arch 544 Slide 28 of 38 Stud Wall Example University of Michigan, TCAUP Arch 544 Slide 29 of 38 Stud Wall Example Find max allowable stress, F’c Calculate max load per stud Determine max stud spacing University of Michigan, TCAUP Arch 544 Slide 30 of 38 Stud Wall Example Check bearing on sill plate University of Michigan, TCAUP Arch 544 Slide 31 of 38 Stud Wall Example Check bearing on sill plate • determine Cb •calculate F’ c┴ • calculate f c┴ • check stress University of Michigan, TCAUP Arch 544 Slide 32 of 38 Timber Column Design Given: • Lumber species, grade • Conditions of use • Load Required: • column size 1. Find adjustment factors (all except CP) CD CM Ct CF Ci 2. Guess CP 3. Estimate Area and d (based on bracing) 4. Calculate slenderness ratio le/d largest ratio governs. Must be < 50 5. Calculate CP 6. Determine F’c by multiplying the tabulated Fc by all the above factors 7. Revise Area: A = P/F’c 8. Revise CP University of Michigan, TCAUP Arch 544 Slide 33 of 38 Timber Column Design Given: • White Oak, No.1 • dry use, normal temp., not incised • Load: D+L=55 psf Required: • column size 1. Find adjustment factors (all except CP) CD CM Ct CF Ci 2. Guess CP try 0.5 University of Michigan, TCAUP Arch 544 Slide 34 of 38 Timber Column Design Given: • White Oak, No.1 • dry use, normal temp., not incised • Load: D+L=55 psf Required: • column size 1. Find adjustment factors (all except CP) CD CM Ct CF Ci 2. Guess CP try 0.5 3. Estimate Area and d (based on bracing) 4. Calculate slenderness ratio le/d largest ratio governs. Must be < 50 University of Michigan, TCAUP Arch 544 Slide 35 of 38 Timber Column Design Given: • White Oak, No.1 • dry use, normal temp., not incised • Load: D+L=55 psf Required: • column size 5. Calculate CP University of Michigan, TCAUP Arch 544 Slide 36 of 38 Timber Column Design Given: • White Oak, No.1 • dry use, normal temp., not incised • Load: D+L=55 psf Required: • column size 6. Determine F’c by multiplying the tabulated Fc by all the above factors 7. Revise Area: A = P/F’c 8. Revise CP University of Michigan, TCAUP Arch 544 Slide 37 of 38 Timber Column Design Design Aids example of a column chart from AWC Manual for Engineered Wood Construction – 2005 University of Michigan, TCAUP Arch 544 Slide 38 of 38.