ASD/LRFD Manual - American Wood Council

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ASD/LRFD MANUAL FOR ENGINEERED Wood Construction 155 Table M16.2-9 Design Load Ratios for Exposed Timber Decks Double and Single Tongue & Groove Decking (Structural Calculations at Standard Reference Conditions: C D = 1.0, C M = 1.0, C t = 1.0, C i = 1.0) Rating 1-HOUR 1.5-HOUR 2-HOUR Deck Thickness Design Load Ratio, R s 2.5 0.22 - - 3 0.46 0.08 - 3.5 0.67 0.23 0.03 4 0.86 0.40 0.12 4.5 1.00 0.56 0.25 5 1.00 0.71 0.38 5.5 1.00 0.85 0.51 Table M16.2-10 Design Load Ratios for Exposed Timber Decks Butt-Joint Timber Decking (Structural Calculations at Standard Reference Conditions: C D = 1.0, C M = 1.0, C t = 1.0, C i = 1.0) Rating 1-HOUR 1.5-HOUR 2-HOUR Decking Width 1.5 2.5 3.5 5.5 2.5 3.5 5.5 3.5 5.5 Decking Depth Design Load Ratio, R s 2.5 0.05 0.12 0.15 0.18 - - - - - 3 0.09 0.24 0.30 0.36 0.03 0.04 0.05 - - 3.5 0.14 0.35 0.44 0.53 0.08 0.12 0.16 - - 4 0.18 0.45 0.57 0.68 0.14 0.21 0.28 0.02 0.08 4.5 0.21 0.54 0.68 0.80 0.19 0.30 0.39 0.04 0.16 5 0.24 0.61 0.77 0.92 0.24 0.38 0.50 0.06 0.24 5.5 0.27 0.68 0.85 1.00 0.29 0.45 0.59 0.09 0.32 M16: FIRE DESIGN 16 American Forest & paper association
156 M16: FIRE DESIGN Example 16.2-1 Exposed Beam Example - Allowable Stress Design Douglas-fir glulam beams span L = 18' and are spaced at s = 6'. The design loads are q live = 100 psf and q dead = 25 psf. Timber decking nailed to the compression edge of the beams provides lateral bracing. Calculate the required section dimensions for a 1-hour fire resistance time. For the structural design of the wood beam, calculate the maximum induced moment. Calculate beam load: w total = s (q dead + q live ) = (6)(25 + 100) = 750 plf Calculate maximum induced moment: M max = w total L²/8 = (750)(18)²/8 = 30,375 ft-lbs Select a 6-3/4" x 13-1/2" 24F visually-graded Douglas-fir glulam beam with a tabulated bending stress, F b , equal to 2,400 psi. Calculate beam section modulus: S s = bd 2 /6 = (6.75)(13.5) 2 /6 = 205.0 in. 3 Calculate the adjusted allowable bending stress (assuming C D = 1.0; C M = 1.0; C t = 1.0; C L = 1.0; C V = 0.98) F′ b = F b (C D )(C M )(C t )(lesser of C L or C V ) = 2,400 (1.0)(1.0)(1.0)(0.98) = 2,343 psi (NDS 5.3.1) Calculate design resisting moment: M′ =F′ b S s = (2,343)(205.0)/12 = 40,032 ft-lbs Structural Check: M′ ≥ M max 40,032 ft-lbs ≥ 30,375 ft-lbs OK For the fire design of the wood beam, the loading is unchanged. Therefore, the maximum induced moment is unchanged. The fire resistance must be calculated. Calculate beam section modulus exposed on three sides: S f = (b – 2a)(d – a) 2 /6 = (6.75 – 3.6)(13.5 – 1.8) 2 /6 = 71.9 in. 3 Calculate the adjusted allowable bending stress (assuming C D = N/A; C M = N/A; C t = N/A; C L = 1.0; C V = 0.98) F′ b = F b (lesser of C L or C V ) = 2,400 (0.98) = 2,343 psi (NDS 5.3.1) Calculate strength resisting moment: M′ = (2.85) F′ b S f = (2.85)(2,343)(71.9)/12 = 40,010 ft-lbs (NDS 16.2.2) Fire Check: M′ ≥ M max 40,010 ft-lbs ≥ 30,375 ft-lbs OK Design Aid Calculate structural design load ratio: r s = M max /M′ = 30,375/40,032 = 0.76 Select the maximum design load ratio limit from Table M16.2-1B or calculate using the following equation: R s 2. 85S f ( 2. 85)( 71. 9) = = = 1. 00 S C C C ( 205)( 1. 0)( 1. 0)( 1. 0) s D M t Fire Check: R s ≥ r s 1.00 ≥ 0.76 OK Example 16.2-2 Exposed Column Example - Allowable Stress Design A southern pine glulam column with an effective column length, < e = 168". The design loads are P snow = 16,000 lbs and P dead = 6,000 lbs. Calculate the required section dimensions for a 1-hour fire resistance time. For the structural design of the wood column, calculate the maximum induced compression stress, f c . Calculate column load: P total = P dead + P snow = 8,000 + 16,000 = 22,000 lbs Select a 8-1/2" x 9-5/8" Combination #48 southern pine glulam column with a tabulated compression parallel-to-grain stress, F c , equal to 2,200 psi and a tabulated modulus of elasticity, E min , equal to 880,000 psi. Calculate column area: A s = bd = (9.625)(8.5) = 81.81 in. 2 I s = bd 3 /12 = (9.625)(8.5) 3 /12 = 492.6 in. 4 Calculate the adjusted allowable compression stress (assuming C D = 1.15; C M = 1.0; C t = 1.0): E min ′ = E min (C M )(C t ) = 880,000 (1.0)(1.0) = 880,000 psi (NDS 5.3.1) F cE = 0.822 E min ′ / (< e /d) 2 = 0.822 (880,000) / (168/8.5) 2 = 1,852 psi (NDS 3.7.1.5) F* c = F c (C D )(C M )(C t ) = 2,200 (1.15)(1.0)(1.0) = 2,530 psi (NDS 3.7.1.5) American Wood Council
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2005 EDITION ASD/LRFD MANUAL MANUAL
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Updates and Errata While every prec
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ii ASD/LRFD MANUAL FOR ENGINEERED W
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iv ASD/LRFD MANUAL FOR ENGINEERED W
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viii ASD/LRFD MANUAL FOR ENGINEERED
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ASD/LRFD MANUAL FOR ENGINEERED Wood
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M1: general requirements for struct
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M2: DESIGN VALUES FOR STRUCTURAL ME
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M3: Design Provisions and Equations
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M3: Design Provisions and Equations
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10 M3: Design Provisions and Equati
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14 M4: SAWN LUMBER M4.4 Special Des
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16 M4: SAWN LUMBER Table M4.4-3 Coe
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18 M4: SAWN LUMBER Table M4.5-1a AS
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20 M4: SAWN LUMBER Table M4.5-2b AS
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22 M4: SAWN LUMBER Table M4.5-3a AS
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24 M4: SAWN LUMBER Table M4.5-4c AS
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26 M4: SAWN LUMBER Column Capacity
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28 M5: structural glued laminated t
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34 M6: ROUND TIMBER POLES AND PILES
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36 M6: ROUND TIMBER POLES AND PILES
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38 M7: PREFABRICATED WOOD I-JOISTS
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54 M8: STRUCTURAL COMPOSITE LUMBER
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60 M9: WOOD STRUCTURAL PANELS M9.1
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62 M9: WOOD STRUCTURAL PANELS Table
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64 M9: WOOD STRUCTURAL PANELS Compr
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66 M9: WOOD STRUCTURAL PANELS M9.3
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68 M9: WOOD STRUCTURAL PANELS Table
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70 M10: MECHANICAL CONNECTIONS M10.
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74 M10: MECHANICAL CONNECTIONS 9. W
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76 M10: MECHANICAL CONNECTIONS 20A.
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78 M10: MECHANICAL CONNECTIONS 31A.
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84 M10: MECHANICAL CONNECTIONS •
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86 M11: DOWEL-TYPE FASTENERS M11.1
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88 M11: DOWEL-TYPE FASTENERS Americ
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90 M12: SPLIT RING AND SHEAR PLATE
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92 M13: TIMBER RIVETS M13.1 General
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94 M14: SHEAR WALLS AND DIAPHRAGMS
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100 M15: SPECIAL LOADING CONDITIONS
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102 M16: FIRE DESIGN M16.1 General
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