ASD/LRFD Manual - American Wood Council

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ASD/LRFD MANUAL FOR ENGINEERED Wood Construction 95 Figure M14.2-2 Shear Wall Special Case Drag Strut V L O Elevation Unit shear above opening = V = v a L Unit shear below opening = V L Maximum force in drag strut = L v L 1 = o a Force in drag strut from L 4 structure = W L 2 2 ( L5 ) 2L4 Figure M14.2-3 Diaphragm Drag Strut (Drag strut parallel to loads) L 1 W1 L1 Unit shear along L3 = 2L3 Force in drag strut from L 3 structure = W L 1 1 ( L L 5) 2 3 v b W2L2 Unit shear along L4 = 2L4 Maximum force in drag strut = L 2 ⎛ ⎜ W L ⎝ L3 5 1 1 W L ⎞ 2 2 + ⎟ L4 ⎠ M14: SHEAR WALLS AND DIAPHRAGMS 14 American Forest & paper association
96 M14: SHEAR WALLS AND DIAPHRAGMS Chords Diaphragms are assumed to act like long deep beams. This model assumes that shear forces are accommodated by the structural-use panel web of the “beam” and that moment forces are carried by the tension or compression forces in the flanges, or chords of the “beam.” These chord forces are often assumed to be carried by the double top-plate of the supporting perimeter walls. Given the magnitude of forces involved in most light-frame wood construction projects, the double top-plate has sufficient capacity to resist tensile and compressive forces assuming adequate detailing at splice locations. However, offset wall lines and other factors sometimes make a continuous diaphragm chord impossible. Because shear walls act as blocked, cantilevered diaphragms, they too develop chord forces and require chords. The chords in a shear wall are the double studs that are required at the end of each shear wall. Just as chords need to be continuous in a diaphragm, chords in a shear wall also need to maintain their continuity. This is accomplished by tension ties (holddowns) that are required at each end of each shear wall and between chords of stacked shear walls to provide overturning restraint. Figure M14.2-4 and the accompanying generalized equations provide a method to calculate chord forces. Figure M14.2-4 Diaphragm Chord Forces L Unit Load (wind or seismic) w Plan F(x) F(x) x L 2 Diaphragm reaction = Lw 2 Diaphragm unit shear = Lw L 2 2 Diaphragm moment = wL2 8 Maximum chord force = wL 8L Chord force at point x, F(x) = wLx wx − 2L 2L 2 2 2 2 2 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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vi 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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12 M4: SAWN LUMBER M4.1 General Pro
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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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Page 103 and 104: 90 M12: SPLIT RING AND SHEAR PLATE
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146 M16: FIRE DESIGN Compression Me
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148 M16: FIRE DESIGN Table M16.2-2
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156 M16: FIRE DESIGN Example 16.2-1
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158 M16: FIRE DESIGN For the fire d
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160 M16: FIRE DESIGN Figure M16.3-5
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American Forest & Paper Association
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ASD/LRFD Manual - American Wood Council

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