wood bolt connection

PROJECT : PAGE : CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY : Bolt Connection Design Based on NDS 2005 INPUT DATA & DESIGN SUMMARY AXIAL TENSILE FORCE (ASD T = 5.3 k NUMBER OF BOLTS n = 3 BOLT DIAMETER = 3/4 BOLT SPACING S = 3 in END DISTANCE OF WOOD E n = 4 in END DISTANCE OF STEEL E n,s = 1.5 in LUMBER TYPE ( 0=Douglas Fir-Larch, 1=Douglas Fir-Larch(N), 0 Douglas Fir-Larch, G=0.5 2=Hem-Fir(N), 3=Hem-Fir, 4=Spruce-Pine-Fir) LUMBER SIZE ( 2 ) - 2 thk. x 6 width STRAP SIZE 5 width x 1/4 thk. LOAD DURATION FACTOR ( Tab 2.3.2, NDS 2005, Page 8 ) C D = 1.6 WET SERVICE FACTOR ( Tab 10.3.3, NDS 2005, Page 58 ) C M = 1.0 TEMPERATURE FACTOR ( Tab 10.3.4, NDS 2005, Page 58 ) C t = 1.0 THE CONNECTION DESIGN IS ADEQUATE. ANALYSIS CHECK STEEL STRAP CAPACITIES (AISC 360-05, ASD) A g = 1.25 in 2 , yielding critierion F y = 36.00 ksi T allow = 0.6 F y A g = 27.00 k > T [Satisfactory] (0.6 from 1/ t , Typ.) A n = 1.03 in 2 , fracture criterion F u = 58.00 ksi T allow = 0.5 F u A n = 29.91 k > T [Satisfactory] A v = 1.33 in 2 , block shear T allow = 0.3 F u A v + 0.5 F u (0.5 A n ) = 38.06 k > T [Satisfactory] r min = t / (12) 0.5 = 0.072 in L = Max (E n , S ) = 4 in L / r min = 55 < 300 [Satisfactory] (AISC 360-05 D1) CHECK EDGE, END, & SPACING DISTANCE REQUIREMENTS (NDS 2005, Table 11.5.1A, Table 11.5.1B, & Table 11.5.1C E g = 2.75 in > 1.5 D [Satisfactory] E n = 4 in > 3.5 D [Satisfactory] S = 3 in > 3 D [Satisfactory] CHECK WOOD CAPACITY C = Min (C 1 , C 2 , C 3 ) = 0.762 , (geometry factor, NDS 2005, 11.5.1, page 76) where C 1 = (actual end distance) / (min end distance for full design value) = E n / 7D = 0.762 C 2 = (actual shear area) / (min shear area for full design value) = 1.000 C 3 = (actual spacing) / (min spacing for full design value) = S / 4D = 1.000 0.988 , (group action factor, NDS 2005, 10.3.6, page 60) where n = 3 R EA = Min [(E s A s /E m A m ) , (E m A m /E s A s )] = 0.616 E s A s = 3.8E+07 lbs, (NDS 2005, Table 10.3.6C) = 180000 D 1.5 = 116913 t m = 3 in u = 1+ S/2 [1 / E m A m + 1 / E s A s ] = 1.012 E m A m = 2.3E+07 lbs, (NDS 2005, Table 10.3.6C) m = u - (u 2 - 1) 0.5 = 0.855 Z' II = n Z II C D C M C t C g C = 5.309 kips > T [Satisfactory] where Z II = 1470 lbs / bolt, (interpolated from NDS 2005, Table 11B, Page 82) Technical References: 2 2 1 1 1 1 1 1 EA EA n n n m m R C g m n m m m R

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Wood Bolt Connection

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PROJECT : PAGE : CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

Bolt Connection Design Based on NDS 2005

INPUT DATA & DESIGN SUMMARYAXIAL TENSILE FORCE (ASD T = 5.3 kNUMBER OF BOLTS n = 3BOLT DIAMETER = 3/4BOLT SPACING S = 3 in

END DISTANCE OF WOOD En = 4 in

END DISTANCE OF STEEL En,s = 1.5 inLUMBER TYPE ( 0=Douglas Fir-Larch, 1=Douglas Fir-Larch(N), 0 Douglas Fir-Larch, G=0.5

2=Hem-Fir(N), 3=Hem-Fir, 4=Spruce-Pine-Fir) LUMBER SIZE ( 2 ) - 2 thk. x 6 widthSTRAP SIZE 5 width x 1/4 thk.LOAD DURATION FACTOR ( Tab 2.3.2, NDS 2005, Page 8 ) CD = 1.6WET SERVICE FACTOR ( Tab 10.3.3, NDS 2005, Page 58 ) CM = 1.0

TEMPERATURE FACTOR ( Tab 10.3.4, NDS 2005, Page 58 ) Ct = 1.0

THE CONNECTION DESIGN IS ADEQUATE.

6

1/4

ANALYSISCHECK STEEL STRAP CAPACITIES (AISC 360-05, ASD)

Ag = 1.25 in2, yielding critierion Fy = 36.00 ksi

Tallow = 0.6 Fy Ag = 27.00 k > T [Satisfactory](0.6 from 1/t, Typ.)

An = 1.03 in2, fracture criterion Fu = 58.00 ksi

Tallow = 0.5 Fu An = 29.91 k > T [Satisfactory]

Av = 1.33 in2, block shear

Tallow = 0.3 Fu Av + 0.5 Fu (0.5 An) = 38.06 k > T [Satisfactory]

rmin = t / (12)0.5 = 0.072 in L = Max (En , S ) = 4 in

L / rmin = 55 < 300 [Satisfactory] (AISC 360-05 D1)

CHECK EDGE, END, & SPACING DISTANCE REQUIREMENTS (NDS 2005, Table 11.5.1A, Table 11.5.1B, & Table 11.5.1C

Eg = 2.75 in > 1.5 D [Satisfactory]En = 4 in > 3.5 D [Satisfactory]S = 3 in > 3 D [Satisfactory]

CHECK WOOD CAPACITYC = Min (C1 , C2 , C3) = 0.762 , (geometry factor, NDS 2005, 11.5.1, page 76)

where C1 = (actual end distance) / (min end distance for full design value) = En / 7D = 0.762C2 = (actual shear area) / (min shear area for full design value) = 1.000C3 = (actual spacing) / (min spacing for full design value) = S / 4D = 1.000

0.988 , (group action factor, NDS 2005, 10.3.6, page 60)

where n = 3 REA = Min [(EsAs/EmAm) , (EmAm/EsAs)] = 0.616

EsAs = 3.8E+07 lbs, (NDS 2005, Table 10.3.6C) = 180000 D1.5 = 116913

tm = 3 in u = 1+ S/2 [1 / EmAm + 1 / EsAs] = 1.012

EmAm = 2.3E+07 lbs, (NDS 2005, Table 10.3.6C) m = u - (u2 - 1)0.5 = 0.855

Z'II = n ZII CD CM Ct Cg C = 5.309 kips > T [Satisfactory]where ZII = 1470 lbs / bolt, (interpolated from NDS 2005, Table 11B, Page 82)

Technical References:

2

2

1 111 1 1

EA

EA

n

n n

m m RC g mn mm mR

PC12Rectangle