Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 3 Timber Column_3 POINT: 3
COORDINATE: x = 1.00 L = 13.50 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /9/ 1*1.20 + 2*1.20 + 3*1.00 + 4*1.60 + 5*0.50
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH WCLIB DenseSelectStructural BS 5+
Visually Graded Timber - Tab.4D
Fb=273600.00 lb/ft2
Ft=158400.00 lb/ft2
Fv=24480.00 lb/ft2
Fcp=105120.00 lb/ft2
Fc=187200.00 lb/ft2
E=244800000.00 lb/ft2
Emin=89280000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: TMBR 6x6
d=5.50 in
b=5.50 in
Ay=20.167 in2
Az=20.167 in2
A=30.250 in2
Iy=76.260 in4
Iz=76.260 in4
Ix=128.654 in4
Sy=27.731 in3
Sz=27.731 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
BUCKLING Z
LT BUCKLING
Key = 1.00
Kez = 1.00
lu = 13.50 ft
ley = 13.50 ft
lez = 13.50 ft
let = 24.84 ft
ley/d = 29.45
lez/b = 29.45
RB = 7.36
CPy = 0.35
CPz = 0.35
CL = 0.99
FcEy = 126547.32 lb/ft2
FcEz = 126547.32 lb/ft2
FbE = 2957310.14 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = 3.87 kip
My = 0.55 kip*ft
Mz = -2.07 kip*ft
Vy = 0.07 kip
Vz = 0.23 kip
fc = 18435.71 lb/ft2 fby = 33990.80 lb/ft2 fbz = -129218.80 lb/ft2 fvy = 499.28 lb/ft2
fvz = 1674.27 lb/ft2
Mx = 0.02 kip*ft
fvty = 897.15 lb/ft2
fvtz = 897.15 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Fc' = Fc(187200.00)*CM(1.00)*Ct(1.00)*CF(1.00)*CP(0.35)*KF(2.40)*Fi(0.90)*Lam(0.80) = 114109.26 lb/ft2
Fb' = Fb(273600.00)*CM(1.00)*Ct(1.00)*CL(0.99)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(0.80) = 468159.35 lb/ft2
Fv' = Fv(24480.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(0.80) = 42301.44 lb/ft2
Emin' = Emin(89280000.00)*CM(1.00)*Ct(1.00)*KF(1.76)*Fi(0.85) = 133562880.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
(fc/Fc')^2 + fby/(Fb'*(1-fc/FcEy) + fbz/(Fb'*(1-fc/FcEz-(fby/FbE)^2) = 0.43 < 1.00 [3.9-3] OK!
fc/FcEz + (fby/FbE)^2 = 0.15 < 1.00 [3.9-4] OK!, fc/FcEy = 0.15 < 1.00 [3.9.2] OK!,
(fvy + 3/2*fvty)/Fv' = 0.04 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.07 < 1.00 [3.4.1] OK!
Date : 13/02/23
Page : 1
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
ley/d = 29.45 < 50.00 STABLE, lez/b = 29.45 < 50.00 STABLE, Rb = 7.36 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 4 Timber Beam_4 POINT: 3
COORDINATE: x = 1.00 L = 16.49 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /20/ 1*1.20 + 2*1.20 + 5*1.00
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH SelectStructural 2+
Visually Graded Dimension Lumber - Tab.4A
Fb=216000.00 lb/ft2
Ft=144000.00 lb/ft2
Fv=25920.00 lb/ft2
Fcp=90000.00 lb/ft2
Fc=244800.00 lb/ft2
E=273600000.00 lb/ft2
Emin=99360000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: LMBR 4x8
d=7.25 in
b=3.50 in
Ay=16.920 in2
Az=16.920 in2
A=25.380 in2
Iy=111.100 in4
Iz=25.900 in4
Ix=72.185 in4
Sy=30.648 in3
Sz=14.800 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
LT BUCKLING
lu = 16.49 ft
leb = 28.70 ft
RB = 14.28
CL = 0.95
FbE = 875251.33 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = -0.15 kip
My = -0.85 kip*ft
Mz = 0.21 kip*ft
Vy = -0.20 kip
Vz = -0.04 kip
ft = -839.79 lb/ft2 fby = -48099.79 lb/ft2 fbz = 24779.72 lb/ft2 fvy = -1677.79 lb/ft2 fvz = -306.95 lb/ft2
Mx = 0.01 kip*ft
fvty = 749.47 lb/ft2
fvtz = 594.17 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Ft' = Ft(144000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.70)*Fi(0.80)*Lam(1.00) = 311040.00 lb/ft2
Fb' = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.95)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 443556.76 lb/ft2
Fv' = Fv(25920.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 55987.20 lb/ft2
Fb* = Fb(216000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 466344.00 lb/ft2
Fb** = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.95)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 443556.76 lb/ft2
Date : 13/02/23
BUCKLING Z
Page : 2
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
ft/Ft' + fby/Fb* + fbz/Fb' = 0.16 < 1.00 [3.9-1] OK!
(fby-ft)/Fb** + fbz/Fb' = 0.16 < 1.00 [3.9-2] OK!
(fvy + 3/2*fvty)/Fv' = 0.05 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.02 < 1.00 [3.4.1] OK!
Rb = 14.28 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 5 Timber Column_5 POINT: 1
COORDINATE: x = 0.00 L = 0.00 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /2/ 1*1.20 + 2*1.20 + 4*1.60 + 5*1.00
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH WCLIB DenseSelectStructural BS 5+
Visually Graded Timber - Tab.4D
Fb=273600.00 lb/ft2
Ft=158400.00 lb/ft2
Fv=24480.00 lb/ft2
Fcp=105120.00 lb/ft2
Fc=187200.00 lb/ft2
E=244800000.00 lb/ft2
Emin=89280000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: TMBR 6x6
d=5.50 in
b=5.50 in
Ay=20.167 in2
Az=20.167 in2
A=30.250 in2
Iy=76.260 in4
Iz=76.260 in4
Ix=128.654 in4
Sy=27.731 in3
Sz=27.731 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
BUCKLING Z
LT BUCKLING
Key = 1.00
Kez = 1.00
lu = 9.50 ft
ley = 9.50 ft
lez = 9.50 ft
leb = 17.48 ft
ley/d = 20.73
lez/b = 20.73
RB = 6.18
CPy = 0.52
CPz = 0.52
CL = 0.99
FcEy = 255548.46 lb/ft2
FcEz = 255548.46 lb/ft2
FbE = 4202493.36 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = 4.05 kip
My = -2.21 kip*ft
Mz = -0.03 kip*ft
Vy = 0.41 kip
Vz = -0.01 kip
fc = 19274.29 lb/ft2 fby = -137823.54 lb/ft2 fbz = -1876.58 lb/ft2 fvy = 2912.94 lb/ft2
fvz = -48.42 lb/ft2
Mx = 0.00 kip*ft
fvty = 30.32 lb/ft2
fvtz = 30.32 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------Date : 13/02/23
Page : 3
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
DESIGN WOOD STRENGHTS:
Fc' = Fc(187200.00)*CM(1.00)*Ct(1.00)*CF(1.00)*CP(0.52)*KF(2.40)*Fi(0.90)*Lam(1.00) = 210100.16 lb/ft2
Fb' = Fb(273600.00)*CM(1.00)*Ct(1.00)*CL(0.99)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 585955.54 lb/ft2
Fv' = Fv(24480.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 52876.80 lb/ft2
Emin' = Emin(89280000.00)*CM(1.00)*Ct(1.00)*KF(1.76)*Fi(0.85) = 133562880.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
(fc/Fc')^2 + fby/(Fb'*(1-fc/FcEy) + fbz/(Fb'*(1-fc/FcEz-(fby/FbE)^2) = 0.27 < 1.00 [3.9-3] OK!
fc/FcEz + (fby/FbE)^2 = 0.08 < 1.00 [3.9-4] OK!, fc/FcEy = 0.08 < 1.00 [3.9.2] OK!,
(fvy + 3/2*fvty)/Fv' = 0.06 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.00 < 1.00 [3.4.1] OK!
ley/d = 20.73 < 50.00 STABLE, lez/b = 20.73 < 50.00 STABLE, Rb = 6.18 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 7 Timber Beam_7 POINT: 3
COORDINATE: x = 1.00 L = 16.49 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /2/ 1*1.20 + 2*1.20 + 4*1.60 + 5*1.00
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH SelectStructural 2+
Visually Graded Dimension Lumber - Tab.4A
Fb=216000.00 lb/ft2
Ft=144000.00 lb/ft2
Fv=25920.00 lb/ft2
Fcp=90000.00 lb/ft2
Fc=244800.00 lb/ft2
E=273600000.00 lb/ft2
Emin=99360000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: LMBR 2x8
d=7.25 in
b=1.50 in
Ay=7.253 in2
Az=7.253 in2
A=10.880 in2
Iy=47.630 in4
Iz=2.039 in4
Ix=7.093 in4
Sy=13.139 in3
Sz=2.719 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
LT BUCKLING
lu = 16.49 ft
leb = 28.70 ft
RB = 33.31
CL = 0.34
FbE = 160760.45 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
Date : 13/02/23
BUCKLING Z
Page : 4
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
N = -0.29 kip
My = -0.75 kip*ft
Mz = -0.01 kip*ft
Vy = -0.15 kip
Vz = 0.00 kip
ft = -3798.31 lb/ft2 fby = -98745.79 lb/ft2 fbz = -7738.46 lb/ft2 fvy = -2949.46 lb/ft2 fvz = 91.96 lb/ft2
Mx = 0.00 kip*ft
fvty = 0.09 lb/ft2
fvtz = 0.07 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Ft' = Ft(144000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.70)*Fi(0.80)*Lam(1.00) = 311040.00 lb/ft2
Fb' = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.34)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 156789.74 lb/ft2
Fv' = Fv(25920.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 55987.20 lb/ft2
Fb* = Fb(216000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 466344.00 lb/ft2
Fb** = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.34)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 156789.74 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
ft/Ft' + fby/Fb* + fbz/Fb' = 0.24 < 1.00 [3.9-1] OK!
(fby-ft)/Fb** + fbz/Fb' = 0.62 < 1.00 [3.9-2] OK!
(fvy + 3/2*fvty)/Fv' = 0.05 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.00 < 1.00 [3.4.1] OK!
Rb = 33.31 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 10 Timber Beam_10 POINT: 1
COORDINATE: x = 0.00 L = 0.00 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 5 Wind X+ 176 ft/s (f =1.00) Simulation
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH SelectStructural 2+
Visually Graded Dimension Lumber - Tab.4A
Fb=216000.00 lb/ft2
Ft=144000.00 lb/ft2
Fv=25920.00 lb/ft2
Fcp=90000.00 lb/ft2
Fc=244800.00 lb/ft2
E=273600000.00 lb/ft2
Emin=99360000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: LMBR 2x8
d=7.25 in
b=1.50 in
Ay=7.253 in2
Az=7.253 in2
A=10.880 in2
Iy=47.630 in4
Iz=2.039 in4
Ix=7.093 in4
Sy=13.139 in3
Sz=2.719 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
Date : 13/02/23
BUCKLING Z
LT BUCKLING
lu = 16.49 ft
let = 28.70 ft
Page : 5
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
RB = 33.31
CL = 0.34
FbE = 160760.45 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = -0.04 kip
My = 0.40 kip*ft
Mz = -0.02 kip*ft
Vy = 0.00 kip
Vz = -0.00 kip
ft = -494.77 lb/ft2 fby = 52309.65 lb/ft2 fbz = -13550.48 lb/ft2 fvy = 69.69 lb/ft2
fvz = -94.76 lb/ft2
Mx = -0.00 kip*ft
fvty = 25.65 lb/ft2
fvtz = 19.01 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Ft' = Ft(144000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.70)*Fi(0.80)*Lam(1.00) = 311040.00 lb/ft2
Fb' = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.34)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 156789.74 lb/ft2
Fv' = Fv(25920.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 55987.20 lb/ft2
Fb* = Fb(216000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 466344.00 lb/ft2
Fb** = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.34)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 156789.74 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
ft/Ft' + fby/Fb* + fbz/Fb' = 0.14 < 1.00 [3.9-1] OK!
(fby-ft)/Fb** + fbz/Fb' = 0.36 < 1.00 [3.9-2] OK!
(fvy + 3/2*fvty)/Fv' = 0.00 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.00 < 1.00 [3.4.1] OK!
Rb = 33.31 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 13 Timber Beam_13 POINT: 1
COORDINATE: x = 0.00 L = 0.00 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 5 Wind X+ 176 ft/s (f =1.00) Simulation
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH SelectStructural 2+
Visually Graded Dimension Lumber - Tab.4A
Fb=216000.00 lb/ft2
Ft=144000.00 lb/ft2
Fv=25920.00 lb/ft2
Fcp=90000.00 lb/ft2
Fc=244800.00 lb/ft2
E=273600000.00 lb/ft2
Emin=99360000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: LMBR 2x8
d=7.25 in
b=1.50 in
Ay=7.253 in2
Az=7.253 in2
A=10.880 in2
Iy=47.630 in4
Iz=2.039 in4
Ix=7.093 in4
Sy=13.139 in3
Sz=2.719 in3
----------------------------------------------------------------------------------------------------------------------------------------------
Date : 13/02/23
Page : 6
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
MEMBER PARAMETERS:
BUCKLING Y
BUCKLING Z
LT BUCKLING
lu = 16.49 ft
let = 28.70 ft
RB = 33.31
CL = 0.34
FbE = 160760.45 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = -0.04 kip
My = 0.41 kip*ft
Mz = -0.00 kip*ft
Vy = 0.00 kip
Vz = -0.00 kip
ft = -515.75 lb/ft2 fby = 53512.12 lb/ft2 fbz = -770.55 lb/ft2
fvy = 48.99 lb/ft2
fvz = -60.15 lb/ft2
Mx = 0.00 kip*ft
fvty = 22.67 lb/ft2
fvtz = 16.80 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Ft' = Ft(144000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.70)*Fi(0.80)*Lam(1.00) = 311040.00 lb/ft2
Fb' = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.34)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 156789.74 lb/ft2
Fv' = Fv(25920.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 55987.20 lb/ft2
Fb* = Fb(216000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 466344.00 lb/ft2
Fb** = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.34)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 156789.74 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
ft/Ft' + fby/Fb* + fbz/Fb' = 0.12 < 1.00 [3.9-1] OK!
(fby-ft)/Fb** + fbz/Fb' = 0.34 < 1.00 [3.9-2] OK!
(fvy + 3/2*fvty)/Fv' = 0.00 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.00 < 1.00 [3.4.1] OK!
Rb = 33.31 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 14 Timber Column_14
POINT: 1 COORDINATE: x = 0.00 L =
0.00 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /2/ 1*1.20 + 2*1.20 + 4*1.60 + 5*1.00
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH WCLIB DenseSelectStructural BS 5+
Visually Graded Timber - Tab.4D
Fb=273600.00 lb/ft2
Ft=158400.00 lb/ft2
Fv=24480.00 lb/ft2
Fcp=105120.00 lb/ft2
Fc=187200.00 lb/ft2
E=244800000.00 lb/ft2
Emin=89280000.00 lb/ft2
----------------------------------------------------------------------------------------------------------------------------------------------
Date : 13/02/23
Page : 7
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
SECTION PARAMETERS: TMBR 6x6
d=5.50 in
b=5.50 in
Ay=20.167 in2
Az=20.167 in2
A=30.250 in2
Iy=76.260 in4
Iz=76.260 in4
Ix=128.654 in4
Sy=27.731 in3
Sz=27.731 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
BUCKLING Z
LT BUCKLING
Key = 1.00
Kez = 1.00
lu = 9.50 ft
ley = 9.50 ft
lez = 9.50 ft
leb = 17.48 ft
ley/d = 20.73
lez/b = 20.73
RB = 6.18
CPy = 0.52
CPz = 0.52
CL = 0.99
FcEy = 255548.46 lb/ft2
FcEz = 255548.46 lb/ft2
FbE = 4202493.36 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = 1.81 kip
My = -1.69 kip*ft
Mz = -0.27 kip*ft
Vy = 0.31 kip
Vz = -0.09 kip
fc = 8594.51 lb/ft2 fby = -105130.53 lb/ft2 fbz = -16984.11 lb/ft2 fvy = 2198.31 lb/ft2
fvz = -632.99 lb/ft2
Mx = -0.02 kip*ft
fvty = 760.31 lb/ft2
fvtz = 760.31 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Fc' = Fc(187200.00)*CM(1.00)*Ct(1.00)*CF(1.00)*CP(0.52)*KF(2.40)*Fi(0.90)*Lam(1.00) = 210100.16 lb/ft2
Fb' = Fb(273600.00)*CM(1.00)*Ct(1.00)*CL(0.99)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 585955.54 lb/ft2
Fv' = Fv(24480.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 52876.80 lb/ft2
Emin' = Emin(89280000.00)*CM(1.00)*Ct(1.00)*KF(1.76)*Fi(0.85) = 133562880.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
(fc/Fc')^2 + fby/(Fb'*(1-fc/FcEy) + fbz/(Fb'*(1-fc/FcEz-(fby/FbE)^2) = 0.22 < 1.00 [3.9-3] OK!
fc/FcEz + (fby/FbE)^2 = 0.03 < 1.00 [3.9-4] OK!, fc/FcEy = 0.03 < 1.00 [3.9.2] OK!,
(fvy + 3/2*fvty)/Fv' = 0.06 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.03 < 1.00 [3.4.1] OK!
ley/d = 20.73 < 50.00 STABLE, lez/b = 20.73 < 50.00 STABLE, Rb = 6.18 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 15 Timber Column_15
POINT: 3 COORDINATE: x = 1.00 L =
13.50 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /9/ 1*1.20 + 2*1.20 + 3*1.00 + 4*1.60 + 5*0.50
---------------------------------------------------------------------------------------------------------------------------------------------Date : 13/02/23
Page : 8
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
MATERIAL: DOUGLAS FIR-LARCH WCLIB DenseSelectStructural BS 5+
Visually Graded Timber - Tab.4D
Fb=273600.00 lb/ft2
Ft=158400.00 lb/ft2
Fv=24480.00 lb/ft2
Fcp=105120.00 lb/ft2
Fc=187200.00 lb/ft2
E=244800000.00 lb/ft2
Emin=89280000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: TMBR 6x6
d=5.50 in
b=5.50 in
Ay=20.167 in2
Az=20.167 in2
A=30.250 in2
Iy=76.260 in4
Iz=76.260 in4
Ix=128.654 in4
Sy=27.731 in3
Sz=27.731 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
BUCKLING Z
LT BUCKLING
Key = 1.00
Kez = 1.00
lu = 13.50 ft
ley = 13.50 ft
lez = 13.50 ft
let = 24.84 ft
ley/d = 29.45
lez/b = 29.45
RB = 7.36
CPy = 0.35
CPz = 0.35
CL = 0.99
FcEy = 126547.32 lb/ft2
FcEz = 126547.32 lb/ft2
FbE = 2957310.14 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = 3.88 kip
My = 0.56 kip*ft
Mz = 2.10 kip*ft
Vy = 0.07 kip
Vz = -0.24 kip
fc = 18451.45 lb/ft2 fby = 34665.44 lb/ft2 fbz = 130606.12 lb/ft2 fvy = 512.58 lb/ft2
fvz = -1699.30 lb/ft2
Mx = -0.02 kip*ft
fvty = 880.13 lb/ft2
fvtz = 880.13 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Fc' = Fc(187200.00)*CM(1.00)*Ct(1.00)*CF(1.00)*CP(0.35)*KF(2.40)*Fi(0.90)*Lam(0.80) = 114109.26 lb/ft2
Fb' = Fb(273600.00)*CM(1.00)*Ct(1.00)*CL(0.99)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(0.80) = 468159.35 lb/ft2
Fv' = Fv(24480.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(0.80) = 42301.44 lb/ft2
Emin' = Emin(89280000.00)*CM(1.00)*Ct(1.00)*KF(1.76)*Fi(0.85) = 133562880.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
(fc/Fc')^2 + fby/(Fb'*(1-fc/FcEy) + fbz/(Fb'*(1-fc/FcEz-(fby/FbE)^2) = 0.44 < 1.00 [3.9-3] OK!
fc/FcEz + (fby/FbE)^2 = 0.15 < 1.00 [3.9-4] OK!, fc/FcEy = 0.15 < 1.00 [3.9.2] OK!,
(fvy + 3/2*fvty)/Fv' = 0.04 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.07 < 1.00 [3.4.1] OK!
ley/d = 29.45 < 50.00 STABLE, lez/b = 29.45 < 50.00 STABLE, Rb = 7.36 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 16 Timber Beam_16 POINT: 3
COORDINATE: x = 1.00 L = 16.49 ft
---------------------------------------------------------------------------------------------------------------------------------------------Date : 13/02/23
Page : 9
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
LOADS:
Governing Load Case: 10 ULS /20/ 1*1.20 + 2*1.20 + 5*1.00
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH SelectStructural 2+
Visually Graded Dimension Lumber - Tab.4A
Fb=216000.00 lb/ft2
Ft=144000.00 lb/ft2
Fv=25920.00 lb/ft2
Fcp=90000.00 lb/ft2
Fc=244800.00 lb/ft2
E=273600000.00 lb/ft2
Emin=99360000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: LMBR 4x8
d=7.25 in
b=3.50 in
Ay=16.920 in2
Az=16.920 in2
A=25.380 in2
Iy=111.100 in4
Iz=25.900 in4
Ix=72.185 in4
Sy=30.648 in3
Sz=14.800 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
BUCKLING Z
LT BUCKLING
lu = 16.49 ft
leb = 28.70 ft
RB = 14.28
CL = 0.95
FbE = 875251.33 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = -0.14 kip
My = -0.87 kip*ft
Mz = -0.20 kip*ft
Vy = -0.20 kip
Vz = 0.04 kip
ft = -807.48 lb/ft2 fby = -49068.26 lb/ft2 fbz = -23513.60 lb/ft2 fvy = -1681.97 lb/ft2 fvz = 306.27 lb/ft2
Mx = -0.01 kip*ft
fvty = 740.40 lb/ft2
fvtz = 586.98 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Ft' = Ft(144000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.70)*Fi(0.80)*Lam(1.00) = 311040.00 lb/ft2
Fb' = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.95)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 443556.76 lb/ft2
Fv' = Fv(25920.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 55987.20 lb/ft2
Fb* = Fb(216000.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 466344.00 lb/ft2
Fb** = Fb(216000.00)*CM(1.00)*Ct(1.00)*CL(0.95)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 443556.76 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
ft/Ft' + fby/Fb* + fbz/Fb' = 0.16 < 1.00 [3.9-1] OK!
(fby-ft)/Fb** + fbz/Fb' = 0.16 < 1.00 [3.9-2] OK!
(fvy + 3/2*fvty)/Fv' = 0.05 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.02 < 1.00 [3.4.1] OK!
Rb = 14.28 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
Date : 13/02/23
Page : 10
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 17 Timber Beam_17 POINT: 1
COORDINATE: x = 0.50 L = 11.00 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /25/ 1*1.20 + 2*1.20 + 3*1.00 + 4*1.60 + 6*0.50
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH WCLIB DenseSelectStructural BS 5+
Visually Graded Timber - Tab.4D
Fb=273600.00 lb/ft2
Ft=158400.00 lb/ft2
Fv=24480.00 lb/ft2
Fcp=105120.00 lb/ft2
Fc=187200.00 lb/ft2
E=244800000.00 lb/ft2
Emin=89280000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: TMBR 6x8
d=7.50 in
b=5.50 in
Ay=27.500 in2
Az=27.500 in2
A=41.250 in2
Iy=193.400 in4
Iz=104.000 in4
Ix=228.921 in4
Sy=51.573 in3
Sz=37.818 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
BUCKLING Z
LT BUCKLING
lu = 22.00 ft
leb = 37.73 ft
RB = 10.60
CL = 0.98
FcEy = INF lb/ft2
FcEz = INF lb/ft2
FbE = 1427596.70 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = 0.24 kip
My = -11.43 kip*ft
Mz = -0.44 kip*ft
Vy = 5.43 kip
Vz = -0.19 kip
fc = 837.22 lb/ft2 fby = -383081.14 lb/ft2 fbz = -19931.44 lb/ft2 fvy = 28432.85 lb/ft2 fvz = -981.81 lb/ft2
Mx = -0.08 kip*ft
fvty = 2762.88 lb/ft2
fvtz = 2461.73 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Fc' = Fc(187200.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.40)*Fi(0.90)*Lam(0.80) = 323481.60 lb/ft2
Fb' = Fb(273600.00)*CM(1.00)*Ct(1.00)*CL(0.98)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(0.80) = 461536.91 lb/ft2
Fv' = Fv(24480.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(0.80) = 42301.44 lb/ft2
Emin' = Emin(89280000.00)*CM(1.00)*Ct(1.00)*KF(1.76)*Fi(0.85) = 133562880.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
(fc/Fc')^2 + fby/(Fb'*(1-fc/FcEy) + fbz/(Fb'*(1-fc/FcEz-(fby/FbE)^2) = 0.88 < 1.00 [3.9-3] OK!
fc/FcEz + (fby/FbE)^2 = 0.07 < 1.00 [3.9-4] OK!,
(fvy + 3/2*fvty)/Fv' = 0.77 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.11 < 1.00 [3.4.1] OK!
Rb = 10.60 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
Date : 13/02/23
Page : 11
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 18 Timber Beam_18 POINT: 3
COORDINATE: x = 0.25 L = 5.50 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /2/ 1*1.20 + 2*1.20 + 4*1.60 + 5*1.00
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH WCLIB DenseSelectStructural BS 5+
Visually Graded Timber - Tab.4D
Fb=273600.00 lb/ft2
Ft=158400.00 lb/ft2
Fv=24480.00 lb/ft2
Fcp=105120.00 lb/ft2
Fc=187200.00 lb/ft2
E=244800000.00 lb/ft2
Emin=89280000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: TMBR 6x8
d=7.50 in
b=5.50 in
Ay=27.500 in2
Az=27.500 in2
A=41.250 in2
Iy=193.400 in4
Iz=104.000 in4
Ix=228.921 in4
Sy=51.573 in3
Sz=37.818 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
LT BUCKLING
lu = 22.00 ft
leb = 37.73 ft
RB = 10.60
CL = 0.97
FcEy = INF lb/ft2
FcEz = INF lb/ft2
FbE = 1427596.70 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = 0.07 kip
My = -2.32 kip*ft
Mz = -0.18 kip*ft
Vy = -2.48 kip
Vz = 0.11 kip
fc = 233.33 lb/ft2 fby = -77819.05 lb/ft2 fbz = -8045.44 lb/ft2 fvy = -13000.65 lb/ft2 fvz = 598.56 lb/ft2
Mx = 0.21 kip*ft
fvty = 6943.08 lb/ft2
fvtz = 6186.28 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Fc' = Fc(187200.00)*CM(1.00)*Ct(1.00)*CF(1.00)*KF(2.40)*Fi(0.90)*Lam(1.00) = 404352.00 lb/ft2
Fb' = Fb(273600.00)*CM(1.00)*Ct(1.00)*CL(0.97)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 571616.37 lb/ft2
Fv' = Fv(24480.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 52876.80 lb/ft2
Emin' = Emin(89280000.00)*CM(1.00)*Ct(1.00)*KF(1.76)*Fi(0.85) = 133562880.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
(fc/Fc')^2 + fby/(Fb'*(1-fc/FcEy) + fbz/(Fb'*(1-fc/FcEz-(fby/FbE)^2) = 0.15 < 1.00 [3.9-3] OK!
fc/FcEz + (fby/FbE)^2 = 0.00 < 1.00 [3.9-4] OK!,
Date : 13/02/23
BUCKLING Z
Page : 12
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
(fvy + 3/2*fvty)/Fv' = 0.44 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.19 < 1.00 [3.4.1] OK!
Rb = 10.60 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 28 Timber Column_28
POINT: 3 COORDINATE: x = 1.00 L =
9.50 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /2/ 1*1.20 + 2*1.20 + 4*1.60 + 5*1.00
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH WCLIB DenseSelectStructural BS 5+
Visually Graded Timber - Tab.4D
Fb=273600.00 lb/ft2
Ft=158400.00 lb/ft2
Fv=24480.00 lb/ft2
Fcp=105120.00 lb/ft2
Fc=187200.00 lb/ft2
E=244800000.00 lb/ft2
Emin=89280000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: TMBR 6x6
d=5.50 in
b=5.50 in
Ay=20.167 in2
Az=20.167 in2
A=30.250 in2
Iy=76.260 in4
Iz=76.260 in4
Ix=128.654 in4
Sy=27.731 in3
Sz=27.731 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
BUCKLING Z
LT BUCKLING
Key = 1.00
Kez = 1.00
lu = 9.50 ft
ley = 9.50 ft
lez = 9.50 ft
let = 17.48 ft
ley/d = 20.73
lez/b = 20.73
RB = 6.18
CPy = 0.52
CPz = 0.52
CL = 0.99
FcEy = 255548.46 lb/ft2
FcEz = 255548.46 lb/ft2
FbE = 4202493.36 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = 4.85 kip
My = 1.66 kip*ft
Mz = -0.07 kip*ft
Vy = 0.24 kip
Vz = 0.02 kip
fc = 23105.55 lb/ft2 fby = 103548.38 lb/ft2 fbz = -4443.04 lb/ft2 fvy = 1740.36 lb/ft2
fvz = 138.08 lb/ft2
Mx = 0.02 kip*ft
fvty = 895.89 lb/ft2
fvtz = 895.89 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Fc' = Fc(187200.00)*CM(1.00)*Ct(1.00)*CF(1.00)*CP(0.52)*KF(2.40)*Fi(0.90)*Lam(1.00) = 210100.16 lb/ft2
Fb' = Fb(273600.00)*CM(1.00)*Ct(1.00)*CL(0.99)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 585955.54 lb/ft2
Fv' = Fv(24480.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 52876.80 lb/ft2
Date : 13/02/23
Page : 13
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
Emin' = Emin(89280000.00)*CM(1.00)*Ct(1.00)*KF(1.76)*Fi(0.85) = 133562880.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
(fc/Fc')^2 + fby/(Fb'*(1-fc/FcEy) + fbz/(Fb'*(1-fc/FcEz-(fby/FbE)^2) = 0.21 < 1.00 [3.9-3] OK!
fc/FcEz + (fby/FbE)^2 = 0.09 < 1.00 [3.9-4] OK!, fc/FcEy = 0.09 < 1.00 [3.9.2] OK!,
(fvy + 3/2*fvty)/Fv' = 0.06 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.03 < 1.00 [3.4.1] OK!
ley/d = 20.73 < 50.00 STABLE, lez/b = 20.73 < 50.00 STABLE, Rb = 6.18 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
TIMBER STRUCTURE CALCULATIONS
---------------------------------------------------------------------------------------------------------------------------------------------CODE:
ANSI/AWC NDS-2012 LRFD
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 29 Timber Column_29
POINT: 3 COORDINATE: x = 1.00 L =
9.50 ft
---------------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 10 ULS /2/ 1*1.20 + 2*1.20 + 4*1.60 + 5*1.00
---------------------------------------------------------------------------------------------------------------------------------------------MATERIAL: DOUGLAS FIR-LARCH WCLIB DenseSelectStructural BS 5+
Visually Graded Timber - Tab.4D
Fb=273600.00 lb/ft2
Ft=158400.00 lb/ft2
Fv=24480.00 lb/ft2
Fcp=105120.00 lb/ft2
Fc=187200.00 lb/ft2
E=244800000.00 lb/ft2
Emin=89280000.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: TMBR 6x6
d=5.50 in
b=5.50 in
Ay=20.167 in2
Az=20.167 in2
A=30.250 in2
Iy=76.260 in4
Iz=76.260 in4
Ix=128.654 in4
Sy=27.731 in3
Sz=27.731 in3
---------------------------------------------------------------------------------------------------------------------------------------------MEMBER PARAMETERS:
BUCKLING Y
BUCKLING Z
LT BUCKLING
Key = 1.00
Kez = 1.00
lu = 9.50 ft
ley = 9.50 ft
lez = 9.50 ft
let = 17.48 ft
ley/d = 20.73
lez/b = 20.73
RB = 6.18
CPy = 0.52
CPz = 0.52
CL = 0.99
FcEy = 255548.46 lb/ft2
FcEz = 255548.46 lb/ft2
FbE = 4202493.36 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND ACTUAL STRESSES:
N = 4.85 kip
My = 1.69 kip*ft
Mz = 0.01 kip*ft
Vy = 0.25 kip
Vz = -0.01 kip
fc = 23080.34 lb/ft2 fby = 105272.97 lb/ft2 fbz = 835.13 lb/ft2
fvy = 1773.94 lb/ft2
fvz = -52.50 lb/ft2
Mx = -0.02 kip*ft
fvty = 833.87 lb/ft2
fvtz = 833.87 lb/ft2
Date : 13/02/23
Page : 14
Robot Structural Analysis Professional 2023
Author:
Address:
File: wood frame final itr 2.rtd
Project: wood frame final itr 2
---------------------------------------------------------------------------------------------------------------------------------------------DESIGN WOOD STRENGHTS:
Fc' = Fc(187200.00)*CM(1.00)*Ct(1.00)*CF(1.00)*CP(0.52)*KF(2.40)*Fi(0.90)*Lam(1.00) = 210100.16 lb/ft2
Fb' = Fb(273600.00)*CM(1.00)*Ct(1.00)*CL(0.99)*CF(1.00)*KF(2.54)*Fi(0.85)*Lam(1.00) = 585955.54 lb/ft2
Fv' = Fv(24480.00)*CM(1.00)*Ct(1.00)*KF(2.88)*Fi(0.75)*Lam(1.00) = 52876.80 lb/ft2
Emin' = Emin(89280000.00)*CM(1.00)*Ct(1.00)*KF(1.76)*Fi(0.85) = 133562880.00 lb/ft2
---------------------------------------------------------------------------------------------------------------------------------------------RESULTS:
(fc/Fc')^2 + fby/(Fb'*(1-fc/FcEy) + fbz/(Fb'*(1-fc/FcEz-(fby/FbE)^2) = 0.21 < 1.00 [3.9-3] OK!
fc/FcEz + (fby/FbE)^2 = 0.09 < 1.00 [3.9-4] OK!, fc/FcEy = 0.09 < 1.00 [3.9.2] OK!,
(fvy + 3/2*fvty)/Fv' = 0.06 < 1.00 [3.4.1] OK!, (fvz + 3/2*fvtz)/Fv' = 0.02 < 1.00 [3.4.1] OK!
ley/d = 20.73 < 50.00 STABLE, lez/b = 20.73 < 50.00 STABLE, Rb = 6.18 < 50.00 STABLE
----------------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
Date : 13/02/23
Page : 15