## Welded Built-up Slender Compressive WF Beam per AISC 2005 E7

## Welded Built-up Slender Compressive WF Beam per AISC 2005 E7

(OP)

Dear friends.

I have checked some beam that needed for the truss structure.

The structure needs the height of the beam 450mm.

Since the truss wide enough, the beam is laying down, means the the minor axis is perpendicular to gravity axis.

I've read the AISC LRFD 1999 and AISC LRFD 2005. I think the LRFD 2005 is missing the equation AISC-LRFD 1999 Specification Appendix B5.3d. (A-B5-15) ] that compute Flexural buckling stress based on K L and lamda_c.

This equation gave more lower stress buckling capabitity of the slender compressive member.Other equation on slender moved in AISC 2005 including Torsional and flexural-torsional buckling stress.

I need your opinion..

This the example:

UNIT SYSTEM : kN, m

SECTION PROPERTIES : Designation = H.450.320.8.16 in mm

Shape = I - Section. (Built-up)

Depth = 0.450,

Top F Width = 0.320,

Bot.F Width = 0.320,

Web Thick = 0.008,

Top F Thick = 0.016,

Bot.F Thick = 0.016,

Area = 1.35840e-002,

Asy = 6.82667e-003,

Asz = 3.60000e-003,

Ybar = 1.60000e-001,

Zbar = 2.25000e-001,

Qyb = 1.60720e-001,

Qzb = 1.28000e-002,

Syy = 2.36044e-003,

Szz = 5.46245e-004,

Zyy = 2.57153e-003,

Zzz = 8.25888e-004

Iyy = 5.31100e-004,

Izz = 8.73992e-005,

Iyz = 0.00000e+000,

ry = 1.97731e-001,

rz = 8.02121e-002,

rp = 0.00000e+000

J = 9.47883e-007,

Cwp = 4.11470e-006

DESIGN PARAMETERS FOR STRENGTH EVALUATION :

Ly = 5.01646e+000,

Lz = 5.01646e+000,

Lu = 5.01646e+000,

Ky = 1.00000e+000,

Kz = 1.00000e+000

MATERIAL PROPERTIES :

Fy = 3.65000e+005, Es = 2.00000e+008,

MATERIAL NAME = JIS SMA490W

Factored force/moments caused by unit load case.

Load combination ID = 34-

Pu = 2793.46 kN.

Muy = -7.27 kN-m.

Muz = -15.73 kN-m.

Vuy = 3.52 kN.

Vuz = -7.74 kN.

Check slenderness ratio of axial compression member (Kl/r).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification B7. ]

Kl/r = 62.5 < 200.0 ---> O.K.

Check width-thickness ratio of flange (BTR).

[ AISC-LRFD2K Specification B5.1 ]

kc = 4/SQRT[h/tw] = 0.553 (for Built-up I-Section)

Lambda_r = 0.64*SQRT[Es/(Fy/kc)] = 11.14

BTR = bf/2tf =10.00 < Lambda_r-> NON-COMPACT SECTION !

Check depth-thickness ratio of web (DTR).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification B5.1 ]

Lambda_r = 1.49*SQRT[Es/Fy] = 34.88

Dweb = H-tf1-tf2 = 0.42 m.

DTR= Dweb/tw = 52.25 > Lambda_r ---> SLENDER SECTION !

Compute elastic compressive stress.

NOT VERY DETAIL on AISC 2005, suggesting Fy for not repetition

Resistance factor for compression : phi = 0.85

f_lim = phi * Fcr = 295173.6487 KPa.

f_cal = Pu/Area - Muy*Ccom/Iyy = 208721.5030 KPa.

f = f_cal (if f_cal < f_lim)

Calculate effective depth of stiffened web (Bew).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification Appendix B5.3b (A-B5-12) ]

Be1 = (1.49*tw)/SQRT[Es/f] = 0.369 m.

1.91*tw [ 0.34 ]

Be2 = ---------- * [ 1 - ---------------- ] = 0.378 m.

SQRT[f/Es] [ (DTR)*SQRT[f/Es] ]

Bew = MAX(Be1,Be2) = 0.378 m.

Calculate reduction factor of stiffened elements (Qa).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification Appendix B5.3c. (A-B5-14) ]

A_eff = Area - (Dweb-Bew)*tw = 0.013 m^2.

Qa = A_eff/Area = 0.976

Full reduction factor for slender section (QsQa).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification Appendix B5.3d. (A-B5-17) ]

Q = Qs*Qa = 0.976

Calculate column slendness parameter (Lambda_c).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification E2. (E2-4) ]

(Kl/r) Fy

Lambda_c = -------- * SQRT[----] = 0.850

pi Es

Calculate critical stress (Fcr1, Fcr2).

Flexural buckling stress (Fcr1)

THIS EQUATION NOT MOVED AND GONE ON AISC 2005 E7

[ AISC-LRFD2K Specification Appendix B5.3d. (A-B5-15) ] ]

Lambda_c*SQRT[Q] = 0.840 < 1.5

Odr = Q*Lambda_c^2 = 0.706

Fcr1 = Q*(0.658^Odr)*Fy = 265167.1822 KPa.

Torsional and flexural-torsional buckling stress (Fcr2)

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification Appendix E3 (A-E3-5) ]

Doubly symmetric shape

Gs = Es/(2(1+Poisson Ratio)) = 7.6923e+007 KPa.

Kx = 1.00 (Conservatively taken value).

L = 5.016 m.

Fe= [Pi^2*E*Cwp/(Kx*L)^2 + Gs*J]/(Iyy+Izz)=639725.8674 KPa.

Lambda_e = SQRT[Fy/Fe] = 0.755

Lambda_e*SQRT[Q] = 0.746 < 1.5

Odr = Q*Lambda_e^2 = 0.557

Fcr2 = Q*(0.658^Odr)*Fy = 282236.8864 KPa.

Calculate axial compressive strength (phiPn).

SAME AS SUCCESSOR AISC 2005

[ AISC-LRFD2K Specification E2. (E2-1), Appendix E3. (A-E3-1) ]

Fcr = MIN[ Fcr1,Fcr2 ] = 265167.1822 KPa.

Resistance factor for compression : phi = 0.85

phiPn = phi*Area*Fcr = 3061.73 kN.

Check ratio of axial strength (Pu/phiPn).

Pu 2793.46

------- = --------------- = 0.912 < 1.000 ---> O.K.

phiPn 3061.73

I have checked some beam that needed for the truss structure.

The structure needs the height of the beam 450mm.

Since the truss wide enough, the beam is laying down, means the the minor axis is perpendicular to gravity axis.

I've read the AISC LRFD 1999 and AISC LRFD 2005. I think the LRFD 2005 is missing the equation AISC-LRFD 1999 Specification Appendix B5.3d. (A-B5-15) ] that compute Flexural buckling stress based on K L and lamda_c.

This equation gave more lower stress buckling capabitity of the slender compressive member.Other equation on slender moved in AISC 2005 including Torsional and flexural-torsional buckling stress.

I need your opinion..

This the example:

UNIT SYSTEM : kN, m

SECTION PROPERTIES : Designation = H.450.320.8.16 in mm

Shape = I - Section. (Built-up)

Depth = 0.450,

Top F Width = 0.320,

Bot.F Width = 0.320,

Web Thick = 0.008,

Top F Thick = 0.016,

Bot.F Thick = 0.016,

Area = 1.35840e-002,

Asy = 6.82667e-003,

Asz = 3.60000e-003,

Ybar = 1.60000e-001,

Zbar = 2.25000e-001,

Qyb = 1.60720e-001,

Qzb = 1.28000e-002,

Syy = 2.36044e-003,

Szz = 5.46245e-004,

Zyy = 2.57153e-003,

Zzz = 8.25888e-004

Iyy = 5.31100e-004,

Izz = 8.73992e-005,

Iyz = 0.00000e+000,

ry = 1.97731e-001,

rz = 8.02121e-002,

rp = 0.00000e+000

J = 9.47883e-007,

Cwp = 4.11470e-006

DESIGN PARAMETERS FOR STRENGTH EVALUATION :

Ly = 5.01646e+000,

Lz = 5.01646e+000,

Lu = 5.01646e+000,

Ky = 1.00000e+000,

Kz = 1.00000e+000

MATERIAL PROPERTIES :

Fy = 3.65000e+005, Es = 2.00000e+008,

MATERIAL NAME = JIS SMA490W

Factored force/moments caused by unit load case.

Load combination ID = 34-

Pu = 2793.46 kN.

Muy = -7.27 kN-m.

Muz = -15.73 kN-m.

Vuy = 3.52 kN.

Vuz = -7.74 kN.

Check slenderness ratio of axial compression member (Kl/r).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification B7. ]

Kl/r = 62.5 < 200.0 ---> O.K.

Check width-thickness ratio of flange (BTR).

[ AISC-LRFD2K Specification B5.1 ]

kc = 4/SQRT[h/tw] = 0.553 (for Built-up I-Section)

Lambda_r = 0.64*SQRT[Es/(Fy/kc)] = 11.14

BTR = bf/2tf =10.00 < Lambda_r-> NON-COMPACT SECTION !

Check depth-thickness ratio of web (DTR).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification B5.1 ]

Lambda_r = 1.49*SQRT[Es/Fy] = 34.88

Dweb = H-tf1-tf2 = 0.42 m.

DTR= Dweb/tw = 52.25 > Lambda_r ---> SLENDER SECTION !

Compute elastic compressive stress.

NOT VERY DETAIL on AISC 2005, suggesting Fy for not repetition

Resistance factor for compression : phi = 0.85

f_lim = phi * Fcr = 295173.6487 KPa.

f_cal = Pu/Area - Muy*Ccom/Iyy = 208721.5030 KPa.

f = f_cal (if f_cal < f_lim)

Calculate effective depth of stiffened web (Bew).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification Appendix B5.3b (A-B5-12) ]

Be1 = (1.49*tw)/SQRT[Es/f] = 0.369 m.

1.91*tw [ 0.34 ]

Be2 = ---------- * [ 1 - ---------------- ] = 0.378 m.

SQRT[f/Es] [ (DTR)*SQRT[f/Es] ]

Bew = MAX(Be1,Be2) = 0.378 m.

Calculate reduction factor of stiffened elements (Qa).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification Appendix B5.3c. (A-B5-14) ]

A_eff = Area - (Dweb-Bew)*tw = 0.013 m^2.

Qa = A_eff/Area = 0.976

Full reduction factor for slender section (QsQa).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification Appendix B5.3d. (A-B5-17) ]

Q = Qs*Qa = 0.976

Calculate column slendness parameter (Lambda_c).

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification E2. (E2-4) ]

(Kl/r) Fy

Lambda_c = -------- * SQRT[----] = 0.850

pi Es

Calculate critical stress (Fcr1, Fcr2).

Flexural buckling stress (Fcr1)

THIS EQUATION NOT MOVED AND GONE ON AISC 2005 E7

[ AISC-LRFD2K Specification Appendix B5.3d. (A-B5-15) ] ]

Lambda_c*SQRT[Q] = 0.840 < 1.5

Odr = Q*Lambda_c^2 = 0.706

Fcr1 = Q*(0.658^Odr)*Fy = 265167.1822 KPa.

Torsional and flexural-torsional buckling stress (Fcr2)

SAME AS SUCCESSOR AISC 2005 E7

[ AISC-LRFD2K Specification Appendix E3 (A-E3-5) ]

Doubly symmetric shape

Gs = Es/(2(1+Poisson Ratio)) = 7.6923e+007 KPa.

Kx = 1.00 (Conservatively taken value).

L = 5.016 m.

Fe= [Pi^2*E*Cwp/(Kx*L)^2 + Gs*J]/(Iyy+Izz)=639725.8674 KPa.

Lambda_e = SQRT[Fy/Fe] = 0.755

Lambda_e*SQRT[Q] = 0.746 < 1.5

Odr = Q*Lambda_e^2 = 0.557

Fcr2 = Q*(0.658^Odr)*Fy = 282236.8864 KPa.

Calculate axial compressive strength (phiPn).

SAME AS SUCCESSOR AISC 2005

[ AISC-LRFD2K Specification E2. (E2-1), Appendix E3. (A-E3-1) ]

Fcr = MIN[ Fcr1,Fcr2 ] = 265167.1822 KPa.

Resistance factor for compression : phi = 0.85

phiPn = phi*Area*Fcr = 3061.73 kN.

Check ratio of axial strength (Pu/phiPn).

Pu 2793.46

------- = --------------- = 0.912 < 1.000 ---> O.K.

phiPn 3061.73