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DIRECT ANALYSIS METHOD Note: Change Cells in Yellow Only Input: E = 200000 MPa Ix = 113214948 mm4 Iy = 38876015 mm4 Fy = 345 MPa number of columns 2 M1(x) = -25.083 kN-m M2(x) = 50.311 kN-m Mnt(x) = 50.311 kN-m M1(y) = -25.083 kN-m M2(y) = 50.311 kN-m Mnt(y) = 50.311 kN-m Pnt = 1275.6 kN Mlt(x) = 13.841 kN-m Mlt(y) = 13.841 kN-m Plt(x) = 10.743 kN Plt(y) = 10.743 kN Pstory = 2551.2 kN Hx = 15.102 kN Hy = 15.102 kN L = 4.268 m ΔHx = 0.00386 m ΔHy = 0.00386 m Pmf = 2551.2 kN KLx = 4.268 m KLy = 4.268 m

W10X49

(non-sway) (non-sway) (non-sway) (non-sway) (non-sway) (non-sway) (non-sway) (sway) major axis (sway) minor axis (sway) major axis (sway) minor axis total load at story story shear (x) story shear (y) height of story drift (x) drift (y) total load carried by moment frames

Cmx Cmy

= =

0.401 0.401

B2(x) B2(y)

= =

1.219 �_2=1/(1−(𝛼𝑃_𝑠𝑡𝑜𝑟𝑦)/𝑃_(𝑒 𝑠𝑡𝑜𝑟𝑦) )≥1.0 1.609

Pe1(x) Pe1(y)

= =

9814.647 kN 3370.177 kN

𝐶_𝑚=0.6−0.4(�_1/�_2 )

𝑃_𝑒1=(𝜋^2 〖𝐸𝐼〗 ^∗)/(𝐾𝐿)^2

�_1=𝐶_𝑚/(1−(𝛼𝑃_𝑟)/𝑃_𝑒1 )≥1.0

�_1=𝐶_𝑚/(1−(𝛼𝑃_𝑟)/𝑃_𝑒1 )≥1.0

B1(x) B1(y)

= =

1.000 1.000

Pr

=

1305.98 kN

Mrx

=

67.185 kN-m

Mry

=

72.581 kN-m

Pc

=

2094.37 kN

Mcx

=

282.198 kN-m

Mcy

=

143.996 kN-m

𝑃_𝑟/𝑃_𝑐 𝑃_𝑟/𝑃_𝑐

≥

+

𝑃_𝑟=𝑃_𝑛𝑡+�_2 𝑃_𝑙𝑡

�_𝑟=�_1 �_𝑛𝑡+�_2 �_𝑙𝑡

0.2 (�_��/�_�� +�_�𝑦/�_�𝑦 ) 8 9 1.283

Section is inadequate

≤

1.0

>

1.0

Description:

Date:

moment frames

𝑃_(𝑒 𝑠𝑡𝑜𝑟𝑦) )≥1.0

∗)/(𝐾𝐿)^2

)/𝑃_𝑒1 )≥1.0

M1

M2

𝑃_(𝑒 𝑠𝑡𝑜𝑟𝑦)=𝑅_� 𝐻𝐿/∆_𝐻

)/𝑃_𝑒1 )≥1.0

+�_2 �_𝑙𝑡

Project:

R.R. Rojo and Associates Prepared by: Checked by:

Description:

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AISC LRFD DESIGN OF COMPRESSION MEMBER WITH AND WITHOUT SLENDER ELEME (W, S, M, HP Shapes Only) Note: Change Cells in Yellow Only PDL = PLL = PU = 1.20PDL + 1.60PLL PU = Select Section for Design W10X49

143 kN 143 kN 400.4 kN

Overall Conclusion:

The Designed Column Is Okay! unsupported length at x axis (meters) 4.268 m unsupported length at y axis (meters) 4.268 m End Restraints Condition Description Fixed at both ends 1

Fixed at one end, free at the other

2 3

Pinned at both Ends Fixed at one end, Pinned at

the other end 4 Enter End Restraints Condition for x-axis

Enter End Restraints Condition for y-axis

3 3

Efective length factor for x axis, Kx 1 Efective length factor for y axis, Ky 1 Properties : Fy = E= A= bf = tf = d= tw = h=

345 Mpa 200000 Mpa 9290.304 mm^2 254 mm 14.224 mm 254 mm 8.636 mm 200.152 mm

k(des) = bf/(2tf) = h/tw = Sx = Zx = rx = ry = rts = Cw = Iy = ho = J=

26.924 mm 8.93 23.1 894733.6944 mm^3 989778.6656 mm^3 110.49 mm 64.516 mm 72.136 mm 555869243737.999 mm^4 38876015.15104 mm^4 239.776 mm 578561.681584 mm^4

Slenderness Ratio Slenderness Ratio about x-axis KxLx/rx KyLy/ry Use Slenderness Ratio (KL/r)max

38.6279301294 66.1541323083

66.1541323083

Type of Flexural Buckling (KL/r)max 66.154132 Therefore,

VS.
λ_r √(Fy/Fcr)

Used h: he

mm

Used Area: A Ag

= =

Nominal Axial Capacity: Pn = Pn = ØPn = ØPn = Pu 400.400

vs


0 50 100 150 200 250 300 350 400 450

100

LOCAL BUCKLING: λp

λ

λf

8.93

λpf

9.15

λrf

24.08

λw

23.1

λpw

90.53

λrw

137.24

Flange is compact Web is compact

Therefore, section is compact

Mn for Local Buckling: Mn

341.47 kN-m

LATERAL TORSIONAL BUCKLING:

Mn

λr

Lb

Lp

Lr

4268

2733.92

9623.90

313.554 kN-m

R.R. Rojo and Associates Prepared by:

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AISC LRFD BEAM FLEXURAL DESIGN

Select Section W10X49 Unbraced Length, Lb

4268 mm

Properties: E bf tf d tw k(des) k1 c bf/(2tf) h/tw Sy Zy ry Cw Iy ho J

200000 MPa 254 mm 14.224 mm 254 mm 8.636 mm 26.924 mm 20.6375 mm 1 8.93 23.1 306438.0968 mm^3 463753.9112 mm^3 64.516 mm

Fy

345 MPa 1200 1100 1000 900 800 700 600 500 400 300 200 100 0

0 50 100 150 200 250 300 350 400 450

555869243738 mm^4 38876015.151 mm^4 239.776 mm 578561.68158 mm^4

Moment Capacity of the section Section is adequate

144.00

>

100

LOCAL BUCKLING: λp

λ

λf

8.93

λpf

9.15

λrf

24.08

λw

23.1

λpw

90.53

λrw

137.24

Flange is compact Web is compact

Therefore, section is compact

Mn for Local Buckling: Mn

not applicable kN-m

Yielding

Mn

λr

159.995 kN-m