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H-3

Example H.1b W-shape Column Subjected to Combined Compression andBending Moment About Both Axes (braced frame)

Verify if an ASTM A992 W1499 has sufficient available strength to support the axial forces and momentslisted below, obtained from a second order analysis that includes second-order effects. The unbraced length is

14 ft and the member has pinned ends. KLx = KLy =Lb = 14.0 ft

LRFD ASD

Pu = 400 kipsMux = 250 kip-ftMuy = 80.0 kip-ft

Pa = 267 kipsMax = 167 kip-ftMay = 53.3 kip-ft

Solution:

Material Properties:ASTM A992 Fy = 50 ksi Fu = 65 ksi

ManualTable 2-3

Take the available axial and flexural strengths from the Manual Tables

LRFD ASD

at KLy = 14.0 ft,

Pc = cPn = 1130 kips

atLb = 14.0 ft,

Mcx = Mnx = 642 kip-ft

Mcy = Mny= 311 kip-ft

u

c n

P

P =400 kips

1,130 kips = 0.354

Since u

c n

P

P> 0.2, use Eqn. H1.1a

8+ +

9

ryrxr

c cx cy

MMP

P M M

M 1.0

400 kips 8 250 kip-ft 80.0 kip-ft+ +

1130 kips 9 642 kip-ft 311 kip-ft

=

( )

80.354 + 0.389 + 0.257

9= 0.929 < 1.0

o.k.

at KLy = 14.0 ft,

Pc = Pn/c= 751 kips

atLb = 14.0 ft,

Mcx =Mnx / = 428 kip-ft

Mcy =Mny / = 207 kip-ft

/

a

n c

P

P =267 kips

751 kips = 0.356

Since/

a

n c

P

P > 0.2, use Eqn. H1.1a

8+ +

9

ryrxr

c cx cy

MMP

P M M

M 1.0

267 kips 8 167 kip-ft 53.3 kip-ft+ +

751 kips 9 428 kip-ft 207 kip-ft

=

( )

80.356 + 0.390 + 0.257

9= 0.931 < 1.0

o.k.

ManualTable 4-1

ManualTable 3-10

ManualTable 3-2

Eq. H1.1a

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H-9

Example H.4 W-Shape Subject to Combined Axial Compression andFlexure

Given: Select an ASTM A992 W-shape with a 10 in. nominal depth to carry nominal axial compressionforces of 5 kips from dead load and 15 kips from live load. The unbraced length is 14 ft and the ends are

pinned. The member also has the following nominal required moment strengths, not including second-order

effects:

MxD = 15 kip-ft MxL = 45 kip-ft

MyD = 2 kip-ft MyL = 6 kip-ft

The member is not subject to sidesway.

Solution:

Material Properties:

ASTM A992 Fy = 50 ksi Fu = 65 ksi

Calculate the required strength, not considering second-order effects

LRFD ASD

Pu = 1.2(5.00 kips) + 1.6(15.0 kips)

= 30.0 kipsMux = 1.2(15.0 kip-ft) + 1.6(45.0 kip-ft)

= 90.0 kip-ft

Muy = 1.2(2.00 kip-ft) + 1.6(6.00 kip-ft)= 12.0 kip-ft

Pa = 5.00 kips + 15.0 kips

= 20.0 kipsMax = 15.0 kip-ft + 45.0 kip-ft

= 60.0 kip-ft

May = 2.00 kip-ft + 6.00 kip-ft= 8.00 kip-ft

Try aW1033

Geometric Properties:

W1033 A = 9.71 in.2 Sx= 35.0 in.3

Zx = 38.8 in.3 Ix = 171 in.

4Sy= 9.20 in.

3Zy = 14.0 in.

3Iy = 36.6 in.

4

Lp = 6.85 ft Lr= 12.8 ft

Manual

Table 1-1Table 3-1

Calculate the available axial strength

For a pinned-pinned condition, K= 1.0.

Since KLx = KLy = 14.0 ft andrx > ry, the y-y axis will govern.

Commentary

TableC-C2.2

LRFD ASD Manual

Pc = cPn = 253 kips Pc = Pn/c = 168 kips Table 4-1

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H-10

Calculate the required flexural strengths including second order amplification

Use Amplified First-Order Elastic Analysis procedure from Section C2.1b. Since the

member is not subject to sidesway, only P- amplifiers need to be added.

1

11 /

m

r e

CB

P P=

Cm = 1.0

X-X axis flexural magnifier

( )

( )( )

( )( )( )( )

2 42

1 2 2

1

29,000ksi 171in.

1.0 14.0 ft 12 in./ft

x

e

x

EIP

K L

= = = 1730 kips

Eqn. C2-2

Eqn. C2-5

LRFD ASD

= 1.0

( )11.0

1 1.0 30.0kips /1730 kipsB =

= 1.02

Mux= 1.02(90.0 kip-ft) = 91.8 kip-ft

= 1.6

( )11.0

1 1.6 20.0kips/1730kipsB =

= 1.02

Max= 1.02(60.0 kip-ft) = 61.2 kip-ft

Eqn. C2-2

Y-Y axis flexural magnifier

( )

( )( )

( )( )( )( )

2 42

1 2 2

1

29,000 ksi 36.6in.

1.0 14.0 ft 12 in./ft

y

e

y

EIP

K L

= = = 371 kips Eqn. C2-5

LRFD ASD

= 1.0

( )11.0

1 1.0 30.0kips/ 371kipsB =

= 1.09

Muy= 1.09(12.0 kip-ft) = 13.1 kip-ft

= 1.6

( )11.0

1 1.6 20.0kips/ 371kipsB =

= 1.09

May= 1.09 (8.00 kip-ft) = 8.76 kip-ft

Eqn. C2-2

Calculate the nominal bending strength about the x-x axis

Yielding limit state

Mnx =Mp = FyZx = 50 ksi(38.8 in.3) = 1940 kip-in or 162 kip-ft

Lateral-torsional buckling limit state

SinceLp

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H-11

Mnx = ( )-

- -0.7-

b p

b p p y x

r p

L LC M M F S

L L

MMp

Mnx= ( )( )( )3 14.0ft - 6.85ft1.14 1940kip-in. - 1940kip-in. - 0.7 50ksi 35.0 in.21.8 ft - 6.85ft

= 1820 kip-in.M

1940 kip-in., therefore use:

Mnx= 1820 kip-in. or 152 kip-ft controls

Local buckling limit state

Per Manual Table 1-1, the member is compact forFy = 50 ksi, so the local buckling limit state

does not apply

Calculate the nominal bending strength about the y-y axis

Since aW1033 has compact flanges, only the yielding limit state applies.

Mny =Mp = FyZyM 1.6FySy

= 50 ksi(14.0 in.3)M 1.6(50 ksi)(9.20 in.3)= 700 kip-in < 736 kip-in., therefore

UseMny = 700 kip-in. or 58.3 kip-ft

Eqn. F2-2

ManualTable 1-1

Section F6.2

Eqn. F6-1

LRFD ASD

b = 0.90Mcx= bMnx = 0.90(152 kip-ft) = 137 kip-ftMcy= bMny = 0.90(58.3 kip-ft) = 52.5 kip-ft

b = 1.67Mcx=Mnx/b = 152 kip-ft/1.67 = 91.0 kip-ftMcy=Mny/b = 58.3 kip-ft/1.67 = 34.9 kip-ft

Section F2

Check limit for Equation H1-1a

LRFD ASD

ur

c c n

PP

P P=

=

30.0 kips

253 kips= 0.119, therefore,

use Specification Equation H1.1b

+ +2

ryrxr

c cx cy

MMP

P M M

M 1.0

30.0 kips 91.8 kip-ft 13.1 kip-ft+ +

2(253 kips) 137 kip-ft 52.5 kip-ft

0.0593 + 0.920 = 0.979 M 1.0 o.k.

/

ar

c n c

PP

P P=

=

20.0 kips

168 kips= 0.119, therefore

use Specification Equation H1.1b

+ +2

ryrxr

c cx cy

MMP

P M M

M 1.0

20.0 kips 61.2 kip-ft 8.76 kip-ft+ +

2(168 kips) 91.0 kip-ft 34.9 kip-ft

0.0595 + 0.924 = 0.983 M 1.0 o.k.

Section H1.1

Eqn. H1.1b