Contents AISC LRFD 93 Example 002

User Manual: AISC LRFD-93 Example 002

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AISC LRFD-93 Example 002

WIDE FLANGE MEMBER UNDER COMBINED COMPRESSION & BIAXIAL BENDING

EXAMPLE DESCRIPTION

A check of the column adequacy is checked for combined axial compression and

flexural loads. The column is 14 feet tall and loaded with an axial load,

1400

u

P=

kips and bending,

ux uy

M ,M

= 200k-ft and 70k-ft, respectively. It is

assumed that there is reverse-curvature bending with equal end moments about

both axes and no loads along the member. The column demand/capacity ratio is

checked against the results of Example 6.2 in the 3rd Edition, LRFD Manual of

Steel Construction, pages 6-6 to 6-8.

GEOMETRY, PROPERTIES AND LOADING

TECHNICAL FEATURES TESTED

 Section compactness check (compression)

 Member compression capacity

 Member bending capacity

 Demand/capacity ratio, D/C

Member Properties

W14X176

E = 29000 ksi

Fy = 50 ksi

Loading

P

u

= 1,400 kips

Mux = 200 kip-ft

Muy = 70 kip-ft

Geometry

H = 14.0 ft

AISC LRFD-93 Example 002 - 1

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RESULTS COMPARISON

Independent results are hand calculated and compared with the results from

Example 6.2 in the 3rd Edition, LRFD Manual of Steel Construction, pages 6-6 to

6-8.

Output Parameter SAP2000 Independent Percent

Difference

Compactness Compact Compact 0.00%

cn

P

φ

(kips) 1937.84 1937.84 0.00%

b nx

M

φ

(k-ft) 1200 1200 0.00%

b ny

M

φ

(k-ft) 600.478 600.478 0.00%

D/C 0.974 0.974 0.00%

COMPUTER FILE: AISC LRFD-93 EX002

CONCLUSION

The results show an exact comparison with the independent results.

AISC LRFD-93 Example 002 - 2

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HAND CALCULATION

Properties:

Material: ASTM A992 Grade 50 Steel

Fy = 50 ksi, E = 29,000 ksi

Section: W14x176

A = 51.8 in2,

bf = 15.7 in, tf = 1.31 in, d = 15.2 in, tw = 0.83 in

2 15.2 2 1.31 12.58

cf

h d t in= − = −• =

Ix = 2,140 in4, Iy = 838 in4, rx = 6.4275 in, ry = 4.0221 in

Sx = 281.579 in3, Sy = 106.7516 in3, Zx = 320.0 in3, Zy = 163.0 in3.

Member:

Kx = Ky = 1.0

L = Lb = 14 ft

Other

0.85φ=

c

0.9φ=

b

Loadings:

Pu = 1400 kips

Mux = 200 k-ft

Muy = 70 k-ft

Section Compactness:

Localized Buckling for Flange:

( )

( )

/2 15.7 / 2 5.99

1.31

λ= = =

f

f

b

t

65 65 9.19

50

λ= = =

p

y

F

λ<λ

p

, No localized flange buckling

Flange is Compact.

AISC LRFD-93 Example 002 - 3

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Localized Buckling for Web:

12.58 15.16

0.83

λ= = =

c

w

h

t

0.9 51.8 50 2331 kipsφ =φ = • •=

by b g y

P AF

1400 0.601

2331

= =

φ

u

by

P

P

Since

0.601 0.125= >

φ

u

by

P

P

191 253

2.33



λ= − ≥





φ



u

p

by

yy

P

P

FF

( )

191 253

2.33 0.601 46.714 35.780

50 50

λ= − = ≥ =

p

λ<λ

p

, No localized web buckling

Web is Compact.

Section is Compact.

Member Compression Capacity:

For braced frames, K = 1.0 and KxLx = KyLy = 14.0 ft, From AISC Table 4-2,

1940 kipsφ=

cn

P

Or by hand,

1.0 14 12 50 0.552

4.022 29000

••

λ= = =

π •π

yy

c

y

KL F

rE

Since

1.5,λ<

c

( )

22

0.552

0.658 50 0.658 44.012ksi= =•=

λ

c

cr y

FF

0.85 44.012 51.8φ=φ = • •

c n c cr g

P FA

1937.84 kipsφ=

cn

P

AISC LRFD-93 Example 002 - 4

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From LRFD Specification Section H1.2,

1400 0.722 0.2

1937.84

= = >

φ

u

cn

P

P

Therefore, LRFD Specification Equation H1-1a governs.

Section Bending Capacity

50 310 1333.333 k-ft

12

•

= = =

px y x

M FZ

py y y

M FZ=

However,

163 1.527 1.5,

106.7516

y

y

Z

S= = >

So

3

1.5 1.5 106.7516 160.1274in==•=

yy

ZS

50 160.1274 667.198k-ft

12

•

= =

py

M

Member Bending Capacity

From LRFD Specification Equation F1-4,

yf

y

pF

E

r

L76

.1

=

29000 1

1.76•4.02 • 14.2ft 14ft

12

50

= = >=

pb

LL

φ=φ

b nx b px

MM

0.9 1333.333φ=•

b nx

M

1200k-ftφ=

b nx

M

φ=φ

b ny b py

MM

0.9•667.198φ=

b ny

M

600.478 k-ft

φ=

b ny

M

AISC LRFD-93 Example 002 - 5

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Interaction Capacity: Compression & Bending

From LRFD Specification section C1.2, for a braced frame, Mlt = 0.

,

1ntxxux

MBM =

where

200=

ntx

M

kip-ft; and

,

1ntyyuy MBM =

where

70=

nty

M

kip-ft

1

1

1

1≥















−

=

e

u

m

P

P

C

B

For reverse curvature bending and equal end moments:

0.1

2

1+=

M

M

















−=

2

1

4.0

6.0 M

M

Cm

( )

2.00.14.06.0 =−=

m

C

( )

2

12

π

=

e

EI

pKL

( )

2

12

29000 2140 21,702

14.0 12

π• •

= =

•

ex

p kips

( )

2

12

29000 838 8,498

14.0 12

π• •

= =

•

ey

p

1

1

1

1≥















−

=

xe

u

mx

x

P

P

C

B

1

0.2 0.214 1

1400

121702

x

B= = ≥



−





AISC LRFD-93 Example 002 - 6

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1

1=

x

B

1

1

1

1

≥















−

=

ye

u

my

y

P

P

C

B

1

0.2 0.239 1

1400

18498

y

B= = ≥



−





1

1

=

y

B

200

2000

.

1=•

=

ux

M

kip-ft;

and

70

700

.1 =

•=

uy

M

kip-ft

From LRFD Specification Equation H1-1a,

1400 8 200 70 0.974 1.0

1940 9 1200 600.478



++ =<





, OK

0.974

D

C=

AISC LRFD-93 Example 002 - 7