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AISC 360-05 Example 002 - 1

AISC 360-05 Example 002

BUILT UP WIDE FLANGE MEMBER UNDER COMPRESSION

EXAMPLE DESCRIPTION

A demand capacity ratio is calculated for the built-up, ASTM A572 grade 50,

column shown below. An axial load of 70 kips (D) and 210 kips (L) is applied toa simply supported column with a height of 15 ft.

GEOMETRY,PROPERTIES AND LOADING

TECHNICAL FEATURES TESTED

Section compactness check (compression) Warping constant calculation, Cw Member compression capacity with slenderness reduction

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AISC 360-05 Example 002 - 2

RESULTS COMPARISON

Independent results are hand calculated and compared with the results from

Example E.2 AISCDesign Examples, Volume 13.0 on the application of the 2005AISC Specification for Structural Steel Buildings (ANSI/AISC 360-05).

Output Parameter ETABS IndependentPercent

Difference

Compactness Slender Slender 0.00%

cPn(kips) 506.1 506.1 0.00 %

COMPUTER FILE: AISC360-05EX002

CONCLUSION

The results show an exact comparison with the independent results.

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AISC 360-05 Example 002 - 3

HAND CALCULATION

Properties:

Material:ASTM A572 Grade 50E = 29,000 ksi,Fy= 50 ksi

Section: Built-Up Wide Flange

d = 17.0 in, bf= 8.00 in, tf= 1.00 in, h = 15.0 in, tw= 0.250 in.

Ignoring fillet welds:

A = 2(8.00)(1.00) + (15.0)(0.250) = 19.75 in2

3 3

32 (1 . 0)(8 . 0) (1 5 . 0)(0. 2 5)85.35 in

12 12

yI

85.42 . 0 8 i n .

19.8

y

y

Ir

A

xx IAdI 2

3 3

2 4(0 .2 5 0 )(1 5 . 0) 2 (8 . 0 )(1 . 0 )

2(8.0)(8.0) 1095 .65 in12 1 2

xI

1 2 1 1

' 1 7 1 6 in

2 2

t td d

2 2

4' (8 5.3 5) (1 6.0 )5462.583 in

4 4

y

w

I dC

3 3 3

42(8.0)(1.0) (15.0)(0.250)5.41 in

3 3

btJ

Member:

K = 1.0 for a pinned-pinned conditionL = 15ft

Loadings:

Pu= 1.2(70.0) + 1.6(210) = 420 kips

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AISC 360-05 Example 002 - 4

Section Compactness:

Check for slender elements using Specification Section E7

Localized Buckling for Flange:

4.04.0

1.0

b

t

2 9 0 0 00 .3 8 0 .3 8 9 .1 5 2

5 0p

y

E

F

p , No localized flange buckling

Flange is Compact.

Localized Buckling for Web:

15.060.0

0.250

h

t

,

2 9 0 0 01 .4 9 1 .4 9 3 5 .9

5 0r

y

E

F

r , Localized web buckling

Web is Slender.

Section is Slender

Member Compression Capacity:

Elastic Flexural Buckling StressSince the unbraced length is the same for both axes, they-yaxis will govern by

inspection.

6.86

08.2

12150.1

y

y

r

KL

2

2

2

2

6.86

29000

r

KL

EF

e= 38.18 ksi

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AISC 360-05 Example 002 - 5

Elastic Critical Torsional Buckling StressNote: Torsional buckling will not govern ifKLy>KLz, however, the check is included

here to illustrate the calculation.

yxz

w

eII

GJLK

ECF

1

2

2

2

2

2 9 0 0 0 5 4 6 2 .4 11 12 00 5 .4 1

1 10 0 8 5.41 8 0e

F

= 91.8 ksi > 38.18 ksi

Therefore, the flexural buckling limit state controls.

Fe= 38.18 ksi

Section Reduction Factors

Since the flange is not slender,

Qs= 1.0

Since the web is slender,

For equation E7-17, takef asFcrwith Q = 1.0

2 9 0 0 04 .7 1 4 .7 1 1 1 3 8 6 .6

1 .0 5 0

y

y y

K LE

Q F r

So

1 . 0 5 0

38.20 .6 5 8 1 .0 0 .6 5 8 5 0 2 8 .9 k si

y

e

QF

F

cr yf F Q F

0.341.92 1 , w h ere

e

E Eb t b b h

f b t f

29000 0.34 290 001 .92 0 .250 1 15.0 in

28.9 15.0 0 .250 28.9

e

b

1 2.5 i n 1 5.0 i n e

b

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AISC 360-05 Example 002 - 6

therefore computeAeffwith reduced effective web width.

2

2 12.5 0 .250 2 8.0 1.0 19.1 in eff e w f f

A b t b t

whereAeffis effective area based on the reduced effective width of the web, be.

19.10.968

19.75

eff

a

AQ

A

1 .0 0 0 .9 6 8 0 .9 6 8s aQ Q Q

Critical Buckling Stress

Determine whether Specification Equation E7-2 or E7-3 applies

290004.71 4.7 1 115.4 86 .6

0 .9 6 6 5 0

y

y y

K LE

Q F r

Therefore, Specification Equation E7-2 applies.

When 4.71y

E K L

Q F r

1 . 0 5 0

38.180 .6 5 8 0 .9 6 6 0 .6 5 8 5 0 2 8 .4 7 k s i

y

e

QF

Fcr y

F Q F

Nominal Compressive Strength

2 8 .5 1 9 .7 5 5 6 2 .3 k ip s n cr g

P F A

0.90 c

0 .9 0 5 6 2. 3 5 0 6. 1 kip s c n cr g P F A > 420 kips

506.1kips c n

P