Contents AISC 360 05 Example 002

User Manual: AISC 360-05 Example 002

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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 to
a 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

AISC 360-05 Example 002 - 1

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RESULTS COMPARISON
Independent results are hand calculated and compared with the results from
Example E.2 AISC Design Examples, Volume 13.0 on the application of the 2005
AISC Specification for Structural Steel Buildings (ANSI/AISC 360-05).

Output Parameter
Compactness
φcPn (kips)

SAP2000

Independent

Percent
Difference

Slender

Slender

0.00%

506.1

506.1

0.00 %

COMPUTER FILE: AISC 360-05 EX002
CONCLUSION
The results show an exact comparison with the independent results.

AISC 360-05 Example 002 - 2

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HAND CALCULATION
Properties:
Material: ASTM A572 Grade 50
E = 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
2(1.0)(8.0)3 (15.0)(0.25)3
+
=85.35 in 3
Iy =
12
12
Iy
85.4
ry =
=
= 2.08 in.
A
19.8
I x = ∑ Ad 2 + ∑ I x
(0.250)(15.0)3 2(8.0)(1.0)3
+
= 1095.65 in 4
12
12
t +t
1+1
d ' =−
d 1 2 =
17 −
=
16 in
2
2
Iy • d '2 (85.35)(16.0) 2
Cw =
=
= 5462.583 in 4
4
4
bt 3 2(8.0)(1.0)3 + (15.0)(0.250)3
=
J ∑
=
= 5.41 in 4
3
3
Member:
K = 1.0 for a pinned-pinned condition
L = 15 ft
I x = 2(8.0)(8.0) 2 +

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

AISC 360-05 Example 002 - 3

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Section Compactness:
Check for slender elements using Specification Section E7
Localized Buckling for Flange:
b 4.0
=
= 4.0
t 1.0
E
29000
=
= 0.38 = 9.152
λ p 0.38
Fy
50

λ=

λ < λ p , No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
h 15.0
=
= 60.0 ,
t 0.250
E
29000
=
λr 1.49
= 1.49 = 35.9
Fy
50

λ=

λ > λr , Localized web buckling
Web is Slender.
Section is Slender
Member Compression Capacity:
Elastic Flexural Buckling Stress
Since the unbraced length is the same for both axes, the y-y axis will govern by
inspection.
KL y
ry

Fe =

=

1.0(15 • 12 )
= 86.6
2.08

π 2E
 KL 


 r 

2

=

π 2 • 29000

(86.6)2

= 38.18 ksi

AISC 360-05 Example 002 - 4

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Elastic Critical Torsional Buckling Stress
Note: Torsional buckling will not govern if KLy > KLz, however, the check is included
here to illustrate the calculation.
 π 2 EC w
 1
Fe = 
+
GJ

2
 (K z L )
 Ix + Iy
 π 2 • 29000 • 5462.4

1
= 91.8 ksi > 38.18 ksi
=
+
•
11200
5.41
Fe 

2

 1100 + 85.4
(180 )

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, take f as Fcr with Q = 1.0
4.71

KLy
E
29000
=4.71
=113 >
=86.6
QFy
ry
1.0 ( 50 )

So
QFy
1.0( 50 )




f = Fcr = Q 0.658 Fe  Fy = 1.0 0.658 38.2  • 50 = 28.9 ksi





 0.34 E 
1 −
 ≤ b, where b = h
b
t
f
(
)


29000 
0.34
29000 
be = 1.92 ( 0.250 )
1 −
 ≤ 15.0in
28.9  (15.0 0.250 ) 28.9 
be 12.5in ≤ 15.0in
=
be = 1.92t

E
f

AISC 360-05 Example 002 - 5

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therefore compute Aeff with reduced effective web width.
Aeff =
betw + 2b f t f =
(12.5)( 0.250 ) + 2 (8.0 )(1.0 ) =19.1 in 2
where Aeff is effective area based on the reduced effective width of the web, be.
Aeff
19.1
=
= 0.968
A 19.75
Q Q=
=
=
(1.00 )( 0.968
) 0.968
s Qa
=
Qa

Critical Buckling Stress
Determine whether Specification Equation E7-2 or E7-3 applies
4.71

KLy
E
29000
= 4.71
= 115.4 >
= 86.6
QFy
ry
0.966 ( 50 )

Therefore, Specification Equation E7-2 applies.
When 4.71

E
KL
≥
QFy
r

QFy
1.0( 50 )




Fe
38.18
 Fy 0.966 0.658
Fcr Q 0.658
=
=
=
 • 50 28.47 ksi





Nominal Compressive Strength
Pn =Fcr Ag =28.5 • 19.75 =562.3kips

φc =0.90

φc P=
Fcr Ag= 0.90 ( 562.3=) 506.1kips > 420 kips
n
φc Pn =
506.1kips

AISC 360-05 Example 002 - 6



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