Contents IS 800 2007 Example 001

User Manual: IS 800-2007 Example 001

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IS 800-2007 Example 001
WIDE FLANGE MEMBER UNDER COMPRESSION
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example.
A continuous column is subjected to factored load N = 1 kN. This example was
tested using the Indian IS 800:2007 steel frame design code. The design
capacities are compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING

NEd

L
A

A

Section A-A
L = 3m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923 MPa

Loading
N = 1 kN

Design Properties
fy = 250 MPa
fu = 410 MPa
Section: ISMB 350

TECHNICAL FEATURES TESTED
 Section compactness check (column)
 Member compression capacity

IS 800-2007 Example 001 - 1

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RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-IS-800-2007.pdf,” which is
available through the program “Help” menu. The example was taken from
Example 9.2 on pp. 765-766 in “Design of Steel Structures” by N. Subramanian.

ETABS

Independent

Percent
Difference

Compactness

Plastic

Plastic

0.00%

Design Axial Strength, Ncrd

733.85

734.07

-0.03%

Output Parameter

COMPUTER FILE: IS 800-2007 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.

IS 800-2007 Example 001 - 2

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HAND CALCULATION
Properties:
Material: Fe 250
E = 200,000 MPa
fy = 250 MPa
Section: ISMB 350
A = 6670 mm2
b = 140 mm, tf = 14.2 mm, d = 350 mm, tw = 8.1 mm, r = 1.8 mm
h =d − 2 ( t f + r ) =350 − 2 (14.2 + 1.8 ) =318 mm

ry = 28.4 mm, rz = 143 mm
Member:
KLy = KLz = 3,000 mm (unbraced length)
γM 0 =
1.1
Loadings:
N Ed = 1 kN
Section Compactness:
=
ε

250
=
fy

250
= 1
250

Localized Buckling for Flange:
λ=
8.4ε= 8.4 •=
1 8.4
p

λe =

b
70
=
= 4.93
t f 14.2

=
λ e 4.93 < λ
=
8.40
p

So Flange is Plastic in compression

IS 800-2007 Example 001 - 3

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Localized Buckling for Web:
λ p= N / A & λ s= 42ε= 42 for compression

λe =

d 318
=
= 39.26
tw 8.1

=
λ e 39.26 < =
λ s 42
So Web is Plastic in compression
Since Flange & Web are Plastic, Section is Plastic.
Member Compression Capacity:
Non-Dimensional Slenderness Ratio:

h 350
= = 2.5 > 1.2
b f 140
and
=
t f 14.2 mm < 40 mm
So we should use the Buckling Curve ‘a’ for the z-z axis and Buckling Curve ‘b’
for the y-y axis (IS 7.1.1, 7.1.2.1, Table 7).
Z-Z Axis Parameters:
For buckling curve a, α =0.21 (IS 7.1.1, 7.1.2.1, Table 7)
Euler Buckling Stress:
=
f cc

=
λz

fy
=
f cc

π2 E
π2 200, 000
=
=
4485 MPa
2
2
 K z Lz 
 3, 000 




 143 
 rz 

250
= 0.2361
4485

IS 800-2007 Example 001 - 4

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=
φ 0.5 1 + α ( λ − 0.2 ) + =
λ 2  0.5 1 + 0.21( 0.2361 − 0.2 ) + 0.23612 
φ =0.532

Stress Reduction
Factor: χ
=

1
1
=
= 0.9920
2
2
φ+ φ −λ
0.532 + 0.5322 − 0.23612

fy
250
f cd , z =
χ
=
0.992 •
=
255.5 MPa
γM 0
1.1
Y-Y Axis Parameters:
For buckling curve b, α =0.34 (IS 7.1.1, 7.1.2.1, Table 7)
Euler Buckling Stress:
=
f cc

λ
=
y

fy
=
f cc

π2 E
π2 200, 000
=
=
177 MPa
2
2
 K z Lz 
 3, 000 




 28.4 
 rz 

250
= 1.189
177

=
φ 0.5 1 + α ( λ − 0.2 ) + =
λ 2  0.5 1 + 0.34 (1.189 − 0.2 ) + 1.1892 

φ =1.375

Stress Reduction
Factor: χ
=

1
1
=
= 0.4842
2
2
φ+ φ −λ
1.375 + 1.3752 − 1.1892

fy
250
f cd , y =
χ
=
=
0.4842 •
110.1MPa Governs
γM 0
1.1
=
Pd Af=
6670 • 110.1
cd , y

Pd = 734.07 kN

IS 800-2007 Example 001 - 5
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