Contents NTC 2008 Example 001

User Manual: NTC 2008 Example 001

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NTC 2008 Example 001
WIDE FLANGE SECTION UNDER COMBINED COMPRESSION & BENDING
EXAMPLE DESCRIPTION
In this example a continuous beam-column is subjected to factored axial load P =
1400 kN and major-axis bending moment M = 200 kN-m. The beam is
continuously braced to avoid any buckling effects. This example was tested using
the Italian NTC-2008 steel frame design code. The design capacities are
compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING

A

M
P

A

L

Section A-A

L = 0.4 m
Material Properties
E = 210x103 MPa
v = 0.3
G = 80769 MPa

Loading
P = 1400 kN
M = 200 kN-m

Design Properties
fy = 235 MPa
Section: 457x191x98 UB

TECHNICAL FEATURES TESTED
 Section compactness check (beam-column)
 Section compression capacity
 Section shear capacity
 Section bending capacity with compression & shear reductions
 Interaction capacity, D/C

NTC 2008 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-NTC-2008.pdf,” which is
available through the program “Help” menu.

Output Parameter
Compactness
Section Compression Resistance,
Nc,Rd (kN)
Section Shear Resistance,
Vc,Rd,y (kN)
Section Plastic Bending Resistance,
Mc,y,Rd (kN-m)
Section Bending Resistance Axially
Reduced,
MN,y,Rd (kN-m)
Section Bending Resistance Shear
Reduced,
MV,y,Rd (kN-m)
Interaction Capacity, D/C

SAP2000

Independent

Percent
Difference

Class 2

Class 2

0.00%

2797.6

2797.6

0.00%

719.2

719.2

0.00%

499.1

499.1

0.00%

310.8

310.8

0.00%

481.3

481.3

0.00%

0.741

0.741

0.00%

COMPUTER FILE: NTC 2008 EX001
CONCLUSION
The results show an exact comparison with the independent results.

NTC 2008 Example 001 - 2

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HAND CALCULATION
Properties:
Material: S275 Steel
E = 210000 MPa
fy = 235 MPa
Section: 457x191x98 UB
A = 12,500 mm2
b = 192.8 mm, tf = 19.6 mm, h = 467.2 mm, tw = 11.4 mm, r = 10.2 mm

hw  h  2t f  467.2  2 19.6  428mm

d  h  2  t f  r   467.2  2  19.6  10.2  407.6mm
c

b  tw  2r 192.8  11.4  2  10.2

 80.5 mm
2
2

Wpl,y = 2,230,000 mm3
Other:
 M 0  1.05

Loadings:

P  1400kN axial load

M y  200 kN-m bending load at one end
Results in the following internal forces:

NEd  1400 kN
VEd  500 kN

M y , Ed  200 kN-m

NTC 2008 Example 001 - 3

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Section Compactness:

235
235

1
fy
235



1   


N Ed
1
1 
2  2htw f y


  1


1
1, 400, 000

1 
  2.7818  1, so
2  2  467.2  11.4  235 

  1.0

Localized Buckling for Flange:
For the tip in compression under combined bending & compression
 cl .1 

e 

9 9  1

9

1

c 80.5

 4.11
t f 19.6

e  4.11 cl .1  9
So Flange is Class 1 in combined bending and compression
Localized Buckling for Web:

  0.5, so
 cl .1 

e 

396
396  1

 33.00 for combined bending & compression
13  1 13  1  1

d 407.6

 35.75
tw 11.4

e  35.75  cl .1  33.00
 cl .2 

456
456  1

 38.00
13  1 13  1  1

e  35.75  cl.2  38.00

NTC 2008 Example 001 - 4

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So Web is Class 2 in combined bending & compression.
Since Web is Class 2, Section is Class 2 in combined bending & compression.
Section Compression Capacity
N c , Rd  N pl , Rd 

Af y
M 0



12,500  235
1.05

Nc, Rd  2797.6 kN

Section Shear Capacity
AV , y  A  2bt f  t f  tw  2r   12,500  2  192.8  19.6  19.6 11.4  2  10.2 
AV , y  5,565.5mm2
Vc , Rd , y 

fy
M 0 3

Avy 

235
 5,565.5
1.05 3

Vc, Rd , y  719.2 kN
  1.0

h 467.1
72 235 72 235

 37.5 

 72
tw 11.4

fy
1.0 235
So no shear buckling needs to be checked.
Section Bending Capacity
M c , y , Rd  M pl , y , Rd 

Wpl , y f y
M 0



2, 230, 000  235
1.05

M c, y , Rd  499.1kN-m

NTC 2008 Example 001 - 5

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with Shear Reduction

VEd  500kN  0.5  Vc, Rd  359.6kN Shear Reduction is needed
Av  htw  467.2 11.4  4,879.2mm2
2

 2V
  2  500 2
   Ed  1  
 1  0.1525
V
  719.2

c
,
Rd



M y ,V , Rd


Av 2 
W

 pl , y 4t  f yk
w 


M 0


0.1525  4879.22 
2,
230,
000


  235
4  11.4


 M y ,c , Rd
1.05

MV ,r , Rd  481.3kN-m
with Compression Reduction

n

N Ed
1400

 0.50
N pl , Rd 2797.6
A  2bt f

12,500  2 192.8 19.6
 0.40  0.5
A
12,500
1 n
1  0.5
M N , y , Rd  M pl , y , Rd
 499.1 
1  0.5a
1  0.5  0.4

a



M N , y , Rd  310.8kN-m

Interaction Capacity: Compression & Bending
Section Bending & Compression Capacity
Formula NTC 4.2.39
2

D  M y , Ed   M z , Ed

 
C  M N , y , Rd   M N , z , Rd

5n

2

M y , Ed
 200 


0

0.414

 0.644

 310.8 
M N , y , Rd




D
 0.644
C

NTC 2008 Example 001 - 6

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Member Bending & Compression Capacity: Method B
k factors used are taken from the software, and determined from Method 2 in
Annex B of Eurocode 3.

k yy  0.540
k yz  0.420

kzy  0.600

kzz  0.700
Formula NTC 4.2.37
M y , Ed
M z , Ed
N Ed
D

 k yy
 k yz
W f
Wpl , z f yk
C  y Af yk
 LT pl , y yk
M1
M1
M1
D
1, 400
200

 0.54 
0
2, 230, 000  235
C 112,500  235
1
1.05
1.05
D
 0.5  0.22  0
C
D
 0.717
C

Formula NTC 4.2.38
M y , Ed
M z , Ed
N Ed
D

 k zy
 k zz
W f
Wpl , z f yk
C  z Af yk
 LT pl , y yk
M1
M1
M1
D
1, 400
200

 0.600 
0
2, 230, 000  235
C 112,500  235
1
1.05
1.05
D
 0.5  0.24  0
C

NTC 2008 Example 001 - 7

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D
 0.741 (Governs)
C

NTC 2008 Example 001 - 8
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Title                           : Contents
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