Exercise 5 Solution NTNU Norwegian University of Science and

NTNU
Exercise 5
Norwegian University of Science
and Technology.
Faculty of Engineering Science
Department of Marine Technology.
Solution
TMR4205 Buckling and Collapse of Structures
Buckling of module truss-work
________________________________________________________________________________________________
Date: March 2008
Signature: JAm
Distributed
Due Date:
The given column can be idealised as shown in Figure 1 below. The weak and strong
axis of the cross-section are also indicated in the figure. Since the element’s
slenderness, , uniquely defines the Euler buckling stress, the governing buckling
mode can be determined by examining the slenderness ratios; the highest slenderness
gives the lowest buckling stress. Therefore, it is natural to design using the highest
slenderness ratio.
y-y
Strong axis (x-x)
L
x-x
Weak axis (y-y)
L/3
Figure 1.
The following calculations has been performed using a simple spreadsheet algorithim
in EXCEL.
a) Design using hot rolled IPE profile.
If we randomly start with IPE240 and neglect the curved shape at the corners, we
obtain the following cross-sectional properties:A  3718.5 mm 2
I x  3.67  10 7 mm 4
I y  2.83  10 6 mm 4


 i


I
A
i x  99.4 mm
 
i y  27.6 mm
The effective buckling length is given by,
lk  k  L

 lkx  1  9000  9000 mm

lky  1  9000 3  3000 mm
where the effective length factor k = 1 has been used for a simple supported
column.
The slenderness ratio is given by,
1

lk
i

 x  90.6

y  109.0
y  x
This result implies that buckling about the y-axis (weak-axis) is the critical one.
Taking the Young’s modulus for steel E = 2.1105 MPa, the Euler buckling stress can
is calculated as,
 2E
 E  2  175MPa
y

 
Y
240

 117
.
E
175
From Figure 4.10 in the compendium or Figure 2.3 (The column selection chart) in
the classification note 30.1, we select column curve b for I and H hot rolled sections,
i.e.,
h b  240 120  2.0  12
. 
  column curve b
Weak  axis buckling

and from Table 4-2 in the compendium we obtain for, column curve b;
column curve b 
  35
o  0.2
Then we proceed as follows,
     o   0.34
 cr   Y 
1    2 
1     
2 2
2 2
 4 2
 118 MPa
The column capacity and the acting load on the column (design load) are given
respectively as,
c 
 cr 118

 102 MPa
 m 115
.
N 390000
d 

 105 MPa
A
3718.5


  d  c


not OK
Since the capacity of the column is less than the design load, we conclude that IPE240
can not carry the given load. The difference in values here is small, and it is likely
that taking the true values (considering the curved corners) for the area and other
cross-sectional properties could make IPE240 satisfy the design criteria. However,
since we have considered the idealized case (sharp corner shape) we should,
theoretically, repeat the design check using the nearest larger cross-section, or recheck
the existing IPE240 by using the true cross-sectional parameters.
The design was repeated using IPE270 and the requirement was satisfied in a way
that,
2
c 
 cr 135.4

 118 MPa
m
115
.
N 390000
d 

 88.6 MPa
A
4401


  d  c


OK
1,6E+02
1,4E+02
stress
critical stress
1,2E+02
1,0E+02
acting stress
8,0E+01
6,0E+01
3,0E+03
4,0E+03
5,0E+03
6,0E+03
area
Figure 2. Acting stress and Critical stress as a function of Area.
b) Design using Aluminium profile
If the aluminium profile is to be used instead of steel, the same calculations as
done in part (a) but using the profiles in Table 2 with aluminium material
properties yield the following results for I300;
c 
 cr 62.34

 56.7 MPa
m
110
.
N 390000
d 

 49.0 MPa
A
7964


  d  c


OK
The weight ratio between aluminium and steel sections which satisfy the design
criteria is calculated from,
r
a Aa 2700  7964

 0.622
s As 7860  4401
oo0O§O0oo
3

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