NTNU
Exercise
Norwegian University of Science
and Technology.
Faculty of Marine Technology
Department of Marine Structures.
Solution
TMR4205 Buckling and Collapse of Structures
Shape factor and plastic analysis of beam
________________________________________________________________________________________________
Date:January 2008
Signature: JAm
Distributed Date:
Due Date:
Problem One
a) The stress distribution over the cross-section is sketched in Figure 1. Since this is
an unsymmetrical cross-section, it is important to note that the position of the
neutral axis in a fully plastic state will not be the same as at first yield state. The
neutral axis will move towards the side of smaller stress value.
Y
Y
Y
Y
N.A
Y
Y
Y
Top fibre yield
Bottom fibre yield
Elasto-plastic
Y
Fully Platic
Figure 1. Stress Distributions.
A plastic hinge can be introduced when a fully plastic state has been reached. This
is simply because at that point the section can not carry any more moment.
b) I order to obtain the elastic and plastic section modulus, we need to calculate the
neutral axis for the elastic and fully plastic state. With reference to Figure 2, the
neutral axis at the elastic state can be calculated as follows.
ye 
 240  24  12   240  24  144  160  16  272
 113.3  113mm
 240  24  240  24  160  16
On the other hand, the neutral axis at fully plastic state, is calculated such that the
area between the two sides of the cross-section (compression and tension sides) are
equal. Therefore,





160  16  240  y p  24  24  y p  24  24  240  24
 y p  77.33  77 mm
1
160
16
24
N.A, elastic state
240
N.A, plastic state
ye
yp
24
240
Dimensions in mm
Figure 2. The Neutral Axis at Elastic and Plastic States.
The elastic section modulus is given by,
W
I nn
y
(1)
where Inn is the moment of inertia about the neutral axis, and y is the longest distance
perpendicular to the neutral axis which in this case is obviously to be towards the
upper flange. By using Figure 2, we obtain,
 1
  1

I nn    240  24 3  240  24  1012     24  240 3  24  240  312 
 12
  12

 1

   160  16 3  160  16  159 2 
 12

 157
.  108 mm 4
157
.  108
W
 9.4  105 mm 3
280  y e
By definition, the plastic section modulus is given by,
Z
Mp
Y
(2)
where Y is the yield stress and Mp is the plastic moment. From Figure 3, we can
write
M p  P y  y'
(3)
where,
2
 24  187  935
.   160  16  195
 130.4  130mm
 24  187  160  16
 24  53  26.5   240  24  65
y' 
 58.0mm
 24  53  240  24
y
P
Y A
2
Therefore,
Z
Mp
Y


Y
2
160  16  24  240  240  24  7040 Y
P y  y '
Y

7040 Y 130  58
Y
 13.2  105 mm 3
Y
160
16
24
P
y
240
77
y'
24
P
Y
240
Dimensions in mm
Figure 3. Plastic Section Modulus.
c) The shape factor is given by,

Z
 1.4
W
This is a higher value as compared to those of typical I-profiles given in the
Søreide’s book, (i.e 1.1 It can be inferred that the unsymmetrical I-profiles
have higher shape factors compared to symmetric ones. Considering the given
profile in this exercise, the factor will decrease with increasing size of the top
flange until symmetry has been reached.
d) The shape factor of a cross-section represents the reserve strength of that section
after first yield.
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Problem Two
a) There are three possible collapse mechanisms for the given structure as shown in
Figure 4. The associated collapse loads can be calculated from the principle of
virtual work, Equation (4).
We  Wi
 We  Wi
P
(4)
0.5P
Loaded Structure
0.5Mp
Mp
2L/3
L/3
L/2
L/2
2L
w
Mechanism I



w
Mechanism II


w2
Mechanism III
w1

Figure 4. Possible Collapse Mechanism.
(i) Mechanism I:
Pw  M p  3   0.5 M p  2 
P

3w
2L
6M p
L
(ii) Mechanism II:
0.5 Pw  0.5 M p  0.5 M p  2 
P

2w
L
6M p
L
(iii) Mechanism III:
4
Pw1  0.5 Pw2  M p  3   0.5 M p  4 
P

3w1 w2

2L
L
60 M p
L
b) From the above results, it is natural to select the true collapse load from
mechanisms I and II as,
Pc 
6M p
L
Drawing the moment diagram will prove that this is the true collapse load if the
moment values do not exceed the plastic moment capacity on any span. That is,
M  Mp
for 0  x  L
M  0.5 M p
for
L  x  2L
Consider mechanism I and assume that the moment diagram has the shape shown
in Figure 5. Then,
R1 
R2 
Mp
2L 3
M
p

3 Mp
2 L
 0.5 M p
L3
  0.5 M
p
 xM p
L2
  11 M
p
2 L
 2x
Mp
L
The moment equilibrium of the left span (about point A) yields,
0.5 M p  3
Mp L
  R3  L  0
L 2
 R3 
Mp
L
and the moment equilibrium of span BC (about point B) yields,
xM p  R3 
L
2

x  0.5
. Pc
Static equilibrium is satisfied by R1  R2  R3  15
5
3Mp/L
6Mp/L
C
A
R1
B
R2
R3
0.5 Mp
x Mp
Mp
Figure 5. Moment Diagram.
6