Takehome

E

() = 0. The values that compose the equation are:

2 2 2

= ∗ ( ) = ∗ ( ) = ∗ ∗ ( ) + ];

[Γ

- ; ;

Considering the symmetry (y = 0), one can simplify the equation, obtaining:

C

2

( ( ( )

() = − ) ∗ ∗ − ) ∗ − ) − ∗ = 0

[ ( ]

The roots of the equation are: =

1

4

2

√(

= ∗ + ± + − ∗ ∗

[( ) ) ]

2,3

2

{

∗

},

= min{ , , = min { , = ∗ [( +

Knowing that one obtains:

1 2 3

2

2 4

√(

− + − ∗ ∗ ].

) ) Considering this, it is possible to evaluate the critical

load as a function of the length of the structure. Such evaluation is presented on the

following table, in which the stability of the structure is analyzed based on the admissible



stress = 200 N/mm².

0  (M

P (kN) P (kN) Pa)

L (m) P (kN) P (kN) P* (kN) Condition

q E

x y

0.10 3.11E+07 2.63E+07 2.91E+06 2.68E+06 2.68E+06 1.67E+05 Unstable

0.25 4.97E+06 4.21E+06 4.67E+05 4.30E+05 4.30E+05 2.69E+04 Unstable

0.50 1.24E+06 1.05E+06 1.18E+05 1.09E+05 1.09E+05 6.80E+03 Unstable

0.75 5.53E+05 4.68E+05 5.38E+04 4.93E+04 4.93E+04 3.08E+03 Unstable

1.00 3.11E+05 2.63E+05 3.12E+04 2.85E+04 2.85E+04 1.78E+03 Unstable

1.25 1.99E+05 1.68E+05 2.08E+04 1.89E+04 1.89E+04 1.18E+03 Unstable

1.50 1.38E+05 1.17E+05 1.51E+04 1.37E+04 1.37E+04 8.55E+02 Unstable

1.75 1.02E+05 8.59E+04 1.16E+04 1.05E+04 1.05E+04 6.57E+02 Unstable

2.00 7.77E+04 6.58E+04 9.42E+03 8.46E+03 8.46E+03 5.28E+02 Unstable

2.25 6.14E+04 5.20E+04 7.89E+03 7.04E+03 7.04E+03 4.40E+02 Unstable

2.50 4.97E+04 4.21E+04 6.80E+03 6.02E+03 6.02E+03 3.77E+02 Unstable

2.75 4.11E+04 3.48E+04 5.99E+03 5.27E+03 5.27E+03 3.29E+02 Unstable

3.00 3.45E+04 2.92E+04 5.38E+03 4.69E+03 4.69E+03 2.93E+02 Unstable

3.25 2.94E+04 2.49E+04 4.90E+03 4.23E+03 4.23E+03 2.65E+02 Unstable

3.50 2.54E+04 2.15E+04 4.52E+03 3.87E+03 3.87E+03 2.42E+02 Unstable

3.75 2.21E+04 1.87E+04 4.22E+03 3.57E+03 3.57E+03 2.23E+02 Unstable

4.00 1.94E+04 1.64E+04 3.97E+03 3.32E+03 3.32E+03 2.07E+02 Unstable

4.10 1.85E+04 1.57E+04 3.88E+03 3.23E+03 3.23E+03 2.02E+02 Unstable

4.11 1.84E+04 1.56E+04 3.87E+03 3.22E+03 3.22E+03 2.01E+02 Unstable

4.12 1.83E+04 1.55E+04 3.86E+03 3.21E+03 3.21E+03 2.01E+02 Unstable

4.13 1.82E+04 1.54E+04 3.85E+03 3.21E+03 3.21E+03 2.00E+02 Unstable

4.14 1.81E+04 1.54E+04 3.84E+03 3.20E+03 3.20E+03 2.00E+02 Stable

4.25 1.72E+04 1.46E+04 3.76E+03 3.11E+03 3.11E+03 1.94E+02 Stable

4.50 1.54E+04 1.30E+04 3.58E+03 2.93E+03 2.93E+03 1.83E+02 Stable

4.75 1.38E+04 1.17E+04 3.44E+03 2.77E+03 2.77E+03 1.73E+02 Stable

5.00 1.24E+04 1.05E+04 3.31E+03 2.63E+03 2.63E+03 1.65E+02 Stable

Observing the table, which was more discretized in the region of transition from

instability to stability, one can conclude that the minimum length of the structure in order

to obtain stability is 4.14 meters. It is also noticeable that the Eulerian critical load

∗

always coincides with , since its values are inferior to the values of . A plot of the

∗

values of , , and as functions of the length of the structure was made to make

the previous conclusion more visible.

1,50E+04

1,25E+04

1,00E+04

(kN) 7,50E+03

P 5,00E+03

2,50E+03

0,00E+00 0,00 5,00 10,00 15,00 20,00 25,00

L (m)

Px (kN) Py (kN) Ptheta (kN) P* (kN)

Problem S2

For the variable section beam depicted below, the student must evaluate the

critical load and the associated critical mode by adopting two cubic finite elements and

critically discuss the results.

Initial considerations

For the solution of the problem, the following finite elements and degrees of

freedom were proposed:

Global system:

Local system:

The degrees of freedom of each element are represented by the vector

{ }

= 3 4

1 2

The incidence vectors, that connect global and local degrees of freedom, for the

problem will then be: [0 ]

= 0 1 2

[1 ]

= 2 0 0

Considering cubic finite elements, the local displacement model can be written

() () () () ()

= ∗ + ∗ + ∗ + ∗ = () ∗

as: , with the

1 1 2 2 3 3 4 4

following shape functions: 2 3

()

= 1 − 3 ∗ ( ) + 2 ∗ ( )

1

2 3

()

= ∗ [( ) − 2 ∗ ( ) + ( ) ]

2

2 3

()

= 3 ∗ ( ) − 2 ∗ ( )

3

2 3

()

= ∗ [− ( ) + ( ) ]

4

{

The total potential energy is described by the following expression:

1 2 2

′′ ′

( ) ()

= ∗ [∫ ∗ − ∫ ∗ ]

2,

2 0 0

1 2 2

′′ ′

[() ] [() ]

()

= ∗ [∫ ∗ ∗ − ∫ ∗ ∗ ]

2 0 0

1 ( )

= ∗ ∗ − ∗ ∗

2

With the previous shape functions, one obtains:

12 6 ∗ −12 6 ∗

2 2

6 ∗ 4 ∗ −6 ∗ 2 ∗

= ∗

−12 −6 ∗ 12 −6 ∗

3

2 2

6 ∗ 2 ∗ −6 ∗ 4 ∗

[ ]

36 3 ∗ −36 3 ∗

2 2

1 3 ∗ 4 ∗ −3 ∗

= ∗

−36 −3 ∗ 36 −3 ∗

30 ∗

2 2

3 ∗ −3 ∗ 4 ∗

[ ]

Assembling procedure

Connecting the local degrees of freedom to the global ones, the global stiffness

matrix will assume the following format:

(1) (2) (1) (2)

+ +

33 11 34 12

= [ ]

(1) (2) (1) (2)

+ +

43 21 44 22

= = = 5 ∗ ; = ,

Considering that and one obtains:

1 2 1 2

(−6

5 ∗ 12 + 12 5 ∗ ∗ ) + 6 ∗ 72 −24 ∗

= ∗[ = ∗

] [ ]

2

2 2 3

(−6 −24 ∗ 24 ∗

5 ∗ ∗ ) + 6 ∗ 5∗4∗ +4∗

³ 1 1

36 + 36 −3 ∗ + 3 ∗ 72 0

= ∗[ = ∗[

] ]

2 2 2

−3 ∗ + 3 ∗ 4∗ +4∗ 0 8 ∗

30 ∗ 30 ∗

The total potential energy for the global system can be obtained summing the

contributions of each finite element, as presented next.

1 1

() ( ) ( )

= ∗ ∑ ∗ − ∗ ∗ = ∗ ∗ − ∗ ∗

2

2 2

2

∗

=

The parameter is proposed so one can obtain the global stiffness matrix

30∗

= − ∗ .

∗ 30 ∗ 1

72 −24 ∗ 72 0

= ∗[ ]− ∗ ∗[ ]

2 2

3 2

−24 ∗ 24 ∗ 0 8 ∗

30 ∗

72 −24 ∗ 72 0

= ∗ ]−∗[

{[ ]}

2 2

3 −24 ∗ 24 ∗ 0 8 ∗

72 ∗ (1 − ) −24 ∗

= ∗[ ]

2

3 −24 ∗ 8 ∗ ∗ (3 − )

Critical load

Considering the Energy Method, the condition of instability, and therefore the

critical load, is associated with a null second variation of the total potential energy, and

det() = 0.

therefore can be determined by the analysis of the condition

2

2 2

[72 ]

(1 (3 (−24

det() = ( ) ∗ ∗ − ) ∗ 8 ∗ ∗ − ) − ∗ ) = 0

3

(1 (3

− ) ∗ − ) − 1 = 0

= 0.5858

1

2

− 4 ∗ + 2 = 0 → {

= 3.4142

2

=

Since must be the minimum of the two computed values, one obtains

= 0.5858, and therefore:

1 0.5858 ∗ 30 ∗ 17.5736 ∗

= =

2 2

Critical mode

As the only load applied on the system is an axial force, the analysis of the

displacements and, with it, of the critical mode can be done by proposing the following

system: (1

72 ∗ − ) −24 ∗ 0

1

∗= → ∗[ ∗ =

] [ ] [ ]

2

3 (3 0

−24 ∗ 8 ∗ ∗ − )

2

(1

72 ∗ − ) ∗ − 24 ∗ ∗ = 0

1 2

→{ 2 (3

−24 ∗ + 8 ∗ ∗ − ) = 0

1

= 0.5858,

Using the computed value one obtains from the first line of the

previous system: 29.823 ∗ = 24 ∗ ∗ → = 0.805 ∗ ∗

1 2 1 2

The same result can be obtained from the second line of the system. Considering

the result, one obtains the expressions for the transversal displacements o