Takehome problems

A

Stability of Structures 3

1. Problem S1

Where:

2 2 2

π A π

π

P = E I P = + GJ

E Γ

P = E I y x θ

x y 2 l 2 l I 2 l

C

The determinant of the matrix, considering the fact that the coordinate y is null, is:

C

I C 2

− − − −

det M = (P P ) (P P ) (P P ) (P x ) = 0

x y θ C

A

The critical load is equal to min(P ; P ), where:

x 1

" #

r

k 4

2

∗ − −

P = P + P (P + P ) P P

y θ y θ y θ

2 k

In the following table we report the values of the critical load varying with the length of the beam.

Comparing the values of the axial stress with the admissible one σ we can observe that the beam is

0

stable if its length is greater than 1.98 m.

Table 1.2: Values of the critical load

∗

L [m] P [kN ] P [kN ] P [kN ] P [kN ] P [kN ] σ [M P a]

x y θ E

0.10 7.28e+06 2.96e+06 8.78e+05 8.08e+05 8.08e+05 8.98e+04

0.50 2.91e+05 1.18e+05 3.59e+04 3.30e+04 3.30e+04 3661.42

1.00 7.28e+04 2.96e+04 9572.17 8728.77 8728.77 969.86

1.50 3.24e+04 1.32e+04 4698.70 4234.38 4234.38 470.49

1.80 2.25e+04 9142.33 3507.41 3130.44 3130.44 347.83

1.90 2.02e+04 8205.30 3229.91 2872.18 2872.18 319.13

1.92 1.98e+04 8035.25 3179.55 2825.24 2825.24 313.92

1.94 1.93e+04 7870.43 3130.74 2779.72 2779.72 308.86

1.96 1.90e+04 7710.63 3083.42 2735.55 2735.55 303.95

1.97 1.88e+04 7632.55 3060.29 2713.97 2713.97 301.55

1.98 1.86e+04 7555.64 3037.52 2692.70 2692.70 299.19

2.00 1.82e+04 7405.29 2992.99 2651.10 2651.10 294.57

3.00 8090.33 3291.24 1774.63 1489.73 1489.73 165.53

5.00 2912.52 1184.85 1150.82 788.34 788.34 87.59

10.00 728.13 296.21 887.65 269.13 269.13 29.90

15.00 323.61 131.65 838.92 126.46 126.46 14.05

Looking at the graph below we immediately see that the beam has flexural-torsional buckling for any

∗

length, since the critical load always coincides with P .

Figure 1.3: Critical load varying with length of the beam

Stability of Structures 4

2

Problem S2

Determine the critical load and the critical mode of the following variable section beam

adopting both the FEM and the Ritz method, the latter by adopting a suitable polynomial

model continuous along the beam length; discuss and compare the results.

Figure 2.1: The beam and its degrees of freedom

Finite element method

The different flexural stiffness helps us in identifying the two finite elements: the first is the one with

stiffness 6 EI and the second EI. Both elements have 4 local DoFs, two vertical displacements and two

rotations, which are linked to the global DoFs by the connectivity matrix:

1 2 3 4

1 / 1 2 3

2 2 3 / 4

With the discretization of the finite element method we get a total potential energy equal to:

1

T −

∆V = U K P K U

E G

2

The stability condition translates into the fact that the stiffness matrix must be positive definite. There-

fore, the eulerian critical load can be computed as the minimum root for which the determinant of the

matrix is null. For a generic finite element, the stiffness matrices are equal to:

   

−12 −36

12 6 L 6 L 36 3 L 3 L

e e e e

2 2 2 2

E I 1

−6 −3

6 L 4 L L 2 L 3 L 4 L L L

e e e e e

(e) (e)

   

e e e e

= =

K K

   

−12 −6 −6 −36 −3 −3

L 12 L L 36 L

3

E G

L 30 L

e e e e

   

e

e 2 2 2 2

−6 −3

6 L 2 L L 4 L 3 L L L 4 L

e e e e

e e e e

Substituting the values of stiffness and length for the two elements we get:

   

−12 −144

12 3 L 3 L 144 6 L 6 L

2 2 2 2

EI 1

−3 −6

3 L L L L /2 6 L 4 L L L

(1) (1)

   

= 48 =

K

K   



−12 −3 −3 −144 −6 −6

L 12 L L 144 L

3

E G

L 60 L 

  

2 2 2 2

−3 −6

3 L L /2 L L 6 L L L 4 L

   

−12 −144

12 3 L 3 L 144 6 L 6 L

2 2 2 2

E I 1

−3 −6

3 L L L L /2 6 L 4 L L L

(2) (2)

   

K = 8 K =

   

−12 −3 −3 −144 −6 −6

L 12 L L 144 L

3

E G

L 60 L

   

2 2 2 2

−3 −6

3 L L /2 L L 6 L L L 4 L

Stability of Structures 5

2. Problem S2

With reference to the connectivity matrix the global stiffness matrices are:

(1) (1) (1)

 

K K K 0 2 2

 

−36

12 L L 6 L 0

22 23 24

(1) (1) (2) (1) (2) (2) EI −36 −30

L 168 L 6 L

K K + K K + K K

   

32 33 11 34 12 14

K = = 4

   

2 2 2

(1) (1) (2) (1) (2) (2) −30

6 L L 14 L L

3

E L

 

K K + K K + K K  

42 43 21 44 22 24

  2 2

0 6 L L 2 L

(2) (2) (2)

0 K K K

41 42 44

2 2

 

−6

4 L L L 0

1 −6 L 288 0 6 L

 

K =  

2 2 2

L 0 8 L L

G 60 L  

2 2

0 6 L L 4 L

The total stiffness matrix is: 2 2

 

− −6 − −

4 L (3 p) L (6 p) L (6 p) 0

EI −6 − − −30 −

L (6 p) 24 (7 12 p) L 6 L (1 p)

 

−

K P K =4  

2 2 2

− −30 − −

L (6 p) L 2 L (7 4 p) L (1 p)

3

E G L  

2 2

− − −

0 6 L (1 p) L (1 p) 2 L (1 2 p)

2

Where p is a dimensionless parameter equal to p = P L /(240 EI). Imposing the condition for stability

we get: 4 3 2

− −

det K = 675 p 3570 p + 4984 p 1932 p + 108 = 0

The solutions of the polynomial are:

≃ ≃ ≃ ≃ −→

p 0.0669 p 0.5222 p 1.3795 p 3.3204 p = 0.0669

1 2 3 4 E

Inverting the relationship we get the eulerian critical load for the beam:

EI

240 EI ≃ 16.0560

P = p

E E 2 2

L L

To compute the critical mode we impose the following condition:

2 2  

 

 

− −6 − − U

4 L (3 p ) L (6 p ) L (6 p ) 0 0

1

E E E

−6 − − −30 −

L (6 p ) 24 (7 12 p ) L 6 L (1 p ) U 0

E E E 2

 

 

 

−→

K U = 0 =

 

 

 

2 2 2

− −30 − −

L (6 p ) L 2 L (7 4 p ) L (1 p ) U 0

E E E 3

 

 

 

2 2

− − −

0 6 L (1 p ) L (1 p ) 2 L (1 2 p ) 0

U

E E E 4

Now we solve the linear system:

−28 p + 168 U U

E 2 2

≃

U = 2.2721

1 2 −

15 p 74 p + 78 L L

E

E

2 − 332 p + 132

90 p U U

E 2 2

E ≃

U = 1.5071

3 2 −

15 p 74 p + 78 L L

E

E

3 2

− − U U

720 p 3494 p + 3820 p 516

E 2 2

E E ≃ −4.0434

U =

4 3 2

−15 −

p + 89 p 152 p + 78 L L

E

E E

Once we know the entity of the displacements we can reconstruct the critical mode by means of the shape

functions: " #

3 2 3 2

x x x x x

−2 − − −

N (x) = +3 1 N (x) = L +2

2 e

1 L L L L L

e e e e e

" #

3 2 3 2

x x x x

− −

N (x) = 2 3 N (x) = L +

3 4 e

L L L L

e e e e

For the first element we have:

v (x) =N U = N (x) U + N (x) U + N (x) U =

E,1 2 1 3 2 4 3

3 2 3 2

x x x U x x

L

2

−8 − −

= +8 2 2.2721 + 16 12 U +

2

2 L L L L L L

3 2 3 2

L x x U x x x

2

−8 −

+ +4 1.5071 = 0.8834 + 0.1025 2.2721 U

2

2 L L L L L L

Stability of Structures 6

2. Problem S2

And for the second we have:

v (x) =N U = N (x) U + N (x) U + N (x) U =

E,2 1 2 2 3 4 4

3 2 3 2

x L x x U

x x

2

− −

−16 −8

+ 12 1 U + +8 2 1.5071

= +

2

L L 2 L L L L

3 2 3 2

L x x U x x x

2 −

− −8 −5.8545 − 1 U

+4 4.0434 = + 9.9414 1.5071 2

2 L L L L L L

Where the coordinate x is in between 0 and L/2.

Figure 2.2: Critical mode associated to the FEM discretization

As we can see from the graph above, the displacement in the beam is more or less symmetric, due to the

symmetric boundary conditions. Anyway, we immediately see that the left finite element is practically

undeformed while the right element shows an evident deformation. Also this characteristics is coherent

with the model, since the left part of the beam has a flexural stiffness six times the one of the right part.

Moreover, for this reason, the stationary point of the displacement function is shifted from the center

and located in the right part of the beam.

Ritz method

Given the boundary conditions, we choose the solution in the form of a sum of harmonics:

πx 2π x

∗

v (x) = A sin + B sin = N (x) A

L L

Substituting in the expression of the total potential energy we get:

Z Z

′′t ′′ ′t ′

K = EI(x) N N dx K = N N dx

E G

x x

Where: i

h 2 2

′ ′′

π π x 2π 2π x

π x 2π 2π x

π

−

cos cos

N = N = sin sin

L L L L L L L L

Since the stiffness is not constant all over the length of the beam, the elastic stiffness matrix will be the

sum of the two traits. 2 π x π x 2π x L 8 L

" # " #

L/2 sin 4 sin sin

Z

4 4

π π

L L L 4 3π

1

K = 6 EI = 6 EI

E L L

2

2π x π x 2π x 8 L

4 sin sin 16 sin 4 L

0 L L L 3π

2 πx π x 2π x L 8 L

" # " #

−

L sin 4 sin sin

Z

4 4

π π

L L L 4 3π

2

K = EI = EI

E L L

2

2π x π x 2π x 8 L

−

4 sin sin 16 sin 4 L

L/2 L L L 3π

Stability of Structures 7

2. Problem S2 2

7 40 L

" #

2

L

4

π

EI 4 3π

1 2

K = K + K =

E E E L L 2

40 L 2

28 L

3π

πx π x 2π x L

" "

# #

2

l cos 2 cos cos 0

Z

2 2

π

π

L L L 2

=

K =