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BITS Pilani
Hyderabad Campus
Eigenvalue and Time
Dependent Problems
Prof. Srinivasa Prakash Regalla
BITS Pilani, Hyderabad Campus
 Eigenvalue Problems
Formulation of eigenvalue problems for parabolic,
hyperbolic equations
Finite element formulation
Natural vibration of a beam
Euler beam
Timoshenko beam
Stability and buckling of beams
 Time dependent problems
Learning Outcomes
BITS Pilani, Hyderabad Campus
It is posing of a BV/IV/BIV problem in a form:
What is an eigenvalue
problem?
)
(
)
( u
B
u
A 

and seeking a solution for eigenvalues  through it for non-
trivial eigenvectors, u.
The  and u have different physical meanings for different
mechanical problems. For example, the natural axial
vibration of a bar are transverse vibration of a cable are
represented by:
and here  is the square of the natural frequency of vibration, .
BITS Pilani, Hyderabad Campus
Different meanings of
eigenvalues in different problems
Problem Area Physical meaning of 
Structural
deformation/stress problems
Square of natural
frequency of vibration
Buckling problems Buckling load
Heat transfer Amplitude of the Fourier
components
Fluid mechanics Amplitude of the Fourier
components
BITS Pilani, Hyderabad Campus
One example application is the 1-D transient heat transfer:
Formulation of eigenvalue
problems: parabolic equation
Here:
u = temperature
k = thermal conductivity
 = density
c = specific heat
A = cross-sectional area
q = heat generation rate per unit length
(1)
BITS Pilani, Hyderabad Campus
Proposing the homogeneous solution as product of two
functions:
Substituting into the homogeneous form of diff. equ.:
Separation of variables gives:
Equating the two terms to a single constant gives:
The minus sign signifies
that the Temperature
decreases with time.
BITS Pilani, Hyderabad Campus
Writing it as two equations:
(2)
(3)
Considering equation (2) and integrating:
t
Ke
T
K
t
T
dt
T
dT
dt
T
dT 


 










 
 )
ln(
)
ln(
For the complete solution to the governing equation (1), we need
solution to (3) also. Solving equation (3) for  (eigenvalues) and
U(x) (eigenvectors) is called as eigenvalue problem.
BITS Pilani, Hyderabad Campus
For constant A, c, k and , the general solution of (2) is:
Substitution of boundary conditions:
0
cos
0
sin
cos
1
0




L
L
L






 cannot be zero hence cosL has to be equal to zero.
BITS Pilani, Hyderabad Campus
For nontrivial solution, the determinant of the coefficient matrix
zero:


 .
,
2
5
,
2
3
,
2
etc
L




The constant K is absorbed into Cn. The constants Cn can be
determined from the initial condition:
If we multiply by sinmx both sides and integrate over the
domain (0,L):

  









L
m
n
n
L
m
n
n
dx
x
u
dx
x
x
C
0
0
1 0
sin
sin
sin 


BITS Pilani, Hyderabad Campus
For the non-trivial condition of m=n, we obtain:
Notice the orthogonality condition:


L
m
n
dx
x
x
u
L
C
0
0
sin
)
(
2

That way we complete the homogeneous solution:
The complete solution of equation (1) can then be obtained
as sum of homogeneous solution and particular solution:
BITS Pilani, Hyderabad Campus
The homogeneous solution and separation of variables gives
Formulation of eigenvalue
problems: hyperbolic equation
BITS Pilani, Hyderabad Campus
BITS Pilani, Hyderabad Campus
Natural Vibration of a Beam – Euler-
Bernoulli Beam Theory (EBT)
For EBT, the governing equation is:
Two example boundary
conditions:
(1) Both ends fixed
(2) One end fixed and the
other end spring
supported
BITS Pilani, Hyderabad Campus
w(x, t) = frequency of natural vibrations
W(x) = Mode shapes
Substituting:
Natural Vibration of a Beam – Euler-
Bernoulli Beam Theory (EBT)
For EBT, the governing equation is:
Where =2
BITS Pilani, Hyderabad Campus
j = Hermite cubic polynomial functions
Week Formulation
 = weight function
BITS Pilani, Hyderabad Campus
The Hermite Shape Functions
BITS Pilani, Hyderabad Campus
Hermite Shape Functions
Using Local Coordinates
The interpolation functions i can be expressed in terms of local
coordinate 𝑥 = 𝑥 − 𝑥𝑒 as:
BITS Pilani, Hyderabad Campus
Elemental level FE Equation
BITS Pilani, Hyderabad Campus
For constant EI and A the stiffness and mass matrices are:
Elemental Matrices
BITS Pilani, Hyderabad Campus
EBT
Example: Cantilever Beam
TBT
How many elements should we consider?
Once the type of element is decided by the weak form, the number of elements is decided by
the number of eigenvalues we want to determine. The number of eigenvalues, and hence
number of natural frequencies is equal to the net number of unconstrained degrees of
freedom in the meshed system.
BITS Pilani, Hyderabad Campus
Example:Cantilever Beam (continued): One 2-nodedlinear1-D
element(with Hermite ShapeFunctions)for EBT:
EBCs
NBCs:
Q3 = Q4 = 0
BITS Pilani, Hyderabad Campus
Setting the determinant of the above coefficient matrix, we
obtain the characteristic equation and solving it we get
eigenvalues.
Two cases can be examined:
(1) Neglecting the inertia forces
(2) Taking inertia forces into account
The characteristic equation
BITS Pilani, Hyderabad Campus
If we write 𝜔𝑖 = 𝜔𝑖𝐿2 𝜌𝐴 𝐸𝐼, the exact solution is
𝜔1 = 3.516 and 𝜔2 = 22.035 comparing which with above
values, it is clear that translational vibration frequency is
accurate but rotational vibration frequency is not accurate.
Characteristic equation neglecting
inertia forces – Eigen values
BITS Pilani, Hyderabad Campus
The eigenvector can be computed from the first of the reduced
equation:
Eigen vectors without inertia
forces
Thus we get one eigenvector for each of two eigenvalues (i):
BITS Pilani, Hyderabad Campus
Writing 𝜌𝐼 =
𝜌𝐵𝐻3
12
= 𝜌 𝐵 × 𝐻 × 𝐻2 12 = 𝜌𝐴 𝐻2 12 and
take H/L=0.01, the characteristic equation is:
Eigenvalues and eigenvectors
without inertia forces
If we write 𝜔𝑖 = 𝜔𝑖𝐿2 𝜌𝐴 𝐸𝐼, the exact solution is
𝜔1 = 3.5158 and 𝜔2 = 22.0226.
BITS Pilani, Hyderabad Campus
BITS Pilani, Hyderabad Campus
Determine the smallest natural frequency of a beam with
clamped ends, and of constant cross-sectional area A,
moment of inertia I, and length L. Use the symmetry and
two Euler—Bernoulli beam elements in the half beam.
Exercise-1
BITS Pilani, Hyderabad Campus
Note that the beam problem is a hyperbolic equation, hence the
eigenvalue is the square of the natural frequency of flexural
vibration, ω. For a mesh of two Euler—Bernoulli elements in a half
beam (i.e., h = L/4), the assembled equations are given by
Solution:
BITS Pilani, Hyderabad Campus
The determinant of the coefficient matrix yields a cubic
polynomial in ω2. Note that by considering the half beam
we restricted the natural frequencies to those of
symmetric modes. The antisymmetric modes (only) can
be obtained by using U5 = 0 instead of U6 = 0.
BITS Pilani, Hyderabad Campus
The Governing Equations are:
Timoshenko Beam Theory
Displacement approximation is:
BITS Pilani, Hyderabad Campus
BITS Pilani, Hyderabad Campus
For linear interpolation functions:
BITS Pilani, Hyderabad Campus
BITS Pilani, Hyderabad Campus
BITS Pilani, Hyderabad Campus
Determine the smallest natural frequency of a beam with
clamped ends, and of constant cross-sectional area A,
moment of inertia I, and length L. Use the symmetry and
two Timoshenko beam elements in the half beam.
Exercise-2
BITS Pilani, Hyderabad Campus
Solution:
BITS Pilani, Hyderabad Campus
Consider a beam (of Young’s modulus E, shear modulus G, area of
cross section A, second moment area about the axis of bending I,
and length L) with its left end (x = 0) clamped and its right end (x =
L) is supported vertically by a linear elastic spring (see Figure P6.5).
Determine the fundamental natural frequency using (a) one Euler-
Bernoulli beam element and (b) one Timoshenko beam (IIE) element
(use the same mass matrix in both elements).
Exercise-3:
BITS Pilani, Hyderabad Campus
One-element mesh is used. The boundary conditions are:
U1 = U2 = 0 and Q1
3 = −kU3. The eigenvalue problems
are formulated below.
Solution:
BITS Pilani, Hyderabad Campus
The equation governing the onset of buckling of a column with
an axial compressive load of No is given by:
Stability in Buckling of Beams,
Eigenvalue Problem Modeling using EBT
This represents an eigenvalue
problem with =No where the
smallest value of No is called
as the critical buckling load.
BITS Pilani, Hyderabad Campus
The finite element model of the above equation is:
The finite element model:
where
BITS Pilani, Hyderabad Campus
The [G] matrix is called as stability matrix.
BITS Pilani, Hyderabad Campus
Where W(x) and S(x) represent transverse deflection and
rotation, respectively, in the onset of buckling. The finite
element model with equal interpolation of W and S is:
Stability in Buckling of Beams,
Eigenvalue Problem Modeling using TBT
For the TBT, the equation are:
BITS Pilani, Hyderabad Campus
BITS Pilani, Hyderabad Campus
BITS Pilani, Hyderabad Campus
BITS Pilani, Hyderabad Campus
Determine the critical buckling load of a cantilever beam (A, I,
L, E) using (a) one Euler—Bernoulli beam element and (b)
one Timoshenko beam element (RIE).
Solution: One element mesh is used. The boundary conditions
are: U1 = U2 = 0.The eigenvalue problems are formulated
below.
Exercise:
BITS Pilani, Hyderabad Campus

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Eigenvalue Problems in BITS Pilani Hyderabad

  • 1. BITS Pilani Hyderabad Campus Eigenvalue and Time Dependent Problems Prof. Srinivasa Prakash Regalla
  • 2. BITS Pilani, Hyderabad Campus  Eigenvalue Problems Formulation of eigenvalue problems for parabolic, hyperbolic equations Finite element formulation Natural vibration of a beam Euler beam Timoshenko beam Stability and buckling of beams  Time dependent problems Learning Outcomes
  • 3. BITS Pilani, Hyderabad Campus It is posing of a BV/IV/BIV problem in a form: What is an eigenvalue problem? ) ( ) ( u B u A   and seeking a solution for eigenvalues  through it for non- trivial eigenvectors, u. The  and u have different physical meanings for different mechanical problems. For example, the natural axial vibration of a bar are transverse vibration of a cable are represented by: and here  is the square of the natural frequency of vibration, .
  • 4. BITS Pilani, Hyderabad Campus Different meanings of eigenvalues in different problems Problem Area Physical meaning of  Structural deformation/stress problems Square of natural frequency of vibration Buckling problems Buckling load Heat transfer Amplitude of the Fourier components Fluid mechanics Amplitude of the Fourier components
  • 5. BITS Pilani, Hyderabad Campus One example application is the 1-D transient heat transfer: Formulation of eigenvalue problems: parabolic equation Here: u = temperature k = thermal conductivity  = density c = specific heat A = cross-sectional area q = heat generation rate per unit length (1)
  • 6. BITS Pilani, Hyderabad Campus Proposing the homogeneous solution as product of two functions: Substituting into the homogeneous form of diff. equ.: Separation of variables gives: Equating the two terms to a single constant gives: The minus sign signifies that the Temperature decreases with time.
  • 7. BITS Pilani, Hyderabad Campus Writing it as two equations: (2) (3) Considering equation (2) and integrating: t Ke T K t T dt T dT dt T dT                   ) ln( ) ln( For the complete solution to the governing equation (1), we need solution to (3) also. Solving equation (3) for  (eigenvalues) and U(x) (eigenvectors) is called as eigenvalue problem.
  • 8. BITS Pilani, Hyderabad Campus For constant A, c, k and , the general solution of (2) is: Substitution of boundary conditions: 0 cos 0 sin cos 1 0     L L L        cannot be zero hence cosL has to be equal to zero.
  • 9. BITS Pilani, Hyderabad Campus For nontrivial solution, the determinant of the coefficient matrix zero:    . , 2 5 , 2 3 , 2 etc L     The constant K is absorbed into Cn. The constants Cn can be determined from the initial condition: If we multiply by sinmx both sides and integrate over the domain (0,L):              L m n n L m n n dx x u dx x x C 0 0 1 0 sin sin sin   
  • 10. BITS Pilani, Hyderabad Campus For the non-trivial condition of m=n, we obtain: Notice the orthogonality condition:   L m n dx x x u L C 0 0 sin ) ( 2  That way we complete the homogeneous solution: The complete solution of equation (1) can then be obtained as sum of homogeneous solution and particular solution:
  • 11. BITS Pilani, Hyderabad Campus The homogeneous solution and separation of variables gives Formulation of eigenvalue problems: hyperbolic equation
  • 13. BITS Pilani, Hyderabad Campus Natural Vibration of a Beam – Euler- Bernoulli Beam Theory (EBT) For EBT, the governing equation is: Two example boundary conditions: (1) Both ends fixed (2) One end fixed and the other end spring supported
  • 14. BITS Pilani, Hyderabad Campus w(x, t) = frequency of natural vibrations W(x) = Mode shapes Substituting: Natural Vibration of a Beam – Euler- Bernoulli Beam Theory (EBT) For EBT, the governing equation is: Where =2
  • 15. BITS Pilani, Hyderabad Campus j = Hermite cubic polynomial functions Week Formulation  = weight function
  • 16. BITS Pilani, Hyderabad Campus The Hermite Shape Functions
  • 17. BITS Pilani, Hyderabad Campus Hermite Shape Functions Using Local Coordinates The interpolation functions i can be expressed in terms of local coordinate 𝑥 = 𝑥 − 𝑥𝑒 as:
  • 18. BITS Pilani, Hyderabad Campus Elemental level FE Equation
  • 19. BITS Pilani, Hyderabad Campus For constant EI and A the stiffness and mass matrices are: Elemental Matrices
  • 20. BITS Pilani, Hyderabad Campus EBT Example: Cantilever Beam TBT How many elements should we consider? Once the type of element is decided by the weak form, the number of elements is decided by the number of eigenvalues we want to determine. The number of eigenvalues, and hence number of natural frequencies is equal to the net number of unconstrained degrees of freedom in the meshed system.
  • 21. BITS Pilani, Hyderabad Campus Example:Cantilever Beam (continued): One 2-nodedlinear1-D element(with Hermite ShapeFunctions)for EBT: EBCs NBCs: Q3 = Q4 = 0
  • 22. BITS Pilani, Hyderabad Campus Setting the determinant of the above coefficient matrix, we obtain the characteristic equation and solving it we get eigenvalues. Two cases can be examined: (1) Neglecting the inertia forces (2) Taking inertia forces into account The characteristic equation
  • 23. BITS Pilani, Hyderabad Campus If we write 𝜔𝑖 = 𝜔𝑖𝐿2 𝜌𝐴 𝐸𝐼, the exact solution is 𝜔1 = 3.516 and 𝜔2 = 22.035 comparing which with above values, it is clear that translational vibration frequency is accurate but rotational vibration frequency is not accurate. Characteristic equation neglecting inertia forces – Eigen values
  • 24. BITS Pilani, Hyderabad Campus The eigenvector can be computed from the first of the reduced equation: Eigen vectors without inertia forces Thus we get one eigenvector for each of two eigenvalues (i):
  • 25. BITS Pilani, Hyderabad Campus Writing 𝜌𝐼 = 𝜌𝐵𝐻3 12 = 𝜌 𝐵 × 𝐻 × 𝐻2 12 = 𝜌𝐴 𝐻2 12 and take H/L=0.01, the characteristic equation is: Eigenvalues and eigenvectors without inertia forces If we write 𝜔𝑖 = 𝜔𝑖𝐿2 𝜌𝐴 𝐸𝐼, the exact solution is 𝜔1 = 3.5158 and 𝜔2 = 22.0226.
  • 27. BITS Pilani, Hyderabad Campus Determine the smallest natural frequency of a beam with clamped ends, and of constant cross-sectional area A, moment of inertia I, and length L. Use the symmetry and two Euler—Bernoulli beam elements in the half beam. Exercise-1
  • 28. BITS Pilani, Hyderabad Campus Note that the beam problem is a hyperbolic equation, hence the eigenvalue is the square of the natural frequency of flexural vibration, ω. For a mesh of two Euler—Bernoulli elements in a half beam (i.e., h = L/4), the assembled equations are given by Solution:
  • 29. BITS Pilani, Hyderabad Campus The determinant of the coefficient matrix yields a cubic polynomial in ω2. Note that by considering the half beam we restricted the natural frequencies to those of symmetric modes. The antisymmetric modes (only) can be obtained by using U5 = 0 instead of U6 = 0.
  • 30. BITS Pilani, Hyderabad Campus The Governing Equations are: Timoshenko Beam Theory Displacement approximation is:
  • 32. BITS Pilani, Hyderabad Campus For linear interpolation functions:
  • 35. BITS Pilani, Hyderabad Campus Determine the smallest natural frequency of a beam with clamped ends, and of constant cross-sectional area A, moment of inertia I, and length L. Use the symmetry and two Timoshenko beam elements in the half beam. Exercise-2
  • 36. BITS Pilani, Hyderabad Campus Solution:
  • 37. BITS Pilani, Hyderabad Campus Consider a beam (of Young’s modulus E, shear modulus G, area of cross section A, second moment area about the axis of bending I, and length L) with its left end (x = 0) clamped and its right end (x = L) is supported vertically by a linear elastic spring (see Figure P6.5). Determine the fundamental natural frequency using (a) one Euler- Bernoulli beam element and (b) one Timoshenko beam (IIE) element (use the same mass matrix in both elements). Exercise-3:
  • 38. BITS Pilani, Hyderabad Campus One-element mesh is used. The boundary conditions are: U1 = U2 = 0 and Q1 3 = −kU3. The eigenvalue problems are formulated below. Solution:
  • 39. BITS Pilani, Hyderabad Campus The equation governing the onset of buckling of a column with an axial compressive load of No is given by: Stability in Buckling of Beams, Eigenvalue Problem Modeling using EBT This represents an eigenvalue problem with =No where the smallest value of No is called as the critical buckling load.
  • 40. BITS Pilani, Hyderabad Campus The finite element model of the above equation is: The finite element model: where
  • 41. BITS Pilani, Hyderabad Campus The [G] matrix is called as stability matrix.
  • 42. BITS Pilani, Hyderabad Campus Where W(x) and S(x) represent transverse deflection and rotation, respectively, in the onset of buckling. The finite element model with equal interpolation of W and S is: Stability in Buckling of Beams, Eigenvalue Problem Modeling using TBT For the TBT, the equation are:
  • 46. BITS Pilani, Hyderabad Campus Determine the critical buckling load of a cantilever beam (A, I, L, E) using (a) one Euler—Bernoulli beam element and (b) one Timoshenko beam element (RIE). Solution: One element mesh is used. The boundary conditions are: U1 = U2 = 0.The eigenvalue problems are formulated below. Exercise: