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Presented By
SARVESH SURESHRAO CHIKTE
Finite Difference Method For The Vibration
Analysis Of Beams
Methods of Analysis
• Classical or exact solution of GDE of motion
• Rayleigh's method
• Modified Rayleigh’s Method
• Rayleigh’s Ritz Method
• Approximate Method
– Finite difference method
– By lumping masses
– FEM
• The FDM is the approximate method for the
solution of vibration problem.
• The differential equation is the starting point
of the method.
• The continuum is divided in the form of mesh
& unknowns in the problem are taken at the
nodes.
• The derivatives of the equation are expressed
in the finite difference form.
• The differential equation, split in this discrete
form is applied at each node.
FINITE DIFFERENCE METHOD
• Let, variable ‘w’ is the function of x which is as
shown in the fig.
• The function is divided into equally spaced
interval of ‘h’.
• Station to the right are indicated as i+1,i+2 ….
and corresponding values of the function are
Wi+1, Wi+2 ….
• Station to the left are indicated as i-1,i-2 ….
and corresponding values of the function are
Wi-1, Wi-2 ….
• The first derivative is given by,
……. (1)
Above equation suggest that the slope of the curve between i+1 and
i-1 considered to be constant.
• Differentiated equation (1) w.r.t. x
we get,
……. (2)
it can be seen from fig,
……. (3)
• Put equation (3) into equation (2)
we get,
…….. (4)
again diff. above equation 2 times w.r.t ‘x’
We get,
……. (5)
Free vibration of Beams
• The differential equation for the free vibration of
a beam is given by,
………(6)
• From equation (5) we get,
……(7)
Where m-mass per unit length
• For a uniform beam above equation becomes
……(8)
When it is expressed in finite difference form
…..(9)
Let,
λ =
we get,
• In order to study the free vibration of a beam, it is
divided into a number of segments.
• The finite difference equation is applied at each
node
• Boundary condition are then applied
• The resulting values are in the form of Eigen value
problem which on solution gives natural
frequencies and the mode shapes
 Example
determine the fundamental frequency of the beam
of the given fig. by dividing it into four equal parts
and draw the first mode shape.
Solution:
The beam has to be divided into 4 equal parts and
the node numbers have been indicated in fig,
The deflection at nodes 1,2 and 3 are w1,w2,w3
resp.
Two ends are indicated as 0 and 4
The imaginary node beyond 0 has been shown as -1
and node beyond 4 shown as 5.
• The boundary condition at node 0 are,
W0 = 0 and = 0 …..(a)
Writing in the finite difference form,
w1 + w-1 = 0 ……(b)
• The boundary condition at node 4 are,
W4 = 0 and = 0 …..(c)
Writing above equation in finite difference form
W3=W5 …..(d)
Equation (b) & (d) relate the external nodes to internal
nodes.
Apply equation (7) to nodes 1,2 & 3 resp.
λ
Above equation is an Eigen value problem..
Solve the above determinant and find out the
lowest root is,
λ=0.72
Therefore
Assume w1=1 and substituting the value λ=0.72 in
the 2nd and 3rd eq. of (e)
5.26w2-4w3=4
-4w2+6.28w3=-1
On solving,
w2 =1.231
w3 =0.625
The first mode shape has been shown in fig,
Thank you

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Finite Difference method in Strucutral Dynamics

  • 1. Presented By SARVESH SURESHRAO CHIKTE Finite Difference Method For The Vibration Analysis Of Beams
  • 2. Methods of Analysis • Classical or exact solution of GDE of motion • Rayleigh's method • Modified Rayleigh’s Method • Rayleigh’s Ritz Method • Approximate Method – Finite difference method – By lumping masses – FEM
  • 3. • The FDM is the approximate method for the solution of vibration problem. • The differential equation is the starting point of the method. • The continuum is divided in the form of mesh & unknowns in the problem are taken at the nodes. • The derivatives of the equation are expressed in the finite difference form. • The differential equation, split in this discrete form is applied at each node.
  • 4. FINITE DIFFERENCE METHOD • Let, variable ‘w’ is the function of x which is as shown in the fig. • The function is divided into equally spaced interval of ‘h’. • Station to the right are indicated as i+1,i+2 …. and corresponding values of the function are Wi+1, Wi+2 …. • Station to the left are indicated as i-1,i-2 …. and corresponding values of the function are Wi-1, Wi-2 ….
  • 5. • The first derivative is given by, ……. (1) Above equation suggest that the slope of the curve between i+1 and i-1 considered to be constant.
  • 6. • Differentiated equation (1) w.r.t. x we get, ……. (2) it can be seen from fig, ……. (3)
  • 7. • Put equation (3) into equation (2) we get, …….. (4) again diff. above equation 2 times w.r.t ‘x’ We get, ……. (5)
  • 8. Free vibration of Beams • The differential equation for the free vibration of a beam is given by, ………(6) • From equation (5) we get, ……(7) Where m-mass per unit length
  • 9. • For a uniform beam above equation becomes ……(8) When it is expressed in finite difference form …..(9) Let, λ = we get,
  • 10. • In order to study the free vibration of a beam, it is divided into a number of segments. • The finite difference equation is applied at each node • Boundary condition are then applied • The resulting values are in the form of Eigen value problem which on solution gives natural frequencies and the mode shapes
  • 11.  Example determine the fundamental frequency of the beam of the given fig. by dividing it into four equal parts and draw the first mode shape. Solution: The beam has to be divided into 4 equal parts and the node numbers have been indicated in fig, The deflection at nodes 1,2 and 3 are w1,w2,w3 resp. Two ends are indicated as 0 and 4 The imaginary node beyond 0 has been shown as -1 and node beyond 4 shown as 5.
  • 12. • The boundary condition at node 0 are, W0 = 0 and = 0 …..(a) Writing in the finite difference form, w1 + w-1 = 0 ……(b)
  • 13. • The boundary condition at node 4 are, W4 = 0 and = 0 …..(c) Writing above equation in finite difference form W3=W5 …..(d) Equation (b) & (d) relate the external nodes to internal nodes. Apply equation (7) to nodes 1,2 & 3 resp. λ
  • 14. Above equation is an Eigen value problem.. Solve the above determinant and find out the lowest root is, λ=0.72 Therefore
  • 15. Assume w1=1 and substituting the value λ=0.72 in the 2nd and 3rd eq. of (e) 5.26w2-4w3=4 -4w2+6.28w3=-1 On solving, w2 =1.231 w3 =0.625 The first mode shape has been shown in fig,