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.
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,