Question

How to calculate natural frequency of beam element of varying youngs modulus by fem

Answer 1

function of quadratic equation as,
( ) 3 2
1 2 34 v x ax ax ax a = + ++ . (1)
The above equation can be rewritten in matrix form of ( ) [ ]{ }e
vx N = δ as,
( ) [ ]
1
1
1234
2
2
.
y
y
u
vx N N N N u
φ
φ
­ ½ ° ° ° ° = ® ¾ ° ° ° ° ¯ ¿
(2)
Where,
123 NNN , , and N4 are shape functions for beam element.
( ) 3 23
1 3
1 N x xl l 23 , l
= −+ ( ) 3 22 3
2 3
1 N x l x l xl 2 , l
= −+ ( ) 3 2
3 3
1 N x xl 2 3, l
=−+ ( ) 3 22
4 2 3
1 N N xl xl . l == −
2.2. The Stiffness Matrix for Beam Element
The element stiffness matrix of the beam element can be derived as,
2 2
2 2
12 6 12 6
6 6 4 2 [] . 12 6 12 6
6 6 2 4
e
l l
l l l l K
l l
l l l l
ª º −
« » − = −− −
¬ ¼ −
(3)
2.3. Mass Matrix for the Beam Element
The shape function is used as the weight functions in Galerkin’s method. Substituting the shape functions in the
weak form equation which is given as below,
3 22 2
2
3 22 2
0 0 0 0
( ) ( ) ( ) 0,
l l l l d v dW d v d v d W W x EI EI EI dx A W x v x dx
dxdx dx dx dx ρ ω
ª ºª º
« »« » −+ − =
¬ ¼¬ ¼ ³ ³ (4)
and evaluating the integrals with respect to each of the weighing functions one can obtain the governing equations at
a time and rewrite those terms in matrix, one can obtain the element mass matrix of an Euler Bernoulli beam as,
2 2
2 2
156 22 54 13
22 13 4 3 [] . 420 54 13 156 22
13 22 3 4
e
l l
Al l l l l M
l l
l l l l
ρ
ª − º « » « − » = « » − « » «− − » ¬ − ¼
(5)
Where, l =Length of the beam element in m, ρ =Density of the beam material in 3 kg m and A =Area of cross
section in 2 m . Only a transverse crack under bending has been used to establish the element stiffness matrix.
3. Modelling of Cracks
A cantilever beam of length L , of uniform rectangular cross section b h × with a crack located at positions
1
l is considered (Fig. 1). The crack is assumed to be open and close periodically. The depth o