74 -Finite Element Methods
Solution
beam iny- direction.
ey
displacement or deflcction of the
be the transverse
i) Formulate differential equation of equilibrium
Differential equation of equilibrium for the beam
El P =0 (A)
ii) Assunme the polynomial displacement function
a,x' + a -(1)
, t a,x + a,x +
coordinates are to be determined and the
where, a , , a, and a, are the generalized
displacement function should satisfy the following boundary conditions:
dv a'v 0
0; (d) at L.
(c) at _x =L.
=
x
(a) at x
=0, v =
0;:(b) at x =0, =0: =
= a , + a , X + ( a , x* + 0 , X"+ 1, X
dv --- (2)
= a, + 2a,x + 3a,x+ 4a,
- (3)
= 2a, + 6a, x + 12 a, **
24 -(4)
d3
=
6a, +
a
Applying Boundary condition. i.e
Using BC (a) i.e., at x =
0;
From Eq.(1) we get.
dv
Using BC (b) i.e., at x = 0: = 0
From Eq.(2), we get, = ()
= 0
Using BC (c) i.e., at x =L:
From Eq. (3). we get 0 = 24, + 6a,L + 12 a, L? (5)
Using BC (d) at x =
L;
From Eq.(4). we get 0 = oa, + 24 a, I.
Substituting value of a, inlo Eq.(5). we get
2a,+6L(-4L a,) +12a, L' =0
Basic Procedure -
75
, = 6L a
Thus.
0 : a, =61
Substituting these values into Eq.(1) a, a, = -
4L a,
and Eq.(3), get we
= 0+ 0 (6L' a,)x' +
Thus, displacement function. +
(4 La,) x' +4 Cu
, (r- 4x'L + 6
=
ii) Substitute the L) -- (6)
displacement function into differential equation
But
R El P
=
( - 4/r + 6L'x*) a,
dv
=
(4r- 12/r+ I2L r) a,
=
(12 -
24Lr + l2L) a,
(24r - 24L) a,
244
dr"
R 24El a,-
iv) Use the Galerkin's formula P%
From weighted residual definition. w,(x)R(x )dx =0
Where.i 4 (given in the problem) R is the Residual
=
w,(x)R(x Jdx =0
1.e. (-4/r'+61x*N24 Ela, - P,)dt = 0
NOTE: value of i depends on the number ol unknowns
ifunknown is only a, then
(generaiised coordinates) suppose
i =
2, W, = W2nknoWns are a, a,, ihen i =2, 4 w =
w,
and w.
(24 Ela,-P -4.6 = ()
- 12)
(24Ela,-P (
(L+5L
(24Ela, P
76 -Finite Element Methods
Po
24El
(iv) To find displacement
Substituting the value a, into Eq.(6),
we get
V
( 4Lr +6L*)
Deflection of the beam 24E
Example 17
cantilever beam subjected to point
Galerkin's method, find the expression for a
Using
I; Young's modulus = E and length of the
load P at the end having moment of inertia =
beam = L
Solution:
X P
A ,E, I
L
Fig. E2.17
Basic Procedure 79
Solution:
() Formulate differential
equation of equilibrium
Differential equation of
equilibrium for the bar subjected to point load is
AE =0
dx
(ii) Assume the
polynomial displacement funetion
whee, (1)
a, , and a, are the
conditions. i.e., generalized coordinates are to be determined from the boundary
(a) at x = 0:
l = (0
(b) at x L; du P
AE
Applying Boundary condition, i.e.
Using BC (a) i.e.. at x =
0; u = 0
From Eq.() we
get. 0
.
lt=
ax +a,x - (2)
and du
= (1, + (1,2x
dx - (3)
Using BC (b) i.e.. atx = L; du P
AE
P
From Eq.(3), we
get = +2la,
AE
=
a,AEP24
214
Substituting the value of a, and a, into Eq.(2). we get
-2a,r
AE +a, x?
P
u
+AE +(r2/r) a, (4)
(iii) Substitute the displacement function into diflerential equation
R AE
80
-Finite Element Methods
Since Px
+ (r-21t) a,
AE
P
+(2 2L) a,
AE
= 20
R = 2AEa,
(iv) Use the Galerkin's
formula
'From weighted residual definition.
w,(x)R(x )dx = 0
Where. i =2
(given in the
problem) R is the Residual
L
(x)R.dr =0
Where. w,(x) =
(x -
2Lr) because a, is the only unknown
-2Lx) (AE2a,)dx = 0
2AEa = 0
2AEd
a, = 0
(v) Find the unknown
displacement
Substituting the value of a, =0 into Eq.(4), we get
Therefore. Displacement at any point x
Px
AE
