Revision E By Tom Irvine - VibrationdataRevision E By Tom Irvine Email: [email protected] November...
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1 TRANSVERSE VIBRATION OF A BEAM VIA THE FINITE ELEMENT METHOD Revision E By Tom Irvine Email: [email protected] November 18, 2008 _______________________________________________________________________ Introduction Many structures are too complex for analysis via classical method. Closed-form solutions are thus unavailable for these structures. For example, a structure may be composed of several different materials. Some of the materials may be anisotropic. Furthermore, the structure may be an assembly of plates, beams, and other components. Consider three examples: 1. A circuit board has numerous chips, crystal oscillators, diodes, connectors, capacitors, jump wires, and other piece parts. 2. A large aircraft consisting of a fuselage, wing sections, tail section, engines, etc. 3. A building has plies, foundation, beams, floor sections, and load-bearing walls. The finite element method is a numerical method that can be used to analyze complex structures, such as the three examples. The purpose of this tutorial is to derive for a method for analyzing beam vibration using the finite element method. The method is based on Reference 1. Theory Consider a beam, such as the cantilever beam in Figure 1. Figure 1. where E is the modulus of elasticity. I is the area moment of inertia. EI, ρ L
Transcript of Revision E By Tom Irvine - VibrationdataRevision E By Tom Irvine Email: [email protected] November...
1
TRANSVERSE VIBRATION OF A BEAM VIA THE FINITE ELEMENT METHOD Revision E By Tom Irvine Email: [email protected] November 18, 2008 _______________________________________________________________________ Introduction Many structures are too complex for analysis via classical method. Closed-form solutions are thus unavailable for these structures. For example, a structure may be composed of several different materials. Some of the materials may be anisotropic. Furthermore, the structure may be an assembly of plates, beams, and other components. Consider three examples:
1. A circuit board has numerous chips, crystal oscillators, diodes, connectors, capacitors, jump wires, and other piece parts.
2. A large aircraft consisting of a fuselage, wing sections, tail section, engines, etc. 3. A building has plies, foundation, beams, floor sections, and load-bearing walls.
The finite element method is a numerical method that can be used to analyze complex structures, such as the three examples. The purpose of this tutorial is to derive for a method for analyzing beam vibration using the finite element method. The method is based on Reference 1.
Theory Consider a beam, such as the cantilever beam in Figure 1.
Figure 1.
where
E is the modulus of elasticity. I is the area moment of inertia.
EI, ρ
L
2
L is the length. ρ is mass per length.
The product EI is the bending stiffness. The vibration modes of the cantilever beam can be found by classical methods. Specifically, the fundamental frequency is
ωρ1
187510 2=⎡⎣⎢
⎤⎦⎥
.L
EI (1)
This problem presents a good opportunity to compare the accuracy of the finite element method to the classical solution. Let y(x,t) represent the displacement of the beam as a function of space and time. The free, transverse vibration of the beam is governed by the equation:
( ) ( ) 2
2
2
2
2
2
t
)t,x(yx)t,x(yx
dxEIx ∂
∂ρ−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
∂∂
∂ (2)
Equation (2) neglects rotary inertia and shear deformation. Note that it is also independent of the boundary conditions, which are applied as constraint equations. Assume that the solution of equation (1) is separable in time and space.
)t(f)x(Y)t,x(y = (3)
( ) ( ) 2
2
2
2
2
2
t
)t(f)x(Yx)t(f)x(Yx
xEIx ∂
∂ρ−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
∂
∂
∂
∂ (4a)
( ) ( ) 2
2
2
2
2
2
t
)t(fx)x(Y)x(Yx
xEIx
)t(f∂
∂ρ−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
∂
∂
∂
∂ (4b)
The partial derivatives change to ordinary derivatives.
( ) 2
2
2
2
2
2
td
)t(fdx)x(Y)x(Ydx
dEIdx
d)t(f ρ−=⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
(5)
3
( ) 2
2
2
2
2
2
td
)t(fd)t(f
1)x(Ydx
dEIdx
dx)x(Y
1−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
ρ (6)
The left-hand side of equation (6) depends on x only. The right hand side depends on t only. Both x and t are independent variables. Thus equation (6) only has a solution if both sides are constant. Let 2ω be the constant.
( )2
2
2
2
2
2
2
td
)t(fd)t(f
1)x(Ydx
dEIdx
dx)x(Y
1ω=−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
ρ (7)
Equation (7) yields two independent equations.
0)x(Y)x()x(Ydx
d)x(EIdx
d 22
2
2
2=ωρ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
(8)
0)t(f)t(ftd
d 22
2=ω+ (9)
Equation (8) is a homogeneous, forth order, ordinary differential equation. The weighted residual method is applied to equation (8). This method is suitable for boundary value problems. An alternative method would be the energy method. The energy method is introduced in Appendix A. There are numerous techniques for applying the weighted residual method. Specifically, the Galerkin approach is used in this tutorial. The differential equation (8) is multiplied by a test function )x(φ . Note that the test function )x(φ must satisfy the homogeneous essential boundary conditions. The essential boundary conditions are the prescribed values of Y and its first derivative. The test function is not required to satisfy the differential equation, however. The product of the test function and the differential equation is integrated over the domain. The integral is set equation to zero.
0dx)x(Y)x()x(Ydx
d)x(EIdx
d)x( 22
2
2
2=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
ωρ−⎥⎥⎦
⎤
⎢⎢⎣
⎡φ∫ (10)
4
The test function )x(φ can be regarded as a virtual displacement. The differential equation in the brackets represents an internal force. This term is also regarded as the residual. Thus, the integral represents virtual work, which should vanish at the equilibrium condition. Define the domain over the limits from a to b. These limits represent the boundary points of the entire beam.
0dx)x(Y)x()x(Ydx
d)x(EIdx
d)x(b
a2
2
2
2
2=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
ωρ−⎥⎥⎦
⎤
⎢⎢⎣
⎡φ∫ (11)
{ } 0dx)x(Y)x()x(dx)x(Ydx
d)x(EIdx
d)x(b
a2b
a 2
2
2
2=ωρφ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ ∫∫
(12)
Integrate the first integral by parts.
{ } 0dx)x(Y)x()x(
dx)x(Ydx
d)x(EIdxd)x(
dxddx)x(Y
dx
d)x(EIdxd)x(
dxd
b
a2
b
a 2
2b
a 2
2
=ωρφ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎭⎬⎫
⎩⎨⎧ φ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ
∫
∫∫
(13)
{ } 0dx)x(Y)x()x(
dx)x(Ydx
d)x(EIdxd)x(
dxd)x(Y
dx
d)x(EIdxd)x(
b
a2
b
a 2
2b
a2
2
=ωρφ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎭⎬⎫
⎩⎨⎧ φ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ
∫
∫
(14)
5
{ } 0dx)x(Y)x()x(
dx)x(Ydx
d)x(EIdxd)x(
dxd)x(Y
dx
d)x(EIdxd)x(
b
a2
b
a 2
2b
a2
2
=ωρφ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎭⎬⎫
⎩⎨⎧ φ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ
∫
∫
(15)
Integrate by parts again.
{ } 0dx)x(Y)x()x(dx)x(Ydx
d)x(EI)x(dx
d
dx)x(Ydx
d)x(EI)x(dxd
dxd)x(Y
dx
d)x(EIdxd)x(
b
a2b
a 2
2
2
2
b
a 2
2b
a2
2
=ωρφ−⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ+
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎥⎦⎤
⎢⎣⎡ φ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ
∫∫
∫
(16)
{ } 0dx)x(Y)x()x(dx)x(Ydx
d)x(EI)x(dx
d
)x(Ydx
d)x(EI)x(dxd)x(Y
dx
d)x(EIdxd)x(
b
a2b
a 2
2
2
2
b
a2
2b
a2
2
=ωρφ−⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ+
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎥⎦⎤
⎢⎣⎡ φ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ
∫∫
(17)
The essential boundary conditions for a cantilever beam are
0)a(Y = (18)
0dxdY
ax=
= (19)
6
Thus, the test functions must satisfy
0)a( =φ (20)
0dxd
ax=
φ
= (21)
The natural boundary conditions are
0)x(Ydx
d)x(EIdxd
bx2
2=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
=
(22)
0)x(Ydx
d)x(EIbx
2
2=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
=
(23)
Equation (23) requires
0)x(dx
d)x(EIbx
2
2=
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ
=
(24)
Apply equations (20), (21), and (24) to equation (17). The result is
{ } 0dx)x(Y)x()x(dx)x(Ydx
d)x(EI)x(dx
d b
a2b
a 2
2
2
2=ωρφ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ ∫∫
(25)
Note that equation (25) would also be obtained for other simple boundary condition cases. Now consider that the beam consists of number of segments, or elements. The elements are arranged geometrically in series form. Furthermore, the endpoints of each element are called nodes.
7
The following equation must be satisfied for each element.
{ } 0dx)x(Y)x()x(dx)x(Ydx
d)x(EI)x(dx
d 22
2
2
2=ωρφ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ ∫∫
(26) Furthermore, consider that the stiffness and mass properties are constant for a given element.
0dx)x(Y)x(dx)x(Ydx
d)x(dx
dEI 22
2
2
2=φωρ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ ∫∫
(27)
Now express the displacement function Y(x) in terms of nodal displacements 1jy − and jy
as well as the rotations 1j−θ and jθ .
hjxh)1j(,hLhLyLyL)x(Y 1j41j3j21j1 <<−θ+θ++= −−− (28)
Note that h is the element length. In addition, each L coefficients is a function of x.
Now introduce a nondimensional natural coordinate ξ .
h/xj−=ξ (29)
Note that h is the segment length.
The displacement function becomes.
10,hLyLhLyL)(Y 1j4j31j21j1 <ξ<θ++θ+=ξ −−− (30) The slope equation is
10,h'Ly'Lh'Ly'L)('Y 1j4j31j21j1 <ξ<θ++θ+=ξ −−− (31)
The displacement function is represented terms of natural coordinates in Figure 2.
8
Figure 2.
Represent each L coefficient in terms of a cubic polynomial.
34i
23i2i1ii ccccL ξ+ξ+ξ+= , i =1, 2, 3, 4
(32)
{ }{ }{ }{ } 10,hcccc
ycccc
hcccc
ycccc)(Y
j3
442
342414
j3
432
332313
1j3
422
232212
1j3
412
312111
<ξ<θξ+ξ+ξ++
ξ+ξ+ξ++
θξ+ξ+ξ++
ξ+ξ+ξ+=ξ
−
−
(33)
x (j-1) h j h
1jy − jyY(x)
ξ
1j−θ
jθ
h
9
{ }{ }{ }{ } 10,hc3c2c
yc3c2c
hc3c2c
yc3c2c)('Y
j2
443424
j2
433323
1j2
422322
1j2
413121
<ξ<θξ+ξ++
ξ+ξ++
θξ+ξ++
ξ+ξ+=ξ
−
−
(34)
Solve for the coefficients jic . The constraint equations are
jy)0(Y = (35)
1jy)1(Y −= (36)
jh)0('Y θ−= (37)
1jh)1('Y −θ−= (38)
Evaluate the displacement at .0=ξ
{ } { } { } { } j14j131j121j11 hcychcyc)0(Y θ++θ+= −− (39)
Boundary condition (35) requires
{ } { } { } { } jj14j131j121j11 yhcychcyc =θ++θ+ −− (39)
0c 11 = (40)
0c 12 = (41)
1c 13 = (42)
0c 14 = (43)
10
The displacement equations becomes
{ }{ }{ }{ } 10,hccc
yccc1
hccc
yccc)(Y
j3
442
3424
j3
432
3323
1j3
422
2322
1j3
412
3121
<ξ<θξ+ξ+ξ+
ξ+ξ+ξ++
θξ+ξ+ξ+
ξ+ξ+ξ+=ξ
−
−
(44)
The slope equations becomes
{ }
{ }{ }{ } 10,hc3c2c
yc3c2c
hc3c2c
yc3c2c)('Y
j2
443424
j2
433323
1j2
422322
1j2
413121
<ξ<θξ+ξ++
ξ+ξ++
θξ+ξ++
ξ+ξ+=ξ
−
−
(45)
Evaluate the slope at .0=ξ
{ } { } { } { } j24j231j221j21 hcychcyc)0('Y θ++θ+= −− (46)
Boundary condition (37) requires.
{ } { } { } { } jj24j231j221j21 hhcychcyc θ−=θ++θ+ −− (47)
0c12 = (48)
0c 22 = (49)
0c32 = (50)
1c 24 −= (51)
11
The displacement equations becomes
{ }
{ }{ }{ } 10,hcc
ycc1
hcc
ycc)(Y
j3
442
34
j3
432
33
1j3
422
23
1j3
412
31
<ξ<θξ+ξ+ξ−+
ξ+ξ++
θξ+ξ+
ξ+ξ+=ξ
−
−
(52)
The slope equations becomes
{ } { }{ } { }
10
,hc3c21yc3c2
hc3c2yc3c2)('Y
j2
4434j2
4333
1j2
42321j2
1413
<ξ<
θξ+ξ+−+ξ+ξ+
θξ+ξ+ξ+ξ=ξ −−
(53)
{ } { }{ } { } j4434j4333
1j42321j4113hcc1ycc1
hccycc)1(Y
θ++−++++
θ+++= −−
(54)
Boundary condition (36) requires
{ } { }{ } { } 1jj4434j4333
1j42321j4113yhcc1ycc1
hccycc
−
−−
=θ++−++++
θ+++
(55)
1cc 4131 =+ (56)
1cc 4131 +−= (57)
0cc 4232 =+ (58)
4232 cc −= (59)
12
0cc1 4333 =++ (60)
4333 c1c −−= (61)
0cc1 4434 =++− (62)
4434 c1c −= (63)
The displacement equation becomes
[ ]{ }[ ]{ }
[ ]{ }[ ]{ } 10,hcc1
ycc11
hcc
ycc1)(Y
j3
442
44
j3
432
43
1j3
422
42
1j3
412
41
<ξ<θξ+ξ−+ξ−+
ξ+ξ−−++
θξ+ξ−+
ξ+ξ−+=ξ
−
−
(64)
The slope equation becomes
[ ]{ }{ }[ ]{ }
[ ]{ } 10,hc3c121
yc3c12
hc3c2
yc3c12)('Y
j2
4444
j2
4343
1j2
4242
1j2
4141
<ξ<θξ+ξ−+−+
ξ+ξ−−+
θξ+ξ−+
ξ+ξ−+=ξ
−
−
(65)
13
The slope equation becomes
[ ]{ }{ }
[ ]{ }[ ]{ } j4444
j4343
1j4242
1j4141
hc3c121
yc3c12
hc3c2
yc3c12)1('Y
θ+−+−+
+−−+
θ+−+
+−+=
−
−
(66) Boundary condition (38) requires
[ ]{ }
{ }[ ]{ }
[ ]{ } 1jj4444
j4343
1j4242
1j4141
hhc3c121
yc3c12
hc3c2
yc3c12
−
−
−
θ−=θ+−+−+
+−−+
θ+−+
+−+
(67)
[ ]{ }
{ }[ ]{ }
[ ]{ } 1jj4444
j4343
1j4242
1j4141
hhc3c221
yc3c22
hc3c2
yc3c22
−
−
−
θ−=θ+−+−+
+−−+
θ+−+
+−+
(68)
0c2 41 =+ (69)
2c 41 −= (70)
1c 42 −= (71)
0c2 43 =+− (72)
2c 43 = (73)
14
0c1 44 =+ (74)
1c44 −= (75)
The displacement equation becomes
( )[ ]{ }
( )[ ]{ }[ ]{ }
( )[ ] ( ){ } 10,h111
y2211
h11
y221)(Y
j32
j32
1j32
1j32
<ξ<θξ−+ξ−−+ξ−+
ξ+ξ−−++
θξ−ξ−−+
ξ−ξ−−+=ξ
−
−
(76)
{ } { }{ } { } 10,h2y231
hy23)(Y
j32
j32
1j32
1j32
<ξ<θξ−ξ+ξ−+ξ+ξ−+
θξ−ξ+ξ−ξ+=ξ −−
(77)
Recall
h/xj−=ξ (78) Thus
h/dxd −=ξ (79a)
dxdh =ξ− (79b)
h/1dxd
−=ξ (80)
Note
ξξ
=dd
dxd
dxd (81)
15
{ } { }{ } { }
10,h/xj,jhxh)1j(
,h2y231
hy23)x(Y
j32
j32
1j32
1j32
<ξ<−=ξ≤≤−
θξ−ξ+ξ−+ξ+ξ−+
θξ−ξ+ξ−ξ+= −−
(82)
{ } [ ] [ ]{[ ] [ ] }
10,h/xj,jhxh)1j(
,h341y661
h32y66h/1)x(Ydxd
j2
j2
1j2
1j2
<ξ<−=ξ≤≤−
θξ−ξ+−+ξ+ξ−+
θξ−ξ+ξ−ξ−= −−
(83)
{ } [ ] [ ]{[ ] [ ] }
10,h/xj,jhxh)1j(
,h64y126
h62y126h/1)x(Ydx
d
jj
1j1j2
2
2
<ξ<−=ξ≤≤−
θξ−+ξ+−+
θξ−+ξ−= −−
(84)
Now Let
h/xj,jhxh)1j(,aL)x(Y T −=ξ≤≤−= (85)
where
321 23L ξ−ξ= (86)
32
2L ξ−ξ= (87)
323 231L ξ+ξ−= (88)
32
4 2L ξ−ξ+ξ−= (89)
16
[ ]Tjj1j1j hyhya θθ= −− (90)
The derivative terms are
h/xj,jhxh)1j(,a'Lh1)x(Y
dxd T −=ξ≤≤−⎟⎟
⎠
⎞⎜⎜⎝
⎛ −= (91)
h/xj,jhxh)1j(,a"L
h
1)x(Ydx
d T22
2−=ξ≤≤−⎟⎟
⎠
⎞⎜⎜⎝
⎛= (92)
Note that primes indicate derivatives with respect to .ξ In summary.
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
ξ−ξ+ξ−
ξ+ξ−ξ−ξ
ξ−ξ
=
32
32
32
32
2231
23
L
(90)
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
ξ−ξ+−
ξ+ξ−ξ−ξ
ξ−ξ
=
2
2
2
2
34166
3266
'L
(91)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ξ−ξ+−
ξ−ξ−
=
6412662
126
"L
(92)
17
Recall
0dx)x(Y)x(dx)x(Ydx
d)x(dx
dEI 22
2
2
2=φωρ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡φ ∫∫
(93)
The essence of the Galerkin method is that the test function is chosen as
)x(Y)x( =φ (94) Thus
[ ] 0dx)x(Ydx)x(Ydx
d)x(Ydx
dEI 222
2
2
2=ωρ−
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡∫∫
(95)
Change the integration variable using equation (79b). Also, apply the integration limits.
[ ] 0d)x(Yhd)x(Ydx
d)x(Ydx
dEIh1
0221
0 2
2
2
2=ξωρ−ξ
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡∫∫
(96)
[ ][ ] 0daLaLh
da"Lh
1a"Lh
1EIh
1
0TT2
1
0T
2T
2
=ξωρ−
ξ⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛
∫
∫
(97)
18
[ ] [ ]{ }
[ ][ ] 0daLaLh
da"La"LEIh
1
1
0TT2
1
0TT
3
=ξωρ−
ξ⎟⎟⎠
⎞⎜⎜⎝
⎛
∫
∫
(98)
[ ] [ ]{ }
[ ][ ] 0daLLah
da"L"LaEIh
1
1
0TT2
1
0TT
3
=ξωρ−
ξ⎟⎟⎠
⎞⎜⎜⎝
⎛
∫
∫
(99)
{ } { } 0daLLahda"L"LaEIh
1 1
0TT21
0TT
3 =ξωρ−ξ⎟⎟⎠
⎞⎜⎜⎝
⎛∫∫
(100)
{ } { } 0adLLhd"L"Lh
EIa1
0T21
0T
3T =
⎭⎬⎫
⎩⎨⎧
ξωρ−ξ⎟⎟⎠
⎞⎜⎜⎝
⎛∫∫ (101)
{ } { } 0dLLhd"L"Lh
EI 1
0T21
0T
3 =ξωρ−ξ⎟⎟⎠
⎞⎜⎜⎝
⎛∫∫ (102)
For a system of n elements,
n,...,2,1j,0MK j2
j ==ω− (103) where
{ } ξ⎟⎟⎠
⎞⎜⎜⎝
⎛= ∫ d"L"L
h
EIK1
0T
3j (104)
19
{ }∫ ξρ=1
0T
j dLLhM (105)
[ ]ξ−ξ+−ξ−ξ−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ξ−ξ+−
ξ−ξ−
=
6412662126
6412662
126
"L"L T
(106)
( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )
( )( ) ( )( )( )( ) ⎥
⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ξ−ξ−ξ−ξ+−ξ+−ξ+−
ξ−ξ−ξ+−ξ−ξ−ξ−ξ−ξ−ξ+−ξ−ξ−ξ−ξ−ξ−
=
646464126126126
64621266262626412612612662126126126
"L"L T
(107)
Note that only the upper triangular components are shown due to symmetry.
( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )
( )( ) ( )( )( )( ) ⎥
⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ξ−ξ−ξ−ξ+−ξ+−ξ+−
ξ−ξ−ξ+−ξ−ξ−ξ−ξ−ξ−ξ+−ξ−ξ−ξ−ξ−ξ−
=
646464126126126
64621266262626412612612662126126126
"L"L T
(108)
20
( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )
( )( ) ( )( )( )( ) ⎥
⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ξ−ξ−ξ−ξ+−ξ+−ξ+−
ξ−ξ−ξ+−ξ−ξ−ξ−ξ−ξ−ξ+−ξ−ξ−ξ−ξ−ξ−
=
646464126126126
64621266262626412612612662126126126
"L"L T
(109)
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
ξ+ξ−
ξ−ξ+−ξ+ξ−
ξ+ξ−ξ−ξ+−ξ+ξ−
ξ+ξ−ξ−ξ+−ξ+ξ−ξ+ξ−
=
2
22
222
2222
T
36481672842414414436
36368726012362447284241441443672601214414436
"L"L
(110)
ξ
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
ξ+ξ−
ξ−ξ+−ξ+ξ−
ξ+ξ−ξ−ξ+−ξ+ξ−
ξ+ξ−ξ−ξ+−ξ+ξ−ξ+ξ−
⎟⎟⎠
⎞⎜⎜⎝
⎛
=
∫ d
36481672842414414436
36368726012362447284241441443672601214414436
h
EI
K
1
02
22
222
2222
3
j
(111)
21
1
032
3232
323232
32323232
3
j
122416244224487236
1218824301212124244224487236243012487236
h
EI
K
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
ξ+ξ−ξ
ξ−ξ+ξ−ξ+ξ−ξ
ξ+ξ−ξξ−ξ+ξ−ξ+ξ−ξ
ξ+ξ−ξξ−ξ+ξ−ξ+ξ−ξξ+ξ−ξ
⎟⎟⎠
⎞⎜⎜⎝
⎛
=
(112)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+−−+−+−
+−−+−+−+−−+−+−+−
⎟⎟⎠
⎞⎜⎜⎝
⎛
=
122416244224487236
1218824301212124244224487236243012487236
h
EI
K
3
j
(113)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
4612
264612612
h
EIK 3j
(114)
22
[ ]32323232
32
32
32
32
T
223123
2231
23
LL
ξ−ξ+ξ−ξ+ξ−ξ−ξξ−ξ
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
ξ−ξ+ξ−
ξ+ξ−ξ−ξ
ξ−ξ
=
(115)
( ) ( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )
( ) ( )( )( ) ⎥
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
ξ−ξ+ξ−
ξ−ξ+ξ−ξ+ξ−ξ+ξ−
ξ−ξ+ξ−ξ−ξξ+ξ−ξ−ξξ−ξ
ξ−ξ+ξ−ξ−ξξ+ξ−ξ−ξξ−ξξ−ξξ−ξ
=
232
3232232
32323232232
323232323232232
T
2
2231231
2231
223231232323
LL
(116)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
44a34a33a24a23a22a14a13a12a11a
LL T
(117)
654 412911a ξ+ξ−ξ= (118)
654 25312a ξ+ξ−ξ= (119)
65432 41292313a ξ−ξ+ξ−ξ−ξ= (120)
23
6543 278314a ξ+ξ−ξ+ξ−= (121)
654 222a ξ+ξ−ξ= (122)
65432 25323a ξ−ξ+ξ−ξ−ξ= (123)
6543 3324a ξ+ξ−ξ+ξ−= (124)
65432 412946133a ξ+ξ−ξ+ξ+ξ−= (125)
65432 2782234a ξ−ξ+ξ−ξ+ξ+ξ−= (126)
65432 46444a ξ+ξ−ξ+ξ−ξ= (127) Recall
{ }∫ ξρ=1
0T
j dLLhM (128)
{ } ξξ+ξ−ξρ= ∫ d4129h,M1
0654
11j (129)
1
0
76511j 7
4259h,M ⎥⎦
⎤⎢⎣⎡ ξ+ξ−ξρ= (130)
⎥⎦⎤
⎢⎣⎡ +−ρ=
742
59h,M 11j
(131)
⎟⎠⎞
⎜⎝⎛ρ=
3513h,M 11j
(132)
⎟⎠⎞
⎜⎝⎛ρ=
420156h,M 11j
(133)
24
{ } ξξ+ξ−ξρ= ∫ d253h,M1
0654
12j (134)
1
0
76512j 7
265
53h,M ⎥⎦
⎤⎢⎣⎡ ξ+ξ−ξρ= (135)
⎥⎦⎤
⎢⎣⎡ +−ρ=
73
65
53h,M 12j
(136)
⎟⎠⎞
⎜⎝⎛ρ=
21011h,M 12j
(137)
⎟⎠⎞
⎜⎝⎛ρ=
42022h,M 12j
(138)
{ } ξξ−ξ+ξ−ξ−ξρ= ∫ d412923h,M1
065432
13j (139)
1
0
7654313j 7
4259
21h,M ⎥
⎦
⎤⎢⎣
⎡ξ⎟⎠⎞
⎜⎝⎛−ξ+ξ⎟
⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛−ξρ= (140)
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛−ρ=
742
59
211h,M 13j
(141)
⎟⎠⎞
⎜⎝⎛ρ=
709h,M 13j
(142)
⎟⎠⎞
⎜⎝⎛ρ=
42054h,M 13j
(143)
25
{ } ξξ+ξ−ξ+ξ−ρ= ∫ d2783h,M1
06543
14j (139)
1
0
765414j 7
267
58
43h,M ⎥
⎦
⎤⎢⎣
⎡ξ⎟⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛−ρ= (140)
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−ρ=
72
67
58
43h,M 14j
(141)
⎟⎠⎞
⎜⎝⎛ −ρ=
42013h,M 14j
(142)
{ } ξξ+ξ−ξρ= ∫ d2h,M1
0654
22j (143)
1
0
76522j 7
131
51h,M ⎥
⎦
⎤⎢⎣
⎡ξ⎟⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛ρ= (144)
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛ρ=
71
31
51h,M 22j
(145)
⎟⎠⎞
⎜⎝⎛ρ=105
1h,M 22j (146)
⎟⎠⎞
⎜⎝⎛ρ=
4204h,M 22j
(147)
{ } ξξ−ξ+ξ−ξ−ξρ= ∫ d253h,M1
065432
23j (148)
26
1
0
7654323j 7
265
53
41
31h,M ⎥
⎦
⎤⎢⎣
⎡ξ⎟⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛ρ=
(149)
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛ρ=
72
65
53
41
31h,M 23j
(150)
⎟⎠⎞
⎜⎝⎛ρ=
42013h,M 23j
(151)
{ } ξξ+ξ−ξ+ξ−ρ= ∫ d33h,M1
06543
24j (152)
1
0
765424j 7
121
53
41h,M ⎥
⎦
⎤⎢⎣
⎡ξ⎟⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛−ρ= (153)
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−ρ=
71
21
53
41h,M 24j
(154)
⎟⎠⎞
⎜⎝⎛ −ρ=
4203h,M 24j
(155)
{ } ξξ+ξ−ξ+ξ+ξ−ρ= ∫ d4129461h,M1
065432
33j (156)
1
0
7654333j 7
42592h,M ⎥
⎦
⎤⎢⎣
⎡ξ⎟⎠⎞
⎜⎝⎛+ξ−ξ⎟
⎠⎞
⎜⎝⎛+ξ+ξ−ξρ= (157)
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+−⎟
⎠⎞
⎜⎝⎛++−ρ=
742
59121h,M 33j
(158)
27
⎟⎠⎞
⎜⎝⎛ρ=
420156h,M 33j
(159)
{ } ξξ−ξ+ξ−ξ+ξ+ξ−ρ= ∫ d27822h,M1
065432
34j (160)
1
0
76543234j 7
267
58
21
32
21h,M ⎥
⎦
⎤⎢⎣
⎡ξ⎟⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛−ρ=
(161)
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−ρ=
72
67
58
21
32
21h,M 34j
(162)
⎟⎠⎞
⎜⎝⎛ −ρ=
42022h,M 34j
(163)
{ } ξξ+ξ−ξ+ξ−ξρ= ∫ d464h,M1
065432
44j (164)
1
0
7654344j 7
132
56
31h,M ⎥
⎦
⎤⎢⎣
⎡ξ⎟⎠⎞
⎜⎝⎛+ξ⎟
⎠⎞
⎜⎝⎛−ξ⎟
⎠⎞
⎜⎝⎛+ξ−ξ⎟
⎠⎞
⎜⎝⎛ρ= (165)
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+−⎟
⎠⎞
⎜⎝⎛ρ=
71
32
561
31h,M 44j
(166)
⎟⎠⎞
⎜⎝⎛ρ=
4204h,M 44j
(167)
28
Recall
{ } ξ⎟⎟⎠
⎞⎜⎜⎝
⎛= ∫ d"L"L
h
EIK1
0T
3j (168)
{ }∫ ξρ=1
0T
j dLLhM (169)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
4612
264612612
h
EIK 3j
(170)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
⎟⎠⎞
⎜⎝⎛ ρ
=
4221563134
135422156
420hM j
(171) Example 1 Model the cantilever beam in Figure 1 as a single element using the mass and stiffness matrices in equations (170) and (171). The model consists of one element and two nodes as shown in Figure 3.
Figure 3. Note that h=L.
N1 N2
E1
29
The eigen problem is
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−−−
ω⎟⎠⎞
⎜⎝⎛ ρ
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−
−−
⎟⎟⎠
⎞⎜⎜⎝
⎛
2
2
1
1
2
2
2
1
1
3
hyhy
422313221561354313422
135422156
420L
hyhy
4626612612
2646612612
LEI
(172)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−−−
λ=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−
−−
2
21
1
2
21
1
hyhy
422313221561354313422
135422156
hyhy
4626612612
2646612612
(173)
where
24
EI420L
ω⎟⎟⎠
⎞⎜⎜⎝
⎛ ρ=λ (174a)
λ⎥⎥⎦
⎤
⎢⎢⎣
⎡
ρ=ω 4L
EI420 (174b)
The boundary conditions at node 1 are
0y1 = (175)
01 =θ (176)
The first two columns and the first two rows of each matrix in equation (173) can thus be struck out.
30
The resulting eigen equation is thus
⎥⎦
⎤⎢⎣
⎡θ⎥
⎦
⎤⎢⎣
⎡−
−λ=⎥
⎦
⎤⎢⎣
⎡θ⎥
⎦
⎤⎢⎣
⎡−
−
22
22
hy
42222156
hy
46612
(177) The eigenvalues are found using the method in Reference 2.
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡λλ
8846.2029715.0
21 (178a)
⎥⎦
⎤⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
λλ
6984.11724.0
2
1 (178b)
The finite element results for the natural frequencies are thus
⎥⎦
⎤⎢⎣
⎡
ρ=⎥
⎦
⎤⎢⎣
⎡ωω
6984.11724.0
LEI420
42
1 (179)
⎥⎦
⎤⎢⎣
⎡
ρ=⎥
⎦
⎤⎢⎣
⎡ωω
807.345331.3
L
EI42
1 (180)
The finite element results are compared to the classical results in Table 1.
Table 1. Natural Frequency Comparison, 1 Element
Index
Finite Element Model
EIL4ρ
ω
Classical Solution
EIL4ρ
ω
1 3.5331 3.5160 2 34.807 22.034
31
The classical results are taken from Reference 3. The finite element results thus over-predicted the natural frequencies. Nevertheless, good agreement is obtained for the first frequency. Example 2 Model the cantilever beam in Figure 1 with two elements using the mass and stiffness matrices in equations (170) and (171). Let each element have equal length. The model consists of two elements and three nodes as shown in Figure 4.
Figure 4.
There are several keys to this problem. One is that h=L/2. The other is that node N2 receives mass and stiffness contributions from both elements E1 and E2. Thus, the resulting global matrices have dimension 6 x 6 prior to the application of the boundary conditions. The local stiffness matrix for element 1 is
( )⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−
−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
2h2y1h
1y
4626612612
2646612612
32/L
EI
(181) The displacement vector is also shown in equation (181) for reference.
N1 N2 N3
E1 E2
32
The local stiffness matrix for element 2 is
( )⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−
−−
⎟⎟⎠
⎞⎜⎜⎝
⎛
3
3
2
2
3
hyhy
4626612612
2646612612
2/LEI (182)
The local mass matrix for element 1 is
( )
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−−−
⎟⎠⎞
⎜⎝⎛ ρ
2
2
1
1
hyhy
422313221561354313422
135422156
4202/L (183)
The local mass matrix for element 2 is
( )
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−−−
⎟⎠⎞
⎜⎝⎛ ρ
3
3
2
2
hyhy
422313221561354313422
135422156
4202/L (184)
The global eigen problem assembled from the local matrices is
( )
( )
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
θ
θ
θ
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−−−−−−−
−−
ω⎟⎠⎞
⎜⎝⎛ ρ
=
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
θ
θ
θ
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−−−
−−−−
−−
⎟⎟⎠
⎞⎜⎜⎝
⎛
3
3
2
2
1
1
2
3
3
2
2
1
1
3
hy
hyhy
422313002215613540031380313
1354031213540031342200135422156
4202/L
hy
hyhy
46260061261200
26802661202461200264600612612
2/LEI
(185)
33
Again, the boundary conditions at node 1 are
0y1 = (186)
01 =θ (187)
The first two columns and the first two rows of each matrix in equation (185) can thus be struck out.
( )( )
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−−−
ω⎟⎠⎞
⎜⎝⎛ ρ
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−
−−
⎟⎟⎠
⎞⎜⎜⎝
⎛
3
3
2
2
2
3
3
2
2
3
hy
hy
42231322156135431380
13540312
4202/L
hy
hy
4626612612
2680612024
2/LEI
(188)
Let
( )
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛
ω⎟⎠⎞
⎜⎝⎛ ρ
=λ
3
2
2/LEI
4202/L
(189)
( )( ) 23
EI4202/L2/L
ω⎟⎟⎠
⎞⎜⎜⎝
⎛ ρ=λ (190)
24
EI6720L
ω⎟⎟⎠
⎞⎜⎜⎝
⎛ ρ=λ (191a)
λ⎥⎥⎦
⎤
⎢⎢⎣
⎡
ρ=ω 4L
EI6720 (191b)
34
The eigen problem in equation (188) becomes
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−−−
λ=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−
−−
3
3
2
2
3
3
2
2
hy
hy
42231322156135431380
13540312
hy
hy
4626612612
2680612024
(192)
The eigenvalues are found using the method in Reference 2. Equation (192) yields four eigenvalues.
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
λλλλ
0810.784056.0073481.0
0018414.0
4
3
2
1
(193a)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
λλλλ
6610.291682.027107.0042912.0
4
3
2
1
(193b)
The finite element results for the first two natural frequencies are thus
⎥⎦
⎤⎢⎣
⎡
ρ=⎥
⎦
⎤⎢⎣
⎡ωω
27107.0042912.0
LEI6720
42
1 (194)
⎥⎦
⎤⎢⎣
⎡
ρ=⎥
⎦
⎤⎢⎣
⎡ωω
221.225177.3
LEI
42
1 (195)
35
The finite element results are compared to the classical results in Table 2.
Table 2. Natural Frequency Comparison, 2 Elements
Index
Finite Element Model
EIL4ρ
ω
Classical Solution
EIL4ρ
ω
1 3.5177 3.5160 2 22.221 22.034
Excellent agreement is obtained for the first two roots. The next step would be to solve for the eigenvectors, which represent the mode shapes. A greater number of elements would be required to obtain accurate mode shapes, however. Example 3 Repeat example 1 with 16 elements. Let each element have equal length. The global stiffness and mass matrix are omitted for brevity. The eigenvalue scale factor is
λ⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
ρ=ω 4
4
LEI)16(420 (196)
The model yields 32 eigenvalues. The first four are
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
λλλλ
0.00053121 0.0001383
005-1.7639e 007-4.4913e
4
3
2
1
(197)
36
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
λλλλ
0.023048 0.01176
0.0041999 0.00067017
4
3
2
1
(198)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ρ=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ωωωω
0.023048 0.01176
0.0041999 0.00067017
LEI)16(420
43 4
42
1
(199)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ρ=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ωωωω
92.120698.61035.22
5160.3
4L
EI
4321
(200)
The finite element results are compared to the classical results in Table 3.
Table 3. Natural Frequency Comparison, 16 Elements
Index
Finite Element Model
EIL4ρ
ω
Classical Solution
EIL4ρ
ω
1 3.5160 3.5160 2 22.035 22.034 3 61.698 61.697 4 120.92 120.90
Excellent agreement is obtained for the first four roots. The corresponding mode shapes are shown in Figures 5 through 8. Note that each mode shape is multiplied by an arbitrary amplitude scale factor. The absolute amplitude scale is thus omitted from the plots.
37
0
0 L
x
Dis
plac
emen
tCANTILEVER BEAM, 16 ELEMENT MODEL, MODE 1
Figure 5.
0
0 L
x
Dis
plac
emen
t
CANTILEVER BEAM, 16 ELEMENT MODEL, MODE 2
Figure 6.
38
0
0 L
x
Dis
plac
emen
tCANTILEVER BEAM, 16 ELEMENT MODEL, MODE 3
Figure 7.
0
0 L
x
Dis
plac
emen
t
CANTILEVER BEAM, 16 ELEMENT MODEL, MODE 4
Figure 8.
39
Additional examples are given in Appendix C.
Reference 1. L. Meirovitch, Computational Methods in Structural Dynamics, Sijthoff & Noordhoff,
The Netherlands, 1980. 2. T. Irvine, The Generalized Eigenvalue Problem, 1999. 3. W. Thomson, Theory of Vibration with Applications, Second Edition, Prentice-Hall,
New Jersey, 1981. 4. K. Bathe, Finite Element Procedures in Engineering Analysis, Prentice-Hall, New
Jersey, 1982.
40
APPENDIX A
Energy Method
The total strain energy P of a beam is
∫ ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
L0
dx2
2dx
y2dEI21P (A-1)
The total kinetric energy T of a beam is
[ ]∫ ρω=L0
22n dxy
21T (A-2)
Again let
h/xj,jhxh)1j(,aL)x(Y T −=ξ≤≤−= (A-3)
h/xj,jhxh)1j(,a'Lh1)x(Y
dxd T −=ξ≤≤−⎟⎟
⎠
⎞⎜⎜⎝
⎛ −= (A-4)
h/xj,jhxh)1j(,a"L
h
1)x(Ydx
d T22
2−=ξ≤≤−⎟⎟
⎠
⎞⎜⎜⎝
⎛= (A-5)
h/dxd −=ξ (A-6)
Assume constant mass density and stiffness. The strain energy is converted to a localized stiffness matrix as
{ } ξ⎟⎟⎠
⎞⎜⎜⎝
⎛= ∫ d"L"L
h
EIK1
0T
3j (A-7)
41
The kinetic energy is converted to a localized mass matrix as
{ }∫ ξρ=10
Tj dLLhM (A-8)
The total strain energy is set equal to the total kinetic energy per the Rayleigh method. The result is a generalized eigenvalue problem. For a system of n elements,
n,...,2,1j,0MK j2
j ==ω− (A-9) where
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
4612
264612612
h
EIK 3j
(A-10)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
⎟⎠⎞
⎜⎝⎛ ρ
=
4221563134
135422156
420hM j
(A-11)
42
APPENDIX B Beam Bending - Alternate Matrix Format The displacement matrix for beam bending is
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
22y11y
(B-1)
The stiffness matrix for beam bending is
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
2h4h612
2h2h62h4h612h612
3h
EIjK (B-2)
The mass matrix for beam bending is
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
⎟⎠⎞
⎜⎝⎛ ρ
=
2h4h22156
2h3132h4h1354h22156
420h
jM (B-3)
43
APPENDIX C
Free-Free Beam Repeat example 1 from the main text with a single element but with free-free boundary conditions. The eigen problem is
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−−−
ω⎟⎠⎞
⎜⎝⎛ ρ
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−
−−
⎟⎟⎠
⎞⎜⎜⎝
⎛
2
2
1
1
2
2
2
1
1
3
hyhy
422313221561354313422
135422156
420L
hyhy
4626612612
2646612612
LEI
(C-1)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−−−
λ=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θ
θ
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−−
−−
221
1
221
1
hyhy
422313221561354313422
135422156
hyhy
4626612612
2646612612
(C-2)
where
24
EI420L
ω⎟⎟⎠
⎞⎜⎜⎝
⎛ ρ=λ (C-3)
λ⎥⎥⎦
⎤
⎢⎢⎣
⎡
ρ=ω 4L
EI420 (C-4)
44
The eigenvalues are found using the method in Reference 2. Equation (C-1) yields four eigenvalues.
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
λλλλ
201.7143
00
4321
(C-5)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
λλλλ
4.47211.3093
00
4321
(C-6)
The finite element results for the first four natural frequencies are thus
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ρ=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ωωωω
4.47211.3093
00
4L
EI420
4321
(C-7)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ρ=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
ωωωω
91.65226.833
00
4L
EI
4321
(C-8)
45
The finite element results are compared to the classical results in Table C-1.
Table C-1. Natural Frequency Comparison, 2 Elements
Index
Finite Element Model
EIL4ρ
ω
Classical Solution
EIL4ρ
ω
1 0 0 2 0 0 3 26.833 22.373 4 91.652 61.673
