Turk J Engin Environ Sci
24 (2000) , 105 – 110.
c TÜBİTAK
Transverse Vibration of a Beam Under Constant Stress
İsmail YÜKSEK, Ahmet ÇELİK
Yıldız Technical University, Mechanical Engineering Department,
Yıldız, İstanbul-TURKEY
Received 30.04.1999
Abstract
In this study, transverse vibrations of a beam made of two materials and with a variable cross section
were investigated. Dimensionless natural frequency values of the system were found by the Rayleigh-Ritz
approach. Moreover, the energy amounts of the system accumulated per unit mass were calculated. The
results were given in tables for comparison.
Key Words: Transverse vibration, Dimensionless natural frequency, Isotropic material.
Sabit Gerilmeli Bir Kirişin Enine Titreşimlerinin İncelenmesi
Özet
Bu çalışmada değişken kesitli izotrop malzemeden yapılmış bir kirişin enine titreşimleri incelendi. RayleighRitz yaklaşımı kullanılarak, sistemin boyutsuz doğal frekans değerleri bulundu. Ayrıca sistemin birim
kütle için biriktirdiği enerji miktarları hesaplandı. Değişik malzeme ve hız değerleri için simulasyonlar
gerçekleştirildi. Karşılaştırma yapabilmek için sonuçlar tablolar halinde verildi.
Anahtar Sözcükler: Enine titreşimler, Boyutsuz doğal frekans, İzotrop malzeme
1.
Introduction
Many investigations were carried out on vibrations
of beams by Prescott (1961) and Meirovitch (1971).
Thickness function for a constant stress beam made
of isotropic material and subjected only to bending
stress was obtained by Georgian (1989). A study
for the optimum design of a flywheel was done by
Georgian (1989). Natural frequencies about this flywheel were obtained for various modes by Güven
and Çelik (1995), and Çelik (1988). The flywheel
was considered to have a constant stress as the design criteria and the stress values about the flywheel were obtained. In this study, transverse vibrations of a rotating cantilever beam with a rectangular cross section under constant stress were in-
vestigated. The Rayleigh-Ritz method was used for
analysis (Meirovitch, 1971). First of all, the thickness function and the stress equations satisfying the
constant stress condition was obtained. To provide
the boundary conditions, a mass was fixed to the
beam. To the beam to accumulate maximum energy,
the outer isotropic material should have high density
and the inner material should have low density and
high strength.
The aim of this study was to investigate the effect
of designing the beam sections stressed uniformly under centrifugal force on lateral vibrations of beams.
105
YÜKSEK, ÇELİK
2.
2.1.
Mathematical Model
σ1
Beam with constant stress
A variable crossbeam with constant stress consists
of two parts, namely, the constant cross section part
and the variable cross section part as can be seen in
Figures 1 and 2.
The following equation can be obtained by force
equilibrium
−σ1 h(x)b1 +P (x)+(σ1 +dσ1 )(h+dh)b1 = 0(1)
Here, h, b1 , m, σ1 , and P are thickness fraction,
constant cross section width, mass, radial stress and
force respectively,
P (x) = m(x)xΩ2
h
P(x)
b1
dx
x
Figure 2. Variable cross section beam with constant
stress
2.2.
Beam with constant cross section
A constant cross section part is added to the beam
in order to achieve the boundary condition in Figure
3.
σ2
m(x) = ρa b1 h(x)dx
σ1+dσ1
σ2+dσ2
Second order terms i.e. in equation [2.1] are ignored
and dividing by b1 dx the following equation is obtained.
x
dx
2
ρa Ω
1 dh
=−
x
h dx
σ1
(2)
Thickness function can be derived by integration of
equation [2.2] with boundary condition
x = a and h = ha
Ea ρa
b1
ha
σ2 =
a 2 ρb Ω2 b2
ρa Ω2 a2
1
−
=
2
b
2ln hha0
Figure 1. Beam with constant stress
2
[1−( x
a) ]
(3)
In equation [2.3], x=0 and h=h0 are taken as boundary conditions, stress value can be found as follows.
2 2
σ1 =
106
ρa Ω a
2ln hha0
x i
ρb Ω2 b2 h
1 − ( )2
2
b
(6)
Therefore, σ1 = σ2 at x=a
b
ρa Ω2 a2
2σ1
(5)
Here, σ2 , Ab and ha are radial stress, cross section
area and thickness respectively
Ab = b1 ha
a
h = ha e
Using equilibrium at force effected on added part,
the following equation is found:
−σ2 Ab + ρb Ω2 Ab xdx + (σ2 + dσ2)Ab = 0
Eb ρb
h0
Figure 3. Beam with constant cross section
(4)
(7)
A relation for a/b ratio can be found as follows
for the optimum designed beam.
 12

1
a 

= ρa 1
b
+
1
h
0
ρb
ha
(8)
YÜKSEK, ÇELİK
3.
Energy Equations
Maximum potential and kinetic energy accumulated
in the system are calculated. Transverse displacement can be calculated for harmonic vibration as
follows:
w = W (x) cos ωn t
(9)
W(x) is a function which provides geometric
boundary conditions for transverse displacement.
Maximum value of potential energy due to bending
(Vemax ) can be written as follows:
Vemax =
E
2
Z
I(x)
∂2W
∂x2
2
dx
(10)
b1 h3a 3ln hh0 [1−( xa )2 ]
b1 h(x)3
a
=
e
12
12
0 < x < a → I(x) =
b1 h3a
12
Maximum potential energy is found for the whole
beam due to bending as follows:
VGmax =
VGmax
2
h0
x 2
∂2W
e3ln ha [1−( a ) ] dx
2
∂x
0
Z b 2 2
∂ W
Eb Ib
dx
(11)
+
2
∂x2
a
Ea Ib
2
Z
a
Eb = Young elasticity module of second part
Potential energy due to rotation (VG ) is written
[2] as follows:
∂w
∂x
2
1
σAdx =
2
Z 2
σA cos ωn tdx
∂W
∂x
Tmax
Tmax
VGmax
1
=
2
2
h0
x 2
eln ha [1−( a ) ] σ1 dx +
Z
a
h0
x 2
W 2 eln ha [1−( a ) ] dx
0
Z b
1
2
W 2 dx
+ ρb ωn b1 ha
2
a
Z a
h0
x 2
Aa
2
ρa ωn
=
W 2 eln ha [1−( a ) ] dx
2
0
Z b
Ab
W 2 dx
+ ρb ωn2
2
a
1 2
JΩ
2
Z
(16)
(17)
a
h0
x 2
Z
b
x2 eln ha [1−( a ) ] dx+ρb b1 ha
0
x2 dx(18)
a
Total mass m is,
Z a
Z
h0
x 2
eln ha [1−( a ) ] dx + ρb b1 ha
m = ρa b1 ha
2
σAdx
Z
1
= ρa ωn2 b1 ha
2
J = ρa b1 ha
(12)
∂W
∂x
0
∂W
∂x
Mass inertia moment can be expressed as follows:
Z
dm = ρAdx
J = x2 dm
2
Maximum potential energy is
Z Kinetic energy of the system can be found as follows:
2
1
1 dw
dm = (−ωn W sin ωn t)2 ρAdx (15)
dT =
2 dt
2
Z
1
Tmax = ρωn2 W 2 Adx
2
Maximum kinetic energy is
T =
Z a
At the same time, kinetic energy can be rewritten as
follows:
Here,
Ea = Young elasticity module of first part
1
VG =
2
Z
2
b1 ha b ∂W
σ2 dx
2
∂x
a
2
Z h0
x 2
Aa a ∂W
=
eln ha [1−( a ) ] σ1 dx +
2 0
∂x
2
Z b
∂W
Aa
σ2 dx
(14)
2 a
∂x
a < x < b → Ib =
Vemax =
b 1 ha
2
(13)
Total potential energy can be calculated by using
the following equations:
0
b
dx (19)
a
Inertia moment per unit mass is
σ1max
T
=
K
m
ρa
(20)
107
YÜKSEK, ÇELİK
Here, K and are the shape function and maximum
stress limit for constant stress area, respectively.
ln ho
K = ha
(t)2
ρa
ρb
R1
2
ho
ξ 2 eln ha [1−(ξ/t) ] .dξ + t ξ 2 dξ
(21)
R1
R t ln ho [1−(/t)2]
ha
.d + t d
0 e
R2
0
ρa
ρb
Non-dimensional natural frequency of beam with
constant stress [Tables 1-3]
Table 1. Ea =29 GPa ρa =1550kg/m3
ρb =2000kg/m3
Ω̄
a
x
t= ,
b
b
Formulation of transverse displacement can be
chosen as follows under geometric boundary conditions; the following matrix form can be obtained
ξ=
W = a 0 x 2 + a1 x 3 + a2 x 4 + . . .
(22)
from solution of the energy equation by the RayleighRitz approach:


a0 






 a1 

2
.
([A] − ωn [B])
=0
(23)




.






an
The non-dimensional natural frequencies can be
found as follows:
s
s
Ab b4
Ab b4
ω̄n = ωn ρb
(24)
Ω̄ = Ω ρb
Eb Ib
Eb Ib
4.
Numerical Results
After the investigation of the vibrations about the
beam with a variable cross section, dimensionless
natural frequency values were obtained according to
dimensionless rotating velocity, while dimensionless
rotating velocity and ln hha0 value increase [Tables 13].
The outer part was made of high density and
isotropical composite material [Eb =85 GPa] having
high elasticity module. The frequency investigation
was performed by changing the material [Ea=2955-65GPa] of the inner part. While increasing the
elasticity of the inner material, the frequencies increase. When comparing the frequencies of the constant cross section, the frequencies of the constant
stress beam were observed to be higher [Tables 1,4].
When choosing a constant cross-sectioned outer part
made of a material with high elasticity module and
high density, and an inner part made of low density, the energy amounts of the constant stress beam
accumulated per unit mass increases [Tables 5-7].
108
0
5
10
ln hha0
0.5
1
1.5
2
0.5
1
1.5
2
0.5
1
1.5
2
¯
ω1
3.92
7.45
13.5
23.4
6.79
9.45
14.9
24.4
11.5
13.6
18.2
27.0
¯
ω2
21.9
30.03
42.76
61.27
25.61
32.46
44.37
62.36
34.38
38.83
48.87
65.54
¯
ω3
63.77
84.08
111.1
156.3
67.17
86.22
112.6
157.2
80.51
92.35
116.9
160.0
¯
ω4
135.9
168.8
234.4
354.4
138.9
170.7
235.6
355.1
147.4
176.4
239.1
357.0
Table 2. Ea =55 GPa ρa =1550kg/m3
ρb =2000kg/m3
Ω̄
0
5
10
ln hh0a
0.5
1
1.5
2
0.5
1
1.5
2
0.5
1
1.5
2
¯
ω1
5.23
9.81
17.6
30.1
7.67
11.4
18.7
30.9
12.1
15.1
21.6
33.2
¯
ω2
26.5
37.1
52.6
76.3
29.4
39.0
54.0
77.2
36.9
44.3
57.7
79.7
¯
ω3
73.8
98.6
133
192
76.5
100
134.4
193.1
84.11
105.4
137.9
195.3
0
5
10
ln hha0
0.5
1
1.5
2
0.5
1
1.5
2
0.5
1
1.5
2
¯
ω1
5.66
10.6
19.1
32.7
7.99
12.1
20.1
33.4
12.4
15.7
22.8
35.5
¯
ω2
28.12
39.81
56.61
82.41
30.93
41.64
57.85
83.23
38.13
46.69
61.41
85.64
¯
ω3
77.59
104
142.3
207.0
80.13
105.6
143.4
207.7
87.31
110.4
140.8
209.7
GPa
¯
ω5
285.2
441.3
649.7
804.5
287.5
442.5
650.3
805.0
294.2
445.8
652.1
806.4
Eb =85
¯
ω4
157.9
208.2
308.1
457.9
160.3
209.7
309.0
458.4
167.2
214.2
311.5
460.0
GPa
¯
ω5
240.7
355.9
549.2
734.54
243.5
357.4
549.9
735.0
252.0
361.9
558.2
736.6
Eb =85
¯
ω4
151.5
196.0
286.9
429.9
154.0
197.6
287.
430.4
161.2
202.3
290.5
432.0
Table 3. Ea =65 GPa ρa =1550kg/m3
ρb =2000kg/m3
Ω̄
Eb =85
GPa
¯
ω5
304.4
474.2
682.7
836.5
306.5
475.2
683.2
836.9
312.8
478.3
685.0
838.3
YÜKSEK, ÇELİK
While the ln hha0 ratio increases the energy amount
per unit mass increases [Tables 5-11]. When a material with high strength and low density is used in
the inner and outer part, the accumulated energy
reaches the maximum value [Table 6].
Non-dimensional cantilever beam with constant
stress
Table 4. Ea =Eb =29 GPa
Ω̄
0
5
10
¯
ω1
3.51
6.44
11.19
¯
ω2
21.99
25.40
33.58
¯
ω3
61.59
65.11
74.58
¯
ω4
127.92
131.2
140.54
¯
ω5
222.59
225.67
234.75
Variation energy value per unit mass for beam
with constant stress Tables 5-11
Table 5. Ea =29 GPa σamax =115 MPa Eb =85 GPa
Table 8. Ea =Eb =29 GPa σamax =115 MPa ρb =1550
kg/m3
ln hha0
0.5
1
1.5
2
K
0.42
0.45
0.48
0.49
T/M
12.17
13.27
13.81
14.09
Table 9. Ea =Eb =55 GPa σamax =480 MPa ρb =1550
kg/m3
ln hha0
0.5
1
1.5
2
K
0.42
0.46
0.48
0.49
T/M
43.6
47.53
49.46
50.46
ln hha0
0.5
1
1.5
2
K
0.38
0.44
0.47
0.48
T/M
11.08
12.75
13.95
13.95
Table 6. Ea =55 GPa σamax =480 MPa Eb =85 GPa
ln hha0
0.5
1
1.5
2
K
0.4
0.45
0.47
0.49
T/M
41.01
46.30
48.84
50.13
Table 7. Ea =65 GPa σamax =290 MPa Eb =85 GPa
ln hha0
0.5
1
1.5
2
K
0.4
0.45
0.47
0.49
T/M
24.78
27.97
29.51
30.29
Table 10. Ea =Eb =65 GPa σamax =290 MPa ρb =1550
kg/m3
ln hha0
0.5
1
1.5
2
K
0.42
0.46
0.48
0.49
T/M
26.34
28.71
29.88
30.49
Table 11. Ea =Eb =85 GPa σamax =480 MPa ρb =2000
kg/m3
ln hha0
0.5
1
1.5
2
K
0.42
0.46
0.48
0.49
T/M
33.79
36.83
38.33
39.11
109
YÜKSEK, ÇELİK
References
Çelik, A., “On Transverse Vibration of a Flywheel
with Optimum Design” PhD Thesis, Istanbul, 1998.
posite Flywheel” Journal of Composite Materials
Vol.23, pp. 2-10, 1989.
Güven, U., and Çelik, A., “On Transverse Vibration
of a Composite Flywheel with Optimum Design” Z.
Angew. Math. Mech. Vol. 75, pp. 879-880, 1995.
Meirovitch, L., “Elements of Vibration Analysis”
New York, McGraw Hill Book Comp. 1975.
Georgian, J. C., “Optimum Design of Variable Com-
110
Prescott, J., “Applied Elasticity” Dover Publications Inc. 1961.