Upload
feelingsofly
View
217
Download
3
Embed Size (px)
DESCRIPTION
hi
Citation preview
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 1/35
TORSIONAL VIBRATION OF BEAMS:
Torsional vibrations of aircraft wings can be closely modeled as torsional vibrations of cantilevered beams.
y
x
y
z
l
y
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 2/35
y
y
yy
TT
T
O Elemental strip
2
2
Pt
yITyy
TT
(1)
2
2
Pt
Iy
T
dAzxIA
22P
Mass Polar Moment of Inertia Ip:
(2)y
GIT P
Equations of Motion:
(3)2
2
P2
2
Pt
Iy
GI
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 3/35
(3)2
2
P2
2
Pt
Iy
GI
Solution of Torsional Vibrations:
Based on Separation of Variables method, the solution of satisfies (3) can be written as,
tYyXty ,
Then,
(4)
YXtYyXtYy
yXtYyX
yy 2
2
2
2
2
2
YXtYyXt
tYyXtYyX
tt 2
2
2
2
2
2
)/:( 3mkg
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 4/35
(5)YXIYXGI PP
Substitute them into (3), we have
Equation (5) can be further written as,
(6)Y
Y
GX
X
Examine equation (6), the LHS is a function of y only while the RHS is a function of t only. If they can be equal, both sides must be equal to a CONSTANT:
2
Y
Y
GX
X
(7)
0YG
Y
0XX2
2
(8a)
(8b)
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 5/35
The general solutions of (8a) and (8b) are,
yByAyX cossin (9a)
(9b)
Constants A and B can be determined from the boundary conditions at the ends of the beam, and C and D can be found as functions of the given initial beam torsional deflection and initial torsional velocity, as will be shown in the examples followed.
t
GDt
GCtY
cossin
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 6/35
EX – Clamped–Free: A uniform circular beam of length l=1m and diameter d=10cm is clamped at one end and is subjected to a torque T=9000Nm at the other end. Shear modulus G is assumed to be G=91010 N/m2 and density =7800kg/m3. The applied torque is then suddenly released (at t=0), determine the subsequent torsional vibration of the beam.
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 7/35
The boundary condition for clamped end becomes:
(10a) 0t0 ,
For the free end, there is no torque applied (Free Vibration Analysis),
Free Vibration Analysis – No Force/Torque Considered!
yIGT P
0
y
tlIGlT P
,
0tl , (10b)
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 8/35
Recall the general solutions,
yByAyX cossin (9a)
(9b)
So, general solution for becomes,
tYyXty ,
t
GDt
GCtY
cossin
t
GDt
GCyByA
cossincossin
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 9/35
Apply boundary condition at y=0:
0tY0Xt0 ,
00B0A cossin
00X
0B
0tYlXtl ,
0lA cos
0lX
Apply boundary condition at y=l:
Update X(y) to become,
yAyX sin yAyX cos
0l cos
2
1i2l
,, 21il2
1i2
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 10/35
Hence, the natural frequencies of torsional vibration of a clamped-free beam become,
,, 21iG
l2
1i2Gii
Substitute known parameters, i becomes
,,. 21i1i275335i
Since X(y) represents the vibration mode shape,
yAyX sin
Hence mode shapes are,
,,sin 21il2
y1i2y
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 11/35
Clamped-Free Beam Torsional Vibration Mode Shapes
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 12/35
The general solution for can now be updated to become,
tYyXty ,
t1i275335Ft1i275335El2
y1i2ii
1i
.cos.sinsin
The initial displacement condition:
rad0102010141593
321
109
19000
GI
lTl
410P
l0 .
..
The initial displacement under constant torque becomes,
rady010200y ., Hence,
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 13/35
Substitute initial displacement condition,
0YyX0y ,
01i275335F01i275335El2
y1i2ii
1i
.cos.sinsin
l2
y1i2F
1ii
sin y01020.
The initial velocity condition:
The initial velocity is assumed to be zero,
00y ,
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 14/35
So, the initial displacement condition leads us to,
y01020
2
y1i2F
1ii .sin
Multiply (11) by sin[(2j-1)y/2] and integrate over the length,
(11)
dy
2
y1j2y01020dy
2
y1j2
2
y1i2F
1
01i
i1
0
sin.sinsin
,,.
sin. 21i1i2
081601dy
2
y1i2y02040F
221i1
i 0
Based on the orthognality properties of mode shapes,
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 15/35
Torsional velocity becomes,
tYyXty ,
t1i275335Ft1i275335E
1i275335l2
y1i2
ii
1i
.sin.cos
.sin
Substitute initial velocity condition,
0
2
y1i2E1i275335 i
1i
sin. (12)
Multiply (12) by sin[(2j-1)y/2] and integrate over the length,
,, 21i0Ei
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 16/35
Hence, the final solution becomes,
t1i2753352
y1i2
1i2
108160ty
22
1i
1i
.cossin
.,
0.0816
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 17/35
Natural Frequencies and Mode Shapes of Other Boundary Conditions - Clamped-Clamped:
The boundary conditions for Clamped-Clamped case,
0tlt0 ,,
00B0A cossin
0tY0Xt0 ,
0B
(13a, 13b)
Applying the first boundary condition at y=0 leads to,
00X
Update the solution of X,
yAyX sin
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 18/35
0lA sin
0tYlXtl ,
,, 21iG
l
iGii
Applying the second boundary condition at y=l leads to,
0lX
0l sin
,, 21iil
Therefore, natural frequencies are given by,
Mode shapes are given by,
,,sinsin 21il
yiyyi
)?( 0AnotWhy
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 19/35
Clamped-Clamped Beam Torsional Vibration Mode Shapes
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 20/35
Free-Free Boundary Conditions:
The boundary conditions for Free-Free case,
0tlt0 ,,
yByAyX cossin
0tY0Xt0 ,
(14a, 14b)
Applying the first boundary condition at y=0 leads to,
00X
yByAyX sincos 00X 00B0A sincos 0A
Update the solution of X, yByX cos yByX sin
(No torques)
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 21/35
0lB sin
0tYlXtl ,
,,, 210iG
l
iGii
Applying the second boundary condition at y=l leads to,
0lX
0OR0l sin
,,, 210iil
Therefore, natural frequencies are given by,
Mode shapes are given by,
,,,coscos 210il
yiyyi
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 22/35
When i=0 (=0), 0=0 is called Rigid Body mode. The corresponding mode shape of a rigid body mode is a constant:
10y0 cos
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 23/35
Free-Free Beam Torsional Vibration Mode Shapes
0
1 32
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 24/35
EX – Clamped–Free: A uniform circular beam of length l=1m and diameter d=10cm is clamped at one end and is subjected to a torque Tl(t)=5000sin2000t Nm. Also, a continuously distributed torque q(y, t)=8000sin2000t N is applied along the beam. Shear modulus G is assumed to be G=91010 N/m2 and density =7800kg/m3. Determine the torsional vibration of the beam.
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 25/35
The generalized torque becomes
dyytyFt il
0i , dy
l2
y1i2lytT0
l
0
sinsin
dy
l2
y1i2tq0
l
0
sinsin
tT1 01i sin t
1i2
lq2 0
sin
Where, .,/,, m1lsrad2000N8000qNm5000T 00
Recall the equation for i-th generalized coordinate,
tt2
lt
2
lii
2ii
(Vibration of Strings)
tt2
Ilt
2
Ilii
2i
Pi
P
(Beam Torsional Vib)
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 26/35
To simplify, let
Then, tiTtt i
2ii sin
The corresponding general solution becomes,
t
iTtBtAt
22i
iiiii
sincossin
So, the solution can be written,
tyty ii1i
,
t
iTtBtA
l2
y1i222
iiiii
1i
sincossinsin
1i2
lq2T1
Il
2iT 0
01i
P
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 27/35
Consider the initial displacement,
00y , ,,21i0Bi
(Previously discussed)
0B
l2
y1i2i
1i
sin
Consider the initial velocity,
t
iTtBtA
l2
y1i2ty
22i
iiiiii1i
cossincossin,
00y ,
0iT
Al2
y1i222
iii
1i
sin
Multiply both sides by sin((2j-1)y/2l) and integrate over [0, l],
0dy
l2
y1j2
l2
y1i2iFA
l
022i
ii1i
sinsin
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 28/35
,, 21i02
liTA
22i
ii
,,)(
21iiT
A22
iii
tt
l2
y1i2iTty i
i22
i1i
sinsinsin,
Substitute known values,
,,. 21i1i275335i
srad2000 /
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 29/35
1i2
lq2T1
Il
2iT 0
01i
P
1i2
509315000
1082917800
2 1i6.
1i2
1033011103061
51i5
..
t2000
t20002
y1i2
1i2
1033011103061
104
1ty
ii
51i5
62i1i
sinsinsin
..,
where, ,,. 21i1i275335i
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 30/35
Natural Frequencies and Mode Shapes of Torsional Beam on Elastic Support:
The boundary condition at y=0 becomes,
0t0 , (15a)
At y=l, apply the moment equation,
tlktlIG TP ,, (15b)
tYyXty , tYyXty ,
lXklXIG TP
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 31/35
Applying the first boundary condition at y=0 leads to,
00B0A cossin
0tY0Xt0 ,
0B 00X
Update the solution of X,
yAyX sin yAyX cosApplying the second boundary condition at y=l leads to,
lXklXIG TP lAklAIG TP sincos
lIG
lkall
P
T sin)cos( (16)
Define a dimensionless support parameter as,
lIG
k PT
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 32/35
Then, equation (16) becomes,
This transcendental equation has infinite set of roots which cannot be found in closed form. However, for any given value of support parameter , can be solved from (17) numerically. The first 4 il (i=1,2,3,4) are shown:
0lll cossin (17)
versusl
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 33/35
When = 0, it is the case of Clamped-Free boundary conditions and when , it becomes a case of Clamped-Clamped boundary conditions we have discussed. Hence, the curves start with il values corresponding to Clamped-Free case and approach values of the Clamped-Clamped case as becomes large.
Having computed the values of i (i=1,2,…), the corresponding natural frequencies of the elastically supported beam become,
,, 21iG
ii
(18)
And the mode shapes become,
,,sin 21iyy ii (19)
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 34/35
Mode Shapes of Elastically Supported Beam =1.0
NB: These mode shapes are no longer orthogonal !!
NANYANG TECHNOLOGICALUNIVERSITY, SINGAPORE
Tuesday, April 18, 2023
SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING
Torsional Vibration of BeamsLectures 7 & 8 35/35
SUMMARY
1. Governing differential equation for torsional vibration of beam has been developed;
2. Torsional vibration natural frequencies and mode shapes have been derived based on Separation of Variable;
3. Various boundary conditions have been investigated to establish vibration properties;
4. Free and forced torsional vibration analyses have been studied.