beam torsion vib 7&8 2015

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





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





(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





(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)





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





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.





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)





Recall the general solutions,

yByAyX cossin (9a)

(9b)

So, general solution for becomes,

tYyXty ,

t

GDt

GCtY

cossin

t

GDt

GCyByA

cossincossin





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





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





Clamped-Free Beam Torsional Vibration Mode Shapes





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,





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 ,





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,





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





Hence, the final solution becomes,

t1i2753352

y1i2

1i2

108160ty

22

1i

1i

.cossin

.,

0.0816





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





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





Clamped-Clamped Beam Torsional Vibration Mode Shapes





Free-Free Boundary Conditions:

The boundary conditions for Free-Free case,

0tlt0 ,,

yByAyX cossin

0tY0Xt0 ,

(14a, 14b)


00X

yByAyX sincos 00X 00B0A sincos 0A

Update the solution of X, yByX cos yByX sin

(No torques)





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





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





Free-Free Beam Torsional Vibration Mode Shapes

0

1 32





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.





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)





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





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





,, 21i02

liTA

22i

ii

,,)(

21iiT

A22

iii

tt

l2

y1i2iTty i

i22

i1i

sinsinsin,

Substitute known values,

,,. 21i1i275335i

srad2000 /





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





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






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





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





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)





Mode Shapes of Elastically Supported Beam =1.0

NB: These mode shapes are no longer orthogonal !!





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.

beam torsion vib 7&8 2015

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