Exact & Numerical Solution of Pure Torsion Shaft NAE 2007
nae2007
The 5th International Conference on Numerical Analysis in Engineering
EXACT AND NUMERICAL SOLUTION OF
PURE TORSION SHAFT
Ismail Thamrin and Hasan Basri
Department of Mechanical Engineering, Sriwijaya University
Kampus Unsri Inderalaya Jalan Raya Palembang-Prabumulih km.32 Inderalaya, Ogan Ilir
Phone : (0711-580272)
E-mail: irin72@plasa.com ; hasanbasri@unsri.ac.id; hasan_bas@yahoo.com
Abstract
Exact analysis on complex structure often meets with problems since it needs a long complicated
mathematical defferential solution. Instead of using this, another method called finite element
method is introduced i.e. a numerical solution undergone by discretion structure of infinite in to
finite element that continuously build mesh. Application of pure torsion on prismatic can be done to
a certain part of cross sectional of shaft loaded by torque subject to couple transmission, therefore in
its manufacturing and fabrication process, stress analysis is significant factor to take into
consideration since one of shaft failures may be caused by excessive stress distribution on some area.
The comparison of exact and numeric solution ( FEM ) on pure torsion shaft which holds torsion 2,5
Nm and whose dimension is major axis ( a ) and minor axis ( b ) is 1,2375 x 10-2 m and 1,05 x 10–2
m respectively, and prismatic length ( l ) = 9,845 x 10-2 m. Mechanical Properties i.e. shear modulus
( G ), Young modulus ( E ), yield point ( Yield ), each 8,02 x 1011 Pa ; 2,07 x 10 11 Pa ; 4,14 x
108Pa, respectively and Poisson and Hardening ratio ; ( = 0,29 ) and 800. Exact and Finite
Element analysis have the same characteristic of maximum shear stress on boundary cross sectional
that is closest from centre point of torsion (Gravity Centre). Comparative exact result to FEM has
divergent deviation to maximum shear stress
Keywords: FEM, Pure torsion shaft, Exact solution, Fast
1. Introduction
Exact analysis on complex structure often
meets with problems since it needs a long
complicated mathematical deferential solution.
Instead of using this, another method called
finite element method is introduced i.e. a
numerical solution undergone by discretion
structure of infinite in to finite element that
continuously build mesh.
Application of pure torsion on prismatic
can be done to a certain part of cross sectional
of shaft loaded by torque subject to couple
transmission, therefore in its manufacturing
and fabrication process, stress analysis is
significant factor to take into consideration
since one of shaft failures may be caused by
excessive stress distribution on some area.
2. Elasticity of Torsion
Elasticity of single crystal is unequal in
the different direction and random. To achieve
a usable homogenous assumption at high
accuration, the elasticity of geometric element
to be average property of crystal and when the
direction of the single crystal is different, the
material can be considered as isotropic
material.
Cartesian Coordinated–displacement of
cross sectional prismatic loaded by torsion
Static and Dynamic Problem
3 - 24
nae2007
The 5th International Conference on Numerical Analysis in Engineering
approach by A.J.C. Barre Venant – Saint
(figure 1).
= -yz
= zx
= ( x,y )
u
v
w
(1)
y
Considering stress and elasticity constant
of material on pure torque is determined by
strain component, hence only influential shear
stress existed perpendicular to prismatic axis.
This is possible in consideration of both strain
to displacement value and occurred shear
strain can be neglected
y
y
x
z
xx
= yy = zz = xy = 0
xz
= Gβ
- y
x
x
A
C
( a )
ny
n
ds
y
dy
Without considering weight force aspect
of prismatic and then eliminating stress
function on element equilibrium hence
boundary condition on cross section area ( A )
and cross sectional boundary ( C ).
nx
-d x
x
2
( b )
K e te ra n g a n :
A
ds
= b a t a n g p r is m a t i k y a n g d ib e r i p u n t i r a n
= elem e n ta s i d a e ra h b a ta s te p ia n
= m o m e n p u n t ir
+
π
=
,
2
C
dy
nx = co s =
,
ds
=
=
d a e r a h b a ta s te p ia n
t e g a n g a n p a d a p e n a m p a n g p r is m a t ik
n y = c o s = s in =
Fig. 1: Twisting of Prismatic
i, j
i,j
=
2
i, j
(Ref. 2, page 159)
(2)
u (a) u j (a
i
a
a
i
j
i,j
=
- j,
0
(4)
d
dn
Relationship between twisting moment
and twist angle is
=
x
=
D
A
yz
y xz dA
(5)
3. Exact Analysis
Based on the definition of stress function
with complex variable.
Linearization of Lagrangian rotation tensor on
rotation component is
1 u i u j
i,j =
a
2 a
i
j
n . =
dx
ds
Eulerian and Lagrangian illustration of
linearization characteristic of strained
component to displacement value ( ) and
strained component to shear strain ( ).
2 L i, j
=
(3)
F(z) = +i
Hence shear strain becomes
(6)
xz =
y
G
y
(7)
yz =
G x
x
Stress function according to Ludwig Prandth
is
Static and Dynamic Problem
3 - 25
nae2007
The 5th International Conference on Numerical Analysis in Engineering
=
1 2
x y2
2
A
=
(8)
b2x 2 a 2y2
a 2 b2
Elliptical cross sectional stiffness torsion
Hence boundary condition of cross sectional
area ( A ) and boundary of cross sectional (C)
D
2
= -2
= K
(9)
G
A
2 2
dA
x y
(10)
While for stress analysis of elliptical prismatic
( figure 2 ) with boundary cross sectional area
2
G π a 3b3
a 2 b2
(12)
Hence it is found that coordinated tangential
stress on cross sectional boundary (C) is the
result of the Pythagoras sum of both occurred
shear stresses in square root, and its resultant
angle is as equal as the result of both shear
stresses
Stiffness torsion of elliptical cross sectional
D=
=
=
2 G β a b b2x 2 a 2 y2
2
a 2 b 2 a 2
b
1
2
(13)
4. FEM Analysis
2
x
y
2 1
2
b
a
Finite element technique involves
element-modeling discretion, which is defined
through a displacement function of each node.
y
F k D
a
T
x
b
Modeling used is rectangular trilinear
element ( fig. 3 ) which has 27 nodes. AS the
result, when the pure torsion is occurred to the
prismatic, then the out coming strain and
stiffness matrix are
{ ( e ) }
=
[ K(e) ]
=
[ N1 N2 . . . Nn ] { ( e ) }
N
Fig. 2 : Cross Sectional of Elliptical Prismatic
Stress function on elliptical cross sectional
boundary according to Ludwig Pradath
x
K 2
a
2
c
=
y
2
b
2
(14)
i 1
Next, we can determine the possible out
coming stress by
{ (e) }
(11)
[k ]
=
[ B ] { (e) }
Finite element analysis is supported by FAST
Software and structure analysis, by FEM
3dat.C. The boundary conditions of prismatic,
both are clamped and are torque. This loading
Static and Dynamic Problem
3 - 26
nae2007
The 5th International Conference on Numerical Analysis in Engineering
characteristic torsion is transformed into
concentrated forces; where each has equal
torsion on each node, therefore result
desirable resultant torsion (figure 3)
Fig. 5: Shear Stress of xz Direction (xz )
Fig. 3: Boundary Condition of
Elliptical Prismatic
Table 1 is the comparison of exact and
numeric solution (FEM) on elliptical
prismatic which holds torsion 2,5 Nm and
whose dimension is major axis (a) and minor
axis (b) : 1,2375 x 10-2 m and 1,05 x 10–2 m
respectively, and prismatic length (l) = 9,845
x 10-2 m. Mechanical Properties i.e. shear
modulus (G), Young modulus (E), yield point
(Yield), each 8,02 x 1011 Pa ; 2,07 x 10 11 Pa ;
4,14 x 108Pa, respectively and Poisson and
Hardening ratio ; ( = 0,29) and 800.
Table 1 Comparison of Exact and Finite
Element result
MERIDIONAL
ANGLE( o )
0,000000
7,469698
15,413007
23,380027
31,652126
40,314084
49,427157
59,014652
69,046075
79,426391
90,000000
SHEAR STRESS ( MPa )
FEM
EXACT
yz
xy
0,989787 -0,020708 0,1554892 0,156862
0,993678 -0,202317 0,1575184 0,256407
1,008002 -0,378269 0,1450308 0,405119
1,028703 -0,544416 0,1193671 0,557348
1,054208 -0,696980 0,0830607 0,701912
1,081777 -0,832578 0,0393264 0,833506
1,108660 -0,948171 -0,0082670 0,948207
1,132364 -1,041099 -0,0559881 1,042600
1,150829 -1,109095 -0,0999039 1,113590
1,162527 -1,150319 -0,1357099 1,158300
1,166535 -1,163527 -0,1589426 1,174330
Appendix III shows the example of exact
solution and appendix IV numeric of Finite
Element analysis supported by FAST software.
Conclusions
Exact and Finite Element analysis have the
same characteristic of maximum shear stress
on boundary cross sectional that is closest
from centre point of torsion (Gravity Centre)
Comparative exact result to FEM has
divergent deviation to maximum shear stress
Fig. 4 : Shear Stress of yz Direction (yz )
Static and Dynamic Problem
3 - 27
nae2007
The 5th International Conference on Numerical Analysis in Engineering
1,4
K
=
Shear Stress ( MPa )
1,2
1
0,8
Node displacement
0,6
0,4
a1 a 2 x a 3 y a 4 z
0,2
0
0
0,2
0,4
0,6
0,8
1
1,2
1,4
1,6
1,8
Meridional Angle ( Radian )
Exact
FEM
Figure 6 : Relationship between Exact
and Numeric
a 5 xy a 6 yz a 7 xz a 8 xyz
a 9 a10 x a11y a12 z
a13 xy a14 yz a15 xz a16 xyz
w a17 a18 x a19 y a 20 z
a 21 xy a 22 yz a 23 xz a 24 xyz
References
[1]
[2]
[3]
[4]
[5]
DaryL.Loogan;”First Course in the
Finite Element Method”, Second
Edition, PWS-KENT Pub.Co. Boston,
1992
Herbert Reisman and Peter S. Pawlik,
“Elasticity”, First Edition, Buffalo Inc.
New York, 1980
Kikuci, M, “ FAST-FEM Analysis
Support
Tools “, Update 2001, Tokyo 106
Larry J. Segerlind, “Applied Finite
Element
Analysis “, Second Edition, John Wiley
and Son Inc. New York, 1984
Robert D. Cook, “ Concepts and
Application of Finite Element Analysis
“,
[6]
Second Edition, John Willey and Sons
Inc. Canada, 1974
S.P. Timoshenko, “ Teori Elastisitas “,
Edisi Ketiga, Penerbit Erlangga,
Jakarta, 1986.
Appendix I
z = x+iy
x
y
a 2b2
a 2 b 2
=
y
=
x
{ ( e ) } =
[ K-1 ] { f (e) }
{ f (e) }
=
[ k(e) ]
=
2G ( N ) dA
[b] [ D]dA
[B]
=
[D]
=
T
( ref. 4, page. 104 )
( ref. 4, page. 91 )
A
T
(N) i , j ,k
x
(N) i , j ,k
y
Dx 0
0 D
y
( ref. 4, page. 91 )
( ref. 4, page. 91 )
( ref. 4, page. 91 )
Appendix II ( List of Symbol )
u, x direction displacement
v, y direction displacement
w, z direction displacement
, twist angle
( x,y ), warping function
Li,j , Lagrangian tensor
, shear stress
G, shear modulus
, differential function
d
, gradient of
dn
w on prismatic cross
sectional
K, stress constant
, twisting moment
Static and Dynamic Problem
3 - 28
nae2007
The 5th International Conference on Numerical Analysis in Engineering
D, torsion stiffness of prismatic cross
sectional
a, major axes
b, minor axes
{ ( e ) }, strain matric
[ N1 N2 . . . Nn ], shape function matric
{ ( e ) }, displacement matric
k , stiffness matric local
[ K(e) ], stiffness matric global
Appendix III (Example of exact analysis)
On ( 1,2375 x 10-2 m, 0 m )
D =
=
x
π a 3 b3
G 2
a b2
8,02 x1 011 ( Nm-2 )
( m ) 1,05 x 10
3
( m )
3
π (1,2375x10 2 ( m ) x 1,05 x 10 -2 ( m )
1,2375 x 10
2
2
=
2,0985 x 104 Nm2
=
T
D
=
2,5 ( N.m )
2,0985 x 10 4 ( Nm 2 )
=
1,19128 x 10-4 ( rad/m )
b2x 2 a 2 y2
2 2
=
b
a
2 x 8,02 x 1011 ( N/m 2 )
2Gβa b
a 2 b2
2
1
2
x 1,19128 x 10- 4 ( rad/m ) x
=
1,2375 x 10 -2 ( m ) x 1,05 x 10 -2 ( m )
1,2375 x 10
=
2
2
( m ) 2 1,05 x 10 -2 ( m )
9,89786 x 10-1 ( MPa )
-2
2
Static and Dynamic Problem
3 - 29
The 5th International Conference on Numerical Analysis in Engineering
EXACT AND NUMERICAL SOLUTION OF
PURE TORSION SHAFT
Ismail Thamrin and Hasan Basri
Department of Mechanical Engineering, Sriwijaya University
Kampus Unsri Inderalaya Jalan Raya Palembang-Prabumulih km.32 Inderalaya, Ogan Ilir
Phone : (0711-580272)
E-mail: irin72@plasa.com ; hasanbasri@unsri.ac.id; hasan_bas@yahoo.com
Abstract
Exact analysis on complex structure often meets with problems since it needs a long complicated
mathematical defferential solution. Instead of using this, another method called finite element
method is introduced i.e. a numerical solution undergone by discretion structure of infinite in to
finite element that continuously build mesh. Application of pure torsion on prismatic can be done to
a certain part of cross sectional of shaft loaded by torque subject to couple transmission, therefore in
its manufacturing and fabrication process, stress analysis is significant factor to take into
consideration since one of shaft failures may be caused by excessive stress distribution on some area.
The comparison of exact and numeric solution ( FEM ) on pure torsion shaft which holds torsion 2,5
Nm and whose dimension is major axis ( a ) and minor axis ( b ) is 1,2375 x 10-2 m and 1,05 x 10–2
m respectively, and prismatic length ( l ) = 9,845 x 10-2 m. Mechanical Properties i.e. shear modulus
( G ), Young modulus ( E ), yield point ( Yield ), each 8,02 x 1011 Pa ; 2,07 x 10 11 Pa ; 4,14 x
108Pa, respectively and Poisson and Hardening ratio ; ( = 0,29 ) and 800. Exact and Finite
Element analysis have the same characteristic of maximum shear stress on boundary cross sectional
that is closest from centre point of torsion (Gravity Centre). Comparative exact result to FEM has
divergent deviation to maximum shear stress
Keywords: FEM, Pure torsion shaft, Exact solution, Fast
1. Introduction
Exact analysis on complex structure often
meets with problems since it needs a long
complicated mathematical deferential solution.
Instead of using this, another method called
finite element method is introduced i.e. a
numerical solution undergone by discretion
structure of infinite in to finite element that
continuously build mesh.
Application of pure torsion on prismatic
can be done to a certain part of cross sectional
of shaft loaded by torque subject to couple
transmission, therefore in its manufacturing
and fabrication process, stress analysis is
significant factor to take into consideration
since one of shaft failures may be caused by
excessive stress distribution on some area.
2. Elasticity of Torsion
Elasticity of single crystal is unequal in
the different direction and random. To achieve
a usable homogenous assumption at high
accuration, the elasticity of geometric element
to be average property of crystal and when the
direction of the single crystal is different, the
material can be considered as isotropic
material.
Cartesian Coordinated–displacement of
cross sectional prismatic loaded by torsion
Static and Dynamic Problem
3 - 24
nae2007
The 5th International Conference on Numerical Analysis in Engineering
approach by A.J.C. Barre Venant – Saint
(figure 1).
= -yz
= zx
= ( x,y )
u
v
w
(1)
y
Considering stress and elasticity constant
of material on pure torque is determined by
strain component, hence only influential shear
stress existed perpendicular to prismatic axis.
This is possible in consideration of both strain
to displacement value and occurred shear
strain can be neglected
y
y
x
z
xx
= yy = zz = xy = 0
xz
= Gβ
- y
x
x
A
C
( a )
ny
n
ds
y
dy
Without considering weight force aspect
of prismatic and then eliminating stress
function on element equilibrium hence
boundary condition on cross section area ( A )
and cross sectional boundary ( C ).
nx
-d x
x
2
( b )
K e te ra n g a n :
A
ds
= b a t a n g p r is m a t i k y a n g d ib e r i p u n t i r a n
= elem e n ta s i d a e ra h b a ta s te p ia n
= m o m e n p u n t ir
+
π
=
,
2
C
dy
nx = co s =
,
ds
=
=
d a e r a h b a ta s te p ia n
t e g a n g a n p a d a p e n a m p a n g p r is m a t ik
n y = c o s = s in =
Fig. 1: Twisting of Prismatic
i, j
i,j
=
2
i, j
(Ref. 2, page 159)
(2)
u (a) u j (a
i
a
a
i
j
i,j
=
- j,
0
(4)
d
dn
Relationship between twisting moment
and twist angle is
=
x
=
D
A
yz
y xz dA
(5)
3. Exact Analysis
Based on the definition of stress function
with complex variable.
Linearization of Lagrangian rotation tensor on
rotation component is
1 u i u j
i,j =
a
2 a
i
j
n . =
dx
ds
Eulerian and Lagrangian illustration of
linearization characteristic of strained
component to displacement value ( ) and
strained component to shear strain ( ).
2 L i, j
=
(3)
F(z) = +i
Hence shear strain becomes
(6)
xz =
y
G
y
(7)
yz =
G x
x
Stress function according to Ludwig Prandth
is
Static and Dynamic Problem
3 - 25
nae2007
The 5th International Conference on Numerical Analysis in Engineering
=
1 2
x y2
2
A
=
(8)
b2x 2 a 2y2
a 2 b2
Elliptical cross sectional stiffness torsion
Hence boundary condition of cross sectional
area ( A ) and boundary of cross sectional (C)
D
2
= -2
= K
(9)
G
A
2 2
dA
x y
(10)
While for stress analysis of elliptical prismatic
( figure 2 ) with boundary cross sectional area
2
G π a 3b3
a 2 b2
(12)
Hence it is found that coordinated tangential
stress on cross sectional boundary (C) is the
result of the Pythagoras sum of both occurred
shear stresses in square root, and its resultant
angle is as equal as the result of both shear
stresses
Stiffness torsion of elliptical cross sectional
D=
=
=
2 G β a b b2x 2 a 2 y2
2
a 2 b 2 a 2
b
1
2
(13)
4. FEM Analysis
2
x
y
2 1
2
b
a
Finite element technique involves
element-modeling discretion, which is defined
through a displacement function of each node.
y
F k D
a
T
x
b
Modeling used is rectangular trilinear
element ( fig. 3 ) which has 27 nodes. AS the
result, when the pure torsion is occurred to the
prismatic, then the out coming strain and
stiffness matrix are
{ ( e ) }
=
[ K(e) ]
=
[ N1 N2 . . . Nn ] { ( e ) }
N
Fig. 2 : Cross Sectional of Elliptical Prismatic
Stress function on elliptical cross sectional
boundary according to Ludwig Pradath
x
K 2
a
2
c
=
y
2
b
2
(14)
i 1
Next, we can determine the possible out
coming stress by
{ (e) }
(11)
[k ]
=
[ B ] { (e) }
Finite element analysis is supported by FAST
Software and structure analysis, by FEM
3dat.C. The boundary conditions of prismatic,
both are clamped and are torque. This loading
Static and Dynamic Problem
3 - 26
nae2007
The 5th International Conference on Numerical Analysis in Engineering
characteristic torsion is transformed into
concentrated forces; where each has equal
torsion on each node, therefore result
desirable resultant torsion (figure 3)
Fig. 5: Shear Stress of xz Direction (xz )
Fig. 3: Boundary Condition of
Elliptical Prismatic
Table 1 is the comparison of exact and
numeric solution (FEM) on elliptical
prismatic which holds torsion 2,5 Nm and
whose dimension is major axis (a) and minor
axis (b) : 1,2375 x 10-2 m and 1,05 x 10–2 m
respectively, and prismatic length (l) = 9,845
x 10-2 m. Mechanical Properties i.e. shear
modulus (G), Young modulus (E), yield point
(Yield), each 8,02 x 1011 Pa ; 2,07 x 10 11 Pa ;
4,14 x 108Pa, respectively and Poisson and
Hardening ratio ; ( = 0,29) and 800.
Table 1 Comparison of Exact and Finite
Element result
MERIDIONAL
ANGLE( o )
0,000000
7,469698
15,413007
23,380027
31,652126
40,314084
49,427157
59,014652
69,046075
79,426391
90,000000
SHEAR STRESS ( MPa )
FEM
EXACT
yz
xy
0,989787 -0,020708 0,1554892 0,156862
0,993678 -0,202317 0,1575184 0,256407
1,008002 -0,378269 0,1450308 0,405119
1,028703 -0,544416 0,1193671 0,557348
1,054208 -0,696980 0,0830607 0,701912
1,081777 -0,832578 0,0393264 0,833506
1,108660 -0,948171 -0,0082670 0,948207
1,132364 -1,041099 -0,0559881 1,042600
1,150829 -1,109095 -0,0999039 1,113590
1,162527 -1,150319 -0,1357099 1,158300
1,166535 -1,163527 -0,1589426 1,174330
Appendix III shows the example of exact
solution and appendix IV numeric of Finite
Element analysis supported by FAST software.
Conclusions
Exact and Finite Element analysis have the
same characteristic of maximum shear stress
on boundary cross sectional that is closest
from centre point of torsion (Gravity Centre)
Comparative exact result to FEM has
divergent deviation to maximum shear stress
Fig. 4 : Shear Stress of yz Direction (yz )
Static and Dynamic Problem
3 - 27
nae2007
The 5th International Conference on Numerical Analysis in Engineering
1,4
K
=
Shear Stress ( MPa )
1,2
1
0,8
Node displacement
0,6
0,4
a1 a 2 x a 3 y a 4 z
0,2
0
0
0,2
0,4
0,6
0,8
1
1,2
1,4
1,6
1,8
Meridional Angle ( Radian )
Exact
FEM
Figure 6 : Relationship between Exact
and Numeric
a 5 xy a 6 yz a 7 xz a 8 xyz
a 9 a10 x a11y a12 z
a13 xy a14 yz a15 xz a16 xyz
w a17 a18 x a19 y a 20 z
a 21 xy a 22 yz a 23 xz a 24 xyz
References
[1]
[2]
[3]
[4]
[5]
DaryL.Loogan;”First Course in the
Finite Element Method”, Second
Edition, PWS-KENT Pub.Co. Boston,
1992
Herbert Reisman and Peter S. Pawlik,
“Elasticity”, First Edition, Buffalo Inc.
New York, 1980
Kikuci, M, “ FAST-FEM Analysis
Support
Tools “, Update 2001, Tokyo 106
Larry J. Segerlind, “Applied Finite
Element
Analysis “, Second Edition, John Wiley
and Son Inc. New York, 1984
Robert D. Cook, “ Concepts and
Application of Finite Element Analysis
“,
[6]
Second Edition, John Willey and Sons
Inc. Canada, 1974
S.P. Timoshenko, “ Teori Elastisitas “,
Edisi Ketiga, Penerbit Erlangga,
Jakarta, 1986.
Appendix I
z = x+iy
x
y
a 2b2
a 2 b 2
=
y
=
x
{ ( e ) } =
[ K-1 ] { f (e) }
{ f (e) }
=
[ k(e) ]
=
2G ( N ) dA
[b] [ D]dA
[B]
=
[D]
=
T
( ref. 4, page. 104 )
( ref. 4, page. 91 )
A
T
(N) i , j ,k
x
(N) i , j ,k
y
Dx 0
0 D
y
( ref. 4, page. 91 )
( ref. 4, page. 91 )
( ref. 4, page. 91 )
Appendix II ( List of Symbol )
u, x direction displacement
v, y direction displacement
w, z direction displacement
, twist angle
( x,y ), warping function
Li,j , Lagrangian tensor
, shear stress
G, shear modulus
, differential function
d
, gradient of
dn
w on prismatic cross
sectional
K, stress constant
, twisting moment
Static and Dynamic Problem
3 - 28
nae2007
The 5th International Conference on Numerical Analysis in Engineering
D, torsion stiffness of prismatic cross
sectional
a, major axes
b, minor axes
{ ( e ) }, strain matric
[ N1 N2 . . . Nn ], shape function matric
{ ( e ) }, displacement matric
k , stiffness matric local
[ K(e) ], stiffness matric global
Appendix III (Example of exact analysis)
On ( 1,2375 x 10-2 m, 0 m )
D =
=
x
π a 3 b3
G 2
a b2
8,02 x1 011 ( Nm-2 )
( m ) 1,05 x 10
3
( m )
3
π (1,2375x10 2 ( m ) x 1,05 x 10 -2 ( m )
1,2375 x 10
2
2
=
2,0985 x 104 Nm2
=
T
D
=
2,5 ( N.m )
2,0985 x 10 4 ( Nm 2 )
=
1,19128 x 10-4 ( rad/m )
b2x 2 a 2 y2
2 2
=
b
a
2 x 8,02 x 1011 ( N/m 2 )
2Gβa b
a 2 b2
2
1
2
x 1,19128 x 10- 4 ( rad/m ) x
=
1,2375 x 10 -2 ( m ) x 1,05 x 10 -2 ( m )
1,2375 x 10
=
2
2
( m ) 2 1,05 x 10 -2 ( m )
9,89786 x 10-1 ( MPa )
-2
2
Static and Dynamic Problem
3 - 29