MIT - 16.20
Fall, 2002
Unit 10
St.
Venant
Torsion Theory
Readings
:
Rivello
8.1, 8.2, 8.4
T & G
101, 104, 105, 106
Paul A. Lagace, Ph.D.
Professor of Aeronautics & Astronautics
and Engineering Systems
Paul A. Lagace
?
2001
MIT - 16.20
Fall, 2002
III.
Torsion
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
MIT - 16.20
Fall, 2002
We have looked at basic in-plane loading.
Let
’
s
now consider
a second
“building block
”
of types of loading:
basic
torsion.
There are 3 basic types of behavior depending on the type of cross-section:
1.
Solid cross-sections
“classical
”
solution technique
via stress functions
2.
Open, thin-walled sections
Membrane Analogy
Paul A. Lagace
?
2001
Unit 10 -
p
. 3
MIT - 16.20
Fall, 2002
3.
Closed, thin-walled sections
Bredt’s Formula
In Unified you developed the basic equations based on some broad assumptions.
Let
’
s
…
?
B
e a bit more rigorous
?
Explore the limitations for the various approaches
?
Better understand how a structure
“resists”
torsion and
the resulting deformation
?
Learn how to model general structures by these three basic approaches
Look first at
Paul A. Lagace
?
2001
Unit 10 -
p
. 4
MIT - 16.20
Fall, 2002
Classical (St.
Venant
’
s
) Torsion Theory
Consider a long prismatic rod twisted by end torques:
T
[in - lbs]
[m - n]
Figure 10.1
Representation of general long prismatic rod
Length (
l
) >> dimensions
in x and y directions
Do not consider
how
end torque is applied (St.
Venant
’
s
principle)
Paul A. Lagace
?
2001
Unit 10 -
p
. 5
MIT - 16.20
Fall, 2002
Assume the following
geometrical
behavior:
a)
Each cross-section (@ each z) rotates as a rigid body (No “distortion
”
of cross-section shape in x, y)
b)
Rate of twist, k = constant
c
)
Cross-sections are free to warp in the z-direction but the
This is the
“
St.
Venant
Hypothesis
”
warping is the same for all cross-sections
“warping”
= extensional deformation in the direction of the axis
about which the torque is applied
Given these assumptions, we see if we can satisfy the equations of elasticity and B.C.
’s.
?
SEMI-INVERSE METHOD
Consider the
deflections
:
Assumptions imply that at any cross-section location z:
?
d
α
?
α
=
?
dz
?
z
=
k
z
(
careful
!
a constant
Rivello
uses
φ
!)
rate of twist
(define as 0 @ z = 0)
Paul A. Lagace
?
2001
Unit 10 -
p
. 6
ααα
???
MIT - 16.20
Fall, 2002
Figure 10.2
Representation of deformation of cross-section due to
torsion
(for small
α
)
undeformed
position
This results in:
consider direction of + u
u (x, y, z) = r
α
(-sin
β
)
v (x, y, z) = r
α
(
cos
β
)
w (x, y, z) = w (x, y)
?
independent of z!
Paul A. Lagace
?
2001
Unit 10 -
p
. 7
MIT - 16.20
Fall, 2002
We can see that:
r
=
y
2
x
+
2
sin
β
=
y
r
x
cos
β
=
r
This gives:
u (x, y, z) =
-y
k
z
(10 - 1)
v (x, y, z) =
x
k
z
(10 - 2)
w (x, y, z) =
w (x, y)
(10 - 3)
Paul A. Lagace
?
2001
Unit 10 -
p
. 8
MIT - 16.20
Fall, 2002
Next look at the
Strain-Displacement
equations
:
?
u
ε
=
=
0
xx
?
x
?
v
ε
=
=
0
yy
?
y
?
w
ε
=
=
0
zz
?
z
?
u
(
consider
: u exists, but
?
x
=
0
v exists, but
?
v
=
0
)
?
y
?
No extensional strains in torsion
if
cross-sections are free
to warp
Paul A. Lagace
?
2001
Unit 10 -
p
. 9
MIT - 16.20
Fall, 2002
?
u
?
v
ε
=
+
=
?
zk
+
zk
=
0
xy
?
y
?
x
?
cross
-
section does not change shape (as assumed!
)
?
v
?
w
?
w
ε
yz
=
?
z
+
?
y
=
kx
+
?
y
(10
-
4)
?
w
?
u
?
w
ε
zx
=
?
x
+
?
z
=
?
ky
+
?
x
(10
-
5)
Now the
Stress-Strain equations
:
let
’
s
first do
isotropic
1
ε
xx
=
E
[
σ
xx
?
ν
(
σ
yy
+
σ
zz
)
]
=
0
ε
yy
=
1
[
σ
yy
?
ν
(
σ
xx
+
σ
zz
)
]
=
0
E
1
ε
zz
=
E
[
σ
zz
?
ν
(
σ
xx
+
σ
yy
)
]
=
0
?
σ
xx
,
σ
yy
,
σ
zz
=
0
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
0
MIT - 16.20
Fall, 2002
(
ε
=
21
+
ν
)
σ
=
0
?
σ
=
0
xy
E
xy
xy
ε
=
21
+
ν
)
σ
yz
(
E
yz
(10
-
6)
1
/
G
(
ε
xz
=
21
+
ν
)
σ
xz
(10
-
7)
E
?
only
σ
xz
and
σ
yz
stresses exist
Look at
orthotropic
case:
1
ε
xx
=
E
11
[
σ
xx
?
ν
12
σ
yy
?
ν
13
σ
zz
]
=
0
1
ε
yy
=
E
22
[
σ
yy
?
ν
21
σ
xx
?
ν
23
σ
zz
]
=
0
1
ε
zz
=
E
33
[
σ
zz
?
ν
31
σ
xx
?
ν
32
σ
yy
]
=
0
?
σ
xx
,
σ
yy
,
σ
zz
=
0
still
equal zero
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
1
MIT - 16.20
Fall, 2002
1
ε
=
σ
yz
G
23
yz
1
ε
=
σ
xz
xz
G
13
Differences are in
ε
yz
and
ε
xz
here as there are two
different
shear moduli
(
G
23
and G
13
) which enter in here.
for
anisotropic
material
:
coefficients of mutual influence and
Chentsov
coefficients
foul everything up (no longer
“simple
”
torsion theory).
[can
’
t
separate torsion from extension]
Back to general case
…
Look at the
Equilibrium
Equations
:
?
σ
xz
?
z
=
0
?
σ
xz
=
σ
xz
(,
xy
)
?
σ
yz
?
z
=
0
?
σ
yz
=
σ
yz
(,
xy
)
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
2
MIT - 16.20
Fall, 2002
So,
σ
xz
and
σ
yz
are only functions of x and y
?
σ
xz
+
?
σ
yz
=
0
(10
-
8)
?
x
?
y
We satisfy equation (10 - 8) by introducing a Torsion (
Prandtl
)
Stress Function
φ
(x, y) where:
?
φ
?
y
=?
σ
xz
(10
-
9a)
?
φ
?
x
=
σ
yz
(10
-
9b)
Using these in equation (10 - 8) gives:
?
? ? ?
?
?
φ
?
+
?
?
?
φ
?
≡
0
?
x
?
y
? ?
?
y
?
?
x
?
?
Automatically satisfies equilibrium
(as a stress function is
supposed to do)
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
3
MIT - 16.20
Fall, 2002
Now consider the
Boundary
Conditions
:
(a)
Along the contour of the cross-section
Figure 10.3
Representation of stress state along edge of solid cross-
section
under
torsion
outer contour is stress-free surface (away from load introduction)
Figure 10.4
Close-up view of edge element from Figure 10.3
σσσσ
xz
(into page)
σσσσ
yz
(out of page)
x
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
4
MIT - 16.20
Fall, 2002
Using equilibrium:
∑
F
z
=
0
(out of page is positive)
gives:
?
σ
xz
dydz
+
σ
yz
dxdz
=
0
Using equation (10 - 9) results in
?
??
?
φ
dy
? ? ?
+
?
?
φ
?
dx
=
0
? ?
?
y
?
?
x
?
?
?
φ
dy
? ? ?
+
?
?
φ
dx
??
=
d
φ
? ?
?
y
?
?
x
And this means:
d
φ
= 0
?
φ
= constant
We take:
φ
= 0
along
contour
(10
-
10)
Note
:
addition of an arbitrary constant does not affect
the stresses, so choose a convenient one (0!)
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
5
MIT - 16.20
Fall, 2002
Boundary condition (b) on edge
z =
l
Figure 10.5
Representation of stress state at top cross-section of rod
under
torsion
Equilibrium tells us the force in each direction:
F
x
=
∫∫
σ
z
x
dxdy
using equation (10 - 9):
=
y
R
?
φ
dxdy
∫∫
y
L
?
y
where y
R
and
y
L
are the geometrical limits of the cross-
section in the y direction
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
6
MIT - 16.20
Fall, 2002
=?
∫
φ
y
R
dx
[]
y
L
and since
φ
= 0 on contour
F
x
= 0
O.K.
(since no force is applied in x-direction)
Similarly
:
F
y
=
∫∫
σ
zy
dxdy
=
0
O.K.
Look at one more case via equilibrium:
Torque
=
Τ
=
∫∫
[
x
σ
zy
?
y
σ
zx
]
dxdy
?
φ
?
φ
=
∫∫
x
B
x
?
x
dxdy
+
∫∫
y
y
L
R
y
?
y
dydx
x
T
where x
T
and
x
B
are geometrical limits of the cross-section
in the x-direction
Integrate each term by parts:
∫
AdB
=
AB
?
∫
BdA
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
7
MIT - 16.20
Fall, 2002
Set:
A
=
x
?
d
A
=
d
x
?
φ
d
B
=
d
x
?
B
=
φ
?
x
and similarly for y
x
Τ
=
∫
[
]
x
φ
B
?∫
φ
dx
]
dy
+
∫
[
]
y
φ
y
R
?∫
φ
dy
]
dx
x
T
y
L
= 0
= 0
since
φ
= 0 in contour
since
φ
= 0 in contour
?
Τ
=?
2
∫∫
φ
dxdy
(10
-
11)
Up to this point, all the equations [with the slight difference in stress-strain of equations (10 - 6) and (10 - 7)] are also valid for orthotropic
materials.
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
8
MIT - 16.20
Fall, 2002
Summarizing
?
Long, prismatic bar under torsion
?
Rate of twist,
k = constant
?
w
?
ε
yz
= kx
+
?
y
?
w
?
ε
xz
= -
k
y
+
?
x
?
φ
?
φ
?
y
=?
σ
xz
?
x
=
σ
yz
? ?
Boundary conditions
φ
= 0
on contour (free boundary)
Τ
=?
2
∫∫
φ
dxdy
Paul A. Lagace
?
2001
Unit 10 -
p
. 1
9
MIT - 16.20
Fall, 2002
Solution of Equations
(now let
’s go back to
isotropic
)
Place equations (10 - 4) and (10 - 5) into equations (10 - 6) and (10 - 7) to get:
?
?
w
?
σ
yz
=
G
ε
yz
=
G
? ?
k
x
+
?
y
? ?
(10
-
12)
σ
xz
=
G
ε
xz
=
G
??
?
k
y
+
?
w
?
(10
-
13)
?
x
?
We want to eliminate w.
We do this via:
?
?
x
{
Eq. (10
-
12)
}
?
?
?
y
{
Eq. (10
-
13)
}
to get:
?
σ
yz
?
σ
xz
?
?
2
w
?
2
w
?
?
x
?
?
y
=
Gk
+
??
+
k
?
yx
? ?
? ?
xy
??
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
0
φφφ
MIT - 16.20
Fall, 2002
and using the definition of the stress function of equation (10 - 9) we get:
?
2
φ
?
2
φ
?
x
2
+
?
y
2
=
2Gk
(10
-
14)
Poisson
’s Equation for
φ
(Nonhomogeneous Laplace
Equation)
Note for
orthotropic
material
We do
not
have a common shear modulus, so we would get:
?
?
+
?
?
=
(
)
+
(
)
?
??
2
2
2
2
2
φ
x
k
G
w
xy
G
G
xz
yz
yz
xz
?
We cannot eliminate w unless
G
xz
and
G
yz
are virtually the
same
+
?
φ
y
G
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
1
MIT - 16.20
Fall, 2002
Overall solution procedure
:
?
Solve Poisson equation (10 - 14) subject to the boundary condition of
φ
= 0 on the contour
?
Get
T
- k relation from equation (10 - 11)
?
Get stresses (
σ
xz
,
σ
yz
) from equation (10 - 9)
?
Get w from equations (10 - 12) and (10 - 13)
?
Get u, v from equations (10 - 1) and (10 - 2)
?
Can also get
ε
xz
,
ε
yz
from equations (10 - 6) and (10 - 7)
This is
“St.
Venant Theory of Torsion
”
Application to a Circular Rod
Figure 10.6
Representation of circular rod under torsion cross-section
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
2
MIT - 16.20
Fall, 2002
“Let”:
φ
=
C
1
(
x
2
+
y
2
?
R
2
)
This satisfies
φ
= 0 on contour since x
2
+ y
2
= R
2
on contour
This gives:
?
2
φ
?
2
φ
?
x
2
=
2
C
1
?
y
2
=
2
C
1
Place these into equation (10-14):
2
C
1
+
2
C
1
=
2
G
k
Gk
?
C
1
=
2
Note
:
(10-14) is satisfied
exactly
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
3
MIT - 16.20
Fall, 2002
Thus:
φ
=
Gk
(
x
2
+
y
2
?
R
2
)
2
Satisfies boundary conditions and partial
differential equation exactly
Now place this into equation (10-11):
Τ
=?
2
∫∫
φ
dxdy
Figure 10.7
Representation of integration strip for circular cross-section
R
+
Τ
=
Gk
∫∫
-
R -
R -
2
2
y
y
2
2
(
R
2
?
y
2
?
x
2
)
dxdy
-R
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
4
1
?
?
MIT - 16.20
Fall, 2002
+
Τ
=
Gk
∫
?
R
R
?? ?
(
R
2
?
y
2
)
x
?
x
3
3
?? ?
y y
2 2
R R
2 2
?
?
dy
?
4
R
/
=
Gk
3
∫
?
R
(
R
2
?
y
2
)
32
dy
+
R
/
=
Gk
4
1
?
y
(
R
2
?
y
2
)
32
+
3
R
2
y
R
2
?
y
2
+
3
R
4
sin
?
1
y
?
3
4
?
2
2
R
?
?
R
= 0
=
0
=
3
R
4
π
2
This finally results in
π
R
4
Τ
=
Gk
2
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
5
MIT - 16.20
Fall, 2002
Since k is the rate of twist:
k
=
d
α
, we can rewrite this as:
dz
d
α
Τ
=
dz
GJ
where:
?
π
R
4
?
J = torsion constant
? ?
=
2
for a circle
? ?
α
= amount of twist
and:
GJ = torsional
rigidity
Note
similarity to:
dw
dx
M
EI
2
2
=
where:
(I) J - geometric part (E) G - material part
EI = bending rigidity
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
6
τττ
MIT - 16.20
Fall, 2002
To get the stresses, use equation (10 - 9):
?
φ
Τ
σ
=
=
G
kx
=
x
yz
?
x
J
Τ
σ
xz
=?
?
φ
=?
G
ky
=
?
J
y
?
y
Figure 10.8
Representation of
resultant shear stress,
τ
res
, as defined
Define
a resultant stress:
τ
=
σ
zx
zy
2
σ
+
2
Τ
=
J
x
2
+
y
2
= r
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
7
τττ
MIT - 16.20
Fall, 2002
The final result is:
τ
=
Τ
r
J
for a circle
Note
: similarity to
??
σ
x
=?
Mz
?
Ι
?
τ
always
acts along the contour (shape)
resultant
Figure 10.9
Representation of
shear resultant stress for circular
cross-section
Cross-Section
No shear stress on surface
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
8
τττ
MIT - 16.20
Fall, 2002
Also note:
1.
Contours of
φ
:
close together near edge
?
higher
τ
Figure 10.10
Representation of contours of
torsional
shear function
2.
Stress pattern
(
τ
) creates twisting
Figure 10.11
Representation of shear stresses acting perpendicular to
radial lines
magnitudes of
τ
res
Paul A. Lagace
?
2001
Unit 10 -
p
. 2
9
MIT - 16.20
Fall, 2002
To get the deflections, first find
α
:
d
α
Τ
=
dz
GJ
(pure rotation of cross-section)
integration yields:
Τ
z
α
=
+
C
1
GJ
Let C
1
= 0 by saying
α
= 0 @ z = 0
Use equations (10 - 1) and (10 - 2) to get:
Τ
z
u
=?
yzk
=?
y
GJ
Τ
z
v
=
xzk
=
x
GJ
Go to equations (10 - 12) and (10 - 13) to find w(x, y):
Equation (10 - 12) gives:
?
w
σ
yz
=
?
kx
?
y
G
Paul A. Lagace
?
2001
Unit 10 -
p
. 3
0
MIT - 16.20
Fall, 2002
using the result for
σ
yz
:
?
w
Gkx
=
?
kx
=
0
?
y
G
integration of this says
w(x, y) = g
1
(x)
(not a function of y)
In a similar manner
…
Equation (10 -13) gives:
?
w
=
σ
xz
+
ky
?
x
G
Using
σ
xz
= -Gky
gives:
?
w
=?
Gky
+
ky
=
0
?
x
G
integration tells us that:
w(x, y) = g
2
(y)
(not a function of x)
Using these two results we see that if w(x, y) is neither a function of x nor y, then it must be a
constant
.
Might as well take this as
zero
Paul A. Lagace
?
2001
Unit 10 -
p
. 3
1
MIT - 16.20
Fall, 2002
(other constants just show a rigid displacement in z which is
trivial)
?
w(x, y) = 0
No
warping
for
circular
cross-sections
(this is the only cross-section that has no warping)
Other Cross-Sections
In other cross-sections, warping is
“the ability of the cross-section to
resist torsion by differential bending
”.
2 parts for torsional
rigidity
?
Rotation
?
Warping
Ellipse
Paul A. Lagace
?
2001
Unit 10 -
p
. 3
2
MIT - 16.20
Fall, 2002
?
x
2
y
2
?
φ
=
C
1
? ?
a
2
+
b
2
?
1
? ?
Equilateral Triangle
φ
=
C
1
? ?
x
?
3y
+
2
?
?
3y
ax
+
1
a
?
ax
+
?
2
?
?
3
?
?
3
?
?
3
?
Rectangle
Paul A. Lagace
?
2001
Unit 10 -
p
. 3
3
MIT - 16.20
Fall, 2002
π
π
φ
=∑
?
C
n
+
D
n
cosh
ny
?
cos
nx
n odd
?
b
?
a
Series:
(the more terms you take, the better the
solution)
These all give solutions to
?
2
φ
= 2GK subject to
φ
= 0 on the boundary.
In
general, there
will
be warping
see Timoshenko
for other relations (Ch. 11)
Note
:
there are also solutions via
“warping
functions
”.
This is a
displacement formulation
see Rivello
8.4
Next we
’
ll look at an analogy used to
“solve”
the general
torsion problem
Paul A. Lagace
?
2001
Unit 10 -
p
. 3
4