MIT - 16.20
Fall, 2002
Unit 12
Torsion of (Thin) Closed Sections
Readings
:
Megson
8.5
Rivello
8.7 (only single cell material),
8.8 (Review)
T
&
G
115, 116
Paul A. Lagace, Ph.D.
Professor of Aeronautics & Astronautics
and Engineering Systems
Paul A. Lagace
? 2001
MIT - 16.20
Fall, 2002
Before we look specifically at thin-walled sections, let us consider the general case (i.e., thick-walled).
Hollow, thick-walled sections:
Figure 12.1
Representation of a general thick-walled cross-section
φφφφ
= C
2
on one boundary
φφφφ
= C
1
on one boundary
This has more than one boundary (multiply-connected)
?
d
φ
= 0 on each boundary
?
However,
φ
= C
1
on one boundary and C
2
on the other (they
cannot be the same constants for a general solution [there’
s
no reason they should be])
=>
Must somehow be able to relate C
1
to C
2
Paul A. Lagace
? 2001
Unit 12 -
2
MIT - 16.20
Fall, 2002
It can be shown that around any
closed
boundary:
∫
=
τ
ds
AGk
2
(12-1)
Figure 12.2
Representation of general closed area
ττττ
where:
τ
= resultant shear stress at boundary
A = Area inside boundary k
= twist rate =
d
α
dz
Paul A. Lagace
? 2001
Unit 12 -
3
MIT - 16.20
Fall, 2002
Notes
:
1.
The resultant shear stresses at the boundary must be in the direction of the tangents to the boundary
2.
The surface traction at the boundary is zero (stress free), but the resultant shear stress is not
Figure 12.3
Representation of a 3-D element cut with one face at the
surface
of
the
body
To prove Equation (12 - 1), begin by considering a small 3-D element from the previous figure
Paul A. Lagace
? 2001
Unit 12 -
4
σσσ
MIT - 16.20
Fall, 2002
Figure 12.4
Exploded view of cut-out 3-D elements
this face is stress free, thus
σ
normal
= 0
Look at a 2-D cross-section in the x-y plane:
Figure 12.5
Stress field at boundary of cross-section
Paul A. Lagace
? 2001
σσσσ
tan
=
σσσσ
res
since
σσσσ
normal
= 0
Unit 12 -
5
MIT - 16.20
Fall, 2002
τ
resul
tan
t
=
σ
zy
cos
γ
+
σ
zx
sin
γ
geometrically
:
cos
γ
=
dy ds
dx
sin
γ
=
ds
Thus:
dx
?
τ
ds
=
∫
?
?
dyds
zy
σ
ds
+
σ
zx
ds
?
∫
?
ds
?
=
∫
σ
zy
dy
+
σ
zx
dx
We know that:
?
?
w
?
σ
zy
=
G
??
k
x
+
?
y
? ?
σ
zx
=
G
?
?
k
y
+
?
w
?
?
?
x
?
Paul A. Lagace
? 2001
Unit 12 -
6
?
MIT - 16.20
Fall, 2002
?
=
∫
τ
ds
∫
G
??
k
x
+
?
+
∫
dy
?
w
?
G
?
?
k
y
+
?
w
?
dx
?
?
y
?
?
?
x
?
+
?
?
+
∫
dx
G
dy
k
∫
?
?
w
?
w
?
=
G
{
xdy
?
ydx
}
?
?
x
?
y
?
= dw
We further
know that:
∫
dw
=
w
?
=
0
around closed contour
?
So we
’
re left with:
∫
τ
ds
=
Gk
∫
{
xdy
?
ydx
}
Paul A. Lagace
? 2001
Unit 12 -
7
MIT - 16.20
Fall, 2002
Use Stoke
’
s
Theorem for the right-hand side integral:
∫
?
?
N
?
M
?
{
Mdx
+
Ndy
}
=
∫∫
? ?
?
x
?
?
y
? ?
dxdy
In this case we have
?
M
M
=
?
y
?
=
?
1
?
y
?
N
N
=
x
?
=
1
?
x
We thus get:
Gk
∫
{
xdy
?
ydx
}
=
Gk
∫
∫
[
1
??
1
)
]
dxdy
(
=
Gk
∫∫
2
dxdy
Paul A. Lagace
? 2001
Unit 12 -
8
MIT - 16.20
Fall, 2002
We furthermore know that the double integral of
dxdy is the planar area:
∫
∫
d
dxdy = Area = A
Putting all this together brings us back to Equation (12 - 1):
∫
=
τ
ds
AGk
2
Q.E.D.
Hence, in the general case we use equation (12 - 1) to relate C
1
and C
2
.
This is rather complicated and we will not do the general case here.
For
further
information
(See
Timoshenko
, Sec. 115)
We can however consider and do the
…
Paul A. Lagace
? 2001
Unit 12 -
9
τττ
MIT - 16.20
Fall, 2002
Special Case of a Circular Tube
Consider the case of a circular tube with inner diameter
R
i
and outer
diameter R
o
Figure 12.6
Representation of cross-section of circular tube
For a solid section, the stress distribution is thus:
Figure 12.7
Representation of stress
“flow
” in circular tube
τ
res
is directed along circles
Paul A. Lagace
? 2001
Unit 12 -
1
0
MIT - 16.20
Fall, 2002
The resultant shear stress,
τ
res
, is always tangent to the boundaries of the
cross-section So, we can
“cut out” a circular piece (around same origin) without violating
the boundary conditions
(of
τ
res
acting tangent to the boundaries)
Using the solution for a solid section, we subtract the
torsional
stiffness of
the
“
removed
piece
”
(radius of
R
i
) from that for the solid section (radius of
R
o
)
π
R
4
π
R
4
J
=
o
?
i
2
2
Exact solution for thick-walled circular tube
let us now consider:
Paul A. Lagace
? 2001
Unit 12 -
1
1
MIT - 16.20
Fall, 2002
Thin-Walled Closed Sections
Figure 12.7
Representation of cross-section of thin-walled closed section
outer
inner
Here, the inner and outer boundaries are nearly parallel
?
resultant
shear stresses throughout wall are tangent to the median line.
Basic assumption for thin, closed section:
τ
resultant
is approximately constant through the thickness t.
Paul A. Lagace
? 2001
Unit 12 -
1
2
MIT - 16.20
Fall, 2002
For such cases:
A
outer
≈
A
inner
≈
A
Hence:
∫
τ
ds
≈
∫
τ
ds
≈
2
GkA
outer
inner
Note
:
basic difference from singly-connected
boundaries (open sections).
Figure 12.9
Representation of stress distribution through thickness in
open
cross-section
under
torsion
very
important difference
ττττ
res
varies
linearly
through- the-thickness
Now, we need to find the boundary conditions:
Paul A. Lagace
? 2001
Unit 12 -
1
3
MIT - 16.20
Fall, 2002
Figure 12.10
Representation of forces on thin closed cross-section
under
torsion
Force:
dF =
ττττ
t
ds
contribution to torque:
dT
=
h
τ
t
d
s
(h = moment arm)
Note
: h,
τ
, t vary with s (around section)
Total torque
=
∫
dT
=
∫
τ
th
d
s
Paul A. Lagace
? 2001
Unit 12 -
1
4
MIT - 16.20
Fall, 2002
But
τ
t is constant around the section.
This can be seen by cutting out a
piece of the wall AB. Figure 12.11
Representation of infinitesimal piece of wall of thin
closed section under torsion
z
x-y plane
Use
∑
F
z
=
0
to give:
?
τ
td
z
+
τ
B
t
B
d
z
=
0
AA
?
τ
t
AA
=
τ
B
t
B
in general:
τ
t = constant
Paul A. Lagace
? 2001
Unit 12 -
1
5
MIT - 16.20
Fall, 2002
Define:
“shear flow” = q =
τ
t
= constant
(we will use the concept of
“shear
flow
”
Analogy:
single 1-D pipe flow
when we deal with shell beams)
uh = constant
velocity
Returning to
∫
dT
=
∫
τ
th
d
s
since
τ
t = constant gives:
∫
dT
=
τ
t
∫
h
d
s
But,
hds = 2dA via geometric argument:
Paul A. Lagace
? 2001
Unit 12 -
1
6
MIT - 16.20
Fall, 2002
hds
dA
=
2
?
height
2
x
base
? ?
=
Area
T
of
Triangle
?
Finally:
?
dT
t
∫
=
τ
∫
2
d
A
?
T
=
2
τ
t
A
T
?
τ
resultant
=
2
At
Bredt
’
s
(12 - 2)
formula
Paul A. Lagace
? 2001
Unit 12 -
1
7
MIT - 16.20
Fall, 2002
Now to find the angle of twist, place (12 - 2) into (12 - 1):
T
ds
=
G
k
2
A
2
At
∫
T
ds
?
k
=
∫
2
4
AG
t
This can be rewritten in the standard form:
d
α
T
k
=
=
dz
GJ
4
A
2
?
J
=
t
ds
∫
(
Note
:
use midline for calculation)
valid for any shape
…..
Paul A. Lagace
? 2001
Unit 12 -
1
8
MIT - 16.20
Fall, 2002
Figure 12.12
Representation of general thin closed cross-section
How good is this approximation?
It will depend on the ratio of the thickness to the overall dimensions of the cross-section (a radius to the center of torsion) Can explore this by considering the case of a circular case since we have an exact solution:
π
R
4
?
π
R
4
J
=
o
i
2
versus approximation:
4
A
2
J
≈
(will explore in home assignment)
t
ds
∫
Paul A. Lagace
? 2001
Unit 12 -
1
9
MIT - 16.20
Fall, 2002
Final note on St.
Venant
Torsion:
When we look at the end constraint (e.g., rod attached at boundary):
Figure 12.13
Overall view of rod under torsion
Here, St.
Venant
theory
is good
in this local region, violation of assumption of St.
Venant
theory
Built-in end
At the base, w = 0.
This is a violation of the
“
free to warp
”
assumption.
Thus,
σ
zz
will be present.
?
resort to complex variables
(See Timoshenko
& Rivello
)
Paul A. Lagace
? 2001
Unit 12 -
2
0