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Department of Aerospace Engineering
AERSP 301Torsion of closed and open section
beams
Jose L. Palacios
July 2008
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Department of Aerospace Engineering
Torsion of closed section beams
To simultaneously satisfy these, q = constant
Thus, pure torque const. shear flow in beam wall
A closed section beam
subjected to a pure
torque T does not in the
absence of axial
constraint, develop anydirect stress, z
Now look at pure torsion of closed c/s
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Department of Aerospace Engineering
Torsion of closed section beams
Torque produced by
shear flow acting on
element s is pqs
[Bredt-Batho formula]
Since q = const. &
Hw # 3, problem 3
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Department of Aerospace Engineering
Torsion of closed section beams
Already derived warping distribution for a shear loaded closed c/s
(combined shear and torsion)
Now determine warping distribution from pure torsion load
Displacements associated with Bredt-Batho shear flow (w & vt):
0 = Normal Strain
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Department of Aerospace Engineering
Torsion of closed section beams
In absence of direct stress,
Recall
No axial
restraint
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Torsion of closed section beams
To hold for all points around the c/s
(all values of)
c/s displacements have a
linear relationship with
distance along the beam, z
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Torsion of closed section beams
Earlier,
For const. q
Twist and Warping of closed
section beams Lecture
Also Needed for HW #5 problem 3
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Torsion of closed section beams
Starting with warping expression:
For const. q
Using
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Twisting / Warping sample problem
Determine warping distribution in doubly symmetrical,closed section beam shown subjected to anticlockwise
torque, T.
From symmetry, center oftwist R coincides with
mid-point of the c/s.
When an axis ofsymmetry crosses a wall,
that wall will be a point ofzero warping.
Take that point as theorigin of S.
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Sample Problem
Assume G is constant
abAt
a
t
bw
tds
tds
A
A
AG
Tww
ab
s
ssos
and,2,0
and
2
0
00s
00
From 0 to 1, 0 S1 b/2 and
4and, 10
1
0
10
1 asA
t
s
t
dss
b
s
s Find Warping Distribution
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Sample Problem
Warping Distribution 0-1 is:
abo t
a
t
b
b
s
abG
T
w
1
1 4
ab t
a
t
b
abG
Tw
bs
8
2/@
1
1
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Department of Aerospace Engineering
Sample Problem
The warping distribution can be deducedfrom symmetry and the fact that wmust be
zero where axes of symmetry intersect the
walls. Follows that: w2= -w1, w3 = w1, w4 = -w1
What would be warping fora square cross-section?
What about a circle?
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Department of Aerospace Engineering
Sample Problem
Resolve the problem choosing the point 1as the origin fors.
In this case, we are choosing an arbitrary
point rather than a point where WE KNEWthat wowas zero.
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Department of Aerospace Engineering
Sample Problem
In the wall 1-2
ab
a
ab
a
s
ss
t
a
t
b
abG
Tw
as
ts
abGTw
abAt
a
t
bw
t
s
t
ds
t
ds
AA
AGTww
42@
42'
and,2,0setting
and
2
2
1112
0
1
00s
00012
a
s
tt
a
t
bs
abGt
a
t
bT
w
a
ab
ab
42
2
2
' 1112
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Department of Aerospace Engineering
Sample Problem
Similarly, it can be show that
2
2
23 4
11
2' sbbt
s
t
a
abG
T
wba
22
1
22
12
2
0
0s
as
baA
t
s
t
a
t
ds
os
ba
s
b
a
s2
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Department of Aerospace Engineering
Thus warping displacement varies linearlyalong wall 2, with a value w2at point 2,
going to zero at point 3.
Distribution in walls 34 and 41 follows fromsymmetry, and the total distribution is
shown below:
Sample Problem
Now, we calculate w0which we
had arbitrary set to zero
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Department of Aerospace Engineering
Sample Problem
We use the condition that for no axial restraint,
the resultant axial load is zero:
0 dstz
tds
dstww
dstww
dswt
s
o
os 0)(
0
zw
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Department of Aerospace Engineering
Sample Problem
Substituting forw12andw23 and evaluating theintegral:
a b
sba
ba
o dstwdstwbtat
w0 0
23112 ''2
2
abo t
a
t
b
abG
Tw
8
Offset that need to be added to previously found warping distributions
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Department of Aerospace Engineering
Torsion / Warping of thin-walled OPEN section
beams
Torsion of open sections creates a different type of shear distribution Creates shear lines that follow boundary of c/s
This is why we must consider it separately
Maximum shear located
along walls, zero in center
of member
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Department of Aerospace Engineering
Torsion / Warping of thin-walled OPEN
section beams
Now determine warping distribution, Recall:
Referring tangential displacement, vt, to center or twist,
R:
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Department of Aerospace Engineering
Torsion / Warping of thin-walled OPEN
section beams
On the mid-line of the
section wall zs = 0,
Integrate to get warping displacement:
where
AR, the area swept by agenerator rotating about
the center of twist from
the point of zero
warping
Distance from wall to shear center
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Department of Aerospace Engineering
Torsion / Warping of thin-walled OPEN
section beams
S = 0 (W = 0)
ARR
R
The sign ofwsis dependent on the
direction of positive torque
(anticlockwise) for closed section
beams.
For open section beams, pris positive if
the movement of the foot ofpralong the
tangent of the direction of the assumed
positive sprovides a anticlockwise areasweeping
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Department of Aerospace Engineering
Torsion / Warping Sample Problem
Determine the warping
distribution when the thin-
walled c-channel section
is subjected to an anti-
clockwise torque of 10 Nm
SideNote:
G = 25 000 N/mm2
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Department of Aerospace Engineering
BEGINNING SIDENOTE
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Department of Aerospace Engineering
SideNote: Calculation of torsional
constant J
(Chapter N, pp 367 Donaldson,
Chapter 4 Megson)
Torsional Constants Examples and
Solutions
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Department of Aerospace Engineering
Stresses for Uniform Torsion
z
x
y
MtMt
Assumptions:
1) Constant Torque Applied
2) Isotropic, Linearly Elastic
3) No Warping Restraint
All Sections Have Identical Twist per Unit Length:
No Elongation
No Shape Change
z
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Department of Aerospace Engineering
St. Venants Constant For Uniform Torsion
(or Torsion Constant)
A
t dA
dz
dG
MJ
2
4
FuEA
MGJ t
F
Mt
z
y
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Department of Aerospace Engineering
Torsion Constant
Jis varies for different cross-sections
#1 #2
#3
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Department of Aerospace Engineering
EXAMPLE #1 (ELLIPSE)
Find S. Torsion Constant For Ellipse:
Find Stress Distribution (xyxz)
01
22
b
z
a
y
2b
2a
1) Eq. Boundary:
2) = 0on Boundary:
22
1),(b
z
a
yCzy o3) Substitute into GDE:
22
222 2),(babaGC
dzdGzy o
22
22
22
1),(
b
z
a
y
ba
baGzy
y
z
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Department of Aerospace Engineering
EXAMPLE # 1
ab
pIba
baJ
22
33
y
zy
z
zy
xz
xy
),(
),(
2b
2a
4) J:
22
33),(2
ba
ba
G
dzdyzy
G
MJ t
5) Substitute into (y,z)J
M
Gt
22
1),(b
z
a
y
ab
Mzy t
y
z
Area Ellipse:
6) Differentiate 5)
Polar Moment of Inertia: 22
4
1baabIp
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Department of Aerospace Engineering
EXAMPLE #2 (RECTANGLE)
m n
mnb
znCos
a
ymCosCzy
),(
b
a
1) Eq. Boundary: Simple Formulas
Do Not Satisfy GDE and BCs
NEED TO USE SERIES
For Orthogonality use Odd COS Series
(n & m odd)
2) Following the procedure in pp 391 and 3923abJ
y
z
2max
1
ab
Mtxs
))/((
1256222226 nabmnmb
af
Find S. Torsion Constant For Ellipse:
Find Stress Distribution (xyxz)
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Department of Aerospace Engineering
31.0
10/
ba
3
1
3
1
/
ba
Stress and Stiffness Parameters
for Rectangular Cross-Sections (pp 393)
,
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Department of Aerospace Engineering
a>>b Rectangle
0)2/( b
GzGz 22)( 2
22
2
2
2)(
bzGz
by
z
No variation in in y
BCs:
G
MJ
dAM
t
A
t 2
J
bMtxs max
3
3
1
abJ
Integrating
Differentiating
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Department of Aerospace Engineering
Similarly: Open Thin Cross-Sections
t
S
J
1
3St
3
S is the Contour Perimeter
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Department of Aerospace Engineering
Extension to Thin Sections with Varying
Thickness (pp 409)
2
2
2
)()(
bGz
ddbGdydzzyM
a b
bA
t0
2/)(
2/)(
22 )(4
12),(2
GJdbGM
a
t0
3 )(3
1
Thickness b()
z
y
By analogy to thin section
J
bM
dbJ
txs
a
max
max
0
3 )(31
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Torsional Constantsfor an Open and Closed CS
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Department of Aerospace Engineering
END SIDENOTE
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Torsion / Warping Sample Problem
Determine the warping
distribution when the thin-
walled c-channel section
is subjected to an anti-
clockwise torque of 10 Nm
Side Note:
G = 25 000 N/mm2
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Department of Aerospace Engineering
Torsion / Warping Sample Problem
433 mm7.316)5.2505.1252(31 J
Origin for s (and AR) taken at intersection of web and axis of symmetry,
where warping is zero
Center of twist = Shear Center, which is located at:(See torsion of beam open cross-section lecture)
42
mm04.8
1
3
h
bh
bs
In wall 0-2:104.8
2
1sAR
Since pRis positive
Positive
pR
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Department of Aerospace Engineering
Torsion / Warping Sample Problem
mm01.07.31625000
1010
04.82
1
2 1
3
102 ssw
Warping distribution is linear in 0-2 and:
mm25.02501.02 w
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Department of Aerospace Engineering
Torsion / Warping Sample Problem
In wall 2-1:
21
252
104.8
2
1ssA
R
dspdspA RRR 2102 21
2
1
??
mm04.8
21
02
R
R
p
p pR21
-25 mm
NegativepR
The are Swept by the generator
in wall 2-1 provides negative
contribution to AR
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Torsion / Warping Sample Problem
mm04.803.07.31625000
101025
2
12504.8
2
12
2
3
221
s
sw
Again, warping distribution is linear in wall 2-1,
going from -0.25 mm at pt.2 to 0.54 mm at pt.1
The warping in the lower half of the web and
lower flange are obtained from symmetry
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