AERSP 301 Torsion of closed and open section beams

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AERSP 301Torsion of closed and open section beams

• Jose L. Palacios
• July 2008

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REMINDERS

• IF YOU HAVE NOT TURN IN HW 4 PLEASE DO SO ASAP
  TO AVOID FURTHER POINT PENALTIES.
• HW 5 DUE FRIDAY, OCTOBER 3
• HW 6 (FINAL HW from me) DUE FRIDAY OCTOBER 10
• EXAM OCTOBER 20 26 HOSLER 815 1015 PM
• REVIEW SESSION OCTOBER 19 220 HAMMOND 6 9
  PM

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Torsion of closed section beams

• Now look at pure torsion of closed c/s

• A closed section beam subjected to a pure torque
  T does not in the absence of axial constraint,
  develop any direct stress, ?z

• To simultaneously satisfy these, q constant
• Thus, pure torque ? const. shear flow in beam wall

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Torsion of closed section beams

• Torque produced by shear flow acting on element
  ?s is pq?s

• Since q const.

Bredt-Batho formula
Hw 3, problem 3
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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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Torsion of closed section beams
No axial restraint

• In absence of direct stress,
• Recall

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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
Twist and Warping of closed section beams Lecture

• Earlier,
• For const. q

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 of twist R coincides with
  mid-point of the c/s.
• When an axis of symmetry crosses a wall, that
  wall will be a point of zero warping.
• Take that point as the origin of S.

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Sample Problem

• Assume G is constant

From 0 to 1, 0 S1 b/2 and
Find Warping Distribution
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Sample Problem

• Warping Distribution 0-1 is

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Sample Problem

• The warping distribution can be deduced from
  symmetry and the fact that w must be zero where
  axes of symmetry intersect the walls.
• Follows that w2 -w1, w3 w1, w4 -w1

What would be warping for a square cross-section?
What about a circle?
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Sample Problem

• Resolve the problem choosing the point 1 as the
  origin for s.
• In this case, we are choosing an arbitrary point
  rather than a point where WE KNEW that wo was
  zero.

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Sample Problem

• In the wall 1-2

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Sample Problem

• Similarly, it can be show that

a
s2
b
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Sample Problem

• Thus warping displacement varies linearly along
  wall 2, with a value w2 at point 2, going to
  zero at point 3.
• Distribution in walls 34 and 41 follows from
  symmetry, and the total distribution is shown
  below

Now, we calculate w0 which we had arbitrary set
to zero
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Sample Problem
We use the condition that for no axial restraint,
the resultant axial load is zero
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Sample Problem

• Substituting for w12 and w23 and evaluating the
  integral

Offset that need to be added to previously found
warping distributions
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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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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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Torsion / Warping of thin-walled OPEN section
beams

• On the mid-line of the section wall ?zs 0,

Distance from wall to shear center

• Integrate to get warping displacement

AR, the area swept by a generator rotating about
the center of twist from the point of zero warping
where
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Torsion / Warping of thin-walled OPEN section
beams
The sign of ws is dependent on the direction of
positive torque (anticlockwise) for closed
section beams. For open section beams, pr is
positive if the movement of the foot of pr along
the tangent of the direction of the assumed
positive s provides a anticlockwise area
sweeping
AR
R
S 0 (W 0)
?R
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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

G 25 000 N/mm2
SideNote
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BEGINNING SIDENOTE
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SideNote Calculation of torsional constant
J(Chapter N, pp 367 Donaldson, Chapter 4 Megson)

• Torsional Constants Examples and Solutions

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Stresses for Uniform Torsion

• Assumptions
• Constant Torque Applied
• Isotropic, Linearly Elastic
• No Warping Restraint

y
Mt
Mt
x
z
All Sections Have Identical Twist per Unit
Length
No Elongation No Shape Change
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St. Venants Constant For Uniform Torsion (or
Torsion Constant)
z
y
Mt
F
F
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Torsion Constant

• J is varies for different cross-sections

1
2
3
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EXAMPLE 1 (ELLIPSE)

• Find S. Torsion Constant For Ellipse
• Find Stress Distribution (sxy sxz)

z
1) Eq. Boundary
2b
y
2) ? 0 on Boundary
2a
3) Substitute ? into GDE
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EXAMPLE 1
Area Ellipse
4) J
Polar Moment of Inertia
z
2b
y
2a
5) Substitute into ?(y,z)
6) Differentiate 5)
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EXAMPLE 2 (RECTANGLE)

• Find S. Torsion Constant For Ellipse
• Find Stress Distribution (sxy sxz)

z

• Eq. Boundary Simple Formulas
• Do Not Satisfy GDE and BCs
• NEED TO USE SERIES
• For Orthogonality use Odd COS Series
• (n m odd)

b
y
a
2) Following the procedure in pp 391 and 392
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Stress and Stiffness Parametersfor Rectangular
Cross-Sections (pp 393)
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agtgtb Rectangle
z
b
y
No variation in ? in y
BCs
Integrating
Differentiating ?
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Similarly Open Thin Cross-Sections
S
t
S is the Contour Perimeter
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Extension to Thin Sections with Varying Thickness
(pp 409)
?
Thickness b(?)
z
?
By analogy to thin section
y
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Torsional Constants for an Open and Closed CS
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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

G 25 000 N/mm2
Side Note
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Torsion / Warping Sample Problem
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)
Positive pR
In wall 0-2
Since pR is positive
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Torsion / Warping Sample Problem
Warping distribution is linear in 0-2 and
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Torsion / Warping Sample Problem
In wall 2-1
pR21
-25 mm
Negative pR
The are Swept by the generator in wall 2-1
provides negative contribution to AR
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Torsion / Warping Sample Problem
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