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**AERSP 301 Torsion 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 – 8:15 – 10:15 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 pqs 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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**Find Warping Distribution**

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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**What would be warping for a square cross-section?**

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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**Now, we calculate w0 which we had arbitrary set to zero**

Sample Problem Thus warping displacement varies linearly along wall 2, with a value w’2 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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**Offset that need to be added to previously found warping distributions**

Sample Problem Substituting for w’12 and w’23 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) ρR

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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 = N/mm2 SideNote:

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BEGINNING SIDENOTE

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**Torsional Constants Examples and Solutions**

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. Venant’s Constant For Uniform Torsion (or Torsion Constant)**

z y Mt Φ F

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**J is varies for different cross-sections**

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 (σxy σxz) z 1) Eq. Boundary: 2b y 2) Ψ = 0 on Boundary: 2a 3) Substitute Ψ into GDE:

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**EXAMPLE # 1 Area Ellipse: Polar Moment of Inertia: 4) J: z 2b y 2a**

5) Substitute into Ψ(y,z) 6) Differentiate 5)

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**EXAMPLE #2 (RECTANGLE) Find S. Torsion Constant For Ellipse:**

Find Stress Distribution (σxy σxz) z Eq. Boundary: Simple Formulas Do Not Satisfy GDE and BC’s 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 Parameters for Rectangular Cross-Sections (pp 393)**

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**a>>b Rectangle z b y No variation in Ψ in y BC’s: Integrating**

Differentiating Ψ

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**Similarly: Open Thin Cross-Sections**

S is the Contour Perimeter

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**Extension to Thin Sections with Varying Thickness (pp 409)**

η Thickness b(ξ) z ξ 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 = 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 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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