# AERSP 301 Torsion of closed and open section beams

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AERSP 301 Torsion of closed and open section beams
Jose L. Palacios July 2008

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

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

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

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

Torsion of closed section beams
No axial restraint In absence of direct stress, Recall

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

Torsion of closed section beams
Twist and Warping of closed section beams Lecture Earlier, For const. q Also Needed for HW #5 problem 3

Torsion of closed section beams
Starting with warping expression: For const. q Using

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.

Find Warping Distribution
Sample Problem Assume G is constant From 0 to 1, 0 ≤ S1 ≤ b/2 and Find Warping Distribution

Sample Problem Warping Distribution 0-1 is:

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?

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.

Sample Problem In the wall 1-2

Sample Problem Similarly, it can be show that a s2 b

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

Sample Problem We use the condition that for no axial restraint, the resultant axial load is zero:

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

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

Torsion / Warping of thin-walled OPEN section beams
Now determine warping distribution, Recall: Referring tangential displacement, vt, to center or twist, R:

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

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

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:

BEGINNING SIDENOTE

Torsional Constants Examples and Solutions
SideNote: Calculation of torsional constant J (Chapter N, pp 367 Donaldson, Chapter 4 Megson) Torsional Constants Examples and Solutions

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

St. Venant’s Constant For Uniform Torsion (or Torsion Constant)
z y Mt Φ F

J is varies for different cross-sections
Torsion Constant J is varies for different cross-sections #1 #2 #3

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:

EXAMPLE # 1 Area Ellipse: Polar Moment of Inertia: 4) J: z 2b y 2a
5) Substitute into Ψ(y,z) 6) Differentiate 5)

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

Stress and Stiffness Parameters for Rectangular Cross-Sections (pp 393)

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

Similarly: Open Thin Cross-Sections
S is the Contour Perimeter

Extension to Thin Sections with Varying Thickness (pp 409)
η Thickness b(ξ) z ξ By analogy to thin section y

Torsional Constants for an Open and Closed CS

END SIDENOTE

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:

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

Torsion / Warping Sample Problem
Warping distribution is linear in 0-2 and:

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

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