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Solving the torsion problem for isotropic matrial with a rectangular cross section using the FEM and FVM methods with triangular elements

Nasser M. Abbasi

June 20, 2014

Contents

1 Introduction
 1.1 Problem setup
  1.1.1 What are the assumptions?
  1.1.2 What is the input and what is the output?
  1.1.3 The output from the problem (the things we need to calculate)
2 Analytical solution using Prandtl stress function
 2.1 Stress components
 2.2 Strain components
 2.3 Determining the twist angle α
 2.4 Displacement calculations
3 References

For MAE 207, Computational methods. UCI. Fall 2006

1 Introduction

We consider bar made of isotropic martial with rectangular cross section subjected to twisting torque T  .  The following diagram illustrate the basic geometry.

PIC

Experiments show that rectangular cross sections do wrap and that cross sections do not remain plane as shown in this diagram (in the case of a circular cross section, cross section do NOT wrap).

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This is another diagram showing a bar under torsion

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1.1 Problem setup

1.1.1 What are the assumptions?
1.
The twist rate (called k  in this problem) and defined as dα
dz  where α  is the twist angle is assumed to be constant.
2.
Cross section can wrap also in the z  direction (i.e. the cross section does not have to remain in the xy  plane) but if this happens, all cross sections will wrap in the z  section by the same amount.
3.
Material is isotropic
1.1.2 What is the input and what is the output?

The input to the problem are the following (these are the known or given):

1.
The width b  and height a  of the cross section.
2.
Material Modulus of rigidity or sheer modulus G  which is the ratio of the shearing stress τ  to the shearing strain γ
3.
The applied torque T
4.
J  the torsion constant for the a rectangular cross section. For a rectangular section of dimensions a,b  it is given by

          (              ∞           (     ) )
     16-3       192--a  ∑    -1-       n-πb-
J =  3 a b  1 −  bπ5         n5 tanh   2 a
                      n=1,3,5⋅⋅⋅
(1)

Hence the torsional rigidity GJ  is known since G  is given (material) and J  is from above (geometry).

1.1.3 The output from the problem (the things we need to calculate)
1.
The stress distribution in the cross section (stress tensor field). Once this is found then using the material constitutive relation we can the strain tensor field.
2.
The angle of twist α  as a function of z  (the length of the beam).

2 Analytical solution using Prandtl stress function

First we solve for the Prandtl stress function Φ (x,y)  by solving the Poisson equation

  2
∇  Φ (x,y) = − 2GK

Where G  is the sheer modulus and k  is the twist rate (which was assumed to be constant).

The boundary conditions (Φ (x,y)  at any point on the edge of the cross section and at the ends of the beam) is an arbitrary constant. We take this constant to be zero. Hence at the cross section boundary we have

Φ =  0

The analytical solution to the above equation is from book Theory of elasticity by S. P. Timoshenko and J. N. Goodier

                                        (          (   ))
           32 Gk  a2  ∑∞    1      (n−1)      cosh  nπy       ( nπx )
Φ (x, y) = ----3----       --3 (− 1) 2    1 − -----(n2aπb)-  cos  ----
              π     n=1,3,5,⋅⋅⋅n                  cosh  -2a-         2a
(2)

where the linear twist k

k = -T--
    GJ

Hence (2) becomes

                                      (                )
                 2  ∑∞            (n−1)      cosh (nπy)      (     )
Φ (x,y) =  32-T-a--       1--(− 1)-2--- 1 − -----(-2a)-  cos  nπx-
            J π3  n=1,3,5,⋅⋅⋅n3                cosh  nπ2ba          2a

Where J  is given by (1)

2.1 Stress components

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Hence

                                         (         (    ))
        ∂Φ    16Gka    ∑∞    1      (n−1)-      cosh  nπy       (nπx )
τyz = − ---=  ---2---       --2 (− 1) 2    1 − -----(2naπb)- sin  ----
        ∂x      π    n=1,3,5,⋅⋅⋅n                  cosh  2a          2a

and

      ∂Φ    16 T a   ∑∞    1      (n−1)(      ( nπx )      ( bnπ )     ( nπy ))
τxz = --- = ----2--        -2 (− 1) 2    − cos  ----  sech   ----  sinh   ----
      ∂y      Jπ   n=1,3,5,⋅⋅⋅n                     2a          2a          2a

Timoshenko gives the maximum sheer stress, which is ∘ ---------
  τy2z + τ2xz  as

                        (               )
                ∞∑
τmax = 16Gka--       -1-  1 − ----1(---)
         π2   n=1,3,5,⋅⋅⋅n2       cosh  n2πab

2.2 Strain components

Given that E  is Young's modulus for the material, υ  is Poisson's ratio for the material, and G  = --E---
     2(1+υ)   we can now obtain the strain components from the constitutive equations (stress-strain equations) since we have determined the stress components from the above solution.

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Hence only γyz  and γxz  are non-zero.

2.3 Determining the twist angle α

If we look at a cross section of the bar at some distance z  from the end of the bar, the angle that this specific cross section has twisted due to the torque is α  .

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This angle is given by the solution to the equation

dα (z)
------=  k
  dz

But k  is the linear twist and is given by     -T-
k = GJ  hence the above equation becomes

dα (z )    T
------=  ----
 dz      GJ

Hence

         T
α (z) = ---z + C1
        GJ

Where C1   is the constant of integration. Assuming α = 0  at z =  0  we obtain that

α(z) =  T--z
        GJ

and using the expression J  given in equation (1) above we can determine α  for each z.

2.4 Displacement calculations

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pict

we see that

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Hence

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Where α = GTJ-z

3 References

1.
Mathematica Structural Mechanics help page
2.
MIT course 16.20 lecture notes. MIT open course website.
3.
Theory of elasticity by S. P. Timoshenko and J. N. Goodier. chapter 10