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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

December 31, 2017

### Contents

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 .  The following diagram illustrate the basic geometry. 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). This is another diagram showing a bar under torsion #### 1.1 Problem setup

##### 1.1.1 What are the assumptions?

1. The twist rate (called in this problem) and defined as where is the twist angle is assumed to be constant.
2. Cross section can wrap also in the direction (i.e. the cross section does not have to remain in the plane) but if this happens, all cross sections will wrap in the 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 and height of the cross section.
2. Material Modulus of rigidity or sheer modulus which is the ratio of the shearing stress to the shearing strain 3. The applied torque 4. the torsion constant for the a rectangular cross section. For a rectangular section of dimensions it is given by (1)

Hence the torsional rigidity is known since is given (material) and 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 as a function of (the length of the beam).

### 2 Analytical solution using Prandtl stress function

First we solve for the Prandtl stress function by solving the Poisson equation Where is the sheer modulus and is the twist rate (which was assumed to be constant).

The boundary conditions ( 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 The analytical solution to the above equation is from book Theory of elasticity by S. P. Timoshenko and J. N. Goodier (2)

where the linear twist  Hence (2) becomes Where is given by (1)

#### 2.1 Stress components Hence and Timoshenko gives the maximum sheer stress, which is as #### 2.2 Strain components

Given that is Young’s modulus for the material, is Poisson’s ratio for the material, and 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. Hence only and are non-zero.

#### 2.3 Determining the twist angle If we look at a cross section of the bar at some distance from the end of the bar, the angle that this specific cross section has twisted due to the torque is . This angle is given by the solution to the equation But is the linear twist and is given by hence the above equation becomes Hence Where is the constant of integration. Assuming at we obtain that and using the expression given in equation (1) above we can determine for each #### 2.4 Displacement calculations  we see that Hence Where ### 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