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CITY AND GUILDS 9210

UNIT 135

MECHANICS OF SOLIDS Level 6

TUTORIAL 15 - FINITE ELEMENT ANALYSIS - PART 2

This tutorial covers some more advanced aspects of FEA theory and may not be required for the

examination.

MODELLING BEAM ELEMENTS

A beam is a commonly used element in FEA. We will consider a simple beam element that deflects

in 2 dimensions only on the x, y plane. We will also assume that:

The beam is loaded only in the y direction.

The beam material is uniform and elastic.

The stiffness is large so the deflections are small.

The beam has a uniform cross section symmetrical about the neutral axis

You should understand that a beam differs from a truss because a beam bends and a truss does not.

Beams carry transverse loads and transmit moments along its length. It seems normal in FEA to use

'v' for the transverse deflection, 'f ' for the transverse force, 'm' for the bending moment and 'θ' for

the rotation angle which is normally the slope of the beam in traditional solid mechanics.

If a beam has a varying cross-section then it should be divided into short but different elements

although it is possible to develop a model for varying cross-sections. The theory given here is based

on the Euler-Bernoulli beam theory. The beam shown has four DOFs.

Without derivation, one model of the stiffness matrix for a beam is:

Iz is the second moment of area of the beam section about the neutral axis. The finite element

equation is:

f is the force, m the moment, θ is the slope or rotation and v is the displacement in the y direction.

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WORKED EXAMPLE No. 1

A cantilever beam is 2 m long and fixed at one end. It has a uniform rectangular cross section

100 mm x 60 mm. The beam undergoes static deflection by a downward load of F = 5000 N applied

at the free end. The modulus of elasticity E is 200 GPa. Solve the slope and deflection at the free

end.

SOLUTION

Because the beam is uniform only one element is needed. Calculate the second moment of area

about the neutral axis.

Iz = BD3/12 = 100 x 60

3 = 1.8 x 10

6 mm

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The stiffness matrix is:

The finite element equation becomes

At the fixed end v1 = 0 and θ1 = 0 At the free end f2 = -5000 N m2 = 0 L = 2 m

Eliminate that shown in red and put

v2 = -37 x 10-3

m θ2 = -0.0278 radian

Just to see how accurate this is lets solve with the standard formulae used in solid mechanics.

y = FL3/3EI = -5000 x 2

3/(3 x 200 x10

9 x 1.8 x 10

-6) = -37 x 10

-3 m

θ2 = FL2/2EI = -5000 x 2

2/(2 x 200 x10

9 x 1.8 x 10

-6) = -0.0277 radian

The bending moment and shear force at the fixed end may be evaluated easily.

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SELF ASSESSMENT EXERCISE No. 1

A cantilever beam is 6 m long and fixed at one end. The flexural stiffness (EI) is 110 MNm2. The

beam undergoes static deflection by a downward load of F = 20 kN applied at the free end. Solve

the slope and deflection at the free end.

(Answers v2 = -13.1 x 10-3

m θ2 = -0.0033 radian)

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The solution in the previous example gives an accurate answer for the end of the beam but it does

not give an accurate answer for points along the length. To do this we need a blending function or

alternatively break the length up into more elements.

BLENDING FUNCTION

The beam will be length L with the x, y origin at the middle as shown. There are two nodes 1 and 2

being the left and right ends respectively. We need to produce four suitable shape functions for the

interpolation of the variables between the nodes one for each DOF.

The 'x' coordinates at the nodes are -a and a where L is the length of the beam. For the shape

functions it is easier to use a special set of dimensionless local coordinates called the natural

coordinate system with the origin at the centre of the element as shown.

Define ξ = x/a and the natural coordinates at the nodes are ξ = -1 and ξ = +1.

Each of the four shape functions is assumed to be a third order polynomial with four unknown

constants α0 to α3. The third order polynomial is chosen because there are four nodal DOFs. You

can find reason and descriptions for the best functions with a web search. The displacement ν at any

value of ξ is:

Note the units of α must have linear dimensions (meter). The rotation θ at any value of ξ is the

gradient of the beam so:

Note the units of α/a are dimensionless for angle or gradient. The displacements at the nodes are:

In matrix form this is:

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Next we must solve the constants α0 to α3 by using the four boundary conditions from the nodal

values. At node 1 ξ1 = -1 and at node 2 ξ2 = 1 so we get:

[d]is the displacement matrix for the nodes. To find the solutions for α we use the inverse matrix

Remember the coefficients α are known values and the displacements at the nodes will be known so

the displacements at the other point can be evaluated.

The displacements in between the nodes are given by:

Substitute

=

[N] is the blending function matrix.

The easiest point to evaluate N is at the centre of the beam where ξ = 0.

Next consider the rotation between the nodes. We had for the rotation

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Substitute

=

The easiest point to evaluate N is at the centre of the beam where ξ = 0.

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WORKED EXAMPLE 2

Find the deflection and rotation at the middle of the cantilever in worked example No. 1

SOLUTION

For the deflection at the middle of the cantilever put ξ = 0

Put a = 1

ν = (1/2) (-37 x 10-3

) - (1/4)(-0.0278) = -11.55 x 10-3

m or -11.55 mm

From classical beam theory

F = 5000 N L = 2 m x = 1 m E = 200 x 109 N/m

2 I = 1.8 x 10

-6 m

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Evaluate and

y = -0.01157 m or 11.57 mm

Rotation at the middle of the cantilever ξ = 0

Put a = 1

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θ = (3/4) (-37 x 10-3

) - (1/4)(-0.0278) = -0.0208 radian

From classical beam theory

F = 5000 N L = 2 m x 1 m E = 200 x 109 N/m

2 I = 1.8 x 10

-6 m

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SELF ASSESSMENT EXERCISE No. 2

A cantilever beam is 6 m long and fixed at one end. The flexural stiffness (EI) is 110 MNm2. The

beam undergoes static deflection by a downward load of F = 20 kN applied at the free end. Solve

the slope and deflection at the middle. (This is a continuation of Self Assessment Exercise No. 1)

(Answers y = -4.091 mm and θ = -0.00245 radian)

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3D STRESS AND STRAIN

Consider a small cube of solid isotropic material with x, y and z axis as shown. In this analysis the

face of the cube is defined by the direction of the axis. Each face has a direct stress (σ) and two

shear stresses (). The first letter of the subscript indicates the face and the second letter the direction.

For a static cube there must be equal and opposite stresses on the reverse faces.

For equilibrium of moments xy = yx, xz = zx and zy = yz. This is commonly known as

complementary shear stress.

What this comes down to is that if we shrink our cube to a point there are 6 stresses three direct

stresses σxx, σyy and σzz and 3 shear stresses yz, xz and xy.

These are written in vector/matrix form [σ ] = [σxx σyy σzz yz xz xy]

The strains are similarly written [ε] = [εxx εyy εzz yz xz xy]

Strain is the derivative of displacement. The displacements in the x, y and z directions are normally

written u, v and w respectively.

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For an isotropic material the elastic constants are related by the following:

E -- Modulus of Elasticity G -- Modulus of Rigidity ν -- Poisson's Ratio

The matrix relating stress and strain can be written:

Using the 3D stress/strain relationships for an isotropic material it can be shown that this reduces to:

Where:

All this is getting well and truly complicated so let's simplify it by seeing what happens in 2

dimensional cases.

2D STRESS AND STRAIN

In this case we will consider models in the x y plane only and there are no forces or stresses in the z

direction. Our equation simplify to:

[σ ] = [σxx σyy xy] and [ε] = [εxx εyy xy]

Strain is the derivative of displacement. The displacements in the x, y and z directions are normally

written u, v and w respectively.

The strain matrix is:

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For an isotropic material the elastic constants are related by the following:

From which we get the relationships:

The matrix relating stress and strain can be written:

This may be altered to:

Now we have the tools to convert displacement into strains and strains into stress.