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Last course

Bar structureEquations from the theory of elasticityStrain energyPrinciple of virtual work

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Today

Bar elementPrinciple of virtual workFormulation of bar element using the principle

of virtual work

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Please submit the 4th homework after the class

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The principle of the virtual work

δU: the virtual strain energy of the internal stresses

δW: the virtual work of external forces on the body

X, u

Z, w

w

δw

δw is the virtual displacement

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The virtual strain energy for a bar

δε : virtual strain, which is the strain associated with the virtual displacement, i.e.

The external virtual work:

0( ) ( )

LU x x Adx

( ) ( )d

x u xdx

x, u

b=b(x)

L

X=0 X=L

P

0( ) ( ) ( )

LW u x b x dx P u L

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The virtual strain energy for a bar

δε : virtual strain, which is the strain associated with the virtual displacement, i.e.

Notice that

Why is it not equal to the definition of actual strain energy?

0( ) ( )

LU x x Adx

( ) ( )d

x u xdx

x, u

b=b(x)

L

X=0 X=L

P

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( ) ( ) L

U x x Adx

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The external virtual work:

The principle of virtual work:

x, u

b=b(x)

L

X=0 X=L

P

0 0( ) ( ) ( ) ( ) ( )

L Lx x Adx u x b x dx P u L

0( ) ( ) ( )

LW u x b x dx P u L

0 0( , ) , ( )

L L

x xE u T u Adx ub dx P u L

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Problem statements of the bar

Strong or classical form Given geometrical

and material properties L, E, A, α, and external actions P, b(x), T, and support displacement u0, find u(x) such that

x, u

b=b(x)

L

X=0 X=L

P

( , ) , 0x xEA u T b on 0 x L

0(0)u u

,x x LEAu P

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Weak or variational form Given geometrical and

material properties L, E, A, α, and external actions P, b(x), T, and support displacement u0, find compatible u(x) such that for all δu(x) satisfying homogeneous boundary condition,

Finite element methods are based upon the weak statement of the problem.

x, u

b=b(x)

L

X=0 X=L

P

0

0

( , ) ,

( ) ( ) ( )

L

x x

L

E u T u Adx

u x b x dx P u L

The basis of the FEM

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Let us go back to finite element!

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Formulation of bar element Consider the two-node bar element as follows

b(x) : body force along the bar element in the unit of force per unit length (e.g. kN/m)

u=u(x) : displacement field within the element, i.e. displacement as a function of point x within the element

0 x L

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Displacements at nodes 1 and 2:

Forces at nodes 1 and 2:

The first and fundamental step in the finite element formulation is to assume the displacement field within the element in terms of its nodal displacements

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Here it is assumed that the displacement field is a linear function

Let us now express the assumed function in terms of nodal displacements d1 and d2

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As a result,

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Written in matrix form

Thus we can write

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Where

N is also called the matrix of interpolation functions, because it interpolates the displacement field u=u(x) from the nodal displacements

L x x

L L

N

1

2

d

d

d

Matrix of shape functions

Vector of nodal displacements

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The graphs of the shape functions

The important property of the shape functions is the partition of unity, i.e.

1

L xN

L

2

xN

L

1

( ) 1nnode

ii

N x

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The strain field

We can write

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where

The stress field

1 1

L L

B

Strain-displacement matrix

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Application of the principle of virtual work Imagine that the bar undergoes a linear virtual nodal

displacement δd1 and δd2

The virtual nodal displacement

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The virtual displacement

The virtual strain

The principle of virtual work

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The virtual strain energy

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The virtual work

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Thus, using the principle of virtual work we obtain

The stiffness matrix

The equivalent nodal force vector

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It can be concluded that

Let us rename the vector of applied nodal forces

The equivalent nodal force vector due to body force

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The stiffness matrix of the bar element with cross-sectional area A (may vary along the bar)

If the bar is prismatic and made up from homogeneous material, i.e. E and A are constant, then we can evaluate the stiffness matrix as follows

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What if the bar is not prismatic, e.g. if the cross-sectional area varies linearly along the bar? Can you evaluate the stiffness matrix?

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The equivalent nodal force

Now suppose that the body force varies linearly along the element as shown here

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The equivalent nodal force

L/6 (2b1+b2)

L/6 (b1+2b2)

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The effect of temperature change

Suppose that now there is no mechanical loads and the bar undergoes temperature change T0C.

Assume that the coefficient of thermal expansion of the bar is constant α /0C.

Using the principle of virtual work and assuming the displacement field is linear as before, show that the equivalent nodal force vector due to the temperature change is

1

1T EA T

f

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Example of the DSM including ‘element load’ Given a steel bar of the

diameter 16 mm. Axial distributed force 5 kN/m acts along the bar and a concentrated force 10 kN acts at the right end.

If the bar is discretized as in the figure, determine the internal stress along the bar.

Compare the finite element solutions to the exact solutions.

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See Lecture 5 OHP Slides