FEM-6. Lecture_0.pdf

Bojtr-Gspr: The finite element method: The basis Lecture 6

07.06.13.1

Lecture 6: Matrices and vectors of an element (stiffness matrix, loadvector)

In this lecture we analyze only one finite element. We assume that we know at this elementthe vectors 0p, u , , , , the matrices L , D , and

a./ we know the exact shape of the element, assigned the nodes, choose a localcoordinate system (if it is a parametric local system, than we defined already theJacobian matrix J and its inverse), we have determined with these coordinates thenecessary shape functions, and we have generated also the interpolation equation

eu Nv (6.1)too. According to these preparations in the following operations:

b./ we generate the strain matrix B of the element, then

c./ we determine the stiffness matrix eK of the element, and finally

d./ we reduce the external loads of the element to the nodes, namely we express thereduced load vector

eq .

Calculation of the strain matrices

The geometrical equation of any structure is the following one:Lu (6.2)

If we substitute here the interpolation of the displacements (equation (6.1)) we obtain theequation:

e eLNv Bv . (6.3)We introduced here a new symbol B LN , where B is called the strain matrix of theelement in the forthcoming, because it yields the connection between strain-vector and thenodal displacement vector.

If we can write the shape functions in a global coordinate-system, then the calculation of Bis relatively simple, because we must derive the polynoms by their any variables. If we haverotated the coordinate-system, then this statement is not true at first sight, although atisotropic material we can take off the directional derivatives in the operator L by the localcoordinate-system. In these cases we shall rotate the elementary stiffness matrix and thereduced load vector into the global coordinate system.

We illustrate the calculation of B with some examples.

Example 6.1

Determine the matrix B at two and three-node bar element and illustrate their physicalmeanings!


07.06.13.2

The operator L and the matrix N are given from our earlier calculations. So at a two-node element:

B LN

lll

xxlxx

dxd 1112 ,

and a three-node element:

2 3 1 3 1 21 2 1 3 2 1 2 3 3 1 3 2

x x x x x x x x x x x xdBdx x x x x x x x x x x x x

2313

21

3212

31

3121

32 222xxxx

xxxxxxx

xxxxxxx

xxx .

We show in figure 6.1 that at a two-node element we obtain zero strains if weprescribe the same dislocations in both nodes, namely Tev u u , since

eBv 011

uu

ll.

If the dislocations are different, the strain will be constant on the whole element:

eBv luu

uu

ll12

2

111

.

Figure 6.1: The strain-functions at two different bar element

According to the figure 6.1/b at the three-node element with rigid body motion thestrain vector will be

eBv

2313

21

3212

31

3121

32 222xxxx

xxxxxxx

xxxxxxx

xxx 0

uuu

,

and with nodal displacements originated from a linear displacement function:


07.06.13.3

B

axxauxxau

u

131

121

1

,

so the strain will be constant. In a general case we obtain a linear strain function,because the elements of the matrix are linear functions and we describe the linearcombinations of them.

We can find from this example that the matrix B is the function of the variable x (this is thebasic variable), but its elements might be constants in a specific case.

In case of parametric coordinates we must apply the chain rule at the differentiation of theshape functions. For instance at a two-variable shape function:

xN

xN

xN iii

, . (6.4)

Example 6.2

Determine the strain matrix of a triangular plate element (see the example of figure 4.8)!

We have seen already in the example 4.2 that with the help of the shape functions 11N , 2N , 3N we can calculate the Jacobian-matrix and its inverse:

1

4 234 34

7 534 34

x xJ

y y

.

So the derivatives of the shape functions by the chain rule: 346

3421

344111

xxN ,

342

3451

34711

yN ,

344

3420

34412

xxN ,

347

3450

34712

yN ,

342

3421

34403

xxN ,

345

3451

34703

yN .

Applying the operator L of a plate:

1 2 3

1 2 3

xN N N

BN N Ny

y x

254762572

246

341 .

We can find two conclusions to a plate with n nodes: on the one hand the derivatives of theshape functions can be expressed from the general relationship:


07.06.13.4

1

11

1

n

nn

N Nx x J N NN Ny y

, (6.5)

on the other hand the matrix B contains n blocks with similar internal structure:

1 nB B B , (6.6)where

i

ii

i i

Nx

NBy

N Ny x

. (6.7)

At the application of the natural coordinates we must use the chain rule. For instance at afour-variable shape function the necessary expression is the following one:

xL

LN

xL

LN

xLLLLN iii

22

1

1

4321 ,,,xL

LN

xL

LN ii

4

4

3

3. (6.8)

The derivatives of the natural coordinates by x, y and z we can generate from the previousmentioned equation. Because it shows linear connection, the desired derivatives will be theelements of the last three columns of the inverse matrix 1A (the first column shows thenatural coordinates of the origin):

1 1 110

2 2 220

1

3 3 330

4 4 440

L L LLx y zL L LLx y z

AL L LLx y zL L LLx y z

. (6.9)

Example 6.3

Determine the strain matrix of this tetrahedron element!

Figure 6.2:Tetrahedron element.


07.06.13.5

Because we can write the shape functions of a four-node tetrahedron element in theexpression ii LN , in the inverse matrix only the i-th term will be different fromzero, actually the first parameter of this term is 1. In this case the last three columns ofthe inverse matrix 1A give immediately the derivatives of the shape functions:

1

1

1 1 1 1 6 3 2 00 2 0 0 0 3 0 010 0 3 3 0 0 2 660 0 0 1 0 0 0 6

A

.

The strain matrix with the help of the vector u and the operator L :

1 2 3 4

1 2 3 46 12

1 2 3 4

x

y

L L L LzB L L L L

L L L Ly x

z x

z y

0626002006063030

000230326600

02020033

61 .

We can find that at a tetrahedron this matrix contains so many blocks as the numberof the nodes and the structure of these blocks is also similar. We wrote here thenontrivial zero elements to emphasize the structure of blocks.

If we express the shape functions so1T Tn x B , (6.10)

so the matrix B is easier to determine from multiplication of two matrices. If the vector ucontains only one function, then TN n , so

1 1ToB LN Lx B B B

. (6.11)The calculation 0B needs fewer calculation, because the vector x contains only mononoms,and the size of the matrix 0B does not depend on the size of the elements.


07.06.13.6

Example 6.4

Determine the 0B strain matrix of a classical non-conform triangular plate bendingelement with 15 degrees of freedom!

We use total fourth-order polynom to the element with 15 degrees of freedom, so2

2

22 2 3 2 2 3 4 3 2 2 3 4

2

2

1

2

o

y

B x y x xy y x x y xy y x x y x y xy yx

x y

0686004400200000026120026002000

1262006200200000

22

22

22

yxyxyxyxyxyx

yxyxyx.

This principle is applicable in those cases, if the vector u has more elements. We illustratethis with a new example.

Example 6.5

Let us analyze a Timoshenko beam, where we approximate the two shape functions withsecond-order C(o) continuous functions.

We suppose in the forthcoming that the three shape functions can be expressed in thisway:

1T T n x B .

The two functions can be approximated also in this expression1

212

312

1

2

3

11

vv

B vv x xx x B

,

because we wrote under each other the two well-known interpolation equations. Wenote that the displacements of the nodal points are written mostly in different order,but returning to a conventional method we must change the columns of the middleparameter:


07.06.13.7

1

12

1 2 3 22

21 2 3

3

3

11

v

g g g vv x xg g gx x

v

,

wherei

g means the i-th column of he matrix1

B

. If the we write the previousequation as

1Te

u X B v ,

then1 1T

o B LX B B B

.So finally

1o

ddxB

ddx

xxxx

xxxx

2101210

11 2

2

2.

Stiffness matrices

We have seen in the previous example that the stiffness matrix of an element can beexpressed by the general equation

e

TeK B DB d

1. (6.12)

The different types of the material stiffness matrix D are given for example in the book 1 ,and the calculation of the matrix B was discussed already in the previous points. Now as asummary we explain the different specific cases of the calculation of the stiffness matrices.

From the analysis of the equation (6.12) we can find that the stiffness matrix of the elementsis always symmetric, because the matrix D is symmetric, and the transposed of amultiplication will be the multiplied value of the individual transposed terms in reversedorder, namely:

TT TB DB B DB , (6.13)and the integration does not destroy the symmetry of course.

The matrices with constant elements

In linear finite element solution we take generally the elements with constant materialparameters in one element, namely the material stiffness matrix D is constant. We have seenin the previous example that in the case of specific approximations the matrix B is alsoconstant. In that case the matrix multiplication we can pick up the matrix-product before the

1 We note the integration will always depend in concrete 1, 2D or 3D cases on the type of the givenexample: practically we must integrate over the length, area or volume of the element.


07.06.13.8

symbol of the integral, and the residuum integral will exactly be the volume of the element(or its area or length). So in the case of matrices with constant elements we can express thestiffness matrix by the equations:

TeeK B DBV , T eeK B DB A , T eeK B DBl (6.14)

Example 6.6

Determine the stiffness matrix of a two-node bar element! It has the length l.With standard relationships:

1

1 11 11 1 1e

EAlK EA ll l l

l

.

In the case of rigid body motion the nodal forces will be exactly zero:

eq 1 1 0

1 1 0eeuEAK vul

,

and the forces will opposite with applications of the general dislocations:

eq 1 1 2

2

1 1 11 1 1ee

u EA u uEAK vul l

.

Example 6.7

Determine the stiffness matrix of a three-node triangular element in the case of plane stressstate (see the example 6.2)! The Poissons ratio is 0,2.

The area of the triangle is 17 (the half of the Jacobian-determinant).

eK =

255247746226

341

2

201120201

201 2 ,,

,

,Eh

254762572

246

341 17=

=96068 ,

Eh

6266831612544614256211883125455816623611616816635418629252116234184182744861162927637

,,,,,,,

,,,,,,,,,,,,,,,,,,,,,,

.

Example 6.8

Let us write the details of the calculation of a stiffness matrix of a tetrahedron element (seethe example 6.3)! The Poissons ratio is 0,3.

The volume of a tetrahedron is exactly 1 (the Jacobian-determinant/6), so:


07.06.13.9

eK =

006600

600206602

620030030

003230032

023

61

2020

20703030307030303070

4031

,,

,,,,,,,,,,

,,E

1

0626002006063030

000230326600

02020033

61

.

Integration with global coordinates

If the matrix B (or perhaps D ) is the function of the global coordinates, then in simpler casesthe elements of the product of TB DB can be expressed with simple expressions and they areintegrable one by one.

Example 6.9

Determine the stiffness matrix of a three-node bar element!

Figure 6.3: Three-node bar element

eK

10

6

7216941

72169

41 dxxxxEA

xx

x

2 2 210

2 2

6 2

18 81 2 34 144 16 63

4 64 256 2 30 11216

symm. 14 49

x x x x x xEA x x x x dx

x x


07.06.13.10

=

2832432643243228

48EA =

1416216321621614

24EA .

Example 6.10

Introduce the steps of the calculation of the stiffness matrix of a four-node rectangle withplane strain state (see figure 6.4)!

Figure 6.4: Four-node rectangular element

The matrix B contains four blocks with similar structures:

1 2 3 4B B B B B .The matrix D is:

D =

221

11

211

Eh .

The stiffness matrix of the element:

1

21 2 3 4

0 0 3

4

T

Tb a

e T

T

B

BK D B B B B dxdy

B

B

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

e e e e

e e e e

e e e e

e e e e

K K K KK K K KK K K KK K K K

,

where a general block is:

0 0

b aT

eij i jK B DB dxdy .In the calculation of the blocks iB we determine the shape function as in the example6.2:

by

axN

11 , by

axN 2 ,

by

axN 13 ,

by

axN 114 .

Let us determine as an example the block 12eK !


07.06.13.11

120 0

1b ae

y a xK

a x yab

221

11

211

Eh dxdy

yxx

y

ab

1 =

= 21122 baEh

dxdy

yxaxxyxyay

xyayxyxaxyb a

0 0 22

22

2211

221

221

2211

a

bba

ba

ab

Eh

621

61

814

841

1221

31

211

.

We note that at a general rectangle or triangle the boundaries of the integrations might befunctions too, for this reason we should divide the region of the elements into different parts.In these cases the numerical integration is more practical.

Integration with parametric coordinates

If the matrix B (or perhaps D ) is the function of parametric coordinates, then to the directcalculation of the matrix Ke is not enough the knowledge the matrix-product

TB DB ,because this value is the function of the local coordinates and we must integrate by theglobal parameters. We can step over to the local variables with the help of the Jacobiandeterminant:

dV dxdydz J d d d , (6.15/a)dA dxdy J d d , (6.15/b)dl dx J d . (6.15/c)

In the case of parametric local coordinate-system the boundaries of the integrals areregulars:

- at line element:

1

1

fd , (rarely: 1

0

fd ), (6.16/a)

- at triangular element: 1

0

1

0

dfd , (6.16/b)

- at rectangular element:

1

1

1

1

dfd , (rarely: 1

0

1

0

dfd ), (6.16/c)

- at tetrahedron element: 1

0

1

0

1

0

ddfd , (6.16/d)

- at hexahedron element:

1

1

1

1

1

1

ddfd , (rarely: 1

0

1

0

1

0

ddfd ). (6.16/e)

In the case of curvilinear coordinate system the elements of the Jacobian-matrix will befunctions of ,, , so the determinant of J might be a higher-order polynom, and its


07.06.13.12

inverse (which is necessary to calculation of B ) is a fraction-function, so the analyticintegration is mostly very difficult. We suggest here also the numerical integration.

Example 6.11

Determine the stiffness matrix of a three-node bar element (see the example 6.9) withanalytic and numerical integrations!

121

1 N , 22 1 N , 121

3 N ,

3

128

iiiNxx ,

2dxJd

,1 0 5J , , D EA ,

1 10 5 2 0 5 0 52 2

B , , , ,

eK 1 1

1 1

12 4 1 1 2

2 4 2 41

2 4

TB DB J d EA d

1416216321621614

24EA .

With two-point numerical integration (the exact places of the integration points2 are:3121 /, , both weights are: 1):

eK 1 31T

/B DB J 1 31 T /B DB J =

= 241

321

31

41

321

41

321

31

41

321

EA +

+ 241

321

31

41

321

41

321

31

41

321

EA =

1416216321621614

24EA .

We emphasize that we have the same results in both solutions (we have substituted the exactvalues of the Gauss points at the numerical integration to show this fact), the stiffness matrixdoes not depend on the coordinate-system. We obtained a correct result with the numericalintegration because the matrix B is first-order one, the Jacobian matrix J is constant, so the

2 See the book 1 , Appendix C.


07.06.13.13

final matrix at integration was second-order, and in this case the Gauss-Legendre methodyields exact result with two integration points.

It is an important question, how many points are necessary to a good numerical integration?The increasing of the integration points increases (generally) the exactness, but of course thecomputation time also will increase. The radical decreasing of the number of points my leylead to a very dangerous case, the so-called mechanisms3 may arise in the finite elementcalculation and we have a wrong result. The mechanisms are such a specific deformationmodes of the element, where the values of the displacement functions will be zero at allintegration points, so we cannot express the effect of deformations. Irons suggested anempirical equation to the number of possible mechanisms:

nrRNdM , (6.17)where

d the degree of freedom of nodes,N the number of nodes,R the number of independent rigid body motions,r the rank of the matrix D ,n the number of integration points.

We have to choose the value of n that will not cause mechanisms, namely 0M .Example 6.12

Analyze the possibility wonder we use at an 8-node serendipity plate element a numericalintegration with 2 x 2 points?

The Irons formula: 1223382 M .Because M > 0, mechanism may arise at the element, so it is not enough to apply 4integration points.

Integration by natural coordinates

We can use natural coordinates only with simplex elements. In the mathematics there arewell-known expressions for the integrations of functions with natural coordinates:

lqp l

qpqpdxLL

!!!121

, Arqp A

rqprqpdxLLL 2

2321 !!!! , (6.18)

Vsrqp V

srqpsrqpdxLLLL 6

34321 !!!!! .

Do not forget the calculation of the factorials: 0!=1 and n!=n(n-1)! . We emphasize that theseexpressions are valid for simplexes with general shape.

Example 6.13

Determine the stiffness matrix of a two-node Timoshenko-beam in the plane xy!

We have generated already the shape functions. Because we approximated here alsotwo displacement functions with C(0) continuous shape functions, the structure of the

3 The English scientist Irons published this phenomenon in 1970, so it is called Irons-mechanisms.


07.06.13.14

matrix N will be similar to matrices of plate elements, only we have here two nodes.With these approximations we have:

11 LN , 22 LN , 1 21 2

N NN

N N

,

1 1

1 1A

x x l

, 1 11

111

x lA

xl

,

1ddxB LN

ddx

21

21

LLLL

=

ll

Ll

Ll

11

1121

,

1

2

1

1

1

1

el

l

LlK

l

Ll

z

y

EIAG

ll

Ll

Ll

11

1121

dx =

1 22 2

21 1 1 22 2

22

222symm.

y y y y

yz zy y

l y y

zy

GA GA GA GAL L

l ll lGAEI EIGA L L GA L L

ll l dxGA GA

Lll

EI GA Ll

=

=

2 2

3 2 6

2

symm.3

y y y y

y y yz z

y y

yz

GA GA GA GAl l

GA l GA GAEI EIl l

GA GAl

GA lEIl

.

Example 6.14

Determine the stiffness matrix of a two-node classical beam element in the plane xy!

,, 3121211 23 LLLLN 3121212 LLlLLN , , ,, 3222213 23 LLLLN 3222214 LLlLLN , ,


07.06.13.15

1 1 2 1 3 2 4 2N N L N L N L N L .

1 1

1 1A

x x l

, 1 11

111

x lA

xl

,

L

22

dxd , D zEI .

All shape functions depend on one natural coordinate. The natural coordinates arelinear functions of x, so their second derivatives by x will be zero, namely the secondderivatives of the shape functions can be expressed easier. For instance:

212

121

12

211

2 1126

l

LdxdL

dLNd

dxLNd .

So

B LN

l

Ll

LlL

lL 2

221

21 6212662126 ,

and the stiffness matrix:1

2

1

22

2

6 12

2 6

6 12

2 6

el

Ll

LlK

Ll

Ll

zEI l

Ll

LlL

lL 2

221

21 6212662126 dx=

2 21 1 1 1 1 2 1 2 1 2 1 24 3 4 3

21 1 1 2 1 2 1 2 1 22 3 2

2 22 2 2 24 3

22 22

1 4 4 1 5 6 1 2 2 4 1 2 3 636 12 36 12

1 6 9 1 3 2 6 1 3 3 94 12 4

1 4 4 1 5 636 12

1 6 9symm. 4

zl

L L L L L L L L L L L Ll l l l

L L L L L L L L L Ll l lEI dx

L L L Ll l

L Ll

=

llll

llll

llll

llll

EI z

4626

612612

2646

612612

22

2323

22

2323

.

Integration without the calculation of matrix B

The calculation is easier if we have to integrate more simple functions. If we describe thematrix B in the equation


07.06.13.16

1

oB B B

, (6.19/a)where the elements of 1B are constants, then the stiffness matrix can be expressed as

1 1 1 1 1 1e e e

T T TT T T

e o o o o o K B DB d B B DB B d B B DB d B B K B

(6.19/b)

so the matrixo

K is:

e

To o oK B DB d

, (6.19/c)

and the functions will be more simple in the integrals. We note that there is a disadvantage ofthis process of course: we have to multiply this matrix with other two to obtain the finalversion.

Example 6.15

Determine the 0K matrix of a four-node plate element with plane stress state (see theexample 6.10)!

Now we approximate two displacement functions by bilinear shape functions, so1

1T x y xyX

x y xy

.

0B =

xy

y

x

xyyx

xyyx1

1=

yxx

y

010100100

010.

By the symbol

21

the multiplication will be:

To oB DB

22

22

2

szimm.1

0

10

1

yxxy

xyyxxyyxxy

Eh

,

so

0 0

b aT

o o oK B DB dxdy =


07.06.13.17

=

2 22

2 2

0

12 2

2 2

3 2 2 401

2

12

symm.3

b a

a b

abb a a bEhab

b

a

a b

.

We note once more, that we have to multiply this matrix from both right and left with1

B

(see the example 6.5) and with the transposed matrix to obtain the final stiffnessmatrix.

The reduced load vectors

We have seen in the third lecture that in 3D problems we can determine the loads reduced tothe nodes of the element by different expressions. From the kinematical effects (fabric error,variation of the temperature) the load vector is:

0

e

Te

V

q B D dV, (6.20)and from distributed loads:

e

Te

V

q N p dV . (6.21)The

eq is called the reduced load vector4. The matrices and vectors of these expressions can

be the functions of either the global, or parametric or natural coordinates. For thecalculation of the integrals we should use the previous mentioned methods.

If the external load is not volume dependent (gravity load, for instance), but is distributed ona surface or on a line, perhaps it is a concentrated force, than we must integrate over the areaor line (in the case of concentrated forces we must not integrate).

The general equation for the integration of the reduced load vector:

e

Te

V

q N p dV TA AA

N p dV T Ts is iis

N p ds N p . (6.22)The vectors

A s ip, p , p , p have always as much elements as the vector u , namely the

multiplications of the correct pairs of the vectors should give (after the integration) the workdone by the loads. The number of the variables of functions of the vectors coincides with thenumber of the necessary integrals.

4 The word reduced means here the reduction for the nodes. We note that the effect of the motionof supports will take into account later.


07.06.13.18

We have to multiply the loads always with the displacement of its active point, so the numberof the variables of matrix N and the load vectors must be the same. For example theconcentrated load has no any variable, so we must substitute into N the coordinates of theactive points of this force (for this reason in the expression (6.22) the matrix N has an indexi ).

Example 6.16

Determine the reduced load vector of a three-node bar element (see the example 6.9) fromconstant variation of the temperature!

0T

el

q B D dx

101

72169

4110

6ss tEAdxtEA

xx

x .

We note that from the rules of the elementary strength of materials comes that if weapply at two ends of the bar such tension forces; we can determine the sameelongation in the bar as was determined from the temperature effect.

Example 6.17

Reduce the constant distributed loads for the nodes in a Timoshenko beam according to thetwo-noded and three-noded elements!

The load vector is in both cases: 0Tp p . We shall use parametric coordinates,where 2J l / .The load vector is at a two-noded element:

1 1

1 1

12

2102

1 0 22 2

1 02

Te

pl

p lq N p J d dpl

,

The load vector is at a three-noded element:

eq

06

06

406

20

21

21

11

21

21

1

12

2

pl

pl

pl

dlp

.


07.06.13.19

In figure 6.5 we can see the summary of the results: at the two-node element the halfof the external load is acting on the individual nodes, but at the three-node elementthe greater part of the load is reduced on the middle point. We note that at applicationof the classical beam theory see the example 6.19 from the external loads wemust take into consideration bending moments too.

Figure 6.5: Reduction of loads in a three-node element

Example 6.18

The plate element in figure 6.6 there is in plane stress state. We must reduce to the nodes thefollowing loads:

- Constant variation of the temperature st ,- gravity load ( is the density), the direction of the gravity is opposite to y,- linear distributed load during a line (figure 6.6/a),- concentrated force (figure 6.6/b) (active point 2, 1F Fx y ).

Figure 6.6:Loads of aplate element

The shape functions in the global coordinate-system are the following ones:

3411

yxN ,42xN ,

33yN .

andN B :

1 2 3

1 2 3

N N NN

N N N

,


07.06.13.20

1 1 04 4

1 103 3

1 1 1 10 03 4 4 3

B

.

The reduced load vector generated from the different effect is:

- In the calculation of thermal effect we can take into account, that the elements ofthe matrix B are constant:

therm.0

e

T

eA

q B D dA= 0TB D Ae=

=

031

310410

041

41

31

31

41

2

11

1

1 2

Eh

0s

s

tt

234 =

400343

12 stEh .

- In the calculation of the gravity load effect we avoid the integrals, because it isenough to work with the volume of the cubes of the shape functions:

1

1 1

grav. 2

2 2

3

3 3

0 0 01 1

00 0 02

1 130 0 0

1 1

e e

T ee

A A

NN N

N Aq N p dA dA h dA h hN Nh

NN N

eA

.

We obtain a result from this calculation that the third part of the gravity load acts toevery nodes. This is not true already at a six-node element.

- At the line-load we transform everything with the help of the s coordinate:

2syx ,

oyx s

spp 123 ,

7212os ,

so


07.06.13.21

line

0

14 2 3 2

14 2 3 2

134 2 112

4 2

3 2

3 2

os

eo

s s

s s

ssq ds

s s

s

s

4924492449184918712712

.

- At the concentrated load we write the coordinates of the load into the matrix N :

conc.

2 1 514 3 6

2 1 714 3 6

2 554 2

2 7 74 2

1 53 3

1 73 3

eq

.

The reduced components from the concentrated load can be seen in figure 6.7:

Figure 6.7:Reduction of aconcentrated load

Example 6.19

We analyze a classical beam element in the plane xy. We reduce the following loads usingnatural coordinates to the nodes:


07.06.13.22

- t linear variation of the temperature,- gravity load ( is the density), the direction of the gravity is opposite to y,- linear distributed load,- concentrated force F.

Figure 6.8: Reduction of external loads

We apply the previous mentioned expressions:

1 k ldxLlki .

We use for description of the thermal-effect the vector o :

therm.0

T

el

q B D dx= dxh tEI

lL

lL

lL

lL

lz

2

22

1

21

62

126

62

126

=h

tEI z

1010

,

So at the two ends of the beam the reduced loads are bending moments.

For reduction of the gravity load we apply the vector p :

2 31 1

2 31 1grav.

2 32 2

2 32 2

123 2

1213 22

12

T

el l

L Ll

l L Lq N p dx dx l

L L

l L Ll

.

At the integral of the partial distributed load we apply a rigid body motion at thecoordinate-system by transformation lxlL /1 and lxL /2 . The load function is

1 2 /y op p x l , so


07.06.13.23

2 3

2 3

distr.2 3

/ 2

2 3

1 3 2 14072

2 96019

3 2 4023960

l

o oel

x xl l

x x x lll l l xq p dx p l

lx xl l

lx xll l

.

The natural coordinates of the concentrated load are: 431 /L , 412 /L .Finally the load will be:

2 3

2 3

conc.2 3

2 3

3 33 2 274 432

3 3 94 4 64

51 13 2 324 43

1 1 644 4

e

l

q F F

l

.

References:

1./ Bojtr I. Gspr Zs. : The finite element method for civil engineers (in Hungarian), Terc,2003.2./ Rao, S. S. : The finite element method in engineering, Pergamon Press, 1989.

FEM-6. Lecture_0.pdf

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