beam element using fem

7/29/2019 BEAM ELEMENT USING FEM

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CHAPTER 4 BEAM ELEMENT

Introduction Beam element has six degrees of freedom at each node

y

vjy

jy vjxi x

jz j jx

z vjz

Beam element is a slender structure Has uniform cross section. The element is unsuitable for structures that have complex

geometry, holes, and points of stress concentration.

The stiffness constant of a beam element is derived bycombining the stiffness constants of a beam under pure

bending, a truss element, and a torsion bar.

A beam element can represent a beam in bending, a trusselement, and a torsion bar.

In FEA its a common practice to use beam elements torepresent all or any of these three loads.


2/23

ME 273 Lecture Notes by R. B. Agarwal 2

Derivation of Stiffness Equation for a Beam

Element Under Pure Bending in 2-D

A beam, under pure bending (without axial loads or torsionloads), has two-degrees of freedom at any point.

F

v

A beam element in pure bending has a total of four degreesof freedom, two at each node.

The size of the stiffness matrix of a beam element has the size4 x 4.

Stiffness matrix equation is derived using the StiffnessInfluence Coefficient Method.

For a two-node beam element, there are two deflections andtwo rotations, namely, v1, 1, v2, and 2.

Force and influence coefficient relationship is established bysetting each of the four deflection values to unity, with the

remaining deflection values equal to zero. The procedure

follows.


3/23


Consider a beam element, loaded in such a way that it has the

deflection values: vi = 1, i = 0, vj = 0, j = 0

i j

vi, i vj, j

The above deflections can be produced by a combination of load

conditions, shown in figure 4.4.

The deflection relationships for loading can be found in any

Machine Design Handbook, and is given as,

vmax

vmax = (FL3)/(3EI)

y

= - (FL2)/(2EI)i L j x

F

(a)


4/23


y

Mi x

i L Mj vmax = - (ML2)/(2EI)

vmax j = (ML)/(EI)

(b)

Applying these relationships to the beam, we get,

1 = vi = (vi)F + (vi)M

1 = vi = (Fi L3)/3 EI - (Mi L

2)/2EI

(4.1)

and = 0 = ()F + ()M

0 = - (Fi L

2

)/2EI + (Mi L)/EI(4.2)

Solving Equations (4.1) and (4.2), we get,

Fi = (12EI)/L3

(A)

Fj = - Fi = -(12EI)/L3

(B)

Mi = (6EI)/L2 (C)


5/23


From the above Figures,

Mj = Fi L - Mi= (12EI)/L

2- (6EI)/ L

2

= (6EI)/ L2 (D)

Writing equations (A) through (D) in a matrix form we get,

Fi (12EI)/L3

1 (12EI)/ L3

0 0 0 1

Mi (6EI)/ L

2

0

(6EI)/ L

2

0 0 0 0= =

Fj -(12EI)/ L3

0

-(12EI)/ L3

0 0 0 0

Mj (6EI)/ L2

0

(6EI)/ L2

0 0 0 0

Using a similar procedure and setting the following deflection

values:

vi = 0, i = 1, vj = 0, j = 0, we get,

Fi (6EI)/L2

0 0 (6EI)/ L2

0 0 0

Mi (4EI)/ L 1 0

(4EI)/ L

0 0 1

= =Fj -(6EI)/ L

20 0 -(6EI)/ L

20 0 0

Mj (2EI)/ L

0 0 (2EI)/ L

0 0 0


6/23


Similarly, setting vj = 1 and , j = 1, respectively, and keeping allother deflection values to zero, we get the final matrix as,

Fi (12EI)/L3

(6EI)/ L2

-(12EI)/ L3

(6EI)/ L2

1

Mi (6EI)/ L2 (4EI)/ L

-(6EI)/ L2

(2EI)/ L 1

= (4.7)

Fj -(12EI)/ L3

-(6EI)/ L2

(12EI)/L3

-(6EI)/ L2

1

Mj (6EI)/ L2

(2EI)/ L -(6EI)/ L2

(4EI)/ L

1

Note that, the first term on the RHS of the above equation is the

stiffness matrix and the second term is the deflection. In the case

where deflections are other than unity, the above equation will

provide an element equation for a beam (in bending), which can be

written as,

Fi (12EI)/L3

(6EI)/ L2

-(12EI)/ L3

(6EI)/ L2

vi

Mi (6EI)/ L2 (4EI)/ L

-(6EI)/ L2

(2EI)/ L i=

Fj -(12EI)/ L3

-(6EI)/ L2

(12EI)/L3

-(6EI)/ L2

vj

Mj (6EI)/ L2

(2EI)/ L -(6EI)/ L2

(4EI)/ L j

Where Fi, Mi, Fj, Mj are the loads corresponding to the deflections

vi, i, vj, j.


7/23


The above equation can be written in a more solution friendly form

as,

Fi 12 6L

-12 6L vi

Mi 6L 4L2

-6L 2L2 i

Fj = EI/L3

-12 -6L 12 -6L vj

Mj 6L 2 L2

-6L 4L2 j

The above equation is the equation of a beam element, whichis under pure bending load (no axial or torsion loads). The stiffness matrix is a 4 x 4, symmetric matrix. Using this equation, we can solve problems in which several

beam elements are connected in an uni-axial direction.

The assembly procedure is identical to the truss elements.

However, if the beam elements are oriented in more than onedirection, we will have to first transform the above equation

into a global stiffness matrix equation (analogues to the

procedure used for truss elements).

For a beam element, transformation of a local stiffness matrixinto a global equation involves very complex trigonometric

relations.


8/23


Example 1

For the beam shown, determine the displacements and slopes at the

nodes, forces in each element, and reactions at the supports.

3 m 3 m 50 kNE = 210 GPa, I = 2x10

-4m

4

K = 200 kN/m

Solution

The beam structure is descritized into three elements and 4-nodes,

as shown.

[1] [2]

3

1 2

[3]

4

First, we will find the element stiffness matrix for each element,

next we will assemble the stiffness matrices, apply the boundaryconditions, and finally, solve for node deflection. Internal forces

and reactions are calculated by back-substituting the deflections in

the structural equation.


9/23


Element 1

1 2EI/L

3= (210 x 10

9) x (2x10

-4)/(3)

3= 15.55 x 10

5

The general equation of a stiffness matrix is given by equation 4.7.

Writing this equation by placing the common factor EI/L3

outside

the matrix, we get

Element 1

12 6L -12 6L v1

6L 4L2

-6L 2 L2

1

[Ke](1)

= (EI/L3)

-12 -6L 12 -6L v2

6L 2 L2

-6L 4 L2

2

[2]

Element 2 2 3

12 6L -12 6L v2

6L 4L2

-6L 2 L2

2

[Ke](2) = (EI/L3)-12 -6L 12 -6L v3

6L 2 L2

-6L 4 L2

3


10/23


3

Element 3

[3]

4

[Ke](3)

= K -K v3

-K K v4

To get the global stiffness matrix, we will use the same procedure

used for assembling truss element stiffness equations. In terms of

E, L, and I the assembled global stiffness matrix is,

v1 1 v2 2 v3 3 v4

v1 12 6L -12 6L 0 0 0

1 4L2

-6L 2 L2

0 0 0

v2 24 0 -12 6L 0x (EI) /(L

3)

2 8L2

-6L 2L2

0

v3 12 +K -6L - K

3 4L2

0

v4 SYMMETRY K

Where K = (K) x [L3

/ (EI)] = 200 x 103

/(15.55 x105) = .1286


11/23


Our next step is to write the structural equation; however, we can

reduce the size of the stiffness matrix by applying the given

boundary conditions:

v1= 1 = 0 node 1 is fixed

v2= 0 node 2 has no vertical deflection, but its free to

rotate.

V4 = 0 node 4 is fixed.

The reduced stiffness matrix is

8L2

-6L 2L2

KG = EI / (L3) -6L 12+K -6L

2L2

-6L 4L2

Substituting the values of E, L, and I the structural equation can be

written as,

0 72 -18 18 2-100 = 15.55 x 10

5-18 12.1 -18 v3

0 18 -18 36 3

2 = - 0.0025 radSolving, we get v3 = - 0.0177 m

3 = -0.0076 rad


12/23


Derivation of a Plane (2-D) Beam Element an an

Arbitrary angle

We will derive the 2-D beam element equation that has an arbitraryorientation with axial and bending loads.

F

M

The stiffness equation will be derived in three steps

1.Pure Beam element arbitrarily oriented in space2.Local Beam stiffness with axial and bending loads3.Arbitrarily oriented beam with axial and bending loads.


13/23


Arbitrarily Oriented 2-D Beam Element

The stiffness equation for an arbitrarily oriented beam element can

be derived with a procedure similar to the truss element.

d2yx

y

2y

d1y d1y

1 d1x x

d1y = d1y cos - d1x sin = d1y C - d1x S = - d1x S + d1y C

d2y = d2y cos d2x sin = d2y C d2x S = d2x S + d2y C

and 1

= 1,

2=

2

Note: The underscored terms represent local coordinate values.

Thus, x and y are local coordinates and x and y are global

coordinates.

The above equations can be written in a compatible matrix form,

by introducing 0s where necessary,

d1xd1y -s c 0 0 0 0 d1y

1 = 0 0 1 0 0 0 1d2y 0 0 0 -s c 0 d2x

2 0 0 0 0 0 1 d2y2


14/23


-s c 0 0 0 0

Let T = 0 0 1 0 0 0 (A)

0 0 0 -s c 00 0 0 0 0 1 , the transformation matrix.

Thus, {d} = [T] {d}

Global

Local

Note that angle is independent of the coordinate systems, and 1= 1, 2 = 2

As derived in the case of the truss element, relationship between

local and global stiffness matrices is given as

[kg] = [T]T

[k] [T]

Where, [kg] = Global stiffness matrix of an element[T] = Transformation matrix

[k] = Local stiffness matrix of the element

Substituting the values of [T] and [k], we get the global equation of

a beam element oriented arbitrarily at an angle as,

12S2

-12SC -6LS -12S2

-12SC -6LS

12C2

6LC 12SC -12C2

6LC

k = EI/L3

4L2

6LS -6LC 2L2

12S2

-12SC 6LSSymmetry 12C

2-6L

4L2


15/23


This is the equation of a beam element (without axial or torsional

load, and oriented at an angle .

Also, S = sin , C = cos in the above equation.


16/23


Stiffness Equation of a Beam Element with

Combined Bending and Axial loads

First, we will derive the stiffness matrix in local coordinates and

then convert it in to global coordinates.

The stiffness equation for the combined bending and axial load can

be written by superimposing the axial stiffness terms over the

bending stiffness.

For axial loading, the structural equation is,

f1x 1 -1 d1x f1x f2x= AE/L

f2x -1 1 d2x Truss Element

And for bending, the structural equation is,

f1y 12 6L -12 6L d1y

m1 6L 4L2

-6L 2L2

1= AE/L

3

f2y -12 -6L 12 -6L d2y

m2 6L 2L2

-6L 4L2

2

f1y f2y

m1 m2Beam Element


17/23


Therefore, the combined loading equation is

d1x d1y 1 d2x d2y 2

f1x C1 0 0 - C1 0 0 d1x

f1y 0 12 C2 6C2L 0 -12 C2 6C2L d1y

m1 0 6 C2L 4C2L2

0 -6C2L 2C2L2

1

=f2x -C1 0 0 C1 0 0 d2x

f2y 0 -12 C2 -6C2L 0 12 C2 -6C2L d2y

m2 0 6 C2L 2C2L

2

0 -6C2L 4C2L

2

2

Where,

C1 = AE/L

C2 = EI/L3

The first matrix on the RHS in the above equation gives the local

Stiffness matrix k as

d1x d1y 1 d2x d2y 2

C1 0 0 - C1 0 0

0 12 C2 6C2L 0 -12 C2 6C2L

0 6 C2L 4C2L2

0 -6C2L 2C2L2

k =

-C1 0 0 C1 0 0

0 -12 C2 -6C2L 0 12 C2 -6C2L

0 6 C2L 2C2L2

0 -6C2L 4C2L2


18/23


The local stiffness matrix k has 3-DOF at each node, representing

bending and axial loads.

For axial loading:

u2y

u2

u1y 2u2x

1 u1x

u1

u1 = u1x cos + u1y sin = c u1x + s u1y

u2 = u2x cos + u2y sin = c u2x + s u2y

where, cos = c, and sin = s

Which can be written as,u1x

u1 c s 0 0 u1y

= u2x (3.2)u2 0 0 c s u2y

Where

c s 0 0

T = (3.2)0 0 c s is the transformation Matrix

Since the transformation matrix T is not a square matrix, it cant be

inverted, which will be incompatible with the matrix operations.


19/23


Let us convert the matrix T into a compatible form by placing 0s as

necessary for compatibility.

u1x u1y u2x u2y

u1 c s 0 0 u1x

v1 = -s c 0 0 u1y

u2 0 0 c s u2xv2 0 0 -s c u2y

Where,u1x u1y u2x u2yc s 0 0

T = -s c 0 00 0 c s

0 0 -s c

For bending loading:

The stiffness equation for an arbitrarily oriented beam element can

be derived with a procedure similar to the truss element.

u2y

y xy

2u1y

u1y 1 u1x x

u1y = u1y cos - u1x sin = u1y C - u1x S = - u1x S + u1y C

u2y = u2y cos u2x sin = u2y C u2x S = u2x S + u2y C


20/23


and 1 = 1, 2 = 2

Note: The underscored terms represent local coordinate values.

Thus, x and y are local coordinates and x and y are globalcoordinates.

The above equations can be written in a compatible matrix form,

by introducing 0s where necessary,

u1x

u1y -s c 0 0 0 0 u1y

1 = 0 0 1 0 0 0 1u2y 0 0 0 -s c 0 u2x

2 0 0 0 0 0 1 u2y2

-s c 0 0 0 0

Let T = 0 0 1 0 0 0

0 0 0 -s c 00 0 0 0 0 1 ,

the transformation matrix.

Combining the axial and bending transformations, we get,

C S 0 0 0 0

-S C 0 0 0 0T = 0 0 1 0 0 0

0 0 0 C S 0

0 0 0 -S C 00 0 0 0 0 1


21/23


Using the relation, [k] = [T]T[k][T] and simplifying, we get,

I

CL

IC

L

IASsymmetry

S

L

ICS

L

IAS

L

IAC

ICL

IS

L

II

CL

IC

L

IASCS

L

IAC

L

IC

L

IAS

SL

ICS

L

IAS

L

IACS

L

ICS

L

IAS

L

IAC

L

EK

4

612

61212

266

4

61212612

6121261212

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

This is the stiffness equation for a 2-D beam element at an

arbitrary orientation and under axial and bending loads.


22/23


2-D Beam Element with combined loading

Bending, Axial, and Torsion ( = 0)

A similar procedure can be used to find the equation of a beam

under general loading of axial, bending and torsion loads.

The torsional loads are m1x and m2x, and the corresponding

deflections are,

x1 and x2

The torsional structural equation is:

m1x 1 -1 1x

= JG/Lm1x -1 1 2x

These terms can be superimposed on the stiffness equation derived

previously for the combined bending and axial loads.

dy

y

3-D Beam Element: dx

z x

dz

A 3-D beam element has 6 DOF at each node, and 12 DOF for

each element. The stiffness matrix can be derived by super-

imposing the axial, bending, and torsion loadings in the XY, XZ,

and YZ planes. The equation is,


23/23

The stiffness equation in local coordinate is:

^

2

^

2

^

2

^

2

^

2

^

2

^

1

^

1

^

1

^

1

^

1

^

1 yxzyxzyxzyx dddddd

L

EI

L

EI

L

EI

L

EI L

EI

L

EI

L

EI

L

EIL

GJ

L

GJL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AE

L

AEL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

GJ

L

GJL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

AEL

AE

K

zzzz

yyyy

yyyy

zzzz

zzzz

yyyy

yyyy

zzzz

4000

60

2000

60

04

06

0002

06

00

0000000000

06

012

0006

012

00

6000

120

6000

120

0000000000

2

000

6

0

4

000

6

0

02

06

0004

06

00

0000000000

06

012

0006

012

00

6000

120

6000

120

0000000000

22

22

2323

2323

22

22

2323

2323

Similarly, we can find the transformation matrix for Axial,

bending, and torsional loading and then use the equation,

]][[][^

TKTKT

The equation gets very complicated, so we will not go in to the

derivation any further.

beam element using fem

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