CHAPTER 4 – BEAM ELEMENT

Introduction

Beam element has six degrees of freedom at each node

y

v

jy

jy

v

jx

i x

jz

j

jx

z v

jz

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 by

combining 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 truss

element, and a torsion bar.

In FEA it’s a common practice to use beam elements to

represent all or any of these three loads.

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 torsion

loads), has two-degrees of freedom at any point.

F

v

A beam element in pure bending has a total of four degrees

of freedom, two at each node.

The size of the stiffness matrix of a beam element has the size

4 x 4.

Stiffness matrix equation is derived using the Stiffness

Influence Coefficient Method.

For a two-node beam element, there are two deflections and

two rotations, namely, v

1

,

1

, v

2

, and

2

.

Force and influence coefficient relationship is established by

setting each of the four deflection values to unity, with the

remaining deflection values equal to zero. The procedure

follows.

ME 273 Lecture Notes © by R. B. Agarwal

3

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

deflection values: v

i

= 1,

i

= 0, v

j

= 0,

j

= 0

i j

v

i

,

i

v

j

,

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,

v

max

v

max

= (FL

3

)/(3EI)

y

= - (FL

2

)/(2EI)

i L j x

F

(a)

ME 273 Lecture Notes © by R. B. Agarwal

4

y

M

i

x

i L M

j

v

max

= - (ML

2

)/(2EI)

v

max

j

= (ML)/(EI)

(b)

Applying these relationships to the beam, we get,

1 = v

i

= (v

i

)

F

+ (v

i

)

M

1 = v

i

= (F

i

L

3

)/3 EI - (M

i

L

2

)/2EI

(4.1)

and = 0 = ()

F

+ ()

M

0 = - (F

i

L

2

)/2EI + (M

i

L)/EI

(4.2)

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

F

i

= (12EI)/L

3

(A)

F

j

= - F

i

= -(12EI)/L

3

(B)

M

i

= (6EI)/L

2

(C)

ME 273 Lecture Notes © by R. B. Agarwal

5

From the above Figures,

M

j

= F

i

L - M

i

= (12EI)/L

2

- (6EI)/ L

2

= (6EI)/ L

2

(D)

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

F

i

(12EI)/L

3

1 (12EI)/ L

3

0 0 0 1

M

i

(6EI)/ L

2

0

(6EI)/ L

2

0 0 0 0

= =

F

j

-(12EI)/ L

3

0

-(12EI)/ L

3

0 0 0 0

M

j

(6EI)/ L

2

0

(6EI)/ L

2

0 0 0 0

Using a similar procedure and setting the following deflection

values:

v

i

= 0,

i

= 1, v

j

= 0,

j

= 0, we get,

F

i

(6EI)/L

2

0 0 (6EI)/ L

2

0 0 0

M

i

(4EI)/ L 1

0

(4EI)/ L

0 0 1

= =

F

j

-(6EI)/ L

2

0

0 -(6EI)/ L

2

0 0 0

M

j

(2EI)/ L

0 0 (2EI)/ L

0 0 0

ME 273 Lecture Notes © by R. B. Agarwal

6

Similarly, setting v

j

= 1 and ,

j

= 1, respectively, and keeping all

other deflection values to zero, we get the final matrix as,

F

i

(12EI)/L

3

(6EI)/ L

2

-(12EI)/ L

3

(6EI)/ L

2

1

M

i

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

-(6EI)/ L

2

(2EI)/ L 1

= (4.7)

F

j

-(12EI)/ L

3

-(6EI)/ L

2

(12EI)/L

3

-(6EI)/ L

2

1

M

j

(6EI)/ L

2

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

2

(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,

F

i

(12EI)/L

3

(6EI)/ L

2

-(12EI)/ L

3

(6EI)/ L

2

v

i

M

i

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

-(6EI)/ L

2

(2EI)/ L

i

=

F

j

-(12EI)/ L

3

-(6EI)/ L

2

(12EI)/L

3

-(6EI)/ L

2

v

j

M

j

(6EI)/ L

2

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

2

(4EI)/ L

j

Where F

i

, M

i

, F

j

, M

j

are the loads corresponding to the deflections

v

i

,

i

, v

j

,

j

.

ME 273 Lecture Notes © by R. B. Agarwal

7

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

as,

F

i

12 6L

-12 6L v

i

M

i

6L 4L

2

-6L 2L

2

i

F

j

= EI/L

3

-12 -6L 12 -6L v

j

M

j

6L 2 L

2

-6L 4L

2

j

The above equation is the equation of a beam element, which

is 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 one

direction, 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 matrix

into a global equation involves very complex trigonometric

relations.

ME 273 Lecture Notes © by R. B. Agarwal

8

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 kN

E = 210 GPa, I = 2x10

-4

m

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 boundary

conditions, and finally, solve for node deflection. Internal forces

and reactions are calculated by back-substituting the deflections in

the structural equation.

ME 273 Lecture Notes © by R. B. Agarwal

9

Element 1

1 2

EI/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/L

3

outside

the matrix, we get

Element 1

12 6L -12 6L v

1

6L 4L

2

-6L 2 L

2

1

[K

e

]

(1)

= (EI/L

3

)

-12 -6L 12 -6L v

2

6L 2 L

2

-6L 4 L

2

2

[2]

Element 2 2 3

12 6L -12 6L v

2

6L 4L

2

-6L 2 L

2

2

[K

e

]

(2)

= (EI/L

3

)

-12 -6L 12 -6L v

3

6L 2 L

2

-6L 4 L

2

3

ME 273 Lecture Notes © by R. B. Agarwal

10

3

Element 3

[3]

4

[K

e

]

(3)

= K -K v

3

-K K v

4

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,

v

1

1

v

2

2

v

3

3

v

4

v

1

12 6L -12 6L 0 0 0

1

4L

2

-6L 2 L

2

0 0 0

v

2

24 0 -12 6L 0

x (EI) /(L

3

)

2

8L

2

-6L 2L

2

0

v

3

12 +K’

-6L - K’

3

4L

2

0

v

4

SYMMETRY

K’

Where K’ = (K) x [L

3

/ (EI)] = 200 x 10

3

/(15.55 x10

5

) = .1286

ME 273 Lecture Notes © by R. B. Agarwal

11

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:

v

1

=

1

= 0 node 1 is fixed

v

2

= 0

node 2 has no vertical deflection, but it’s free to

rotate.

V

4

= 0 node 4 is fixed.

The reduced stiffness matrix is

8L

2

-6L 2L

2

K

G

= EI / (L

3

) -6L 12+K’ -6L

2L

2

-6L 4L

2

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 v

3

0 18 -18 36

3

2

= - 0.0025 rad

Solving, we get v

3

= - 0.0177 m

3

= -0.0076 rad

ME 273 Lecture Notes © by R. B. Agarwal

12

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

Arbitrary angle

We will derive the 2-D beam element equation that has an arbitrary

orientation with axial and bending loads.

F

M

θ

The stiffness equation will be derived in three steps

1. Pure Beam element arbitrarily oriented in space

2. Local Beam stiffness with axial and bending loads

3. Arbitrarily oriented beam with axial and bending loads.

ME 273 Lecture Notes © by R. B. Agarwal

13

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.

d

2y

x

y

2

y

d

1y

d

1y

1

d

1x

x

d

1y

= d

1y

cosθ - d

1x

sinθ = d

1y

C - d

1x

S = - d

1x

S + d

1y

C

d

2y

= d

2y

cosθ – d

2x

sinθ = d

2y

C – d

2x

S = – d

2x

S + d

2y

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,

d

1x

d

1y

-s c 0 0 0 0 d

1y

1

= 0 0 1 0 0 0

1

d

2y

0 0 0 -s c 0 d

2x

2

0 0 0 0 0 1 d

2y

2

ME 273 Lecture Notes © by R. B. Agarwal

14

-s c 0 0 0 0

Let T

= 0 0 1 0 0 0 (A)

0 0 0 -s c 0

0 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

[k

g

] = [T]

T

[k] [T]

Where, [k

g

] = 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

猬

†ㄲ

2

ㄲ千 ⴶ䱓-ⴱ㉓

2

ㄲ千 ⴶ䱓-

† ‱2C

2

‶䱃 ㄲ千1ⴱ㉃

2

‶䱃

欠㴠䕉⽌

3

‴

2

‶䱓 ⴶ䱃-㉌

2

‱㉓

2

ㄲ千 㙌匠

⁓祭 整特e ㄲ1

2

㙌

† † L

2

ME 273 Lecture Notes © by R. B. Agarwal

15

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

load, and oriented at an angle

⸠

䅬獯Ⱐ匠㴠獩A

Ⱐ䌠㴠捯,

渠瑨攠慢潶攠敱畡瑩潮⸠

ME 273 Lecture Notes © by R. B. Agarwal

16

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,

f

1x

1 -1 d

1x

f

1x

f

2x

= AE/L

f

2x

-1 1 d

2x

Truss Element

And for bending, the structural equation is,

f

1y

12 6L -12 6L d

1y

m

1

6L 4L

2

-6L 2L

2

φ

1

= AE/L

3

f

2y

-12 -6L 12 -6L d

2y

m

2

6L 2L

2

-6L 4L

2

φ

2

f

1y

f

2y

m

1

m

2

Beam Element

ME 273 Lecture Notes © by R. B. Agarwal

17

Therefore, the combined loading equation is

d

1x

d

1y

φ

1

d

2x

d

2y

φ

2

f

1x

C

1

0 0 - C

1

0 0 d

1x

f

1y

0 12 C

2

6C

2

L 0 -12 C

2

6C

2

L d

1y

m

1

0 6 C

2

L 4C

2

L

2

0 -6C

2

L 2C

2

L

2

φ

1

=

f

2x

-C

1

0 0 C

1

0 0 d

2x

f

2y

0 -12 C

2

-6C

2

L 0 12 C

2

-6C

2

L d

2y

m

2

0 6 C

2

L 2C

2

L

2

0 -6C

2

L 4C

2

L

2

φ

2

Where,

C

1

= AE/L

C

2

= EI/L

3

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

Stiffness matrix k as

d

1x

d

1y

φ

1

d

2x

d

2y

φ

2

C

1

0 0 - C

1

0 0

0 12 C

2

6C

2

L 0 -12 C

2

6C

2

L

0 6 C

2

L 4C

2

L

2

0 -6C

2

L 2C

2

L

2

k =

-C

1

0 0 C

1

0 0

0 -12 C

2

-6C

2

L 0 12 C

2

-6C

2

L

0 6 C

2

L 2C

2

L

2

0 -6C

2

L 4C

2

L

2

ME 273 Lecture Notes © by R. B. Agarwal

18

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

bending and axial loads.

For axial loading:

u

2y

u

2

u

1y

2

u

2x

1

u

1x

u

1

u

1

= u

1x

cos + u

1y

sin = c u

1x

+ s u

1y

u

2

= u

2x

cos + u

2y

sin = c u

2x

+ s u

2y

where, cos = c, and sin = s

Which can be written as,

u

1x

u

1

c s 0 0 u

1y

= u

2x

(3.2)

u

2

0 0 c s u

2y

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 can’t be

inverted, which will be incompatible with the matrix operations.

ME 273 Lecture Notes © by R. B. Agarwal

19

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

necessary for compatibility.

u

1x

u

1y

u

2x

u

2y

u

1

c s 0 0 u

1x

v

1

= -s c 0 0 u

1y

u

2

0 0 c s u

2x

v

2

0 0 -s c u

2y

Where

,

u

1x

u

1y

u

2x

u

2y

c s 0 0

T

= -s c 0 0

0 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.

u

2y

y x

y

2

u

1y

u

1y

1

u

1x

x

u

1y

= u

1y

cosθ - u

1x

sinθ = u

1y

C - u

1x

S = - u

1x

S + u

1y

C

u

2y

= u

2y

cosθ – u

2x

sinθ = u

2y

C – u

2x

S = – u

2x

S + u

2y

C

ME 273 Lecture Notes © by R. B. Agarwal

20

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,

u

1x

u

1y

-s c 0 0 0 0 u

1y

1

= 0 0 1 0 0 0

1

u

2y

0 0 0 -s c 0 u

2x

2

0 0 0 0 0 1 u

2y

2

-s c 0 0 0 0

Let T

= 0 0 1 0 0 0

0 0 0 -s c 0

0 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 0

T = 0 0 1 0 0 0

0 0 0 C S 0

0 0 0 -S C 0

0 0 0 0 0 1

ME 273 Lecture Notes © by R. B. Agarwal

21

Using the relation, [k] = [T]

T

[k][T] and simplifying, we get,

I

C

L

I

C

L

I

ASsymmetry

S

L

I

CS

L

I

AS

L

I

AC

IC

L

I

S

L

I

I

C

L

I

C

L

I

ASCS

L

I

AC

L

I

C

L

I

AS

S

L

I

CS

L

I

AS

L

I

ACS

L

I

CS

L

I

AS

L

I

AC

L

E

K

4

612

61212

2

66

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.

ME 273 Lecture Notes © by R. B. Agarwal

22

2-D Beam Element with combined loading

Bending, Axial, and Torsion (θ = 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 m

1x

and m

2x

, and the corresponding

deflections are,

x1

湤

x2

周攠瑯牳楯湡氠獴牵捴畲慬煵慴楯渠楳㨠

ㅸ

†ㄠ ⴱ- φ

1x

= JG/L

m

1x

-1 1 φ

2x

These terms can be superimposed on the stiffness equation derived

previously for the combined bending and axial loads.

d

y

φ

y

3-D Beam Element: d

x

φ

z

φ

x

d

z

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,

ME 273 Lecture Notes © by R. B. Agarwal

23

The stiffness equation in local coordinate is:

^

2

^

2

^

2

^

2

^

2

^

2

^

1

^

1

^

1

^

1

^

1

^

1 zyxzyxzyxzyx

dddddd

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

GJ

L

GJ

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

AE

L

AE

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

GJ

L

GJ

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

AE

L

AE

K

zzzz

yyyy

yyyy

zzzz

zzzz

yyyy

yyyy

zzzz

4

000

6

0

2

000

6

0

0

4

0

6

000

2

0

6

00

0000000000

0

6

0

12

000

6

0

12

00

6

000

12

0

6

000

12

0

0000000000

2

000

6

0

4

000

6

0

0

2

0

6

000

4

0

6

00

0000000000

0

6

0

12

000

6

0

12

00

6

000

12

0

6

000

12

0

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,

]][[][

^

TKTK

T

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

derivation any further.

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