DirectMethod Beams

7/30/2019 DirectMethod Beams

1/49

CCOORRNNEELLLLU N I V E R S I T Y 1MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/13/2011)

Finite Element Analysis forMechanical and Aerospace Design

Prof. Nicholas ZabarasMaterials Process Design and Control Laboratory

Sibley School of Mechanical and Aerospace Engineering

101 Rhodes Hall

Cornell University

Ithaca, NY 14853-3801

[email protected]

http://mpdc.mae.cornell.edu/
mailto:[email protected]://mpdc.mae.cornell.edu/http://mpdc.mae.cornell.edu/mailto:[email protected]


2/49


A refresher on beam bending

Beams are different from truss structuresin that they are designed to resisttransverse loads.

The tranverse loads are transported tothe supports of the beam viaextensional action.

There are several beam models

depending on the assumptionsemployed. We herein consider theBernoulli-Euler beam theoryusuallyintroduced in introductory staticscourses.

We assume thatnormals to the middle line of

the beam remain straight and normal.

This will allow us to approximate thedisplacements at a given point.,x yu u


3/49


Bernoulli-Euler beam theory

The x-component of the displacementthrough the depth of the beam is given as:

xu

sin ( ) ( : 0 0)x x

u y x Note y u= < >

where (x) is the rotation of the middle line(positive counterclockwise) at x and y isthe distance from the middle line.

We assume that

For (x) small, sin =,

( )ydu x

dx =

y

x

duu y

dx=

Thus we conclude:

2

2

yxx

d uduy y

dx dx = = =

y

curvature

y( , ) ( ).y yu x y u x=


4/49


Displacements im Bernoulli-Euler Beam model

In summary, the following displacements are considered:

( )( , )( , ) ( )

( )

yx

y yy

du xu x y yy

dxu x y u x

u x

= =

( )ydu xdx

=,y

y u

( , ) ( , 0)y y

u x y u x=

,x

x u

'( , )x yP x u y u+ +


5/49


Stress and moment calculation

From Hookes law, we can compute theaxial stress as:

2

2

y

x

d uy

dx

=

2

2

y

x x

d uE Ey

dx = =

With integration, we can now computethe bending moment as follows:

2 2 2

2

2 2 2( ) y y y

x

A A A

d u d u d uM x ydA Ey ydA E y dA EIdx dx dx

= = = = Here I denotes the moment of inertia of

the cross section:

2

A

I y dA=

: 0,

0

xNote For

M

>

2

2.y

d uM EI M

dx =

2

2 .y

d u

M EI Mdx =

0M >

.V

2

2

2

2

,

,

,

.

yy u

y

y

M

y

V

u u on

duon

dx

d uMn EI n M on

dxd ud

Vn EI n V ondx dx

=

=

=

=


11/49


Boundary conditions for beams

Free end with an applied load:

Simple support:

Clamped support:

,

.

M

V

Mn M on

Vn V on

=

=

0 ,

0 .

y u

M

u on

M on

=

=

0 ,

0 .

y uu on

on

=

=


12/49


Potential energy of a beam element: Pe=Ue-We

Using these, let us compute the potential energy of a beamelement of length Le (to be defined shortly in more detail).

Strain energy:

External work:

2 22 2( )

2 2 2e ee e

e

e e ee x

L A

I

E E y EU dV dAdx y dA dx

= = =

2

2 2

2( )

2 2

y

e e

ee e e ee

d uE I E IU dx dx

dx

= =

Notes:

For this strain energy to make sense as anintegral, we need to introduce an approximation(interpolation) for such that issquare integrable (C1 continuity)

From now-on, we simply denote with thex-dimensional domain of a beam element e oflength Le (to be defined later in more detail)

e

yu 2 2/e

yd u dx

2

2( ) ,

yd u

M x EI

dx

=2

2,

yd udM d

V EI

dx dx dx

= =

22

2 2

yd udV dq EI

dx dx dx

= =

( ) ( ) ( ) | ( ) |e eMV

e

e e e

y y y

e

W q x u x dx Vu x M x

= + +

e


13/49


C1 continuity

Displacement that is C1

continuous: Both andare continuous.

yu

yu

yu ydu

dx=

yu

yuy

du

dx =

Displacement that is C0continuous: iscontinuous but

is discontinuous.y

u
http://../FiniteElementClass/CarlosFelippa/Introduction%20to%20Finite%20Element%20Methods%20(ASEN%205007)%20Course%20Material.mhthttp://../FiniteElementClass/CarlosFelippa/Introduction%20to%20Finite%20Element%20Methods%20(ASEN%205007)%20Course%20Material.mhthttp://../FiniteElementClass/CarlosFelippa/Introduction%20to%20Finite%20Element%20Methods%20(ASEN%205007)%20Course%20Material.mhthttp://../FiniteElementClass/CarlosFelippa/Introduction%20to%20Finite%20Element%20Methods%20(ASEN%205007)%20Course%20Material.mhthttp://../FiniteElementClass/CarlosFelippa/Introduction%20to%20Finite%20Element%20Methods%20(ASEN%205007)%20Course%20Material.mhthttp://../FiniteElementClass/CarlosFelippa/Introduction%20to%20Finite%20Element%20Methods%20(ASEN%205007)%20Course%20Material.mhthttp://../FiniteElementClass/CarlosFelippa/Introduction%20to%20Finite%20Element%20Methods%20(ASEN%205007)%20Course%20Material.mhthttp://../FiniteElementClass/CarlosFelippa/Introduction%20to%20Finite%20Element%20Methods%20(ASEN%205007)%20Course%20Material.mhthttp://../FiniteElementClass/CarlosFelippa/Introduction%20to%20Finite%20Element%20Methods%20(ASEN%205007)%20Course%20Material.mht


14/49


Two-noded beam element

We need interpolation forboth displacements and

slopes at the ends of thebeam (C1 continuity)

Our element degrees offreedomare taken as the

following:

1

1

2

2

{ }

y

e

y

u

d u

=

1

1

2

2

{ }

y

e

y

F

M

f F

M

=

The corresponding

conjugate nodal forcesare:

Note that:( ),

1, 2

I IM M X

I

=

el

2yu1y

u


15/49


Two-noded beam element: Sign conventions

Please note the convention forpositive nodal

displacements and slopesat both ends (posit ive y-axis- upwards andcounterclockwise,respectively)

As a result, the notation for the conjugate nodal forces is asfollows:

1 2,y yF F are positive when they point in the positive y-axis

1 2,y yM M are positive when they are anticlockwise

1V2

V2

M1

M

1x x= 2x x=2

1 2yu1y

u


16/49


Interpolation functions for 2-noded beam element

We interpolate thedeflection along the beam

using the followingHermite interpolationfunctions for beamelements.

21

21

22

22

1(1 ) (2 )

4

(1 ) (1 )

81

(1 ) (2 )4

(1 ) ( 1)8

u

e

u

e

N

LN

N

LN

= +

= +

= +

= +

12( )

1, 1 1

e e

e

x x

L

=

1

1

1 1 2 2

2

2

( ) [ ]

y

e

y u u

y

u

u x N N N N u

=

( ) [ ] { }

e e e

yrow vector column vector

u x N d =

1e

x x= 2e

x x=

1(1 ) , 1 12

eeL

x x = + +


17/49


How the interpolation functions are computed?

We are looking for anapproximation of the form:

We compute byinterpolation using thevalues of deflection andslope at the ends of theelement:

1x x= 2x x=

0

12 3 2 3

0 1 2 3

2

3

( ) 1y

aa

u x a a a aa

a

= + + + =

1

2

1 1 1 0 1 2 3

1 1 1 1 2 3

1 2 2 0 1 2 3

1 2 2 1 2 3

1: | ,

1: | | , 2 32 2 2

1: | ,

1: | | , 2 32 2 2

y y y

e e e

y yx x

y y y

e e ey y

x x

u u u a a a a

du du L L L a a ad dx

u u u a a a a

du du L L La a a

d dx

=

= =

=

= =

= = = +

= = = = +

= = = + + + = = = = + +

1= 1= +

12( ) 1, 1 1e

e

x x

L

=

0 1

1 1

2 2

3 2

1 / 2 / 8 1 / 2 / 83 / 4 / 8 3 / 4 / 8

0 / 8 0 / 8

1 / 4 / 8 1 / 4 / 8

e e

y

e e

e ey

e e

a uL La L L

a uL L

a L L

=

in :

2

ey y y

du du dudx LCha rule

d dx d dx

= =

1(1 ) , 1 12

eeL

x x = + +

2

1 2 3

( )2 3

ydu xa a a

d

= + +


18/49


How the interpolation functions are computed?

With substitution, weobtain:

This can finally besimplified as:

1x x= 2x x=

0

12 3 2 3

0 1 2 3

2

3

( ) 1y

a

a

u x a a a a a

a

= + + + =

1= 1= +

12( ) 1, 1 1e

e

x x

L

=

0 1

1 1

2 2

3 2

1 / 2 / 8 1 / 2 / 8

3 / 4 / 8 3 / 4 / 8

0 / 8 0 / 8

1 / 4 / 8 1 / 4 / 8

e ey

e e

e ey

e e

a uL L

a L L

a uL L

a L L

=

1

12 3

2

2

1/ 2 / 8 1/ 2 / 8

3 / 4 / 8 3 / 4 / 8( ) 1

0 / 8 0 / 8

1/ 4 / 8 1/ 4 / 8

e ey

e e

y e ey

e e

uL L

L Lu x

uL L

L L

=

1 21 2

1

13 2 3 3 2 3

2

2

1 3 1 1 1 1 1 1 3 1 1 1 1 1( )

2 4 4 2 4 4 4 4 2 4 4 2 4 4 4 4u u

y

e e

y

y

N NN N

u

L Lu x

u

= + + + + +

1(1 ) , 1 12

eeL

x x = + +


19/49


Interpolation functions for 2-noded beam element

Use change rule to findderivatives with respect to

x, e.g.

12( ) 1, 1 1e

e

x x

L

=

1 1u udN dN d

dx d dx

=

21 3 (1 )4

udN

d

=

2eddx L

=

21

2

1

22

22

1(1 ) (2 )

4

(1 ) (1 )8

1(1 ) (2 )

4

(1 ) ( 1)8

u

e

u

e

N

L

N

N

LN

= +

= +

= +

= +

1x x= 2x x=

1(1 ) , 1 12

eeL

x x = + +


20/49


Continuity of displacements

Note that at =-1 (left node), . Similarly at

the right node.

1

11 1 2 2

2

2

( ) [ ]

y

ey u u

y

u

u x N N N N u

=

1 1 2 2( 1) 1, ( 1) ( 1) ( 1) 0u uN N N N = = = = = = = =

1( )e

y yu left node u=

1 =


21/49


Continuity of slopes

Note that at =-1 (left node), . Similarly at

the right node.

1

11 1 2 2

2

2

( )( ) [ ]

y

e

y u u

y

u

du x dN dN dN dN x

udx dx dx dx dx

=

1 1 2 2( 1) 1, ( 1) ( 1) ( 1) 0u udN dN dN dN

dx dx dx dx

= = = = = = = =

1( )e

left node =


22/49


Curvature calculation

12( ) 1, 1 1e

e

x x

L

=

1 1 2 2( ) [ ]{ }ye eu u

du dN dN dN dN x d

dx dx dx dx dx

= =

2 2 2 2 21 1 2 2

2 2 2 2 2[ ]{ }

ey eu ud u d N d N d N d N d

dx dx dx dx dx

= =

2

2 2 2 2 21 1 2 2

2 2 2 2 2

4[ ]{ }

e

y eu u

e

d u d N d N d N d N d

dx d d d d L

=

2

2

1 6 6[ 3 1 3 1]{ }

e

e

y e

e e e

B

d u

ddx L L L

= +

2

2[ ]{ }

e

y e ed u

B ddx

=

2

2

e

yd u

dx

=

We will need to alsocompute the curvature

to compute the strainenergy. Starting from:

21

21

22

22

1(1 ) (2 )4

(1 ) (1 )8

1(1 ) (2 )

4

(1 ) ( 1)8

u

e

u

e

N

LN

N

LN

= +

= +

= +

= +

1(1 ) , 1 12

eeLx x = + +


23/49


Strain energy calculation

Recall that:

Thus with our Hermite interpolation, we can write:

Recall that (from global to local degrees offreedom), so the assembly of the strain energy term will

give:

2

2

2( )

2e

ee eye

d uE IU dx

dx

=

1{ } [ ] [ ] { }

2 e

e e T e T e e e e e

Beam element stiffness

U d B E I B dx d

=

{ } [ ]{ }e ed L d=

1

1

1 { } ([ ] [ ] [ ] [ ]){ }2 2

eT e T e T e e e e

e

Jacobian

dx

LU d L B E I B d L d

=

2

2[ ]{ }

e

y e ed u

B ddx

=


24/49


Element stiffness

For constant on the element, the

stiffness is given as (use directintegration):

e eE I

[ ]

e

K

22 2

3 3

2 2

2 2

1 2 2

2 21

2 2

12 6 12 636 6 (3 1) 36 6 (3 1)

6 4 66 (3 1) (3 1) 6 (3 1) (9 1)

36 6 (3 1) 36 6 (3 1)2

6 (3 1) (9 1) 6 (3 1) (3 1)

e ee e

e ee e e ee e e ee

e ee e

e e e

L LL L

L LL L L LE I E IK d

L LL L

L L L L

+

= =

+ + + +

2

2 2

2

12 6 12 6

6 2 6 4

e e

e e

e e e e

L L

L L

L L L L

1

1 2

Te

e e e e e LK B E I B d

=


25/49


The external

work is: Upon assembly all terms with V and M cancel at each nodeunless external force and/or moments are applied there.

We can thus write:

The boundary is non-zero only if the element e has at oneof its ends a prescribed shear force .

Similarly, the boundary is non-zero only if the element e

has at one of its ends a prescribed moment . Note that the sign notation for and is consistent with the

sign convention for , respectively.

External work calculation

( ) ( ) ( ) | ( ) |e eMVe

e e e

y y yeW q x u x dx Vu x M x

= + + e

V

e

M

1 1 2 1 1 1 2 2( ) ( ) ( ) ( ) ( ) ( )e

e e e e e e e e e e e e e

y y y y y

e

W q x u x dx V u x V u x M x M u x

= + + + +

V M

V

M

1 2 1 2, ,e e e e

V V and M M


26/49


Recall that:

For example, if we only account for prescribedmoments and shear force at both ends, the

above equation is written explicitly as:


( ) ( ) ( ) | ( ) |e eMV

e

e e e

y y y

e


= + +

2 12 12 1 2 1( ) ( )e

e

y

e

W q x u x dx V u V u M M

= + + + +

1V2V

2M1M

1x x= 2x x=2

1 2yu1y

u

E l k l l i


27/49


We will keep the general notation here as it will alsoallow us to account for shear force and moments

applied even inside the element!

We will consider these cases in our followingderivations but please note that if you apply amoment or shear force at a point, you will be betterserved to make that point a finite element node!!


( ) ( ) ( ) | ( ) |e eMV

e

e e e

y y y

e


= + +

1V2V

2M1M

1x x= 2x x=2

1 2yu1y

u

M

V

E t l k l l ti


28/49

CCOORRNNEELLLLU N I V E R S I T Y 28MAE 4700 FE Analysis for Mechanical & Aerospace Design

N. Zabaras (9/13/2011)

Recall that:

We assume known distributed load, applied concentratedload and applied concentrated moments. Applying theHermite interpolation:

Note that in this expression, only the elements e that haveboundaries (i.e. one of their end points) on the boundaries

contribute to the last two terms.


( ) ( ) ( ) | ( ) |e eMV

e

e e e

y y y

e


= + +

[ ]

{ } ( )[ ] [ ] | |e eMVe

e Te e T e T e T d N

W d q x N dx N V M dx

= + +

,V M

E t l k l l ti


29/49


N. Zabaras (9/13/2011)

Also note that with our notation, and are row vectorsand is a column vector. Assembling (transform fromlocal to global degrees of freedom) then gives:


[ ]eN [ ]e

B

{ }ed

[ ]{ } [ ] ( )[ ] [ ] | |

[ ]

{ } [ ] ( )[ ] [ ] | |

e eV M

e

e eV M

e

e T

T e T e T e T

e

e TT e T e T e T

e columnvector column

vector

d NW d L q x N dx N V M dx

d N

d L q x N dx N V M dx

= + +

+ +

[ ]{ } ( )[ ] [ ] | |e e

MVe

e Te e T e T e T d N

W d q x N dx N V M dx

= + +

Mi i i ti f t t l t ti l


30/49


N. Zabaras (9/13/2011)

This minimization problem results in:

Minimization of total potential energy

{ }{ }

1min min { } ([ ] [ ] [ ] [ ]){ }

2[ ]

{ } [ ] ( )[ ] [ ] | |

e

e

e eV M

e

T e T e T e e e e e

dde

e TT e T e T e T

e

d L B E I B dx L d

d Nd L q x N dx N V M

dx

=

+ +

[ ] { }

[ ]([ ] [ ] [ ] [ ]){ } [ ] ( )[ ] [ ] | |e e

V Me e

e Te T e T e e e e e e T e T e T

e e

Global stiffness K Global force vector F

d NL B E I B dx L d L q x N dx N V M

dx

= + +

Fi l fi it l t ti


31/49


N. Zabaras (9/13/2011)

Note that the above minimization process is with respect tothe nodal displacements (i.e. excluding the DOF withprescribed displacement or rotation).

As was done in earlier lectures, by splitting the stiffnessmatrix, we finally obtain:

Final finite element equations

[ ]([ ] [ ] [ ] [ ]){ } [ ] ( )[ ] [ ] | |e e

V Me e

e Te T e T e e e e e e T e T e T

e e

d NL B E I B dx L d L q x N dx N V M

dx

= + +

{ }Fd

[ ]{ } [ ] { }T EF F F F EF K d f f K d = +

Distributedloads

BoundaryLoads (forces,moments)

U if di t ib t d l d


32/49


N. Zabaras (9/13/2011)

For uniform over the element load q the first term gives:

Uniform distributed load

1

612

6

e

e

e

e

L

qLf

L

=

( )[ ]e

e e Tf q x N dx

=

1

11

21int

2 tan

( )[ ] ( )

2e

u

ee e T

performuegration

forcons t q

N

NLf q x N dx q x d

NN

= =

2

eqL

2

eqL2

12

eqL2

12

eqL

1 2

C t t d l d i id th l t


33/49


N. Zabaras (9/13/2011)

Assume a contentrated load at . We write this load asa distributed load using a delta function (nice trick!):

Substitution into the first term of the formula for gives:

Concentrated load inside the element

( )[ ]e

e e Tf q x N dx

=

e

L

1F1x x=

1x x=

1 1( ) ( )q x F x x=

1 1( )[ ]e

e e Tf F x x N dx

=

ef

1 1[ ( )]e e Tf F N x =

If the applied load is at a

node e.g. , then:2e

x1

00

1

0

ef F

=

E l bl


34/49


N. Zabaras (9/13/2011)

Example problem

The beam ABC is clamped at the left side

and simply supported at the right side.Dimensions are in m, forces in N andloading q in N/m . Also, EI = 104 Nm2. Atx

=12m, and . Find thedeflection, shear forces and moments.20V Nt

= 20 .M N m

=

20M = 20V =

Fi it l t di ti ti


35/49


N. Zabaras (9/13/2011)

Finite element discretization

We consider 2 beam elements as follows:

1 2

1,2 3,4 5,6Degrees

of freedom

20M = 20V =

Stiffness of element 1


36/49


N. Zabaras (9/13/2011)


2 2

3

3

2 2

12 6 12 6 0.23 0.94 0.23 0.94

6 4 6 2 0.94 5.00 0.94 2.5010

12 6 12 6 0.23 0.94 0.23 0.94

0.94 2.50 0.94 5.006 2 6 4

e

L L

L L L LEIK

L LL

L L L L

= =

4

10 , 8EI L= =

1 2 3 41

2

3

4



37/49


N. Zabaras (9/13/2011)


2 2

3

3

2 2

12 6 12 6 1.88 3.75 1.88 3.75

6 4 6 2 3.75 10.00 3.75 5.0010

12 6 12 6 1.88 3.75 1.88 3.75

3.75 5.00 3.75 10.006 2 6 4

e

L L

L L L LEIK

L LL

L L L L

= =

3 4 5 63

4

5

6

410 , 4EI L= =

Global stiffness


38/49


N. Zabaras (9/13/2011)

Global stiffness

3

0.23 0.94 0.23 0.94 0 0

0.94 5.00 0.94 2.50 0 0

0.23 0.94 2.11 2.81 1.88 3.75

10 0.94 2.50 2.81 15.00 3.75 5.00

0 0 1.88 3.75 1.88 3.75

0 0 3.75 5.00 3.75 10.00

K

=

1 2 3 4 5 61

2

3

4

5

6

Boundary moments and forces


39/49


N. Zabaras (9/13/2011)

Boundary moments and forces

Element 1 has no boundary with applied V or M.

Element 2 has and at its right end.

Assembly of this vector gives:

[ ][ ] | |e e

V M

e Te e T d N

f N V M

dx

= +

20V Nt= 20 .M N m=

410 , 4EI L= =

2

0 0 0

0 0 0

1 0 20

0 1 20

f V MN

N

= + =

3

4

5

6

0

0

0

0

20

20

f

N

N

=

1

2

3

4

5

6

Distributed load and concentrated load inside the element


40/49


N. Zabaras (9/13/2011)

Distributed load and concentrated load inside the element

Element 1 has distributed load q=-1 andconcentrated load P1=-10Nt.

11 0

1 1

[ ( )] |

1 1 1928

1 15.3( 1)86 6( 0) ( 10)1 1 1 92 2

28 15.316 6

T

T

A

N P

L

qL

f N P

L

=

= + = = + =

1 2

1

2

3

4

410 , 8EI L= =

( ) 1q x = 20M = 20V =

21

21

22

22

1(1 ) (2 )

4

(1 ) (1 )8

1(1 ) (2 )

4

(1 ) ( 1)8

u

e

u

e

N

LN

N

LN

= +

= +

= +

= + 0=

Distributed load


41/49


N. Zabaras (9/13/2011)

Distributed load

Element 2 has concentrated load P2 at an end point

2 22

5

0( 1)

00

T

f N P

= = =

1 2

3

4

56

410 , 8EI L= = 2 2

2 1[ ( )] |T

f N P ==

20M = 20V =

1=

Assembled load


42/49


N. Zabaras (9/13/2011)

Assembled load

Assembling all load contributions gives:

1 2

1 2

0 9 0 9

0 15.3 0 15.3

0 9 5 4

0 15.3 0 15.3

20 0 0 20

20 0 0 20

Boundary From element From elementconditions

f

N

N

= + + =

1

2

3

4

5

6

20M = 20V =

Solution step


43/49


N. Zabaras (9/13/2011)

Solution step

1

1

23

2

3

3

0 90.23 0.94 0.23 0.94 0 000.94 5.00 0.94 2.50 0 0

0.23 0.94 2.11 2.81 1.88 3.7510

0.94 2.50 2.81 15.00 3.75 5.00

0 0 1.88 3.75 1.88 3.75

0 0 3.75 5.00 3.75 10.00

y

y

y

y

y

y

u

u

u

= =

=

1

115.3

4

15.3

20

20

uR

R

+ +

1 2

1,2 3,4 5,6

Note that we applied the essential boundary conditions and included thecorresponding reaction force and moment at this location

20M = 20V =

Solution step


44/49


N. Zabaras (9/13/2011)

Solution step

2

23

3

3

2.11 2.81 1.88 3.75 4

2.81 15.00 3.75 5.00 15.310

1.88 3.75 1.88 3.75 20

3.75 5.00 3.75 10.00 20

y

y

y

y

u

u

=

1 2

1,2 3,4 5,6

2

2

3

3

0.55

0.11

1.03

0.12

y

y

y

y

u

u

=

2

2 13

3 1

3

90.23 0.94 0 010

0.94 2.50 0 0 15.3

y

y u

y

y

uR

u R

+

= +

1

1

33

252 .

uR N

R N m

=

20M = 20V =

Displacement field


45/49


N. Zabaras (9/13/2011)

Displacement field

For element 1:

For element 2:

1 1 1 1 1 1 11 1 2 2 2 2 2 2

2

2

00

( ) [ ]y u u u y y

y

y

u x N N N N N u N u

= = +

2

22 2 2 2 2 2 2 2 3

1 1 2 2 1 2 1 2 2 3 2 33

3

( ) [ ]

y

y

y u u u y y u y yy

y

u

u x N N N N N u N N u N u

= = + + +

Moments and shear forces


46/49


N. Zabaras (9/13/2011)

Moments and shear forces

1

2 12 1 2 1 2 1 2 11 11 2 2

2 2 2 2 2[ ]{ } 236.67 23.76y u u

d Nd u d N d N d N M EI EI d x

dx dx dx dx dx

= = = +

1

3 13 1 3 1 3 1 3 11 11 2 2

2 3 3 3 3[ ]{ } 23.76y u u

d Nd u d N d N d N V EI EI d

dx dx dx dx dx= = =

1

2 22 2 2 2 2 2 2 22 21 2 2

2 2 2 2 2[ ]{ } 222 20.25y u u

d Nd u d N d N d N M EI EI d x

dx dx dx dx dx= = = +

1

3 23 2 3 2 3 2 3 22 21 2 2

2 3 3 3 3[ ]{ } 20.25y u u

d Nd u d N d N d N V EI EI d

dx dx dx dx dx

= = =

Place nodes on location of concentrated load


47/49


N. Zabaras (9/13/2011)

Place nodes on location of concentrated load

To compute an accurate shear force, youneed to split element 1 in more elements.For sure place a node at the application of

load P1.

0 2 4 6 8 10 12-1.5

-1

-0.5

0

x

displacement

Displacements: FE versus analytical beam solutions

FE

Exact S olution

0 2 4 6 8 10 12-300

-250

-200

-150

-100

-50

0

50

100

x

moment

Moments: FE versus analytical beam solutions

FE

Exact Solution

0 2 4 6 8 10 12-35

-30

-25

-20

-15

-10

x

shear

Shear: FE versus analytical beam solutions

FE

Exact Solution

Results with refined grids: 3 elements


48/49


N. Zabaras (9/13/2011)


0 2 4 6 8 10 12

-1.5

-1

-0.5

0

x

displacement


FE

Exact Solution

0 2 4 6 8 10 12-300

-250

-200

-150

-100

-50

0

50

100

x

moment


FE

Exact Solution

0 2 4 6 8 10 12-35

-30

-25

-20

-15

-10

x

shear


FE

Exact Solution



49/49

MAE 4700 FE Analysis for Mechanical & Aerospace Design


0 2 4 6 8 10 12-1.5

-1

-0.5

0

x

displacement


FE

Exact Solution

0 2 4 6 8 10 12-300

-250

-200

-150

-100

-50

0

50

100

x

moment


FE

Exact Solution

0 2 4 6 8 10 12-35

-30

-25

-20

-15

-10

x

shear


FE

Exact Solution

Documents

DirectMethod Beams