Chapter4 [Modo de Compatibilidad]

7/28/2019 Chapter4 [Modo de Compatibilidad]

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Beams and Frames


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beam theory can be used to solve simple

beamscomplex beams with many cross section

many 2-d and 3-d frame structures are

better modeled by beam theory


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One Dimensional Problems

x

y

P

p

u(x)

y

px

StretchContract

u(x)

kx + P(x=l)=0dudx

The geometry of the problem is three dimensional, but the

variation of variable is one dimensional

x v(x)v(x)

kx + P(x)=0d4v

dx4

Variable can be scalar field liketemperature, or vector field like

displacement.

For dynamic loading, the variable can

be time dependent


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Element Formulation assume the displacement w is a cubic polynomial in

L = Length

a1, a2, a3, a4 are the undetermined coefficients

2 3

1 2 3 4v(x) a a x a x a x= + + +

the cross sectional areaE = Modulus of Elsaticity

v = v(x) deflection of the

neutral axis

= dv/dx slope of theelastic curve (rotation of

the section

F = F(x) = shear force

M= M(x) = Bending

moment about Z-axis

i j

L, EI

viFi,Fj,vj

Mi,i

Mj,j

j vjivi

y

x

x

v(x)

(x)i

j


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`

1 1

x 0

dvx 0, v(0) v ;

dx

dv

=

= = =

{ } { }

1 1

2 22 3 2

3 3

4 4

a a

a av(x) 1 x x x ; (x) 0 1 2x 3x

a a

a a

= =

2 2x L

x , v v ;dx =

= = =

i 1

i 2

2 3j 3

2j 4

v 1 0 0 0 a

0 1 0 0 a

v a1 L L L

a0 1 2L 3L

=


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Applying these boundary conditions, we get

1

1 1 2 1

{d} [P(x)]{a}

{a} [P(x)] {d}

a v ; a

1a 3v 2L 3v L

=

=

= =

= +

Substituting coefficients ai back into the original equation

for v(x) and rearranging terms gives

L

{ }

1

22 3

3

4

a

av(x) 1 x x x

a

a

=


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The interpolation function or shape function is given by

2 3 2 3

1 12 3 2

2 3 2 3

3x 2x 2x xv(x) (1 )v (x )

L L L L

3x 2x x x( )v ( )

= + + +

+ + +

N1=1

dN1dx

(x=0) = 0

N1(x=L) =1

dN1dx

(x=L) = 0

N2=1

[ ]

1

1

1 2 3 4

2

2

v

Lv N (x) N (x) N (x) N (x) [N]{d}

vL

= =


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strain for a beam in bending is defined by the curvature, so

Hence

N A

dv

dx

y

x

dvdx

u(x) = y2 2

2 2

du d v d [N]y {d} y[B]{d}

dx dx dx = = = =

3 2 3 2 2 3 3 2

12x 6 6x 4 6 12x 6x 2[B]

L L L L L L L h

=

{ } { }

{ } [ ]{ }

{ } [ ]{ }

{ } [ ] { }

{ } [ ] [ ][ ]{ }

e

e

e

Te

v

Te

v

T Te 2

v

Internal virtual energy U = dv

substitute E in above eqn.

U = E dv

= y B d

U = d B E B d y dv

=


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e eFrom virtual work principle U W =

{ } { } { } [ ] { }

{ } { } { } [ ] { }

{ } { } { }

e e

T T Te

b y

v v

T T Te

s y

s s

TTe e e

c

External virtual workdue to body force

w = d(x) b dv d N b dv

External virtual work due to surface force

w = d(x) p dv d N p ds

External virtual work due to nodal forces

w d P , P

=

=

=

{ }yi i yj= P , M , P ,....

{ } [ ] [ ][ ] { } { } [ ] { } [ ] { } { }

{ } { }

[ ] [ ][ ]

{ } [ ] { } [ ] { } { }

T TT T T2 ed ( B E B y dv d d N b dv N p dv Py y

e e sv v

eK U Fe e

where

T 2K B D B y dv Element stiffness matrixe

ev

T T eF N b dv N p ds P Total nodal force vectore y y

e sv

= + +

=

= =

= + + =


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the stiffness matrix [k] is defined

dAy

To compute equivalent nodal force vector

( )L

T 2 T

VA 0

2 2

3

2 2

[k] [B] E[B]dV dAy E [B] [B]dx

12 6L 12 6L

6L 4L 6L 2LEI

12 6L 12 6LL

6L 2L 6L 4L

= =

=

for the loading shown

{ } [ ] { }

[ ]

T

e y

s

y

y

2 3 2 3 2 3 2 3

2 3 2 2 3 2

F N p ds

From similar triangles

p w w; p x; ds = 1 dxx L L

3x 2x 2x x 3x 2x x xN (1 ) (x ) ( ) ( )

L L L L L L L L

=

= =

= + + +

w

x

py

L


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{ } [ ] { }

{ }

T

e y

s

2 3

2 3

22 3

2

e 2 3L

2 3

F N p ds

3wL3x 2x(1 )

20L L

wL2x x(x )

wx 30L LF dx

7wLL3x 2x( )

20L L

=

+

+

= =

+ve directions

vi vj

i j

22 3

2wLx x( )20L L

+

w

Equivalent nodal force due toUniformly distributed load w


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v1 v2v3

1 23

v4

v3


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1

1

2 5

2

3

3

V 12 6 -12 6

M 6 4 -6 2

V -12 -6 12+12 -6+6 -12 68x10

M 6 2 -6+6 4+4 -6 2

-12 -6V

M

=

1

1

2

2

3

3

1 1 2 3

v

v

12 -6 v

6 2 -6 4

Boundary condition

v , , v , v 0

=

2 3

5

M 1000; M 1000

8 28x10

2

= =

{ } [ ]{ }

2

3

42

5 43

1000

4 1000.0

4 -2 1000 2.679x101

-2 8 1000.028*8x10 4.464x10

Final member end forces

f k d {FEMS}

=

= =

= +


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Find slope at joint 2 and deflection at

joint 3. Also find member end forces

v

vv

EI EI

20 KN 20kN/m Guided Support

2m 2m 6m

EI=2 x 10

4

kN-m

2

Global coordinates

20 KN 20kN/m

m mm

m

m

m

m

Fixed end reactions (FERs)

Action/loads at global

coordinates


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1 14

1 1

32 2

2 2

1 1 2 2

For element 1

f v12 24 -12 24

m 24 64 -24 321X10

f -12 -24 12 -24 v4

24 32 -24 64m

v v

For el

=

1 1

41 1

ement 2

f v12 36 -12 36

m 36 144 -36 721X10

=2 2

2 2

2 2 3 3

f -12 -3 12 -3 v6

36 72 -36 144m

v v


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1

1

2

2

3

3

F 1875 3750 -1875 3750

M 3750 10000 -3750 5000

F -1875 -3750 1875+555.56 -3750+1666.67 -555.56 1666.67

M

F

M

=

3750 5000 -3750+1666.67 10000+6666.67 -1666.67 3333.33

-555.56 -1666.67 555.56 -1666.67

1666.67

1

1

2

2

3

3

1 1 2 3

v

v

v

3333.332 -1666.67 6666.67

Boundary condition

v , , v , 0

=

6/16/2013 26

2 3

M 50; F 60

16666.67 -1666.67

-1666.67 555.56

= =

{ } [ ]{ }

2

3

2

3

50

v 60

555.56 1666.67 50 0.0197141

v 1666.67 16666.67 60 0.167146481481.5

Final member end forces

f k d {FEMS}

=

= =

= +


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x

y

q3

q2

q1

q5

q6

q4

displacement in local coordinates


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3 3 3 3

3 3 3 3

AE AE0 0 0 0

L L

12EI 6EI 12EI 6EI0 0

L L L L6EI 4EI 6EI 2EI

0 0L L L L

[k]AE AE

0 0 0 0L L

=

{ } [ ]{ }

If f ' member end forces in local coordinates then

f' k ' q '=

3 3 3 3

3 3 3 3

12EI 6EI 12EI 6EI

0 0L L L L

6EI 2EI 6EI 4EI0 0

L L L L

{ }q {q q q q q q }


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'

1 1 2

'

2 1 2

'

At node i

q q cos q sin

q q sin q cos

= +

= +

{ } 1 2 3 4 5 6q {q ,q ,q ,q ,q ,q }

are forces in global coordinate direction

=

3 3q q

l cos ; m sin

=

= =


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{ } { }

[ ] [ ] [ ][ ]T

using conditions q' [L]{q}; and f' [L]{f}

Stiffness matrix for an arbitrarily oriented beam element is given by

k L k ' L

= =

=


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Grid Elements

a

a

f = GJ/l

qxi fi qxj fj

xi i

xj j

JG JGq fL L

q fJG JG

L L

=


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3 3 3 3

3 3 3 3

GJ GJ0 0 0 0L L

12EI 6EI 12EI 6EI0 0

L L L L

6EI 4EI 6EI 2EI0 0

L L L LGJ GJ

0 0 0 0L L

12EI 6EI 12EI 6EI

{ } [ ]{ }

If f ' member end forces in local coordinates then

f' k ' q '=

3 3 3 3

3 3 3 3

L L L L

6EI 2EI 6EI 4EI0 0

L L L L


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[ ]

C 0 -s 0 0 0

0 1 0 0 0 0

-s 0 c 0 0 0L

0 0 0 c 0 -s

=

0 0 0 0 1 0

0 0 0 --s 0 c

[ ] [ ] [ ][ ]T

k L k ' L

=


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Beam element for 3D analysis

x

y

q2

q1

q5q7

q8

q11

z

q3

q6q4

displacement in local coordinatesq9

q10

q12


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if axial load is tensile, results from beam

elements are higher than actual

resultsare conservative

than actual size of error is small until load is about 25% of

Euler buckling load


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for 2-d, can use rotation matrices to get

stiffness matrix for beams in anyorientation

,

add capability for torsional loads about theaxis of the element, and flexural loading in

x-z plane


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to derive the 3-d beam element, set up the

beam with thex axis along its length, andyandz axes as lateral directions

of simple strength of materials solution

JG

L

JG

LJG

L

JG

L

T

Txi

xj

i

j

=


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J = torsional moment aboutx axis

G = shear modulusL = length

xi, xj are nodal degrees of freedom ofangle of twist at each end

Ti, Tj are torques about thex axis at each

end


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flexure in x-z plane adds another stiffness

matrix like the first one derivedsuperposition of all these matrices gives a

to orient a beam element in 3-d, use 3-d

rotation matrices


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for beams long compared to their crosssection, displacement is almost all due toflexure of beam

for short beams there is an additional lateral

displacement due to transverse shearsome FE programs take this into account,

but you then need to input a shear

deformation constant (value associated withgeometry of cross section)


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limitations:

same assumptions as in conventional beam andtorsion theories

no better than beam anal sis

axial load capability allows frame analysis, butformulation does not couple axial and lateral

loading which are coupled nonlinearly


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Finite Element Model

Element formulation exact for beam spans

with no intermediate loads need only 1 element to model any such

member that has constant cross section

for distributed load, subdivide into severalelements

need a node everywhere a point load isapplied


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need nodes where frame members connect,

where they change direction, or where thecross section properties change

,

have the same linear and rotationaldisplacement

boundary conditions can be restraints onlinear displacements or rotation


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simple supports restrain only linear

displacementsbuilt in supports restrain rotation also


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restrain vertical and horizontal displacements

of nodes 1 and 3

no restraint on rotation of nodes 1 and 3

need a restraint inx direction to revent ri id

body motion, even if all forces are inydirection


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cantilever beam

has x and y linear displacements and rotation of node 1

fixed


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point loads are idealized loads

structure away from area of applicationbehaves as though point loads are applied


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only an exact formulation when there are no

loads along the span for distributed loads, can get exact solution

ever where else b re lacin the distributed

load by equivalent loads and moments at thenodes


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C t I t A i t


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Computer Input Assistance

preprocessor will usually have the same

capabilities as for trussesa beam element consists of two node

physical properties


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material properties:

modulus of elasticity

if dynamic or thermal analysis, mass densityand thermal coefficient of expansion

physical properties: cross sectional area

2 area moments of inertia

torsion constant location of stress calculation point


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boundary conditions:

specify node numbers and displacementcomponents that are restrained

specify by node number and load components

most commercial FE programs allows

application of distributed loads but they use

and equivalent load/moment set internally


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Output Processing and Evaluation


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Output Processing and Evaluation

graphical output of deformed shape usually

uses only straight lines to representmembers

constraints on the deformed shape of eachmember

to check these, subdivide each member andredo the analysis


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most FE codes do not make graphical

presentations of beam stress results

user must calculate some of these from the stress

values returned

for 2- beams, you get a norma stress norma to

the cross section and a transverse shear acting on

the face of the cross section

normal stress has 2 components

axial stress

bending stress due to moment


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3-d beams

normal stress is combination of axial stress,flexural stress from localy- andz- moments

stress due to moment is linear across a section

the combination is usually highest at theextreme corners of the cross section

may also have to include the effects of torsion

get a 2-d stress state which must be evaluated

also need to check for column buckling

Chapter4 [Modo de Compatibilidad]

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