Stiffness method for 2D frames

EG-225

Structural Mechanics II(b)

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Dr. Antonio Martinez

28/02/2012

Stiffness Method (I: frames)

1

Stiffness method for 2D frames

l  Introduction and nomenclature l  Element internal forces and displacements l  Element local stiffness matrix l  Transformation of internal forces and displacements l  Element global stiffness matrix l  Nodal equilibrium and compatibility l  Direct assembly of the global stiffness matrix l  Introduction of support/reactions l  Solution of the equilibrium equations l  Calculation of internal forces l  Analysis of beams l  Analysis of continuous beams l  Intermediate loading

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Examples of frames in Civil Engineering

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2

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3

Introduction

X

Y

OGlobal coordinate axes

3 Degrees Of Freedom (DOF) per node

External force 2

3

1

2 3

4

Element

Node

1

l  2D frames will be analysed (3 DOF per node).

l  Consider initially (forces and moments) to be applied at nodes

l  Divide and conquer approach:

l  Every frame member will be analysed sequentially

l  Finally, all the element information will be assembled

External moment

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4

Basic nomenclature

2

3

1

2 3

4

1

X

Y

O Global coordinate axes 1

2

1

x

yo

Local coordinate axes

l  Define global coordinate axes

l  Define nodes (joints) ; elements (members)

l  Define arbitrarily element orientation

l  Define local coordinate axes per element

OXY

2 3

2oxy

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l  Contrary to a truss element (which only translates along its local axis ), a frame element translates and bends due to the shear force and the bending moment

l  Notice that local displacements now involve translations and per node as well as rotation

l  These element internal forces can be arranged in a vector format

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5

Frame element kinematics

XO

Y

Initial shape

Displaced shape

aiaiaiaajajaj

uv

uv

θ

θ

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

u

Local displacements

i

j

ax

y !i

!j

aiu

aju

aiv

ajq

ajv

aiq

ox

uq

v

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6

Frame element internal forces

XO

Y

l  Consider a single frame element extracted from a general frame structure and establish its Free Body Diagram (FBD)

l  The frame element will be subjected to the standard internal forces: axial force, shear force and bending moment

l  Recall the classical sign convention

NN SS BMBM

Initial shape Displaced shape

i

j

ax

y!i

!j

ajBM

ajS

ajN

aiN

aiS

aiBM

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7

Frame element kinematics analysis

XO

Y

l  The kinematics of the frame element can be split into two consecutive stages:

l  STAGE I: due to axial deformation

l  STAGE II: due to shear and bending deformation

i

j

ax

y !i

!j

aiu

ajua

iv

ajq

ajv

aiq

!!j

!!i

Initial shape

Final displaced shape

Intermediate shape STAGE I

STAGE II

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Stiffness Method (I: trusses)

8

Axial deformation: constitutive law l  Similarly as it was carried out for a truss element, the following

relationship can be derived:

( )a a

a a a a a a a aj ia

E AN A E A u uL

σ ε= = = −

a aj iaa

u uL

ε−

=

a a aEσ ε=

Linear strain

Direct stress

Axial force Local displacements

Element geometric and mechanical properties

i

j

ax

XO

Y

y!i

j 'aN

aN

aiu

aju

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l  Recall that distributed loads are not considered yet, thus:

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9

Shear and bending deformation: constitutive law (I)

XO

Y

a!i

j ¢

aiv

ajq

ajv

aiq

j ¢¢

i¢¢

xy

BM (x) = Ea (x)I a (x)R(x)

! Ea (x)I a (x)d2y(x)dx2

l  Following Euler beam theory, the following relationship can be established between the bending moment and the radius of curvature :

BMR

l  For a frame element with constant section and same material throughout,

BM (x) = EaI a d2y(x)dx2

l  From the equilibrium of a beam section, it follows:

dBM (x)dx

= S(x)

dS(x)dx

= !w(x)

dx

S S dS+

BM dBM+BM

w

w(x) = 0

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10

Shear and bending deformation: constitutive law (II)

l  Combining the relationships previously derived, it yields:

l  This is the classical Ordinary Differential Equation (ODE) for the Euler beam theory, whose solution is the cubic polynomial:

EaI a d4y(x)dx4

= 0

3 2( ) ; , , ,y x ax bx cx d a b c d= + + + Œ°

l  Boundary Conditions (BCs) are required to calculate the unknown constants , , ,a b c d

1

11

1

Case 1 Case 3

Case 4 Case 2

l  Let us analyse the following four deformation cases:

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11

Shear and bending deformation: constitutive law (III)

l  For the deformation case 1, the BCs are as follows:

(0) 1y =1

0

( ) 0x

dy xdx =

=

( ) 0ay L =

( ) 0ax L

dy xdx =

=

3 2( )y x ax bx cx d= + + +l  Substituting the BCs into the cubic polynomial

it yields: y(x) = 2

La( )3x3 ! 3

La( )2x2 +1

l  The internal shear force and bending moment result in:


= EaI a 12

La( )3x ! 6

La( )2"

#

$$

%

&

''

S(x) = EaI a d3y(x)dx3

= EaI a 12

La( )3!

"

##

$

%

&&

1 ( )26( )

a aa

aE IBM LL

=

( )312( )

a aa

aE IS LL

=

a

( )26(0)

a a

aE IBML

= -‐

( )312(0)

a a

aE ISL

=

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12

Shear and bending deformation: constitutive law (III)

l  For the deformation case 2, the BCs are as follows:

(0) 0y =

0

( ) 1x

dy xdx =

=

( ) 0ay L =

( ) 0ax L

dy xdx =

=

3 2( )y x ax bx cx d= + + +l  Substituting the BCs into the cubic polynomial

it yields: ( )

3 22

1 2( ) aay x x x xLL

= -‐ +

l  The internal shear force and bending moment result in:


= EaI a 6

La( )2x ! 4

La

"

#

$$

%

&

''

S(x) = EaI a d3y(x)dx3

= EaI a 6

La( )2!

"

##

$

%

&&

BM (La ) = 2EaI a

La

S(La ) = 6EaI a

La( )2

a

BM (0) = ! 4EaI a

La

S(0) = 6EaI a

La( )2

1

1

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13

Frame element local internal forces

XO

i

j

ax

Y y

Initial shape

Displaced shape

!i

!j

l  Consider a single frame element extracted from a general frame structure and establish its Free Body Diagram (FBD)

l  Element local internal forces (notice the sign convention) can be introduced representing the axial, shear and bending moment effects

l  These element internal forces can be arranged in a vector format

axif

axjf

ayif

ajM

ayjf

aiM

a axi ia ayi ia ai iaa axj ja ayj ja aj j

f Nf SM BMf Nf SM BM

⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−

= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

f

Local internal forces

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14

Element local forces-local displacements relationship

l  We must establish a relationship between the local internal forces and the local displacements

l  This relationship will define the so called element local stiffness matrix

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( )

a a a a a aii ii ii ij ij ijxx xy x xx xy x

aa a a a a axiii ii ii ij ij ijyx yy y yx yy ya

yia a a a a a

a ii ii ii ij ij ijx y x yia a a a a a axj ji ji ji jj jj jjxx xy x xx xy xayj a aa ji jiyx yyj

k k k k k kf k k k k k kf

k k k k k kMf k k k k k kf

k k kM

θ θ

θ θ

θ θ θθ θ θ θθ

θ θ

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

=⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

aiaiaiajaja a a aaji jj jj jjy yx yy yj

a a a a a aji ji ji jj jj jjx y x y

uv

uv

k k k

k k k k k kθ θ

θ θ θθ θ θ θθ

θ

θ

⎡ ⎤⎢ ⎥

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦

⎢ ⎥⎢ ⎥⎣ ⎦

a a a=f k ua a a ai ii ij ia a a aj ji jj j

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

f k k uf k k u

l  The same relationship can be presented in a more compact matrix notation as:

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Stiffness coefficients (I)

000100

aiaiaiaajajaj

uv

uv

θ

θ

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

u

a

a a

aE AL

1

00

00

a a

a axiayiaiaa a axja ayjaj

E Af LfMf E Af LM

⎡ ⎤−⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥= =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦

f

( )ajj xxk

a a

aE AL

l  The term represents the horizontal force at end caused by a unit displacement at end and zero displacement/rotation at the rest of the degrees of freedom

jj

( )ajj xxk

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16

Stiffness coefficients (II)

000010

aiaiaiaajajaj

uv

uv

θ

θ

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

u

a

( )26 a a

aE IL( )3

12 a a

aE IL

( )26 a a

aE IL

( )312 a a

aE IL

1

( )

( )

( )

( )

3

2

3

2

012

6

012

6

a a

aaxi

a aayi

aaiaaxja a ayja aj

a a

a

E I

Lf

E IfLM

ff E IM L

E I

L

⎡ ⎤⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥= =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥−⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

f( )aij yyk

l  The term represents the vertical force at end caused by a unit displacement at end and zero displacement/rotation at the rest of the degrees of freedom

ij

( )aij yyk

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Stiffness coefficients (III)

001000

aiaiaiaajajaj

uv

uv

θ

θ

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

u

( )

( )

( )

( )

2

2

06

4

06

2

a a

aaxi

a aayi

aaiaaxja a ayja aj

a a

a

E I

LfE IfLM

ff E IM L

E IL

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

f( )aji yk

θ

a

2 a a

aE IL

( )26 a a

aE IL

4 a a

aE IL

( )26 a a

aE IL

1

l  The term represents the vertical force at end caused by a unit rotation at end and zero displacement/rotation at the rest of the degrees of freedom

ji

( )aji yk

θ

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Local internal forces-local displacements

3 2 3 2

2 2

3 2 3 2

2 2

0 0 0 0

12 6 12 60 0

6 4 6 20 0

0 0 0 0

12 6 12 60 0

6 2 6 40 0

a axi ia ayi iaiaxjayjaj

EA EAL L

EI EI EI EIf uL L L Lf vEI EI EI EIM L L L Lf EA EA

L LfEI EI EI EIML L L LEI EI EI EIL L L L

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ − − −⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

aiajajaj

uv

θ

θ

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

a aa aii iji ia aa aji jjj j

⎡ ⎤⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

k kf uk kf u

a a a=f k u

Element local stiffness matrix Local

forces

Local

displacements

l  The element local stiffness matrix is a 6x6 symmetric matrix

( as ): ( )Ta ak k=

ak

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Transformation of displacements l  The displacements/rotations at each joint of the member are

transformed from local to global coordinate axes

l  Analogously for node j:

00

0 0 1

a a a ai i

a a a a a a ai i i i

a ai i

u c s Uv s c Vθ θ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥

= = − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

u T U

a a aj j=u T U

XOi

xY

y

i¢

aiU

aiV

aiu

aaaa

ai

a ai i

ai

uvθ

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎣ ⎦

u

ai

a ai i

ai

UVθ

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎣ ⎦

U

From global to local displacements

From local to global displacements

l  For node i:

l  where is the so-called transformation matrix aT

sina as α=

cosa ac α=

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Transformation of internal forces

( )00

0 0 1

a a a aXi xi

Ta a a a a a ai Yi yi i

a ai i

F c s fF s c fM M

⎡ ⎤ ⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥

= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

F T f

l  The local forces at each joint of the member are transformed from local to global coordinate axes

l  is the transpose of the transformation matrix

xy

axif

ayif

aiM

aY iF

aX iF

aiM

Y

X

a

a

a

( )TaT

l  Analogously for node j:

axi

a axi yi

ai

ffM

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎣ ⎦

f

aXi

a ai Yi

ai

FFM

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎣ ⎦

F

From global to local forces

From local to global forces

l  For node i:

( )Ta a aj j=F T f

sina as α=

cosa ac α=

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From local to global element stiffness matrix

F a =Fia

Fja

!

"

##

$

%

&&=

T a( )T

03x3

03x3 T a( )T

!

"

####

$

%

&&&&

fia

f ja

!

"

##

$

%

&&

F a =T a( )

T03x3

03x3 T a( )T

!

"

####

$

%

&&&&

kiia kij

a

k jia k jj

a

!

"

###

$

%

&&&

uia

u ja

!

"

##

$

%

&&

F a =T a( )

T03x3

03x3 T a( )T

!

"

####

$

%

&&&&

kiia kij

a

k jia k jj

a

!

"

###

$

%

&&&

T a 03x303x3 T a

!

"

##

$

%

&&

Uia

U ja

!

"

##

$

%

&&

F a =T a( )

Tkiia T a( ) T a( )

Tkija T a( )

T a( )Tk jia T a( ) T a( )

Tk jja T a( )

!

"

####

$

%

&&&&

Uia

U ja

!

"

##

$

%

&&=

Kiia Kij

a

K jia K jj

a

!

"

###

$

%

&&&

Uia

U ja

!

"

##

$

%

&&

Step 1: Global forces-local forces relationship

Step 2: Global forces-local displacements relationship

Step 3: Global forces-global displacements relationship

Step 4: Element global stiffness matrix

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22

Element global stiffness matrix

i

jaY iF

aX iF

ajq

aiM

aY jF

aX jFajM

ajV

aiq

aiV

aiU

F a = K aU a =Fia

Fja

!

"

##

$

%

&&=

Kiia Kij

a

K jia K jj

a

!

"

###

$

%

&&&

Uia

U ja

!

"

##

$

%

&&

Element Global

Stiffness Submatrices

l  and represent the element global force and displacement vector, respectively

l  The matrix is a 6x6 symmetric ( as ) matrix called the element global stiffness matrix

aK ( )Ta aΚ K=

aF aU

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23

Nomenclature

i

j

ax

y

,a axi if u

,a axj jf u

XO

i

j

aY

,a aY j jF V

,a aX j jF U,a a

Y i iF V

,a aX i iF U

Global axes

Local axes

Member global internal forces

Member global displacements

Member local displacements

Member local internal forces

l  The Free Body Diagram (FBD) of any frame member can then be expressed both in global and in local axes:

,a ayj jf v

,a aj jM q

,a ayi if v

,a aj jM q

,a ai iM q

,a ai iM q

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24

Element global stiffness matrix for a general frame member

K a =

EALc2 +12EI

L3s2 EA

L!12EIL3

"

#$

%

&'cs !

6EIL2s

EAL!12EIL3

"

#$

%

&'cs

EALs2 +12EI

L3c2 6EI

L2c

!6EIL2s 6EI

L2c 4EI

L

!EALc2 !12EI

L3s2 !

EAL+12EIL3

"

#$

%

&'cs !

6EIL2s

!EAL+12EIL3

"

#$

%

&'cs !

EALs2 !12EI

L3c2 6EI

L2c

6EIL2s !

6EIL2c 2EI

L

!EALc2 !12EI

L3s2 !

EAL+12EIL3

"

#$

%

&'cs

6EIL2s

!EAL+12EIL3

"

#$

%

&'cs !

EALs2 !12EI

L3c2 !

6EIL2c

!6EIL2s 6EI

L2c 2EI

L

EALc2 +12EI

L3s2 EA

L!12EIL3

"

#$

%

&'cs

6EIL2s

EAL!12EIL3

"

#$

%

&'cs

EALs2 +12EI

L3c2 !

6EIL2c

6EIL2s !

6EIL2c 4EI

L

(

)

******************

+

,

------------------

l  Step 1: Define the global axes

l  Step 2: Label nodes, members, element orientation, local axes per element

l  Step 3: Calculate the angle between the local axes and the global axes

l  Step 4: Compute transformation matrix

l  Step 5: Compute element global stiffness matrix

EG-225


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25

Element global stiffness matrix for a beam

l  A particular case of frame elements are continuous beams

l  As local and global axes coincide, the global and local stiffness matrices coincide too, thus:

3 2 3 2

2 2

3 2 3 2

2 2

0 0 0 0

12 6 12 60 0

6 4 6 20 0

0 0 0 0

12 6 12 60 0

6 2 6 40 0

a

EA EAL L

EI EI EI EIL L L LEI EI EI EIL L L L

EA EAL L

EI EI EI EIL L L LEI EI EI EIL L L L

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥= ⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥

− − −⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

K

1

1

2

3

45 2 3 41M 2M 3M

2YF 4YF

EG-225


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26

Nodal equilibrium and compatibility (I)

3XP

2

1

2

31 3M

11YF

11XF

11M

13YF1

3XF

13M

11YF1

1XF11M1XR

1YR

1M

1

1

3YP

RX1 = FX11

RY1 = FY11

Equilibrium

U1 = 0 =U11Compatibility

V1 = 0 = V11!1 = 0 = !11

M1 = M11

l  Each node and member of the structure must be in equilibrium

l  We analyse the FBD of frame members and joints

Node 1 FBD

EG-225


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27

Nodal equilibrium and compatibility (II)

11YF

11XF

11M

13YF1

3XF

13M1

23YF

23XF

23M

22YF2

2XF 22M

223YF

23XF 2

3M

13YF1

3XF

13M 3XP

3M

33YP

3XP

2

1

2

31 3M

3YP

PX3 = FX31 + FX32

PY 3 = FY 31 + FY 32

Equilibrium

U3 =U31 =U32Compatibility

V3 = V31 = V32!3 = !3

1 = !32

M3 = M31 +M3

2

Node 3 FBD

EG-225


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28

Nodal equilibrium and compatibility (III)

23YF2

3XF 23M

22YF2

2XF 22M

22XR2YR

2M

22YF2

2XF 22M

2

RX2 = FX22RY 2 = FY 22

Equilibrium U2 = 0 =U22

Compatibility

V2 = 0 = V22

!2 = 0 = !22M2 = M22

3YP

2

1

2

31

3XP

3M

Node 2 FBD

EG-225


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29

Assembling equilibrium equations (I)

F11

F31

!

"

##

$

%

&&=

K111 K131

K311 K331

!

"

##

$

%

&&

U1U3

!

"##

$

%&&

11YF

11XF

11M

13YF1

3XF

13M

1 23YF2

3XF 23M

22YF2

2XF 22M

2F22

F32

!

"

##

$

%

&&=

K222 K23

2

K322 K332

!

"

##

$

%

&&

U2U3

!

"##

$

%&&

R1 = F11 = K111 U1 +K131 U3

R2 = F22 = K222 U2 +K23

2 U3

P3 = F31 + F32 = K311 U1 +K331 U3 +K322 U2 +K332 U3

l  Recall the element global stiffness matrix relationships:

l  Combine above formulae with the nodal equilibrium and compatibility relationships derived previously:

EG-225


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30

Assembling equilibrium equations (II)

R1R2P3

!

"

###

$

%

&&&=

K111 0 K131

0 K222 K23

2

K311 K322 K331 +K332

!

"

####

$

%

&&&&

U1U2U3

!

"

###

$

%

&&&

ndgof ndgof x ndgof ndgof= F K U

where ndgof: No. of degrees of freedom 2

1

2

31

3XP

3M 3YP

R1 = F11 = K111 U1 +K131 U3R2 = F22 = K22

2 U2 +K232 U3

P3 = F31 + F32 = K311 U1 +K331 U3 +K322 U2 +K332 U3

l  This is the system of nodal global equilibrium equations expressed in terms of the nodal global displacements and global stiffness matrix coefficients:

l  The above expressions can be re-written in matrix format as follows:

EG-225


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31

l  Relationship between member global internal forces and member global displacements (element equilibrium and constitutive behaviour)

l  Relationship between member global displacements and nodal global displacements (compatibility equations)

l  Relationship between member global internal forces and external global forces or reactions (equilibrium equations)

Assembled global stiffness matrix (I)


Assembled Global Displacement Vector

Assembled Global Force

Vector

Assembled Global Stiffness Matrix

l  The matrix is a ndgofxndgof symmetric ( as ) matrix called the assembled global stiffness matrix

K TK=K

EG-225


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32

l  A close examination of the assembled equilibrium equations reveals that the assembled global stiffness matrix can be obtained directly

l  The member global stiffness submatrices must be inserted into the assembled global stiffness matrix in the appropriate joint block row and block column according to the indices and

l  This direct procedure avoids having to consider joint equilibrium and compatibility

Assembled global stiffness matrix (II)

K =

K111 03!3 K131

03!3 K222 K23

2

K311 K322 K331 +K332

"

#

$$$$

%

&

''''

aijK

i j

2

1

2

31

3XP

3M 3YP

EG-225


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33

Direct assembly of the Global Stiffness Matrix (GSM)

3 3 3 3 3 3

3 3 3 3 3 3

3 3 3 3 3 3

¥ ¥ ¥

¥ ¥ ¥

¥ ¥ ¥

È ˘Í ˙Í ˙Í ˙Í ˙Í ˙Í ˙Î ˚

0 0 00 0 00 0 0

=K

K=K111 03!3 K131

03!3 03!3 03!3K311 03!3 K331

"

#

$$$$

%

&

''''

K=K111 03!3 K131

03!3 K222 K23

2

K311 K322 K331 +K332

"

#

$$$$

%

&

''''

GSM after member 1

GSM after members 1 & 2

Initial GSM

2

1

2

31

EG-225


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Assembled GSM coefficients (I)

28/02/2012


34

RX1RY1M1RX2RY 2M2PX3PY3M3

!

"

##############

$

%

&&&&&&&&&&&&&&

=

K111( )XX K111( )XY K111( )X! 0 0 0 K131( )XX K131( )XY K131( )X!K111( )YX K111( )YY K111( )Y! 0 0 0 K131( )YX K131( )YY K131( )Y!K111( )!X K111( )!Y K111( )!! 0 0 0 K131( )!X K131( )!Y K131( )!!0 0 0 K222( )XX K222( )XY K222( )XX K232( )XX K232( )XY K232( )X!0 0 0 K222( )YX K222( )YY K222( )Y! K232( )YX K232( )YY K232( )Y!0 0 0 K222( )!X K222( )!Y K222( )!! K232( )!X K232( )!Y K232( )!!

K311( )XX K311( )XY K311( )X! K322( )XX K322( )XY K322( )X! K331+2( )XX K331+2( )XY K331+2( )X!K311( )YX K311( )YY K311( )Y! K322( )YX K322( )YY K322( )Y! K331+2( )YX K331+2( )YY K331+2( )Y!K311( )!X K311( )!Y K311( )!! K322( )!X K322( )!Y K322( )!! K331+2( )!X K331+2( )!Y K331+2( )!!

!

"

#################

$

%

&&&&&&&&&&&&&&&&&

000000001

!

"

###########

$

%

&&&&&&&&&&&

1

2

3

1

2

11

( )2133 XK

q+

( )2133 YK

q+ ( )21

33K qq+

l  Apply unit rotation at joint 3 and fix the other dgofs. Thus, the assembled global forces are the 9th column of : K

l  The assembled global equilibrium equations can be expanded:

EG-225


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Assembled GSM coefficients (II)

28/02/2012


35


!

"

##############

$

%

&&&&&&&&&&&&&&

=



!

"

#################

$

%

&&&&&&&&&&&&&&&&&

000000100

!

"

###########

$

%

&&&&&&&&&&&

l  Apply unit OX displacement at joint 3 and fix the other dgofs. Thus, the assembled global forces are the 7th column of : K

l  The assembled global equilibrium equations can be expanded:

1

2

3

1

2

1

( )133 XX2K +

( )133 YX2K + ( )21

33 XK

q+

EG-225


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Symmetry of the GSM: Reciprocity theorem

28/02/2012


36

l  Consider two possible deformed configurations:


!

"

##############

$

%

&&&&&&&&&&&&&&

=



!

"

#################

$

%

&&&&&&&&&&&&&&&&&

U1V1!1U2V2!2U3V3!3

!

"

##############

$

%

&&&&&&&&&&&&&&

1

2

3

1

2

1

( )133 XX2K +

( )133 YX2K + ( )21

33 XK

q+

1

2

3

1

2

11

( )2133 XK

q+

( )2133 YK

q+ ( )21

33K qq+

l  The force at node 3 along OX due to a rotation at node 3 is equal to the moment at node 3 due to a displacement at node 3 along OX [Maxwell’s reciprocity theorem (1864)]

EG-225


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37

Introduction of loads and support conditions

F=

FX1FY1M1

FX2FY 2M2

FX3FY3M3

!

"

##############

$

%

&&&&&&&&&&&&&&

=

RX1RY1M1

Rx2RY 2M2

PX3PY3M3

!

"

############

$

%

&&&&&&&&&&&&

U =

U1V1!1U2V2!2U3V3!3

!

"

############

$

%

&&&&&&&&&&&&

=

000000U3V3!3

!

"

###########

$

%

&&&&&&&&&&&

2

1

2

31

3XP

3M 3YP

Known values

Unknown values

Notice that known loads and known displacement cannot coincide for the same node and same degree of freedom

l  Known support conditions can be introduced into the assembled global displacement vector

l  Suppose initially zero displacement wherever there is a support (no settlement allowed)

l  Loads can be introduced into the assembled global force vector

U

F

EG-225


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28/02/2012


38

Internal forces calculation

U a =Uia

U ja

!

"

##

$

%

&&= Ui

a Via !i

a U ja Vj

a ! ja!

"#$%&

T

ua =uia

u ja

!

"

##

$

%

&&= ui

a via !i

a u ja v j

a ! ja!

"#$%&

T

=T a 03x303x3 T a

!

"

##

$

%

&&

Uia

U ja

!

"

##

$

%

&&

Step 1: Extract element global displacements from assembled global displacement vector

Step 2: Compute element local displacements

Step 3: Compute element internal forces

U

i

jaY iF

aX iF

ajq

aiM

aY jF

aX jFajM

ajU

ajV

aiq

aiV

aiU

f a =

f xia

f yia

Mia

f xja

f yja

M ja

!

"

##########

$

%

&&&&&&&&&&

=

'Nia

Sia

'BMia

N ja

'S ja

BM ja

!

"

##########

$

%

&&&&&&&&&&

=kiia kij

a

k jia k jj

a

!

"

###

$

%

&&&

T a 03x303x3 T a

!

"

##

$

%

&&

Uia

U ja

!

"

##

$

%

&&=

kiiaT aUi

a + kijaT aU j

a

k jiaT aUi

a + k jjaT aU j

a

!

"

###

$

%

&&&

EG-225


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28/02/2012


39

Step 1: Establish global axes

Step 2: Label nodes and elements

Step 3: Establish element local axes

Step 4: Calculate element geometrical and mechanical properties

Step 5: Compute element GSM

Step 6: Formulate the assembled GSM

Step 7: Assemble global force and displacement vectors

Step 8: Substitute known nodal loads and known support conditions

Step 9: Solve for unknown displacements/rotations

Step 10: Compute unknown reactions

Step 11: Verify overall translational and rotational equilibrium

Step 12: Extract element global displacements

Step 13: Compute element local displacements

Step 14: Compute element local forces

Step 15: Deduce element axial force, shear force and bending moment

The “road to non-perdition” in the stiffness method

EG-225


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28/02/2012


40

Prescribed support displacements l  Generally, the equilibrium equations can be partitioned by

rearranging the rows and columns:

FfR s

!

"

##

$

%

&&=

K ff K fs

Ksf Kss

!

"##

$

%&&

U f

Us

!

"##

$

%&&

s: prescribed support displacements

f: displacements free to move

Ff = K ffU f +K fsUs ! U f = K ff"1(Ff "K fsUs )

R s = KsfU f +KssUs !

R s = KsfK ff"1(Ff "K fsUs )+KssUs

l  The unknown displacements can be obtained as:

l  The unknown reactions are calculated subsequently:

l  In general, non-zero prescribed displacements can be imposed in some support as a result of possible settlement:

1

2

3

1'

EG-225


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41

l  For beam elements, the rotations and vertical displacements at the joints are the only degrees of freedom

l  The local and global axes coincide so no transformation is required

l  The global and local stiffness matrices are identical

l  The axial components of the stiffness matrix are not required

Analysis of beams: introduction

1

1

2

3

45 2 3 41M 1M 1M

2YF 4YF

1

1 23

45 2 3 4

2 Degrees Of Freedom (DOF) per node

EG-225


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28/02/2012


42

Analysis of beams: the simplified element GSM

K a =Kiia Kij

a

K jia K jj

a

!

"

###

$

%

&&&=

12EIL3

6EIL2

'12EIL3

6EIL2

6EIL2

4EIL

'6EIL2

2EIL

'12EIL3

'6EIL2

12EIL3

'6EIL2

6EIL2

2EIL

'6EIL2

4EIL

!

"

##########

$

%

&&&&&&&&&&

K a =

EAL

0 0 !EAL

0 0

0 12EIL3

6EIL2

0 !12EIL3

6EIL2

0 6EIL2

4EIL

0 !6EIL2

2EIL

!EAL

0 0 EAL

0 0

0 !12EIL3

!6EIL2

0 12EIL3

!6EIL2

0 6EIL2

2EIL

0 !6EIL2

4EIL

"

#

$$$$$$$$$$$$$$$$

%

&

''''''''''''''''

Rotational and vertical components

1

1 23

45 2 3 4

EG-225


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43

Analysis of beams: the assembled GSM

FY1M1

FY 2M 2

FY 3M 3

!

"

#########

$

%

&&&&&&&&&

=

K111+2( )YY K11

1+2( )Y! K122( )YY K12

2( )Y! K131( )YY K13

1( )Y!K111+2( )

!YK111+2( )

!!K122( )

!YK122( )

!!K131( )

!YK131( )

!!

K212( )YY K21

2( )Y! K222( )YY K22

2( )Y! 0 0

K212( )

!YK212( )

!!K222( )

!YK222( )

!!0 0

K311( )YY K31

1( )Y! 0 0 K331( )YY K33

1( )Y!K311( )

!YK311( )

!!0 0 K33

1( )!Y

K331( )

!!

!

"

############

$

%

&&&&&&&&&&&&

0!1V2!200

!

"

########

$

%

&&&&&&&&


l  Select rows and columns of the assembled global stiffness matrix

l  Solve for the unknown rotations

l  Substitute computed rotations to obtain unknown moments

1

1

23 21M 1M

2YF

EG-225


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M1

FY 2M 2

!

"

####

$

%

&&&&

=

K111+2( )

!!K122( )

!YK122( )

!!

K212( )

!YK222( )YY K22

2( )Y!K212( )

!!K222( )

!YK222( )

!!

!

"

#####

$

%

&&&&&

!1V2!2

!

"

####

$

%

&&&&

28/02/2012


44

Analysis of beams: solving the equilibrium equations

FY1M1

FY 2M 2

FY 3M 3

!

"

#########

$

%

&&&&&&&&&

=

K111+2( )YY K11

1+2( )Y! K122( )YY K12

2( )Y! K131( )YY K13

1( )Y!K111+2( )

!YK111+2( )

!!K122( )

!YK122( )

!!K131( )

!YK131( )

!!

K212( )YY K21

2( )Y! K222( )YY K22

2( )Y! 0 0

K212( )

!YK212( )

!!K222( )

!YK222( )

!!0 0

K311( )YY K31

1( )Y! 0 0 K331( )YY K33

1( )Y!K311( )

!YK311( )

!!0 0 K33

1( )!Y

K331( )

!!

!

"

############

$

%

&&&&&&&&&&&&

0!1V2!200

!

"

########

$

%

&&&&&&&&

( ) ( ) ( )( )( )

111

1 11

3 3

2 2 212 12

1 2

13 231

0 0

0 0

YYYY

Y Y

YKF

F K V

K

K

M

Kθθ

θ

θθ

θ

θ

+⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

⎢ ⎥⎣ ⎦

EG-225


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45

Continuous beams (I)

l  Most structural beams will be so-called continuous beams.

l  They have multiple supports and the rotations at the joints are the only degrees of freedom (the vertical displacements at the joints are all constrained). Then, a reduced stiffness matrix can be employed.

l  Only the rotational components of the stiffness matrix are required

1 2 5

15 62

3 4

43

Rotational degrees of freedom

1 Degree Of Freedom (DOF) per node

EG-225


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46

Continuous beams (II)

K a =Kiia Kij

a

K jia K jj

a

!

"

###

$

%

&&&=

4EIL

2EIL

2EIL

4EIL

!

"

####

$

%

&&&&

K a =

EAL

0 0 !EAL

0 0

0 12EIL3

6EIL2

0 !12EIL3

6EIL2

0 6EIL2

4EIL

0 !6EIL2

2EIL

!EAL

0 0 EAL

0 0

0 !12EIL3

!6EIL2

0 12EIL3

!6EIL2

0 6EIL2

2EIL

0 !6EIL2

4EIL

"

#

$$$$$$$$$$$$$$$$

%

&

''''''''''''''''

Rotational components

1 2 5

15 62

3 4

43

1M 2M 3M 4M

EG-225


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28/02/2012


47

Continuous beams (III)

M1

M 2

M 3

M 4

M5

M 6

!

"

#########

$

%

&&&&&&&&&

=

K111 + K11

2 K122 0 0 K15

1 0

K212 K22

2 + K223 K23

3 0 0 0

0 K323 K33

3 + K334 K34

4 0 0

0 0 K434 K44

4 + K445 0 K46

5

K511 0 0 0 K55

1 0

0 0 0 K645 0 K66

5

!

"

#########

$

%

&&&&&&&&&

!1!2!3!400

!

"

########

$

%

&&&&&&&&

Fndgof =Kndgof x ndgof Undgof

1 2 5

15 62

3 4

43

1M 2M 3M 4M

l  Select rows and columns of the assembled global stiffness matrix

l  Solve for the unknown rotations

l  Substitute computed rotations to obtain unknown moments

EG-225


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28/02/2012


48

Continuous beams (IV)

3 322 23

3 332 3

4 433 43

4

2 211 121

2

34

43

2 212

4

1111

2

544

22

4

3

4

3

4

0 00

00 0

K K

K

K K

KK K

K KK K

M K

MK

M

M

θ

θ

θ

θ

⎡ ⎤+⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥+⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥+⎢ ⎥⎢ ⎥ ⎢ ⎥

+⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

1 2 5

15 62

3 4

43

1M 2M 3M 4M

564

15 151

6 4

00

M KM K

θ

θ

⎡ ⎤⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦ ⎣ ⎦

2 211

1 1111 15

2

3

4

3 322 23

3 332 33

1 15 5

1

2

3

4

1

5 544 46

5 5

122 221 2

4 433 34

5

4 443 4

6

5

6

2

4

64 6

0 0 00 0 0

0 0 0

0 0 00 0 0 0

0 00

0 00

K K

K K

K KK KK

MMM

M K

K K

KM

M

K

K

K KK K

K

θ

θ

θ

θ

⎡ ⎤+⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥+⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥+

= ⎢ ⎥⎢ ⎥ ⎢ ⎥+⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

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l  So far we have considered that beam elements are loaded at their joints.

l  In general, apart from nodal loads and prescribed displacements at supports, a frame structure can undergo other effects

l  In a general case, point loads or distributed loads can be acting along the span

l  How can we account for these new effects?

l  Consider a general beam structure subjected to multiple loads such as the ones shown below

Intermediate loading (I)

w

P

1 21

2 3

3M

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50

Intermediate loading (II)

P2P

8PL

2P

8PL2 32

w2wL

2

12wL

2wL

2

12wL 1 21

l  Establish a STAGE I, where every frame member is analysed independently with its ends fixed. Consider only the non-nodal loads. Compute the necessary fixed end forces

STAGE I 2

24wL

2

12wL

2wL

2wL

8PL

8PL

2P

2P

STAGE I

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51

Intermediate loading (III)

l  Establish a STAGE II, where the overall frame structure is subjected to forces at the nodes opposite to the previously calculated fixed end forces (plus any other already applied nodal loads)

l  Notice that forces at supports can be removed

l  The solution of the problem is the addition of the results in STAGE I and STAGE II

This force is absorbed by the support: it does not affect the

frame structure

2wL

2

12wL

2wL 2

12wL

2P

8PL

2P

8PL

1 22 31

3M

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28/02/2012


52

Intermediate loading (IV) 2

12 8wL PL-‐ 38

PL M+

1 22 31

M1

wL2

12!PL8

PL8+M3

"

#

$$$$

%

&

''''

=

K111 K12

1 0

K211 K22

1 + K222 K23

2

0 K322 K33

2

"

#

$$$$

%

&

''''

0!2!3

"

#

$$$$

%

&

''''

M1 = K121!2

3

1 2 222 23

2 232 33

222 8 22

8 3

1wL PL

PL M

K K KK K

θ

θ

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

−

+

⎡ ⎤+ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦

l  Recall that the shear force diagram and bending moment diagram corresponding to STAGE I must be added to obtain the final solution

STAGE II

Documents

Stiffness method for 2D frames