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Stiffness method for 2D frames
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EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
Examples of frames in Civil Engineering
28/02/2012
Stiffness Method (I: frames)
2
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
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Dr. Antonio Martinez
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
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
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Dr. Antonio Martinez
l Recall that distributed loads are not considered yet, thus:
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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= + + + Ρ
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:
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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:
BM (x) = EaI a d2y(x)dx2
= EaI a 12
La( )3x ! 6
La( )2"
#
$$
%
&
''
S(x) = EaI a d3y(x)dx3
= EaI a 12
La( )3!
"
##
$
%
&&
1 ( )26( )
a aa
aE IBM LL
=
( )312( )
a aa
aE IS LL
=
a
( )26(0)
a a
aE IBML
= -‐
( )312(0)
a a
aE ISL
=
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
12
Shear and bending deformation: constitutive law (III)
l For the deformation case 2, the BCs are as follows:
(0) 0y =
0
( ) 1x
dy xdx =
=
( ) 0ay L =
( ) 0ax L
dy xdx =
=
3 2( )y x ax bx cx d= + + +l Substituting the BCs into the cubic polynomial
it yields: ( )
3 22
1 2( ) aay x x x xLL
= -‐ +
l The internal shear force and bending moment result in:
BM (x) = EaI a d2y(x)dx2
= EaI a 6
La( )2x ! 4
La
"
#
$$
%
&
''
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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:
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
15
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
17
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
18
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
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Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
19
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=
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
20
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=
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
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Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
21
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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)
ndgof ndgof x ndgof ndgof= F K U
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
"
#
$$$$
%
&
''''
aijK
i j
2
1
2
31
3XP
3M 3YP
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
33
Direct assembly of the Global Stiffness Matrix (GSM)
3 3 3 3 3 3
3 3 3 3 3 3
3 3 3 3 3 3
¥ ¥ ¥
¥ ¥ ¥
¥ ¥ ¥
È ˘Í ˙Í ˙Í ˙Í ˙Í ˙Í ˙Î ˚
0 0 00 0 00 0 0
=K
K=K111 03!3 K131
03!3 03!3 03!3K311 03!3 K331
"
#
$$$$
%
&
''''
K=K111 03!3 K131
03!3 K222 K23
2
K311 K322 K331 +K332
"
#
$$$$
%
&
''''
GSM after member 1
GSM after members 1 & 2
Initial GSM
2
1
2
31
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
Assembled GSM coefficients (I)
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
Assembled GSM coefficients (II)
28/02/2012
Stiffness Method (I: frames)
35
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( )!!
!
"
#################
$
%
&&&&&&&&&&&&&&&&&
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
Symmetry of the GSM: Reciprocity theorem
28/02/2012
Stiffness Method (I: frames)
36
l Consider two possible deformed configurations:
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( )!!
!
"
#################
$
%
&&&&&&&&&&&&&&&&&
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: trusses)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: trusses)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: trusses)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
!
"
########
$
%
&&&&&&&&
ndgof ndgof x ndgof ndgof= F K U
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
M1
FY 2M 2
!
"
####
$
%
&&&&
=
K111+2( )
!!K122( )
!YK122( )
!!
K212( )
!YK222( )YY K22
2( )Y!K212( )
!!K222( )
!YK222( )
!!
!
"
#####
$
%
&&&&&
!1V2!2
!
"
####
$
%
&&&&
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
⎡ ⎤+⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥+⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥+
= ⎢ ⎥⎢ ⎥ ⎢ ⎥+⎢ ⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
49
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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
EG-225
Structural Mechanics II(b)
SWANSEA UNIVERSITY!
!School of
Engineering
Dr. Antonio Martinez
28/02/2012
Stiffness Method (I: frames)
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