1. Structural Analysis - III
Di t Stiff M th dDirect Stiffness Method
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKNDept. of CE, GCE Kannur Dr.RajeshKN
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2. Module IIIModule III
Direct stiffness method
• Introduction – element stiffness matrix – rotation transformation
Direct stiffness method
matrix – transformation of displacement and load vectors and
stiffness matrix – equivalent nodal forces and load vectors –
assembly of stiffness matrix and load vector – determination ofassembly of stiffness matrix and load vector determination of
nodal displacement and element forces – analysis of plane truss
beam and plane frame (with numerical examples) – analysis of
grid space frame (without numerical examples)grid – space frame (without numerical examples)
Dept. of CE, GCE Kannur Dr.RajeshKN
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3. Introduction
• The formalised stiffness method involves evaluating the
displacement transformation matrix CMJ correctlyp y
• Generation of matrix CMJ is not suitable for computer
programming
H th l ti f di t tiff th d• Hence the evolution of direct stiffness method
Dept. of CE, GCE Kannur Dr.RajeshKN
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4. Direct stiffness method
• We need to simplify the assembling process of SJ , theJ
assembled structure stiffness matrix
• The key to this is to use member stiffness matrices for actionsThe key to this is to use member stiffness matrices for actions
and displacements at BOTH ends of each member
If b di l d i h f• If member displacements are expressed with reference to
global co-ordinates, the process of assembling SJ can be made
simplesimple
Dept. of CE, GCE Kannur Dr.RajeshKN
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5. Member oriented axes (local coordinates)
d t t i t d ( l b l di t )and structure oriented axes (global coordinates)
Lδ
x
y
L
Local axes
LδL
Lδ
sinLδ θ
Lδ θ
x
YYYY
cosLδ θ
θ
Global axes yGlobal axesGlobal axes
θ
XXXX
Dept. of CE, GCE Kannur Dr.RajeshKN
Global axes
6. 1. Plane truss member
Stiffness coefficients in local coordinates
1
3
2 4
y
0 0
Degrees of freedom
1
⎛ ⎞
⎜ ⎟
Unit displacement
xEA
L
EA
L
⎜ ⎟
⎝ ⎠corr. to DOF 1
0 0
EA EA⎡ ⎤
−⎢ ⎥
[ ]
0 0
0 0 0 0
0 0
M
L L
S
EA EA
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
−⎢ ⎥
=
Member stiffness matrix in local
coordinates
Dept. of CE, GCE Kannur Dr.RajeshKN
0 0 0 0
L L
⎢ ⎥
⎢
⎢⎣ ⎦
⎥
⎥
7. Transformation of displacement vector
θ
Displacements in global
22 2 cosU D θ=
p g
coordinates: U1 and U2
Displacements in local
di D d D2D
DY iU D θ
12 2 sinU D θ= − θ
1 11 12 1 2cos sinU U U D Dθ θ= + = −
coordinates: D1 and D2
1DY
11 1 cosU D θ=
21 1 sinU D θ=
2 21 22 1 2sin cosU U U D Dθ θ= + = +
1 1cos sinU Dθ θ−⎧ ⎫ ⎧ ⎫⎡ ⎤
⎧ ⎫ ⎧ ⎫⎡ ⎤
1 1
2 2
cos sin
sin cos
U D
U D
θ θ
θ θ
⎧ ⎫ ⎧ ⎫⎡ ⎤
=⎨ ⎬ ⎨ ⎬⎢ ⎥
⎣ ⎦⎩ ⎭ ⎩ ⎭
X 1 1
2 2
cos sin
sin cos
D U
D U
θ θ
θ θ
⎧ ⎫ ⎧ ⎫⎡ ⎤
∴ =⎨ ⎬ ⎨ ⎬⎢ ⎥−⎣ ⎦⎩ ⎭ ⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
{ } [ ]{ }D R U=
8. C id i b th d
⎧ ⎫ ⎧ ⎫
Considering both ends,
1 1
2 2
cos sin 0 0
sin cos 0 0
0 0 i
D U
D U
D U
θ θ
θ θ
θ θ
⎧ ⎫ ⎧ ⎫⎡ ⎤
⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬
⎢ ⎥3 3
4 4
0 0 cos sin
0 0 sin cos
D U
D U
θ θ
θ θ
⎨ ⎬ ⎨ ⎬
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢ ⎥⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎣ ⎦ ⎩ ⎭
{ } [ ]{ }LOCAL T GLOBALD R D=
[ ] [ ]R O⎡ ⎤
[ ]
[ ] [ ]
[ ] [ ]T
R O
R
O R
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
Rotation matrix
Dept. of CE, GCE Kannur Dr.RajeshKN
9. Transformation of load vector Actions in global
θ
coordinates:
F1 and F2
22 2 cosF A θ= 1 11 12 1 2cos sinF F F A Aθ θ= + = −
22 2
12 2 sinF A θ= − θ 2A 2 21 22 1 2sin cosF F F A Aθ θ= + = +
Y
11 1 cosF A θ=
21 1 sinF A θ=1A 1 1
2 2
cos sin
sin cos
F A
F A
θ θ
θ θ
−⎧ ⎫ ⎧ ⎫⎡ ⎤
=⎨ ⎬ ⎨ ⎬⎢ ⎥
⎣ ⎦⎩ ⎭ ⎩ ⎭
11 1
1 1cos sinA Fθ θ⎧ ⎫ ⎧ ⎫⎡ ⎤
=⎨ ⎬ ⎨ ⎬⎢ ⎥
X
2 2sin cosA Fθ θ
⎨ ⎬ ⎨ ⎬⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭
{ } [ ]{ }A R F=
Dept. of CE, GCE Kannur Dr.RajeshKN
{ } [ ]{ }A R F=
10. C id i b th dConsidering both ends,
1 1
2 2
cos sin 0 0
sin cos 0 0
A F
A F
θ θ
θ θ
⎧ ⎫ ⎧ ⎫⎡ ⎤
⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬
3 3
4 4
0 0 cos sin
0 0 sin cos
A F
A F
θ θ
θ θ
⎢ ⎥=⎨ ⎬ ⎨ ⎬
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢ ⎥⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎣ ⎦ ⎩ ⎭
{ } [ ]{ }LOCAL T GLOBALi.e., A R A=
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11. Transformation of stiffness matrix
{ } [ ]{ }LOCAL M LOCALA S D=
[ ]{ } [ ][ ]{ }T GLOBAL M T GLOBALR A S R D=
{ } [ ] [ ][ ]{ }
1
GLOBAL T M T GLOBALA R S R D
−
=
[ ]{ } [ ][ ]{ }T GLOBAL M T GLOBAL
[ ] [ ]1 T
R R
−{ } [ ] [ ][ ]{ }GLOBAL T M T GLOBAL
{ } [ ]{ }A S D
[ ] [ ]T TR R=
{ } [ ]{ }GLOBAL MS GLOBALA S D=
[ ] [ ] [ ][ ]
T
MS T M TS R S R=where,
Member stiffness matrix in
global coordinates
Dept. of CE, GCE Kannur Dr.RajeshKN
12. [ ] [ ] [ ][ ]
T
S R S R=
T
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
[ ] [ ] [ ][ ]MS T M TS R S R=
0 0 1 0 1 0 0 0
0 0 0 0 0 0 0 0
T
c s c s
s c s cEA
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥=
⎢ ⎥ ⎢ ⎥ ⎢ ⎥0 0 1 0 1 0 0 0
0 0 0 0 0 0 0 0
c s c sL
s c s c
−⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
2 2
c cs c cs⎡ ⎤− −
⎢ ⎥ M b tiff t i2 2
2 2
cs s cs sEA
L c cs c cs
⎢ ⎥
− −⎢ ⎥=
⎢ ⎥− −
⎢ ⎥
Member stiffness matrix
in global coordinates
for a plane truss member
2 2
cs s cs s
⎢ ⎥
− −⎣ ⎦
p
Dept. of CE, GCE Kannur Dr.RajeshKN
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13. 2. Plane frame member
Stiffness coefficients in local coordinates
2 52
4
5
Degrees of freedom0 0 0 0
EA EA⎡ ⎤
1
3
4
6
Degrees of freedom
3 2 3 2
0 0 0 0
12 6 12 6
0 0
L L
EI EI EI EI
L L L L
⎡ ⎤
−⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎢ ⎥
[ ]
2 2
6 4 6 2
0 0
0 0 0 0
Mi
EI EI EI EI
L L L L
S
EA EA
⎢ ⎥
⎢ ⎥−
⎢ ⎥=
⎢ ⎥
−⎢ ⎥
Member stiffness matrix
in local coordinates
3 2 3 2
0 0 0 0
12 6 12 6
0 0
L L
EI EI EI EI
L L L L
−⎢ ⎥
⎢ ⎥
⎢ ⎥
− − −⎢ ⎥
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
2 2
6 2 6 4
0 0
EI EI EI EI
L L L L
⎢ ⎥
⎢ ⎥−
⎢ ⎥⎣ ⎦
14. Transformation of displacement vector
1 1
2 2
cos sin 0 0 0 0
sin cos 0 0 0 0
D U
D U
θ θ
θ θ
⎧ ⎫ ⎧ ⎫⎡ ⎤
⎪ ⎪ ⎪ ⎪⎢ ⎥−
⎪ ⎪ ⎪ ⎪⎢ ⎥
3 3
4 4
0 0 1 0 0 0
0 0 0 cos sin 0
D U
D Uθ θ
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
=⎨ ⎬ ⎨ ⎬⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥5 5
6 6
0 0 0 sin cos 0
0 0 0 0 0 1
D U
D U
θ θ⎪ ⎪ ⎪ ⎪⎢ ⎥−
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎣ ⎦⎩ ⎭ ⎩ ⎭
{ } [ ]{ }LOCAL T GLOBALD R D=
[ ] [ ]R O⎡ ⎤
{ } [ ]{ }LOCAL T GLOBAL
[ ]
[ ] [ ]
[ ] [ ]T
R O
R
O R
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
Rotation matrix
Dept. of CE, GCE Kannur Dr.RajeshKN
15. Transformation of load vector
{ } [ ]{ }LOCAL T GLOBALA R A={ } [ ]{ }
cos sin 0 0 0 0θ θ⎡ ⎤
⎢ ⎥
[ ]
sin cos 0 0 0 0
0 0 1 0 0 0
R
θ θ⎢ ⎥−
⎢ ⎥
⎢ ⎥
= ⎢ ⎥[ ]
0 0 0 cos sin 0
0 0 0 sin cos 0
TR
θ θ
θ θ
= ⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥
0 0 0 0 0 1
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
16. Transformation of stiffness matrix
{ } [ ]{ }GLOBAL MS GLOBALA S D=
[ ] [ ] [ ][ ]T
MS T M TS R S R= Member stiffness matrix in
global coordinatesglobal coordinates
EA EA⎡ ⎤
0 0 0 0
0 0 0 0
c s
s c
⎡ ⎤
⎢ ⎥−
⎢ ⎥
3 2 3 2
0 0 0 0
12 6 12 6
0 0
EA EA
L L
EI EI EI EI
L L L L
⎡ ⎤
−⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎢ ⎥
[ ]
0 0 1 0 0 0
0 0 0 0
0 0 0 0
T
c s
s c
R
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥
[ ]
2 2
6 4 6 2
0 0
0 0 0 0
M
EI EI EI EI
L L L LS
EA EA
L L
⎢ ⎥
⎢ ⎥−
⎢ ⎥=
⎢ ⎥
−⎢ ⎥
Where,
,
0 0 0 0 0 1
⎢ ⎥
⎣ ⎦
3 2 3 2
12 6 12 6
0 0
6 2 6 4
0 0
L L
EI EI EI EI
L L L L
EI EI EI EI
⎢ ⎥
⎢ ⎥
⎢ ⎥− − −⎢ ⎥
⎢ ⎥
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
2 2
0 0
L L L L
⎢ ⎥−
⎢ ⎥⎣ ⎦
17. Assembling global stiffness matrixg g
Plane truss
2
Force
21
3
31
3
2 4
1
3
Action/displacement components in
local coordinates of members
Dept. of CE, GCE Kannur Dr.RajeshKN
19. 1 2 3 4
Global DOF
11 12 13 14
1 1 1 1
21 22 23 24
2 2
2 2
1
2
M M M M
s s s s
s s s s
c cs c cs
cs s cs sEA
− −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥
1 2 3 4
[ ] 1 1 1 1
31 32 33 34
1 1 1 1
41 42 43 44
1 2 2
2 2
2
3
4
M M M M
M M M M
M
s s s s
s s s s
cs s cs sEA
S
c cs c csL
− −
⎢ ⎥ ⎢ ⎥= =
− −⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦⎣ ⎦
41 42 43 44
1 1 1 1
2 2
4M M M M
s s s scs s cs s
⎢ ⎥ ⎢ ⎥− − ⎣ ⎦⎣ ⎦
1 2 3 4 5 6
1
2
× × × ×⎡ ⎤
⎢ ⎥× × × ×
⎢ ⎥ C t ib ti f
[ ]
2
3
4
JS
× × × ×
⎢ ⎥
× × × ×⎢ ⎥
= ⎢ ⎥
Contribution of
Member 1 to global
stiffness matrix[ ]
4
5
J ⎢ ⎥
× × × ×⎢ ⎥
⎢ ⎥
⎢ ⎥
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⎢ ⎥
⎣ ⎦
20. 3 4 5 6
Global DOF
11 12 13 14
2 2 2 2
21 22 23 24
3
4
M M M M
s s s s
s s s s
⎡ ⎤
⎢ ⎥
⎢ ⎥
3 4 5 6
[ ] 2 2
2
2 2
31 32 33 34
2 2 2 2
41 42 43 44
4
5
6
M M M M
M M M
M
M
s s s s
s s s s
s s s s
S = ⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦2 2 2 2
6M M M M
s s s s⎣ ⎦
1 2 3 4 5 6
1
2
⎡ ⎤
⎢ ⎥
⎢ ⎥ C t ib ti f
[ ]
2
3
4
JS
⎢ ⎥
× × × ×⎢ ⎥
= ⎢ ⎥
× × × ×⎢ ⎥
Contribution of
Member 2 to global
stiffness matrix
4
5
6
× × × ×⎢ ⎥
⎢ ⎥× × × ×
⎢ ⎥
⎣ ⎦
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⎢ ⎥
× × × ×⎣ ⎦
21. 11 12 13 14
1⎡ ⎤
1 2 5 6
Global DOF
[ ]
11 12 13 14
3 3 3 3
21 22 23 24
3 3 3 3
1
2
M M M M
M M M M
s s s s
s s s s
S
⎡ ⎤
⎢ ⎥
⎢ ⎥[ ] 3 3
3
3 3
31 32 33 34
3 3 3 3
41 42 43 44
5
6
M M M M
M M M
M
M
s s s s
S = ⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
41 42 43 44
3 3 3 3
6M M M M
s s s s
⎢ ⎥
⎣ ⎦
1 2 3 4 5 6
1
2
× × × ×⎡ ⎤
⎢ ⎥
1 2 3 4 5 6
[ ]
2
3
JS
⎢ ⎥× × × ×
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
Contribution of
Member 3 to global
stiffness matrix[ ]
4
5
JS ⎢ ⎥
⎢ ⎥
⎢ ⎥× × × ×
⎢ ⎥
stiffness matrix
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⎢ ⎥
× × × ×⎣ ⎦
22. Assembled global stiffness matrix
11 12 13 1411 12 13 14
s s s ss s s s+ + 1⎡ ⎤
1 2 3 4 5 6
1 1 1 1
21 22 23 24
1 1 1 1
3 3 3 3
21 22 23 24
3 3 3 3
M M M M
M M M M
M M M M
M M M M
s s s s
s s s s
s s s s
s s s s
+ +
+ +
1
2
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
[ ]
31 32 3 11 12 13 14
2 2 2 2
21 22 23 24
3 34
1 1 1 1
41 42 43 44
M M M MM M M M
J
s s s s
s s s s
s s s s
s
S
ss s
+ +
=
+ +
3
4
⎢ ⎥
⎢ ⎥
⎢ ⎥
31 3
2 2 2 2
31 32 33
1 1
2
2
1 1
23 3 32
M M M M
M MM M
M M M M
MM
s s
s s s s
s s
s s
s
s s
s
+ +
+
34
2
33 34
3
4
5MM
ss +
⎢ ⎥
⎢ ⎥
⎢ ⎥41 42 43 44
2
41 42 43 44
3 2 23 32 3
6M M M MM M M M
s s s ss s s s+ +
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
23. Imposing boundary conditions 2
p g y
Plane truss example
21
1
2
0U U U U= = = =Boundary conditions are:
3
3
11 2 5 6
0U U U UBoundary conditions are:
{ }GLOBALD
11 12 13 1411 12 13 14
s s s ssF s s s+ +⎧ ⎫ U⎧ ⎫⎡ ⎤
{ }GLOBALD
3 3 3 31 1 1 1
2 21 221 22 23 24
1 1 1 1
23 24
1
2 3 3 3 3
M MM M M M
M M M M
M M
M M M M
s s s s
s s s s
sF
F
s s s
s s s s
+ +
+ +
⎧ ⎫
⎪ ⎪
⎪ ⎪
⎪ ⎪
1
2
U
U
⎧ ⎫⎡ ⎤
⎪ ⎪⎢ ⎥
⎪ ⎪⎢ ⎥
⎪ ⎪⎢ ⎥11 12 13 14
2 2 2 2
21 22 23 24
2
31 32 33 34
1 1 1 1
41 42 43 44
1
3
1 1 1 24 2 2
M M M M M M M
M M M
M
M M M M M
s s s s
s s s
s s s s
s s s s
F
F s
+ +
=
+ +
⎪ ⎪
⎨ ⎬
⎪ ⎪ 4
3
U
U
⎪ ⎪⎢ ⎥
⎨ ⎬⎢ ⎥
⎪ ⎪⎢ ⎥21 1 1 1 24
5
2 2M M MM M M M M
F
F
s
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎩ ⎭
31 32 33 34
2 2 2 2
41 42 43 44
31 32 33 34
3 3 3 3
41 42 43 44
4
5M M M MM M M M
s ss s s s s U
U
+ +
⎪ ⎪⎢ ⎥
⎪ ⎪⎢ ⎥
⎪ ⎪⎢ ⎥
⎣ ⎦⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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F⎩ ⎭
41 42 43 44
2 2 2 2
41 42 43 44
3 3 3 3 6M MM M M MM M
ss s s s ss s U+ +⎣ ⎦⎩ ⎭
24. Reduced equation systemq y
(after imposing boundary conditions)
33 34
1 1
43 44
11 12
2 2
21 22
3 3MM M M
F U
F U
s ss s+ +⎧ ⎫ ⎧ ⎫⎡ ⎤
=⎨ ⎬ ⎨ ⎬⎢ ⎥+ +⎩ ⎭ ⎣ ⎦⎩ ⎭1 2 1 24 4M MM M
F Us sss⎢ ⎥+ +⎩ ⎭ ⎣ ⎦⎩ ⎭
•This reduced equation system can be solved to get the unknown
displacement components 3 4
,U U
{ } [ ]{ }LOCAL T GLOBALD R D=•From
{ }LOCALD can be found out.
{ } { }D D=F h b
Dept. of CE, GCE Kannur Dr.RajeshKN
{ } { }LOCAL MiD D=•For each member,
25. { } { } [ ]{ }A A S D{ } { } [ ]{ }Mi MLi Mi MiA A S D= +Member end actions
Where,
Fixed end actions on the member,{ }MLiA
Member stiffness matrix,[ ]MiS
[ ]
in local
coordinates
Displacement components of the member,[ ]MiD
{ } { } [ ]{ }LOCAL T GLOBALMi D R DD ==As we know,
{ } { } [ ][ ]{ }i i iM ML M i iT GLOBALA A S R D∴ = +
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25
26. Direct Stiffness Method: Procedure
STEP 1: Get member stiffness matrices for all members [ ]MiS
STEP 2: Get rotation matrices for all members
[ ]TiR
STEP 3: Transform member stiffness matrices from local coordinates
into global coordinates to get [ ]MSiS[ ]MSi
STEP 4: Assemble global stiffness matrix [ ]JS
STEP 5: Impose boundary conditions to get the reduced stiffness
matrix [ ]S[ ]FFS
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26
27. STEP 6: Find equivalent joint loads from applied loads on eachq j pp
member (loads other than those applied at joints directly)
STEP 7 T f b ti f l l di t i t l b lSTEP 7: Transform member actions from local coordinates into global
coordinates to get the transformed load vector
STEP 8: Find combined load vector by adding the above
transformed load vector and the loads applied directly at joints
[ ]CA
STEP 9: Find the reduced load vector by removing members in
h l d di b d di i
[ ]FCA
the load vector corresponding to boundary conditions
STEP 10: Get displacement components of the structure in global
coordinates { } [ ] { }
1
F FF FCD S A
−
=
Dept. of CE, GCE Kannur Dr.RajeshKN
27
28. { } [ ]{ }LOCAL T GLOBALD R D=
STEP 11: Get displacement components of each member in local
coordinates
STEP 12: Get member end actions from
{ } { } [ ][ ]{ }Mi MLi Mi iT GLOB iALR DA A S∴ = +
{ } { } [ ][ ]R RC RF FA A S D= − +STEP 13: Get reactions from
{ }RCA represents combined joint loads (actual and
equivalent) applied directly to the supports.q ) pp y pp
Dept. of CE, GCE Kannur Dr.RajeshKN
28
29. • Problem 1:
1 Member stiffness matrices in local co ordinates
1EA ⎡ ⎤⎡ ⎤
1. Member stiffness matrices in local co-ordinates
(without considering restraint DOF)
[ ]1
1
00
1.155
0 0 0 0
M
EA
S L
⎡ ⎤⎡ ⎤
⎢ ⎥⎢ ⎥= =
⎢ ⎥⎢ ⎥
⎣ ⎦ ⎣ ⎦0 0 0 0⎣ ⎦ ⎣ ⎦
[ ]
1 0⎡ ⎤
[ ]
1 2 0⎡ ⎤
Dept. of CE, GCE Kannur Dr.RajeshKN
[ ]2
1 0
0 0
MS
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
[ ]3
1 2 0
0 0MS
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
37. • Problem 2 :
10kN m
5
4
8
4m
0kN m
C
B
20kN 2
4
6
7
94m
1
9
4m 1
2
A EI is constant.
1
2
3
Displacements in
global co-ordinates
Dept. of CE, GCE Kannur Dr.RajeshKN
37
global co ordinates
38. 2 5
6
9 2
1
4
3 6
4
6
2
6
9 2
5
1 1
Ki ti 1Kinematic
indeterminacy
DOF in local co-
ordinates
1
2
33
Dept. of CE, GCE Kannur Dr.RajeshKN
38
39. 0 0 0 0
12 6 12 6
EA EA
L L
EI EI EI EI
⎡ ⎤
−⎢ ⎥
⎢ ⎥
⎢ ⎥
3 2 3 2
2 2
12 6 12 6
0 0
6 4 6 2
0 0
EI EI EI EI
L L L L
EI EI EI EI
L L L L
⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥
Member stiffness matrix
[ ]
0 0 0 0
12 6 12 6
Mi
L L L LS
EA EA
L L
EI EI EI EI
⎢ ⎥=
⎢ ⎥
−⎢ ⎥
⎢ ⎥
⎢ ⎥
of a 2D frame member in
local coordinates
3 2 3 2
2 2
12 6 12 6
0 0
6 2 6 4
0 0
EI EI EI EI
L L L L
EI EI EI EI
L L L L
⎢ ⎥− − −⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥⎣ ⎦L L L L⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
39
40. 1. Member stiffness matrices in local co-ordinates
( ith t id i t i t DOF)
6Local DOFMember 1
(without considering restraint DOF)
[ ] 4EI
S ⎡ ⎤=
⎢ ⎥ [ ]1EI EI= =
6Global DOF
[ ]1MS
L
=
⎢ ⎥⎣ ⎦
[ ]1EI EI
Member 2
6 9Global DOF
3 6Local DOF
Member 2
[ ]
4 2EI EI
L LS
⎡ ⎤
⎢ ⎥
⎢ ⎥
6 9Global DOF
0.5EI EI⎡ ⎤
[ ]2
2 4
M
L LS
EI EI
L L
= ⎢ ⎥
⎢ ⎥
⎢ ⎥⎣ ⎦
0.5EI EI
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
40
41. 2. Rotation (transformation) matrices
In this case, transformation matrices are:
[ ] [ ]1 1TR = corresponding to local DOF 6
[ ]2
1 0
0 1
TR
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
corresponding to local DOFs 3 & 6
⎣ ⎦
3. Member stiffness matrices in global co-ordinates
(Transformed member stiffness matrices)
[ ]1MSS EI= [ ]2
0.5
0.5
MS
EI EI
S
EI EI
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
⎣ ⎦
42. 4 A bl d ( d d d) l b l tiff t i
6 9Gl b l DOF
4. Assembled (and reduced) global stiffness matrix
[ ]
0.5EI EI
S
IE⎡ + ⎤
= ⎢ ⎥
6 9Global DOF
2 0.5
EI
⎡ ⎤
= ⎢ ⎥[ ]
0.5FFS
EI EI
= ⎢ ⎥
⎣ ⎦ 0.5 1
EI= ⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
43. 5. Loads
20
13.33 13.33
B
C
20
13.33 20
13.33B
C20
B
20 20
C C20
Fi d d ti A
Combined (Eqlt.+
actual) joint loads
A
Fixed end actions A (Loads in global
co-ordinates)
{ }
13.33
13 33FCA
−⎧ ⎫
= ⎨ ⎬
⎩ ⎭
Loads corresponding to global DOF 6, 9:
Dept. of CE, GCE Kannur Dr.RajeshKN
43
{ }
13.33FC ⎨ ⎬
⎩ ⎭
Loads corresponding to global DOF 6, 9:
44. 6 Joint displacements
11 431 ⎧ ⎫
6. Joint displacements
{ } [ ] { }
1
F FF FCD S A
−
=
11.43
19.04
1
EI
−⎧
=
⎫
⎨ ⎬
⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
44
45. 7. Member end actions
{ } { } [ ][ ]{ }Mi MLi Mi iT GLOB iALR DA A S∴ = +
: fixed end actions for member i{ }MLiA
Member 1
{ } { } [ ][ ]{ }1 1 1 1 1TM M G LM L BAL OR DA A S= +
{ } [ ]{ }DA S { } { }
4 1
0 11 43
EI⎡ ⎤
{ } [ ]{ }1 1 1ML M GLOBALDA S= + { } { }0 11.43
11.43
L EI
⎡ ⎤
= + −⎢ ⎥⎣ ⎦
= −
This is the member end action corresponding to local DOF 6
f M b 1 i b d h d f
Dept. of CE, GCE Kannur Dr.RajeshKN
45
of Member 1. i.e., member end moment at the top edge of
Member 1.
46. Member 2
{ } { } [ ]{ }2 2 2 2M ML M GLOBALA A DS= +
Member 2
13.33 11.43
13.33 1
0.5
9.04
1
0.5
EI EI
EI EI EI
−⎧ ⎫ ⎧ ⎫
+
⎡ ⎤
= ⎢ ⎥
⎣
⎨ ⎬ ⎨
−⎩ ⎭ ⎩⎦
⎬
⎭
13.33 1.91−⎧ ⎫ ⎧ ⎫
+⎨ ⎬ ⎨= ⎬
11.42
=
⎧ ⎫
⎨ ⎬
Th th b ti di t l l DOF 3
13.33 13.325
+⎨ ⎬ ⎨
⎩ ⎭
⎬
− ⎩ ⎭ 0
⎨ ⎬
⎩ ⎭
These are the member actions corresponding to local DOFs 3
and 6 of Member 2. i.e., member end moments of Member 2.
Dept. of CE, GCE Kannur Dr.RajeshKN
46
52. 1. Member stiffness matrices in local co-ordinates
( ith t id i t i t DOF)(without considering restraint DOF)
0 0
12 6
XEA
L
EI EI
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
1000 0 0⎡ ⎤
⎢ ⎥[ ]1 3 2
12 6
0
6 4
0
Z Z
M
Z Z
EI EI
S
L L
EI EI
⎢ ⎥= −
⎢ ⎥
⎢ ⎥
⎢ ⎥−
5
0 120 6000
0 6000 4 10
⎢ ⎥= −
⎢ ⎥
− ×⎢ ⎥⎣ ⎦
2
0
L L
⎢ ⎥−
⎣ ⎦
0 0
12 6
XEA
L
EI EI
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
800 0 0⎡ ⎤
⎢ ⎥
[ ]2 3 2
12 6
0
6 4
0
Z Z
M
Z Z
EI EI
S
L L
EI EI
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎢ ⎥
5
0 61.44 3840
0 3840 3.2 10
⎢ ⎥=
⎢ ⎥
×⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
52
2
0
L L
⎢ ⎥
⎣ ⎦
63. 1. Member stiffness matrices in local co-ordinates
( ith t id i t i t DOF)(without considering restraint DOF)
5
0 0
3.5 10 0 0
12 6
XEA
L
EI EI
⎡ ⎤
⎢ ⎥
⎡ ⎤×⎢ ⎥
⎢ ⎥⎢ ⎥[ ]1 3 2
12 6
0 0 6562.5 13125
0 13125 35000
6 4
0
Z Z
M
Z Z
EI EI
S
L L
EI EI
⎢ ⎥⎢ ⎥= − = −⎢ ⎥⎢ ⎥
⎢ ⎥−⎢ ⎥ ⎣ ⎦
⎢ ⎥− 2
0
L L
⎢ ⎥
⎣ ⎦
⎡ ⎤
[ ]
5
0 0
3.5 10 0 0
12 6
X
Z Z
EA
L
EI EI
⎡ ⎤
⎢ ⎥
⎡ ⎤×⎢ ⎥
⎢ ⎥⎢ ⎥[ ]2 3 2
2
12 6
0 0 3888.89 11666.67
0 11666.67 46666.67
6 4
0
Z Z
M
Z Z
EI EI
S
L L
EI EI
⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥
⎢ ⎥⎢ ⎥ ⎣ ⎦
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
63
2
L L
⎢ ⎥
⎣ ⎦
64. ⎡ ⎤
[ ]
5
0 0
3.5 10 0 0
12 6
0 0 6 62 1312
X
Z Z
EA
L
EI EI
S
⎡ ⎤
⎢ ⎥
⎡ ⎤×⎢ ⎥
⎢ ⎥⎢ ⎥[ ]3 3 2
2
12 6
0 0 6562.5 13125
0 13125 35000
6 4
0
Z Z
M
Z Z
EI EI
S
L L
EI EI
⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥
⎢ ⎥⎢ ⎥ ⎣ ⎦
⎢ ⎥
⎣ ⎦
2
L L
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
64
74. 7. Member end actions
{ } { } [ ][ ]{ }Mi MLi Mi iT GLOB iALR DA A S∴ = +
Dept. of CE, GCE Kannur Dr.RajeshKN
75. Summary
Direct stiffness method
Summary
• Introduction – element stiffness matrix – rotation transformation
matrix – transformation of displacement and load vectors and
stiffness matrix – equivalent nodal forces and load vectors –
assembly of stiffness matrix and load vector – determination ofassembly of stiffness matrix and load vector determination of
nodal displacement and element forces – analysis of plane truss
beam and plane frame (with numerical examples) – analysis of
grid space frame (without numerical examples)grid – space frame (without numerical examples)
Dept. of CE, GCE Kannur Dr.RajeshKN
75