Module3 direct stiffness- rajesh sir

GCE Kannur

1. Structural Analysis-III
Di t Stiff Mth d Di
rec
t Stiff
ness
M
e
th
o
d
D
r
. Rajesh K. N.
Assistant
Professor in Civil Engineering
Assistant
Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE KannurDr.RajeshKN Dept. of CE, GCE KannurDr.RajeshKN
1

2. Module III Module 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 of
assembly 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 KannurDr.RajeshKN
2

3. Introduction •The formalised stiffness method involves evaluating the
dis
p
lacement transformation matrix C
MJ
correctl
y
p
y
•Generation of matrix C
MJ
is not suitable for computer
programming H th l ti f di t
tiff th d
•
H
ence
th
e
evo
l
u
ti
on
o
f di
rec
t
s
tiff
ness
me
th
o
d
Dept. of CE, GCE KannurDr.RajeshKN
3

4. Direct stiffness method
• We need to simplify the assembling process of S
J
, the
J
assembled structure stiffness matrix
•
The key to this is to use member stiffness matrices for actions The key to this is to use member stiffness matrices for actions
and displacements at BOTHends of each member
If b di l d i h f
•
If
mem
b
er
di
sp
l
acements
are
expresse
d
w
i
t
h
re
f
erence
to

global co-ordinates, the process of assembling S
J
can be made
simple simple
Dept. of CE, GCE KannurDr.RajeshKN
4

5. Member oriented axes (local coordinates)
d t t i t d ( l b l di t )
an
d
s
t
ruc
t
ure

or
i
en
t
e
d
axes
(
g
l
o
b
a
l
coor
di
na
t
es
)
L
δ
x
y
L
Local axes
L
δ
L
L
δ
sin L
δ
θ
L
δθ
x
Y Y Y Y
cos
L
δθ
θ
Global axes
y
Global axes Global axes
θ
XX X X
Dept. of CE, GCE KannurDr.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
x
E
A
L
E
A
L
⎜⎟⎝⎠corr. to DOF 1
00
EA EA
⎡
⎤
−
⎢⎥
[]
00
0000
00
M
LL
S
EA EA
⎢⎥⎢
⎥
⎢
⎥
⎢
⎥
−
⎢⎥
=
Member stiffness matrix in local
coordinates
Dept. of CE, GCE KannurDr.RajeshKN
0000
L
L
⎢⎥⎢
⎢
⎣
⎦⎥⎥

7. Transformation of displacement vector
θ
Dis
p
lacements in
g
lobal
22 2
cos UD
θ
=
pg
coordinates: U
1
and U
2
Displacements in local
di
D
d
D
2
D
D
Y
i
UD
θ
12 2
sin
UD
θ
=
−
θ
111121 2
cos sin UUU D D
θ
θ
=+= −
coor
di
nates:

D
1
an
d
D
2
1
D
Y
11 1
cos
UD
θ
=
21 1
s
i
n
UD
θ
=
221221 2
sin cos UUU D D
θ
θ
=
+= +
11
cos sin
UD
θθ
−
⎧⎫ ⎧⎫
⎡⎤
⎧⎫ ⎧⎫
⎡⎤
11 22
cos sin sin cos
UD UD
θθ θθ
⎧⎫ ⎧⎫
⎡⎤
=
⎨⎬ ⎨⎬⎢⎥
⎣⎦ ⎩⎭ ⎩⎭
X
11
22
cos sin
sin cos
D
U
D
U
θ
θ
θθ
⎧⎫ ⎧⎫
⎡⎤
∴
= ⎨⎬ ⎨⎬⎢⎥
−⎣⎦ ⎩⎭ ⎩⎭
Dept. of CE, GCE KannurDr.RajeshKN
{
}
[
]
{
}
D
RU =

8. Cidi bth d
⎧⎫ ⎧⎫
C
ons
id
er
i
ng
b
o
th
en
d
s,
11
22
cos sin 0 0
sin cos 0 0
00 i
D
U
D
U
DU
θ
θ
θθ
θθ
⎧⎫ ⎧⎫
⎡⎤
⎪
⎪⎪⎪⎢⎥
−
⎪
⎪⎪⎪
⎢⎥=
⎨⎬ ⎨⎬
⎢⎥
33
44
00
cos s
i
n
00sincos
DU D
U
θθ θθ
⎨⎬ ⎨⎬
⎢⎥
⎪
⎪⎪⎪
⎢⎥
⎪
⎪⎪⎪− ⎩⎭⎣ ⎦⎩⎭
{
}
[]
{
}
LOCAL T GLOBAL
DRD=
[
]
[
]
RO
⎡⎤
[
]
[
]
[
]
[][]
T
RO
R
OR
⎡⎤
=
⎢
⎥
⎣
⎦
Rotation matrix
Dept. of CE, GCE KannurDr.RajeshKN

9. Transformation of load vector
Actions in global
θ
coordinates:
F
1
and F
2
22 2
cos
F
A
θ
=
111121 2
cos sin
FFF A A
θ
θ
=
+= −
22 2 12 2
sin
FA
θ
=
−
θ
2
A
221221 2
sin cos
FFF A A
θ
θ
=
+= +
Y
11 1
cos
F
A
θ
=
21 1
sin
FA
θ
=
1
A
11
22
cos sin
sin cos
FA
FA
θ
θ
θθ−
⎧⎫ ⎧⎫
⎡
⎤
= ⎨⎬ ⎨⎬⎢⎥
⎣⎦ ⎩⎭ ⎩⎭
11 1
11
cos sin
A
F
θθ
⎧⎫ ⎧⎫⎡⎤
=
⎨⎬ ⎨⎬
⎢⎥
X
22
sin cos
A
F
θ
θ
⎨⎬ ⎨⎬
⎢⎥
−
⎩⎭⎣ ⎦⎩⎭
{
}
[
]
{
}
ARF
=
Dept. of CE, GCE KannurDr.RajeshKN
{
}
[
]
{
}
ARF
=

10. Cidi bth d C
ons
id
er
i
ng
b
o
th
en
d
s,
11
22
cos sin 0 0
sin cos 0 0 A
F
A
F
θθ
θθ
⎧⎫ ⎧⎫⎡⎤
⎪⎪ ⎪⎪⎢⎥
− ⎪⎪ ⎪⎪
⎢⎥
=
⎨⎬ ⎨⎬
33
44
00cossin
00sincos
A
F
A
F
θ
θ
θθ
⎢⎥
=
⎨⎬ ⎨⎬
⎢⎥
⎪⎪ ⎪⎪
⎢⎥
⎪⎪ ⎪⎪− ⎩⎭⎣ ⎦⎩⎭
{}
[
]
{
}
L
OCAL T GLOBAL
i.e., A R A=
Dept. of CE, GCE KannurDr.RajeshKN
10

11. Transformation of stiffness matrix
{
}
[]
{
}
LOCAL M LOCAL
ASD=
[
]
{
}
[
]
[
]
{
}
T GLOBAL M T GLOBAL
RA SRD=
{
}
[
]
[
]
[
]
{
}
1
GLOBAL T M T GLOBAL
A
RSRD
−
=
[
]
{
}
[
]
[
]
{
}
T GLOBAL M T GLOBAL
[
]
[
]
1T
RR
−
{
}
[
]
[
]
[
]
{
}
GLOBAL T M T GLOBAL
{
}
[
]
{
}
ASD
[
]
[
]
TT
RR
=
{
}
[
]
{
}
GLOBAL MS GLOBAL
ASD
=
[
]
[
]
[
]
[
]
T
M
STMT
SRSR
=
where, Member stiffness matrix in
global coordinates
Dept. of CE, GCE KannurDr.RajeshKN

12. [
]
[
]
[
]
[
]
T
SRSR
=
T
⎡⎤⎡⎤⎡⎤
[
]
[
]
[
]
[
]
M
STMT
SRSR
=
00 10 10 00
00 0000 00
T
cs cs
sc scEA
−
⎡⎤⎡⎤⎡⎤ ⎢⎥⎢⎥⎢⎥
−−
⎢⎥⎢⎥⎢⎥ =
⎢⎥⎢⎥⎢⎥
00 101000
00 000000
cs cs
L
sc sc
−
⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥
−− ⎣⎦⎣⎦⎣⎦
22
ccsccs ⎡⎤−−
⎢⎥
Mb tiff ti
22
22
cs s cs s EA
Lccsccs
⎢⎥
−
−
⎢⎥
=
⎢⎥−−
⎢⎥
M
em
b
er
s
tiff
ness
ma
t
r
i
x

in global coordinates
for a
p
lane truss member
22
cs s cs s
⎢⎥
−−⎣⎦
p
Dept. of CE, GCE KannurDr.RajeshKN
12

13. 2. Plane frame member
Stiffness coefficients in local coordinates
2
5
2
4
5
Degrees of freedom
00 00
E
AEA
⎡⎤
1
3
4
6
Degrees of freedom
32 32
00 00
12 6 12 6
00
LLE
IEI EIEI
L
LLL
⎡⎤
−
⎢
⎥
⎢
⎥
⎢
⎥
−
⎢
⎥
⎢⎥
[]
22
64 62
00
00 00
Mi
E
IEI EIEI
LL LL
S
EA EA
⎢⎥⎢
⎥ −
⎢
⎥
=
⎢
⎥
−
⎢⎥
Member stiffness matrix
in local coordinates
32 3 2
00 00
12 6 12 6
00
LLE
IEI EI EI
L
LLL
−
⎢⎥⎢
⎥
⎢
⎥
−− −
⎢
⎥
⎢⎥
Dept. of CE, GCE KannurDr.RajeshKN
22
62 64
00
E
IEI EIEI
LL LL
⎢⎥⎢
⎥ −
⎢
⎥
⎣
⎦

14. Transformation of displacement vector
11
22
cos sin 0 0 0 0
sin cos 0 0 0 0
D
U
D
U
θθ
θθ
⎧⎫ ⎧⎫⎡⎤
⎪⎪ ⎪⎪⎢⎥
−
⎪⎪ ⎪⎪
⎢⎥
33
44
001000
000cossin0
D
U
D
U
θθ
⎪⎪ ⎪⎪
⎢⎥
⎪⎪ ⎪⎪⎢⎥
= ⎨⎬ ⎨⎬⎢⎥
⎪⎪ ⎪⎪
⎢
⎥
⎪⎪ ⎪⎪
⎢⎥
55
66
000sincos0
000001
D
U
D
U
θ
θ
⎪⎪ ⎪⎪
⎢⎥
−
⎪⎪ ⎪⎪⎢⎥
⎣⎦ ⎩⎭ ⎩⎭
{
}
[
]
{
}
LOCAL T GLOBAL
DRD
=
[
]
[
]
RO
⎡⎤
{
}
[
]
{
}
LOCAL T GLOBAL
[
]
[
]
[
]
[][]
T
RO
R
OR
⎡⎤
=⎢⎥
⎣⎦
Rotation

matrix
Dept. of CE, GCE KannurDr.RajeshKN

15. Transformation of load vector
{
}
[
]
{
}
LOCAL T GLOBAL
ARA=
{
}
[
]
{
}
cos sin 0 0 0 0
θθ
⎡⎤⎢⎥
[
]
sin cos 0 0 0 0
001000
R
θθ
⎢⎥
−
⎢⎥
⎢⎥
=
⎢⎥
[
]
000cossin0
000sincos0
T
R
θ
θ
θθ
=
⎢⎥⎢⎥
⎢⎥−
⎢⎥
000001
⎢⎥⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN

16. Transformation of stiffness matrix
{
}
[]
{
}
GLOBAL MS GLOBAL
ASD=
[
]
[
]
[
]
[
]
T
M
STMT
SRSR=
Member stiffness matrix in global coordinates global coordinates
EA EA
⎡⎤
0000
0000
cs
sc
⎡
⎤
⎢
⎥
−
⎢⎥
32 32
00 00
12 6 12 6
00
EA EA
LL
E
IEI EIEI
LL LL
⎡⎤
−
⎢⎥
⎢⎥
⎢⎥
−
⎢⎥
⎢⎥
[]
001000
000 0
000 0
T
cs
sc
R
⎢⎥⎢
⎥
=
⎢
⎥
⎢
⎥
⎢
⎥ −
⎢⎥
[]
22
64 62
00
00 00
M
E
IEI EIEI
LL LL
S
EA EA
LL
⎢⎥⎢⎥−
⎢⎥
=
⎢⎥
−
⎢⎥
Where,
,
000001
⎢⎥⎣
⎦
32 3 2
12 6 12 6
00
62 64
00
LL
E
IEI EI EI
LL L L E
IEI EIEI
⎢⎥⎢⎥
⎢⎥
−− −
⎢⎥
⎢⎥
⎢⎥
Dept. of CE, GCE KannurDr.RajeshKN
22
00
LL LL
⎢⎥
−
⎢⎥⎣⎦

17. Assemblin
g
g
lobal stiffness matrix
gg
Plane truss
2
Force
2
1
3
3
1
3
2
4
1
3
Action/displacement components in
local coordinates of members
Dept. of CE, GCE KannurDr.RajeshKN

18. 4
4
3
4
3
4
2
6
1
5
6
2
5
1
4
3
Action/displacement components in
2
5
6
Action/displacement components in
global coordinates of the structure
Dept. of CE, GCE KannurDr.RajeshKN
1
5

19. 1234
Global DOF
11 12 13 14
1111
21 22 23 24
22
22
1
2
MMMM
ssss ssss
ccsccs cs s cs s
EA
−− ⎡⎤
⎡
⎤
⎢⎥
⎢
⎥
−−
⎢⎥
⎢⎥
1234
[
]
1111
31 32 33 34
1111
41 42 43 44
122
22
2 3 4
MMMM
MMMM
M
ssss ssss
cs s cs s
EA
S
ccsccs L
−−
⎢⎥
⎢⎥
==
−−⎢⎥
⎢
⎥
⎢⎥
⎢⎥⎣⎦
⎣⎦
41 42 43 44
1111
22
4
MMMM
ssss
cs s cs s
⎢⎥
⎢⎥
−−
⎣⎦
⎣⎦
123456
1 2
××××
⎡
⎤
⎢
⎥
××××
⎢⎥
C t ib ti f
[
]
2 34
J
S
××××
⎢⎥
××××
⎢
⎥
=
⎢⎥
C
on
t
r
ib
u
ti
on
o
f
Member 1 to global
stiffness matrix
[
]
4 5
J
⎢⎥
×
×××
⎢
⎥
⎢
⎥
⎢⎥
Dept. of CE, GCE KannurDr.RajeshKN
6
⎢⎥⎣
⎦

20. 3456
Global DOF
11 12 13 14
2222
21 22 23 24
3
4
MMMM
ssss ssss
⎡⎤
⎢⎥
⎢⎥
3456
[
]
22
2
22
31 32 33 34
2222
41 42 43 44
4 5 6
MMMM
MM M
M
M
ssss ssss ssss
S=
⎢⎥⎢⎥
⎢⎥
⎣⎦
2222
6
MMMM
ssss
⎣⎦
123456
1 2
⎡⎤
⎢⎥
⎢⎥
C t ib ti f
[]
2 3 4
J
S
⎢⎥
×
××× ⎢⎥
=⎢⎥
××××
⎢⎥
C
on
t
r
ib
u
ti
on
o
f
Member 2 to global
stiffness matrix
4 5 6
××××
⎢⎥⎢⎥
×
×××
⎢⎥⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
6
⎢⎥
×
×××
⎣⎦

21. 11 12 13 14
1
⎡⎤
1256
Global DOF
[
]
11 12 13 14
3333
21 22 23 24
33
33
1 2
MMMM MMMM
s
sss
ssss
S
⎡⎤⎢⎥⎢⎥
[
]
33
3
33
31 32 33 34
3333
41 42 43 44
5 6
MMMM MM M
M
M
ssss
S
=
⎢⎥⎢⎥⎢⎥⎣⎦
41 42 43 44
3333
6
MMMM
s
sss
⎢⎥⎣⎦
123456
1 2
×
×××
⎡
⎤
⎢⎥
123456
[
]
2 3
J
S
⎢⎥
×
×××
⎢
⎥
⎢
⎥
=
⎢⎥
Contribution of
Member 3 to global
stiffness matrix
[
]
4
5
J
S
⎢⎥⎢
⎥
⎢
⎥
×
×××
⎢⎥
stiffness matrix
Dept. of CE, GCE KannurDr.RajeshKN
6
⎢⎥
×
×××
⎣
⎦

22. Assembled global stiffness matrix
11 12 13 1411 12 13 14
ss ss
ss
ss
++
1
⎡⎤
12 3456
11 11
21 22 23 24
11 11
33 33
21 22 23 24
33 33
MM MM
MM MM
MM MM
MM MM
ss ss ss ss
ss
ss
ss ss
++ ++
1 2
⎡⎤⎢
⎥
⎢
⎥
⎢⎥
[]
31 32 311 12 13 14
2222
21 22 23 24
334
1111
41 42 43 44
MMMM MMMM
J
ssss
ssss
ssss
s
S
s
ss
++
=
++
3 4
⎢⎥⎢
⎥
⎢⎥
31 3
2222
31 32 33
11
2
2
11
2 33 32
MMMM
MM MM
MMMM
M M
ss
ssss
ss
s
s
s
ss
s
++
+
34
2
33 34
3
4 5
M M
s s+
⎢⎥⎢
⎥
⎢⎥
41 42 43 44
2
41 42 43 44
322 3323
6
MMMM MM MM
ss ssssss++
⎢⎥⎣
⎦
Dept. of CE, GCE KannurDr.RajeshKN

23. Im
p
osin
g
boundar
y
conditions
2
pg y
Plane truss example
2
1
1
2
0
UUUU
=
===
Boundary conditions are:
3
3
1
1256
0
UUUU
Boundary conditions are:
{
}
GLOBAL
D
11 12 13 14 11 12 13 14
ss ss
s
F
sss
++
⎧⎫
U
⎧⎫
⎡⎤
{
}
GLOBAL
D
33 33 11 11
221 22 122 2324
11 11
23 24
1
233 33
MM MM MM
MM MM
MM
MM MM
ss ss ss ss
s
F F
sss
ss ss
++ ++
⎧⎫⎪⎪
⎪⎪
⎪⎪
1
2
U U
⎧⎫
⎡⎤
⎪
⎪ ⎢⎥
⎪
⎪
⎢
⎥
⎪⎪
⎢⎥
11 12 13 14
2222
21 22 23 24
2
31 32 33 34
1111
41 42 43 44
1
3
11 1
2
4
2
2
MMMMMMM
MMM
M
MMMM
M
ssss
sss
ssss
ssss
F
F
s
++
=
+
+
⎪⎪⎨⎬⎪⎪
43
U
U
⎪⎪
⎢⎥
⎨
⎬ ⎢⎥
⎪⎪
⎢⎥
2
1
11 1
2
4 5
2
2
MMM
MMMM
M
F F
s
⎪⎪⎪⎪
⎪⎪
⎩⎭
31 32 33 34
2222
41 42 43 44
31 32 33 34
33 33
41 42 43 44
4 5 MMMM MM MM
sssssssU
U
++
⎪⎪
⎢⎥
⎪
⎪ ⎢⎥
⎪
⎪
⎢
⎥
⎣⎦
⎩⎭
Dept. of CE, GCE KannurDr.RajeshKN
6
F
⎩⎭
41 42 43 44
2222
41 42 43 44
33 336 MM MM MMMM
s ssss sss
U
+
+
⎣⎦
⎩⎭

24. Reduced e
q
uation s
y
stem
qy
(after imposing boundary conditions)
33 34
11
43 44
11 12
22
21 22
33
M MMM
FU
FU
ssss++ ⎧⎫ ⎧⎫⎡⎤
= ⎨⎬ ⎨⎬
⎢⎥
++
⎩⎭⎣ ⎦⎩⎭
1212
44
MM MM
FU
sss s
⎢⎥
++
⎩⎭⎣ ⎦⎩⎭
•This reduced equation system can be solved to get the unknown
displacement components
34,UU
{
}
[
]
{
}
LOCAL T GLOBAL
DRD=
•From
{
}
LOCAL
D
can be found out. {
}
{
}
DD
=
F h b
Dept. of CE, GCE KannurDr.RajeshKN
{
}
{
}
LOCAL Mi
DD
=
•
F
or
eac
h
mem
b
er,

25. {
}
{
}
[
]
{
}
AASD
{
}
{
}
[
]
{
}
M
i MLi Mi Mi
AASD
=+
Member end actions
Where,
Fixed end actions on the member,
{
}
M
Li
A
Member stiffness matrix,
[
]
M
i
S
[
]
in local
coordinates
Displacement components of the member,
[
]
M
i
D
{
}
{
}
[
]
{
}
LOCAL T GLOBAL Mi
DRD D= =
As we know,

{
}
{
}
[
]
[
]
{
}
iii MMLMiiTGLOBAL
AASRD ∴=+
Dept. of CE, GCE KannurDr.RajeshKN
25

26. Direct Stiffness Method: Procedure
STEP 1: Get member stiffness matrices for all members
[
]
M
i
S
STEP 2: Get rotation matrices for all members
[
]
Ti
R
STEP 3: Transform member stiffness matrices from local coordinates
into global coordinates to get
[
]
MSi
S
[
]
MSi
STEP 4: Assemble global stiffness matrix
[
]
J
S
STEP 5: Impose boundary conditions to get the reduced stiffness
matrix
[
]
S
[
]
FF
S
Dept. of CE, GCE KannurDr.RajeshKN
26

27. STEP 6: Find e
q
uivalent
joint loads from a
pp
lied loads on each
qj pp
member (loads other than those applied at joints directly) STEP 7
Tf b ti f ll dit it lbl
STEP 7
: T
rans
f
orm
mem
b
er

ac
ti
ons
f
rom
l
oca
l
coor
di
na
t
es
i
n
t
o
g
l
o
b
a
l
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
[
]
C
A
STEP 9: Find the reduced load vector by removing members in
h l d di b d di i
[
]
FC
A
t
h
e
l
oa
d
vector
correspon
di
ng

to
b
oun
d
ary

con
di
t
i
ons
STEP 10: Get displacement components of the structure in global
coordinates
{}
[
]
{
}
1
FFFFC
D
SA
−
=
Dept. of CE, GCE KannurDr.RajeshKN
27

28. {
}
[
]
{
}
LOCAL T GLOBAL
DRD=
STEP 11: Get displacement components of each member in local
coordinates
STEP 12: Get member end actions from
{
}
{
}
[]
[
]
{
}
Mi MLi MiiTGLOBiAL
RD AAS
∴
=+
{
}
{
}
[
]
[
]
R
RC RF F
A
ASD =− +
STEP 13: Get reactions from
{
}
RC
A
represents combined joint loads (actual and
e
q
uivalent
)
a
pp
lied directl
y
to the su
pp
orts.
q)pp y pp
Dept. of CE, GCE KannurDr.RajeshKN
28

29. •Problem 1:
1
Member stiffness matrices in local co
ordinates
1
EA
⎡⎤
⎡⎤
1
.
Member stiffness matrices in local co
-
ordinates
(without considering restraint DOF)
[]
1
1
0 0
1.155
00
00
M
EA
SL
⎡⎤
⎡⎤
⎢
⎥ ⎢⎥
==
⎢
⎥ ⎢⎥
⎣⎦
⎣⎦
00
00
⎣⎦
⎣⎦
[
]
10
⎡⎤
[
]
12 0
⎡⎤
Dept. of CE, GCE KannurDr.RajeshKN
[
]
2
10 00
M
S
⎡⎤
=
⎢⎥
⎣⎦
[
]
3
12 0
00
M
S
⎡⎤
=
⎢
⎥
⎣
⎦

30. 2. Rotation (transformation) matrices
[
]
1
cos sin 0.5 0.866
i086605
T
R
θθ θθ
− ⎡⎤⎡ ⎤
==
⎢⎥⎢ ⎥ ⎣⎦⎣ ⎦
[
]
160
s
i
ncos
0
.
866 0
.
5
T
θ
θθ
=−
⎢⎥⎢ ⎥
−
⎣⎦⎣ ⎦
[]
290
01 10
T
R θ
−
⎡⎤
=
⎢⎥
⎣⎦
90
10
θ
=
−
⎣⎦
[]
3
150
0.866 0.5
0.5 0.866
T
R
θ
=−
−−⎡⎤
=
⎢⎥
−
⎣⎦⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
30

31. 3. Member stiffness matrices in global co-ordinates
(Transformed member stiffness matrices)
[
]
[
]
[
]
[
]
1111
T
MS T M T
SRSR=
0.5 0.866 1 0 0.5 0.866 1
0866 05 0 0 0866 05
1155
T
−−
⎡
⎤⎡⎤⎡ ⎤
=
⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦⎣⎦
0
.
866 0
.
5000
.
866 0
.
5
1
.
155
⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦⎣⎦
0216 0375
−
⎡⎤
0
.
216 0
.
375
0.375 0.649
−
⎡⎤
=
⎢
⎥
−
⎣
⎦
Dept. of CE, GCE KannurDr.RajeshKN

32. []
2
011001
10 0010
T
MS
S
−− ⎡⎤⎡⎤⎡⎤
=
⎢⎥⎢⎥⎢⎥
⎣⎦⎣⎦⎣⎦ ⎣⎦⎣⎦⎣⎦
00⎡⎤
=
⎢⎥
01
=
⎢⎥⎣⎦
[
]
3
0.866 0.5 1 0 0.866 0.5 1
05 0866 0 0 05 0866
2
T
MS
S
−− −− ⎡⎤⎡⎤⎡⎤
=
⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦⎣⎦
[
]
3
0
.
50
.
866 0 0 0
.
50
.
866
2
MS
⎢⎥⎢⎥⎢⎥
−
−
⎣⎦⎣⎦⎣⎦
0.375 0.217 ⎡⎤
⎢⎥
0.217 0.125
=
⎢⎥⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
32

33. 4. Global stiffness matrix
021600375 037500217
⎡⎤
(by assemblingtransformed member stiffness matrices)
[]
0
.
21600
.
375 0
.
375 0 0
.
217
0.375 0 0.217 0.649 1 0.125
FF
S
+
+−++
⎡⎤
=
⎢⎥
−++ ++ ⎣⎦
0.591 0.158
0158 1774
− ⎡⎤
=
⎢⎥⎣⎦
0
.158

1
.
774

⎢⎥
−
⎣⎦
This is the reducedglobal stiffness matrix, since restraint DOF
were not considered. Hence, boundary conditions are
automatically incorporated automatically incorporated
.

Dept. of CE, GCE KannurDr.RajeshKN

34. 5. Loads
{}
0
5
FC
A
⎧
⎫
=
⎨
⎬
−
⎩⎭
6. Joint displacements 6. Joint displacements
0772
⎧⎫
{}
[
]
{
}
1
F
FF FC
D
SA
−
=
0
.
772
2.89
−
⎧⎫
=
⎨
⎬
−⎩⎭
Dept. of CE, GCE KannurDr.RajeshKN
34

35. 7. Member forces
{
}
{
}
[
]
[
]
{
}
M
iMLiMiiTGLOBi
A
L
RD AAS
∴
=+
Member 1
{
}
{
}
[
]
[
]
{
}
{}{}
[
]
[
]
{
}
1111 1T MMGL MLBA LO
RD AAS=+
[
]
[
]
{
}
11 1TGLOBAL M
RD S=
1
0.5 0.866 0
1155
0.772
⎡⎤
− ⎡⎤
⎢⎥
=
⎢⎥
⎢
−⎧
⎥
⎫
⎨⎬
1.833
=
⎧
⎫
⎨⎬
1
.
155
0.866 0.5
00
2.89
⎢⎥
⎢
⎥
⎣⎦
⎣⎦
⎨⎬
−
⎩⎭0
⎨⎬⎩⎭
Dept. of CE, GCE KannurDr.RajeshKN
35

36. {
}
[
]
[
]
{
}
RD
AS
Mb 2
{
}
[
]
[
]
{
}
222 2T GLOBAL MM
RD
AS
=
M
em
b
er
2
0.772
2
100 1
0010.89
−
⎡⎤⎡ ⎤
=
⎢⎥⎢ ⎥
⎣
⎦
−
⎧
⎫
⎨
⎦−⎩
⎣
⎬
⎭
2.89
0
⎧
=
⎫
⎨
⎬
⎩⎭
Member 3
{}
[
]
[
]
{
}
333 3TGLOBAL MM
R
D
A
S
=
12 0 0.866 0.5
00 0.5 0.866
0.772
2.89
−
− ⎡⎤⎡ ⎤
=
⎢⎥⎢ ⎥
−
⎣⎦
−
⎧
⎫
⎨
⎣⎦
⎬
−
⎩⎭
1.057
0
=
⎧
⎫
⎨
⎬
⎩⎭
⎣⎦
⎣⎦
⎩⎭
⎩⎭
Dept. of CE, GCE KannurDr.RajeshKN
36

37. •Problem 2 :
1
0
kN m
5
4
8
4
m
0
kN m
C
B
20kN
2
4
6
7
9
4
m
1
9
4m
1
2
A
EI is constant.
1
2
3
Displacements in global co
-
ordinates
Dept. of CE, GCE KannurDr.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
1
Ki
nema
ti
c

indeterminacy
DOF in local co-
ordinates
1
2
33
Dept. of CE, GCE KannurDr.RajeshKN
38

39. 00 00
12 6 12 6
EA EA
LL
EI EI EI EI
⎡
⎤
−
⎢
⎥
⎢⎥⎢⎥
32 32
22
12 6 12 6
00
64 62
00
EI EI EI EI
LL LL E
IEI EIEI
LL LL
⎢⎥⎢⎥
−
⎢
⎥
⎢
⎥
⎢
⎥ −
⎢⎥
Member stiffness matrix
[]
00 00
12 6 12 6
Mi
LL LL
S
EA EA
LL
EI EI EI EI
⎢⎥
=
⎢
⎥
−
⎢
⎥
⎢
⎥
⎢⎥
of a 2D frame member in
local coordinates
32 3 2
22
12 6 12 6
00
62 64
00
EI EI EI EI
L
LLL
E
IEI EIEI
LL LL
⎢⎥
−− −
⎢
⎥
⎢
⎥
⎢
⎥ −
⎢⎥⎣⎦
LL LL
⎢⎥⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
39

40. 1. Member stiffness matrices in local co-ordinates
( ith t id i t i t DOF)
6
Local DOF
Member 1 (
w
ith
ou
t
cons
id
er
i
ng
res
t
ra
i
n
t DOF)
[
]
4EI
S
⎡⎤
=
⎢⎥
[
]
1
EI EI
=
=
6
Global DOF
[
]
1
M
S
L
=
⎢⎥⎣⎦
[
]
1
EI EI
Member 2
69
Global DOF
3 6
Local DOF
Member 2
[
]
42EI EI
LL
S
⎡
⎤
⎢
⎥
⎢⎥
6

9
Global DOF
0.5 EI EI
⎡⎤
[
]
2
24
M
LL
S
EI EI L
L
=
⎢⎥⎢
⎥
⎢
⎥
⎣
⎦
0.5EI EI
⎡⎤
=
⎢⎥
⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
40

41. 2. Rotation (transformation) matrices
In this case, transformation matrices are:
[
]
[
]
1
1
T
R=
corresponding to local DOF 6
[]
2
10
01
T
R
⎡⎤
=
⎢⎥
⎣⎦
corresponding to local DOFs 3 & 6
⎣⎦
3. Member stiffness matrices in global co-ordinates
(Transformed member stiffness matrices)
[
]
1MS
SEI=
[]
2
0.5
0.5
MS
EI EI
S
EI EI
⎡
⎤
=
⎢
⎥
⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
⎣⎦

42. 4 A bld ( d d d)
lbl
tiff t i
69
Gl b l DOF
4
.
A
ssem
bl
e
d (
an
d
re
d
uce
d)
g
l
o
b
a
l
s
tiff
ness
ma
t
r
i
x
[
]
0.5 EI EI
S
IE ⎡
+
⎤
=
⎢⎥
6

9
Gl
o
b
a
l DOF
20.5
EI
⎡
⎤
=
⎢⎥
[
]
0.5
F
F
S
EI E
I
=
⎢⎥⎣
⎦0.5 1
EI
=
⎢⎥⎣
⎦
Dept. of CE, GCE KannurDr.RajeshKN

43. 5. Loads
20
13.33
13.33
B
C
20
13.33
20
13.33
B
C
20
B
20
20
C
C
20
Fi d d ti
A
Combined (Eqlt.+
actual) joint loads
A
Fi
xe
d
en
d
ac
ti
ons
A
(Loads in global
co-ordinates)
{
}
13.33
1333
FC
A
−
⎧
⎫
=
⎨⎬⎩⎭
Loads corresponding to global DOF
6,9:
Dept. of CE, GCE KannurDr.RajeshKN
43
{
}
13
.
33
FC
⎨⎬⎩⎭
Loads corresponding to global DOF
6,
9:

44. 6 Joint displacements
1143
1
⎧
⎫
6
.
Joint displacements
{}
[]
{
}
1
FFFFC
D
SA
−
=
11
.
43
19.04
1
EI
−
⎧
=
⎫
⎨
⎬
⎩⎭
Dept. of CE, GCE KannurDr.RajeshKN
44

45. 7. Member end actions
{}
{
}
[
]
[
]
{
}
Mi MLi MiiTGLOBiAL
RD AAS
∴
=+
: fixed end actions for member
i
{
}
M
Li
A
Member 1
{
}{ }
[
]
[
]
{
}
1111 1
T MMGL MLBA LO
RD AAS
=+
{
}
[
]
{
}
D
AS
{
}
{
}
41
01143
E
I
⎡⎤
{
}
[
]
{
}
11 1
ML MGLOBAL
D
AS
=+
{
}
{
}
011
.
43
11.43
LEI⎡⎤
=+ −
⎢
⎥
⎣⎦
=−
This is the member end action corresponding to local DOF 6
f
Mb 1
i b d h d f
Dept. of CE, GCE KannurDr.RajeshKN
45
o
f
M
em
b
er
1
.

i
.e.,
mem
b
er
en
d
moment
at
t
h
e
top
e
d
ge
o
f
Member 1.

46. Member 2
{
}
{
}
[
]
{
}
222 2 MMLMGLOBAL
AAD S =+
Member 2
13.33 11.43
13.33 1
0.5
9.04
1
0.5
EI EI
EI EIEI
−
⎧
⎫⎧⎫
+
⎡⎤
=
⎢⎥
⎣
⎨⎬ ⎨
−⎩⎭ ⎩⎦
⎬
⎭
13.33 1.91
−
⎧
⎫⎧ ⎫
+
⎨⎬⎨
=
⎬
11.42
=
⎧
⎫
⎨⎬
Th th b ti di t
ll DOF 3
13.33 13.325
+
⎨⎬⎨ ⎩⎭
⎬
−
⎩⎭0
⎨⎬⎩⎭
Th
ese
are
th
e
mem
b
er
ac
ti
ons
correspon
di
ng
t
o
l
oca
l DOF
s
3
and 6 of Member 2. i.e., member end moments of Member 2.
Dept. of CE, GCE KannurDr.RajeshKN
46

47. 11.43
0
B
0
B
C
11.43
Member end moments
A
Dept. of CE, GCE KannurDr.RajeshKN
47

48. •Problem 3 :
10 kips
0.24 kips/in.
20 kips
1000 kips-in
75 in
100 in
50 in
50 in
100 in
50 in
50 in
Dept. of CE, GCE KannurDr.RajeshKN
48

49. 25
1
3
4
6
8
7
9
8
Global DOF
7
Dept. of CE, GCE KannurDr.RajeshKN
49

50. 2
13
5
6
Local DOF
4
6
Local DOF
Dept. of CE, GCE KannurDr.RajeshKN
50
4

51. Free DOF
Dept. of CE, GCE KannurDr.RajeshKN
51

52. 1. Member stiffness matrices in local co-ordinates
( ith t id i t i t DOF) (
w
ith
ou
t
cons
id
er
i
ng
res
t
ra
i
n
t DOF)
00
12 6
X
EA
L
EI EI
⎡
⎤
⎢
⎥
⎢
⎥
⎢⎥
1000 0 0
⎡
⎤
⎢⎥
[
]
132
12 6
0
64
0
ZZ
M
Z
Z
EI EI
S
LL
EI EI
⎢⎥
=
−
⎢
⎥
⎢
⎥
⎢⎥
−
5
0 120 6000
0 6000 4 10
⎢⎥
=−
⎢
⎥
−×
⎢
⎥
⎣
⎦
2
0
LL
⎢⎥
−
⎣
⎦
00
12 6
X
EA
L
EI EI
⎡
⎤
⎢
⎥
⎢
⎥
⎢⎥
800 0 0
⎡
⎤
⎢⎥
[
]
232
12 6
0
64
0
ZZ
M
Z
Z
EI EI
S
LL
EI EI
⎢⎥
=
⎢
⎥
⎢
⎥
⎢⎥
5
0 61.44 3840
038403.210
⎢⎥
=
⎢
⎥
×
⎢
⎥
⎣
⎦
Dept. of CE, GCE KannurDr.RajeshKN
52
2
0
LL
⎢⎥⎣
⎦

53. 2. Rotation (transformation) matrices
[
]
cos sin 0
i0
R
⎡
⎤
⎢⎥
θθ θθ
[
]
s
i
ncos
0
001
T
R
⎢⎥
=
−
⎢
⎥
⎢
⎥
⎣
⎦
θθ
Member 1
Member 2
[
]
2
0.8 0.6 0
0.6 0.8 0 R
−
⎡
⎤
⎢
⎥
=
⎢⎥
[
]
1
100
010 R
⎡
⎤
⎢
⎥
=
⎢⎥
[
]
001
⎢⎥⎢
⎥
⎣
⎦
[
]
1
001
⎢⎥⎢
⎥
⎣
⎦
Dept. of CE, GCE KannurDr.RajeshKN
53

54. 3. Member stiffness matrices in global co-ordinates
(Transformed member stiffness matrices)
[]
[
][ ]
[
]
1111
T
MS T M T
SRSR=
100 1000 0 0 100 0 1 0 0 120 6000 0 1 0
T
⎡
⎤⎡ ⎤⎡ ⎤
⎢
⎥⎢ ⎥⎢ ⎥
=
⎢⎥⎢ ⎥⎢⎥
5
0 1 0 0 120 6000 0 1 0 0 0 1 0 6000 4 10 0 0 1
=
−
⎢⎥⎢ ⎥⎢⎥
−×
⎢
⎥⎢ ⎥⎢ ⎥
⎣
⎦⎣ ⎦⎣ ⎦
5
1000 0 0
0 120 6000
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢⎥
5
0 6000 4 10 −×
⎢⎥⎣
⎦
Dept. of CE, GCE KannurDr.RajeshKN
54

55. [
]
0.8 0.6 0 800 0 0 0.8 0.6 0 06 08 0 0 6144 3840 06 08 0
T
S
−− ⎡⎤⎡ ⎤⎡⎤
⎢⎥⎢ ⎥⎢⎥
[
]
2
5
0
.
60
.
80 061
.
44 3840 0
.
60
.
80
0 0 1 0 3840 3.2 10 0 0 1
MS
S
⎢⎥⎢ ⎥⎢⎥
=
⎢⎥⎢ ⎥⎢⎥
× ⎢⎥⎢ ⎥⎢⎥
⎣⎦⎣ ⎦⎣⎦
0.8 0.6 0 640 480 0
06 08 0 3686 4915 3840
−
⎡⎤⎡ ⎤
⎢⎥⎢ ⎥ =−
⎢⎥⎢ ⎥
5
0
.
60
.
8036
.
86 49
.
15 3840
0 0 1 2304 3072 3.2 10
=−
⎢⎥⎢ ⎥
× ⎢⎥⎢ ⎥ ⎣⎦⎣ ⎦
534.12 354.51 2304
35451 32732 3072
−
⎡
⎤
⎢
⎥
=
−
⎢⎥
5
354
.
51 327
.
32 3072
2304 3072 3.2 10
⎢⎥
×
⎢
⎥
⎣
⎦
Dept. of CE, GCE KannurDr.RajeshKN

56. 4
.
A
sse
m
b
l
ed (
r
educed)
g
l
oba
l
st
iffn
ess
m
at
ri
x
1000 534.12 0 354.51 0 2304
+
−+
⎡⎤
. sse b ed ( educed)
goba
st ess at x
[]
55
0 354.51 120 327.32 6000 3072
0 2304 6000 3072 4 10 3.2 10
FF
S
⎡⎤⎢⎥
=− + −+
⎢⎥
+
−
+×+×
⎢⎥⎣⎦
0 2304 6000 3072 4 10 3.2 10
++×+×
⎢⎥⎣⎦
153412 35451 2304
⎡⎤
5
1534
.
12 354
.
51 2304
354.51 447.32 2928
−
⎡⎤⎢⎥
−−
⎢
⎢
=
⎥⎥
5
2304 2928 7.2 10
−
×
⎢⎣⎦
⎥
Dept. of CE, GCE KannurDr.RajeshKN
56

57. 5. Loads
⎧⎫
{}
0
10
J
A
⎧⎫⎪⎪
=−⎨⎬
⎪⎪
Actual joint loads
1000
⎪⎪
−
⎩⎭
{
}
022
A
⎧⎫
⎪⎪
⎨⎬
Equivalent joint loads
{
}
22 50
E
A
=−
⎨⎬⎪⎪
−
⎩⎭
Equivalent joint loads
0
⎧
⎫
⎪⎪
{
}
{
}
{
}
32
1050
FC J E
AAA
⎪⎪
=+=−
⎨
⎬
⎪
⎪
−
⎩⎭
Hence, combined joint loads
Dept. of CE, GCE KannurDr.RajeshKN
⎩⎭

58. 6. Joint displacements {
}
[
]
{
}
1
FFFFC
DSA
−
=
{
}
[
]
{
}
FFFFC
DSA
1
{}
1
1534.12 354.51 2304 0
354.51 447.32 2928 32
F
D
−
−
⎧
⎫ ⎡⎤
⎪
⎪ ⎢⎥
∴=− − −
⎨
⎬
⎢⎥
⎪⎪
5
2304 2928 7.2 10 1050
⎢⎥
⎪⎪
−×− ⎢⎥⎩⎭ ⎣⎦
0.0206
0.09936
−⎧⎫
⎪⎪
=−
⎨⎬
0.09936
0.001797
⎨⎬⎪⎪
−
⎩⎭
Dept. of CE, GCE KannurDr.RajeshKN

59. 7. Member end actions
{}{}
[
]
[
]
{
}
Mi MLi MiiTGLOBiAL
RD AAS
∴
=+
1000 0 0 1 0 00.0206
⎡⎤⎡⎤
−
⎧⎫⎪⎪
{}
5
0 120 6000 0 1 00.0993 0 6000 4 10 0
6
01
0.001797
MLi
A
⎡⎤⎡⎤ ⎢⎥⎢⎥
=+ −
⎢⎥⎢⎥
−
×
⎢⎥⎢⎥
⎧⎫⎪⎪
−⎨
⎣⎦⎣
⎪⎩
⎦
⎬⎪
−
⎭
0 6000 4 10 0
01
0.001797
⎢⎥⎢⎥ ⎣⎦⎣
⎩
⎦
⎭
8. Reactions
{
}
{
}
[
]
[
]
RRCRFF
A
ASD =− +
Dept. of CE, GCE KannurDr.RajeshKN

60. •Problem 4:
20kN/
6m
2m
2m
40kN
20kN/
m
16kNm
2m
80kN
2m
I
1
=I
3
=
I
2
=
I
2
A
1
=A
3
=
Dept. of CE, GCE KannurDr.RajeshKN
60
A
2
=

61. 1
2
3
1
2
3
4
Ctiit C
onnec
ti
v
it
y:
Member Node 1 Node 2
112
223
3
2
4
Dept. of CE, GCE KannurDr.RajeshKN
61
3
2
4

62. 4
6
5
4
6
Free DOF
Dept. of CE, GCE KannurDr.RajeshKN
62

63. 1. Member stiffness matrices in local co-ordinates
( ith t id i t i t DOF) (
w
ith
ou
t
cons
id
er
i
ng
res
t
ra
i
n
t DOF)
5
00
3.5 10 0 0 12 6
X
EA
L
EI EI
⎡⎤
⎢⎥
⎡
⎤ × ⎢⎥
⎢⎥
⎢⎥
[]
132
12 6
0 0 6562.5 13125
0 13125 35000
64
0
ZZ
M
ZZ
EI EI
S
LL
EI EI
⎢⎥
⎢⎥
=
−= −
⎢
⎥
⎢⎥
⎢
⎥ − ⎢⎥
⎣
⎦
⎢⎥
−
2
0
LL
⎢⎥⎣⎦ ⎡⎤
[
]
5
00
3.5 10 0 0 12 6
X
ZZ
EA
L
EI EI
⎡⎤⎢⎥
⎡
⎤ ×
⎢⎥
⎢⎥
⎢⎥
[
]
232
2
12 6
0 0 3888.89 11666.67
0 11666.67 46666.67
64
0
ZZ
M
ZZ
EI EI
S
LL
EI EI
⎢⎥
⎢⎥
=
=
⎢
⎥
⎢⎥
⎢
⎥
⎢⎥
⎣
⎦
⎢⎥⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
63
2
LL
⎢⎥⎣⎦

64. ⎡⎤
[
]
5
00
3.5 10 0 0 12 6
0 0 6 62 1312
X
ZZ
EA
L
EI EI
S
⎡⎤⎢⎥
⎡
⎤ × ⎢⎥
⎢⎥
⎢⎥
[
]
332
2
12 6
006
5
62
.5
1312
5
0 13125 35000
64
0
ZZ
M
ZZ
EI EI
S
LL
EI EI
⎢⎥
⎢⎥
==
⎢
⎥
⎢⎥
⎢
⎥
⎢⎥
⎣
⎦
⎢⎥⎣⎦
2
LL
⎢⎥⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
64

65. 2. Rotation (transformation) matrices
010
−
⎡⎤
100
⎡⎤
100
⎡⎤
[]
3
90
010 100 001
R
θ
=−
⎡⎤⎢
⎥
=
⎢
⎥
⎢⎥⎣⎦
[]
1
0
100 010 001
R θ
=
⎡⎤⎢
⎥
=
⎢
⎥
⎢⎥⎣⎦
[]
2
0
010
001
R
θ
=
⎡⎤⎢
⎥
=
⎢
⎥
⎢
⎥
⎣⎦
Member 1
001
⎢⎥⎣⎦
Member 2
001
⎢⎥⎣⎦
Member 3
⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
65

66. 3. Member stiffness matrices in global co-ordinates
(Transformed member stiffness matrices)
[][]
[
]
[
]
1111
T
MS T M T
SRSR
=
5
3.5 10 0 0
0 6562.5 13125
⎡⎤×
⎢⎥
=
−
⎢⎥
0 13125 35000
⎢⎥⎢⎥−
⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
66

67. 5
T
⎡⎤
⎡⎤ ⎡⎤
[]
5
2
100 3.510 0 0 100
0 1 0 0 3888.89 11666.67 0 1 0
T
MS
S
⎡⎤
×
⎡⎤ ⎡⎤
⎢⎥ ⎢⎥ ⎢⎥
=
⎢⎥ ⎢⎥ ⎢⎥
⎢⎥
⎢⎥ ⎢⎥
0 0 1 0 11666.67 46666.67 0 0 1
⎢⎥
⎢⎥ ⎢⎥ ⎣⎦ ⎣⎦⎣⎦
5
3.5 10 0 0
0 3888.89 11666.67
⎡⎤×
⎢⎥
=
⎢⎥
0 11666.67 46666.67
⎢⎥⎢⎥
⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
67

68. []
5
3
0103.510 0 0 010
1 0 0 0 6562.5 13125 1 0 0
T
MS
S
⎡
⎤ −× − ⎡⎤ ⎡⎤
⎢⎥ ⎢⎥ ⎢⎥
=
⎢⎥
⎢⎥ ⎢⎥
0 0 1 0 13125 35000 0 0 1
⎢⎥
⎢⎥ ⎢⎥
⎢
⎥ ⎢⎥ ⎢⎥
⎣⎦ ⎣⎦⎣⎦
5
010 0 3.510 0
1 0 0 65625 0 13125
⎡
⎤ −× ⎡⎤
⎢⎥
⎢⎥
⎢⎥
1006562
.
5 0 13125
0 0 1 13125 0 35000
⎢⎥
⎢⎥
=
−
⎢⎥
⎢⎥
⎢
⎥ ⎢⎥⎣⎦
⎣
⎦
6562.5 0 13125 ⎡⎤
⎢⎥
5
03.5100
13125 0 35000 ⎢⎥
=×
⎢⎥
⎢⎥
⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
68
⎣⎦

69. 4
.
A
sse
m
b
l
ed (a
n
d
r
educed)
g
l
oba
l
st
iffn
ess
m
at
ri
x
. sse b ed (a d educed)
goba
st ess at x
55
3.5 10 3.5 10 6562.5 0 0 0 0 0 13125
⎡
⎤ × + × + ++ ++
⎢⎥
[
]
5
0 0 0 6562.5 3888.89 3.5 10 13125 11666.67 0
0 0 13125 13125 11666.67 0 35000 46666.67 35000
FF
S
⎢⎥
=++ ++×−++
⎢
⎥
⎢
⎥ ++ − + + + +
⎣
⎦
7065625 0 13125
⎡⎤
706562
.
5 0 13125
0 360451.39 1458.33
⎡⎤⎢
⎥
=−
⎢⎥
13125 1458.33 116666.67
⎢⎥
−
⎢
⎥
⎣
⎦
Dept. of CE, GCE KannurDr.RajeshKN

70. 60
5. Loads
20
20 kNm60
60
20
20
60
60
40
4040
40
Fixed end actions
Dept. of CE, GCE KannurDr.RajeshKN
70
40
Fixed end actions

71. 80kN
0
+
16
=
16 kNm
01616 kNm
40kN
Combined
(
e
q
uivalent + actual
)
joint loads
Dept. of CE, GCE KannurDr.RajeshKN
71
(q )j

72. 0
⎧⎫
0
⎧⎫
0
⎧⎫
0
⎧
⎫
⎪⎪
00
0
⎧⎫⎪
⎪
⎪
⎪
⎪⎪
0 20
20
⎧⎫⎪
⎪
−
⎪
⎪
−
⎪⎪
020
20
⎧⎫⎪
⎪
−
⎪
⎪
−
⎪⎪
20
20
0
⎪⎪
−
⎪
⎪
−
⎪
⎪
⎪⎪
0
0
⎪⎪⎪
⎪
⎪
⎪
⎪
⎪
⎪⎪
40
80
⎪⎪⎪
⎪
⎪
⎪
⎪
⎪ −
⎪⎪
40
80
⎪⎪⎪
⎪
⎪
⎪
⎪
⎪ −
⎪⎪
{}
0 60
60
RC
A
⎪⎪⎪
⎪
⎪
⎪
=−
⎨
⎬
⎪
⎪
⎪⎪
{}
16
;
0
0
J
A
⎪⎪
−
⎪
⎪
=
⎨
⎬
⎪
⎪
⎪⎪
{}
0
;
0
60
E
A
⎪⎪⎪
⎪
=
⎨
⎬
⎪
⎪
⎪⎪
−
{}{}
{}
16
;
0
60
CJE
AAA
⎪⎪
−
⎪
⎪
=+=
⎨
⎬
⎪
⎪
⎪⎪
−
40
80
⎪⎪⎪
⎪
⎪
⎪
−
⎪
⎪
⎪⎪
00
0
⎪⎪⎪
⎪
⎪
⎪
⎪
⎪
⎪⎪
60
60
40
⎪⎪⎪
⎪
⎪
⎪
⎪
⎪
⎪⎪
40
⎧
⎫
⎪⎪
60
60
40
⎪⎪⎪
⎪
⎪
⎪
⎪
⎪
⎪⎪
40
⎪⎪
−
⎩⎭
80
0
⎪⎪
−
⎪
⎪
⎪
⎪
⎩⎭
0
40
⎪⎪⎪
⎪
⎪
⎪
−⎩⎭
{
}
80
16
FC
A
⎪⎪
=−
⎨
⎬
⎪
⎪
−⎩⎭
80
40
⎪⎪
−
⎪
⎪
⎪
⎪
−⎩⎭
Dept. of CE, GCE KannurDr.RajeshKN
72

73. [
]
[
]
{
}
1
DS A
−
=
6. Jo
in
t d
i
sp
l
ace
m
e
n
ts
1
706562.5 0 13125 40
−
⎡⎤⎧⎫
⎪⎪
[
]
[
]
{
}
F
FF FC
DS A
=
6. Jo t d sp ace e ts
0 360451.39 1458.33 80
13125 1458.33 116666.67 16
⎡⎤⎧⎫
⎪⎪
⎢⎥
=−−⎨⎬
⎢⎥
⎪⎪
−−
⎢⎥⎣⎦⎩⎭⎢⎥⎣⎦⎩⎭
596
0.142 10 0.646 10 0.160 10
40

−−−
⎡⎤
×−×−×
⎧⎫
⎢⎥
957
675
40
0.646 10 0.277 10 0.348 10 80
16
0160 10 0348 10 0859 10

−−− −−−
⎡⎤
⎧⎫
⎢⎥
⎪
⎪
=− × × × −
⎨
⎬ ⎢⎥
⎪
⎪
⎢⎥
−
×××
⎩⎭
⎣⎦
16
0
.
160 10 0
.
348 10 0
.
859 10

⎢⎥
−× × ×
⎩⎭
⎣⎦
4
0.594 10
−
⎧
⎫ ×
⎪⎪
3
3
0.222 10
0.147 10
−
−
⎪⎪
=− ×
⎨
⎬
⎪
⎪
−
×
⎩⎭
Dept. of CE, GCE KannurDr.RajeshKN
73
⎩⎭

74. 7. Member end actions
{
}{ }
[
]
[
]
{
}
Mi MLi MiiTGLOBiAL
RD AAS
∴
=+
Dept. of CE, GCE KannurDr.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 of
assembly 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 KannurDr.RajeshKN
75

76. • Problem X:
2
6
1
3
5
6
4
5
Dept. of CE, GCE KannurDr.RajeshKN
76

77. 00
EA EA
⎡⎤
[
]
1
00
10 10
0000 00001
M
LL
S
⎡⎤
−
⎢⎥−
⎡
⎤
⎢⎥
⎢
⎥
⎢⎥
⎢
⎥
=
=
⎢⎥
⎢⎥
[
]
1
10 1 0 2
00
0000
0000
M
EA EA
LL⎢⎥
−
⎢⎥
−⎢⎥
⎢
⎥
⎣
⎦ ⎢⎥
⎢⎥⎣⎦
0000
⎢⎥⎣⎦ [
]
[
]
[
]
234
SSS
=
==
[
]
[
]
[
]
234
M
MM
SSS
10 10 0000
1
− ⎡⎤
⎢⎥
⎢⎥
10 10 0000
1
−
⎡
⎤
⎢
⎥
⎢⎥
[
]
5
0000
1
10 1 0 2.83
0000
M
S
⎢⎥
=
−⎢⎥
⎢⎥
⎣⎦
[
]
6
0000
1
10 1 0 2.83
0000
M
S
⎢⎥
=
−
⎢
⎥
⎢
⎥
⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
77
⎣⎦
⎣⎦

78. Rotation matrices
cos sin 0 0 0 1 0 0
sin cos 0 0 1 0 0 0
θθ θθ
⎡⎤⎡⎤ ⎢⎥⎢⎥
[
]
1
90
sin cos 0 0 1 0 0 0
00cossin0001
T
R θ
θθ
θθ
=
⎢⎥⎢⎥
−−
⎢⎥⎢⎥ ==
⎢⎥⎢⎥
⎢⎥⎢⎥
00sincos0010
θ
θ
⎢⎥⎢⎥
−
− ⎣⎦⎣⎦
0100
−
⎡
⎤
⎢⎥
1000 0100
⎡⎤⎢⎥
[]
3
90
1000
000 1
T
R
θ
=−
⎢⎥⎢
⎥ =
−
⎢
⎥
⎢⎥
[
]
2
0
0100 0010 0001
T
R θ
=
⎢⎥⎢⎥=
⎢⎥
⎢⎥⎣⎦
0010
⎢⎥⎣
⎦
0001
⎢⎥⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN

79. [
]
1100
11 0 0
0707
R
⎡
⎤
⎢
⎥
−
⎢⎥
=
[
]
10 0 0
0100
R
−⎡⎤
⎢⎥
−
⎢⎥
[
]
5
45
0
.
707
0011
00 11
T
R θ
=
⎢⎥
=
⎢
⎥
⎢
⎥
−
⎣
⎦
[
]
4
180
00 10
000 1
T
R θ
=
⎢⎥
=
− ⎢⎥
⎢⎥
−
⎣⎦⎣⎦
[
]
1100 1100
−
⎡
⎤
⎢
⎥
⎢⎥
[
]
6
45
1100
0.707
001 1
0011
T
R θ
=−
⎢⎥
=
−
⎢
⎥
⎢
⎥
⎣⎦
0011
⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
79

80. [
]
[
]
[
]
[
]
T
SRSR
[
]
[
]
[
]
[
]
1111 MS T M T
SRSR
=
0100 10 100100
1000 0000 10001
T
− ⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
−−
⎢⎥⎢⎥⎢⎥
=
0001 101000012
00 10 000000 10
⎢⎥⎢⎥⎢⎥
=
−
⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥
−− ⎣⎦⎣⎦⎣⎦
0000
⎡⎤
0 100010 1 −−
⎡⎤⎡⎤
010 1 1
0000 2
⎡⎤⎢
⎥
−
⎢
⎥ =
⎢
⎥
⎢⎥
10000000 1
000 10 101 2
⎡⎤⎡⎤ ⎢⎥⎢⎥
⎢⎥⎢⎥ =
−− ⎢⎥⎢⎥
⎢⎥⎢⎥
0101
⎢⎥
−
⎣
⎦
00100000
⎢⎥⎢⎥ ⎣⎦⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN

81. 10 10
−
⎡⎤
[]
2
10 10 0000 1
10 1 0
2
MS
S
−
⎡⎤⎢⎥
⎢⎥ =
−
⎢⎥
10 1 0
2
0000
⎢⎥⎢⎥
⎣⎦
[
]
0100 10100100
1000 000010001
T
S
−−− ⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥
[
]
3
000 1 1010000 12
0010 00000010
MS
S
⎢⎥⎢⎥⎢⎥
=
−
−− ⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥
⎣⎦⎣⎦⎣⎦
01000 101
10000000
1
− ⎡⎤⎡⎤
⎢⎥⎢⎥
0000 010 1
1
⎡
⎤
⎢⎥
10000000
1
0001010 1 200 100000
⎢⎥⎢⎥
−
⎢⎥⎢⎥ =
− ⎢⎥⎢⎥
⎢⎥⎢⎥
⎣⎦⎣⎦
010 1
1
0000 20101
⎢⎥
−
⎢
⎥ =
⎢
⎥
⎢
⎥
⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
81
00 100000
−
⎣⎦⎣⎦
0101
−
⎣⎦

82. 10 10− ⎡⎤
⎢⎥
[]
4
0000 1
10 1 0 2
0000
MS
S
⎢⎥⎢⎥ =
−⎢⎥
⎢⎥
⎣⎦
0000
⎣⎦
1100 10 10 1100
T
−
⎡⎤⎡⎤⎡⎤ ⎢⎥⎢⎥⎢⎥
[]
5
1100 0000 11001
0.707 0.707
0011 1010 00112.83
MS
S
⎡⎤⎡⎤⎡⎤ ⎢⎥⎢⎥⎢⎥
−−
⎢⎥⎢⎥⎢⎥ =
− ⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥
00 11 0000 00 11
⎢⎥⎢⎥⎢⎥
−
− ⎣⎦⎣⎦⎣⎦
110011 11
1100 0 0 0 0
0.1766
00111111
−
−− ⎡⎤⎡ ⎤
⎢⎥⎢ ⎥
⎢⎥⎢ ⎥
=
⎢⎥⎢ ⎥
11 11
11 11
0.1766
1111
−
−
⎡
⎤
⎢
⎥
−
−
⎢
⎥
=
⎢⎥
00111111 0011 0 0 0 0
−
−−
⎢⎥⎢ ⎥ ⎢⎥⎢ ⎥
⎣⎦⎣ ⎦
1111 1111
−
−
⎢⎥⎢
⎥
−−
⎣
⎦
Dept. of CE, GCE KannurDr.RajeshKN

83. 1100 1010 1100
T
⎡⎤⎡⎤⎡⎤
[]
6
1100 1010 1100 1100 0000 11001
0.707 0.707
001 1 1010 001 1
2.83
MS
S
−
−−
⎡⎤⎡⎤⎡⎤ ⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥ =
−
−−
⎢⎥⎢⎥⎢⎥
001 1 1010 001 1
2.83
0011 0000 0011
⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥
⎣⎦⎣⎦⎣⎦
1111
11 1 1
−−
⎡
⎤
⎢⎥
11001 1 11
11000000
−− ⎡⎤⎡ ⎤
⎢⎥⎢ ⎥
11 1 1
0.177
11 1 1
1111
⎢⎥
−
−
⎢
⎥ =
−
−
⎢
⎥
⎢
⎥
−
−
⎣⎦
11000000
0.177
0011 11 1 1
00110000
⎢⎥⎢ ⎥
−
⎢⎥⎢ ⎥ =
−
− ⎢⎥⎢ ⎥
⎢⎥⎢ ⎥
−
⎣⎦⎣ ⎦
1111
⎣⎦
00110000
⎣⎦⎣ ⎦
Dept. of CE, GCE KannurDr.RajeshKN

84. Assembled
g
lobal stiffness matrix
11
0 0.1766 0 0 0.1766 0 0 0.1766 0.1766 0
22
11
++ ++ − − −
⎡⎤
⎢⎥
⎢⎥
⎢⎥
g
11
0 0 0.1766 0 0.1766 0 0.1766 0.1766 0 0
22
11
0 0 0 0.1766 0 0 0.1766 0 0.1766 0.1766
22
11
++ ++ − − −
++ +− − −
⎢⎥ ⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎢⎥
[]
11
0 0 0 0.1766 0 0.1766 0 0 0.1766 0.1766
22
11
0.1766 0.1766 0 0 0.1766 0 0 0.
22
J
S
−+−++ −
=
−− − ++++1766 0 0
11
0 1766 0 1766 0 0 0 0 0 1766 0 0 1766 0
⎢⎥ ⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎢⎥
11
0
.
1766 0
.
1766 0 0 0 0 0
.
1766 0 0
.
1766 0
22
11
0 0.1766 0.1766 0 0 0 0.1766 0 0 0.1766
22
11
0 0 0 1766 0 1766 0 0 0 0 1766 0 0 1766
⎢⎥
−− ++++ −
⎢⎥
⎢⎥
⎢− − ++ +− ⎥
⎢⎥
⎢⎥
+++
⎢⎥
000
.
1766 0
.
1766 0 0 0 0
.
1766 0 0
.
1766
22
−
−
+
−
++
⎢⎥ ⎣⎦
128
0
DDD
=
==
Boundary conditions are:
Dept. of CE, GCE KannurDr.RajeshKN
128
0
DDD
Boundary conditions are:

85. 0.677 -0.177 -0.500 0.000 -0.177 ⎡⎤
⎢⎥
Reduced stiffness matrix
[
]
-0.177 0.677 0.000 0.000 0.177
-0.500 0.000 0.677 0.177 0.000
FF
S=
⎢⎥⎢⎥
⎢⎥
⎢⎥
[
]
0.000 0.000 0.177 0.677 0.000
-0.177 0.177 0.000 0.000 0.677
⎢⎥⎢⎥
⎢⎥
⎣⎦⎣⎦
Dept. of CE, GCE KannurDr.RajeshKN
85

86. 5
⎧
⎫
⎪⎪
{}
0
0
FC
A
⎪⎪⎪
⎪
⎪
⎪
=
⎨
⎬
⎪⎪
In this problem, the reduced load vector
(in global coords) can be directly written as
0
0
⎪⎪⎪
⎪
⎪
⎪⎩⎭
{}
[
]
{}
1
F
FF FC
D
SA
−
=
24.144
⎧
⎫
⎪⎪
4.829 1.000 3. 829 -1.000 1.000 5 ⎡⎤⎧⎫
⎢⎥⎪⎪
5.000
19.144
⎪⎪⎪
⎪
⎪
⎪
⎨
⎬
⎪
=
⎪
1.000 1.793 0. 793 -0.207 -0.207
3.829 0.793 4. 622 -1.207 0.793 =
0
0
⎢⎥⎪⎪ ⎢⎥⎪⎪
⎪
⎪
⎢⎥
⎨
⎬
⎢⎥
⎪⎪
-5.000
5.000
⎪⎪
⎪
⎪⎪⎪
⎩⎭
-1.000 -0.207 -1. 207 1.793 -0.207 0
1.000 -0.207 0. 793 -0.207 1.793 0
⎢⎥
⎪⎪
⎢⎥
⎪
⎪
⎢⎥
⎪
⎪
⎣⎦⎩⎭ Dept. of CE, GCE KannurDr.RajeshKN ⎣⎦⎩⎭

87. Assignment
Dept. of CE, GCE KannurDr.RajeshKN