1. Dept. of CE, GCE Kannur Dr.RajeshKN
1
Structural Analysis - III
Dr. Rajesh K. N.
Asst. Professor in Civil Engineering
Govt. College of Engineering, Kannur
Stiffness Method
2. Dept. of CE, GCE Kannur Dr.RajeshKN
2
• Development of stiffness matrices by physical approach –
stiffness matrices for truss,beam and frame elements –
displacement transformation matrix – development of total
stiffness matrix - analysis of simple structures – plane
truss beam and plane frame- nodal loads and element
loads – lack of fit and temperature effects.
Stiffness method
Module II
3. Dept. of CE, GCE Kannur Dr.RajeshKN
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• Displacement components are the primary unknowns
• Number of unknowns is equal to the kinematic indeterminacy
• Redundants are the joint displacements, which are
automatically specified
• Choice of redundants is unique
• Conducive to computer programming
FUNDAMENTALS OF STIFFNESS METHOD
Introduction
4. Dept. of CE, GCE Kannur Dr.RajeshKN
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• Stiffness method (displacements of the joints are the primary
unknowns): kinematic indeterminacy
• kinematic indeterminacy
• joints: a) where members meet, b) supports, c) free ends
• joints undergo translations or rotations
• in some cases joint displacements will be known, from the
restraint conditions
• the unknown joint displacements are the kinematically
indeterminate quantities
o degree of kinematic indeterminacy: number of degrees
of freedom
5. Dept. of CE, GCE Kannur Dr.RajeshKN
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• in a truss, the joint rotation is not regarded as a degree
of freedom. joint rotations do not have any physical
significance as they have no effects in the members of the
truss
• in a frame, degrees of freedom due to axial
deformations can be neglected
6. Dept. of CE, GCE Kannur Dr.RajeshKN
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•Example 1: Stiffness and flexibility coefficients of a beam
Stiffness coefficients
Unit displacement
Unit action
Forces due to unit dispts
– stiffness coefficients
11 21 12 22, , ,S S S S Dispts due to unit forces
– flexibility coefficients
11 21 12 22, , ,F F F F
7. Dept. of CE, GCE Kannur Dr.RajeshKN
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•Example 2: Action-displacement equations for a beam subjected to
several loads
1 11 12 13A A A A= + +
1 11 1 12 2 13 3A S D S D S D= + +
2 21 1 22 2 23 3A S D S D S D= + +
3 31 1 32 2 33 3A S D S D S D= + +
8. Dept. of CE, GCE Kannur Dr.RajeshKN
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1 11 1 12 2 13 3 1
2 21 1 22 2 23 3 2
1 1 2 2 3 3
...
...
............................................................
...
n n
n n
n n n n nn n
A S D S D S D S D
A S D S D S D S D
A S D S D S D S D
= + + + +
= + + + +
= + + + +
1 11 12 1 1
2 21 22 2 2
1 2
11
...
...
... ... ... ... ......
...
n
n
n n nn nn
n n nn
A S S S D
A S S S D
S S S DA
× ××
⎧ ⎫ ⎧ ⎫⎧ ⎫
⎪ ⎪ ⎪ ⎪⎪ ⎪
⎪ ⎪ ⎪ ⎪⎪ ⎪
=⎨ ⎬ ⎨ ⎬⎨ ⎬
⎪ ⎪ ⎪ ⎪⎪ ⎪
⎪ ⎪ ⎪ ⎪⎪ ⎪⎩ ⎭⎩ ⎭⎩ ⎭
A = SD
•Action matrix, Stiffness matrix, Displacement matrix
•Stiffness coefficient ijS
{ } [ ] { } [ ] [ ]
1 1
A F D S F
− −
= ⇒ =
oIn matrix form,
{ } [ ]{ }A S D=
[ ] [ ]
1
F S
−
=
Stiffness matrix
9. Dept. of CE, GCE Kannur Dr.RajeshKN
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Example: Propped cantilever (Kinematically indeterminate to
first degree)
Stiffness method (Direct approach: Explanation using principle
of superposition)
•degrees of freedom: one
• Kinematically determinate structure is obtained by restraining
all displacements (all displacement components made zero -
restrained structure)
• Required to get Bθ
10. Dept. of CE, GCE Kannur Dr.RajeshKN
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Restraint at B causes a reaction of MB as shown.
Hence it is required to induce a rotation of Bθ
The actual rotation at B is Bθ
2
12
B
wL
M = −
(Note the sign convention:
anticlockwise positive)
•Apply unit rotation corresponding to Bθ
Let the moment required for this unit rotation be Bm
4
B
EI
m
L
= anticlockwise
11. Dept. of CE, GCE Kannur Dr.RajeshKN
• Moment required to induce a rotation of Bθ B Bm θis
2
4
0
12
B
wL EI
L
θ− + =
3
48
B
B
B
wL
EI
M
m
θ∴ = − =
Bm (Moment required for unit rotation) is the stiffness
coefficient here.
0B B BM m θ+ = (Joint equilibrium equation)
12. Dept. of CE, GCE Kannur Dr.RajeshKN
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Stiffnesses of prismatic members
Stiffness coefficients of a structure are calculated from the
contributions of individual members
Hence it is worthwhile to construct member stiffness
matrices
[ ] [ ] 1
Mi MiS F
−
=
13. Dept. of CE, GCE Kannur Dr.RajeshKN
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Member stiffness matrix for prismatic beam member with
rotations at the ends as degrees of freedom
[ ]
2 12
1 2
Mi
EI
S
L
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
[ ] [ ]
1
1
1 2 13 6
1 26
6 3
Mi Mi
L L
LEI EI
S F
L L EI
EI EI
−
−
−
−⎡ ⎤
⎢ ⎥ −⎛ ⎞⎡ ⎤
= = =⎢ ⎥ ⎜ ⎟⎢ ⎥− −⎣ ⎦⎝ ⎠⎢ ⎥
⎢ ⎥⎣ ⎦
2 1 2 16 2
1 2 1 2(3)
EI EI
L L
⎡ ⎤ ⎡ ⎤
= =⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
Verification:
B
A
1
14. Dept. of CE, GCE Kannur Dr.RajeshKN
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[ ] 11 12
21 22
3 2
2
12 6
6 4
M M
Mi
M M
S S
S
S S
EI EI
L L
EI EI
L L
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
⎡ ⎤
−⎢ ⎥
= ⎢ ⎥
⎢ ⎥−
⎢ ⎥⎣ ⎦
Member stiffness matrix for prismatic beam member with
deflection and rotation at one end as degrees of freedom
15. Dept. of CE, GCE Kannur Dr.RajeshKN
[ ] [ ]
13 2
1
2
1
2
2 33 2
6 3 6
2
Mi Mi
L L
L LLEI EI
S F
EI LL L
EI EI
−
−
−
⎡ ⎤
⎢ ⎥ ⎛ ⎞⎡ ⎤
= = =⎢ ⎥ ⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥ ⎣ ⎦⎝ ⎠
⎢ ⎥⎣ ⎦
( )
2
2
3
2
2
6 36
3
12 6
6 423
EI EI
L L
EI EI
L
LEI
L LL L
L
−⎡ ⎤
= =⎢ ⎥−⎣
⎡ ⎤
−⎢ ⎥
⎢ ⎥
⎢
⎢
⎦ ⎥−
⎥⎣ ⎦
Verification:
[ ]Mi
EA
S
L
=
•Truss member
16. Dept. of CE, GCE Kannur Dr.RajeshKN
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•Plane frame member
[ ]
11 12 13
21 22 23
31 32 33
3 2
2
0 0
12 6
0
6 4
0
M M M
Mi M M M
M M M
S S S
S S S S
S S S
EA
L
EI EI
L L
EI EI
L L
⎡ ⎤
⎢ ⎥=
⎢ ⎥
⎢ ⎥⎣ ⎦
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥= −
⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎣ ⎦
17. Dept. of CE, GCE Kannur Dr.RajeshKN
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•Grid member
[ ]
3 2
2
12 6
0
0 0
6 4
0
Mi
EI EI
L L
GJ
S
L
EI EI
L L
⎡ ⎤
−⎢ ⎥
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎣ ⎦
18. Dept. of CE, GCE Kannur Dr.RajeshKN
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•Space frame member
[ ]
3 2
3 2
2
2
0 0 0 0 0
12 6
0 0 0 0
12 6
0 0 0 0
0 0 0 0 0
6 4
0 0 0 0
6 4
0 0 0 0
Z Z
Y Y
Mi
Y Y
Z Z
EA
L
EI EI
L L
EI EI
L LS
GJ
L
EI EI
L L
EI EI
L L
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥⎣ ⎦
19. Dept. of CE, GCE Kannur Dr.RajeshKN
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(Explanation using principle of complimentary virtual work)
Formalization of the Stiffness method
{ } [ ]{ }Mi Mi MiA S D=
Here{ }MiD contains relative displacements of the k end with respect
to j end of the i-th member
If there are m members in the structure,
{ }
{ }
{ }
{ }
{ }
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]
{ }
{ }
{ }
{ }
{ }
11 1
22 2
33 3
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
MM M
MM M
MM M
MiMi Mi
MmMm Mm
SA D
SA D
SA D
SA D
SA D
⎧ ⎫ ⎧ ⎫⎡ ⎤
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
=⎨ ⎬ ⎨ ⎬⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎩ ⎭ ⎩ ⎭⎣ ⎦
M M MA = S D
20. Dept. of CE, GCE Kannur Dr.RajeshKN
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{ } [ ]{ }M M MA S D=
[ ]MS is the unassembled stiffness matrix of the entire structure
21. Dept. of CE, GCE Kannur Dr.RajeshKN
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• Relative end-displacements in will be related to a vector of joint
displacements for the whole structure,
{ }MD
{ }JD
• If there are no support displacements specified,
{ }RD will be a null matrix
• Hence, { } [ ]{ } [ ] [ ]
{ }
{ }
F
M MJ J MF MR
R
D
D C D C C
D
⎧ ⎫
= = ⎡ ⎤ ⎨ ⎬⎣ ⎦
⎩ ⎭
{ }JD
{ }FDfree (unknown) joint displacements
{ }RDand restraint displacements
consists of:
{ } [ ]{ }M MJ JD C D=
displacement transformation matrix (compatibility matrix)[ ]MJC
22. Dept. of CE, GCE Kannur Dr.RajeshKN
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• Elements in displacement transformation matrix
(compatibility matrix) [ ]MJC are found from compatibility conditions.
•Each column in the submatrix consists of member
displacements caused by a unit value of a support displacement
applied to the restrained structure.
[ ]MRC
[ ]MFC• Each column in the submatrix consists of member
displacements caused by a unit value of an unknown displacement
applied to the restrained structure.
{ }MDrelate to respectively
[ ]MFC
[ ]MRC
and
{ }FD
{ }RD
and
23. Dept. of CE, GCE Kannur Dr.RajeshKN
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{ }MDδ
{ } [ ]{ } [ ] [ ]
{ }
{ }
F
M MJ J MF MR
R
D
D C D C C
D
δ
δ δ
δ
⎧ ⎫
= = ⎡ ⎤ ⎨ ⎬⎣ ⎦
⎩ ⎭
• Suppose an arbitrary set of virtual displacements
is applied on the structure.
{ } { } { } { }
T T T F
J J F R
R
D
W A D A A
D
δ
δ δ δ
δ
⎧ ⎫⎡ ⎤= = ⎨ ⎬⎣ ⎦ ⎩ ⎭
{ }JDδ• External virtual work produced by the virtual displacements
{ }JAand real loads is
24. Dept. of CE, GCE Kannur Dr.RajeshKN
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{ } { }
T
M MU A Dδ δ=
• Internal virtual work produced by the virtual (relative) end
displacements { }MDδ { }MAand actual member end actions is
{ } { } { } { }
T T
J J M MA D A Dδ δ=
• Equating the above two (principle of virtual work),
{ } [ ]{ }M MJ JD C D= { } [ ]{ }M M MA S D=But and
{ } [ ]{ }M MJ JD C Dδ δ=Also,
{ } { } { } [ ] [ ] [ ]{ }
TT T T
J J J MJ M MJ JA D D C S C Dδ δ=Hence,
{ } [ ]{ }J J JA S D=
25. Dept. of CE, GCE Kannur Dr.RajeshKN
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[ ] [ ] [ ][ ]T
J MJ M MJS C S C=Where, , the assembled stiffness matrix for the
entire structure.
• It is useful to partition into submatrices pertaining to free
(unknown) joint displacements
[ ]JS
{ }FD { }RDand restraint displacements
{ } [ ]{ }
{ }
{ }
[ ] [ ]
[ ] [ ]
{ }
{ }
FF FRF F
J J J
RF RRR R
S SA D
A S D
S SA D
⎧ ⎫ ⎧ ⎫⎡ ⎤
= ⇒ =⎨ ⎬ ⎨ ⎬⎢ ⎥
⎩ ⎭ ⎩ ⎭⎣ ⎦
[ ] [ ] [ ][ ]T
FF MF M MFS C S C= [ ] [ ] [ ][ ]T
FR MF M MRS C S C=
[ ] [ ] [ ][ ]T
RF MR M MFS C S C= [ ] [ ] [ ][ ]T
RR MR M MRS C S C=
Where,
26. Dept. of CE, GCE Kannur Dr.RajeshKN
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{ } [ ]{ } [ ]{ }F FF F FR RA S D S D= + { } [ ]{ } [ ]{ }R RF F RR RA S D S D= +
{ } [ ] { } [ ]{ }
1
F FF F FR RD S A S D
−
⇒ = −⎡ ⎤⎣ ⎦
{ } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + +
• Support reactions
{ }RCA
represents combined joint loads (actual and equivalent) applied
directly to the supports.
If actual or equivalent joint loads are applied directly to the supports,
Joint displacements
27. Dept. of CE, GCE Kannur Dr.RajeshKN
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{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
• Member end actions are obtained adding member end actions
calculated as above and initial fixed-end actions
{ } { } [ ][ ]{ }M ML M MJ JA A S C D= +i.e.,
{ }MLAwhere represents fixed end actions
28. Dept. of CE, GCE Kannur Dr.RajeshKN
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Important formulae:
Joint displacements:
Member end actions:
Support reactions:
{ } [ ] { } [ ]{ }
1
F FF F FR RD S A S D
−
= ⎡ − ⎤⎣ ⎦
{ } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + +
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
29. Dept. of CE, GCE Kannur Dr.RajeshKN
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•Problem 1
[ ]
4 2 0 0
2 4 0 02
0 0 2 1
0 0 1 2
M
EI
S
L
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
Unassembled stiffness matrix
[ ]
2 12
1 2
Mi
EI
S
L
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
Member stiffness matrix of beam member
Kinematic indeterminacy = 2
30. Dept. of CE, GCE Kannur Dr.RajeshKN
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Fixed end actions
Equivalent joint loads
32. Dept. of CE, GCE Kannur Dr.RajeshKN
Joint displacements
Free (unknown) joint displacements { }FD { }RDRestraint displacements
•Each column in the submatrix consists of member
displacements caused by a unit value of an unknown displacement
applied to the restrained structure.
[ ]MFC
•Each column in the submatrix consists of member
displacements caused by a unit value of a support displacement
applied to the restrained structure.
[ ]MRC
40. Dept. of CE, GCE Kannur Dr.RajeshKN
2
2
2
6 0
2 0 02
3
3 3 118
3
3 3
P
PL
LEI PL
L EI
P
P
−⎧ ⎫
⎡ ⎤⎪ ⎪− ⎢ ⎥⎪ ⎪ ⎧ ⎫⎢ ⎥= − +⎨ ⎬ ⎨ ⎬
−⎢ ⎥ ⎩ ⎭⎪ ⎪− ⎢ ⎥⎪ ⎪ − −⎣ ⎦−⎩ ⎭
2
2
0
2 2
0 0
02
3 3
318 33 3
3
3
P P
PL PL
PL EI P
EI L
P P
P
P P
⎧ ⎫
−⎧ ⎫ ⎧ ⎫ ⎪ ⎪⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪
= − + = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎩ ⎭− −⎩ ⎭ ⎩ ⎭ ⎪ ⎪
⎩ ⎭
{ } { } [ ]{ }R RC RF FA A S D∴ = − +
2
3
10
3
2
3
P
PL
P
P
⎧ ⎫
⎪ ⎪
⎪ ⎪
⎪ ⎪⎪ ⎪
⎨ ⎬
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪⎩ ⎭
=
41. Dept. of CE, GCE Kannur Dr.RajeshKN
41
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
{ }
2
3 4 2 0 0 0 0
3 2 4 0 0 1 0 02
2 0 0 2 1 1 0 19 18
2 0 0 1 2 0 1
M
PL EI PL
A
L EI
⎧ ⎫ ⎡ ⎤ ⎡ ⎤
⎪ ⎪ ⎢ ⎥ ⎢ ⎥− ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎪ ⎪
⎢ ⎥ ⎢ ⎥⎪ ⎪−⎩ ⎭ ⎣ ⎦ ⎣ ⎦
2
3 4 2 0 0 0
3 2 4 0 0 02
2 0 0 2 1 09 18
2 0 0 1 2 1
PL EI PL
L EI
⎧ ⎫ ⎧ ⎫⎡ ⎤
⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥= +⎨ ⎬ ⎨ ⎬
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢ ⎥⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭⎣ ⎦
3 0
3 0
2 19 9
2 2
PL PL
⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪
= +⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭
1
1
13
0
PL
⎧ ⎫
⎪ ⎪−⎪ ⎪
⎨ ⎬
⎪
⎪
=
⎪
⎪⎩ ⎭
is a null matrix{ }RD
Member end actions
{ } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +
42. Dept. of CE, GCE Kannur Dr.RajeshKN
42
3
0 0
0 0 0 2 0 6 4 6 0
1 1 0 0 4 0 6 2 6 02
0 0 0 2 0 0 3 3
1 1 1 1 2 0 0 3 3
0 0 1 1
L L
L L L
L LEI
L L LL
L L
⎡ ⎤
⎢ ⎥ −⎡ ⎤⎢ ⎥
⎢ ⎥−⎢ ⎥
⎢ ⎥= ⎢ ⎥
−⎢ ⎥⎢ ⎥
⎢ ⎥⎢ ⎥− − −⎣ ⎦
⎢ ⎥
− −⎣ ⎦
[ ] [ ]
[ ] [ ]
2 2 2
2 2
3 2
6 6 2 3 3
2 0 0 3 3
6 0 12 6 12 02
2 0 6 4 6 0
3 3 12 6 18 6
3 3 0 0 6 6
FF FR
RF RR
L L L L L L
L L L L
S SL LEI
S SL L L L L
L L L
L L
⎡ ⎤− −
⎢ ⎥
−⎢ ⎥
⎢ ⎥ ⎡ ⎤−
= =⎢ ⎥ ⎢ ⎥
− ⎣ ⎦⎢ ⎥
⎢ ⎥− − − −
⎢ ⎥
− − −⎣ ⎦
[ ] [ ] [ ][ ]T
J MJ M MJS C S C=
0 0
0 0 0 4 2 0 0 0 0 1 1 0
1 1 0 0 2 4 0 0 0 1 0 1 01 2 1
0 0 0 0 0 2 1 0 0 0 1 1
1 1 1 1 0 0 1 2 0 0 0 1 1
0 0 1 1
L L
L L
LEI
L LL L L
L
⎡ ⎤
⎢ ⎥ −⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎣ ⎦
⎢ ⎥
− −⎣ ⎦
Alternatively, if the entire [SJ] matrix is assembled at a time,
43. Dept. of CE, GCE Kannur Dr.RajeshKN
43
•Problem 2:
A
B C D
40 kN10 kN/m
2 m
4 m 4 m 2 m
100 kN
Kinematic indeterminacy = 3 (Not considering joint D in the
overhanging portion)
Degrees of freedom
A
B C D
DF2DF1 DF3
44. Dept. of CE, GCE Kannur Dr.RajeshKN
44
[ ]
2 1 0 0
1 2 0 02
0 0 2 14
0 0 1 2
M
EI
S
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
Unassembled stiffness matrix
[ ]
2 12
1 24
Mi
EI
S
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
Member stiffness matrix of beam member
45. Dept. of CE, GCE Kannur Dr.RajeshKN
45
Fixed end actions
Equivalent joint loads + actual joint loads
A
B
13.33 kNm 13.33
20 kN
20 kN
B
C
20 20
20 kN
20 kN
A
B
13.33 kNm 6.67
20 kN 20 +20 = 40 kN 20 +100 = 120 kN
2x100 – 20 = 180 kNm
46. Dept. of CE, GCE Kannur Dr.RajeshKN
1 0 0 0
0 1 1 0
DF1 =1
DF2 =1 L
50. Dept. of CE, GCE Kannur Dr.RajeshKN
50
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
{ }
43.435
60.33
210.6
13.33 2 1 0 0 1 0 0
13.33 1 2 0 0 0 1 02 1
20 0 0 2 1 0 1 04
20 0 0 1 2 0 0 1
M
EI
A
EI
⎧ ⎫ ⎡ ⎤ ⎡ ⎤
⎪ ⎪ ⎢ ⎥ ⎢ ⎥−⎪ ⎪ ⎢ ⎥ ⎢ ⎥= +⎨ ⎬
⎢ ⎥ ⎢ ⎥⎪ ⎪
⎢ ⎥ ⎢ ⎥⎪ ⎪−
−⎧ ⎫
⎪ ⎪
⎨ ⎬
⎪ ⎪−
⎣
⎩
⎭ ⎦ ⎣ ⎦
⎭
⎩
0
25
25
200
kNm
⎧ ⎫
⎪ ⎪
⎪ ⎪
⎨ ⎬
−⎪ ⎪
⎪ ⎪−⎩ ⎭
=
is a null matrix{ }RD
Member end actions
{ } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +
51. Dept. of CE, GCE Kannur Dr.RajeshKN
•Problem 3
Analyse the beam. Support B has a downward settlement of 30mm.
EI=5.6×103 kNm2
[ ]
4 2 0 0 0 0
2 4 0 0 0 0
0 0 2 1 0 02
0 0 1 2 0 06
0 0 0 0 4 2
0 0 0 0 2 4
M
EI
S
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
Unassembled stiffness matrix
[ ]
2 12
1 2Mi
EI
S
L
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
Member stiffness matrix of beam member
52. Dept. of CE, GCE Kannur Dr.RajeshKN
Fixed end moments
2525
Equivalent joint loads
2525
Support settlements { }RD(Restraint displacements)
A
B
C D30mm 1RD
Free (unknown) joint displacements { }FD
1FD 2FD
3FD
53. Dept. of CE, GCE Kannur Dr.RajeshKN
53
BA C D
1 1FD =
0 1 1 0 0 0
and consist of member displacements due to unit
displacements on the restrained structure.
[ ]MFC [ ]MRC
75. Dept. of CE, GCE Kannur Dr.RajeshKN
75
2 1 0 0 0 0 0 0 1
1 2 0 0 0 0 1 0 1
0 1 1 0 0 0
0 0 2 1 0 0 1 0 02
0 0 0 1 1 0
0 0 1 2 0 0 0 1 0
1 1 0 0 1 1
0 0 0 0 2 1 0 1 1
0 0 0 0 1 2 0 0 1
L
L
EI
L
L L L L
L
L
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎡ ⎤
⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥
⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
[ ] [ ] [ ][ ]T
FF MF M MFS C S C=
1 0 3
2 0 3
0 1 1 0 0 0
2 1 02
0 0 0 1 1 0
1 2 0
1 1 0 0 1 1
0 2 3
0 1 3
L
L
EI
L
L L L L
L
L
⎡ ⎤
⎢ ⎥
⎢ ⎥⎡ ⎤
⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥
⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥
⎢ ⎥
⎣ ⎦
2
4 1 3
2
1 4 3
3 3 12
L
EI
L
L
L L L
⎡ ⎤
⎢ ⎥=
⎢ ⎥
⎢ ⎥⎣ ⎦
76. Dept. of CE, GCE Kannur Dr.RajeshKN
76
{ } [ ] { } [ ]{ }
1
F FF F FR RD S A S D
−
= −⎡ ⎤⎣ ⎦
{ }
1
2
4 1 3 0
2
1 4 3 0
3 3 12
F
L
EI
D L
L
L L L P
−
⎛ ⎞ ⎧ ⎫⎡ ⎤
⎪ ⎪⎜ ⎟⎢ ⎥= ⎨ ⎬⎜ ⎟⎢ ⎥
⎪ ⎪⎜ ⎟⎢ ⎥ ⎩ ⎭⎣ ⎦⎝ ⎠
{ } [ ] { }
1
F FF FD S A
−
= , since there are no support displacements.
2
13 3 3 0
3 13 3 0
84
3 3 5
L
L
L
EI
L L L P
− − ⎧ ⎫⎡ ⎤
⎪ ⎪⎢ ⎥= − − ⎨ ⎬⎢ ⎥
⎪ ⎪− −⎢ ⎥ ⎩ ⎭⎣ ⎦
2
3
3
84
5
PL
EI
L
−⎧ ⎫
⎪ ⎪
−⎨ ⎬
⎪
⎩
=
⎪
⎭
Joint displacements
77. Dept. of CE, GCE Kannur Dr.RajeshKN
77
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
{ }
2
2 1 0 0 0 0 0 0 1
1 2 0 0 0 0 1 0 1
3
0 0 2 1 0 0 1 0 02
3
0 0 1 2 0 0 0 1 0 84
5
0 0 0 0 2 1 0 1 1
0 0 0 0 1 2 0 0 1
M
L
L
EI PL
A
L EI
L
L
L
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ −⎧ ⎫
⎢ ⎥ ⎢ ⎥ ⎪ ⎪
= −⎨ ⎬⎢ ⎥ ⎢ ⎥
⎪ ⎪⎢ ⎥ ⎢ ⎥
⎩ ⎭⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
2
2 1 0 0 0 0 5
1 2 0 0 0 0 2
0 0 2 1 0 0 32
0 0 1 2 0 0 384
0 0 0 0 2 1 2
0 0 0 0 1 2 5
EI PL
L EI
⎧ ⎫⎡ ⎤
⎪ ⎪⎢ ⎥
⎪ ⎪⎢ ⎥
−⎪ ⎪⎢ ⎥
= ⎨ ⎬⎢ ⎥
−⎪ ⎪⎢ ⎥
⎪ ⎪⎢ ⎥
⎪ ⎪⎢ ⎥
⎩ ⎭⎣ ⎦
{ } [ ][ ]{ }M M MF FA S C D=
2
12
9
92
984
9
12
EI PL
L EI
⎧ ⎫
⎪ ⎪
⎪ ⎪
−⎪ ⎪
= ⎨ ⎬
−⎪ ⎪
⎪ ⎪
⎪ ⎪
⎩ ⎭
4
3
3
314
3
4
PL
⎧ ⎫
⎪ ⎪
⎪ ⎪
−
=
⎪ ⎪
⎨ ⎬
−⎪ ⎪
⎪ ⎪
⎪ ⎪
⎩ ⎭
Member end actions
78. Dept. of CE, GCE Kannur Dr.RajeshKN
78
[ ]
0.2 0.0 0.0
0.0 0.2 0.0
0.0 0.0 0.2
MS
⎡ ⎤
⎢ ⎥=
⎢ ⎥
⎢ ⎥⎣ ⎦
Unassembled stiffness matrix
•Problem 6:
1
2
3
50 kN
80 kN
5 m
5 m
5 m
4 m 4 m
3 m
3 m
88. Dept. of CE, GCE Kannur Dr.RajeshKN
88
{ } [ ] { } [ ]{ }
1
F FF F FR RD S A S D
−
= −⎡ ⎤⎣ ⎦
{ }
0.577 -0.155 5
-0.155 1.732 0
FD
⎡ ⎤ ⎧ ⎫
= ⎨ ⎬⎢ ⎥
⎣ ⎦ ⎩ ⎭
{ } [ ] { }
1
F FF FD S A
−
= , since there are no support displacements.
2.887
-0.773
⎧ ⎫
⎨
⎩
= ⎬
⎭
Joint displacements
[ ] [ ] [ ][ ]T
FF MF M MFS C S C= 1.774 0.158
0.158 0.591
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
89. Dept. of CE, GCE Kannur Dr.RajeshKN
{ }
0.866 0.000 0.000 0.866 0.500
2.887
0.000 1.000 0.000 1.000 0.000
-0.773
0.000 0.000 0.500 0.500 -0.866
MA
⎡ ⎤⎡ ⎤
⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥⎩ ⎭
⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
{ } [ ][ ]{ }M M MF FA S C D=
1.830
2.887
1.057
⎧ ⎫
⎪ ⎪
⎨ ⎬
⎪ ⎪
⎩ ⎭
=
Member Forces:
90. Dept. of CE, GCE Kannur Dr.RajeshKN
90
•Homework 4:
AE is constant.
20 kN
1 2 3
600
450
A B C
D
1m
91. Dept. of CE, GCE Kannur Dr.RajeshKN
91
• Development of stiffness matrices by physical approach –
stiffness matrices for truss, beam and frame elements –
displacement transformation matrix – development of total
stiffness matrix - analysis of simple structures – plane truss
beam and plane frame- nodal loads and element loads – lack of
fit and temperature effects.
Stiffness method
Summary
