Upload
shamjith-km
View
82
Download
7
Tags:
Embed Size (px)
DESCRIPTION
GCE Kannur
Citation preview
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
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
Dept. of CE, GCE Kannur Dr.RajeshKN3
• 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
Dept. of CE, GCE Kannur Dr.RajeshKN4
• 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 kinematicallyindeterminate quantities
o degree of kinematic indeterminacy: number of degrees of freedom
Dept. of CE, GCE Kannur Dr.RajeshKN
5
• 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
Dept. of CE, GCE Kannur Dr.RajeshKN6
•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
Dept. of CE, GCE Kannur Dr.RajeshKN
7
•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= + +
Dept. of CE, GCE Kannur Dr.RajeshKN
8
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 DA 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
1 211
...
...... ... ... ... ......
...
n
n
n n nn nnn n nn
A S S S DA S S S D
S S S DA× ××
⎧ ⎫ ⎧ ⎫⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪=⎨ ⎬ ⎨ ⎬⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪⎩ ⎭⎩ ⎭⎩ ⎭
A = SD
•Action matrix, Stiffness matrix, Displacement matrix
•Stiffness coefficient ijS
{ } [ ] { } [ ] [ ]1 1A F D S F− −= ⇒ =
oIn matrix form,
{ } [ ]{ }A S D=
[ ] [ ] 1F S −=
Stiffness matrix
Dept. of CE, GCE Kannur Dr.RajeshKN
9
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
Dept. of CE, GCE Kannur Dr.RajeshKN
10
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
12BwLM = −
(Note the sign convention: anticlockwise positive)
•Apply unit rotation corresponding to Bθ
Let the moment required for this unit rotation be Bm
4B
EImL
= anticlockwise
Dept. of CE, GCE Kannur Dr.RajeshKN
• Moment required to induce a rotation of Bθ B Bm θis
2 4 012 BwL EI
Lθ− + =
3
48B
BB
wLEI
Mm
θ∴ = − =
Bm (Moment required for unit rotation) is the stiffness coefficient here.
0B B BM m θ+ = (Joint equilibrium equation)
Dept. of CE, GCE Kannur Dr.RajeshKN
12
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
[ ] [ ] 1Mi MiS F −=
Dept. of CE, GCE Kannur Dr.RajeshKN
13
Member stiffness matrix for prismatic beam member with rotations at the ends as degrees of freedom
[ ] 2 121 2Mi
EISL
⎡ ⎤= ⎢ ⎥
⎣ ⎦
[ ] [ ]
1
11 2 13 6
1 266 3
Mi Mi
L LLEI EIS F
L L EIEI EI
−
−−
−⎡ ⎤⎢ ⎥ −⎛ ⎞⎡ ⎤
= = =⎢ ⎥ ⎜ ⎟⎢ ⎥− −⎣ ⎦⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦
2 1 2 16 21 2 1 2(3)
EI EIL L
⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
Verification:
BA
1
Dept. of CE, GCE Kannur Dr.RajeshKN
14
[ ] 11 12
21 22
3 2
2
12 6
6 4
M MMi
M M
S SS
S S
EI EIL LEI EIL L
⎡ ⎤= ⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥
= ⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
Member stiffness matrix for prismatic beam member with deflection and rotation at one end as degrees of freedom
Dept. of CE, GCE Kannur Dr.RajeshKN
[ ] [ ]
13 212
1
2
2 33 26 3 6
2
Mi Mi
L LL LLEI EIS F
EI LL LEI EI
−
−−
⎡ ⎤⎢ ⎥ ⎛ ⎞⎡ ⎤
= = =⎢ ⎥ ⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥ ⎣ ⎦⎝ ⎠⎢ ⎥⎣ ⎦
( )2
2
3
2
2
6 363
12 6
6 423
EI EIL LEI EIL
LEIL LL L
L
−⎡ ⎤= =⎢ ⎥−⎣
⎡ ⎤−⎢ ⎥⎢ ⎥⎢⎢
⎦ ⎥− ⎥⎣ ⎦
Verification:
[ ]MiEASL
=
•Truss member
Dept. of CE, GCE Kannur Dr.RajeshKN
16
•Plane frame member
[ ]11 12 13
21 22 23
31 32 33
3 2
2
0 0
12 60
6 40
M M M
Mi M M M
M M M
S S SS S S S
S S S
EAL
EI EIL LEI EIL L
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
17
•Grid member
[ ]
3 2
2
12 60
0 0
6 40
Mi
EI EIL L
GJSL
EI EIL L
⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
18
•Space frame member
[ ]
3 2
3 2
2
2
0 0 0 0 0
12 60 0 0 0
12 60 0 0 0
0 0 0 0 0
6 40 0 0 0
6 40 0 0 0
Z Z
Y Y
Mi
Y Y
Z Z
EAL
EI EIL L
EI EIL LS
GJL
EI EIL L
EI EIL L
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
19
(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 00 0 0 00 0 0 0
0 0 0 0
0 0 0 0
MM M
MM M
MM M
MiMi Mi
MmMm Mm
SA DSA D
SA D
SA D
SA D
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦
M M MA = S D
Dept. of CE, GCE Kannur Dr.RajeshKN
20
{ } [ ]{ }M M MA S D=
[ ]MS is the unassembled stiffness matrix of the entire structure
Dept. of CE, GCE Kannur Dr.RajeshKN
21
• 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, { } [ ]{ } [ ] [ ] { }{ }
FM MJ J MF MR
R
DD 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
Dept. of CE, GCE Kannur Dr.RajeshKN
22
• 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 displacementapplied to the restrained structure.
[ ]MRC
[ ]MFC• Each column in the submatrix consists of member displacements caused by a unit value of an unknown displacementapplied to the restrained structure.
{ }MDrelate to respectively
[ ]MFC
[ ]MRC
and{ }FD
{ }RD
and
Dept. of CE, GCE Kannur Dr.RajeshKN
23
{ }MDδ
{ } [ ]{ } [ ] [ ] { }{ }
FM MJ J MF MR
R
DD C D C C
Dδ
δ δδ
⎧ ⎫= = ⎡ ⎤ ⎨ ⎬⎣ ⎦
⎩ ⎭
• Suppose an arbitrary set of virtual displacements
is applied on the structure.
{ } { } { } { }T T T FJ J F R
R
DW A D A A
Dδ
δ δ δδ⎧ ⎫⎡ ⎤= = ⎨ ⎬⎣ ⎦ ⎩ ⎭
{ }JDδ• External virtual work produced by the virtual displacements
{ }JAand real loads is
Dept. of CE, GCE Kannur Dr.RajeshKN
24
{ } { }TM MU A Dδ δ=
• Internal virtual work produced by the virtual (relative) end displacements { }MDδ { }MAand actual member end actions is
{ } { } { } { }T TJ 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 TJ J J MJ M MJ JA D D C S C Dδ δ=Hence,
{ } [ ]{ }J J JA S D=
Dept. of CE, GCE Kannur Dr.RajeshKN
25
[ ] [ ] [ ][ ]TJ 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 FJ J J
RF RRR R
S SA DA S D
S SA D⎧ ⎫ ⎧ ⎫⎡ ⎤
= ⇒ =⎨ ⎬ ⎨ ⎬⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦
[ ] [ ] [ ][ ]TFF MF M MFS C S C= [ ] [ ] [ ][ ]T
FR MF M MRS C S C=
[ ] [ ] [ ][ ]TRF MR M MFS C S C= [ ] [ ] [ ][ ]T
RR MR M MRS C S C=
Where,
Dept. of CE, GCE Kannur Dr.RajeshKN
26
{ } [ ]{ } [ ]{ }F FF F FR RA S D S D= + { } [ ]{ } [ ]{ }R RF F RR RA S D S D= +
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−⇒ = −⎡ ⎤⎣ ⎦
{ } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + +
• Support reactions
{ }RCArepresents 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
Dept. of CE, GCE Kannur Dr.RajeshKN
27
{ } { } [ ] [ ]{ } [ ]{ }( )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
Dept. of CE, GCE Kannur Dr.RajeshKN
28
Important formulae:
Joint displacements:
Member end actions:
Support reactions:
{ } [ ] { } [ ]{ }1F 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= + +
Dept. of CE, GCE Kannur Dr.RajeshKN
29
•Problem 1
[ ]
4 2 0 02 4 0 020 0 2 10 0 1 2
MEISL
⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦
Unassembled stiffness matrix
[ ] 2 121 2Mi
EISL
⎡ ⎤= ⎢ ⎥
⎣ ⎦Member stiffness matrix of beam member
Kinematic indeterminacy = 2
Dept. of CE, GCE Kannur Dr.RajeshKN
30
Fixed end actions
Equivalent joint loads
Dept. of CE, GCE Kannur Dr.RajeshKN
Joint displacements
Free (unknown) joint displacements { }FD { }RDRestraint displacements
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 displacementapplied to the restrained structure.
[ ]MFC
•Each column in the submatrix consists of member displacements caused by a unit value of a support displacementapplied to the restrained structure.
[ ]MRC
Dept. of CE, GCE Kannur Dr.RajeshKN
0 1 1 0
0 0 0 1
[ ]
0 01 01 00 1
M FC
⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦
DF1 DF2=1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN
1 L 1 L 0 0 1 0 0 0
1 L− 1 L− 1 L 1 L 0 0 1 L− 1 L−
Dept. of CE, GCE Kannur Dr.RajeshKN
35
[ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦
0 0 1 1 00 1 0 1 010 0 0 1 1
0 0 0 1 1
LLLL
L
−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥−⎣ ⎦
[ ]
1 1 1 01 0 1 0
0 0 1 10 0 1 1
MR
L LL L
CL LL L
−⎡ ⎤⎢ ⎥−⎢ ⎥=
−⎢ ⎥⎢ ⎥−⎣ ⎦
DR1 DR2 DR3 DR4=1 =1 =1 =1
DR1 DR2 DR3 DR4=1 =1 =1 =1
DF1 DF2=1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN
36
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦
{ } [ ] { }1F FF FD S A−∴ =
Joint displacements
[ ] [ ] [ ][ ]TFF MF M MFS C S C=
4 2 0 0 0 00 1 1 0 2 4 0 0 1 020 0 0 1 0 0 2 1 1 0
0 0 1 2 0 1
EIL
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
6 121 2
EIL
⎡ ⎤= ⎢ ⎥
⎣ ⎦
is a null matrix, since there are no support displacements
{ }RD
Dept. of CE, GCE Kannur Dr.RajeshKN
{ }1
6 1 121 2 29F
EI PLDL
−⎛ ⎞⎡ ⎤ ⎧ ⎫
= ⎨ ⎬⎜ ⎟⎢ ⎥⎣ ⎦ ⎩ ⎭⎝ ⎠
Free (unknown) joint displacements
2 201.11 818 11
011
PLE EII
PL⎧ ⎫ ⎧= =
⎫⎨⎨ ⎬
⎭ ⎩⎩⎬⎭
2
3
6 2 3 320 0 3 3L L L LEI
L L L⎡ ⎤− −
= ⎢ ⎥−⎣ ⎦
[ ] [ ] [ ][ ]TFR MF M MRS C S C=
4 2 0 0 1 1 00 1 1 0 2 4 0 0 1 0 1 02 10 0 0 1 0 0 2 1 0 0 1 1
0 0 1 2 0 0 1 1
LEIL L
−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
{ } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + +
is a null matrix.{ }RD
Support reactions
[ ] [ ] [ ][ ]TRF MR M MFS C S C=
2
6 02 02
3 33 3
LEIL
⎡ ⎤⎢ ⎥⎢ ⎥=−⎢ ⎥⎢ ⎥− −⎣ ⎦
1 1 0 4 2 0 0 0 01 0 1 0 2 4 0 0 1 01 20 0 1 1 0 0 2 1 1 00 0 1 1 0 0 1 2 0 1
TLEI
L L
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥=
−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
{ } { } [ ]{ }R RC RF FA A S D∴ = − +
Dept. of CE, GCE Kannur Dr.RajeshKN
[ ] [ ] [ ][ ]TRR MR M MRS C S C=
1 1 0 4 2 0 0 1 1 01 0 1 0 2 4 0 0 1 0 1 01 2 10 0 1 1 0 0 2 1 0 0 1 10 0 1 1 0 0 1 2 0 0 1 1
TL LEI
L L L
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥=
− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
3
12 6 12 06 4 6 0212 6 18 60 0 6 6
LL L LEI
LL
−⎡ ⎤⎢ ⎥−⎢ ⎥=− − −⎢ ⎥⎢ ⎥−⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
2
2
2 6 02 0 02
33 3 11833 3
PPL LEI PL
L EIPP
−⎧ ⎫ ⎡ ⎤⎪ ⎪− ⎢ ⎥⎪ ⎪ ⎧ ⎫⎢ ⎥= − +⎨ ⎬ ⎨ ⎬−⎢ ⎥ ⎩ ⎭⎪ ⎪− ⎢ ⎥⎪ ⎪ − −⎣ ⎦−⎩ ⎭
2
2
02 20 002
3 3318 33 33
3
P PPL PL
PL EI PEI LP P
PP P
⎧ ⎫−⎧ ⎫ ⎧ ⎫ ⎪ ⎪⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪= − + = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎩ ⎭− −⎩ ⎭ ⎩ ⎭ ⎪ ⎪⎩ ⎭
{ } { } [ ]{ }R RC RF FA A S D∴ = − +
2
310
323
PPL
P
P
⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
=
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 03 2 4 0 0 1 0 02
2 0 0 2 1 1 0 19 182 0 0 1 2 0 1
MPL EI PLA
L EI
⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥− ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪−⎩ ⎭ ⎣ ⎦ ⎣ ⎦
2
3 4 2 0 0 03 2 4 0 0 02
2 0 0 2 1 09 182 0 0 1 2 1
PL EI PLL EI
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥= +⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭⎣ ⎦
3 03 0
2 19 92 2
PL PL⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪= +⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭
11
130
PL⎧ ⎫⎪ ⎪−⎪ ⎪⎨ ⎬⎪⎪
=⎪⎪⎩ ⎭
is a null matrix{ }RD
Member end actions
{ } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +
Dept. of CE, GCE Kannur Dr.RajeshKN
42
3
0 00 0 0 2 0 6 4 6 01 1 0 0 4 0 6 2 6 02
0 0 0 2 0 0 3 31 1 1 1 2 0 0 3 3
0 0 1 1
L LL L L
L LEIL L LL
L L
⎡ ⎤⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= ⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− − −⎣ ⎦⎢ ⎥− −⎣ ⎦
[ ] [ ][ ] [ ]
2 2 2
2 2
3 2
6 6 2 3 32 0 0 3 3
6 0 12 6 12 022 0 6 4 6 03 3 12 6 18 63 3 0 0 6 6
FF FR
RF RR
L L L L L LL L L L
S SL LEIS SL L L L L
L L LL L
⎡ ⎤− −⎢ ⎥−⎢ ⎥⎢ ⎥ ⎡ ⎤−
= =⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎢ ⎥⎢ ⎥− − − −⎢ ⎥− − −⎣ ⎦
[ ] [ ] [ ][ ]TJ MJ M MJS C S C=
0 00 0 0 4 2 0 0 0 0 1 1 01 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 11 1 1 1 0 0 1 2 0 0 0 1 1
0 0 1 1
L LL L
LEIL LL L L
L
⎡ ⎤⎢ ⎥ −⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎣ ⎦⎢ ⎥− −⎣ ⎦
Alternatively, if the entire [SJ] matrix is assembled at a time,
Dept. of CE, GCE Kannur Dr.RajeshKN
43
•Problem 2:
AB C D
40 kN10 kN/m
2 m4 m 4 m 2 m
100 kN
Kinematic indeterminacy = 3 (Not considering joint D in the overhanging portion)
Degrees of freedom
AB C D
DF2DF1 DF3
Dept. of CE, GCE Kannur Dr.RajeshKN
44
[ ]
2 1 0 01 2 0 020 0 2 140 0 1 2
MEIS
⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦
Unassembled stiffness matrix
[ ] 2 121 24Mi
EIS ⎡ ⎤= ⎢ ⎥
⎣ ⎦Member stiffness matrix of beam member
Dept. of CE, GCE Kannur Dr.RajeshKN
45
Fixed end actions
Equivalent joint loads + actual joint loads
AB
13.33 kNm 13.33
20 kN 20 kN
BC
20 20
20 kN 20 kN
AB
13.33 kNm 6.67
20 kN 20 +20 = 40 kN 20 +100 = 120 kN
2x100 – 20 = 180 kNm
Dept. of CE, GCE Kannur Dr.RajeshKN
1 0 0 0
0 1 1 0
DF1 =1
DF2 =1 L
Dept. of CE, GCE Kannur Dr.RajeshKN
47
[ ]
1 0 00 1 00 1 00 0 1
M FC
⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦
DF1 DF2 DF3=1 =1 =1
0 0 0 1
DF3 =1
Dept. of CE, GCE Kannur Dr.RajeshKN
48
{ } [ ] { }1F FF FD S A−∴ =
Joint displacements
[ ] [ ] [ ][ ]TFF MF M MFS C S C=
1 0 0 2 1 0 0 1 0 00 1 0 1 2 0 0 0 1 020 1 0 0 0 2 1 0 1 040 0 1 0 0 1 2 0 0 1
T
EI⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
1 0.5 00.5 2 0.50 0.5 1
EI⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
is a null matrix.{ }RD∴
Dept. of CE, GCE Kannur Dr.RajeshKN
{ }
11 0.5 0 13.33
1 0.5 2 0.5 6.670 0.5 1 180
FDEI
−−⎛ ⎞⎡ ⎤ ⎧ ⎫⎪ ⎪⎜ ⎟⎢ ⎥= −⎨ ⎬⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟ −⎢ ⎥⎣ ⎦ ⎩ ⎭⎝ ⎠
Joint displacements
43.43560.33210.6
1EI
−⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪−⎩ ⎭
=
Dept. of CE, GCE Kannur Dr.RajeshKN
50
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
{ }43.43560.33210.6
13.33 2 1 0 0 1 0 013.33 1 2 0 0 0 1 02 120 0 0 2 1 0 1 0420 0 0 1 2 0 0 1
MEIA
EI
⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥−⎪ ⎪ ⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪−
−⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪−
⎣⎩
⎭ ⎦ ⎣ ⎦⎭
⎩
02525200
kNm
⎧ ⎫⎪ ⎪⎪ ⎪⎨ ⎬−⎪ ⎪⎪ ⎪−⎩ ⎭
=
is a null matrix{ }RD
Member end actions
{ } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +
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 02 4 0 0 0 00 0 2 1 0 020 0 1 2 0 060 0 0 0 4 20 0 0 0 2 4
MEIS
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
Unassembled stiffness matrix
[ ] 2 121 2Mi
EISL
⎡ ⎤= ⎢ ⎥
⎣ ⎦Member stiffness matrix of beam member
Dept. of CE, GCE Kannur Dr.RajeshKN
Fixed end moments
2525
Equivalent joint loads
2525
Support settlements { }RD(Restraint displacements)
A
BC D30mm 1RD
Free (unknown) joint displacements { }FD
1FD 2FD3FD
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
Dept. of CE, GCE Kannur Dr.RajeshKN54
[ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦
0 0 0 1 31 0 0 1 31 0 0 1 60 1 0 1 60 1 0 00 0 1 0
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
31 2 1
1 1 11FF F RDD D D
= = ==
A
BC D1 1RD =
1 3− 1 3− 1 6 1 6 0 0
Dept. of CE, GCE Kannur Dr.RajeshKN
55
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦
Joint displacements
[ ] [ ] [ ][ ]TFF MF M MFS C S C=
0 0 0 4 2 0 0 0 0 0 0 01 0 0 2 4 0 0 0 0 1 0 01 0 0 0 0 2 1 0 0 1 0 00 1 0 0 0 1 2 0 0 0 1 030 1 0 0 0 0 0 4 2 0 1 00 0 1 0 0 0 0 2 4 0 0 1
T
EI
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥
= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
6 1 01 6 2
30 2 4
EI⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
[ ] [ ] [ ][ ]TFR MF M MRS C S C=
0 0 0 4 2 0 0 0 0 1 31 0 0 2 4 0 0 0 0 1 31 0 0 0 0 2 1 0 0 1 60 1 0 0 0 1 2 0 0 1 630 1 0 0 0 0 0 4 2 00 0 1 0 0 0 0 2 4 0
T
EI
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥
= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
3 21 2
30
EI−⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
{ }
16 1 0 0 3 21 6 2 25 1 2 0.03
3 30 2 4 25 0
EI EI−
−⎛ ⎞ ⎡ ⎤⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − − −⎨ ⎬⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎪ ⎪⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎩ ⎭⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎣ ⎦
20 4 2 0 0.0453 56004 24 12 25 0.015
116 32 12 35 25 0
EI
− ⎡ ⎤⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥= − − − − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎢ ⎥ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦
3
16423 108
116 5.6 106
0.0075830.00050.007 311
−⎧ ⎫⎪ ⎪= =⎨ ⎬× × ⎪ ⎪
−⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩⎩ ⎭⎭
20 4 2 0 843 4 24 12 25 28
1162 12 35 25 0
EI
− ⎡ ⎤⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥= − − − − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎢ ⎥ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
58
{ } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + +
Support reactions
[ ]
4 2 0 0 0 0 0 0 02 4 0 0 0 0 1 0 00 0 2 1 0 0 1 0 0
1 3 1 3 1 6 1 6 0 00 0 1 2 0 0 0 1 030 0 0 0 4 2 0 1 00 0 0 0 2 4 0 0 1
EI
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥
= − − ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
[ ] [ ] [ ][ ]TRF MR M MFS C S C=
[ ]3 2 1 2 03
EI= −
Dept. of CE, GCE Kannur Dr.RajeshKN
{ } { } [ ] { }0.007583
0 3 2 1 2 0 0.0005 0.033 2
0.0031R
EI EIA−⎧ ⎫⎪ ⎪ ⎡ ⎤= − + − + −⎨ ⎬ ⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭
(Support reaction corresponding to DR . i.e., reaction at B)
( )0.0116 0.045 0.0334 62 53 3
.3EI EI= − = − −=
[ ] [ ] [ ][ ]TRR MR M MRS C S C=
[ ]
4 2 0 0 0 0 1 32 4 0 0 0 0 1 30 0 2 1 0 0 1 6
1 3 1 3 1 6 1 6 0 00 0 1 2 0 0 1 630 0 0 0 4 2 00 0 0 0 2 4 0
EI
−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥
= − − ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
2EI⎡ ⎤= ⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN60
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
{ } { }
0 4 2 0 0 0 0 0 0 0 1 30 2 4 0 0 0 0 1 0 0 1 3
0.0075830 0 0 2 1 0 0 1 0 0 1 62 0.0005 0.030 0 0 1 2 0 0 0 1 0 1 66
0.003125 0 0 0 0 4 2 0 1 0 025 0 0 0 0 2 4 0 0 1 0
MEIA
−⎛⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎧ ⎫⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪= + + −⎜⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎩ ⎭⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝
⎞⎟⎟⎟⎟⎟⎟
⎜ ⎟⎜ ⎟⎠
83.755.455.440.3
40.30
⎧ ⎫⎪ ⎪⎪ ⎪−⎪ ⎪
⎨ ⎬−⎪⎪⎪⎩
=⎪⎪⎪⎭
Member end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
61
[ ] [ ][ ] [ ]
FF FR
RF RR
S SS S
⎡ ⎤= ⎢ ⎥⎣ ⎦
6 1 0 3 21 6 2 1 20 2 4 033 2 1 2 0 3 2
E I−⎡ ⎤
⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥−⎣ ⎦
2 0 0 20 1 1 0 0 0 4 0 0 20 0 0 1 1 0 2 1 0 1 20 0 0 0 0 1 1 2 0 1 231 3 1 3 1 6 1 6 0 0 0 4 2 0
0 2 4 0
EI
−⎡ ⎤⎢ ⎥−⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎢ ⎥⎣ ⎦
[ ] [ ] [ ][ ]TJ MJ M MJS C S C=
4 2 0 0 0 0 0 0 0 1 30 1 1 0 0 0 2 4 0 0 0 0 1 0 0 1 30 0 0 1 1 0 0 0 2 1 0 0 1 0 0 1 60 0 0 0 0 1 0 0 1 2 0 0 0 1 0 1 631 3 1 3 1 6 1 6 0 0 0 0 0 0 4 2 0 1 0 0
0 0 0 0 2 4 0 0 1 0
EI
−⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥ −⎡ ⎤ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
Alternatively, if the entire [SJ] matrix is assembled at a time,
Dept. of CE, GCE Kannur Dr.RajeshKN
62
[ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦
0 0 0 1 3 1 1 3 0 01 0 0 1 3 0 1 3 0 01 0 0 0 0 1 6 1 6 00 1 0 0 0 1 6 1 6 00 1 0 0 0 0 1 3 1 30 0 1 0 0 0 1 3 1 3
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−
= ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥
−⎢ ⎥⎣ ⎦[ ] [ ] [ ][ ]TJ MJ M MJS C S C=
0 0 01 3 1 1 3 0 0 4 2 0 0 0 0 0 0 01 3 1 1 3 0 01 0 01 3 0 1 3 0 0 2 4 0 0 0 0 1 0 01 3 0 1 3 0 01 0 0 0 0 1 6 1 6 0 0 0 2 1 0 0 1 0 0 0 0 1 6 1 6 00 1 0 0 0 1 6 1 6 0 0 0 1 2 0 0 0 1 0 0 0 1 6 1 6 030 1 0 0 0 0 1 3 1 3 0 0 0 0 4 2 0 1 0 0 0 0 1 30 0 1 0 0 0 1 3 1 3 0 0 0 0 2 4 0 0 1
T
EI
− −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎢ ⎥
= ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎣ ⎦⎣ ⎦
1 30 0 0 1 3 1 3
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥
−⎢ ⎥⎣ ⎦
Alternatively, if ALL possible support settlements are accounted for,
Dept. of CE, GCE Kannur Dr.RajeshKN
63
[ ] [ ][ ] [ ]
FF FR
RF RR
S SS S
⎡ ⎤= ⎢ ⎥⎣ ⎦
0 1 1 0 0 00 0 0 1 1 0 2 0 0 2 4 2 0 00 0 0 0 0 1 4 0 0 2 2 2 0 0
1 3 1 3 0 0 0 0 2 1 0 0 0 1 2 1 2 01 0 0 0 0 0 1 2 0 0 0 1 2 1 2 031 3 1 3 1 6 1 6 0 0 0 4 2 0 0 0 2 20 0 1 6 1 6 1 3 1 3 0 2 4 0 0 0 2 20 0 0 0 1 3 1 3
EI
⎡ ⎤⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥⎢ ⎥− − −⎣ ⎦⎢ ⎥
− −⎣ ⎦
6 1 0 2 2 3 2 1 2 01 6 2 0 0 1 2 3 2 20 2 4 0 0 0 2 22 0 0 4 3 2 4 3 0 02 0 0 2 4 2 0 033 2 1 2 0 4 3 2 3 2 1 6 01 2 3 2 2 0 0 1 6 3 2 4 30 2 2 0 0 0 4 3 4 3
EI
− −⎡ ⎤⎢ ⎥−⎢ ⎥
−⎢ ⎥⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥− − − −⎢ ⎥⎢ ⎥− − −⎢ ⎥
− − −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
64
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦
10
6 1 0 0 2 2 3 2 1 2 0 01 6 2 25 0 0 1 2 3 2 2 0.03
3 30 2 4 25 0 0 0 2 2 0
0
EI EI−⎡ ⎤⎧ ⎫⎢ ⎥⎪ ⎪− −⎛ ⎞ ⎧ ⎫⎡ ⎤ ⎡ ⎤⎢ ⎥⎪ ⎪⎪ ⎪ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥= − − − −⎨ ⎬ ⎨ ⎬⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥⎪ ⎪ ⎪ ⎪⎜ ⎟ −⎢ ⎥ ⎢ ⎥⎩ ⎭⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦
20 4 2 0 0.0453 56004 24 12 25 0.015
116 32 12 35 25 0
20 4 2 843 4 24 12 3
1162 12 35 25
EI
EI
− ⎡ ⎤⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥= − − − − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎢ ⎥ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦
− −⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥= − − ⎨ ⎬⎢ ⎥ ⎪ ⎪−⎢ ⎥ ⎩ ⎭⎣ ⎦
16423 108
116671
EI
−⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭
0.0075830.00050.0031
−⎧ ⎫⎪ ⎪⎨ ⎬⎪⎩
=⎪⎭
Joint displacements
Dept. of CE, GCE Kannur Dr.RajeshKN
65
{ }
0 2 0 0 4 3 2 4 3 0 0 00.0075830 2 0 0 2 4 2 0 0 00.00050 3 2 1 2 0 4 3 2 3 2 1 6 0 0.03
3 30.003150 1 2 3 2 2 0 0 1 6 3 2 4 3 0
50 0 2 2 0 0 0 4 3 4 3 0
REI EIA
−⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪= − + +− − − − −⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪− − − −⎩ ⎭⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− − − −⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦
{ } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + +
{ }
0 28.373 74.6670 28.373 1120 21.653 8450 19.973 9.333
46.29483.62762.34779.3136.5650 13.44 0
RA
−⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪= − + + =−⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪ ⎪ ⎪− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩
⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪−⎨ ⎬⎪ ⎪⎪ ⎪⎪⎩ ⎭⎭ ⎪
Support reactions
Dept. of CE, GCE Kannur Dr.RajeshKN
66
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
{ }
0 4 2 0 0 0 0 0 0 0 1 3 1 1 3 0 00 2 4 0 0 0 0 1 0 0 1 3 0 1 3 0 0
0.0075830 0 0 2 1 0 0 1 0 0 0 0 1 6 1 6 02 0.00050 0 0 1 2 0 0 0 1 0 0 0 1 6 1 6 06
0.003125 0 0 0 0 4 2 0 1 0 0 0 0 1 3 1 325 0 0 0 0 2 4 0 0 1 0 0 0
MEIA
−⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎧ ⎫
−⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪= + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎢ ⎥ ⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎣ ⎦
00
0.0300
1 3 1 3
⎛ ⎞⎡ ⎤⎧ ⎫⎜ ⎟⎢ ⎥⎪ ⎪⎜ ⎟⎢ ⎥⎪ ⎪⎜ ⎟⎢ ⎥ ⎪ ⎪−⎜ ⎟⎨ ⎬⎢ ⎥
⎜ ⎟⎪ ⎪⎢ ⎥⎜ ⎟⎪ ⎪⎢ ⎥⎜ ⎟⎪ ⎪⎢ ⎥ ⎩ ⎭⎜ ⎟−⎣ ⎦⎝ ⎠
{ }
83.755.455.440.3
40.30
MA
⎧ ⎫⎪ ⎪⎪ ⎪−⎪ ⎪
⎨ ⎬−⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
=
Member end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
[ ]
2 4 1 4 0 0 0 01 4 2 4 0 0 0 00 0 6 4.8 3 4.8 0 0
20 0 3 4.8 6 4.8 0 00 0 0 0 4 8 2 80 0 0 0 2 8 4 8
MS EI
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
•Problem 4
Unassembled stiffness matrix
Equivalent joint loads
85.33
42.67
1
2
3
Dept. of CE, GCE Kannur Dr.RajeshKN
68
• consists of member displacements due to unit displacements on the restrained structure.
[ ]MFC
1 1FD =2 1FD =
[ ]
1 00 10 00 10 10 0
MFC
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
DF1 DF2=1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN
69
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦
{ } [ ] { }1F FF FD S A−= since there are no support displacements.
Joint displacements
[ ] [ ] [ ][ ]TFF MF M MFS C S C=
2 4 1 4 0 0 0 0 1 01 4 2 4 0 0 0 0 0 1
1 0 0 0 0 0 0 0 6 4.8 3 4.8 0 0 0 02
0 1 0 1 1 0 0 0 3 4.8 6 4.8 0 0 0 10 0 0 0 4 8 2 8 0 10 0 0 0 2 8 4 8 0 0
EI
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤
= ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
70
8 44 8
1 0 0 0 0 0 0 200 1 0 1 1 0 0 108
0 80 4
EI
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤
= ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥
⎢ ⎥⎢ ⎥⎣ ⎦
2 11 92
EI ⎡ ⎤= ⎢ ⎥
⎣ ⎦
{ } [ ] { }1
1 2 1 01 9 85.3332F FF F
EID S A−
− ⎛ ⎞⎡ ⎤ ⎧ ⎫∴ = = = ⎨ ⎬⎜ ⎟⎢ ⎥
⎣ ⎦ ⎩ ⎭⎝ ⎠
36 4 0 341.3328 14 8 85.333 682.66
10.034272
91203 . 784 0EI EIEI
− −⎧ ⎫ ⎧ ⎫⎡ ⎤= = =⎨ ⎬ ⎨ ⎬⎢ ⎥− ⎩ ⎭ ⎩ ⎭⎣
−
⎦
⎧ ⎫⎨ ⎬⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
71
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
0 2 4 1 4 0 0 0 0 1 00 1 4 2 4 0 0 0 0 0 1
42.667 0 0 6 4.8 3 4.8 0 0 0 1 10.0391285.333 0 0 3 4.8 6 4.8 0 0 0 0 20.078
0 0 0 0 0 4 8 2 8 0 10 0 0 0 0 2 8 4 8 0 0
EIEI
⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥
−⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎧ ⎫= +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥− ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎣ ⎦
is a null matrix{ }RD
0 00 120.47
42.667 200.78185.333 401.568
0 160.6240 80.312
⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪
= +⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭
015.05967.76535.138
20.07810.039
⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎨= ⎬−⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
Member end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
72
•Homework 2:
32
1
80kNm
3m3m
4m
6m
20kN mC
A
B D
Dept. of CE, GCE Kannur Dr.RajeshKN
•Problem 5
[ ]
2 1 0 0 0 01 2 0 0 0 00 0 2 1 0 020 0 1 2 0 00 0 0 0 2 10 0 0 0 1 2
MEISL
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
Unassembled stiffness matrix
Dept. of CE, GCE Kannur Dr.RajeshKN
74
[ ]MFC =
0 0 11 0 11 0 00 1 00 1 10 0 1
LL
LL
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
12
3
DF1 DF2 DF3=1 =1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN
75
2 1 0 0 0 0 0 0 11 2 0 0 0 0 1 0 1
0 1 1 0 0 00 0 2 1 0 0 1 0 020 0 0 1 1 00 0 1 2 0 0 0 1 0
1 1 0 0 1 10 0 0 0 2 1 0 1 10 0 0 0 1 2 0 0 1
LL
EIL
L L L LLL
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
[ ] [ ] [ ][ ]TFF MF M MFS C S C=
1 0 32 0 3
0 1 1 0 0 02 1 020 0 0 1 1 01 2 0
1 1 0 0 1 10 2 30 1 3
LL
EIL
L L L LLL
⎡ ⎤⎢ ⎥⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦
2
4 1 32 1 4 3
3 3 12
LEI LL
L L L
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
76
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦
{ }
1
2
4 1 3 02 1 4 3 0
3 3 12F
LEID LL
L L L P
−⎛ ⎞ ⎧ ⎫⎡ ⎤
⎪ ⎪⎜ ⎟⎢ ⎥= ⎨ ⎬⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎩ ⎭⎣ ⎦⎝ ⎠
{ } [ ] { }1F FF FD S A−= , since there are no support displacements.
2
13 3 3 03 13 3 0
843 3 5
LL LEI
L L L P
− − ⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥= − − ⎨ ⎬⎢ ⎥ ⎪ ⎪− −⎢ ⎥ ⎩ ⎭⎣ ⎦
233
845
PLEI
L
−⎧ ⎫⎪ ⎪−⎨ ⎬⎪⎩
=⎪⎭
Joint displacements
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 11 2 0 0 0 0 1 0 1
30 0 2 1 0 0 1 0 02 30 0 1 2 0 0 0 1 0 84
50 0 0 0 2 1 0 1 10 0 0 0 1 2 0 0 1
M
LL
EI PLAL EI
LLL
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ −⎧ ⎫⎢ ⎥ ⎢ ⎥ ⎪ ⎪= −⎨ ⎬⎢ ⎥ ⎢ ⎥
⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
2
2 1 0 0 0 0 51 2 0 0 0 0 20 0 2 1 0 0 320 0 1 2 0 0 3840 0 0 0 2 1 20 0 0 0 1 2 5
EI PLL EI
⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥−⎪ ⎪⎢ ⎥
= ⎨ ⎬⎢ ⎥ −⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦
{ } [ ][ ]{ }M M MF FA S C D=
2
12992984
912
EI PLL EI
⎧ ⎫⎪ ⎪⎪ ⎪−⎪ ⎪
= ⎨ ⎬−⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
433314
34
PL
⎧ ⎫⎪ ⎪⎪ ⎪−
=⎪ ⎪⎨ ⎬−⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
Member end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
78
[ ]0.2 0.0 0.00.0 0.2 0.00.0 0.0 0.2
MS⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Unassembled stiffness matrix
•Problem 6:
1
2
3
50 kN
80 kN
5 m
5 m5 m
4 m 4 m
3 m
3 m
Dept. of CE, GCE Kannur Dr.RajeshKN
[ ]MFC =
0.8 -0.6-0.8 0.60.8 0.6
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
1cos 3
6.87=0.8
36.87
1
23
0.8
10.6
53.13
cos 53.13=0.6
DF1 DF2=1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN
80
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦
{ }2.930 1.302 -501.302 5.208 -80FD ⎧ ⎫⎡ ⎤
= ⎨ ⎬⎢ ⎥⎩ ⎭⎣ ⎦
{ } [ ] { }1F FF FD S A−= , since there are no support displacements.
-250.651-481.771⎧
=⎫
⎨ ⎬⎩ ⎭
Joint displacements
[ ] [ ] [ ][ ]TFF MF M MFS C S C= 0.384 -0.096
-0.096 0.216⎡ ⎤
= ⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
0.2 0.0 0.0 0.8 -0.6250.651
0.0 0.2 0.0 -0.8 0.6481.771
0.0 0.0 0.2 0.8 0.6
⎡ ⎤⎡ ⎤−⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥ −⎩ ⎭⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
[ ][ ]{ }M MF FS C D=
17.70817.70897.917
⎧ ⎫⎪ ⎪⎨−⎪⎩
= ⎬− ⎪⎭
Member Forces:
Dept. of CE, GCE Kannur Dr.RajeshKN
82
•Problem 7:
[ ]0.2 0.0 0.00.0 0.5 0.00.0 0.0 0.2
MS⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Unassembled stiffness matrix
1
2
3
Dept. of CE, GCE Kannur Dr.RajeshKN
[ ]MFC =
0.600 0.8000.000 -1.000-0.600 0.800
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
1
53.13
5 m
cos 5
3.13=
0.6
0.6
1
053.13
5 m
036.87
cos 36.87=0.8
cos 36.87=0.8 0.8
DF1 DF2=1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN84
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦
{ }6.944 0.000 15.0000.000 1.323 -10.000FD ⎡ ⎤ ⎧ ⎫
= ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭
{ } [ ] { }1F FF FD S A−= , since there are no support displacements.
104.167-13.228⎧
=⎫
⎨ ⎬⎩ ⎭
Joint displacements
[ ] [ ] [ ][ ]TFF MF M MFS C S C= 0.144 0.000
0.000 0.756⎡ ⎤
=⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
{ }0.2 0.0 0.0 0.600 0.800
104.1670.0 0.5 0.0 0.000 -1.000
-13.2280.0 0.0 0.2 -0.600 0.800
MA⎡ ⎤⎡ ⎤
⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥⎩ ⎭⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦
{ } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + +
{ } [ ][ ]{ }M M MF FA S C D=
10.46.6-14.6
⎧ ⎫⎪ ⎪⎨ ⎬⎪
=⎪⎩ ⎭
Member Forces:
Dept. of CE, GCE Kannur Dr.RajeshKN
[ ] 0.866 0.000 0.000 0.000 1.000 0.000 0.000 0.000 0.500
MS⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Unassembled stiffness matrix
•Problem 8 (Homework 3):
Dept. of CE, GCE Kannur Dr.RajeshKN
1
0.8660.51
0.8660.5
[ ]MFC =
0.866 0.500 1.000 0.0000.500 -0.866
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
DF1 DF2=1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN
88
{ } [ ] { } [ ]{ }1F FF F FR RD S A S D−= −⎡ ⎤⎣ ⎦
{ }0.577 -0.155 5-0.155 1.732 0FD⎡ ⎤ ⎧ ⎫
= ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭
{ } [ ] { }1F FF FD S A−= , since there are no support displacements.
2.887-0.773⎧ ⎫⎨⎩
= ⎬⎭
Joint displacements
[ ] [ ] [ ][ ]TFF MF M MFS C S C= 1.774 0.158
0.158 0.591⎡ ⎤
= ⎢ ⎥⎣ ⎦
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=
1.8302.8871.057
⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭
=
Member Forces:
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
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