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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
1
Example 4.1
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Principle Of Superposition
Mathematically, the principle of superposition is stated in the following fashion
Relative to a linear structural system, the mathematical expression above infers the
displacement at a given point in the system caused by two or more loads is the sum of the
responses which would have been caused by each load individually. Since the addition
function is preserved this is sometimes referred to as an additive map.
Consider a linear spring where
and K is the linear spring constant.
21
2121 ,,
ADAD
AADAAD
DKA
2
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
For the initial load on the spring
Now increase the deflection on the system by an amount DD. The additional load
on the spring is
The final force on the spring is from the additional deflection is
with
and
11 DKA
DKA DD
DDK
DKDKAA
D
DD
1
11
AAA D 12
DDD D 12
3
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
then
This result, although deceptively obvious, indicates that for a linear spring system
• the deflection caused by a force can be added to the deflection caused by
another force to obtain the deflection resulting from both forces being
applied;
• the order of loading is not important (DD or D1 could be applied first);
The is the Principle of Superposition – For a structure with a linear response, the
load effects caused by two or more loads are the sum of the effects caused by each
load applied separately.
For the principle to be applicable in a structural analysis the material the structure
is fabricated from must be linear elastic. To guarantee this we typically require the
structure to undergo small deformations.
22 DKA
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Consider the beam in figure (a) subject to
external actions A1 and A2. These actions
produce various reactions and
displacements throughout the structure.
Reactions are developed at the supports. A
displacement is produced at the mid span.
The effects of A1 and A2 are shown
separately in (b) and (c). A single prime is
associated with A1 and a double prime with
A2.
From the figure it becomes obvious that the
following equations can be developed
through the use of superposition:
DDDRRR
MMMRRR
BBB
AAAAAA
5
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Next consider the same beam subjected
to displacements, i.e., the support at B is
translated down an amount D and rotated
counterclockwise an amount q. Again
various reactions and displacements are
induced in the structure.
Reactions and displacements with single
primes are associated with D. Those
with double primes are associated with
q.
Focusing on the reactions, superposition
can also be invoked for a linear system if
the input variables are displacements,
i.e.,
6 2121 DADADDA
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Action And Displacement Equations
The relationship between actions and displacements play an important role in structural
analysis. A convenient and simplistic way to see this relationship is through the use of a
linear, elastic spring
The action A will compress (translate) the spring an amount D. This can be expressed
through the simple expression:
In this equation F (in Example 4.1 we used d) is the flexibility of the spring. The spring
flexibility is defined as the displacement produced by a unit value of the (force) action A.
FAD
7
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
This relationship can also be expressed as
Here S is the stiffness of the spring and is defined as the action (force) required to produce a
unit displacement in the spring. The flexibility and stiffness of the spring are inverse to one
another. To see this consider
Thus
or
This requires a formal discussion of matrix division which occurs in Section 5 of the class
notes.
11
SS
F
SDA
8
FS
AFSA
DSA
1
11 FF
S
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
3
3 48
48 L
EIDA
EI
ALD
EI
LAF
48
1 3
3
481
L
EIDS
The relationship that holds for a
spring holds for any structural
component. Consider the
simple beam subjected to an
action A that produces a
translation D.
The action and displacement
equation holds if the flexibility
F and stiffness S are determined
as shown.
The action and displacement
equation given on the previous
slide is valid only when one
action is present and we are
looking for one displacement
within the structure. More
than one action and/or more
than one displacement
requires a matrix format.9
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Consider a general example where a beam is subjected to
three actions, i.e., two forces (A1 and A2) and a moment
(A3). The directions for the actions are assumed positive.
The deflected shape is given in figure (b) and displacements
D1, D2 and D3 correspond to A1, A2, and A3.
By using superposition each displacement can be expressed
as the sum of displacements due to actions A1 through A3
In a similar manner expressions for D2 and D3 are
1312111
312111
32113211 ,,
DDDD
ADADAD
AAADAAAD
333231
3323133
232221
3222122
DDD
ADADADD
DDD
ADADADD
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Now it is quite obvious that
and if we note the fact that
We can extend the notation and introduce the concept of a matrix of flexibility coefficients
via the following relationships
3113
2112
1111
AbycausedAatdeflectionD
AbycausedAatdeflectionD
AbycausedAatdeflectionD
onlyAtoalproportiondirectlyisD
onlyAtoalproportiondirectlyisD
onlyAtoalproportiondirectlyisD
313
212
111
31313
21212
11111
AFD
AFD
AFD
1132323
22222
12121
AFD
AFD
AFD
33333
23232
13131
AFD
AFD
AFD
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Example 4.2
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Example 4.3
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
We can express the equations for the deformations D1,
D2 and D3 as
Each term on the right-hand side of the equations is a
displacement written in the form of a coefficient times
the action that produces a deformation represented by
the coefficient. The coefficients are called flexibility
coefficients. The physical significance of the flexibility
coefficients are depicted in figures (c), (d) and (e)
3332321313
3232221212
3132121111
AFAFAFD
AFAFAFD
AFAFAFD
All the flexibility coefficients in the figures have two subscripts (Fij). The first subscript
identifies the displacement (Di) associated with an action (Aj). The second subscript
denotes where the unit action is being applied. Figure (c) is associated with action A1,
figure (d) is associated with action A2, and figure (e) is associated with action A3.
Flexibility coefficients are taken as positive when the deformation represented by the
coefficient is in the same direction as the ith action.14
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Instead of expressing displacements in terms of
actions, it is possible to express actions in terms of
displacements, i.e.,
Here S is a stiffness coefficient and represents an
action due to a unit displacement. To impose these
unit displacements requires that artificial restraints
must be provided. These restraints are shown in the
figure by simple supports corresponding to actions
A1, A2 and A3.
3332321313
3232221212
3132121111
DSDSDSA
DSDSDSA
DSDSDSA
15
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Each stiffness coefficient is shown acting in its assumed positive direction, which is
the same direction as the corresponding action. If the actual direction of one of the
stiffness coefficients is opposite to that assumption, then the stiffness coefficient will
have a negative value.
The calculations of the stiffness coefficients for the beam shown can be quite
lengthy. However, analyzing a beam like the one shown previously by the stiffness
method can be expedited by utilizing a special structure where all the joints of the
structure are restrained. We will get into the details of this in the next section of
notes.
The primary purpose of this discussion is for the student to visualize what flexibility
and stiffness coefficients represent physically.
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Example 4.4
17
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Example 4.5
18
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Example 4.6
19
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Example 4.7
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Flexibility and Stiffness Matrices
We can now generalize the concepts introduced in the preceding section. If the number of
actions applied to a structure is n, the corresponding equations for displacements are:
In matrix format these equations become
or
nnnnnn
nn
nn
AFAFAFD
AFAFAFD
AFAFAFD
2211
22221212
12121111
nnnnn
n
n
n A
A
A
FFF
FFF
FFF
D
D
D
2
1
21
22221
11211
2
1
AFD 21
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
The action equations with n actions applied to the structure with a corresponding n
displacements are
In matrix format these equations become
or
nnnnnn
nn
nn
DSDSDSA
DSDSDSA
DSDSDSA
2211
22221212
12121111
nnnnn
n
n
n D
D
D
SSS
SSS
SSS
A
A
A
2
1
21
22221
11211
2
1
DSA 22
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Since the actions Ai and displacements Di correspond to one another in both formats, it
follows the flexibility matrix Fij and the stiffness matrix Sij are related to each other. Taking
the matrix inverse of
yields
With
then
How to formulate the inverses appearing above is discussed in the next section of the notes.
AFD
DFA 1
DSA
1 FS
23
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
In a similar fashion one can show that
Thus the stiffness matrix is the inverse of the flexibility matrix and vice versa provided that
the same set of actions and displacements are being considered in both equations
Note that a flexibility matrix or stiffness matrix is not an array that is determined by the
geometry of the structure only. The matrices are directly related to the geometry and the set
of actions as well as displacements under consideration.
1 SF
24
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Example 4.8
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Example 4.9
The cantilever beam shown in the figure below is subjected to a force (A1) and moment (A2)
at the free end. Develop the flexibility matrix and the stiffness matrix for assuming
displacements D1 and D2 are of interest.
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Making use of Case #7 and Case #8 from the following table
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
(continued)
28
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Then the flexibility coefficients are as follows:
The displacements are
The flexibility matrix becomes
EI
LF
EI
LFF
EI
LF 22
2
2112
3
1123
21
2
22
2
1
3
1223
AEI
LA
EI
LDA
EI
LA
EI
LD
EI
L
EI
LEI
L
EI
L
F
2
232
23
29
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
In order to develop the stiffness matrix consider the following beam reactions due to
applied displacements:
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Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Then the stiffness coefficients are as follows:
The actions are
The stiffness matrix becomes
L
EIS
L
EISS
L
EIS
46122222112311
1 1 2 2 1 23 2 2
12 6 6 4EI EI EI EIA D D A D D
L L L L
L
EI
L
EIL
EI
L
EI
S46
612
2
23
31
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
When the flexibility matrix and the stiffness matrix are multiplied together, the
result is the identity matrix:
This infers but does not prove that the two matrices are inverses of one another.
10
01
2
2346
612
2
23
2
23
EI
L
EI
LEI
L
EI
L
L
EI
L
EIL
EI
L
EI
FS
32
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
nnDADADADAW 33221121
Symmetry – Flexibility & Stiffness Matrices
If the loads on a structure are zero and gradually increase such that all loadings hit peak
values at the same time, the work done during this period of time will be summation of
the area under each individual load deflection curve, i.e.,
In a matrix format but both {A} and {D} are vectors by definition so to perform this
matrix multiplication we must use the transpose of one or the other column vectors. Thus
Recall that
Now, substitute this in the first equation for W immediately above.
T
T
DA
DAW
21
21
AFD
33
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
This substitution leads to
In addition, the following relationship holds from matrix algebra
Substituting this relationship in the equation from the previous slide yields
AFA
DAW
T
T
2
1
2
1
TT
TT
AF
AFD
TT
T
AFA
DAW
2
1
2
1
34
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Now
Multiplying both sides by ({A}T)-1 and {A}-1 we obtain
TTTAFAAFA 2121
TTTAFAAFA
TTTTTAFAAAAFAAA
1111
TTTTTFAAAAFAAAA
1111
TFIIFII
TFF
35
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Thus the flexibility matrix must be symmetric. To prove the stiffness matrix is symmetric
recall that
Substituting this in the equation for work
In addition, the following relationship holds from matrix algebra
DSA
T
T
DDS
DAW
2
1
2
1
TT
TT
DS
DSA
36
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Substituting this in the above equation for work
Equating these two relationships for work
Multiplying both sides by [D]-1 and [DT]-1 we obtain
DDS
DAW
TT
T
2
1
2
1
DDSDDS
DDSDDS
TTT
TTT
2
1
2
1
TTTTTDSDDDDSDDD
1111
37
Section 4: Matrix Structural Analysis (MSA) -
FundamentalsWashkewicz College of Engineering
Further manipulation yields
Hence the stiffness matrix is symmetric. Of course the fact that the stiffness matrix is
symmetric could have been concluded from the fact that the flexibility matrix is symmetric
and the stiffness matrix is the inverse of the flexibility matrix. But this has not been formally
proven.
TTTTTSDDDDSDDDD
1111
TSIISII
TSS
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