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Transcript of fm_truss_determinate
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8/7/2019 fm_truss_determinate
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FLEXIBILITY METHOD FOR
STATICALLY DETERMINATE TRUSSES
PROCEDURE:
1. Assemble the static matrix (A). Use the same techniques being used inthe matrix implementation of the method of joints.
2. Form the unit load matrix. This matrix contains rows equal to the number
of DOF and columns equal to the number of translation values beingdesired.
3. Compute for the equlibrium (b) matrix. This contains the member forces as
a result of applying unit loads on the locations being desired.
4. Form fm (unassembled flexibility matrix). This is a diagonal matrix of size
NM x NM. It contains the member flexibilities ( LAE
) on its main diagonal.
5. Calculate the Global Flexibility Matrix f. It can be shown that the overall
coupled structure matrix is given by the triple matrix product ( ). The
f matrix is a square matrix of size equal to the number of columns in the bmatrix (number of displacements being desired).
T
Mb f b
6. Solve for member forces using equilibrium equations.
*
where :
Q b P
Q Member Forces Matrixb Equilibrium Matrix
P External Load Matrix
=
=
=
=
7. Solve for displacement vector using the global relationship between
external forces and displacements.*
where :
D f P
D Structure Displacement Matrix
f Global Flexibility Matrix
P External Forces Matrix
=
=
=
=
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ILLUSTRATIVE PROBLEM (Entire Displacement and Member Forces Solution) :
REQUIRED : All Joint Translations and Member Forces
Static Matrix (A)
A 1 2 3 4 5 6 7
1 -0.8 0 -1 0 -0.8 0 0
2 -0.6 0 0 1 0.6 0 0
3 0.8 0 0 0 0 0 0
4 0.6 1 0 0 0 0 0
5 0 0 1 0 0 0 0
6 0 -1 0 0 0 1 07 0 0 0 0 0.8 0 1
Unit Load Matrix
1 1 0 0 0 0 0 0
2 0 1 0 0 0 0 0
3 0 0 1 0 0 0 0
4 0 0 0 1 0 0 0
5 0 0 0 0 1 0 0
6 0 0 0 0 0 1 0
7 0 0 0 0 0 0 1
Member Forces Due to Unit Loads (b)
1 0.00 0.00 1.25 0.00 0.00 0.00 0.00
2 0.00 0.00 -0.75 1.00 0.00 0.00 0.00
3 0.00 0.00 0.00 0.00 1.00 0.00 0.00
4 0.75 1.00 1.50 0.00 0.75 0.00 0.00
5 -1.25 0.00 -1.25 0.00 -1.25 0.00 0.00
6 0.00 0.00 -0.75 1.00 0.00 1.00 0.00
7 1.00 0.00 1.00 0.00 1.00 0.00 1.00
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Sample Results
Unassembled Flexibility Matrix (fm)
fm 1 2 3 4 5 6 7
1 0.00080 0 0 0 0 0 0
2 0 0.00048 0 0 0 0 0
3 0 0 0.00064 0 0 0 0 in/kip
4 0 0 0 0.00048 0 0 0
5 0 0 0 0 0.00080 0 0
6 0 0 0 0 0 0.00048 0
7 0 0 0 0 0 0 0.00064
Global Flexibility Matrix (f)
f 1 2 3 4 5 6 7
1 0.00216 0.00036 0.00243 0.00000 0.00216 0.00000 0.00064
2 0.00036 0.00048 0.00072 0.00000 0.00036 0.00000 0.00000
3 0.00243 0.00072 0.00476 -0.00072 0.00243 -0.00036 0.00064
4 0.00000 0.00000 -0.00072 0.00096 0.00000 0.00048 0.00000 in/kip
5 0.00216 0.00036 0.00243 0.00000 0.00280 0.00000 0.00064
6 0.00000 0.00000 -0.00036 0.00048 0.00000 0.00048 0.00000
7 0.00064 0.00000 0.00064 0.00000 0.00064 0.00000 0.00064
External Load Matrix (P) Member Forces (F) Displacement Matrix (D)
1 0 62.50 0.3375
2 0 -37.50 0.072
3 50 100.00 0.481
4 0 kips 150.00 kips -0.036 in
5 100 -187.50 0.4015
6 0 -37.50 -0.018
7 0 150.00 0.096
`
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ILLUSTRATIVE PROBLEM (Selective Displacement and All Member Forces Solution) :
REQUIRED : horizontal translation at joint 3, vert. & hor. translation at joint 2, all member forces
Static Matrix (A)
A 1 2 3 4 5 6 7
1 -0.8 0 -1 0 -0.8 0 0
2 -0.6 0 0 1 0.6 0 0
3 0.8 0 0 0 0 0 0
4 0.6 1 0 0 0 0 0
5 0 0 1 0 0 0 0
6 0 -1 0 0 0 1 0
7 0 0 0 0 0.8 0 1
Unit Load Matrix `
1 0 0 0
2 0 0 0
3 0 1 0
4 0 0 1
5 1 0 0
6 0 0 0
7 0 0 0
Member Forces Due to Unit Loads (b)
1 0.00 1.25 0.00
2 0.00 -0.75 1.00
3 1.00 0.00 0.00
4 0.75 1.50 0.00
5 -1.25 -1.25 0.00
6 0.00 -0.75 1.00
7 1.00 1.00 0.00
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Sample Results
Unassembled Flexibility Matrix (fm)
fm 1 2 3 4 5 6 7
1 0.00080 0 0 0 0 0 0
2 0 0.00048 0 0 0 0 0
3 0 0 0.00064 0 0 0 0 in/kip
4 0 0 0 0.00048 0 0 0
5 0 0 0 0 0.00080 0 0
6 0 0 0 0 0 0.00048 0
7 0 0 0 0 0 0 0.00064
Global Flexibility Matrix (f)
f 1 2 3
1 0.00280 0.00243 0.00000
2 0.00243 0.00476 -0.00072 in/kip
3 0.00000 -0.00072 0.00096
External Load Matrix (P) Member Forces (F) Displacement Matrix (D)
5 100 62.50 0.4015
3 50 kips -37.50 0.4810 in
4 0 100.00 -0.0360
150.00 kips
-187.50
-37.50
150.00
