Seoul National University Structural Design Laboratory Structural Design Lab.(Prof. Ho-Kyung Kim)
Dept. of Civil & Environmental Eng.
Seoul National University
457.649 Advanced Structural Analysis
Part III:
Direct Stiffness Method
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§ Reverse of force method
§ Also called “stiffness method” or “equilibrium method”
▶ See “Step in the displacement method”
These are “equilibrium equations”. Solve for {r}.
If linear, can use superposition. → greatly simplifies the calculation
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Review of Procedure for Displacement Method
Nodes Elements
{R} {S} Forces
{r} Geometry {v} Displacements
(outside-in) Equilibrium (inside-out) known loads
(function of {r}) Action-Deformation (stiffness form)
(functions of {r}) (functions of {r})
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▶ Kinematics
▶ Equilibrium
▶ Principle of virtual displacement
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Transformation Matrix
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Definition of the “a” Matrix
• Slope-deflection eq.
Therefore,
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Derivation of Stiffness by Direct Stiffness
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Derivation of Stiffness by Direct Stiffness
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▶ Deformed shapes for element stiffness definition
▶ Forces required for unit deformations
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Two-Dimensional Beam Element Stiffness Derivation
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§ The solution process becomes more systematic and mechanical when using this formulation. The only hand work comes from defining the geometric relationship of the transformation matrix a. The process can now be reduced to the following steps:
1. Define the structural (global) DOFs (r).
2. Find the a matrix for each element. This is done by applying a unit displacement to each DOF, holding all others still, and finding the corresponding element
displacements. This fills a column of the a matrix for each displacement applied.
Repeat until all columns of all a matrices are filled.
3. Sum aTKeafor all elements in the structure.
4. Form the load vector R.
5. Solve the set of equations Kr = R.
6. Recover the element force.
§ There are still two steps that need to be explained to complete the solution process.
This first is formation of the load vector R. the load vector can be considered to be composed of two parts, the concentrated load applied directly to global DOFs and the equivalent loads that were calculated previously by equivalent work. The
concentrated loads again are just placed into the corresponding location for the DOF at which they are applied.
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Direct Stiffness Solution Procedure
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Beam Element Loads
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Beam Element Force Recovery
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▶ Final equations
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Beam Element Force Recovery
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Beam Element Force Recovery
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§ The final solution procedure can be summarized as follows 1. Define the global DOF, r, for the structure.
2. Find the a matrix for each element in the structure.
3. Sum aTKea for all elements.
4. Form the load R=Rconcentrated+∑aTS𝒇. 5. Solve Kr=R.
6. Recover the element force, S=Kear−S𝒇.
§ Using this procedure, general structures can be analyzed using the direct stiffness method. Now only the geometric relationship for developing the transformation matrix, a, needs to be done by hand. All the other work is accomplished by linear algebra. Also notice that the element stiffness matrix never has to be rederived-only different properties used.
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Final Direct Stiffness Solution Procedure
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▶ Introduction of CAL-90
▶ Examples in Textbook P166-176
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The a Matrix Revisited
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The a Matrix Revisited
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The a Matrix Revisited
Local Global
v1
v2 v3
v4
*
v1
*
v2
*
v3
*
v4
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▶ Frame element stiffness matrix
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Inclusion of Axial Deformations
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▶ The matrix form for rotation of coordinates has the form
§ Coordinate transformation
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Derivation of Rotation Transformation
§ With rotation axis
§ Element rotation matrix aα
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▶ Location matrix, a"
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Derivation of Location Transformation
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▶ Element stiffness in global coordinate
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Derivation of Location Transformation
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▶ Example Structure
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Stiffness Assembly
§ a" of the element
§ Element’s contribution to the global stiffness
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▶ Location vector (LV)
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Direct Stiffness Assembly
a"=
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Force Recovery Using Transformations
informationLV
Seoul National University Structural Design Laboratory 1. Form all element-related matrices for each element in the structure. [Form the
element stiffness for all elements in global coordinates (K#$), and the location vector (LV) and element contribution to the global load vector (aT%S&).]
a. Form the element stiffness in local coordinates (K').
b. Rotate the stiffness to global coordinates (a%TK
'a% ). [Usually, save K'a% (force transformation).]
c. Form the location vector (LV) for each element. (This is the compact form of the a" matrix.)
d. Form the element load contribution to the global loads (aT%S&). (Note that this will also have to be placed into the proper locations in the global load vector, the opposite of pulling out the proper displacements using the LV vector.)
2. Assemble all element stiffnesses (K#$) into the global stiffness as well as the element load contributions into the global load vector. Use the LV vector pointer method.
3. Add all concentrated loads to the global load vector.
4. Solve for the global displacements (r).
5. Recover the local element forces (S=FT𝜶v∗ − S𝒇). (Again, use the LV vector pointer method.)
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Split Transformation Analysis Procedure
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▶ Section 5.10 in the textbook (p.189~)
§ FRAME: Form elemental stiffness matrix
§ ADDK: Assemble global stiffness matrix
§ ADDL: Assemble load vector
§ MEMFRC: Recover member forces
▶ Examples in the textbook
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Stiffness Analysis Using CAL-90
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Three-Dimensional Beam Elements
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Final Stiffness Matrix
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Rotational Transformation in Three Dimensions
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▶ Necessity of local coordinate definition
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Definition of Local Coordinate
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Definition of Local Coordinate
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Transformation of Global Vector to Local Vector
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Transformation of Global Vector to Local Vector
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Transformation of Global Vector to Local Vector
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Definition of Local Coordinate in SAP2000
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▶ n Simultaneous linear equations
(1) Full unsymmetric 𝒂 (Gauss4)
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Numerical Matrix Solver: Gaussian Elimination 1
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Numerical Matrix Solver: Gaussian Elimination 2
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Numerical Matrix Solver: Gaussian Elimination 3
Seoul National University Structural Design Laboratory (4) Banded symmetric 𝒂in rectangular array (Gauss4)
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Numerical Matrix Solver: Gaussian Elimination 4
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Numerical Matrix Solver: Gaussian Elimination - Subroutines
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Numerical Matrix Solver: Gaussian Elimination - Subroutines
Seoul National University Structural Design Laboratory In problems, we have used “deformation degrees of freedom” for the elements. That is,
§ For a truss bar:
§ For a beam: or
Hence the stiffness matrix for a truss element is 1x1, and for a beam element is 3x3 (or 2x2 for a beam with a hinge).
This is, however, not the only way to define element deformations. In particular, the Direct Stiffness Method requires the use of “displacement degrees of freedom”, so that the element deformations are the some as nodal displacements.
Element stiffnesses based on displacement DOFs can be set up using either “local”
(element) coordinates or “global” (structure) coordinates, as follows:
For the Direct Stiffness Method, the element deformation must correspond exactly to the nodal displacements. Hence, the global coordinates must be used. However, local coordinates may be useful as an intermediate step in forming the element stiffness.
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Beam & Truss Element Stiffnesses
v1
v1 v2 v3
v1 v2
v1
v2
v3
v4
v5
v6
Local v1
v2
v3
v5
v6
v4
Global
Seoul National University Structural Design Laboratory Consider the process of developing the global 6x6 stiffness for beam element, beginning with the 3x3 stiffness in terms of deformation dofs.
, set up the displacement transformation
To get [a’], impose unit value of each v’ in turn, then get {v}. This is COL,i of [a’].
E.g. for v2’=1
Hence is :
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Beam & Truss Element Stiffnesses
1, 1
S v
2, 2
S v
3, 3
S v
1 1
2 2
3 3
4 2 0
2 4 0
0 0
S EI L EI L v
S EI L EI L v
S EA L v
ì ü é ùì ü
ï ï= ê úï ï
í ý ê úí ý
ï ï êë úûï ï
î þ î þ
for a uniform beam with GAʹ=∞
1 , 1
S v¢ ¢
2, 2
S v¢ ¢
3, 3
S v¢ ¢ S v4¢ ¢, 4
5, 5
S v¢ ¢
6, 6
S v¢ ¢
[ ]
1 1
6 6
6 6
S v
S v
ì ü¢ ì ü¢
ï ï ï ï
ï ï= ´ ï ï
í ý í ý
ï ï¢ ï ï¢
ï ï ï ï
î þ î þ
! ! We can get this 6x6 by transforming the 3x3.
{ } 12 3
v
v v
v ì üï ï
=í ý ï ïî þ
{ } 1
6
v v
v ì ü¢
¢ =ï ïí ý ï ï¢ î þ
! { }
( )!
[ ]
( )!{ }
!( )
3 1 3 6 6 1
v a v
´ ´ ´
¢ ¢
=
COL,i={v} when vi’=1
v1 v2
2 1
v¢ =
1 2 3
1 1 0
v L
v L
v
ì ü ì- ü ï ï ï= - ï í ý í ý ï ï ïî ïþ î þ
Similarly for the other vi’
{ }v =
[ ]
a v¢ { }¢ 1 12
3 6
0 1 1 0 1 0
0 1 0 0 1 1
1 0 0 1 0 0
v L L v
v L L
v v
ì ü¢
ì ü é - ù ï ï
ï ï= ê - úï ï
í ý ê úí ý
ï ï êë- úûï ï¢
î þ ï ïî þ
!
Seoul National University Structural Design Laboratory We can also establish a force transformation, by statics as follows:
Hence:
Note that [C’]=[a’]T. This is not surprising - it can be proved by virtual work (Try this as an exercise).
Now:
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Beam & Truss Element Stiffnesses
{ }
!( )
[ ]
!( ){ }
!( )
6 1 6 3 3 1
S C S
´ ´ ´
¢ = ¢
S1
S2
S3
{S} system
= S3
1 2
S S L +
S1
S3
S2
1 2
S S L +
{S’} system
[ ]
1
2
1 3
2
4 3
5
6 '
0 0 1
1 1 0
1 0 0
0 0 1
1 1 0
0 1 0
C
S
S L L
S S
S
S S
L L
S S ì ü¢
ï ï é - ù
ï ï¢ ê- - ú
ï ï ê¢ ú ì ü
ï ï ê ú
ï ï= ï ï
í ý ê úí ý
ï ï¢ ê ú î þï ï
ï ï ê¢ ú
ï ï ê ú
ë û
ï ï¢
ï ïî þ !"""#"""$
{ }
!( )
[ ]
( )
!
{ }
!( )
[ ]
( )
!
[ ]
( )!
{ }
( )!
[ ]
( )
!
[ ]
( )!
[ ]
( )
!
[ ]( )
{ }
!( )
6 6
6 1 6 3 3 1 6 3 3 3 3 1 6 3 3 3 3 6 6 1
T T T
k
S a S a k v a k a v
´
´ ´ ´ ´ ´ ´ ´ ´ ´ ´
¢
¢ = ¢ = ¢ = ¢ ¢ ¢
"#$#%
: obtained by transforming [k](3x3)
Compare with Sʹ values in figure above.
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Element Stiffness Matrices for the Direct Method
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(2) 3D Truss
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Element Stiffness Matrices for the Direct Method
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Element Stiffness Matrices for the Direct Method
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Element Stiffness Matrices for the Direct Method
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Element Stiffness Matrices for the Direct Method
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Element Stiffness Matrices for the Direct Method
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Element Stiffness Matrices for the Direct Method
(5) 3D Frame
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Element Stiffness Matrices for the Direct Method
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Element Stiffness Matrices for the Direct Method
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Element Stiffness Matrices for the Direct Method
Seoul National University Structural Design Laboratory (5) 3D Frame
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