University of California, San Diego SE 201 – Fall 2004 2004 Class Notes Notes
3
3.1
′
F,
Departmen Departmentt of Structural Structural Engineering Instruct Instructor: or: J oel P. Conte
DIRECT STIFFNESS METHOD APPL IED TO 2-D FRAME STRUCTURES
GENERAL
′: ′
′
F = k ⋅
′
(3.1)
∆′ 2
9 0 ° N O T
∆′ 3 ∆′1
Figure 3.1
F′ , ′: F′5 , ∆′5
F′ = k ′ ⋅ ′
(3.2) F′4 , ∆′4
F′6 , ∆′6
j F′2 , ∆′2
i
F′1 , ∆′1
F′3 , ∆′3 Figure 3.2
3-2
F, : F = k ⋅
J
F5 , ∆ 5
(3.3)
F6 , ∆ 6 F4 , ∆ 4
j
F2 , ∆ 2 F3 , ∆ 3 i
F1 , ∆1
I
Figure 3.3
F5 , ∆ 5
F, : F = k ⋅
F6 , ∆ 6
(3.4)
F4 , ∆ 4
J d 2 y
j
, A, E, G F2 , ∆ 2 i F3 , ∆3
Y F1 , ∆1
d 1 y
I d 1 x X
Figure 3.4
, L A s, I φ
d 2 x
3-2
F, : F = k ⋅
J
F5 , ∆ 5
(3.3)
F6 , ∆ 6 F4 , ∆ 4
j
F2 , ∆ 2 F3 , ∆ 3 i
F1 , ∆1
I
Figure 3.3
F5 , ∆ 5
F, : F = k ⋅
F6 , ∆ 6
(3.4)
F4 , ∆ 4
J d 2 y
j
, A, E, G F2 , ∆ 2 i F3 , ∆3
Y F1 , ∆1
d 1 y
I d 1 x X
Figure 3.4
, L A s, I φ
d 2 x
3-3
Structure Stiffness Matrix:
3.1.1
K
100 k a
1
Element a: 2
Node I = node 1 b
6′
Node J = node 2
3
Y
∆x ( a )
= x ( a ) (node J) - x ( a ) (node I) = 8
∆y ( a )
= y( a ) (node J) - y( a ) (node I) = 0 2
( x (a ) )
L( a ) = 8′
cos φ
(a )
sin φ (a ) =
X
⎛ EI ⎞ ⎜ ⎟ ⎝L⎠
Figure 3.5
Element b: Node I = node 3 Node J = node 2 ∆x ( b )
= x ( b ) (node J) - x ( b ) (node I) = 8
∆y ( b )
= y( b ) (node J) - y( b ) (node I) = 0
L( b ) =
82 + 62 = 10 8 4 = = 0.80 cos φ (b ) = 10 5 6 = 0 .6 0 sin φ (b ) = 10 (b)
⎛ EA ⎞ k-in ; ⎜ = 165,000 k-in ⎟ ⎝ L ⎠
(b)
= 4,200
k in
L(a ) ∆y(a ) L(a )
=1 =0
(a)
⎛ EA ⎞ ⎜ ⎟ ⎝ L ⎠
⎛ EI ⎞ ⎜ ⎟ ⎝L⎠
=
∆x (a )
2
+ ( y( a ) ) = 8
= 25,000 k-in (a )
= 3,750
k in
3-4
Member a: EI
= 25,000 [k-in ] L EI = 25, 000 × 8 × 12 = 2, 400, 000 I =
2,400,000 29,000
= 82.759
[ in 4 ]
⎡ k ⎤ ⎢⎣ in ⎥⎦ L EA = 3, 750 × 8 × 12 = 360, 000 EA
A =
[ k-in2 ]
= 3,750
360,000 29,000
= 12.414
[ k]
[ in2 ]
Member b: EI
= 165,000 [k-in] L EI = 165, 000 × 10 × 12 = 19, 800, 000 I =
19,800,000 29,000
= 682.759
[ in4 ]
⎡ k ⎤ ⎢⎣ in ⎥⎦ L EA = 4, 200 × 10 × 12 = 504, 000 EA
A =
= 4,200
504,000 29,000
= 17.379
[ in2 ]
[ k]
[ k-in2 ]
3-5
3.1.2
Struct ure Degrees of Freedom Numberin g:
U2 U5 U3 U4
a
1
2
U1
U6
b U8
Node 2 : free node Node 1 and 3 : restrained nodes
U7
3 U9
Figure 3.6
T
Node displacements at free dof’s: U f = [ U1 U 2 U3 ]
T
Node displacements at restrained dof’s: Ud = [ U 4 U 5 U 6 U 7 U 8 U 9 ]
3-6
3.1.3
Equili briu m Equation s for Free DOF’s onl y:
F2(a)
P5 P6
(a) 1
(a) 1
F
1
P7
a
F
F6
F5(a)
(a)
F4 F4
F3(a) F2
F5(a)
F3(a) F5
(a)
(b)
T
F6(a)
P3
(a)
2
(a)
P1
F4(b) F6(b) F5(b)
F6 (b)
Pf = [ P1 P2 P3 ] : Applied forces
P2
F4 (b)
in free dof’s b
P1 = F4(a ) + F4(b ) P2 = F5(a ) + F5(b ) P3 = F6(a ) + F6(b )
F1(b)
P8
F3(b)
F2(b) F2
P7
3
F3(b)
(b)
(b) 1
F
T
Pd = [ P4 P5 P6 P7 P8 P9 P10 ] :
forces at dof’s with known displacements (support reactions)
P9
Figure 3.7
3.1.4
Direct Assembly/Formation of Equilibrium Matrix
Equilibrium equations for free and restrained dof’s: P1 = F4(a ) + F4(b ) P2 = F5(a ) + F5(b ) P3 = F6(a ) + F6(b ) P4 = F1(a ) P5 = F2(a ) P6 = F3(a ) P7 = F1(b ) P8 = F2(b ) P9 = F3(b )
: Equilibrium between internal (F’s) and external (P’s) forces
3-7
In matrix form:
= A (bf )T
= A (af )T ⎡ P1 ⎤ ⎡0 ⎢P ⎥ ⎢0 2 ⎢ ⎥ ⎢ ⎢ P3 ⎥ ⎢0 ⎢ ⎥ ⎢ ⎢ P4 ⎥ ⎡ P ⎤ ⎢1 f P = ⎢ P5 ⎥ = ⎢ ⎥ = ⎢0 ⎢ ⎥ ⎣ Pd ⎦ ⎢ ⎢ P6 ⎥ ⎢0 ⎢P ⎥ ⎢0 7 ⎢ ⎥ ⎢ ⎢ P8 ⎥ ⎢0 ⎢P ⎥ ⎢⎢0 ⎣ ⎣⎢ 9 ⎦⎥
0 0 1 0 0⎤
0 0 1 0 0⎤
0 0 0 1
0 0 0 1 0
0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
⎡0 ⎥ ⎢0 0 (a ) ⎥ ⎡ F1 ⎤ ⎢ ⎢0 1⎥ ⎢ ⎥ F ⎥ 2 ⎢ 0⎥ ⎢ ⎥ ⎢0 ⎢ F3 ⎥ 0⎥ ⋅ ⎢ ⎥ + ⎢ 0 ⎥ F4 ⎢ 0⎥ ⎢ ⎥ ⎢0 ⎢ F5 ⎥ ⎢1 0⎥ ⎢ ⎥ ⎥ ⎢⎣ F6 ⎥⎦ ⎢ 0⎥ ⎢0 ⎢⎢0 0⎥⎦⎥ ⎣
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0
Pf ( e ) = A (bf e )T ⋅ F (e ) (e) f
= A(bfe ) ⋅ U f
(3.5)
= A (bd)T
= A (ad)T
Notes:
⎥ ⎥ ⎡ F1 ⎤ (b ) 1⎥ ⎢ ⎥ ⎥ F2 0⎥ ⎢ ⎥ ⎢ F3 ⎥ 0⎥ ⋅ ⎢ ⎥ ⎥ F4 0⎥ ⎢ ⎥ ⎢ F5 ⎥ 0⎥ ⎢ ⎥ ⎥ ⎢⎣ F6 ⎥⎦ 0⎥ 0⎥⎦⎥
Pd ( e ) = A (bde )T ⋅ F (e )
(e )
(e) f
+
(e ) d
(e) d
= A (bde ) ⋅ U d
From element equilibrium, we can express the complete set of element end forces in the global reference system (F1 , F2 , F3, F4 , F5 , F6 ) in terms of the basic forces (F1′ , F 2′ , F3′ ) as F = Γ REZ T ⋅ Γ ROT T ⋅ Γ RBM T ⋅ F′
(∆′ = Γ RBM ⋅ Γ ROT ⋅ Γ REZ ⋅ )
3-8
=I
Here
Γ REZ
Γ ROT
⎡ ∆x ⎢ L ⎢ ⎢ − ∆y ⎢ L ⎢ 0 =⎢ ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ ⎣⎢ 0
Γ RBM
⇒
⎡ ⎢0 ⎢ =⎢0 ⎢ ⎢ −1 ⎢ ⎣⎢
ΓT
∆y
L ∆x
1 L 1
=
(identity matrix, since
0
0
0
L 0
0
0
0
1
0
0
0
0
∆x
∆y
L
0
0
∆y
L ∆x
0
0
L 0
L 0
1
0
−
−
1 L 1
)
⎤
0⎥
⎥ 0⎥ ⎥ ⎥ 0 ⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ 1⎦⎥
⎤
0⎥
⎥ 0 0 − 1⎥ ⎥ L L ⎥ 0 0 1 0 0 ⎥ ⎦⎥
⎡ ∆y ⎢ − L2 ⎢ ⎢ ∆x ⎢ L2 ⎢ 1 = Γ REZT ⋅ ΓROT T ⋅ ΓRBM T = ⎢ ⎢ ∆y ⎢ 2 ⎢ L ⎢ − ∆x ⎢ L2 ⎢ ⎢⎣ 0
−
∆y
L2 ∆x L2 0
− −
∆x ⎤
L ⎥ ⎥ ∆y ⎥
L ⎥ ⎥ 0
∆y
∆x
L2 ∆x
L ∆y
L2 1
L 0
−
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
3-9
Thus,
⎡0 ⎢0 ⎢ ⎢0 ⎢ 1 ⎡ Pf ⎤ ⎢ P = ⎢ ⎥ = ⎢0 ⎣ Pd ⎦ ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢⎢0 ⎣
0
0 1
0
0 0
0 1
0 0
0 0
0
0 0
1
0
0 0
0
1
0
0
0
0 0
0
0
0 0
0
0 0
0
0
0
0⎤
⎡0 ⎥ ⎢0 0 ⎥ ⎢ ⎥ ⎢0 1 ⎥ ⎢ 0⎥ ⎢0 ( a ) ′ 0⎥ ⋅ Γ ( a )T ⋅ F + ⎢ 0 ⎥ ⎢ 0⎥ ⎢0 ⎢1 0⎥ ⎥ ⎢ 0⎥ ⎢0 ⎢⎢ 0 0⎥⎥⎦ ⎣
0
0 1
0
0 0
0 1
0 0
0 0
0 0
0
0
0 0
0
0
0
0 0
0
0 0
0
0
1
0
0
0
0
1
0
0
0⎤
⎥ ⎥ 1⎥ ⎥ 0⎥ ′( b ) 0⎥ ⋅ Γ (b )T ⋅ F ⎥ 0⎥ 0⎥ ⎥ 0⎥ 0⎥⎥⎦ 0
3-10
3.1.5
Direct Assembly/Formation of Compatibility Matrix
Compatibility between node displacements in global reference system and element end displacements in global reference system: U1 = ∆ 4 (a) = ∆4 (b)
1
a
2
b
3
U 2 = ∆5 (a) = ∆5 (b)
1
a
2
b
3
1
a
2
U3 = ∆6 (a) = ∆6 (b)
b
U1 = ∆ 4(a) = ∆ 4 (b) U 2 = ∆ 5(a) = ∆ 5 (b)
3
U 3 = ∆ 6 (a) = ∆ 6 (b) Figure 3.8
3-11
(a)
(b )
=
=
(a) f
(b ) f
⎡ ∆1 ⎤ ⎢∆ ⎥ ⎢ 2⎥ ⎢∆ ⎥ = ⎢ 3⎥ ⎢∆ 4 ⎥ ⎢∆5 ⎥ ⎢ ⎥ ⎣⎢∆ 6 ⎦⎥
(a)
⎡ ∆1 ⎤ ⎢∆ ⎥ ⎢ 2⎥ ⎢∆ ⎥ = ⎢ 3⎥ ⎢∆ 4 ⎥ ⎢∆5 ⎥ ⎢ ⎥ ⎢⎣ ∆ 6 ⎥⎦
(b )
⎡0 ⎢0 ⎢ ⎢0 =⎢ ⎢1 ⎢0 ⎢ ⎣⎢0 ⎡0 ⎢0 ⎢ ⎢0 =⎢ ⎢1 ⎢0 ⎢ ⎢⎣0
0⎤
0
0
⎥ ⎥ ⎡ U1 ⎤ 0⎥ ⎢ ⎥ (a ) ⎥ ⋅ ⎢ U 2 ⎥ = A bf ⋅ U f 0⎥ ⎢⎣ U 3 ⎥⎦ 0⎥ ⎥ 1 ⎦⎥
0
0⎤
0
0
0 0 1
⎥ ⎥ ⎡ U1 ⎤ 0⎥ ⎢ ⎥ ⋅ Uf A (bfb ) ⎥ ⋅ ⎢U2 ⎥ = 0⎥ displacement ⎢⎣ U 3 ⎥⎦ Boolean transformation matrix 0⎥ ⎥ 1 ⎥⎦
0
0
0 0
N
1 0
′ ′ ′ Using element compatibility, we can related the element deformations ( ∆1 , ∆ 2 , ∆3 ) to the element end displacements in the global reference system ( ∆1 , ∆ 2 , ... , ∆ 6 ) :
′( a ) = Γ ( a ) ⋅ where
⎡ ∆y ⎢ − L2 ⎢ ⎢ ∆y Γ= − ⎢ L2 ⎢ ∆x ⎢− ⎢⎣ L
⇒
Γ
(a)
′( b ) = Γ ( b ) ⋅
;
= Γ RBM ⋅ Γ ROT ⋅ Γ REZ (here Γ REZ = I) ∆x
L2 ∆x L2
−
∆y
L
1 0 0
∆y
L2 ∆y L2 ∆x L
− −
∆x
L2 ∆x
L2 ∆y L
′( a ) = Γ( a ) ⋅ A (a ) ⋅ U f f ′(b ) = Γ( b ) ⋅ A (b ) ⋅ U f f
⎤
0⎥
⎥ 1⎥ ⎥ ⎥ 0⎥ ⎥⎦
(b )
3-12
⎡ ′( a ) ⎤ ⎥= Form ⎢ ⎢ ′( b ) ⎥ ⎥⎦ ⎣⎢
′=A
where
f
⋅ U f
(a ) ⎡ Γ( a ) ⋅ A bf ⎤ A f = ⎢ ⎥: ( b ) ⋅ A (b ) Γ bf ⎦ ⎣
structure compatibility matrix for free dof’s
3-13
3.1.6
Direct Stiffness Implementatio n
In the direct stiffness implementation of the displacement method, we proceed as follows:
•
We realize that the equilibrium equations at the free dof’s can be written as (see p. 3-9):
′( a ) + A (b )T ⋅ Γ (b )T ⋅ F′(b )
Pf = A (bfa )T ⋅ Γ ( a )T ⋅ F
•
bf
′
′
We express the basic element forces F in terms of basic element deformations according to the force-deformation relation of element “e” :
′( e ) = k ′( e ) ⋅ ′(e )
F
•
Finally, the basic element deformations can be expressed in terms of the displacements at the free global dof’s according to (see p. 3-11):
′( e ) = Γ( e ) ⋅ A (e ) ⋅ U f f •
In the absence of initial forces, we obtain Pf = A (bfa )T ⋅ Γ ( a )T ⋅ k ′
⇒ •
(a)
(a) (b ) (b ) ⋅ ′ + A (bfb )T ⋅ Γ(b )T ⋅ k ′ ⋅ ′
′( a ) ⋅ Γ( a ) ⋅ A (a ) ⋅ U + A (b )T ⋅ Γ (b )T ⋅ k ′(b ) ⋅ Γ(b ) ⋅ A(b ) ⋅ U bf f bf bf f
Pf = A (bfa )T ⋅ Γ ( a )T ⋅ k
We note that the product Γ ( a )T
(a) ⋅ k ′ ⋅ Γ( a ) = k ( a )
is the element stiffness matrix in the global reference system.
•
After factoring out the displacements at the free global dof’s, we obtain Pf = ⎡⎣ A (bfa )T ⋅ k ( a ) ⋅ A (bfa ) + A (bfb )T ⋅ k (b ) ⋅ A (bfb ) ⎤⎦ ⋅ U f
•
We recognize in the square brackets the structure stiffness matrix for the free global dof’s, and generalizing to Nel elements, we have K ff =
Nel
∑A e =1
(e )T bf
⋅ k ( e ) ⋅ A (bfe )
(3.6)
3-14
3.1.7
•
Direct Assemb ly of Struct ure Stiffness Matrix
From K ff =
Nel
∑A
(e )T bf
recognize that the structure stiffness matrix is derived ⋅ k ( e ) ⋅ A (bfe ) , we
e =1
from the summation of element stiffness contributions, as long as these are expressed first in the global reference system.
•
We also recognize that the compatibility matrices A ( f ) are Boolean matrices of 1’s and e
0’s. The 1’s lie at the rows and columns corresponding to the relation between local (element) and global (structure) dof’s (each row and column thus contains at most one non-zero term). Consequently, multiplication by A ( f ) involves the positioning of the e
terms of the element stiffness matrix in the appropriate address of the structure stiffness matrix. Element contributions are summed up in the process.
We illustrate this for
element a in the example:
(a) A bf
⎡0 ⎢0 ⎢ ⎢0 =⎢ ⎢1 ⎢0 ⎢ ⎣⎢ 0
0 0 0 0 1 0
0⎤
⎥ ⎥ which indicates that element dof 4 corresponds to global (structure) dof 1, 0⎥ ⎥ element dof 5 to global dof 2, and element dof 6 to global dof 3. The 0⎥ 0⎥ other element dof’s correspond to restrained global (structure) dof’s. ⎥ 1 ⎦⎥ 0
⇒ “ID” array for element a reads
⎡ 4⎤ ⎡0 ⎤ ⎢5⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢6⎥ ⎢0⎥ ⎢ ⎥→⎢ ⎥ ⎢1 ⎥ ⎢1 ⎥ ⎢ 2⎥ ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ 3⎦⎥ ⎣⎢ 3⎦⎥
3-15
⎡7 ⎤ ⎡0⎤ ⎢8 ⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢9 ⎥ ⎢0⎥ ⎢ ⎥→⎢ ⎥ ⎢ 1 ⎥ ⎢1 ⎥ ⎢ 2⎥ ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ 3⎦⎥ ⎣⎢ 3⎦⎥
Similarly, “ID” array for element b reads
The “ID” array of the element provides the scheme for the proper addressing of the element stiffness coefficients into the structure stiffness matrix as follows:
“ID”
k ( a )
K ff
[
0
(a ) ⎡ 0 ⎤ ⎡ k11 ⎢ 0 ⎥ ⎢ (a ) ⎢ ⎥ ⎢ k 21 (a ) ⎢ 0 ⎥ ⎢ k 31 = ⎢ ⎥ ⎢ (a ) ⎢1 ⎥ ⎢ k 41 (a ) ⎢ 2⎥ ⎢ k 51 ⎢ ⎥ ⎢ (a ) ⎢⎣ 3⎥⎦ ⎣⎢ k 61
⎡ K11 K12 = ⎢⎢ K 21 K 22 ⎢⎣ K 31 K 32
3
]
0
0
1
2
(a ) k12
(a ) k13
(a ) k14
(a ) k15
(a ) ⎤ k16
(a ) k 22
(a ) k 23
(a ) k 24
(a ) k 25
(a ) k 26 ⎥
(a ) k 32
(a ) k 33
(a ) k 34
(a ) k 35
(a ) ⎥ k 36
(a ) k 42
(a ) k 43
(a ) k 44
(a ) k 45
(a ) k 46 ⎥
(a ) k 52
(a ) k 53
(a ) k 54
(a ) k 55
(a ) ⎥ k 56
(a ) k 62
(a ) k 63
(a ) k 64
(a ) k 65
(a ) k 66 ⎦⎥
⎥ ⎥ ⎥
K 13 ⎤
⎥ ⎥ K 33 ⎥⎦
K 23
Note: In Matlab (Fedeaslab), the redirection of the stiffness coefficients of a single element into the appropriate addresses of the structure stiffness matrix can be accomplished with vector indexing in a single command:
K(id , id ) = K(id , id ) + kh
structure stiffness matrix
element stiffness matrix
3-16
3.1.8
Direct Assemb ly of Resisting Force Vector
Equilibrium equations (at the structure level): Pr = Pf :
express static equilibrium between external forces and internal (resisting) forces.
where Pr = Resisting force vector in free global dof's U f Pf = Applied (external) forces in free global dof's U f
Pr = A (bfa )T ⋅ F ( a ) + A (bfb )T ⋅ F (b )
Generalizing to Nel elements in the model: Pr =
Nel
∑A
( e )T bf
⋅ F (e )
(3.7)
e =1
It is important to recall that we have not yet included the initial end forces (fixed-end forces) due to temperature changes, shrinkage, lack-of-fit (unintentional deviations from the reference geometry), prestressing, etc. We’ll include these effects later.
′( e ) = Γ( e)T ⋅ F′( e )
F ( e ) = Γ TREZ ⋅ ΓTROT ⋅ ΓTRBM ⋅ F e T ( ) =Γ
From element force-deformation relation:
′( e ) = k ′(e ) ⋅ ′( e )
F
where
⇒
′( e ) = Γ ( e ) ⋅
(e )
= Γ (e ) ⋅ A (ef) ⋅ U f
′( e ) ⋅ Γ(e ) ⋅ A (e ) ⋅ U
F ( e ) = Γ( e )T ⋅ k
f
f
3-17
3.1.9
Implementatio n of Direct Assemb ly
From the preceding discussion, it is apparent that it is possible to directly assemble the structure resisting force vector (and thus express the structure equilibrium equations) and the structure stiffness matrix. These two operations can be expressed in compact form as follows.
K ff =
Nel
∑A
(e )T bf
⋅ k ( e ) ⋅ A (bfe ) :
structure stiffness matrix
(3.6)
e =1
k ( e )
where
(e ) = ΓTREZ ⋅ Γ TROT ⋅ Γ TRBM ⋅ k ′ ⋅ Γ RBM ⋅ ΓROT ⋅ Γ REZ
= Γ(e )T Pr =
Nel
∑A
( e )T bf
= Γ(e)
⋅ F(e)
(3.7)
e =1
where, for linear elastic material response,
′( e ) ⋅ Γ (e ) ⋅ A (e ) ⋅ U
F ( e ) = Γ (e )T ⋅ k
f
f
= k ( e ) ⋅ A bf(e ) ⋅ U f ⇒
Pr = K ff ⋅ U f :
Structure resisting force vector
The structure equilibrium equations read:
Pf = Pr
3.1.10 Struct ure State Determin ation
The process of determining the structure stiffness matrix K ff and the structure resisting force vector Pr is called structure state determination. During this process, each element “e” in the model is called to supply its stiffness matrix k ( e ) and resisting force vector F ( e ) in global coordinates. This sub-process is known as element state determination.
3-18
Back to the example: Element a: (a)
Γ REZ
units: k-in
=I ⎡c ⎢- s ⎢ ⎢0 =⎢ ⎢0 ⎢0 ⎢ ⎣⎢ 0
(a)
Γ ROT
(no rigid end-offsets) s
0
0
0
c
0
0
0
0
1
0
0
0
0
c
s
0
0
- s
c
0
0
0
0
0⎤
⎡1 ⎥ ⎢0 0 ⎥ ⎢ 0⎥ ⎢ 0 ⎥=⎢ 0⎥ ⎢ 0 0⎥ ⎢ 0 ⎥ ⎢ 1⎦⎥ ⎣⎢ 0
0
0
0
0
1
0
0
0
0 1
0
0
0
0 1
0
0
0
0 1
0 0
0 0
0⎤
⎥ ⎥ 0⎥ ⎥=I 0⎥ 0⎥ ⎥ 1⎦⎥ 0
c = cos φ = 1 s = sin φ = 0
L = 8 ft = 96 in
Γ RBM
′
k
(a)
⎡ ⎢0 ⎢ =⎢0 ⎢ ⎢ −1 ⎢ ⎢⎣
⎡ 4 EI ⎢ L ⎢ 2 EI =⎢ ⎢ L ⎢ ⎢ 0 ⎣⎢
1 1 ⎡ ⎤ 0 1 0 − 0⎥ ⎢ 96 96 L L ⎥ ⎥ ⎢ 1 1 1 1 0 0 − 1⎥ = ⎢ 0 0 0 − 1⎥ ⎢ ⎥ ⎥ L L 96 96 ⎥ ⎥ ⎢ −1 0 0 1 0 0 1 0 0 0 0⎥ ⎥ ⎢ ⎥⎦ ⎣⎢ ⎥⎦ 1
1
0
−
⎤ 0 ⎥ L ⎥ 4 EI 0 ⎥ ⎥ L ⎥ EA ⎥ 0 L ⎦⎥ 2 EI
1
⎤
0⎥
(a )
0 ⎡100, 000(k-in) 50, 000(k-in) ⎤ ⎥ = ⎢⎢ 50, 000(k-in) 100, 000(k-in) 0 ⎥ ⎢⎣ 0 0 3, 750(k/in)⎥⎦
3-19
a )T (a )T a )T k ( a ) = Γ (REZ ⋅ ΓROT ⋅ Γ(RBM ⋅ k ′
“ID” array of element a
[
0
0
(a)
0
a) a) a) ⋅ Γ (RBM ⋅ Γ(ROT ⋅ Γ (REZ
1
2
3
]
0 0 0 0 − 3, 750. ⎡ 0 ⎤ ⎡ 3, 750. ⎤ ⎢0 ⎥ ⎢ 0 ⎥ 35.552 1,562.5 0 35.552 1, 562.5 − ⎢ ⎥⎢ ⎥ ⎢0 ⎥ ⎢ 0 1, 562.5 100, 000. 0 − 1, 562.5 50, 000. ⎥ = ⎢ ⎥⎢ ⎥ 0 0 3, 750. 0 0 ⎢1 ⎥ ⎢ −3, 750. ⎥ ⎢ 2⎥ ⎢ 0 0 32.552 − 1, 562.5⎥ −32.552 −1, 562.5 ⎢ ⎥⎢ ⎥ 1, 562.5 50, 000. 0 100, 000.⎥⎦ − 1, 562.5 100,000. ⎢⎣ 3 ⎥⎦ ⎢⎣ 0 to K 11 “ID” array:
∆1( ) = 0 a
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⇒ “ID” array of element a = ⎢ ⎥ ⎢1 ⎥ ⎢2⎥ ⎢ ⎥ ⎣⎢ 3 ⎦⎥
∆(2 ) = 0 a
∆(3 ) = 0 a
∆(4 ) = U1 a
∆(5 ) = U 2 a
∆(6 ) = U3 a
Element b: (b)
Γ REZ
(b)
Γ ROT
=I ⎡c ⎢- s ⎢ ⎢0 =⎢ ⎢0 ⎢0 ⎢ ⎢⎣ 0
(no rigid end-offsets) s
0
0
0
c
0
0
0
0 1
0
0
0
0
c
s
0
0
- s
c
0 0
0
0
0⎤
0 0 0⎤ ⎡ 0.8 0.6 0 ⎥ ⎢ −0.6 0.8 0 ⎥ 0 0 0 0 ⎥ ⎢ ⎥ 0⎥ ⎢ 0 0 1 0 0 0⎥ ⎥=⎢ ⎥ 0⎥ ⎢ 0 0 0 0.8 0.6 0⎥ 0⎥ ⎢ 0 0 0 −0.6 0.8 0⎥ ⎥ ⎢ ⎥ 1⎥⎦ ⎢⎣ 0 0 0 0 0 1⎥⎦
c = cos φ = 0.80 s = sin φ = 0.60
to K 22
to K 23
3-20
( b )
Γ RBM
′
k
(b)
⎡ ⎢0 ⎢ =⎢0 ⎢ ⎢ −1 ⎢ ⎣⎢
1 L 1 L 0
⎡ 4 EI ⎢ L ⎢ 2 EI =⎢ ⎢ L ⎢ ⎢ 0 ⎢⎣
1 0
0 0
− −
0 1
⎤ 0 ⎥ L ⎥ 4 EI 0 ⎥ ⎥ L ⎥ EA ⎥ 0 L ⎥⎦ 2 EI
L = 10 ft = 120 in 1 ⎤ ⎡ 1 0⎥ ⎢ 0 120 ⎥ ⎢ 1 1⎥ = ⎢ 0 0 ⎥ ⎢ 120 ⎥ ⎢ −1 0 0 0 ⎢
1 L 1
L 0
⎥ ⎦⎥
⎣⎢
0
−
⎤
1
0⎥
120 1
⎥ 0 − 1⎥ ⎥ 120 ⎥ 1 0 0⎥ ⎦⎥
(b)
0 ⎡ 660, 000(k-in) 330, 000(k-in) ⎤ ⎥ 0 = ⎢⎢330, 000(k-in) 660, 000(k-in) ⎥ ⎢⎣ 0 0 4, 200(k/in)⎥⎦
′(b ) ⋅ Γ(b ) ⋅ Γ(b ) ⋅ Γ(b ) RBM ROT REZ
b )T b )T (b )T ⋅ Γ(ROT ⋅ ΓRBM ⋅ k k (b ) = Γ (REZ
“ID” array of element b
[
0
0
0
1
⎡0 ⎤ ⎡ 2, 737.5 1,950. − 4, 950. − 2, 737.5 ⎢0 ⎥ ⎢ 1,950. 1, 600. 6, 600. − 1,950. ⎢ ⎥⎢ ⎢0 ⎥ ⎢ −4,950. 6, 600. 660,000. 4,950. = ⎢ ⎥⎢ 2, 737.5 ⎢1 ⎥ ⎢ −2, 737.5 −1,950. 4,950. ⎢ 2 ⎥ ⎢ −1,950. −1, 600. −6, 600. 1, 950. ⎢ ⎥⎢ ⎣⎢ 3 ⎦⎥ ⎣⎢ −4,950. 6, 600. 330, 000. 4,950. to K 11
2
3
]
− 1,950. − 4, 950. ⎤ − 1, 600. 6, 600. ⎥⎥ −6, 600. 330, 000.⎥ ⎥ 1,950. 4, 950. ⎥ 1, 600. − 6, 600. ⎥ ⎥ − 6, 600. 660, 000.⎦⎥ to K 22
to K 23
3-21
Direct assembly of structure stiffness matrix: K ff =
Nel=2
∑A
(e )T bf
⋅ k ( e ) ⋅ A (bfe )
e =1
1, 950. 4, 950. ⎤ ⎡ 6, 487.5 = ⎢⎢ 1, 950. 1, 632.552 − 8,162.5⎥⎥ ⎢⎣ 4,950. −8,162.5 760, 000.⎥⎦
K f f ⋅ U f = Pf
Equilibrium equations: 1 ⎡ 6, 487.5
⎢
2 ⎢ 1,950. 3 ⎢⎣ 4,950.
⎡ 0 ⎤ ⎥⎢ ⎥ ⎢ ⎥ 1, 632.552 − 8,162.5 U2 = − 100 ⎥⎢ ⎥ ⎢ ⎥ −8,162.5 760, 000.⎥⎦ ⎢⎣ U3 ⎥⎦ ⎢⎣ 0 ⎥⎦ 1,950.
4, 950. ⎤ ⎡ U1 ⎤
Eq. 1
expresses the horizontal equilibrium of node 2 (free node 1).
Eq. 2
expresses the vertical equilibrium of node 2 (free node 1).
Eq. 3
expresses the rotational equilibrium of node 2 (free no de 1).
Solve equilibrium equations for U f (using Matlab) Uf = K ff ⋅ Pf -1
⎡ 0.03361 ⎤ in = ⎢⎢ −0.10831 ⎥⎥ in ⎢⎣ −0.001382⎥⎦ rad
Check that
Pf - K ff ⋅ U f Pf
1
3-22
3.1.10.1 Element State Determination (more generally know n as stress reco very phase in finite element analysis)
Element a:
(a)
′
′
F
(a )
(a)
0 ⎡0⎤ ⎡ ⎤ ⎢0⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢ ⎥ ⎢0⎥ ⎢ ⎥ 0 =⎢ ⎥=⎢ ⎥ ⎢ U1 ⎥ ⎢ 0.03361 ⎥ ⎢ U 2 ⎥ ⎢ −0.10831 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ U3 ⎥⎦ ⎢⎣ −0.001382⎥⎦
⎡ 0.0011282 ⎤ rad a) a) a) = Γ(RBM ⋅ Γ (ROT ⋅ Γ(REZ ⋅ ( a ) = ⎢⎢ −0.00025394⎥⎥ rad ⎢⎣ 0.033610 ⎥⎦ in
= k ′ ⋅ ′ (a)
(a )
0 ⎤ ⎡100, 000. 50, 000. ⎥ ⋅ ′(a ) 0 = ⎢⎢ 50, 000. 100, 000. ⎥ ⎢⎣ 0 0 3, 750.⎥⎦
⎡100.12⎤ ← M(I a ) (k-in) = ⎢⎢ 31.02 ⎥⎥ ← M(Ja ) (k-in) ⎢⎣ 126. ⎥⎦ ← N( a ) (k) Element b:
(b )
0 ⎡0⎤ ⎡ ⎤ ⎢0⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢ ⎥ ⎢0⎥ ⎢ ⎥ 0 =⎢ ⎥=⎢ ⎥ ⎢ U1 ⎥ ⎢ 0.03361 ⎥ ⎢ U 2 ⎥ ⎢ −0.10831 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ U 3 ⎥⎦ ⎢⎣ −0.001382⎥⎦
3-23
′
F′
(b)
(b)
⎡ 0.00089 ⎤ rad b) b) b) = Γ (RBM ⋅ Γ(ROT ⋅ Γ (REZ ⋅ (b ) = ⎢⎢ −0.000492⎥⎥ rad ⎢⎣ −0.038098⎥⎦ in
= k ′ ⋅ ′ (b )
(b )
0 ⎤ ⎡ 660, 000. 330, 000. ⎥ ⋅ ′(b ) = ⎢⎢330, 000. 660, 000. 0 ⎥ ⎢⎣ 0 0 4, 200.⎥⎦
⎡ 425.1 ⎤ ← M(Ib ) (k-in) = ⎢⎢ −31.02⎥⎥ ← M(Jb ) (k-in) ⎢⎣ −160. ⎥⎦ ← N(b ) (k)
3-24
3.1.10.2 Check Nodal Equilibr ium and Determine Supp ort Reactio ns
Support reactions at Node 1
1.366 126
1
100.12
1.366 =
100.12+31.02 8 (12 )
126
1.366 31.02 126 126
8 ft
126
100.12 1.366
31.02
100.12
1.366
0 1 6
Units: k, in
0 1 6
100
2
31.02 3.28
3.28 5
4 5
t f 0 1
3
4
(160 ) − (3.28 )
31.02
Equilibrium Check
∑F
x
5 = 126.03
4
= −126 + (160) 5
3
0 1 6
3.28 =
425 3.28
− (3.28) = +0.032 5 ≈0
425
0 1 6
3
3
425 − 31.02 10 (12 )
Support reactions
∑F
x
3 5
4
= −100 + 1.366 + (3.28) 5
3
+ (160) = − 0.01
at Node 3
425
5 ≈0
4
(160 ) + ( 3.28 ) = 98.624 5
∑M
2
(round-off error)
= −31.02 + 31.02 =0
Figure 3.9
(round-off error)
3-25
3.1.10.3 Check Global Equilibrium
100
1.366 126
1
a
100.12
2
∑F
= −126 + 126.03 = +0.03 ≈ 0 (round-off error)
∑F
= 1.366 − 100 + 98.624 = − 0.01 ≈ 0 (round-off error)
x
y
b 98.624 126.03
∑M
2
= 100.12 + 126(6)(12) + 425 − 100(8)(12) = = −2.88 ≈ 0
(round-off error)
3
425 Figure 3.10
Draw internal forces (M,V, N) diagrams according to sign c onvention:
•
Bending moment diagram M:
−100.12
2 0 . 1 3
−
31.02 5 2 4
−
Figure 3.11
3-26
Shear force diagram V: 8 2 3.
1.37
1.37
8 2 3.
Figure 3.12
•
Axial force diagram N: 126
0 1 6
126
−
0 1 6
−
Figure 3.13
Draw accurate sketch of deflected shape consistent with internal forces, especially bending moment diagram:
+ 0.0011282 rad
1
− 0.00025394 rad 2
c h o r d
inflection point 100 k 2
3
− 0.000492 rad + 0.00089 rad
Figure 3.14
3-27
If the above structure were to resist as a truss: (5.8% difference with frame results)
133.33 4 = (100) 3
133.33
133.33 1
100
2
0 . 7 6 166.67 6 6 1 5
= (100 ) 3
(4% difference with frame results)
7 6 6. 6 1
133.33
7 6 6. 6 1
3 100 Figure 3.15
⇒ Frame action is very modest in this structural system.
3-28
3.2
SLAVING
Slaving is a particular case of elements with Rigid End Zones. In this case, the dof’s of several nodes are “slaved” to those of a Master node. In other words, the displacements of several nodes are rigidly linked to those of a single Master node. The typical example is that of a rigid floor, where the displacements of the story columns are rigidly linked to those of the rigid floor, represented by a Master node. In the figure below, the displacements at the top of T (i) column i, ∆ = [ui vi θi ] , depend on (are rigidly linked to) those of Master node M (typically taken as the center of stiffness or the center of mass of the floor), T vM θM ] . ∆ M = [uM Rigid floor (in- and out-of-plane)
vM Master node
θM
uM
Axially rigid columns
Single story shear building model
xi y
vi
θM
vM
i
θi ui x
yi
uM
For node i, whose coordinates are ( xi, yi):
∆
(i)
=
(i ) Γ REZ
⋅ ∆M
⎡1 0 − yi ⎤ = ⎢⎢0 1 xi ⎥⎥ ⋅ ∆ M ⎢⎣0 0 1 ⎥⎦
Based on the properties of the transformation matrices: (i)
F M =
T
(i)
REZ
⋅ F(i )
3-29
where F
(i)
are the forces from the slab onto the top of column
Master node M , which are in equilibrium with F
F
(i )
T
(i )
i
and F
(i)
represent the forces at
. Thus:
T
T
i) (i ) ( i) ( i) = Γ REZ ⋅ F( i ) = Γ(REZ ⋅ k ( i) ⋅ ∆( i) = ΓREZ ⋅ k ( i) ⋅ ΓREZ ⋅ ∆M = k M( i) ⋅ ∆M
(i ) k M
where
k
(i )
=
T
(i)
REZ
⋅ k ( i ) ⋅
(i) REZ
After assembly of all the column contributions to the equilibrium equations of the slab, we have: K ⋅ ∆ M = FM
where F M denote the external forces applied to Master node M , K is the stiffness of master node M (or of the one-story building structure shown in the figure above), and
are the
displacements of master node M .
General Procedure (for one-story shear building model) :
(1) For each column element, find the element stiffness matrix in coordinates ∆ respectively: (i)
T
(i )
(i )
(i)
(2) Assemble the Master node stiffness matrix: number _ of _ columns
∑
k M =
by solving: F = K M ⋅ ∆ M
(4) Find the top column displacements: ∆
(i )
=
(i) REZ
⋅ ∆M
(5) Find the top column forces: F
(i )
(i)
k M
i =1
(3) Find ∆
( i)
k = Γ REZ ⋅ k ⋅ Γ REZ
k ,
= k (i ) ⋅ ∆( i )
(i)
and
∆
,
