1.571 Structural Analysis and Control
Prof. Connor
Section 4: Non­linear Analysis of Trusses
4,1 Notation and key definitions
4.1.1 Deformation
(+)
F
v
L+e
u
(­)
θ
ψ
L
u = u
v
α = cos θ sin θ
β = cos ψ sin ψ
( L + e ) 2 = ( L sin θ + v ) 2 + ( L cos θ + u ) 2
e
( L 2 + 2 Le + e 2 ) = L 2 + 2 eL ⎛ 1 + ------⎞ ≈ L 2 + 2 eL Assuming e « L
⎝
2 L⎠
Then
where
L 2 + 2 eL ≈ L 2 s 2 + 2 Lsv + v 2 + L 2 c 2 + 2 Lcu + u 2
s = sin θ and c = cos θ
Combine terms
L 2 ( s 2 + c 2 ) + 2 Lsv + v 2 + 2 Lcu + u 2 ≈ L 2 + 2 eL
1
e = cu + sv + ------ ( u 2 + v 2 )
2L
1
e = αu + ------ u T u
2L
β = cos ψ sin ψ
L cos θ + u
cos ψ = ------------------------- =
L+e
L sin θ + v
cos ψ = ----------------------- =
L+e
Then
cos θ + u ⁄ L
u
---------------------------- ≈ cos θ + --1+e⁄L
L
sin θ + v ⁄ L
v
--------------------------- ≈ sin θ + --1+e⁄L
L
β = cos θ + --u- sin θ + --vL
L
1
β = α + --- u T
L
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 1 of 15
4.1.2 Force
AE
F = ------- e
L
F has direction defined by ψ
Define
P = End action matrix
T
P = F cos ψ = Fβ
F sin ψ
At positive end
P
At negative end
P
+ end
-end
= +P
= –P
(+)
(+)
P+
ψ
P-
4.2 Incremental Equations
4.2.1 Deformation
Consider an incremental end displacement that changes u to u + ∆u . Find the corre­
sponding change in elongation from e to e + ∆e .
1
e = αu + ------ u T u
2L
1
(e + ∆e) = α(u + ∆u) + ------ (u + ∆u) T (u + ∆u)
2L
1
(e + ∆e) = αu + α∆u + ------ (u T u + uT ∆u + ∆u T u + ∆u T ∆u)
2L
1
1
∆e = (e + ∆e) – e = α∆u + --- u T ∆u + ------ ∆u T ∆u
L
2L
1
∆e = de + --- d 2 e
2
1
de = α∆u + --- u T ∆u = β∆u
L
1
d 2 e = --- ∆u T ∆u
L
If ∆u is small in comparison to u , it is reasonable to take ∆e ≈ de
Then
∆e ≈ β∆u
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 2 of 15
4.2.2 Force
σ
Et
Non Linear Stress­Strain
σ = σ(ε )
ε
∆ε
σ + ∆σ = σ ( ε + ∆ )
From a Taylor Series expansion
∂σ
1 ∂2 σ
∆σ = ------ ∆ε + --- --------2- ∆ε 2 + …
∂ε
2 ∂ε
∂σ
Note ------ = Et
∂ε
We want the change in stress due to the change in displacement
1
ε = --- e
L
1
1
1
∆ε = --- ∆e = --- ⎛ de + --- d 2 e⎞⎠
⎝
L
L
2
So
∂σ ∆e
1 ∂2 σ
∆σ = ------ ------ + ⎛ --- --------2- ∆ε 2⎞ + …
⎠
∂ε L ⎝ 2 ∂ε
1
∆σ = dσ + --- d 2 σ + …
2
Et
Et
∂σ∆e
dσ = ------ ------ = ----- de = ----- β∆u
∂ε L
L
L
∂σd 2 e ∂ 2 σ ∆e 2 ∂σd 2 e ∂ 2 σ
d 2 σ = -------------- + --------2- ⎛ ------⎞ ∼ -------------- + --------2- ( de ) 2
∂ε L
∂ε L
∂ε ⎝ L ⎠
∂ε
The incremental forces are
F = Aσ
Set
Then
A
∆F = A∆σ = Adσ + --- d 2 σ
2
AE t
AEt
dF = Adσ = --------- de = --------- β∆u
L
L
A
d 2 F = --- d 2 σ
2
AE
---------t = k t
L
dF = k t de = k t β∆u
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 3 of 15
4.2.3 End Actions
P = Fβ
T
Consider the change in u to u + ∆u and get the corresponding
(P + ∆P) = (F + ∆F)(β T + ∆β T )
(P + ∆P) = Fβ T + F∆β T + ∆Fβ T + ∆F∆β T
∆P = (P + ∆P) – P = F∆β T + ∆F∆β T
1
∆P = dP + --- d 2 P + …
2
dP = Fdβ T + dFβ T
2
∂β
1∂ β
β(u + ∆u) = β + ∆β = β + ------ ∆u T + --- --------2- ∆u T ∆u
∂u
2 ∂u
∂β
dβ = ------ ∆u T =
∂u
So
Can write as
where
1
dβ = --- ∆u T
L
⎧ ∂ ⎛
⎨ ------ ⎝ α
⎩ ∂u
⎫
1
+ --- u T⎞ ⎬∆u T
⎠
L
⎭
1
dβ T = --- ∆u
L
F
dP = k T β T β∆u + --- ∆u
L
dP = k T ∆u
k T = Tangent member stiffness matrix
F
k T = k T β T β + --- I
L
AE T
F
k T = ⎛ ----------⎞ β T β + --- I
⎝ L ⎠
L
where
I = 1 0 = Identity matrix
01
So
AE
⎛ ---------T-⎞ β T β
⎝ L ⎠
is the pseudo linear stiffness based on the deformed position defined by β or α .
and
F
--- I is the “geometric” stiffness due to initial forces.
L
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 4 of 15
4.2.4 Illustration ­ Geometric stiffness associated with an initial force
F sin ∆θ
F
F cos ∆θ
y
∆v
F
∆θ
F
L
x
∆u
∆u = ∆u
∆v
--­ 1 0 ∆u
Geometric term of dP = F
L 0 1 ∆v
F
dP = --­ ∆u
L ∆v
This can also be done by inspection
dP = F cos ∆θ – F
F sin ∆θ
Assuming small deformations ( e « L )
Therefore
L + ∆u
∆u
cos ∆θ = ---------------- ≈ 1 + ------L+e
L
∆v
∆v
sin ∆θ = ------------ ≈ ----­L+e L
∆u
F + F ------- – F
F
L
= --­ ∆u
dP =
L ∆v
∆v
F -----L
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 5 of 15
4.3 System of Incremental Equilibrium Equations
P E = External load vector
For Equilibrium
P int = Nodal load vector due to end actions
P E + (– Pint
) = 0
P E =
Pint
Let u* be an estimate of displacement due to P E
P int ( u* ) = Load vector (end actions) due to u* = P* int
Equilibirum imbalance = PE – P* int
Let ∆u be an estimate of the “correction” required to satisfy equilibrium.
The correction is found with: k*∆u = P E – P* int where k* is a “predictor” stiffness matrix.
Once ∆u is found, update P*int and k* and use to determine another correction.
This process is repeated until the equilibrium error satisfies the specified convergence
criteria.
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 6 of 15
Various choices for k*
1) k* = k linear ; the initial stiffness matrix
P
∆P
P1
uo
u 1 u2 u 3
Successive iterations use the initial stiffness
Can also start with an inital displacement
P
uo
2) k* = k t ; the instantaneous tangent stiffness matrix
P
∆P
P1
uo
u1
Update k t at each iteration. Requires less iteration but is computationally more expensive.
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 7 of 15
Example using k* = k t
AE
k = ------L
P
v
L
h
θ
Using symmetry
E = constant
h
sin θ = --L
P int = 2F sin θ
PE = P
P⁄2
θ
Find the tangent stiffness of the system
1
1
β = α + --- u T = α + --- u v
L
L
u = 0
α = cos θ sinθ
1
β = cos θ -­h- + --- 0 v = cos θ --h- + --v- = cos ψ sin ψ
L L
L L
F
dP = kβ T β∆u + --- ∆u
L
∆u =
0
∆v
dP = k
cos 2 ψ
sin ψcos ψ
sin ψ cosψ
sin 2 ψ
0 +F
--­ 0
∆v L ∆v
0
∆P ⁄ 2
Set
dP =
So
F
∆P
ksin 2 ψ∆v + --- ∆v = -----­L
2
h v 2 F
2 ⎛⎝ k ⎛⎝ --- + ---⎠⎞ + ---⎠⎞ ∆v = ∆P
L
L L
2
---⎞ must be updated.
At each interval k t (v, F) = 2 ⎛⎝ k ⎛⎝ --h- + --v-⎠⎞ + F
⎠
L
1.571 Structural Analysis and Control
Prof Connor
L
L
Section 4
Page 8 of 15
For initial step take
F = 0
V = 0
h2
→ k to = 2 k -----2
L
Set
P
k to v o = P → v o = -----­k to
and
P
F o = --------------­
2 sin ψ
Update k t to k t1 by including v o and Fo
Get
P 1 = k t1 v o
P 1 + ∆P1 = P → ∆P1 = P – P 1
Then
k t1 v o + k t1 ∆v = P
P – kt1 v 1
∆v 1 = ---------------------­
k t1
Repeat until ∆v < v tol or ∆Pn < P tol
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 9 of 15
4.4 Linearized Stability Analysis
4.4.1 Bifurcation
The system incremental equation expressed in terms of the instantaneous tangent stiffness matrix
∆P = K t ∆u
∆P = load increment from the equilibrium position (P*, u* )
∆u = first order estimate for displacement increment due to ∆P
Then ∆u can be evaluated using
∆u = ( Kt ) – 1 ∆ P
and update P*, u*, Kt and ∆P
Question: What happens if Kt becomes singular?
If Kt = 0 , Kt is singular and there exists a non­trivial solution of Kt ∆u = 0 . This
implies that there is more than one equilibrium position for a given load.
Suppose Kt = 0 for u* and P* . Then, u* + ∆u is also a solution for the loading P* .
The existence of multiple equilibrium positions for a given loading is called
“bifurcation.” In order to determine the structural behaviour in the neighborhood of a
bifurcation point, we need to include second and higher order terms in the incremental equilibrium equations. Bifurcation is viewed as an intermediate state. The position is neither stable nor unstable. The terminology “neutral equilibrium” is
used to interpret “bifurcation.”
4.4.2 Linearized Analysis
Consider the case where the equilibrium position (i.e. the displaced position) is very close to the initial unloaded position. In this case, it is reasonable to neglect the difference between the direction cosines for the initial and deformed configuration.
The expression for the member tangent stiffness reduces to
F
F
K t = k t β T β + --- I ≈ k t α T α + --- I
L
L
AE­
When the material is linear elastic and the member is prismatic k t = -----L
This can be stated
F
K t = K l + --- I
L
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 10 of 15
Using this result, the system tangent stiffness matrix is expressed as
Kt = Kl + Kg
where
K l = linear system stiffness matrix
K g = “geometric” stiffness matrix containing the terms involving
the memb er forces
K g depends on the loading. As the loading is increased, K t varies from its initial
linear value Kl . Depending on the loading and structural makeup, Kt may decrease
with increasing load and eventually reach 0, at which point a “bifurcation” occurs. To
evaluate whether this is possible, one evaluates Kl + Kg as a function of the loading.
4.4.2 Example 1
L
P, v
k1
A
u
B
1
2 k2
L
C
From equilibrium
F 2 = –P
k1 0
k t1 ≅ k l1 =
0 0
k t2 ≅ k l2 + k g2 =
k t2 =
Kt =
Solve
Giving
F
0 0
+ ----­2- 1 0
L 01
0 k2
0 0
+ –P ⁄ L 0
0 k2
0 –P ⁄ L
k1 – P ⁄ L
0
0
k2 – P ⁄ L
K t = ( k 1 – P ⁄ L)(k 2 – P ⁄ L) = 0
For P = k1 L ; ∆u is arbitrary
For P = k2 L ; ∆v is arbitrary
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 11 of 15
4.4.3 Modified Geometric Stiffness Matrix
1
β = ------------ (∆x + ∆u) T (exect expression)
L+e
For (+) and (­) ends
Y, v
∆x = difference in nodal coordinates
∆u = difference in nodal displacements
∆u
∆v
(+)
∆y
(­)
∆x
X, u
∆x = ∆x
∆y
∆u = ∆u
∆v
Case 1 ∆y ≅ 0 (Element ~ parallel to x direction)
∆u will be small wrt ∆x
1
∆x
β ≅ --L ∆y + ∆v
T
Then
∂β
1
dβ T = ---------- ∆du = --- 0
∂∆u
L ∆dv
and
Fdβ T
F
= --­ 0
L ∆dv
T
T
∆du
= 0 0
0 F ⁄ L ∆dv
Case 2 ∆x ≅ 0 (Element ~ parallel to y direction)
∆v will be small wrt ∆y
1
β ≅ -­ - ∆x + ∆u
L
∆y
Then
and
T
1
dβ T = --- ∆du
L 0
F
Fdβ T = --­ ∆du
L 0
1.571 Structural Analysis and Control
Prof Connor
T
T
= F ⁄ L 0 ∆du
0 0 ∆dv
Section 4
Page 12 of 15
Generalizing
F
k g = --- E
L
Examine
α = αx αy
If αx close to 1, take E = 0 0
01
If αy close to 1, take E = 1 0
00
4.4.4 Example 2
P
L1
R
1
Y, v
F1 = – R
F2 = – P
2 L2
X, u
0 0
k t1 ≅
+
R
0 – ---­0 0
L1
k1 0
P
– ----- 0
0 0
k t2 ≅
+ L2
0 k2
0 0
Kt =
P
k 1 – ----L2
0
0
R
k 2 – ----­
L1
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 13 of 15
4.4.5 Equilibrium Approach
Perturb structure from its initial equilibrium position, holding the loading constant.
Apply equilibrium equations to the perturbed position. Determine whether a non­triv­
ial solution for the “perturbed” displacement is possible. This approach is also seeking
whether Kt = 0 . However, one does not formally establish Kt .
Example 3
∆u P
P
k
B
B
F
A
A
∑M
A
= 0 = FL – P∆u
F = k∆u
Then
k∆uL = P∆u
P cr = kL
Example 4
∆u 2
P
k2
B
F2
∆u 1
L2
P
B
k1
F1
L1
A
A
Equilibrium Equations
∑M
∑M
top of B
A
= P(∆u 2 – ∆u 1 ) = F 2 L2
= P∆u 2 = F2 (L 1 + L 2 ) + F1 L1
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 14 of 15
Member Relations
Then
F 1 = k 1 ∆u 1
F 2 = k 2 ∆u 2
P
k2L2 – P
∆u 1
k 1 L 1 k 2 (L 1 + L 2 ) – P ∆u 2
= 0
0
Solve by setting detrminant = 0
P 2 – P ( k 1 L 1 + k 2 (L 1 + L 2 )) + k 1 k 2 L 1 L 2 = 0
If k 1 = k 2 and L 1 = L 2
P1
----= --- {3 ± 5 } = 0.38, 2.62
kL
2
⎧ P
⎫
∆u 2 = ∆u 1 ⎨ ---------------- ⎬ ← Mode Shapes
⎩ P – kL ⎭
4.4.6 Eigenvalue Approach
For the case where the loading is defined in terms of a single parameter, λ , one can
interpret linearized stability analysis as an eigenvalue problem.
Write
F = λF
F
k t = k l + --- I
L
Set
k t = k l – λk g
and the equilibrium equations can be written as
k l ∆u = λk g ∆u
1.571 Structural Analysis and Control
Prof Connor
Section 4
Page 15 of 15