3
Residual Force
Equations
3–1
Chapter 3: RESIDUAL FORCE EQUATIONS
TABLE OF CONTENTS
Page
§3.1
§3.2
§3.3
§3.4
§3.5
§3.
§3.
Introduction
. . . . . . . . . . . . . . . . . . . . .
Residual Equations
. . . . . . . . . . . . . . . . . .
§3.2.1
Control and State Variables . . . . . . . . . . . . .
§3.2.2
Balanced Force Residual . . . . . . . . . . . . .
§3.2.3
Accounting for Memory Effects
. . . . . . . . . . .
Stiffness and Control Matrix . . . . . . . . . . . . . . .
Parametric Representations and Rate Forms
. . . . . . . . .
Reduction to Single Control Parameter by Staging . . . . . . .
§3.5.1
Explicit Reduction . . . . . . . . . . . . . . . .
§3.5.2
*Implicit Reduction . . . . . . . . . . . . . . .
Notes and Bibliography
. . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . .
3–2
3–3
3–3
3–3
3–4
3–4
3–5
3–6
3–9
3–10
3–12
3–13
3–14
§3.2
RESIDUAL EQUATIONS
§3.1. Introduction
Chapters 3 through 6 discuss basic properties of systems of algebraic nonlinear equations that
depend on one or more control parameters. Algebraic means that these systems contain a finite
number of equations and unknowns.
Algebraic systems result from the discretization of continuum models of nonlinear structures. As
noted in Chapter 1, the most widely used discretization method is the displacement-based Finite
Element Method (FEM). The coverage of FEM discretization schemes is deferred until Chapter 9
and following ones. For the moment it is assumed that the discretization has been carried out.
Physically the algebraic systems represent force equilibrium at the discrete level. Specifically, for
discrete models coming from the displacement FEM, the sum of external and internal node forces
on each degree of freedom vanishes. These are collectively known as force residual equations or
residual equations for short. In the present Chapter, residual and differential forms of use in later
Chapters are presented, and the key concept of staging, through which multiple control parameters
are reduced to one, is introduced.
§3.2. Residual Equations
Discrete equilibrium equations encountered in nonlinear static structural analysis formulated by
the displacement method may be presented in the compact total force residual form
r (u, Λ) = 0.
(3.1)
Here u is the state vector that contains the degrees of freedom that uniquely characterize the state
of the structure, r is the residual vector that contains out-of-balance forces conjugate to u, and Λ
is an array of assignable control parameters. Occasionally an indicial form of (3.1) is convenient:
ri (u j , k ) = 0.
(3.2)
Here ri , u j , and k are indexed entries of vectors r, u, and Λ, respectively, that range over
appropriate dimensions. We will normally restrict consideration to problems where the dimensions
of r and u are the same; if that is the case, the range of indices i and j is identical.
As noted above, the formulation of these discrete equations using FEM is treated in later Chapters.
For the moment it is simply assumed that, given u and Λ, a computational procedure exists that
returns r. This “black box view” is sketched in the block diagram of Figure 3.1(a). In addition,
most solution methods may require residual derivative (rate) information, as discussed later.
§3.2.1. Control and State Variables
In structural mechanics, control parameters are commonly mechanical load levels. They may
also be, however, prescribed physical or generalized displacements, temperature variations, fluid
velocities, imperfection amplitudes, and even (in design and optimization) design variables such as
geometric dimensions or material properties.
The degrees of freedom collected in u are usually physical or generalized unknown displacements.
In this book the term state variables will be used to designate them in general terms.1
1
The names behavior and configuration variables for the entries of u are also found in the literature. However, “state
variable” is more precise, and in line with standard terminology in analytical dynamics. In a mathematical context, u
and Λ are called the active and passive variables, respectively.
3–3
Chapter 3: RESIDUAL FORCE EQUATIONS
(a)
Control: Λ
(b)
Control: Λ
State u
Residual
evaluation
Residual r
Ext force
evaluation
State u
Int force
evaluation
f
−
Residual r
p
Figure 3.1. Black box diagrams for residual evaluation: (a) total force residual form (3.1);
(b) balanced force residual form (3.4).
The dependence of r on u and Λ is assumed to be piecewise smooth so that first and second
derivatives exist except possibly at isolated critical points. If the system is conservative,2 r is the
gradient of the total potential energy (u, Λ) for fixed Λ:
r=
∂
,
∂u
or ri =
∂
.
∂u i
(3.3)
If (3.3) holds, (3.1) expresses that equilibrium is associated with a energy-stationarity condition.
§3.2.2. Balanced Force Residual
An alternative version of (3.1) that displays more physical meaning is the balanced force residual
form:
p(u) = f(u, Λ).
(3.4)
Here p denotes the configuration-dependent internal forces resisted by the structure whereas f are
the control-dependent external or applied forces, which may also be state dependent. This form
says that in static equilibrium, the internal forces p balance the external forces f. A black box view
of (3.4) is sketched in Figure 3.1(b).
The total force residual corresponding to (3.4) is either r = p − f or r = f − p, the two versions
being equivalent except for sign. In the sequel only the first form: r = p − f, is used so as to
conform with the conventional definition of tangent stiffness matrix in §3.3.
If a total potential energy exists, the decomposition associated with (3.4) is
p=
∂U
,
∂u
f=
∂W
,
∂u
(3.5)
in which U and W are the internal energy and external work potential components, respectively, of
= U − W.
§3.2.3. Accounting for Memory Effects
Both the total force residual form (3.1) and its force-balance variant (3.4) are restricted in that no
account for historical or memory effects is made. A more general form would be
r(u, , Λ) = 0,
2
A property studied more carefully in Chapter 6.
3–4
(3.6)
§3.3
STIFFNESS AND CONTROL MATRIX
in which is a functional of the past history of deformation. This generalized form is needed for the
treatment of inelastic, path-dependent materials. (Think, for example, of loading versus unloading
paths in plasticity). The path-independent forms (3.1) or (3.4) are sufficient, however, for the class
of problems considered here. In addition to geometric and force B.C. nonlinearities, these forms
are applicable to nonlinear elasticity (hyperelasticity), as well as several types of displacement B.C.
nonlinearities.
Example 3.1. Consider the following residual equilibrium equations
r1 = 4 u 1 − u 2 + u 2 u 3 − 6 1 = 0,
r2 = 6 u 2 − u 1 + u 1 u 3 − 3 2 = 0,
(3.7)
r3 = 4 u 3 + u 1 u 2 − 3 2 = 0.
The vector form of (3.7) is
r=
r1 (u, Λ)
r2 (u, Λ)
r3 (u, Λ)
=
4 u 1 − u 2 + u 2 u 3 − 6 1
6 u 2 − u 1 + u 1 u 3 − 3 2
4 u 3 + u 1 u 2 − 3 2
= 0,
with
The force-balance vector form of = (3.7) is p = f, in which
p=
p1 (u)
p2 (u)
p3 (u)
=
4 u1 − u2 + u2 u3
6 u2 − u1 + u1 u3
4 u3 + u1 u2
,
f=
u=
f 1 (Λ)
f 2 (Λ)
f 3 (Λ)
u1
u2
u3
,
=3
Λ=
21
2
2
1
.
2
(3.8)
.
(3.9)
In this particular case f does not depend on u.
Remark 3.1. The function u(Λ) characterizes the equilibrium surface of the structure in the space spanned by
u and Λ. In a mathematical context the set of {u,Λ} pairs that satisfies (3.1) is called a solution manifold.
Remark 3.2. The usefulness of the residual equation (3.1) is not restricted to static problems. It is also
applicable to nonlinear dynamical systems
r(u(τ ), Λ(τ )) = 0,
(3.10)
which have been discretized in time τ by implicit methods.3 In this case a system of nonlinear equations
arises at each time station.
§3.3. Stiffness and Control Matrix
Varying the vector r with respect to the components of u while keeping Λ fixed yields the Jacobian
matrix K:
∂r
∂ri
, with entries K i j =
K=
.
(3.11)
∂u
∂u j
This is called the tangent stiffness matrix in structural mechanics applications. The inverse of K, if
it exists, is denoted by F = K−1 ; a notation suggested by the name flexibility matrix traditionally
used in linear structural analysis for the reciprocal of the stiffness. If (3.1) derives from a potential,
both K and F are symmetric matrices (see Exercise 3.5).
3
τ is used throughout this book for real time in lieu of t, which is used more extensively to denote pseudotime.
3–5
Chapter 3: RESIDUAL FORCE EQUATIONS
Varying the negative of r with respect to Λ while keeping u fixed yields
Q=−
∂r
,
∂Λ
with entries
Qi j = −
∂ri
.
∂ j
(3.12)
There is no agreed upon name for Q in the literature. In the sequel it is called the control matrix,
although the name loads matrix could also be appropriate. The specialization of Q to the usual
incremental load vector q is discussed in the next Chapter.
Remark 3.3. Are the foregoing definitions compatible with linear FEM? The master stiffness equations
produced by the linear Direct Stiffness Method (DSM) [263, Chapters 2-3] are
K u = f,
(3.13)
in which K and f are constant. The associated residual can be written as r = K u − f. Evidently K = ∂r/∂u,
which is consistent with (3.11). For the control matrix the verification is less obvious, because the concept of
control parameters is unnecessary in linear FEM analysis. But we can bring them in, somewhat artificially, by
expressing the force vector as a linear combination of the i as
f = ΛT Q,
(3.14)
in which the columns of Q contain the so-called load cases. Then r = K u − ΛT Q, and Q = −∂r/∂Λ = Q.
This is consistent with (3.12).
§3.4. Parametric Representations and Rate Forms
Parametric representations of the state variables u and control parameters Λ are useful in the
description of solution methods as pseudodynamical processes. The general form is
u = u(t),
Λ = Λ(t),
(3.15)
in which t is a dimensionless time-like parameter. Derivatives with respect to t will be denoted by
superposed dots, as in real dynamics. The first two t-derivatives of the residual in component form
are (with summation convention implied):
∂ri
∂ri
˙ j,
u̇ j +
(3.16)
∂u j
∂ j
∂ 2r
∂ 2r
∂ri
∂ 2 ri
∂r
∂ 2 ri
i
i
˙ k u̇ j + i ¨ j+
˙k ˙ j . (3.17)
r̈i =
ü j +
u̇ k +
u̇ k +
∂u j
∂u j ∂u k
∂u j ∂k
∂ j
∂ j ∂u k
∂ j ∂k
ṙi =
In matrix form,
ṙ = Ku̇ − QΛ̇,
(3.18)
r̈ = Kü + K̇u̇ − QΛ̈ − Q̇Λ̇.
(3.19)
Note that both K̇ and Q̇ are matrices. Their (i, j) entries are shown in square brackets in (3.17).
On the other hand, terms such as ∂ 2ri /∂u j ∂u k , are three-dimensional arrays that may be visualized
3–6
§3.4 PARAMETRIC REPRESENTATIONS AND RATE FORMS
(a)
(b)
P = λ EA
C'
E,A constant
B
A
A
θ
B
θ
u
C
k
k = β EA
L
(c)
P = λ EA
C'
FAC
L
L
2L
Fs
FBC
Figure 3.2. Pin-jointed truss structure composed of two aligned elastic bars propped by an
extensional spring: (a) unloaded and undeformed reference state; (b) deformed configuration
under load P; (c) FBD of midspan joint C in the deformed configuration.
as “cubic matrices.” Matrices K̇ and Q̇ are projections of those arrays on the subspace spanned by
the directions u̇ and Λ̇. Often these matrices can be more expediently formed by direct pseudotime
differentiation:
dQ
dK
(3.20)
,
Q̇ =
.
K̇ =
dt
dt
Example 3.2. The pin-jointed truss structure shown in Figure 3.2(a) consists of two aligned, identical elastic
bars of length L, elastic modulus E and cross section area A. The bars are propped at the midspan joint C
by an extensional spring as shown (without the spring, the bars would form an infinitesimal mechanism). The
extensional spring stiffness is expressed as k = β E A/L for convenience, where β is dimensionless.
A vertical point load P = λ E A is applied to the center pin hinge C as shown in Figure 3.2(b). The load
factor λ = P/(E A) is a dimensionless control parameter defined in terms of E A for convenience in eventually
making all equations dimensionless.4 The structure deflects symmetrically as pictured in Figure 3.2(b). The
deformed configuration is fully specified by either the deflection u, which is the movement CC’ of the load
point, or by the rise angle θ defined in that figure. For convenience we introduce the dimensionless deflection
µ = u/L. Deflection
and rise angle are related by tan θ = u/L = µ.The deformed bar lengths are
√
2
2
L d = L + u = L 1 + µ2 . The bar elongations are δ = L d − L = L( 1 + µ2 − 1). We shall assume
that Hooke’s law holds for both bars in terms of the axial engineering strain
=
δ
Ld − L
=
= 1 + µ2 − 1,
L
L
(3.21)
so that the axial bar stress is σ = E . Neglecting cross section change, the internal bar forces are Fb = FAC =
FBC = A σ = E A . The restoring force of the extensional spring is Fs = k u = β (E A/L) µL = β E A µ.
To establish the total force residual equation in terms of µ, it is sufficient to draw the Free Body Diagram (FBD)
of the center hinge in the deformed configuration C . This FBD is depicted in Figure 3.2(c). Horizontal and
moment equilibrium are automatically verified, whereas vertical equilibrium requires 2Fb sin θ + Fs = λP.
4
This λ is labeled the stage parameter in §3.5.1, and further studied in the next Chapter.
3–7
0.4
0.2
0
β=1
β = 1/10
β = 1/100, 1/1000
(indistinguishable
at plot scale)
−0.2
−0.4
Load factor λ = P/EA
−0.6
(c)
(b)
Response using
engineering
strain measure
0.6
0.4
0.2
−1
1
−0.5
0
0.5
Dimensionless displacement µ = u/L
(d)
Response using
Green-Lagrange
strain measure
0
−0.2
β=1
β = 1/10
β = 1/100, 1/1000
(indistinguishable
at plot scale)
−0.4
−0.6
−1
1
−0.5
0
0.5
Dimensionless displacement µ = u/L
Stiffness coefficient K = dr/dµ
0.6
Load factor λ = P/EA
(a)
2.5
Stiffness coefficient K = dr/dµ
Chapter 3: RESIDUAL FORCE EQUATIONS
2.5
Stiffness using
engineering
strain measure
2
1.5
1
0.5
0
−1
1
−0.5
0
0.5
Dimensionless displacement µ = u/L
Stiffness using
Green-Lagrange
strain measure
2
1.5
1
0.5
0
−1
1
−0.5
0
0.5
Dimensionless displacement µ = u/L
Figure 3.3. Results for problems worked out in Examples 3.2 and 3.3: (a,b) load-deflection response λ = λ(µ)
and stiffness coefficient K = K (µ), respectively, using engineering strain measure; (c,d) load-deflection response
λ = λ(µ) and stiffness coefficient K = K (µ), respectively, using GL strain measure.
Since sin θ = tan θ/ (1 + tan2 θ) = µ/ 1 + µ2 , replacing the sine and canceling out E A yields the force
residual in dimensionless form as
r (µ, λ) = 2
1 + µ2 − 1
µ
2
+βµ−λ=µ 2+β − 2
1+µ
1 + µ2
− λ = 0.
(3.22)
Note that µ = λ = 0 give r = 0, so the undeformed configuration is in equilibrium. Equation (3.22) can be
easily solved for λ to get the response λ = λ(µ) in terms of the deflection µ. On the other hand, obtaining a
closed form solution for µ = µ(λ) is not feasible since it requires finding the roots of a sixth order polynomial
in µ in terms of its symbolic coefficients. The associated tangent stiffness matrix is simply 1 × 1 and its only
entry is the dimensionless stiffness coefficient
K =
2(1 + µ2 )3/2 − 1
∂r
=β+
.
∂µ
(1 + µ2 )3/2
(3.23)
Figures 3.3(a,b) display plots of λ = λ(µ) and K = K (µ) for µ ∈ [−1, 1] and four extensional spring ratios:
β = 1, 1/10, 1/100, and 1/1000. The λ(µ) curves are of hardening type, a response flavor introduced in
§2.7. Physical reason: as the deflection grows, both bars develop increasing tension forces, which stiffen the
structure. This effect is qualitatively displayed by the K (µ) plots. Note that if β → 0, the stiffness at µ = 0
approaches zero, and the response curve has a zero-slope tangent at the inflexion point there.
3–8
§3.5
REDUCTION TO SINGLE CONTROL PARAMETER BY STAGING
Example 3.3. Consider again the problem defined in Figure 3.2. Would the response change appreciably if
another bar strain measure is used? More specifically, assume now that the bar responds linearly in terms of
the Green-Lagrange strain5
e = + 12 2 ,
(3.24)
in which is the axial engineering strain given in (3.21). The axial bar stress is now σ = E e,6 and the axial
force Fb = FAC = FBC = A σ = E A e. Repeating the previous derivations one finds, after simplification
µ
r (µ, λ) = µ β + 1 + µ2
2
− λ = 0,
K =
µ2 (3 + 2µ2 )
∂r
=β+
.
∂µ
(1 + µ2 )3/2
(3.25)
Observe that the total residual r is significantly simpler than (3.22) despite (3.24) being a more complicated
strain measure than (3.21). Figures 3.3(c,d) display plots of λ = λ(µ) and K = K (µ) for µ ∈ [−1, 1] and
four spring ratios: β = 1, 1/10, 1/100, and 1/1000. Eyeball comparison to Figures 3.3(a,b) show some
minor discrepancies in the load response diagrams λ(µ) as µ increases, and more significant ones as regard
the stiffness K (µ), since differentiation amplifies differences.
Example 3.4. This is a variation of the previous example that brings into focus the effect of initial forces (also
called prestress) on the nonlinear response. Suppose that in the reference state of Figure 3.2(a), the aligned
bars are under an internal axial force F0 , which is the same in both. This force may be positive or negative.
The reference state µ = λ = 0 is still in equilibrium since the initial forces cancel each other at C.
We will again use the Green-Lagrange (GL) strain measure (3.24) for the bar elasticity to keep the residual
relatively simple. For convenience define the initial GL strain to be e0 = F0 /(E A). The bar constitutive
equation becomes σ = E(e + e0 ), and the bar axial forces are Fb = FAB = FBC = Aσ = E A(e + e0 ). The
total force residual and the associated stiffness coefficient are easily worked out to be
µ2 + 2e0
r (µ, λ) = µ β + 1 + µ2
− λ,
K =
2µ4 + 3µ2 + 2e0
∂r
=β+
.
∂µ
(1 + µ2 )3/2
(3.26)
These reduce to (3.25) if e0 = 0, as may be expected. The reference state λ = µ = 0 is still an equilibrium
configuration for any e0 . However the initial stiffness at µ = 0 is K 0 = β + 2e0 . If e0 is negative and |e0 |
exceeds β/2, K 0 < 0, whence the initial slope of the load-deflection response is negative, and the structure is
unstable in the reference state. It recovers stability after it deflects enough under P, so the bar compressive
forces are sufficiently reduced.
Figures 3.3(c,d) display plots of λ = λ(µ) and K = K (µ) for µ ∈ [−1, 1] the four spring ratios: β = 1,
1/10, 1/100, and 1/1000, and a negative initial strain e0 = −1/5 = −0.2, so aligned bars are in compression.
(This is admittedly a very large initial strain, exaggerated for plot visualization convenience.) The structure is
initially unstable if β < |2e0 | = 2/5 = 0.4. It recovers stability upon traversing a limit point at which K = 0.
For instance, if β = 1/10 those limit points occur at µ = ±3/10.
The benefits of presenting results in dimensionless form for the foregoing examples should be evident. The
load-deflection and stiffness-deflection solutions are valid for any E, A, L and k, once a bar strain measure is
decided upon. One plot chart for each of λ(µ) and K (µ) suffices. Also notice that working out these simple
problems does not require any use of finite elements: simple Mechanics of Materials does it.
5
A strain measure defined generally (for 3D) in Chapter 8. For now assume that (3.24) is correct.
6
Strictly speaking the bar stress-strain equation should be s = E e, in which s is the second Piola-Kirchhoff (PK2) stress
measure that is conjugate to e. For now we will not bother about such nitpicking details.
3–9
0.6
0.4
0.2
(b)
Response using
Green-Lagrange
strain measure
and e0 = −0.2
0
−0.2
β=1
β = 1/10
β = 1/100, 1/1000
(indistinguishable
at plot scale)
−0.4
−0.6
−1
1
−0.5
0
0.5
Dimensionless displacement µ = u/L
Stiffness coefficient K = dr/dµ
(a)
Load factor λ = P/EA
Chapter 3: RESIDUAL FORCE EQUATIONS
2.5
Stiffness using
Green-Lagrange
strain measure
and e0 = −0.2
2
1.5
1
0.5
0
−0.5
−1
1
−0.5
0
0.5
Dimensionless displacement µ = u/L
Figure 3.4. Results for problem worked out in Example 3.4 with initial GL strain e0 = −0.2:
(a) load-deflection response λ = λ(µ); (b) stiffness coefficient K = K (µ).
Figure 3.5. The Brooklyn Bridge in 1876, drawn by F. Hildebrand for the Library of Congress.
§3.5. Reduction to Single Control Parameter by Staging
Multiple control parameters are quite common in real-life nonlinear problems. They are the analog
of multiple load conditions in linear problems. But in the linear world, multiple load conditions can
be processed independently because any load combination is readily handled by superposition. In
nonlinear problems, however, control parameters are not varied independently. This aspect deserves
an explanation, as it is rarely mentioned in the literature.
§3.5.1. Explicit Reduction
Typically the analysis proceeds as follows. The user defines the control parameters to the computer
program during a model preprocessing phase. To illustrate the process with a real problem, let us
assume that for the analysis of a suspension bridge (Figure 3.5) there are six control parameters
Λ = [ 1 2 3 4 5 6 ]T ,
3–10
(3.27)
§3.5 REDUCTION TO SINGLE CONTROL PARAMETER BY STAGING
in which parameter 1 is associated with own weight, 2 and 3 with live loads, 4 with temperature changes, 5 with foundation settlement and 6 with wind velocity.
Suppose that 1 = 10 corresponds to full own weight. The first nonlinear analysis involves going
from the reference state at
(3.28)
Λr e f = [ 0 0 0 0 0 0 ]T
to the full own-weight condition
ΛW = [ 10
0
0
0
0 ]T .
0
(3.29)
Next, assume that the effect of a temperature drop of −20◦ C is to be investigated. If a unit increment
of 4 corresponds to 1◦ C, then the next nonlinear analysis corresponds to going from ΛW to
ΛT = [ 10
0
0
−20
0
0 ]T .
(3.30)
Note that own weight obviously stays on as a fixed value of 1 = 10, once the bridge is assumed
finished and in service. Analysis for a live load parametrized by 2 = 20, combined with a 6◦
temperature rise, would entails going from ΛW to
Λ L L = [ 10
20
0
6
0
0 ]T ,
(3.31)
and so on. Each of these processes is called an analysis stage or simply stage. A stage can be
defined as “advancing the solution” from
ΛA
to
ΛB .
(3.32)
when the solution u A at Λ A is known. Furthermore, if we assume that the components of Λ will
vary proportionally, we can introduce a single control parameter λ that varies from 0 through 1 as
per the linear interpolation
Λ = (1 − λ)Λ A + λΛ B .
(3.33)
This λ is called the stage control parameter. The nonlinear equation to be solved in (A → B) is
r (u, λ) = 0,
(3.34)
with the initial condition u = u A at λ = 0. The solution curve u = u(λ) is called the response of
the structure in the (A → B) stage.
The importance of staging in nonlinear static analysis arises from the inapplicability of the superposition principle of linear analysis. For example, the sequences
Λ A → Λ B → ΛC ,
Λ A → ΛC ,
(3.35)
Λ A → ΛC → Λ A ,
(3.36)
or
Λ A → ΛB → Λ A,
do not necessarily produce the same final solution. Of course, this phenomenon is especially
important for path-dependent problems in which material failure may occur during a stage.
3–11
Chapter 3: RESIDUAL FORCE EQUATIONS
Remark 3.4. In mathematical circles, (3.33) receive names such as linear homotopy imbedding and piecewise
linear imbedding. The vector ur e f for Λr e f = 0 is called the reference or initial configuration. Commonly
ur e f = 0, i.e., the reference state is the origin of displacements. This is not necessarily a physically attainable
configuration. In the suspension bridge example, ur e f is fictitious because bridges are not erected in zero
gravity fields; on the other hand the own-weight configuration uW is physically relevant once the bridge is
contructed.
Remark 3.5. Many staged analysis sequences do start from the same configuration, for example the ownweight solution uW in the case of the suspension bridge. A comprehensive analysis system must therefore
provides facilities for saving selected solutions on a permanent database, and the ability to restart the analysis
from any previously saved solution.
Remark 3.6. The definition (3.33) of λ as a [0 → 1] parametrization of the [A → B] stage is somewhat
artificial for the following reasons. First, there is no guarantee that a solution at Λ B exists, so λ = 1 may not
be in fact attainable. Second, in stability analysis (3.33) defines only a direction in control parameter space;
in this case the user may want to know the smallest λ (or largest −λ) at which a critical point of (3.34) occurs,
and the analysis stops there.
Remark 3.7. As the number of control parameters grows, the number of possible analysis sequences increases
combinatorially. Given the substantial costs usually incurred in these analyses, the experience and ability of
the engineer can play an important role in weeding out unproductive paths. Selective linearization can also
reduce the number of cases substantially, because invoking the superposition principle “factors out” certain
parameters. For example, if the bridge response to live loads is essentially linear about the own-weight solution,
parameters 2 and 3 may be removed from the nonlinear analysis, and the dimension of Λ is reduced from
6 to 4.
§3.5.2. *Implicit Reduction
The most general reduction from multiple control parameters Λ to a single parameter λ can be
formulated as follows. Suppose that over a stage an implicit vector algebraic relation is established:
m(Λ, λ) = 0,
(3.37)
where m is a vector of m equations that uniquely defines Λ if λ is given. Augmenting the residual
(3.1) with (3.37) we obtain the expanded system
r(u, Λ)
0
=
.
(3.38)
m(Λ, λ)
0
This relation implicitly defines u as a function of λ. (It would be explicit if the Λ are eliminated,
but that might not be possible or convenient.) Derivatives of r with respect to pseudotime t are
obtained by the chain rule, for example
0
ṙ(u, Λ)
Ku̇ − QΛ̇
=
,
(3.39)
=
0
ṁ(Λ, λ)
AΛ̇ − aT λ̇
in which A = ∂m/∂Λ and a = −∂m/∂λ. The explicit definition of λ by the linear interpolation
(3.33) is a special case of (3.38), for which
m(Λ, λ) = Λ − (1 − λ)Λ A − λΛ B = 0.
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(3.40)
§3. Notes and Bibliography
In this case the elimination of Λ is trivial. For most applications use of the more general expression
(3.38) is unnecessary. It may come handy, however, in complicated load parameter dependencies.
For example, those associated with hydro or aerodynamic loads: beyond the laminar flow regime
physical loads on the structure are complicated nonlinear functions of relative fluid velocities.
Notes and Bibliography
The general treatment of nonlinear structural problems in terms of force residuals is a natural one, since it
merges physics visualization with standard techniques in numerical analysis and computational mathematics.
In textbooks of that ilk, systems of nonlinear equations are typically presented as
f(x) = 0,
(3.41)
and attention is directed to finding the roots, whether a few, billions or an infinite number. Filtering the root
morass by attaching physical significance to roots requires the introduction of control parameters, as done in
§3.2. No treatment of that nature can be found in FEM textbooks. One needs to go to the literature in system
control to run into comparable formulations.
The use of parametrized residual derivative forms is a useful tool that unifies continuation solution methods.
It also provides a natural link to nonlinear dynamics. Again those rate forms are ignored in most FEM
presentations, reflecting the unease of many authors in dealing with rate equations in pseudotime. A short
historical account of rate forms is provided in the book by Ortega and Rheinboldt [552, pp. 234–235].
No FEM book mentions staging, despite its crucial importance in design verification by nonlinear analysis.
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Chapter 3: RESIDUAL FORCE EQUATIONS
Homework Exercises for Chapter 3
Residual Force Equations
EXERCISE 3.1 [A:15] For the example system (3.7) find K and Q.
EXERCISE 3.2 [A:15] For the example system (3.7), and assuming the parametric representation (3.15) for
u and Λ, write down ṙ, K̇, Q̇ and r̈ in matrix-vector form.
EXERCISE 3.3 [A:15] Continuing the above exercise: if 1 = 2λ and 2 = λ, write down r(u, λ) and
ṙ(u, λ) in matrix-vector form.
EXERCISE 3.4 [A:20] Is the example system (3.7) derivable from a potential energy function so that r
can be represented as r = ∂/∂u? Can you guess by inspection what the potential is?
EXERCISE 3.5 [A:20] Show that both K and F = K−1 (assuming the latter exists) are symmetric if the
residual is derivable from a potential energy function.
EXERCISE 3.6 [A:20] For the example problem defined in Figure 3.2, derive equations for λ(µ) and K (µ)
if the logarithmic (also called Hencky) strain e H = log(1 + ) is used for the bars, so that σ = E e H and
Fb = FAC = FBC = Aσ = E A e H . Provide plots of λ(µ) and K (µ) similar to those shown in Figure 3.3.
Does the change in strain measure makes much difference?
EXERCISE 3.7 [A:15] Show that the total potential energy for the example problem in Figure 3.2, using
(3.21) as bar strain measure, is given by
= U − W,
U = E A L 2 + 12 k u 2 ,
W = P u.
(E3.1)
Then derive the total residual force equation directly as r = ∂/∂µ = 0. Does this method provide the same
result as (3.22)?
EXERCISE 3.8 [D:10] The K (µ) curves plotted in Figures 3.3(b,d) for varying β are identical except for a
translation along the vertical λ axis. Explain the physical reason behind this feature.
EXERCISE 3.9 [A:20] Assume that the total force residual equation for a one-DOF system with state
parameter µ and control parameter µ is r (µ, λ), and that the response satisfying r = 0 can be explicitly solved
as λ = λ(µ). The stiffness matrix is simply 1 × 1, and its only entry is K = ∂r/∂µ. Discuss under which
condition(s) the following relation holds true:
K =
∂r ? ∂λ
.
=
∂µ
∂µ
(E3.2)
(Hint: differentiate both sides of r = 0 wrt µ and apply the chain rule.) Can (E3.2) be extended to the case
where the response λ = λ(µ) satisfies r = rc , where rc is constant?
3–14