ASEN 6107: Nonlinear FEM Take-Home Midterm Exam 1 Spring 2016
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Web posted Monday February 22, 2016 by noon. Due on or before Th February 25 at class time.
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bilinear spring
;;
D
λF
A (λ>0 means load
acts downward)
λ>0 (compression
on column)
B3
y
;;
;;
B2
rigid column
x
B1
will curve down
for large motions
R
u xA
uxA =λ=0
C
Figure Q1.1. Structure for Question 1(a).
QUESTION 1. Conceptual questions, 30 pts, 6+6+4+4+5+5 pts/item Be brief
(a) The structure shown in Figure Q1.1 is an axially loaded rigid column pinned at C and propped
at A by a bilinear spring that is much stiffer in tension than in compression. Load λF stays
vertical. Briefly explain the control-state response: λ versus lateral deflection u x A , sketched
on the right of that Figure (λ > 0 means downward load.) Which path portions do you think
are stable? Note: equilibrium paths go further; they are drawn truncated for visualization
conveniency. (Hint: for path B1-B3 check Exercise E2.7.)
(b) The structure is as in Figure Q1.1, but now subject to a horizontal load λF where λ > 0 means
the load acts toward the right extending the spring. The load stays horizontal, and no vertical
load is applied. Sketch the control-state response of λ versus u x A for λ > 0 (spring extended)
and λ < 0 (spring compressed). Do you expect any critical points?
(c) You crush an empty styrofoam coffee cup flat by pushing down on the lid (or stepping on the
lid.) What kind of nonlinearities do you think are involved in the process? Explain.
(d) In preparation to mail a letter you unstick a stamp glued to the standard stamp book sold by
the Post Office. What kind of nonlinearities are involved in the removal operation? Explain.
(e) A submarine goes too deep. Its hull collapses by buckling and starts to crack. What kind of
nonlinearities are involved? Explain. Note: ignore what happens after cracking starts; that
can be grisly.
(f)
Suppose that the two-bar structure pictured in Figure 3.2(a) of Chapter 3 has no spring, that the
initial bar axial forces vanish, and that only vertical motions of node 2 are allowed. Then the
stiffness coefficient in the reference configuration is zero. Can you explain physically why?
(No equations, please) Is that a critical point? If so, of which kind?
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QUESTION 2. Residual equations: 25 pts, 5+10+10 pts/item
A structure has one degree of freedom: a displacement u, and one dimensionless control parameter
called λ. The total potential energy is
2
u
u4
2
1
1
(u, λ) = U (u) − W (u, λ), U = 2 k u , W = 3 λ F L
+
,
(1)
L2
2L 4
in which k, F and L are given constants. In the reference configuration λ = 0 and u = 0. Note
that (1) is not separable.
(a) Find the residual force equation r = 0. Verify that u = 0 is the fundamental equilibrium path.
(b) Find the λcr in terms of k, L, and F at which the structure first encounters a critical point on
the fundamental equilibrium path. Using the criterion of §5.3, determine whether it is a limit
or bifurcation point.
(c) Take k=F=L=1 for simplicity. Obtain the equation λs = λs (u) of the secondary equilibrium
path. Picture it, along with the fundamental path u = 0, on the {λ, u} plane for λ ∈ [0, 3] and
u ∈ [−1, 1], with λ along the vertical axis. Identify reference and critical points. A computer
draw plot is fine. (This kind of plot is called a “fork” in the applied mathematics literature,
although for this problem it does not resemble a fork that much.)
QUESTION 3. TL element derivation: 45 points, 5+10+10+10+10 pts/item
Upon graduation you join a LA software company that sells a FEM code with name starting with N.
Your first assignment is to develop a 3-node, small-strain, large-deflection TL plane cable element.
This is to be used for modeling suspension bridges, power transmission lines, antennas and guyed
towers, as requested by customers. The element nodal freedoms include two displacements per
node, which makes for a total of six DOF. In the reference configuration C 0 the element is straight
while in the current configuration C it takes up a parabolic shape defined by the 3 nodes. See Figure
Q1.2. In C 0 the element has length L 0 , cross section area A0 , elastic modulus E, and axial force F0 .
It is to be subjected to a uniform, conservative external load wY per unit of cable length, directed
along the Y axis. (An example is own weight.)
But the task of formulating a 6-DOF nonlinear element from scratch seems a big imposition on
your time. After all, this is surf season in Malibu. So you decide to start small, doing a 1-DOF
cable element first, and tackling the more complicated one after your first paycheck. Your laid-back
manager agrees.
The simplified element is shown in Figure Q1.3. The undeformed reference configuration C 0 is a
straight line defined by end nodes 1-2, with node 3 at the midpoint of 1-2. The reference frame
{X, Y } ≡ {x, y} is placed with origin at node 3, with X ≡ x along 1–2. The initial axial force in
C 0 is F0 . The uniform applied load along Y is denoted simply by w; for simplicity it is specified
per unit of reference chord length instead of per unit of deformed cable length.
On applying the load, the element moves to the current configuration C shown in Figure Q1.3. The
end nodes 1-2 are kept fixed and node 3 is forced to move vertically. Consequently the only DOF
is the so-called sag deflection d = u Y 3 of node 3. In C the cable element length becomes L and
the axial force F. The area is assumed unchanged: A = A0 ; also E is kept fixed. For convenience
introduce the following control and state dimensionless parameters:
d
w L0
,
µ=
.
(2)
λ=
E A0
L0
2
wY
Y, y
F
X, x
3
1
Base (same as reference)
configuration C o
F0 1
3
2
Current (deformed)
configuration C
F
F0
2
Figure Q1.2. A general 3-node plane cable element with six DOF.
Current (deformed)
configuration C
Base (same as reference)
configuration C o
F0
F
Displacement uY (X)
1
Elastic modulus E,
cross section area A ~A 0
Y, y
3
3
X, x
2
F0
sag d=uY (0)
F
Applied load w per
unit length of chord
L0
End nodes 1 and 2
stay fixed
Figure Q1.3. Simplification of the 3-node plane cable element of Figure Q1.2 to one DOF.
The state parameter µ is called the element-sag-to-length (ESL) ratio whereas the control parameter
λ is a load factor. It is assumed that µ is fairly small compared to 1. Formulation steps follow.
(a) Show that the deflected cable shape u Y (X ) can be expressed as
(3)
u Y (ξ ) = N3 (ξ ) u Y 3 = N3 (ξ ) d.
in which ξ is the isoparametric natural coordinate ξ = 2X/L 0 , which varies from −1 at node
1 to +1 at node 2, and N3 = 1 − ξ 2 is a quadratic shape function. (Hint: see Exercise
E16.4 in Intro to FEM for the shape functions.) Note that the Jacobian needed below is
J = d X/dξ = 12 L 0 .
(b) Using (3), expand the deflected cable arclength differential ds = d X 1 + (u Y )2 , where
u Y = du Y (X )/d X , as a Taylor series in µ:
(4)
ds = d X (1 + α1 µ + α2 µ2 + . . .)
State what α1 through α4 are, using (2), (3), & chain rule on ξ . Integrate ds over X ∈
[− 12 L 0 , 12 L 0 ] ⇒ ξ ∈ [−1, 1] with J = 12 L 0 , to express the current length as L = L 0 (1 +
β2 µ2 + β4 µ4 + . . .), and find what β2 and β4 are.
(c) Show that for µ << 1, the mean GL axial strain along the cable element can be written
2
2
def L − L 0
= γ 2 µ2 + γ 4 µ4 + . . .
(5)
e =
2
2L 0
where coefficients γ2 and γ4 are just numbers. What are those?
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(d) The total potential energy of the element at C is e = U e − W e , in which
L 0 /2
+1
e
2
e
1
w u Y (X ) d X = w J d
N3 (ξ ) dξ.
U = 2 E A L0 e , W =
−L 0 /2
−1
(6)
Notes: (i) the axial force does no work because the end nodes are fixed; (ii) we have taken the
initial force F0 = 0 so the constitutive equation is s = E e and not s = s0 + E e, to keep U e
simple; (iii) for small strains, you may simply assume A = A0 .
Divide (6) through by (E A L 0 ) to bring up the parameter λ of (2), and convert the rest to
depend only on µ through the Taylor series found in (a)–(c). Retain terms only up to µ4 .
¯ e = e /(E AL 0 ); this should be a function of the
Call the scaled total potential energy dimensionless parameters λ and µ only.
¯ e in the
(e) Obtain the residual force equation r = 0 and tangent stiffness coefficient K from usual manner. Show that the primary control-state equilibrium path λ = λ(µ) is cubic in µ.
(This is typically the dependence assumed in commercial programs.) Sketch λ = λ(µ) for
µ ∈ [0, 1/4]. Does the element stiffen or weaken as the ESL ratio µ increases?
BONUS QUESTION Up to 5 pts.
You have proved your worth to the employer by polishing off the 1-DOF element in one day, and
can take the rest of the week off. On coming back, what steps would be needed to extend it to the
6-DOF case of Figure Q1.2? Accounting for a nonzero initial force F0 is now esssential. (Explain
in words; no equations please.)
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