Muddy Card Responses Lecture M16.
The mud marked * is from the equivalent lecture (also M16) in 2000 .
I seem to have done a
better job of explaining some of the derivations, as there were fewer questions on it this time
through.
*I didn't understand the integral in the first PRS question.
The concept is that strain is the
differential of displacement w.r.t position, so in order to obtain displacement from strain we
need to integrate between the appropriate boundary conditions.
In the PRS example the
strain was one dimensional (just
e
11
) so the tip displacement just result from the integral of
that component of strain along the bar.
In strain elongation can it only have a displacement in one direction at a time in line with
our example?
No, in general there can be u
1
, u
2
and u
3
components of displacement.
The
first PRS question was supposed to be a simple example that you could visualize.
*In your diagrams you consistently label
f
2
as the angle from the x
1
edge and
f
1
as the
angle from the x
2
edge.
Is this arbitrary, (i.e. does it affect the derivation)?
It is arbitrary
(although I try to be consistent).
I was doing it to allow you to keep track of where the terms
came from in the derivation.
*Would you be able to clarify what you mean by "gradient"
u
2
+
∂
u
2
x
1
d
x
1
Ê
Ë
Á
ˆ
¯
˜
?
Since we are
trying to obtain continuum definitions of stress and strain (i.e. they are quantities that can
vary continuously across a structure, we need to allow them to vary!.
The simplest way to do
this is to choose a sufficiently small element size that the variation is linear i.e.
u
2
x
1
Ê
Ë
Á
ˆ
¯
˜
is
constant.
Then the total increase in displacement across the element is given by the gradient
times the length of the element, i.e.
u
2
x
1
Ê
Ë
Á
ˆ
¯
˜
x
1
.
From there we apply the definition of strains
(extensional components of strain are changes in length, shear strains are changes in angle) in
order to obtain straindisplacement relationships.
*I'm still confused on how you get
e
11
=
x
1
+
u
1
+
u
1
x
1
x
1

u
1
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜ 
x
1
x
1
=
u
1
x
1
.
I hope that
the arithmetic is straightforward.
All we are doing is using our basic definition of
extensional strain, namely that it is the ratio of the change in length to the initial
(undeformed) length.
We are applying this to an element which is sufficiently small that
there is a linear variation of strain over the element.
For an alternative view see CDL 4.10.
Note that by subtracting out the u1 terms we are removing the rigid body translations from
our definition of strain (i.e. deformations are distinguished from displacements).
In the
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View Full Documentfigure I drew a dashed square to indicate the undeformed, undisplaced unit element and the
coloured, solid, square was the displaced and deformed equivalent element.
*Why do we want a symmetrical tensor?
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 Winter '12
 CharlesColeman
 Aeronautics, Astronautics

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