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An in-depth analysis of elasticity, focusing on equilibrium, strain-displacement, and stress-strain relations. It covers the concepts of strain compatibility equations, hooke's law, and the elasticity tensor. The document also discusses the importance of static determinacy and the implications for calculating stresses.
Typology: Exercises
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Equilibrium
Strain - Displacement
Stress - Strain Relations (
Constitutive
Relations)
Consider each:
Equilibrium
i = 0,
i = 0
Free body diagrams
yieldsApplying these to an infinitesimal element
equilibrium equations
Figure 4.
elementRepresentation of general infinitesimal
∂σ
11
∂σ
21
∂σ
31
1
=
∂ y 1 ∂ y 2 ∂ y 3
∂σ
12
∂σ
22
∂σ
32
2
=
∂ y 1 ∂ y 2 ∂ y 3
∂σ
13
∂σ
23
∂σ
33
3
=
m mn
n
Strain - Displacement
Based on geometric considerations
Linear considerations
I.e.,
small
strains only --
we will talk about
large strains later
(and
infinitesimal
displacements only)
Stress - Strain
mn
mnpq
pq
Static Determinance Let’s review the “4th important concept”:^ we’ll come back to this …
There are there possibilities
(as noted in U.E.)
a. A structure is not sufficiently restrained
(fewer reactions than d.o.f.)
freedom degrees of
b. Structure is exactly (or “simply”) restrained
(# of reactions = # of d.o.f.)
determinate)STATICS (statically
Implication
: can calculate stresses via
equilibrium (
as done in Unified
c. Structure is overrestrained
(# reactions > # of d.o.f.)
simultaneously …must solve for reactions
with stresses, strains, etc.
in this case, you must employ the stress-strain equations
--> Overall, this yields for elasticity:
15 unknowns
and
15 equations
6 strains =
ε mn
3 equilibrium (
σ )
6 stresses =
σ mn
6 strain-displacements (
ε )
3 displacements = u
m
6 stress-strain (
σ
ε )
equilibrium (equilibrium).material) as they depend on geometry (strain-displacement) and The first two sets of equations are “universal” (independent of the
Only
the stress-strain equations are
dependent on the material.
Note that deformations (u
m ) must be continuous
doesn’t makesingle-valued functions for continuity. (or it
physical sense
Step 2
: Now consider the case where there are gradients in the strain field
∂ε
12
∂ε
12
∂ y 1 ∂ y 2
∂ Take derivatives on both sides:structure This is the most general case and most likely in a general 2 ε 12
1 ∂ 3 u 1 ∂ 3 u 2
2
2
y
y 2
y y 2 y y 2
1
1
1
Step 3
: rearrange slightly and recall other strain-displacement equations
∂ u 1 = ε 1
u 2
ε 2
∂ y 1 ∂ y 2
2 ε 12
2 ε 11
2 ε 22
(^)
2
y
y 2
y 2
y (^12)
1
they are all related to the 3 displacements. So, the gradients in strain are related in certain ways since
Same for other 5 cases …
Law: The basic relation between force and displacement (recall 8.01) is Hooke’sgeneralize it to encompass all cases. In Unified, you saw particular examples of this, but we now want to
F = kx
spring constant (linear case)
2212
(symmetry in switching
last two
indices
(1st law of thermo)
mnpq
pqmn
(symmetry in switching
pairs of
indices
Also note that:
Since
σ mn
=
σ nm
are only 6!
, the apparent 9 equations for stress
With these symmetrics, the resulting equations are: σ 11
(^)
1111
1122
1133
1123
1113
1112
ε 11
(^)
1122
2222
2233
2223
2213
2212
ε 22
^
σ 22
^
σ 33 (^)
1133
2233
3333
3323
3313
3312
ε 33
(^)
σ 23
^
1123
2223
3323
2323
1323
1223
ε 23
^
σ 13 (^)
1113
2213
3313
1323
1313
1213
ε 13 (^)
σ 12 (^)
1112
2212
3312
1223
1213
1212
ε 12 (^)
Results in
independent components of the elasticity tensor
Along diagonal (6)
Upper right half of matrix (15)
[don’t worry about 2’s]
Also note
: 2’s come out automatically…don’t put them in
For example:
σ 12
= … E
1212
ε 12
1221
(^) ε 21 (^) …
1212
ε 12 …
These E
mnpq
can be placed into
3 groups
Extensional strains to extensional stresses
1111
1122
2222
1133
3333
2233
e.g.,
σ 11
= … E
1122
ε 22 …
Shear strains to shear stresses
1212
1213
1313
1323
2323
2312
Isotropic
Cubic
“Transversely Isotropic”*
Tetragonal
Orthotropic
Monoclinic
Anisotropic
# of Independent
Components of E
mnpq
Material Type
Useful
Engineering Materials
LaminatesComposite Basic
Composite
Ply
Metals
(on average)
Good Reference
BMP, Ch. 7
*not in BMP
For
orthotropic material
s (which is as complicated as we usually get),
there are no coupling terms in the
principal axes of the material
When you apply an extensional stress, no shear strains arise
e.g., E
1112
(total of 9 terms are now zero)
When you apply a shear stress, no extensional strains arise
(some terms become zero as for
previous condition)
(stresses) in another planeShear strains (stresses) in one plane do not cause shear strains
1223
1213
1323
With these additional terms zero, we end up with
independent
components:
and the equations are:
