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MIT - 16.20
Fall, 2002
Unit 6
Plane Stress and Plane Strain
Readings:
T & G 8, 9, 10, 11, 12, 14, 15, 16
Paul A. Lagace, Ph.D.
Professor of Aeronautics & Astronautics
and Engineering Systems
Paul A. Lagace © 2001 MIT - 16.20
Fall, 2002
There are many structural configurations where we do not
have to deal with the full 3-D case.
• First let’s consider the models
• Let’s then see under what conditions we can
apply them
A. Plane Stress
This deals with stretching and shearing of thin slabs.
Figure 6.1 Representation of Generic Thin Slab
Paul A. Lagace © 2001
Unit 6 - p. 2 MIT - 16.20
Fall, 2002
The body has dimensions such that
h << a, b
(Key: where are limits to “<<“??? We’ll
consider later)
Thus, the plate is thin enough such that there is no variation of
displacement (and temperature) with respect to y (z).
3
Furthermore, stresses in the z-direction are zero (small order of
magnitude).
Figure 6.2 Representation of Cross-Section of Thin Slab
Paul A. Lagace © 2001
Unit 6 - p. 3 MIT - 16.20
Fall, 2002
Thus, we assume:
σ = 0
zz
σ = 0
yz
σ = 0
xz
∂
= 0
∂z
So the equations of elasticity reduce to:
Equilibrium
∂ σ ∂ σ
11 21
+ + f = 0 (1)
1
∂y ∂y
1 2
∂σ ∂σ
12 22
+ + f = 0 (2)
2
∂y ∂y
1 2
(3rd equation is an identity) 0 = 0
(f = 0)
3
∂σ
βα
In general:
+ f = 0
α
∂y
β
Paul A. Lagace © 2001
Unit 6 - p. 4 MIT - 16.20
Fall, 2002
Stress-Strain (fully anisotropic)
Primary (in-plane) strains
1
ε = σ − ν σ − η σ (3)
[ ]
1 1 12 2 16 6
E
1
1
ε = − ν σ + σ − η σ (4)
[ ]
2 21 1 2 26 6
E
2
1
ε = − η σ − η σ + σ (5)
[ ]
6 61 1 62 2 6
G
6
Invert to get:
*
σ = E ε
αβ αβσγ σγ
Secondary (out-of-plane) strains
⇒ (they exist, but they are not a primary part of the problem)
1
ε = − ν σ − ν σ − η σ
[ ]
3 31 1 32 2 36 6
E
3
Paul A. Lagace © 2001
Unit 6 - p. 5 MIT - 16.20
Fall, 2002
1
ε = − η σ − η σ − η σ
[ ]
4 41 1 42 2 46 6
G
4
1
ε = − η σ − η σ − η σ
[ ]
5 51 1 52 2 56 6
G
5
Note: can reduce these for orthotropic, isotropic
(etc.) as before.
Strain - Displacement
Primary
∂u
1
ε = (6)
11
∂y
1
∂u
2
ε = (7)
22
∂y
2
1 ∂u ∂u
1 2
ε = + (8)
12
2 ∂y ∂y
2 1
Paul A. Lagace © 2001
Unit 6 - p. 6 MIT - 16.20
Fall, 2002
Secondary
∂u
1 ∂u
3
1
ε = +
13
2 ∂y ∂y
3 1
1 ∂u ∂u
2 3
ε = +
23
2 ∂y ∂y
3 2
∂u
3
ε =
33
∂y
3
Note: that for an orthotropic material
( ε ) ( ε )
23 13
(due to stress-strain relations)
ε = ε = 0
4 5
Paul A. Lagace © 2001
Unit 6 - p. 7 MIT - 16.20
Fall, 2002
This further implies from above
∂
(since )
= 0
∂y
3
No in-plane variation
∂u
3
= 0
∂y
α
but this is not exactly true
⇒ INCONSISTENCY
Why? This is an idealized model and thus an approximation. There
are, in actuality, triaxial ( σ , etc.) stresses that we ignore here as
zz
being small relative to the in-plane stresses!
(we will return to try to define “small”)
Final note: for an orthotropic material, write the tensorial
stress-strain equation as:
2-D plane stress
∗
σ = E ε (, αβ, σ, γ = 12, )
σγ
αβ αβσγ
Paul A. Lagace © 2001
Unit 6 - p. 8 MIT - 16.20
Fall, 2002
There is not a 1-to-1 correspondence between the 3-D E and
mnpq
*
the 2-D E . The effect of ε must be incorporated since ε does
αβσ γ 33 33
not appear in these equations by using the ( σ = 0) equation.
33
This gives:
ε = f(ε )
33 αβ
Also, particularly in composites, another “notation” will be used in
the case of plane stress in place of engineering notation:
subscript x = 1 = L (longitudinal)…along major axis
change y = 2 = T (transverse)…along minor axis
The other important “extreme” model is…
B. Plane Strain
This deals with long prismatic bodies:
Paul A. Lagace © 2001
Unit 6 - p. 9 MIT - 16.20
Fall, 2002
Figure 6.3 Representation of Long Prismatic Body
Dimension in z - direction is much, much larger than in
the x and y directions
L >> x, y
Paul A. Lagace © 2001
Unit 6 - p. 10