1.050 Engineering Mechanics

1.050 – Content overview
I. Dimensional analysis
1.
On monsters, mice and mushrooms
2.
Similarity relations: Important engineering tools
II. Stresses and strength
3.
Stresses and equilibrium
4.
Strength models (how to design structures, foundations.. against mechanical failure)
III. Deformation and strain
5.
How strain gages work?
6.
How to measure deformation in a 3D structure/material?
IV. Elasticity
7.
Elasticity model – link stresses and deformation
8.
Variational methods in elasticity
V. How things fail – and how to avoid it
9.
Elastic instabilities
10.
Plasticity (permanent deformation)
11.
Fracture mechanics
Lectures 1-3
Sept.
Lectures 4-15
Sept./Oct.
Lectures 16-19
Oct.
Lectures 20-31
Oct./Nov.
Lectures 32-37
Dec.

1.050 – Content overview
Lecture 20: Introduction to elasticity (thermodynamics)
Lecture 21: Generalization to 3D continuum elasticity
Lecture 22: Special case: isotropic elasticity
Lecture 23: Applications and examples
…

• Isotropic elasticity = elastic properties do not depend on direction
• In terms of the free energy change , this means that the change of the free energy does not depend on the direction of deformation
• Rather, it depends on quantities that are independent on the direction of deformation (i.e., independent of coordinate system)
• Idea: Use invariants of strain tensor to calculate free energy change
– Volume change
– Shape change (shear deformation)
• Note: Invariants are defined as properties of strain tensor that are independent of coordinate system (C.S.)

Important mathematical tools tr
( )
= ε
: 1
= ε
11
+ ε
22
+ ε
33
= d
Ω − d
Ω d 0 d
Ω
0
Trace of a tensor
Relates to the chain of volume of REV
Independent of C.S. – trace of a tensor is an invariant
ε =
2
1
(
ε
:
ε T
)
=
1
ε
2 i j ij
2
‘Magnitude’ of a tensor (2 nd order norm)
Note: Analogy to the ‘magnitude’ of a tensor is the norm of a first order tensor
(=vector), that is, its length

Overview: Approach
Step 1: Calculate change in volume
ε v
= tr
(
ε
)
= ε
: 1
Step 2: Calculate magnitude of angle change
Define strain deviator tensor = tensor that describes deformation without the volume change ( trace of strain deviator tensor is zero!
)
ε e
⎝
ε −
1 tr
3
( )
1 ⎟ d
=
2 e
=
2
2
1
( e : e
T tr
( )
=
)
=
2
ε
11
+ ε
22
1
e
2 i j ij
2
+ ε
33
−
1 (
3
(
3
ε
11
+ ε
22
+ ε
33
))
=
0
Step 3: Define two coefficients to link energy change with deformation
(“spring model”):
Ψ =
1
2
K
ε
2 + v
1
2
G
ε d
2
Shear modulus Bulk modulus

Note
The approach that the free energy under deformation depends only on volume change and overall angle change is not derived from physical principles
Rather, it is an assumption, which is made to ‘model’ the behavior of a solid
( modeling is finding a mathematical representation of a physical phenomenon )
Generally, models must be validated, for instance through experiments
Alternative approach: Calculation of from ‘first principles’ – by explicitly considering the atomistic scale of atomic, molecular etc. interactions
Spring 2008: 1.021J Introduction to Modeling and Simulation (Buehler,
Radovitzky, Marzari) – continuum methods, particle methods, quantum mechanics

Stress-strain relation
Total stress tensor = sum of contribution from volume change and contribution from shape change:
σ = σ + σ v d
σ = v
∂Ψ v
∂ ε
σ = d
∂Ψ d
∂ ε
Next step : Carry out differentiations

Stress-strain relation
Total stress tensor = sum of contribution from volume change and contribution from shape change:
σ = σ + σ v d
σ = v
∂Ψ v
∂ ε
σ d
=
∂Ψ d
∂ ε
1. Calculation of
Ψ v
=
1
K
2
ε v
2
σ v
σ v
=
∂Ψ v
∂ ε
=
∂Ψ v
∂ ε v
:
∂
∂
ε
ε v
=
K
ε v
1
∂Ψ v
∂ ε v
=
K
ε v
∂ ε
∂ ε v
=
∂
(tr(
∂ ε
ε
))
=
∂
(
ε
∂ ε
: 1)
=
1

Stress-strain relation
2. Calculation of
σ d
σ = d
∂ Ψ d
∂ ε
Ψ d
=
1
G
2
ε d
2
=
∂ Ψ d
∂ e
∂
:
∂ e
ε
=
2 Ge :
⎛
⎝
1
−
1
3
1
⊗
1
⎞
⎠
=
2 Ge
−
1
3
( e : 1
)
⊗
1
=0
∂Ψ
∂ e d
=
1
2
G
∂
(
2 e : e
T
∂ e
)
=
2 Ge
Note (definition of
ε d
=
2 e
=
2
ε d
):
1
2
( e : e
T
)
∂ e
∂ ε
=
1
−
1
1
⊗
1
3 tr since:
( e
)
=
0 tensor product
Note (definition of e
): e
= ε −
1
3
ε v
1
= ε −
1
3
(
ε
: 1)
⊗
1

Complete stress-strain relation
3. Putting it all together:
σ = σ + σ v d
σ =
K
ε v
1
+
2 Ge
=
K
ε v
1
+
2 G
⎝
ε −
1
3
ε v
1
⎠ e
= ⎜ ε −
1
3 tr
( )
1
⎟
Deviatoric part of the strain tensor
σ =
⎛
⎝
K
−
2
3
G
⎟ ε v
1
+
2 G
ε
Reorganized…

Complete stress-strain relation
σ = ⎜
K
−
2
3
G
ε v
1
+
2 G
ε = ⎜
K
−
2
3
G
( ε
11
+ ε
22
+ ε
33
)
1
+
2 G
ε
Writing it out in coefficient form:
σ
11
σ
22
σ
33
=
=
=
⎜
⎜
⎜
K
−
K
−
K
−
2
3
2
3
2
3
G
G
G
⎞
⎠
⎟
⎟
( ε
( ε
( ε
11
11
11
+
+
+
ε
22
ε
22
ε
22
+
+
+
ε
33
)
+
2 G
ε
11
ε
ε
33
)
+
2 G
ε
33
)
+
2 G
ε
22
33
σ
12
=
2 G
ε
12
σ
23
=
2 G
ε
23
σ
13
=
2 G
ε
13

Complete stress-strain relation
Rewrite by collecting terms multiplying
ε ii
σ
σ
σ
11
22
33
=
=
=
⎜
⎜
⎜
K
+
K
−
K
−
4
3
2
3
2
3
G
G
G
⎞
⎠
⎟
⎟
ε
ε
ε
11
11
11
+
+
+
⎜
⎜
⎜
K
K
K
−
+
−
2
3
4
3
2
3
G
G
G
⎞
⎠
⎟
⎠
ε
ε
ε
22
22
22
+
+
⎜
⎜
+ ⎜⎛
K
−
K
−
K
+
2
3
2
3
4
3
G
G
G
⎞
⎠
⎠
⎟
ε
ε
ε
33
33
33
… collecting terms multiplying
σ
12
=
2 G
ε
12
ε
12
,
ε
23
,
ε
13
σ
23
=
2 G
ε
23
σ
13
=
2 G
ε
13
(1)
(2)
(3)

Complete stress-strain relation
σ
11
=
⎛
⎝
K
+
4
3
G
⎟ ε
11
+ ⎜
K
−
2
3
G
⎟ ε
22
+
⎛
⎝
K
−
2
3
G
⎟ ε
33
(1) c
1111
=
K
+
4
G
3 c
1122
=
K
−
2
G
= c
3
1133

Complete stress-strain relation
σ
22
=
⎛
⎝
K
−
2
3
G
⎟ ε
11
+ ⎜ K
+
4
3
G ⎟ ε
22
+
⎛
⎝
K
−
2
3
G ⎟ ε
33 (2) c
2211
=
K
−
2
G
= c
3
2233 c
2222
=
K
+
4
G
3

Complete stress-strain relation
σ
33
= ⎜
K
−
2
3
G
⎞
⎠
ε
11
+ ⎜
K
−
2
3
G
⎞
⎠
ε
22
+ ⎜⎛
K
+
4
G
⎞
⎝ 3 ⎠
ε
33
(3) c
3311
=
K
−
2
G
= c
3
3322 c
3333
=
K
+
4
G
3

Complete stress-strain relation
σ
12
=
2 G
ε
12
σ
23
=
2 G
ε
23
σ
13
=
2 G
ε
13 c
1212
=
2 G c
2323
=
2 G c
1313
=
2 G
All other c ijkl are zero

c
1111
= c
2222
= c
3333
=
K
+
4
G
3 c
1122
= c
1133
= c
2233
=
K
−
2
G
3 c
1212
= c
2323
= c
1313
=
2 G

Examples – numerical values
Concrete
K = 14 GPa G = 10 GPa
Quartz (sand, stone..)
K = 27 GPa G = 26 GPa
Steel
K = 200 GPa G = 140 GPa