1.050 – Content overview
I. Dimensional analysis
1.
2.
1.050 Engineering Mechanics I
On monsters, mice and mushrooms
Similarity relations: Important engineering tools
Lectures 1-3
Sept.
II. Stresses and strength
3.
4.
Lecture 28 Introduction: Energy bounds in linear elasticity (cont’d)
Stresses and equilibrium
Strength models (how to design structures,
foundations.. against mechanical failure)
Lectures 4-15
Sept./Oct.
III. Deformation and strain
5.
6.
How strain gages work?
How to measure deformation in a 3D
structure/material?
Lectures 16-19
Oct.
IV. Elasticity
7.
8.
Elasticity model – link stresses and deformation
Variational methods in elasticity
Lectures 20-31
Oct./Nov.
V. How things fail – and how to avoid it
1
9.
10.
11.
Elastic instabilities
Plasticity (permanent deformation)
Fracture mechanics
Lectures 32-37
Dec.
2
Outline and goals
1.050 – Content overview
I. Dimensional analysis
Use concept of concept of convexity to derive conditions that
specify the solutions to elasticity problems
II. Stresses and strength
III. Deformation and strain
IV. Elasticity
…
Lecture 23:
Lecture 24:
Lecture 25:
Lecture 26:
Lecture 27:
Lecture 28:
Lecture 29:
…
Obtain two approaches:
Approach 1: Based on minimizing the potential energy
Applications and examples
Beam elasticity
Applications and examples (beam elasticity)
… cont’d and closure
Introduction: Energy bounds in linear elasticity (1D system)
Introduction: Energy bounds in linear elasticity (1D system), cont’d
Generalization to 3D, examples
Approach 2: Based on minimizing the complementary energy
Last part: Combine the two approaches: Upper/lower bound
V. How things fail – and how to avoid it
Lectures 32 to 37
3
4
1
Reminder: convexity of a function
f (x)
Total external work
∂f
| (b − a) ≤ f (b) − f (a)
∂x x=a
v r
v r
Wd = ξ ⋅ Fd +ξ d ⋅ R
Work done by
prescribed
forces
Displacement
s unknown
secant
f (x)
Work done by
prescribed
displacements,
force unknown
tangent
a
x
b
Free energy and complementary free energy functions
ψ i* ,ψ i
5
are convex!
6
Total internal work
Combining it…
Ni
v r
v r !
W d = ξ ⋅ F d + ξ d ⋅ R =ψ + ψ *
State equations
Complementary free energy
ψ
Ni =
*
i
δi =
ψi
Free energy
∑δ N
i
i
∂ψ
∂δ i
∂ψ
(
*
∂N i
Complementary
energy
=: ε com
δi
i
=ψ * (N i ) + ψ (δ i )
)
v r !
v r
− ψ * − ξ d ⋅ R =ψ − ξ ⋅ F d
Solution to elasticity problem
7
Potential
energy
=: ε pot
− ε com = ε pot
8
2
Minimum potential energy approach
Example system: 1D truss structure
Consider two kinematically admissible (K.A.)
displacement fields
(1)
Rigid boundary
1
N1
N3
N2
δ2
δ1
Approximation
to solution
(K.A.)
3
2
δ3
ξ0
9
Minimum potential energy approach
(2)
ξ0 P = ∑ N iδ i
(1)-(2)
δ2
P
δ1
δ3
Actual solution
Prescribed force
Unknown
displacement
(2)
2
(ξ0 − δ1 )
3
4
δ 3 = δ1 + (ξ0 − δ1 )
3
δ 2 = δ1 +
10
Minimum potential energy approach
ε pot (δ i , ξ0 ) = ψ (δ i ) − Pξ0 ≤ ψ (δ i' ) − Pξ0' = ε pot (δ i' , ξ0' )
P (ξ0' − ξ0 ) = ∑ N i (δ i' − δ i ) = ∑
i
Convexity:
3
N3
ξ0
Rigid bar
ξ0' P = ∑ N iδ i'
2
N2
δ1
P
(1)
1
N1
δ1‘ = δ 2‘ = δ 3‘ = ξ0‘
i
∂ψ
Ni =
∂δ i
∂ψ '
(δ i − δ i ) ≤ ψ (δ i' ) − ψ (δ i )
∂δ i
P (ξ0' − ξ0 ) ≤ ψ (δ i' ) − ψ (δ i )
∂ψ '
(δ i − δ i )
∂δ i
Therefore, the actual solution realizes a minimum of the potential
energy:
ε pot (δ i , ξi ) = min ε pot (δ i' , ξi' )
δ i' K. A.
ε pot (δ i , ξ0 ) = ψ (δ i ) − Pξ0 ≤ ψ (δ ) − Pξ = ε pot (δ , ξ )
'
i
Potential energy of actual solution is always smaller than the solution
to any other displacement field
'
0
'
i
'
011
To find a solution, minimize the potential energy for a selected choice
of kinematically admissible displacement fields
We have not invoked the EQ conditions!
12
3
Minimum complementary energy approach
Conditions for statically
admissible (S.A.)
N1 + N 2 + N 3 = R
Consider two statically admissible force fields
3N1 + N 2 − N 3 = 0
N1'
N 3'
N 2'
N1' , N 2' , N 3'
Approximation
to solution
Still S.A.
1
2
3
N1
N2
N3
δ1
δ2
R
(1)
ξ0d R ' = ∑ N i'δ i
(2)
ξ0d R = ∑ N iδ i
(1)-(2)
ξ0d (R ' − R) = ∑ δ i (N i' − N i ) = ∑
i
i
R'
N1
δ3
ξ 0d
Prescribed
displacement
Unknown force
(1)
Minimum complementary energy approach
(2)
N 3 N = 1/12R
1
N2
Actual
solution
(obtained in lecture 20)
Convexity:
N 2 = 1/ 3R
N 3 = 7 /12R
R
13
Minimum complementary energy approach
ε com ( N i , R ) = ψ * ( N i ) − ξ0d R ≤ ψ * ( N i' ) − ξ0d R ' = ε com ( N i' , R ' )
Combine: Upper/lower bound
Recall that the solution to elasticity problem
(change sign)
δ i' K. A.
'
To find a solution, minimize the complementary energy for a selected
choice of statically admissible force fields
We have not invoked the kinematics of the problem!
(− ε com (N i' , R' ))
− ε com (N i , R) = max
'
ε pot (δ i , ξi ) = min ε pot (δ i' , ξi' )
ε com ( N i , R ) = min ε com ( N , R )
N i' S. A.
− ε com = ε pot
N i S. A.
Therefore, the actual solution realizes a minimum of the
complementary energy:
'
i
∂ψ *
δi =
∂N
i
∂ψ * '
(N i − N i ) ≤ ψ * ( N i' ) − ψ * (N i )
∂N i
ξ0d (R ' − R) ≤ ψ * (N i' ) − ψ * (N i )
ε com ( N i , R ) = ψ * ( N i ) − ξ0d R ≤ ψ * ( N i' ) − ξ0d R ' = ε com ( N i' ,14R ' )
Therefore
Complementary energy of actual solution is always smaller than the
solution to any other displacement field
∂ψ * (N i ) '
(N i − N i )
∂N i
15
⎧ max (− ε com (N i' , R ' ) )⎫
'
⎪⎪ N i S. A.
⎪⎪
'
'
− ε com (N i' , R ' ) ≤ ⎨
is equal to
⎬ ≤ ε pot (δ i , ξi )
⎪ min ε (δ ' , ξ ' ) ⎪
i
i
pot
⎪ '
⎪⎭ Upper bound
Lower bound ⎩ δ i K. A.
16
At the solution to the elasticity problem, the upper and lower bound coincide
4
Approach to approximate/numerical
solution of elasticity problems
• Minimum potential energy approach:
Select a guess for a displacement field; the only condition that must be satisfied is
that it is kinematically admissible. In a numerical solution, this displacement field is
typically a function of some unknown parameters (a1,a2,…)
– Express the potential energy as a function of the unknown parameters a1,a2,…
– Minimize the potential energy by finding the appropriate set of parameters
(a1,a2,…) for the minimum – generally yields approximate solution
– The actual solution is given by the displacement field that yields a total
minimum of the potential energy. Otherwise, an approximate solution is
obtained
• Minimum complementary energy approach:
Select a guess for a force field; the only condition that must be satisfied is that it is
statically admissible. In a numerical solution, this force field is typically a function
of some unknown parameters (b1,b2,…)
– Express the complementary energy as a function of the unknown parameters
b1,b2,…
– Minimize the complementary energy by finding the appropriate set of
parameters (b1,b2,…) for the minimum – generally yields approximate solution
– The actual solution is given by the force field that yields a total minimum of the
complementary energy. Otherwise, an approximate solution is obtained
• At the elastic solution, the minimum potential energy approach solution and
the negative of the solution of the minimum complementary energy
17
approach coincide
5