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 29 Stresses and equilibrium
Strength models (how to design structures,
foundations.. against mechanical failure)
Lectures 4-15
Sept./Oct.
III. Deformation and strain
Energy bounds in 1D systems
5.
6.
Examples and applications
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
Example system: 1D truss structure
1.050 – Content overview
I. Dimensional analysis
II. Stresses and strength
Rigid boundary
III. Deformation and strain
IV. Elasticity
…
Lecture 23:
Lecture 24:
Lecture 25:
Lecture 26:
Lecture 27:
Lecture 28:
Lecture 29:
Lecture 30:
…
1
N1
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
1D examples
Generalization to 3D
3
2
δ1
N3
N2
δ2
δ3
P
Rigid bar
ξ0
V. How things fail – and how to avoid it
Lectures 32 to 37
3
4
1
Minimum potential energy approach
Conditions for
kinematically admissible
(K.A.): Deformation
must be compatible w/
rigid bar
Consider two kinematically admissible (K.A.)
displacement fields
1
2
3
N2
N3
δ1
δ2
P
δ1‘ = δ 2‘ = δ 3‘ = ξ0‘
Actual solution
Prescribed force
Unknown
displacement
Potential energy of actual solution is always smaller than the solution
to any other displacement field
Therefore, the actual solution realizes a minimum of the potential
energy:
δ1
δ3
ξ0
ε pot (δ i , ξ0 ) = ψ (δ i ) − Pξ0 ≤ ψ (δ i' ) − Pξ0' = ε pot (δ i' , ξ0' )
(1)
Approximation
to solution
(K.A.)
N1
Minimum potential energy approach
(2)
2
δ 2 = δ1 + (ξ0 − δ1 )
3
4
δ 3 = δ1 + (ξ0 − δ1 )
3
5
Minimum potential energy approach
ε pot (δ i , ξi ) = min ε pot (δ i' , ξi' )
δ i' K. A.
To find a solution, minimize the potential energy for a selected choice
of kinematically admissible displacement fields
Minimum complementary energy approach
Conditions for statically
admissible (S.A.)
N1 + N 2 + N 3 = R
(1)
Approximation
to solution
(K.A.)
δ1‘ = δ 2‘ = δ 3‘ = ξ0‘
‘
δ1
Actual solution
(2)
2
δ 2 = δ1 + (ξ0 − δ1 )
3
4
δ 3 = δ1 + (ξ0 − δ1 )
3
ε pot (ξ 0' ) = −
1 2
P
6k
is larger than
ε pot (ξ 0 , δ 1 ) = −
11 2
P
48k
7
6
We have not invoked the EQ conditions!
Consider two statically admissible force fields
3N1 + N 2 − N 3 = 0
N1'
(1)
N 2'
N 1' , N 2'
Approximation
to solution
Still S.A.
1
2
3
N1
N2
N3
δ1
δ2
R
R ' ξ 0d
N1
δ3
ξ 0d
Prescribed
displacement
Unknown force
Actual
solution
(obtained in lecture 20)
N2
(2)
N 3 N = 1/12R
1
N 2 = 1/ 3R
d
R ξ0
N 3 = 7 /12R
8
2
Minimum complementary energy approach
Minimum complementary energy approach
ε com ( N i , R ) = ψ * ( N i ) − ξ0d R ≤ ψ * ( N i' ) − ξ0d R ' = ε com ( N i' , R ' )
N1'
N 1' , N 2'
Approximation
to solution
Still S.A.
Complementary energy of actual solution is always smaller than the
solution to any other displacement field
N1
ε com ( N i , R ) = min ε com ( N i' , R ' )
N2
Actual
solution
(obtained in
lecture 20)
N i' S. A.
Combine: Upper/lower bound
R ξ 0d
3
N3
δ1
δ2
P
ε com (N 1 ) = ψ * (N i ) − ξ 0d R
=0
δ3
ξ0
ε com (N 1 ) =
Prescribed force
Unknown displacement
Find min:
∂ε com (N 1 )
=0
∂N 1
1
(12N12 − 2PN1 + P 2 )
4k
1
N1 =
P
12
N 3 = 7 /12R
ε com
2
is larger than
ε com (R, N 1 ) = −
2
12
k (ξ 0d )
11
10
Combine: Upper/lower bound
Treat this BC with complementary energy approach
2
(2)
N 3 N = 1/12 R
1
9
We have not invoked the kinematics of the problem!
N2
1
5
N 2 = 1/ 3R
To find a solution, minimize the complementary energy for a selected
choice of statically admissible force fields
1
ε com (R ' ) = − k (ξ 0d )
d
R' ξ 0
Therefore, the actual solution realizes a minimum of the
complementary energy:
N1
(1)
N 2'
11 2
=
P
48k
11
⎧ max (− ε com (N i' , R ' ) )⎫
⎪⎪ Ni' 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.
ε com =
11 2
P
48k
− ε com = −
ε pot (ξ 0 , δ 1 ) = −
11 2
P
48k
11 2
P
48k
12
At the solution to the elasticity problem, the upper and lower bound coincide
3
Step-by-step approach
• Step 1: Determine K.A. displacement field (for
approximation, find appropriate assumed displacement
field)
Another example
• Step 2: Express work balance – find
Approximate solution for coupled
beam-truss problem
• Step 3: Find min of
ε pot / ε com
ε pot / ε com
• Step 4: Determine displacement field, forces etc.
13
• Solution is approximation to actual solution
14
Minimum potential energy approach
Rigid boundary
Step 1: Assume K.A. displacement field
1
2
3
4
N1
N2
N3
N4
L
δ1
ξz
x
L
δ2
⎛ x ⎞
⎟
⎝ 3L ⎠
(approximation of the actual solution…)
L
δ3
δ4
δ1
P
Elastic beam
2
ξ z (x; α , β ) = β + α ⎜
ξ z (x = 0) = β
ξ0
15
δ2
δ3
δ4
ξ z (x = 3L) = β + α
16
4
Minimum potential energy approach
Minimum potential energy approach
Step 2:
Total free energy of a beam:
Displacement of the four truss members
ψB =
δ 1 = ξ z (x = 0) = β
δ 2 = ξ z (x = L) = β +
(*)
9
4
δ 3 = ξ z (x = 2L) = β + α
9
δ 4 = ξ z (x = 3L) = β + α
2
1
E (ε xx0 + ϑ y0 z ) dydzdx
2
x=0 z=−h / 2 y=−b / 2
3L
h/2
0
⎛ x ⎞
⎟
⎝ 3L ⎠
∫
=0
ϑ y0 = −
2
ξ z ( x; α , β ) = β + α ⎜
ψB =
ξ 0 = δ 4
b/2
∫ ∫
with: ε xx
α
Page 215 in manuscript
(chapter 5)
E 4α 2
2 81L4
3L
(no displacement in the x-direction)
∂ ξz
2α
=− 2
∂x 2
9L
2
h/2
∫ ∫
(curvature can be calculated from
the assumed displacement field)
b/2
2
∫ z dydzdx
x=0 z=−h / 2 y=−b / 2
17
Minimum potential energy approach
1
ψ B (α ) = k Bα 2
2
Total free energy:
Sum of free energies of four trusses…
ψ (α , β ) = ψ B (α ) +
ψ B (α , β ) =
“spring constant”
bh 3 E 2
α
162L3
=: 12 k18B
Minimum potential energy approach
ψi =
1 2
kδ i
2
ε pot (α , β ) =
2
2
α⎞ ⎛
1
1 ⎛
4α ⎞
⎛
2⎞
k Bα 2 + k ⎜ β 2 + ⎜ β + ⎟ + ⎜ β +
⎟ + (β + α ) ⎟⎟ − F (β + α )
⎜
2
2 ⎝
9⎠ ⎝
9 ⎠
⎝
⎠
∑ψ (α , β )
i
i=1..4
ψ (α , β ) =
2
2
⎞
α⎞ ⎛
1
1 ⎛
4α ⎞
⎛
2
k Bα 2 + k ⎜ β 2 + ⎜ β + ⎟ + ⎜ β +
⎟ + (β + α ) ⎟⎟
⎜
2
2 ⎝
9⎠ ⎝
9 ⎠
⎝
⎠
Step 3:
min (ε pot (α , β ) )
α ,β
How to find minimum of this function?
External work
W = F (α + β )
Take partial derivatives, and set each to zero
19
20
5
Minimum potential energy approach
∂
∂α
2
2
⎛1
⎞
⎛
⎞
⎜ k Bα 2 + 1 k ⎜ β 2 + ⎛⎜ β + α ⎞⎟ + ⎛⎜ β + 4α ⎞⎟ + (β + α )2 ⎟ − F (β + α )⎟ = 0
⎟
⎜2
⎟
2 ⎜⎝
9⎠ ⎝
9 ⎠
⎝
⎠
⎠
⎝
∂
∂β
2
2
⎛1
⎞
⎛
⎞
⎜ k Bα 2 + 1 k ⎜ β 2 + ⎛⎜ β + α ⎞⎟ + ⎛⎜ β + 4α ⎞⎟ + (β + α )2 ⎟ − F (β + α )⎟ = 0
⎜
⎟
⎜2
⎟
2
9
9
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
Results in a system of linear equations:
98
⎛
k
⎜ kB +
81
⎜
⎜⎜ 14 k
⎝ 9
14 ⎞
k⎟
9 ⎟⎛⎜ α ⎞⎟ = ⎛⎜ F ⎞⎟
⎜ ⎟ ⎜ ⎟
4k ⎟⎟⎝ β ⎠ ⎝ F ⎠
⎠
Step 4: Based on solution, determine
displacement field δ i (from (*)), then
forces:
N = kδ
i
i
98
⎛
⎛ α ⎞ ⎜ k B + 81 k
⎜⎜ β ⎟⎟ = ⎜ 14
⎝ ⎠ ⎜⎜
k
⎝ 9
−1
14 ⎞
k⎟
9 ⎟ ⎛⎜ F ⎞⎟
⎜ ⎟
4k ⎟⎟ ⎝ F ⎠
⎠
99F
⎛
⎞
⎜
⎟
⎛ α ⎞ ⎜ 2(81k B + 49k ) ⎟
⎜⎜ ⎟⎟ =
⎝ β ⎠ ⎜ F (81k B − 28k ) ⎟
⎜ 4k (81k + 49k ) ⎟
B
⎝
⎠
21
6