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 32
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 beam structures (cont’d) How to solve problems
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 20-32
Oct./Nov.
V. How things fail – and how to avoid it
1
Lectures 33-37
Dec.
2
⎧ max (− ε com (σ ' ) )⎫
⎪⎪σ ' S. A.
⎪⎪
r
− ε com (σ ' ) ≤ ⎨ is equal to ⎬ ≤ ε pot
(
ξ
')
r
r
σ ' S. A.
ξ ' K. A.
⎪ rmin ε (ξ ' ) ⎪
⎭⎪
⎩⎪ ξ ' K . A. pot
I. Dimensional analysis
II. Stresses and strength
III. Deformation and strain
Introduction: Energy bounds in linear elasticity (1D system)
Introduction: Energy bounds in linear elasticity (1D system), cont’d
1D examples
Generalization to 3D
Energy bounds in beam structures
Energy bounds in beam structures (cont’d): How to solve problems
V. How things fail – and how to avoid it
Lecture 33 (Mon): Buckling (loss of convexity)
Lecture 34 (Wed): Fracture mechanics I (and surprise!)
Lecture 35 (Fri): Fracture mechanics II
Lecture 36 (Mon): Plastic yield
Lecture 37 (Wed): Wrap-up plastic yield and closure
Elastic instabilities
Plasticity (permanent deformation)
Fracture mechanics
Review: 3D isotropic elasticity
1.050 – Content overview
IV. Elasticity
…
Lecture 27:
Lecture 28:
Lecture 29:
Lecture 30:
Lecture 31:
Lecture 32:
9.
10.
11.
Solution
Lower bound
Upper bound
Potential energy
approach
Complementary energy
approach
r
ε com (σ ' ) = ψ * (σ ' ) − W * (T ' )
r
3
r
ε pot (ξ ' ) = ψ (ε ' ) − W (ξ ' )
4
1
Beam structures (2D)
Beam structures
Complementary free energy
External work by prescribed displacements
[
⎡ 1 N 2 1 M y2 ⎤
ψ* = ∫ ⎢
+
⎥dx
2 ES 2 EI ⎦
x=0..l ⎣
[
r0 r d
r r
⋅ f (x)dx + ∑ ξ 0 ⋅ F d (xi ) + ω y M yd (xi )
∫ξ
W=
i
x=0..l
=
∫ [ξ
0
x
]
]
[
f xd (xi ) + ξ z0 f zd (xi ) dx + ∑ ξ x0 Fxd (xi ) + ξ z0 Fzd (xi ) + ω y M yd (xi )
]
i
x=0..l
5
6
Clapeyron’s formulas Beam elasticity (
)
r
⎧ max − ε com (FS ', M y ' ) ⎫
⎪⎪
⎪⎪ N ',M y ' S. A.
r
r
is equal to
≤ ε pot (rξ ', ω y ' )
− ε com (FS ', M y ' ) ≤ ⎨
⎬
v
N ',M y 'S. A.
ξ 'K. A.
⎪
⎪
rmin ε pot (ξ ', ω y ' )
ξ ' K . A.
⎭⎪
⎩⎪
1 *
(W + W )
2
1 *
(W − W )
2
1
= (W − W * )
2
ε pot =
ε com
]
i
External work by prescribed force densities/forces/moments
2
2⎤
1
⎡1
ψ = ∫ ⎢ ES (ε xx0 ) + EI (ϑ y0 ) ⎥dx
2
2
⎦
x=0..l ⎣
ψ =ψ * =
[
i
Free energy
Note: For 2D, the only contributions are axial forces & moments and
axial strains and curvatures (general 3D case see manuscript page
263 and following)
]
r
r
W * = ∑ ξ d (xi ) ⋅ R + ω yd (xi )M y,R = ∑ ξ xd (xi )R x + ξ zd (xi )Rz + ω yd (xi )M y,R
Lower bound
Complementary energy
approach
“Stress approach”
Significance: Calculate solution potential/complementary
energy (“target”) from BCs
Work with unknown but
S.A. moments and
forces
7
Solution
r
FS', M y '
that provide
absolute max
of − ε com
r
ξ ', ω y '
that provide
absolute min of ε pot
Upper bound
Potential energy
approach
“Displacement
approach)
Work with unknown
but K.A. displacements
8
2
Step-by-step solution approach
Example
Use complementary energy approach!
• Step 1: Express target solution (Clapeyron’s formulas) – calculate
complementary energy AT solution
P
• Step 2: Determine reaction forces and reaction moments
• Step 3: Determine force and moment distribution, as a function of reaction
forces and reaction moments (need My and N)
l/2
l/2
δ = unknown displacement at point of
• Step 4: Express complementary energy as function of reaction forces and
reaction moments (integrate)
load application
• Step 5: Minimize complementary energy (take partial derivatives w.r.t. all
unknown reaction forces and reaction moments and set to zero); result: set of
unknown reaction forces and moments that minimize the complementary energy
• Step 6: Calculate complementary energy at the minimum (based on resulting
forces and moments obtained in step 5)
• Step 7: Make comparison with target solution = find solution displacement
9
Structure is statically indeterminate to degree 1
Can not be solved by relying on static equilibrium only (too many
unknown forces, ‘hyperstatic’).
Goal: Solve problem using complementary energy approach
Example
Step 1: Target solution ε com =
δ
10
Example
1
Pδ
2
Step 3: Determine force and moment distribution
(as a function of hyperstatic force R’):
Step 2: Determine hyperstatic forces and moments (here: R’)
x
2x
⎧ Pl
⎪ (1 − l ) − R' l(1 − l ) 0 ≤ x ≤ l / 2
M y (x) = ⎨ 2
x
⎪
− R' l(1 − )
l/2 < x ≤ l
l
⎩
P
Concept of
superposition often
helpful
P
M y (x)
+
Note: Only need expression for N and My
l/2
δ
R’
M y (x)
l/2
Step 4: Express complementary energy
R’
ε com
Hyperstatic force
11
=0
⎡ 1 N 2 1 M y2 ⎤
*
=ψ * −W * = ∫ ⎢
+
⎥dx − W
2 ES 2 EI ⎦
x=0..l ⎣
ε com (R' ) =
1 ⎛ l3 2 5 3
1 3 2⎞
⎜ R' − l R' P +
l P ⎟⎟
2EI ⎜⎝ 3
24
24
⎠
=0
No contribution
from prescribed
displacements
12
3
Example
Example
Step 5: Find min of ε com (R' )
Step 7: Compare with target solution
∂ε com (R' )
=0
∂R'
ε com =
1 ⎛ 2l 3 R' 5 3 ⎞
⎜
− l P ⎟⎟ = 0
2EI ⎜⎝ 3
24
⎠
R' =
δ=
5
P
16
7
l 3 P represents a minimum of the complementary energy
768EI
Is it a global minimum, that is, the solution?
Step 6: Minimum complementary energy
ε com (R' =
1
5
7
Pδ ≤ ε com (R' =
P) =
l 3P
2
16
1536EI
7
δ≤
l 3P
768EI
5
7
P) =
l 3P
16
1536EI
13
1. My’ is S.A.
2. R’ is the only hyperstatic reaction force (in other words, the only
source of additional moments)
3. Therefore, the minimum is actually a global minimum, and therefore, it
is the solution
14
Generalization (important)
• For any homogeneous beam problem, the minimization
of the complementary energy with respect to all
hyperstatic forces and moments
X i = {Ri , M y,R;i }
yields the solution of the linear elastic beam problem:
∂
(ε com ( X i )) =! 0
∂X i
1
(W − W * ) ≡ min
ε com ( X i )
Xi
2
15
4