1.050 Engineering Mechanics I
Review session
1
1.050 – Content overview
I. Dimensional analysis
1.
2.
On monsters, mice and mushrooms
Similarity relations: Important engineering tools
Lectures 1-3
Sept.
II. Stresses and strength
3.
4.
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 20-32
Oct./Nov.
V. How things fail – and how to avoid it
9.
10.
11.
Elastic instabilities
Fracture mechanics
Plasticity (permanent deformation)
Lectures 33-37
Dec.
2
1
Notes regarding final exam
• Please contact me or stop by at any time for any questions
• The final will be comprehensive and cover all material discussed in
1.050. Note that the last two p-sets will be important for the final.
– To get an idea about the style of the final, work out old finals and the practice
final
– There will be 2-3 problems with several questions each (e.g. beam problem/
truss problem, continuum problem)
– We will post old final exams from 2005 and 2006 today
– We will post an additional, new practice final exam on or around Wednesday
next week
– Another list of variables and concepts will be posted next week
– Stay calm, read carefully, and practice time management
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Stress, strain and elasticity concepts
4
2
Overview: 3D linear elasticity
r
Stress tensor σ (x )
Strain tensor
Basis: Geometry
Kinematically admissible (K.A.)
Statically admissible (S.A.)
Basis: Physical laws
(Newton’s laws)
Ω
rd r
r
∂ΩTr d : T ( n ) = σ ⋅ n
r
Ω : T (nr ) = σ ⋅ nr
r
div σ + ρg = 0
σ ij = σ ji
BCs on boundary of domain
Elasticity
σ = c:ε
r
ε (x )
Basis: Thermodynamics
⎛
⎝
BCs on boundary of domain
rd
r
∂Ωξ : ξ = ξ
rd
Linear deformation theory
r
Grad ξ << 1
(
(
))
r
rT
1
grad ξ + grad ξ
2
1 ⎛ ∂ξ ∂ξ ⎞
ε ij = ⎜⎜ i + j ⎟⎟
2 ⎝ ∂x j ∂xi ⎠
ε=
2 ⎞
3 ⎠
σ = ⎜ K − G ⎟ε v 1 + 2Gε
σ ij = cijklε kl
Ω
5
Isotropic solid
Isotropic elasticity
⎛
⎝
2 ⎞
3 ⎠
⎛
⎝
2 ⎞
3 ⎠
σ = ⎜ K − G ⎟ε v 1 + 2Gε = ⎜ K − G ⎟(ε11 + ε 22 + ε 33 )1 + 2Gε
2 ⎞
⎛
3 ⎠
⎝
2 ⎞
⎛
σ 22 = ⎜ K − G ⎟(ε11 + ε 22 + ε 33 ) + 2Gε 22
3 ⎠
⎝
2 ⎞
⎛
σ 33 = ⎜ K − G ⎟(ε11 + ε 22 + ε 33 ) + 2Gε 33
3 ⎠
⎝
σ 12 = 2Gε12
σ 11 = ⎜ K − G ⎟(ε11 + ε 22 + ε 33 ) + 2Gε11
Linear isotropic elasticity
Written out for individual
stress tensor coefficients
σ 23 = 2Gε 23
σ 13 = 2Gε13
σ 11 = ⎛⎜ K + G ⎞⎟ε 11 + ⎛⎜ K − G ⎞⎟ε 22 + ⎛⎜ K − G ⎞⎟ε 33
σ 22
σ 33
4
2
2
3 ⎠
3 ⎠
3 ⎠
⎝
⎝
⎝
2 ⎞
4 ⎞
2 ⎞
⎛
⎛
⎛
= ⎜ K − G ⎟ε 11 + ⎜ K + G ⎟ε 22 + ⎜ K − G ⎟ε 33
3 ⎠
3 ⎠
3 ⎠
⎝
⎝
⎝
2 ⎞
2 ⎞
4 ⎞
⎛
⎛
⎛
= ⎜ K − G ⎟ε 11 + ⎜ K − G ⎟ε 22 + ⎜ K + G ⎟ε 33
3 ⎠
3 ⎠
3 ⎠
⎝
⎝
⎝
σ 12 = 2G ε 12
σ 23 = 2G ε 23
σ 13 = 2G ε 13
Linear isotropic elasticity
Written out for individual
stress tensor coefficients,
collect terms that multiply
strain tensor coefficients
4
c1111 = c2222 = c3333 = K + G
3
2
c1122 = c1133 = c2233 = K − G
3
6
c1212 = c2323 = c1313 = 2G
3
Variable
ν
Definition
Notes & comments
ε yy = ε zz = −νε xx ν =
E=
E
1 3K − 2G
2 3K + G
9 KG
3K + G
Poisson’s ratio (lateral
contraction under uniaxial
tension)
Young’s modulus (relates
stresses and strains under
uniaxial tension)
σ xx = Eε xx
z
F
x
Uniaxial beam deformation
F Æ σ= F/A
x Æ ε= x/L
7
Solving problems with
strength approach
Use conditions for S.A. plus
strength criterion (S.C.)
8
4
Variable
Definition
Notes & comments
At any point, σ must be:
Two pillars of
stressstrength
approach
(1) Statically admissible (S.A.)
and
(2) Strength compatible (S.C.)
• Equilibrium conditions “only” specify statically admissible
stress field, without worrying about if the stresses can
actually be sustained by the material – S.A.
From EQ condition for a REV we can integrate up
(upscale) to the structural scale
Examples: Many integrations in homework and in class;
Hoover dam etc.
• Strength compatibility adds the condition that in addition
to S.A., the stress field must be compatible with the
strength capacity of the material – S.C.
In other words, at no point in the domain can the stress
vector exceed the strength capacity of the material
Examples: Sand pile, foundation etc. – Mohr circle
9
Variable
Definition
Notes & comments
DS
Strength domain for beams
For rectangular cross-section
b,h
N x lim = N 0
N x lim = N 0 = bhσ 0
f (M y , N x ) =
f (M y , N x ) =
My
M0
+
Strength capacity for beams
Nx
−1 ≤ 0
Nx
2
⎛N ⎞
+ ⎜ x ⎟ −1 ≤ 0
M 0 ⎜⎝ N x ⎟⎠
My
Moment capacity for beams
M-N interaction (linear)
f (M y , N x ) ≤ 0
1
M-N interaction (actual);
convexity
10
1
5
Variable
Definition
Safe
strength
domain
Notes & comments
Linear combination is safe
(convexity)
: load bearing capacity of
i-th load case
11
Mohr circle
Display 3D strain tensor in 2D
projection – enables us to ‘see’ largest
shear stresses, largest normal
stresses…
Thereby facilitates application of
strength criterion
12
6
Variable
Definition
Notes & comments
Strength domain (general
definition)
Equivalent to condition for S.C.
Dk
Dk ,Tresca
Max. shear stress
c
Tresca criterion
v
v
∀n : f (T ) = σ − c ≤ 0
Max. tensile stress
Dk ,Tension−cutoff
Tension cutoff criterion
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c
Variable
Definition
τ
Dk ,Mohr −Coulomb
Notes & comments
=µ
Max. shear stress
function of σ
Mohr-Coulomb
σ
c cohesion
c=0 dry sand
τ
Angle of repose
σ
14
7
Variable
Definition
Notes & comments
S = ∫ dS
S
Cross-sectional area
S
I = ∫ z 2 dS
I
Second order area moment
S
M y = − EI
EI
d 2ξ z0
= EIϑ y
dx 2
d 2ξ x0
f
=− x
ES
dx 2
d 4ξ z0
f
= z
EI
dx 4
•
•
•
•
Step 1: Write down BCs (stress BCs and
displacement BCs), analyze the problem to be
solved (read carefully!)
Step 2: Write governing equations for ξ z , ξ x ...
Step 3: Solve governing equations (e.g. by
integration), results in expression with unknown
integration constants
Step 4: Apply BCs (determine integration
constants)
Beam bending stiffness
(relates bending moment and
curvature)
Governing differential
equation, axial forces
Governing differential
equation, shear forces
Solution procedure to solve
beam elasticity problems
15
Energy approach
Approximate solution or find exact
solution
16
8
Upper/lower bounds
⎧ 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
⎪⎭
Solution
Lower bound
Upper bound
Potential energy
approach
Complementary energy
approach
r
ε com (σ ' ) = ψ * (σ ' ) − W * (T ' )
r
r
ε pot (ξ ' ) = ψ (ε ' ) − W (ξ ' )
17
Boundary conditions
Important concept in 1.050 and
elsewhere
18
9
Important BCs in beams/frames
Free end
r
F =0
r
M =0
Concentrated force
Qz = −P
ξz = 0
My = 0
ξx = 0
ωy = 0
P
Hinge (bending)
My = 0
r
ξ =0
ωy = 0
19
Buckling of beams in
compression
Elastic instability
20
10
Buckling
Euler beam buckling
Different boundary conditions
21
Fracture mechanics
How to treat cracks in a continuum
22
11
Example: 3D fracture model
Expressions for G can be found for a variety of geometries and structures:
rd
r
T = σ0ez
σ0
For this geometry:
G = 1.12 2
πaσ 02 !
E
= 2γ s
r
ez
a
σ0 =
far away from
crack
r
ex
2γ s E
1.12 2 πa
σ0
1
K Ic
σ0 =
1.12 2 πa
“Surface crack”
r
r
T d = −σ0ez
23
Example application
Detect crack of length
am
Photograph of crack (crack length of a(sub m)
removed due to copyright restrictions.
Concrete beam
am
am
Measure length of crack
E.g. add fluorescent fluid
use UV light
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http://www.amesresearch.com/images/cshst/block_before.jpg
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Example application
Detect crack of length
am
Concrete beam
Question: Will structure fail?
Solution: Solving beam problem provides us with stress distribution in section σ xx (z; x) =
M y (x)
I
z σ xx (z) =
σ0
h/2
z
z
−h/2
h/2
σ xx (z)
Calculate critical fracture stress and compare with stress in beam structure
σ 0,crit =
1
K Ic
1.12 2 πam
h
2
h
σ xx (z = − ) < σ 0,crit
2
σ xx (z = − ) ≥ σ 0,crit failure
no failure
25
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