1.050 Engineering Mechanics I
Summary of variables/concepts
Lecture 16-26
1
Variable
Definition
Notes & comments
Define undeformed position
Deformed position
Displacement vector
r
X
Position vector, underformed
configuration
r
x
Position vector, deformed
configuration
r
ξ
r
v
r
ξ =x−X
r
r
F = Grad( x ) = 1 + Grad(ξ )
r r
F = Fij ei ⊗ e j
∂xi
∂x j
v
r
dx = F ⋅ dX
Fij =
Note: Distinction between
capital “X” and small “x”
Displacement vector
Deformation gradient tensor
Relates position vector of
undeformed configuration with
deformed configuration
Lectures 16 and 17: Introduction to deformation and strain
Key concepts: Undeformed and deformed configuration, displacement vector, the
transformation between the undeformed and deformed configuration is described by
the deformation gradient tensor
Derivation first for general case of large deformation
2
Variable
Definition
Large-deformation theory
J=
Notes & comments
dΩd
= det F
dΩ0
( )
r
T
nda = J F
−1
J = Jacobian volume change
r
⋅ NdA
Surface change (area & normal)
(
)
r
r
r
r
T
E = F F −1 L2d − L20 = dX ⋅ F T F − 1 ⋅ dX = dX ⋅ 2E ⋅ dX
λα =
∆Lα
2Eαα + 1 −1
L0,α
sin θα ,β =
Definition of strain tensor
Relative length variation in the
α-direction
2Eαβ
(1 + λα )(1 + λβ )
Angle change between two
vectors
r
Grad ξ << 1
ε
ε=
(
(
))
r
rT
1
grad ξ + grad ξ
2
1 ⎛ ∂ξ ∂ξ ⎞
ε ij = ⎜⎜ i + j ⎟⎟
2 ⎝ ∂x j ∂xi ⎠
Small deformation strain tensor
For Cartesian coordinate system
Lecture 18: How to calculate change of geometry (angle, volume, length..)
Small deformation theory: The small deformation theory is valid for small
deformations only; for this case the equations simplify. These concepts are most
important for the remainder of 1.050.
3
Small-deformation theory
Variable
Definition
1 r r
θ ( eα , eβ ) = γ αβ = ε αβ
2
r
λ (eα ) = ε αα
r
r
1
θ mr ,nr = m ⋅ ε ⋅ n
2
r
r
λnr = n ⋅ ε ⋅ n
Notes & comments
Angle change
Dilatation
Volume change
Surface change
“The” Mohr circle
r r
r
E ( n ) = ε ⋅ n (strain vector)
ε II
Mohr circle of strain tensor
Lecture 18: Small deformation - Mohr circle for strain tensor. Any strain tensor can
be represented in the Mohr plane; this way, one can display a 3D tensor quantity in
a 2D projection. All concepts are the same as for the stress tensor Mohr plane.
The quantities on the x/y-axes are dilatations and angle change (shear).
4
Variable
Definition
δW
Work done by external forces
dψ
Free energy change
dψ = δW
Non-dissipative deformation=
elastic deformation
All work done on system
stored in free energy
∂ψ
∂ψ
dxi =
dξ j
∂xi
∂ξ j
Notes & comments
Defines thermodynamics of
elastic deformation
Solution approach
1D truss systems
Link between stress and
strain
Also called “generalized
Hooke’s law”
∀dxi , ∀dξ j
σ ij = cijklε kl
σ = c:ε
Lectures 20 and 21: Elasticity, basic definitions. The most important concept of
this lecture is that elastic deformation is a thermodynamic process under which no
energy dissipation occurs. This concept can be generally applied to characterize
any elasticity problem. We derived elasticity for 1D systems (including solution
strategy), and then generalized it to 3D. This led to the link between stress and
strain.
5
Variable
Definition
Elastic properties of material
do NOT depend on direction
Isotropic elasticity
ε =
ε
(
)
1
1
ε :εT =
∑∑ ε ij2
2
2 i j
tr (ε ) = ε :1 = ε11 + ε 22 + ε 33 =
tr (ε )
Ψ (ε v , ε d )
⎛
⎝
⎞
⎠
Ψ=
⎛
⎝
2 ⎞
3 ⎠
1
1
Kε v2 + Gε d2
2
2
σ = ⎜ K − G ⎟ε v 1 + 2Gε = ⎜ K − G ⎟(ε11 + ε 22 + ε 33 )1 + 2Gε
2
3
Notes & comments
dΩ d − dΩ 0
dΩ 0
Isotropic elasticity described
uniquely by 2 parameters, K
and G
“Length” of a tensor
“Volume change” of a tensor
Free energy due to volume
strain and shear strain
(assumption, mathematical
model to describe elastic
behavior of isotropic solids)
Linear isotropic elasticity
Tensor notation
Lecture 22: Isotropic elasticity, basic concepts. The most important equation on
this slide is the one on the bottom, for linear isotropic elasticity. Note that isotropic
elasticity is fully characterized by two constants, K and G. These two parameters
have physical meaning; K describes how the free energy changes under volume
changes, and G describes how the free energy changes under shear (shape)
changes.
6
Variable
Definition
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 ⎠
⎝
Notes & comments
σ 11 = ⎜ K − G ⎟(ε11 + ε 22 + ε 33 ) + 2Gε11
σ 12 = 2Gε12
Linear isotropic elasticity
Written out for individual
stress tensor coefficients
σ 23 = 2Gε 23
σ 13 = 2Gε13
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 11 = ⎜ K + G ⎟ε 11 + ⎜ K − G ⎟ε 22 + ⎜ K − G ⎟ε 33
σ 22
σ 33
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
c1212 = c2323 = c1313 = 2G
Lecture 22: Isotropic elasticity, equations that relate stress and strain. Here we
summarize the equations in different forms. On the bottom, right, you see how to
calculate the elasticity tensor coefficients from K and G.
7
Overview: 3D linear elasticity
r
Stress tensor σ (x )
Strain tensor
Basis: Geometry
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
σ = c:ε
Kinematically admissible (K.A.)
Statically admissible (S.A.)
Basis: Physical laws
(Newton’s laws)
Elasticity
r
ε ( x)
Basis: Thermodynamics
σ ij = cijklε kl
⎛
⎝
BCs on boundary of domain
rd
r
∂Ωξr d : ξ = ξ
Ω
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
Isotropic solid
Summary, 3D linear elasticity. This page may be useful to keep an overview over
the methods and approaches covered here. This summary is valid for any linear
elasticity problem.
8
Variable
Definition
Notes & comments
• Step 1: Write down BCs (stress BCs and
displacement BCs), analyze the problem to
be solved (read carefully!)
• Step 2: Write governing equations for
stress tensor, strain tensor, and constitutive
equations that link stress and strain, simplify
expressions
Solution procedure to solve
3D elasticity problems
• Step 3: Solve governing equations (e.g. by
integration), typically results in expression
with unknown integration constants
• Step 4: Apply BCs (determine integration
constants)
Lecture 23: Solution approach, 3D isotropic elasticity problems. This is a 4-step
solution procedure that guides you through the process.
9
Variable
Definition
Notes & comments
ε xx = ε xx0 + ϑ y0 z
ε xx
ϑ y0 = −
ε xx0 =
ε xx =
d 2ξ z0
dx 2
dξ
dx
0
x
Curvature
Strain
Navier-Bernouilli beam
model; strain distribution in
beam section
dξ x0 d 2ξ z0
−
z
dx
dx 2
z
F
ν
ν=
1 3K − 2G
2 3K + G
ε yy = ε zz = −νε xx
E=
E
9 KG
3K + G
σ xx = Eε xx
x
Uniaxial beam deformation
Poisson’s ratio (lateral
contraction under uniaxial
tension)
Young’s modulus (relates
stresses and strains under
uniaxial tension)
Lecture 19 and 24: Beam deformation and beam elasticity. Here we only review
the beam bending case for 2D systems. Beam elasticity is a special case of 3D
elasticity, adapted for the particular (stretched) geometry of beams. This slide also
reviews the introduction of Young’s modulus E and Poisson’s ratio. Both can be
calculated from K and G.
10
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
dx 2
ES
d 4ξ z0
f
= z
4
dx
EI
•
•
•
•
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
Lecture 25: Beam elasticity, cont’d. Note the two differential equations for axial
load/displacement and shear load/displacement in the z-direction. This slide also
summarizes the 4-step approach to solve beam problems.
11
Beam stretching
(axial forces..)
Beam bending
(shear forces, moments..)
Variable
Definition
Notes & comments
d 4ξ z
f
= z
dx 4
EI
d 3ξ z
Q
=− z
dx 3
EI
M
d 2ξ z
=− y
dx 2
EI
dξ z
= −ω y
dx
d 4ξ z
EI = f z
dx 4
d 3ξ z
−
EI = Qz
dx 3
d 2ξ z
−
EI = M y
dx 2
dξ
− z =ω y
dx
ξz
ξz
d 2ξ x0
f
=− x
dx 2
ES
dξ x0
= ε xx0
dx
ξ x0
− ES
Shear force density
Shear force
Bending moment
Rotation (angle)
Displacement
d 2ξ x0
= fx
dx 2
Axial force density (e.g. gravity)
Axial strain
Axial displacement
Lecture 25: Beam elasticity, governing equations for both beam bending and beam
stretching. This slide reviews the physical meaning of the different derivatives.
12
Variable
Definition
f (x)
function of x
f ' (x) = 0
necessary condition for
min/max
f ' ' (x) < 0
local maximum
f ' ' (x) > 0
local minimum
f ' ' (x) = 0
inflection point
Notes & comments
How to find
min/max of
functions
• Start from f z = EIξ z'''' , then work your way up…
• Note sign changes:
ξ z'''' ~ f z
ξ z''' ~ −Qz
+→−
ξ z'' ~ −M y
ξ z' ~ −ω y
−→+
ξz ~ ξz
• At each level of derivative, first plot extreme cases at ends of
beam
• Then consider zeros of higher derivatives; determine points of
local min/max
• ξ z represents physical shape of the beam (“beam line”)
Drawing/sketching
approach
Lecture 26: Drawing of beam problems. Note the sign changes, as indicated. The
approach is based on the concept of considering min/max values of the functions;
since all physical quantities are derivates of one another, this approach can be
easily applied to plot the solution.
13
Variable
Definition
Notes & comments
f z ( x ) = − p ~ ξ z''''
Example
5 ⎞
⎛
Qz ( x ) = p⎜ x − l ⎟ ~ −ξ z'''
8 ⎠
⎝
⎛1
x2 5 ⎞
M y ( x ) = p⎜⎜ l 2 + − lx ⎟⎟ ~ −ξ z''
8
2 8 ⎠
⎝
ω y( x) =
p
EI
ξz ( x) = −
⎛1 2
⎞
x3 5
⎜⎜ l x + − lx 2 ⎟⎟ ~ −ξ z'
8
6
16
⎝
⎠
p
EI
z
p = force/length
x
l length
EI
⎛ 1 2 2 x4 5 3 ⎞
⎜⎜ l x +
− lx ⎟
24 48 ⎟⎠
⎝ 16
Lecture 26: Example. Remember to clearly indicate the coordinate system when
you draw beam elasticity solutions.
14
Variable
Free end
Definition
r
F =0
r
M =0
Concentrated force
Qz = −P
Notes & comments
ξz = 0
My = 0
ξx = 0
ωy = 0
Commin beam
boundary conditions
P
Hinge (bending)
My = 0
r
ξ =0
ωy = 0
⎛ N (x) M y (x) ⎞ N (x) M y (x)
z ⎟⎟ =
z
+
+
EI
S
I
⎠
⎝ ES
σ xx (z; x) = E ⎜⎜
Stress distribution within
cross-section
Lecture 26: Common boundary conditions in beam problems, plotting of stress
distribution within cross-section.
15