1.050 Engineering Mechanics
Lecture 24: Beam elasticity – derivation of governing equation
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 20-31
Oct./Nov.
V. How things fail – and how to avoid it
9.
10.
11.
Elastic instabilities
Plasticity (permanent deformation)
Fracture mechanics
Lectures 32-37
Dec.
2
1
1.050 – Content overview
I. Dimensional analysis
II. Stresses and strength
III. Deformation and strain
IV. Elasticity
Lecture 20:
Lecture 21:
Lecture 22:
Lecture 23:
Lecture 24:
Lecture 25:
Lecture 26:
…
Introduction to elasticity (thermodynamics)
Generalization to 3D continuum elasticity
Special case: isotropic elasticity
Applications and examples
Beam elasticity
Applications and examples (beam elasticity)
… cont’d and closure
V. How things fail – and how to avoid it
3
Goal of this lecture
• Derive differential equations that can be solved to determine stress,
strain and displacement fields in beam
• Consider 2D beam geometry:
z
x
+ boundary conditions (force, clamped, moments…)
• Approach: Utilize beam stress model, strain model for beams and
combine with isotropic elasticity
4
2
Stress
(σ ) = ⎜⎜
ij
Strain
⎛ σ xx 0 σ xz ⎞
0
⎜σ
⎝ xz
0
0
Navier-Bernouilli beam model
⎟
0 ⎟
0 ⎟
⎠
ε xx = ε xx0 + ϑ y0 z
d 2ξ z0
ϑ =− 2
dx
dξ x0
0
ε xx =
dx
0
y
Shape of stress tensor
for 2D beam problem
N = ∫ σ xx dS Qz = ∫ σ xz dS
S
M y = ∫ zσ xx dS
dx
= Qz
d 2M y
dx 2
dN
= − fx
dx
Axial strain
S
Thus:
dξ x0 d 2ξ z0
−
z
ε xx =
dx
dx 2
S
dM y
Curvature
= − fz
Isotropic elasticity:
Strain completely determined from
displacement of beam reference axis
⎛
⎝
2 ⎞
3 ⎠
σ = ⎜ K − G ⎟ε v 1 + 2Gε
5
Derivation of beam constitutive equation in 3-step
approach
Section number below corresponds to section numbering used in class
Step 1: Consider continuum scale alone (derive a relation between stress and
strain for the particular shape of the stress tensor in beam geometry)
2.1)
Step 2: Link continuum scale with section scale (use reduction formulas)
2.2)
Step 3: Link section scale to structural scale (beam EQ equations)
2.3)
6
3
Overview
Structural
scale
Section
scale
Continuum
scale
N ( x ), Qz ( x ), M y ( x ),
ω y ( x ), ξ x ( x ), ξ z ( x )
N , Qz , M y
σ ,ε
slope
Reminder:
z
x
ξ
Rotation (slope)
ω y0 = −
Curvature (=first derivative of rotation)
dξ
dx
0
d 2ξ z0 dω y
ϑ =− 2 =
dx
dx
0
z
0
y
2.1) Step 1 (continuum scale)
z
0
z
Consider a beam in
uniaxial tension:
F
x
(1)
σ yy = ⎜ K − G ⎟(ε xx + ε yy + ε zz ) + 2Gε yy = 0
(2)
2 ⎞
3 ⎠
2 ⎞
!
⎛
3 ⎠
⎝
2 ⎞
!
⎛
σ zz = ⎜ K − G ⎟(ε xx + ε yy + ε zz ) + 2Gε zz = 0
3 ⎠
⎝
Eqns. (2) and (3) provide relation between
ε yy
0 0⎞
⎟
0 0⎟
0 0 ⎟⎠
σ xx = ⎜ K − G ⎟(ε xx + ε yy + ε zz ) + 2Gε xx
⎛
⎝
3 unknowns, 2
equations; can
eliminate one variable and obtain
relation between 2
remaining ones
⎛ σ xx
(σ ij ) = ⎜⎜ 0
⎜ 0
⎝
7
ε xx
and
ε yy , ε zz
1 3K − 2G
= ε zz = −
ε xx = −νε xx
2 3K + G
=: ν Poisson’s ratio
(3)
:
8
4
Physical meaning “Poisson’s effect”
• The ‘Poisson effect’ refers to the fact that beams
contract in the lateral directions when subjected to
tensile strain
ε yy = ε zz = −νε xx
d1
d2
From eq. (1) (with Poisson relation):
d 2 = (1 + ε yy )d1 = (1 − νε xx )
σ xx =
9
9 KG
ε xx
3K + G
=: E
Young’s modulus
σ xx = Eε xx
This result can be generalized: In bending, the shape of the stress tensor is
identical, for any point in the cross-section (albeit the component σzz typically
varies with the coordinate z)
Thus, the same conditions for the lateral strains applies
Therefore: We can use the same formulas!
⎛ σ xx ( z ) 0 0 ⎞
(σ ij ) = ⎜⎜ 0 0 0 ⎟⎟
⎜ 0
0 100 ⎟⎠
⎝
5
2.2) Step 2 (link to section scale) Now: Plug in relation
Consider that
ε xx =
σ xx = Eε xx
dξ x0 d 2ξ z0
−
z
dx
dx 2
into reduction formulas
and thus
⎛ dξ 0 d 2ξ 0
σ xx = E ⎜⎜ x − 2z
dx
⎝ dx
⎞
z ⎟⎟
⎠
Results in:
Assume: E constant over S
⎛ dξ x0 d 2ξ z0
N = ∫ E ⎜⎜
−
dx 2
⎝ dx
S
⎞
z ⎟⎟dS
⎠
N=E
⎛ dξ 0
d 2ξ z0 2 ⎞
M y = ∫ E ⎜⎜ x z −
z ⎟⎟dS
2
dx
dx
⎠
⎝
S
Finally:
=0
d 2ξ z0
dξ x0
zdS
dS
−
E
dx 2 ∫S
dx ∫S
dξ
My = E
dx
dξ 0
N = ES x
dx
=0
0
x
d 2ξ z0 2
∫S zdS − E dx 2 ∫S z dS
d 2ξ z0
M y = −EI
dx 2
=I
Area
11
moment
of inertia
2.3) Step 4 (link to structural scale)
Beam EQ equations:
d 2M y
dx 2
Beam constitutive equations:
d 4ξ z0
M y = −EI
= − fz
dx 4
= − fz
with:
dN
= − fx
dx
d 2ξ z0
dx 2
dξ 0
N = ES x
dx
M y = −EI
d 4ξ z0
f
= z
4
dx
EI
d 2ξ x0
f
=− x
2
dx
ES
12
6
Beam bending elasticity
Governed by this differential equation:
d 4ξ z0
f
= z
4
dx
EI
Integration provides solution for displacement
Solve integration constants by applying BCs
Note:
E = material parameter (Young’s modulus)
I = geometry parameter (property of cross-section)
fz = distributed shear force
How to solve? Lecture 25
13
7