ADVANCED STRUCTURAL MECHANICS
Transcript of ADVANCED STRUCTURAL MECHANICS
VSB– TECHNICAL UNIVERSITY OF OSTRAVA
FACULTY OF CIVIL ENGINEERING
ADVANCED STRUCTURAL
MECHANICS
Lecture 1
Jirı BrozovskyOffice: LP – H 406/3Phone: 597 321 321
E-mail: [email protected]
WWW: http://fast10.vsb.cz/brozovsky/
Topics
1. Basic equations
2. Plane structures
3. Energetical principles, variational methods
4. Finite element method
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Recommended books
• Timoshenko. S. – Gere, J.: Mechanics of Materials, Van
Nostrand Company, New York, 1972 (or newer edition)
• Boresi, A. – Schmidt, R.: Advanced Mechanics of Materials,
John Wiley & Sons, 2003
• Cook, R. D., Malkus, D. S., Plesha, M. E., Witt, R. J.: Con-
cepts and Applications of Finite Element Analysis, John
Wiley and Sons, 1995
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Topic 1:Basic equations of elasticity
• Basic unknowns
• Geometrical relations
• Equilibrium equations
• Material law
• Compatibility equations
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Basic unknowns
σ
σ
τ
τ
τ
ττ
τ
x
yy
z
yx
yz xy
xz
zx
zy
σz x
Stress vector
σ = {σx, σy, σz, τxy, τyz, τzx}T
(1)
Strain vector
ε = {εx, εy, εz, γxy, γyz, γzx}T
(2)
Displacement vector
u = {ux, uy, uz}T (3)
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Geometrical relations (1)
� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �
� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �0
y, v
x, u
A B
D C
β
αA’
D’
uv
dxux
dx
dy
−
−x
v
B’
− dxx
y
v
u− dy
dy
C’
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Geometrical relations (2)
� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �
� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �
0
y, v
x, u
A B
D C
β
αA’
D’
uv
dxux
dx
dy
−
−x
v
B’
− dxx
y
v
u− dy
dy
C’
εx =A′B′ −AB
AB=(x + dx + u + ∂u
∂xdx)− (x + u)− dx
dx=
∂u
∂x
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Geometrical relations (3)
Normal deformations
εx =∂u
∂xεy =
∂v
∂yεz =
∂w
∂z, (4)
Shear deformations
γyz = γzy =∂v
∂z+
∂w
∂y(5)
γzx = γxz =∂w
∂x+
∂u
∂z(6)
γxy = γyx =∂u
∂y+
∂v
∂x. (7)
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Geometrical relations (4)
Uvedene vztahy obecne neplatı:
γyz = γzy,
γzx = γxz,
γxy = γyx.
The the assumption of equality of shear stresses is based
on approximative fulfillment of moment equilibrium equati-
ons of differential element. Shear deformations are assumed
to be equal in the same way.
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Differential equilibrium conditions(1)
yxyx
zxzx xzxz
yz
yz
zy
τ
τ
τ
ττ
τ
τ
ττ
dx
zy
xy
σ’
y
’
x
’
xy
’τ
dz
dy
’
’
’
’
’τ
σ
σ
σ
σ
x
τ
z
z
y
σ
σx‘ = σx +∂σx
∂xdx, τxy‘ = τxy +
∂τxy
∂xdy, ... (8)
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Diff. equilibrium conditions (2)
σx‘ = σx +∂σx
∂xdx, τxy‘ = τxy +
∂τxy
∂ydy, ...
∑
Fi,y = (σ′x−σx) dx dy+(τxy−τ ′xy) dx dz+(τxz−τxz‘) dx dy = 0
(σx−σx−∂σx
∂xdx) dy dz+(τxy−τxy−
∂τxy
∂ydy) dx dz+(τxz−τxz−
∂τxz
∂zdz) dx dy = 0
After simplification:
∂σx
∂x+
∂τxy
∂y+
∂τxz
∂z= 0 (9)
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Diff. equilibrium conditions (3)
∂σx
∂x+
∂τxy
∂y+
∂τxz
∂z+X = 0
∂τxy
∂x+
∂σy
∂y+
∂τyz
∂z+ Y = 0 (10)
∂τzx
∂x+
∂τzy
∂y+
∂σz
∂z+ Z = 0
where X,Y , Z are volume forces.
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Material law (1)
Defines relations between stresses and strains.
Hooke law for 1D problem (simple tension/compression):
εx =σx
E
F
L L∆
Αx
εx =∆L
L=
∂L
∂x
σx =F
A= E εx
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Material law (2)
Hooke law for 3D problems:
εx =1
E[σx − ν (σy + σz)] , γyz =
τyz
2 G
εy =1
E[σy − ν (σx + σz)] , γxz =
τxz
2 G(11)
εz =1
E[σz − ν (σx + σy)] , γxy =
τxy
2 G
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Conclusion
15 unknowns:
3 displacements u
6 stresses ε
6 strains σ
15 available equations:
6 geometrical relations
6 matrial law equations 3 equilibrium conditions
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Compatibility conditions (1)
They define continuity of deformations – the continuous vo-
lume must remain to be continuous also after deformation.
They can be obtained from geometrical relations by elimination
of displacements.
∂2εx
∂y2+
∂2εy
∂x2+ =
∂2εxy
∂x∂y
∂2εy
∂z2+
∂2εz
∂y2+ =
∂2εyz
∂y∂z(12)
∂2εz
∂x2+
∂2εx
∂z2+ =
∂2εzx
∂z∂x16
Compatibility conditions (2)
∂2εx
∂y∂z=1
2
−∂γyz
∂x+
∂γxz
∂y+
∂γxy
∂z
∂2εy
∂z∂x=1
2
+∂γyz
∂x−
∂γxz
∂y+
∂γxy
∂z
(13)
∂2εz
∂x∂y=1
2
+∂γyz
∂x+
∂γxz
∂y−
∂γxy
∂z
For plane problems (2D, to be discussed later) it can be written:
∂2εx
∂y2+
∂2εy
∂x2=
∂2γxy
∂x∂y(14)
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Topic 2: Plane problems
2D plane structures supported and loaded in their middle plane
will be discussed.
Two main variants of plane problem:
• plane stress
• plane strain
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Plane strain
1
All deformations (strains) must
lie in the x− y plane:
ε = {εx, εy, γxy}T (15)
The structure can not freely move in z direction. As a result,
there is σz stress. σz:
σ = {σx, σy, σz, τxy}T (16)
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Plane strain
y
x
στ
x
yσ
xy
z
All stresses lie in the x−y plane:
σ = {σx, σy, τxy}T (17)
The wall can freely move in z direction, therefore there is on
σz stress but the εz strain is nonzero:
ε = {εx, εy, εz, γxy}T (18)
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Plane problem – conclusions (1)
Equilibrium equations
∂σx
∂x+
τxy
∂y+X = 0 (19)
∂τxy
∂x+
σy
∂y+ Y = 0
Geometrical relations
εx =∂u
∂x, εy =
∂v
∂y, γxy = γyx =
∂u
∂y+
∂v
∂x. (20)
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Plane problem – conclusions (2)
Material law: plane stress
σx =E
1− µ2(εx + µ εy)
σy =E
1− µ2(εy + µ εx) (21)
τx =E
2(1− µ)γxy
(22)
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Plane problem – conclusions (3)
Material law: plane strain
σx =E
(1 + µ)(1− 2µ)[(1− µ) εx + µ εy]
σy =E
(1 + µ)(1− 2µ)[µ εx + (1− µ) εy] (23)
τx =E
(1 + µ)(1− 2µ)γxy1
2(1− µ)
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Airy’s equation for plane problem
Airy function F – describes stress state of plane problem if:
σx =∂2F
∂x2, σy =
∂2F
∂x2, τxz = −
∂2F
∂x∂y. (24)
Airy equation – compatibility relation is (14) writtnen with use
of Airy function:
∂4F
∂x4+ 2
∂4F
∂x2∂y2+
∂4F
∂y4= 0 (25)
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Stress distribution on wall:N
S
N . . . beam theory,
S . . . plane stress theory.
25
Forces in wall
Nx = σx h [N
m]
Ny = σy h [N
m]
Nxy = τxy h [N
m]
x
yx
xyn
z
n
n
n
y
τxy
yx
h
τ
σ
σx
y
y
�����������������������������
�����������������������������
x
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� �� �� �� �� �� �� �� �
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