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Elasticity and Plasticity
1.Basic principles of Elasticity and plasticity
2.Stress and Deformation of Bars in Axial load
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Basic principles of Elasticity and plasticity
Elasticity and plasticity in building engineering – theoretical
basement for the theory of structures (important for steel,
concret, timber structures design) - to be able design safe
structures (to resist mechanical load, temperature load…)
Statics: external forces, internal forces
Elasticity and plasticity new terms:
1) stress
2) strain
3) stability
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Material
Elastic behavior of material – Hooke´s law - Elasticity
Plastic behavior of material - Plasticity
Load
External force load (F, M, q)
Temperature load
Principal terms
Stress
Strain
Stability
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External forces, internal forces, stress
M
F
N
V
AN
V
A
... Normal component of the vector F
... Tangential component of F
... Element of cross section area A
A
N
A
0lim
A
V
A
0lim
stress (intensity of the
internal forces distributed
over a given section)
normal shear
Stress and Strain
Elasticity and plasticity - new terms stress and strain
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Basic principles of Elasticity and plasticity
The initial presumptions of the clasic linear elasticity:
1) continuity of material,
2) homogenity(just one material) and isotropy (properties are the
same in all directions),
3) linear elasticity (valid Hook´s law),
4) the small deformation theory,
5) static loading,
6) no initial state of stress
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1. Continuity of material:
A solid is a continuum, it has got its volume without any holes,
gaps or any interruptions. Stress and strain is a continuous
function.
2. Homogenity and isotropy
3. Linear elasticity
4. Small deformation
5. Static loading
6. No initial state of stress
Basic principles of Elasticity and plasticity
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1. Continuity of material
2. Homogenity a isotropy:
Homogeneous material has got physical characteristics
identical in all places (concret, steel, timber).
Combination of two or more materials ( concret + steel) is not
homogeneous material.
Isotropy means that material has got characteristics
undependent on the direction – (concret, steel – yes,
timber – not).
3. Linear elasticity
4. Small deformation
5. Static loading
6. No initial state of stress
Basic principles of Elasticity and plasticity
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1. Continuity of material
2. Homogenity and isotropy
3. Linear elasticity:
Elasticity is an ability of material to get back after removing the
couses of changes (for example load) into the original state. If
there is a direct relation between stress and strain than we talk
about Hooke´s law = this is called physical linearity.
4. Small deformation
5. Static loading
6. No initial state of stress
Basic principles of Elasticity and plasticity
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Plasticity (on the contrary from linearity): This is an ability of
material to deform without any rupture by non-returnable way.
After removing the load there are staying permanent
deformations.
e
Ideally elastic-plastic material
Basic principles of Elasticity and plasticity
Plastic rangePlastic range
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Stress-strain diagram of an ideal
elastic-plastic material:
+e ∆l/l
+
fy
fy
Yield limit
εelast. εplast.
Plastic rangePlastic range
Tension
Compression
Y A=C
BB
Y
F → N →
xl
Δl
For axial load
- normal stress
e –strain
Linear elastic
range
Exx .e
arctg E = a
Basic principles of Elasticity and plasticity
εelast.εplast
εplast
dx dx
x
xx
d
de
1 e
Small-deformation theory
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Tension
a = arctg E x ... Normal stress [Pa]
ex ... Axial strain [-]
E ... Young´s modulus of elasticity in
tension and compression [Pa]
Hooke´s law - physical relations between stress and strain
it is valid only in the linear elastic range of the stress-stran diagram
Y
+ N/A
fy
Yield limit
εelast.
Linear elastic
range
A
Nx
EA
Nll
l
lx
e
By substituting:
E.xx e
ε
e = ∆l/l
Ee
atan
Hooke´s law
Other version of
Hooke´s law
Basic principles of Elasticity and plasticity
Hooke´s law,
another expression
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Hooke´s law in shear
a = arctg G
xz ... Shear stress [Pa]
xz ... Angle deformation
G ... modulus of elasticity in shear [Pa]
xz
xz
G
atan
Gxzxz
Basic principles of Elasticity and plasticity
zzxxz
d
3
3
dz
Small-deformation theory: 1
Simplyfying: tan
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Physical constants – E, G, n
In case of isotropic materila E, G are not mutual indipendent.
5,00 + 1.2G
E
23
EG
E
Benchmark values of physical constants of some matherials:
E G
Steel 210 000 MPa 81 000 MPa 0,3
Glass 70 000 MPa 28 000 MPa 0,25
Granite 12 000 to 50 000 MPa - 0,2
SoftwoodE|| = 10 000 MPa
E^ = 300 MPa600 MPa -
Physical Relations between Strains and Deformations
Poisson´s ratio
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Stress-strain diagrams
concrete
steel
Plasticity: the ability of material to get permanent deformations without fracture
Ductility: plastic elongation of a broken bar (range /OT/), steel 15%).
fe … Elasticity limit
fy ... Yield limit
fu ... Ultimate limit
Basic principles of Elasticity and plasticity
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1. Continuity of material
2. Homogenity and isotropy
3. Linear elasticity
4. Small deformations theory:
Changes of a shape of a (solid) structure are small
with aspect to its size (dimensions). Than we can use
a lot of mathematical simplifications, which usually
lead to linear dependency.
5. Static loading
6. No initial state of stress
Basic principles of Elasticity and plasticity
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a
F
l
b
a
F
l
b
H H
d
May=H.l May=H.l+F.d
Theory of the
I-st order
Theory of the II-nd order
- Geometric nonlinearity
d << l
Theory of small deformations
d l
Theory of large deformation
(finite deformation)
Basic principles of Elasticity and plasticity
The equilibrium conditions we set on the
deformated construction (buckling of columns).
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1. Continuity of material
2. Homogenity and isotropy
3. Linear elasticity
4. Small deformations theory
5. Static loading:
It means gradually growing of load (not dynamic
effects)
6. No initial state of stress
Basic principles of Elasticity and plasticity
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1. Continuity of material
2. Homogenity and isotropy
3. Linear elasticity
4. Small deformations theory
5. Static loading
6. No initial state of stress:
In the initial state there are all stresses equal zerro.
(Inner tension e.g. from the production).
Basic principles of Elasticity and plasticity
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1. Continuity of material
2. Homogenity and isotropy
3. Linear elasticity
4. Small deformations theory
5. Static loading
6. No initial state of stress
All these assumptions enable to use principal of superposition
which is based on linearity of all mathematic relationship.
Basic principles of Elasticity and plasticity
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Saint - Venant principle of local effect
F
F area of failure
part without affect
F
q
Near surroundings
Makes possible to replace a given load by
a simpler one for the purpose of an easier
calculation of stresses in a member.
• the state of stress is influenced just in near
surroundings of the load
• farther from this load we have nearly
uniform distribution of stress
Used for:
replacement the surface load by the load
statically equivalent but simpler for solution
Jean Claude Saint-Venant
(1797-1886)
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Saint - Venant principle of local is not valid in
these cases:a) concentrated loads on the end of bar:
F
F area of failure
part without affect
part without affect
area of failure
.constx .constx
q
N
b) bars with variable cross-section area: deduced conditions are valid for
bars with gradual changes of cross-section area. Abrupt changes (announce
by holes, nicks or narrowing) lead to no validity of condition.
q
q
d
b
q
q
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Basic Theorems of Statics
1) Principle of action and reaction
2) Principle of superposition
3) Principle of proportionality
Theorems of superposition and proportionality
Issac Newton
(1642 - 1727) These theorems are valid when the basic principles are kept.
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External forces, internal forces, stress
M
F
N
V
AN
V
A
... Normal component of the vector F
... Tangential component of F
... Element of cross section area A
A
N
A
0lim
A
V
A
0lim
stress (intensity of the
internal forces distributed
over a given section)
normal shear
Stress and Strain
Elasticity and plasticity - new terms stress and strain
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Stress: vector, characterised by its components
Format of the unit:
Unit: Pascal ... [Pa]
2m
NPa
22
6
mm
N
m
MNPa10MPa
Stress
2
3
m
kNPa10kPa
The Pascal is a small
quantity, in practise we use
multiplies of this unit
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Basic (simple) types of mechanical stress
1. Axial loading 2. Bending 3. Torsion 4. Shear
ba
F
+
Normal force N 0
ba-
tension
compression
FRax
Rax
NN
NN
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Bending moment My , Mz 0
b
Rbz
a
Raz
FM M
b
Rbz
a
Raz FMM
tension
compression
compression
-
+
Basic (simple) types of mechanical stress
1. Axial loading 3. Torsion 4. Shear2. Bending
tension
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Torsion moment Mx 0
+y
+z+x
1
2 3
F1
F2
F3
nv = 6
Basic (simple) types of mechanical stress
2. Bending 3. Torsion 4. Shear1. Axial loading
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Shear force Vy , Vz 0
ba
VV
RbzRaz
F
VV-+
Basic (simple) types of mechanical stress
2. Bending 3. Torsion 4. Shear1. Axial loading
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Type of the
loading
Internal force Stress
Axial Loading
(tension, compression)N x - normal
Bending My, Mz x- normal
(bending stress)
Shear Vy, Vz xy, xz - shear
Torsion Mx xy, xz - shear
Basic (simple) types of mechanical stress
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Types of loading
a) simple (axial loading, bending, torsion, shear)
b) combined
Combined loading:
• general bending (unsymmetric bending)
• eccentric axial loading
• torsion combined with tension or compression and with
bending
Due to the Principle of superposition, which is valid in a linear
elastic range, we can solve the combined stresses. First by
spliting up to basic stresses and then we can add these results
together.
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Axial loading – tension, compression
Basic principles and condition of solution
The only one inner force in each cross-section is an axial force N.
0N
0N
0 zy VV 0 zy MM
… tension
… compression
ba
F
ba
Tension
compression
FRax
Rax
NN
NN
+
-
0xM
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Conditions of solution
a) deformated cross-sections stay on plane figure and it is
vertically to the axis (Bernoulli hypothesis)
b) longitudinal fibres are not mutually compressed together
dx dx
N N
Character of condition is deformation-
geometrical. Cross sections stay mutually
paralel without tapering.
before and after
deformation
Outcome:
00 xzxyxzxy
.constx … for x = const.
N N0 zy
Daniel Bernoulli(1700 - 1782)
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Stress and strain in external load
Axis x = axis of a member
R
+N
axial force F → normal forces N → normal stress σx
- Intensity of internal forces distributed
over a given section[MPa]
-Tensile stress - positive sign
compressive stress – negative sign
x
F
l
σx
Constant
a) Stress under axial loading
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b) Strain under axial loading - deformation
Dimension´s changes:
l
lx
e
xzy eee
Axial strain
Lateral strain
b´ = b+∆b
h´ = h+∆h
b
by
e
h
hz
e
l´= l + ∆l
50,ν Poisson´s ratio
Circle - diameter d?
x
F
l Δl - elongation
z
b
h
b´
h´y
(dimensionless quantity [-])
Deformations : elongation or contraction
Stress and strain in external load
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TTzTyTxT aeee
Ta - Coefficient of thermal expansion [°C-1]
lTl T ..a
If there is not defended the deformation of a member – doesn´t
come up normal force and stress, later on indeterminate members
+ΔT
εxT = Δl/l = Δb/b = Δh/h = Δd/d
l´= l +
∆lb´ = b+∆b
h´ = h+∆h
a) stress
b) Strain (thermal strain)
ba
Stress and strain in temperature changes
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Example 1
The steel rod (see the picture) has a circle cross-sectional area of a diameter
d = 0,025 m. E=2,1.105MPa. ν =0,3
Determine σx, elongation of the rod, the lateral changes ( in dimensions) and
determine new dimensions of the rod). (Ignore the dead weight).
l=
l0
m
P = 100 kN
+
N
RResults:
A = 490,87.10-6m2
σx = 203,718MPa
Δl = 0,0097m =9,7.10-3m = 9,7mm
εx = 9,7.10-4
l´= 10,0097m
εy = εz = -2,91.10-4
Δd = -7,28.10-6m = -7,28.10-3mm
d´= 0,02499m
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Ocelová tyč kruhového průřezu d = 0,01 m a délky l = 2 m je
namáhána tahovou silou N = 15 kN.
Určete normálové napětí x, celkové prodloužení l a příčné
zkácení prutu.
E = 210 000 MPa, n = 0,3
Example 2
N
-
+
R
Results:
A1 = 314,159.10-6m2
A2 = 78,539.10-6m2
Δl1 = -1,364.10-3m = -1,364mm
Δl2 = 1,212.10-3m = 1,212mm
σ1 = -95,49MPa
σ2 = 127,33MPa
Determine the total deformation of the rod in its length (see the picture).
S Fix = 0: R - F1 + F2 = 0
R = F1 - F2 = 40 – 10 = 30 kN
N1 = -R = -30kN
N2 = -R + F1 = -30+40 = 10 kN
Result: l =
-30
+10
N1
N2
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The concret column of a square cross-section of a side a =0,6m and the hight h =
3,6 m is uniformly warmed by T = 75°C. Determine the changes in dimensions of
the column - cube. αT = 10 ·10-6 °C-1
Example 3
Results: h´= 3,6027m, a´= 0,6005m
h = 3,6m
x
a = 0,6 m
a = 0,6 m
y
z
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Determine the stress in the circle rod and determine the change of the lenght of the rod.
F =20kN, ∆t =15°C, d=0.02m, l =1,5m, l1 = 1 m, l2 = 2 m, E = 210GPa, αT= 0.000012°C-1
1- N force from F
2- normal stress from N
3- elongation of the rod (from N + from temperature)
Solution:
Example 4
l1
l =1,5
l2F
∆t
Results: N=30kN, σx = 95,5MPa (tension), ∆lN =0,682mm, ∆lT =0,27mm