ELASTICITY AND PLASTICITY I Ing. Lenka Lausová LPH 407/1 tel. 59 732 1326 [email protected]http://fast10.vsb.cz/lausova 1 / 28 Literature: Higgeler – Statics and mechanics of material, USA 1992 Beer, Johnston, DeWolf, Mazurek – Mechanics of materials, USA 2009, Fifth Edition
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Higgeler – Statics and mechanics of material, USA 1992
Beer, Johnston, DeWolf, Mazurek – Mechanics of materials,
USA 2009, Fifth Edition
Elasticity and Plasticity
1.Basic principles of Elasticity and plasticity
2.Stress and Deformation of Bars in Axial load
2 / 28
Department of Structural Mechanics
Faculty of Civil Engineering, VSB - Technical University Ostrava
3.Design and assessment of structures
Basic principles of Elasticity and plasticity
Elasticity and plasticity in building engineering – studying the strenght of material, 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
3 / 28
Elasticity and plasticity new terms: 1) stress2) strain3) stability
Material
Elastic behavior of a material – Hooke´s law - Elasticity
Principal terms Principal terms Principal terms Principal terms
Stress
Strain
Stability
4 / 28
Elastic behavior of a material – Hooke´s law - Elasticity
Plastic behavior of a material - Plasticity
Load
External force load (F,M, q)
Temperature load
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,
5 / 28
5) static loading, 6) no initial state of stress
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
Basic principles of Elasticity and plasticity
6 / 28
2. Homogenity and isotropy3. Linear elasticity4. Small deformation5. Static loading6. No initial state of stress
(str. 4 učebnice)
Základní pojmy, výchozí předpoklady
1. Continuity of material
2. Homogeneity 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.
Basic principles of Elasticity and plasticity
7 / 28
steel) is not homogeneous material.
Isotropy means that material has got characteristics
undepended on the direction – (concret, steel – yes,
timber – not).
3. Linear elasticity4. Small deformation5. Static loading6. No initial state of stress (str. 4 učebnice)
Základní pojmy, výchozí předpoklady
1. Continuity of material2. 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 former (original) state. If there is valid direct
Basic principles of Elasticity and plasticity
8 / 28
into the former (original) state. If there is valid direct
proportion between stress and strain than we talk
about Hookes law = this is physical linearity.
4. Small deformation5. Static loading6. No initial state of stress
(str. 4 učebnice)
Základní pojmy, výchozí předpoklady
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.
(It is used at composit steel-concret structures).
Basic principles of Elasticity and plasticity
9 / 28
ε
σ
Ideally elastic-plastic material
Stress-strain diagrams
concrete
steel
fe … Elasticity limitfy ... Yield limitf ... Ultimate limit
Basic principles of Elasticity and plasticity
10 / 28
Plasticity: the ability of material to get permanent deformations without fracture
Ductility: plastic elongation of a broken bar (range /OT/), steel 15%).
fu ... Ultimate limit
Stress-strain diagram of an ideal
elastic-plastic material
+σ
fyYield limit
εelast. εplast.
Tension
Y A=C
F → N → σ
xl
∆l
Derivation for axial load σ - normal stress (axial)
ε – normal strain (axial)
Basic principles of Elasticity and plasticity
11 / 28
+ε = ∆l/l
fy
Elastic-plastic rangeElastic-plastic range
Tension
Compression
BσB
Y
Linear elastic
range
Exx .εσ =
arctg E = α
Tension
Hooke´s law - physical relations between stress and strain
Y
+σ =Ν/Α
fy
Yield limit
εelast.
Linear elastic
range
A
Nx =σ
l
lx
∆ε =
By substituting:
E.xx ε=σ Hooke´s law
Basic principles of Elasticity and plasticity
12 / 28
α = arctg Eσx ... Normal stress [Pa]
εx ... Axial strain [-]
E ... Young´s modulus of elasticity in tension and
compression [Pa]
EA
Nll =∆
ε
σ
ε = ∆l/l
E==ε
σαtan
Other version of Hooke´s law
Hooke´s law in shear
τxz ... Shear stress [Pa]
γxz ... Angle deformation
τxz
G==τ
αtan
Gxzxz γτ = (((( ))))υG
E++++==== 12
Basic principles of Elasticity and plasticity
13 / 28
α = arctg G
G ... modulus of elasticity in shear
[Pa]
γxz
G==γ
ταtan
1. Continuity of material2. Homogenity and isotropy 3. Linear elasticity
4. Small deformations theory: Changes of a shape of a (solid) structure are small
Basic principles of Elasticity and plasticity
14 / 28
Changes of a shape of a (solid) structure are small
with aspect to the its size (dimensions). Than we can
use a lot of mathematical simplifications, which usually
lead to linear dependency.
5. Static loading6. No initial state of stress
(str. 4 učebnice)
F
b
F
b
H H
δTheory I-st order
Theory of the II-nd order
δ << l
Theory of small deformations
δ ≈ l
Theory of large deformation (finite deformation)
Basic principles of Elasticity and plasticity
15 / 28
a
l
a
l
δ
May=H.l May=H.l+F.δ
Theory I-st orderTheory of the II-nd order
- Geometric nonlinearity
The equilibrium conditions we set on the
deformated construction (buckling of columns).
1. Continuity of material2. Homogenity and isotropy 3. Linear elasticity4. Small deformations theory
5. Static loading:
Basic principles of Elasticity and plasticity
16 / 28
5. Static loading:It means gradually growing of load (not dynamic
effects)
6. No initial state of stress
(str. 4 učebnice)
Základní pojmy, výchozí předpoklady
1. Continuity of material2. Homogenity and isotropy 3. Linear elasticity4. Small deformations theory 5. Static loading
Basic principles of Elasticity and plasticity
17 / 28
5. Static loading
6. No initial state of stress:In the initial state there are all stresses equal zerro.
Inner tension (from the production).
(str. 4 učebnice)
Základní pojmy, výchozí předpoklady
1. Continuity of material2. Homogenity and isotropy 3. Linear elasticity4. Small deformations theory 5. Static loading 6. No initial state of stress
Basic principles of Elasticity and plasticity
18 / 28
6. No initial state of stress
All these assumptions enable to use principal of
superposition which is based on linearity of all
mathematic relationship.
(str. 4 učebnice)
Základní pojmy, výchozí předpoklady
Saint - Venant principle of local effect
F
F area of failure
part without affect
F
Makes possible to replace a given load by a simpler one for the purpose of easier calculation the stresses in a member.
• the state of stress is influenced just in near
Jean Claude Saint-Venant
(1797-1886)
19 / 28
q
Near surroundings
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
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
.const=xσ .const≠xσ
q
N
20 / 28
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
Elasticity and Plasticity
Stress and Strain
21 / 28
Department of Structural Mechanics
Faculty of Civil Engineering, VSB - Technical University Ostrava
External forces, stress
M
Fr
∆
Nr
∆
Vr
∆
22 / 28
∆
A∆Nr
∆
Vr
∆
A∆
... Normal component of the vector Fr
∆... Tangential component of F
r∆
... Element of cross section area A
A
N
Ar
r
∆
∆=
→∆ 0limσ
A
V
Ar
r
∆
∆=
→∆ 0limτ
stress (intensity of the
internal forces distributed over a given section)
normal shear
Stress: vector, characterised by its components
Format of the unit:Unit: Pascal ... [Pa]
NPa =
Stress
23 / 28
2m
NPa =
22
6
mm
N
m
MNPa10MPa ===
2
3
m
kNPa10kPa ==
The Pascal is a small
quantity, in practise we use
multiplies of this unit
Basic (simple) types of mechanical stress
1. Axial loading 2. Bending 3. Torsion 4. Shear
Normal force N ≠ 0
24 / 28
ba
F
+
ba-
tension
compression
FRax
Rax
NN
NN
Bending moment My , Mz ≠ 0
F
M M
Basic (simple) types of mechanical loading
1. Axial loading 3. Torsion 4. Shear2. Bending
25 / 28
b
Rbz
a
Raz
M M
b
Rbz
a
Raz FMM
tension
compression
compression
-
+
tension
2 3
F1
F2
F3
Basic (simple) types of mechanical stress
2. Bending 3. Torsion 4. Shear1. Axial loading
26 / 28
Torsion moment Mx ≠ 0
+y
+z+x
1nv = 6
Shear force Vy , Vz ≠ 0
Basic (simple) types of mechanical stress
2. Bending 3. Torsion 4. Shear1. Axial loading
27 / 28
ba
VV
RbzRaz
F
VV-+
Type of the
loading
Internal force Stress
Axial Loading
(tension, compression)N σx - normal
Basic (simple) types of mechanical stress
28 / 28
(tension, compression)
Bending My, Mz σx- normal
Shear Vy, Vz τxy, τxz - shear
Torsion Mx τxy, τxz - shear
Basic Theorems of Statics
1) Principle of action and reaction
2) Principle of superposition3) Principle of proportionality
Theorems of superposition and proportionality
Issac Newton
(1642 - 1727)
29 / 28
3) Principle of proportionality
Types of loading
a) simple (axial loading, bending, torsion, shear)
b) combined
Combined loading:
• general bending (unsymmetric bending)
• eccentric axial loading
30 / 28
• eccentric axial loading
• torsion combined with tension or compression and
with bending
Due to 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.
Axial loading – tension, compression
The only one inner force in each cross-section is an axial force N.
0>N
0== zy VV 0== zy MM
… tensionNN
+
0=xM
31 / 28Basic principles and condition of solution
0<N … compression
ba
F
ba
Tension
compression
FRax
Rax
NN
+
-
Conditions of solution
a) deformated cross-sections stay on plane figure
and it is vertically to the axis (Bernoulli hypothesis)
N N
Character of condition is deformation-
geometrical. Cross sections stay mutually
paralel without tapering.
Outcome:
00 ==→== ττγγ
Daniel Bernoulli(1700 - 1782)
32 / 28
b) longitudinal fibres are not mutually compressed together
dx ∆dx
before and after deformation
00 ==→== xzxyxzxy ττγγ
.const=xσ … for x = const.
N N0== zy σσ
1. External load
Axis x = axis of a member
R
+N
x
F
l
σx
Constant
33 / 28
+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)
a) Normal stress under axial loading
b) Strain under axial loading - deformation
l
lx
∆ε =
Axial strain (changes in length of a member – relative deformations)
x
F
l ∆l - elongation
z
b
h
b´
h´y
(dimensionless quantity [-])
Deformations : elongation or contraction
34 / 28
Dimension changes:
l
xzy υεεε −==
– relative deformations)
Lateral strain
b´ = b+∆b
h´ = h+∆h
b
by
∆ε =
h
hz
∆ε =
l´= l + ∆l
50,≤≤≤≤ν
Poisson´s ratio
Circle - diameter d?
2) Temperature changes
If there is not defended the deformation of a member – doesn´t
come up normal force and stress, later on indeterminate members
+∆T
a) stress
b) Strain (thermal strain)
ba
35 / 28
TTzTyTxT ∆αεεε ===
Tα - Coefficient of thermal expansion [°C-1]
lTl T ..∆α∆ =
εxT = ∆l/l = ∆b/b = ∆h/h = ∆d/d
l´= l +
∆lb´ = b+∆b
h´ = h+∆h
b) Strain (thermal strain)
Examples (it is necessery to keep this rules)
� Construct the diagram of internal forces (N)
� Write general formula, express numerically (make sure to have the right units of quantities)
� Answer will be highlighted
36 / 28
� Answer will be highlighted
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).
N
RResults:
A = 490,87.10-6m2
37 / 28
�=
10
mP = 100 kN
+
NA = 490,87.10-6m2
σx = 203,718MPa
∆l = 0,0097m =9,7.10-3m = 9,7mm
εx = 9,7.10-4
l´= 10,0097m
εy = εz = -2,91.10-4
∆d = -7,28.10-6m = -7,28.10-3mm
d´= 0,02499m
Ocelová tyč kruhového průřezu d = 0,01 m a délky � = 2 m je
namáhána tahovou silou N = 15 kN.
Určete normálové napětí σx, celkové prodloužení ∆� a příčné
zkácení prutu.
E = 210 000 MPa, ν = 0,3
Example 2
R
Determine the total deformation of the rod in its length (see the picture).
38 / 28
N-
+
R
Intermediate data:
A1 = 314,159.10-6m2
A2 = 78,539.10-6m2
∆l1 = -1,364.10-3m = -1,364mm
∆l2 = 1,212.10-3m = 1,212mm
σ1 = -95,49MPa
σ2 = 127,33MPa
Example 3
R
°C-1
The rod is under the temperature load of -20°C. Determine the force F so that the total
deformation of the rod is 0.
39 / 28
+N
…derive
Determine the normal force N and the normal stress σx which
causes elongation 3mm on the steel rod of the circle sross section (diameter d = 0,05 m, length l = 3 m, E = 210 000 MPa).
Example 4
40 / 28
(N = 412,334kN, σx = 210,00MPa)
The concret column of a square cross-section 0,6 x 0,6 m and a 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 6
41 / 28
(h´= 3,6027m, a´= 0,6005m, b´= 0,6005m)
Determine the changes in dimensions and the stress in the steel and concrete part of the column (see the picture).
2) Determine Areq and minimal side of the square cross section areq from δall=1,70mm
3) Determine the real side areal (round up areq)
4) Count the real area A
5) Determine the maximal value of the real deformation ∆lreal
6) Make the assessment ∆lreal,2 < δall,2=1,70mm 22
222lim
AE
lNl k=∆≥δ
Questions for the exam– part 1
1. Elasticity and plasticity in building engineering.The initial presumptions of the clasic linear elasticity. The term of the plasticity, the small deformation theory, the theory of the II.order. Stress, state of the stresses of a member.
2. Relations between stresses and internal forces in a member, diferencial equilibrium conditions. The main types of stresses – simple and combined. Saint - Venant princip of local effect.
61 / 28
3. Deformations and displacement of a member, geometric equations, Hook´s law, linear elastic material, physical constants.
4. Stress-strain diagrams of building material, non-elastic and ideal elastic-plastic material, ductility.
5. Changes Temperature Deformations.
6. Axial Stress – tension, compression
7. Deformations of a member in tension or compression.8. Design and assessment of structures under axial loading