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ELASTICITY AND PLASTICITY I
Ing. Lenka Lausová
LPH 407/1
tel. 59 732 1326
http://fast10.vsb.cz/lausova
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Literature:
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
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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
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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
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ε
σ
Ideally elastic-plastic material
Stress-strain diagrams
concrete
steel
fe … Elasticity limitfy ... Yield limitf ... Ultimate limit
Basic principles of Elasticity and plasticity
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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
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+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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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Department of Structural Mechanics
Faculty of Civil Engineering, VSB - Technical University Ostrava
External forces, stress
M
Fr
∆
Nr
∆
Vr
∆
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∆
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
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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
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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
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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
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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
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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
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(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)
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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
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• 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)
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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
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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
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� 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
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�=
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.
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+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).
Example 7
1
h1
=0
,5m
P1=150kN
P2=165kNP2
d1 = 0,03mE1 = 210 000MPaν1 = 0,3
N
-
ocel
42 / 28
2
h1
=0
,5m
h2
=0
,2m
P2=165kN
a = 0,15m
b =
0,1
m
E2=33 500MPaν2 = 0,2
-
beton
(d´= 30,009mm, a´= 150,029mm, b´= 100,019mm, h1´=499,495mm, h2´=199,809mm, h´= 699,304mmσ1= -212,21MPa, σ2= -32,0MPa)
Elasticity and Plasticity
Design and assessment
of structures
43 / 28
Department of Structural Mechanics
Faculty of Civil Engineering, VSB - Technical University Ostrava
Design and assessment of structuresLimit state design requires the structure to satisfy in two principal criterias: the
ultimate limit state (ULS) and the serviceability limit state (SLS)
In Europe, the Limit State Design is enforced by the Eurocodes.
The ultimate limit state is reached when the applied stresses actually exceed the
allowable strength of the structure or structural elements – it causes to fail or
collapse of the structure. We use magnification factor to get higher the load (=
design load) and reduction factors to get lower strength of the structure ( = design
strenght).
44 / 28
� ultimate limit state – we compare carrying capacity and design inner force
� serviceability limit state – we compare deformation and limit given deformation
The serviceability limit state is the point where a structure can no longer
be used for it's intended purpose (but it would still be structurally all
right). The tolerances for serviceability depend on the intended use of the
structure ( large deformations).
Characteristic and design load
Fk characteristic value of load
EU Czech
γ.kd FF = 1≥γ
Coefficients of reliability – for load: permanent load γG , changeable load γQ
Fd design value of load
45 / 28
EU Czech republic
γG1,35 1,1 (1,35)
γQ1,50 1,50
Design and assessment of tensile beams
� ultimate limit state
� serviceability limit state
EdRd NN ≥
maxall ll ∆≥∆
NEd inner normal force in design value (from Fd)
46 / 28
NRd carrying capacity
∆l – deformation in axial load = alongation or contraction
– load is given in characteristic values
∆lmax = ∆llim - limit given deformation
Strenght of material
M
kd
ff
γ=
γM ... Coefficients of reliability for material
Strenght of material f = resistace of structure R (carrying
capacity) on the contrary of load get lower:
47 / 28
Strenght of material: characteristic = fk and design fd = design
γM ... Coefficients of reliability for material
yk ff =
1≥γ
fy ... Stress at yield limit (from stress- strain diagram)fu ... Stress at ultimate limit
Design and reliability assessment of bar exposed to axial forces
Design of carrying structure
Dimensioning
dreqEd fAN ,,
Adjusted design
d
Edreq
f
NA =
48 / 28
Reliability assessment of design(Limit state of carrying capacity)
Realization
dRdEd fANN .=≤ M
kd
ff
γ=
1≤Rd
Ed
N
N fd = fyd
For steel:
(yield limit)
Rx,aF3 F2 F1
11 FN −=
l3 l2 l1
x
Tension and compression – basics of asssessment
N
Cross sectional characteristic for tension and compression is area
Assessment of members in stress:
M
yk
yd
ff
γ= … Yield limit
From the last lesson:
EdydrealRd NfAN ≥⋅=ULS:
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3213 FFFN −−−=
212 FFN −−=
11 FN −=
A
Nx =σ
∑=∆ii
ii
AE
lNl
-Rx
,a
Normal stress [Pa]
σ = const.
Deformation [m]
A
Nf Ed
allowableyd =≥= maxσσ
Assessment of members in deformations :
∑=∆≥ii
iikallowable
AE
lNlδ
yd
Edreq
f
NA =⇒
...=⇒ reqA
SLS:
Determine stress in both rods (profile I80 – area from tables of one I80:
A=0,757.103mm2). Make the assessment after ULS: fyk = 150MPa,
γM=1,00, γG=1,1.
a
b
FN1
γDetermination N – Two Equilibrium Conditions in the hinge a:
3846.0
ba
bsin
22=
+=γ 9231.0
ba
acos
22=
+=γ
0sin:0 =+−=∑ dFNF γ
b
a
Example 1 – ULS (ultimate limit state)
Conditions of solution:
Rb
50 / 28
a = 12 m,
b = 5 m,
Fk = 39,3 kN,
Fd=43,23kN
FN1
N2
0cos:0
0sin:0
21
1
=−=
=+−=
∑
∑
ddz
ddx
NNF
FNF
γ
γ
kNN
kNN
d
d
68,103
32,112
2
1
=
=
( If you count the reactions, then
c
a
Rb=N1, Rc=N2)
Conditions of solution:
1) Hinges in nodes
2) Load in nodes
Then in rods just N forces
Rc
kNN
kNN
k
k
3,94
1,102
2
1
=
=
FN1
σx = N/A
N1
Diagram of N forces along the rods
Example 1 - ULS
Distribution of the stress in the section
- Normal stress is constant
51 / 28
FN1
N2
N2
Results:
Normal stress σx1 =134,9 Mpa, σx2 =124,5 Mpa, kNNkNfAN EdydRd 32,112 55,113 =≥=⋅=
1) Determine stress in the rod (I80 –profile). The cross-sectional area -tables value -
A=0,757.103mm2 ).
2) SLS - Compare allowable (limit) elongation (∆lall = 10 mm) and real elongation.
a = 3 m, b = 1.5 m, l = 10 m, g =180kNm-1 , E = 210000 MPa
Ra
RN =
1- Stressá
Example 2 - SLS
N=σ
52 / 28
Results:
Normal stress σx =237,78 Mpa, ∆lreal = 11,28mm ≤ ∆lall =10,00mm
a b
gl
ab
Rbx=0
Rbz
aRN =
AE
lNl =∆
2- Deformation
A
Nx =σreactions
Determine the stress in the circle rod, draw its behaviour during the cross section and
determine the change of the lenght of the rod. Calculate in given characteristic values.
Fk =20kN ∆t=15°C
d=0.02m l=1.5m
l1 = 1 m l2 = 2 m
E=210GPa αT=0.000012°C-1
Example 5
l∆t
53 / 28
1- N force from F
2- normal stress from N
3- elongation of the rod (from N + from temperature)
Solving:
l1 l2F
Results: N=30kN, σx = 95,5MPa (tension), ∆lN =0,682mm, ∆lT =0,27mm
Steel bar of the circle cross section area – sida a =16mm. E=2,1.105 MPa.
- determine the elongations of the bar ∆l and compare with δlim= 5mm (assessment after SLS)
- determine normal stress in all parts of a bar.
Don´t forget to construct N forces. Calculate in given characteristic values.
b = 1,7 m P1 P2 P3
Example 6 - SLS
54 / 28
cb d
c = 1,1 m
d = 0,6 m
P1 = 20 kN
P2 = 10 kN
P3 = 20 kN
y
σx = N/A
Results:
∆l = 1,376 mm < δlim = 5 mm
σ1=117,19MPa, σ2=39,06MPa, σ3=78,13MPa – all of them are tensile
Make the assessment after ULS:
F1k=333,3kN, F2k=266,7kN, γQ=1,5, designe value, l1=l2=l3=2m, fyk=200MPa, γΜ = 1,0,
square cross section (A=a2)
l1 l2 l3
F1,d F2
Example 7 - ULS
1) Determine reactions, construct the N forces, determine NEd
55 / 28
Results: NEd=N2=-400kN
Areq=2,0.10-3m2=2,0.103mm2 areq=44,7mm areal = 45,0mm
Areal=2,025.103mm2
1) Determine reactions, construct the N forces, determine NEd
2) Determine Areq and minimal side of the square cross section areq
3) Determine the real side areal (round up areq)
4) Count the real area A
5) Make the assessment EdydrealRd NfAN ≥⋅=
EdydRd NfAN ≥⋅=
Make the assessment after ULS:
F1k=333,3kN, F2k=266,7kN, γQ=1,5, l1=l2=l3=2m, fyk=200MPa, γΜ = 1,0,
circle cross section (A=πd2/4)
l1 l2 l3
F1 F2
Example 8 - ULS
1) Determine reactions, construct the N forces, determine NEd
56 / 28
Results: NEd=N2=400kN
Areq=2,0.10-3m2=2,0.103mm2 dreq=50,463mm use dreal=51,0mm
Areal=2,043.103mm2
1) Determine reactions, construct the N forces, determine NEd
2) Determine Areq and minimal diameter of the circle cross section dreq
3) Determine the real dreal (round up dreq)
4) Count the real area A
5) Make the assessment EdydrealRd NfAN ≥⋅=
EdydrealRd NfAN ≥⋅=
Make the assessment after ULS:
F1k=333,3kN, F2k=266,7kN, γQ=1,5, l1=l2=l3=2m, fyk=200MPa, γΜ = 1,0,
U - cross section (use the tables)
l1 l2 l3
F1 F2
Example 9 - ULS
1) Determine reactions, construct the N forces, determine N
57 / 28
1) Determine reactions, construct the N forces, determine NEd
2) Determine Areq
3) Find in the tables of the U-sections the first section ot fhe higher value
4) Write down real area A and type of U-section
5) Make the assessment
Results:
NEd=400kN Areq=2,0.10-3m2=2,0.103mm2 use U140
Areal=2,04.103mm2EdydrealRd NfAN ≥⋅=
EdydrealRd NfAN ≥⋅=
Make the assessment after ULS:
F1k=333,3kN, F2k=266,7kN, γQ=1,5, l1=l2=l3=2m, fyk= 200MPa, γΜ = 1,0,
2U - cross section (use the tables)
l1 l2 l3
F1 F2
Example 10 - ULS
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1) Determine reactions, construct the N forces, determine NEd
2) Determine Areq (for two profiles U), Areq/2 will be for one profile U
3) Find in the tables of the U-sections the first section ot fhe higher value
4) Write down type of 2U-section and real area A of the 2U
5) Make the assessment
Results: Nmax=N2= - 400kN
Areq=2,0.10-3m2=2,0.103mm2 →Areq for 1U =1.103mm2 use 2U80
AU80=1,1.103mm2 →Areal=2,2.103mm2EdydrealRd NfAN ≥⋅=
EdydrealRd NfAN ≥⋅=
Make the assessment after ULS:
F1k=333,3kN, F2k=266,7kN, l1=l2=l3=2m, fyk=200MPa, γΜ = 1,0,
b=100mm, rectangular cross section (A=ab)
l1 l2 l3
F1 F2
Example 11 - ULS
1) Determine reactions, construct the N forces, determine NEd
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1) Determine reactions, construct the N forces, determine NEd
2) Determine Areq and minimal side of the square cross section areq
3) Determine the real side areal (round up areq)
4) Count the real area A
5) Make the assessment
Results:
NEd=-400kN Areq=2,0.10-3m2=2,0.103mm2 areq=20mm areal=20mm
Areal=2,0.103mm2
EdydrealRd NfAN ≥⋅=
EdydrealRd NfAN ≥⋅=
Make the assessment of the beam , if the limit deformation of the part two (on the lenght l2) is
δlim=1,70mm. F1k=333,3kN, F2k=266,7kN, γQ=1,5, l1=l2=l3=2m, E=210GPa, square cross section
(A=a2).
l1 l2 l3
F1 F2
Example 12 - SLS
1) Determine reactions, construct the N forces
2) Determine Areq and minimal side of the square cross section areq from δall=1,70mm
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Results:
N2k=-266,7kN Areq=1,49.10-3m2=1,49.103mm2 areq=38,7mm a=40,0mm
A=1600mm2=0,0016m2 ∆l2real= 1,587m ∆l2,real < ∆l2,lim
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.
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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