1 / 59
ELASTICITY AND PLASTICITY I
Ing. Lenka Lausová, Ph.D.
LPH 407/1
tel. 59 732 1326
http://fast10.vsb.cz/lausova
Literature:
Higgeler – Statics and mechanics of material, USA 1992
Beer, Johnston, DeWolf, Mazurek – Mechanics of materials,
USA 2009, Fifth Edition
2 / 59
Department of Structural Mechanics
Faculty of Civil Engineering, VSB - Technical University Ostrava
Elasticity and Plasticity
1.Basic principles of Elasticity and plasticity
2.Stress and Deformation of Bars in Axial load
3.Design and assessment of structures
3 / 59
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
Elasticity and plasticity new terms: 1) stress2) strain3) stability
4 / 59
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 Principal terms Principal terms Principal terms
Stress
Strain
Stability
5 / 59
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
6 / 59
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 isotropy3. Linear elasticity4. Small deformation5. Static loading6. No initial state of stress
(str. 4 učebnice)
Základní pojmy, výchozí předpoklady
Basic principles of Elasticity and plasticity
7 / 59
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 elasticity4. Small deformation5. Static loading6. No initial state of stress (str. 4 učebnice)
Základní pojmy, výchozí předpoklady
Basic principles of Elasticity and plasticity
8 / 59
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
proportion between stress and strain than we talk
about Hooke´s law = this is called physical linearity.
4. Small deformation5. Static loading6. No initial state of stress
(str. 4 učebnice)
Základní pojmy, výchozí předpoklady
Basic principles of Elasticity and plasticity
9 / 59
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.
ε
σ
Ideally elastic-plastic material
Basic principles of Elasticity and plasticity
Plastic rangePlastic range
10 / 59
Stress-strain diagram of an ideal
elastic-plastic material:
+ε = ∆l/l
+σ
fy
fy
Yield limit
εelast. εplast.
Plastic rangePlastic range
Tension
Compression
Y A=C
BσB
Y
F → N → σ
xl
∆l
For axial load σ - normal stress
ε – normal strain
Linear elastic
range
Exx .εσ =
arctg E = α
Basic principles of Elasticity and plasticity
εelast.
εelast.εplast
11 / 59
Tension
α = arctg Eσx ... Normal stress [Pa]
εx ... Axial strain [-]
E ... Young´s modulus of elasticity in tension and
compression [Pa]
Hooke´s law - physical relations between stress and strain
Y
+σ =Ν/Α
fy
Yield limit
εelast.
Linear elastic
range
A
Nx =σ
EA
Nll =∆
l
lx
∆ε =
By substituting:
E.xx ε=σ
ε
σ
ε = ∆l/l
E==ε
σαtan
Hooke´s law
Other version of Hooke´s law
Basic principles of Elasticity and plasticity
12 / 59
Hooke´s law in shear
α = arctg G
τxz ... Shear stress [Pa]
γxz ... Angle deformation
G ... modulus of elasticity in shear [Pa]
τxz
γxz
G==γ
ταtan
Gxzxz γτ = (((( ))))υG
E++++==== 12
Basic principles of Elasticity and plasticity
13 / 59
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 limitfy ... Yield limitfu ... Ultimate limit
Basic principles of Elasticity and plasticity
14 / 59
1. Continuity of material2. Homogenity and isotropy3. 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 loading6. No initial state of stress
(str. 4 učebnice)
Basic principles of Elasticity and plasticity
15 / 59
a
F
l
b
a
F
l
b
H H
δ
May=H.l May=H.l+F.δ
Theory of the
I-st order
Theory of the II-nd order
- Geometric nonlinearity
δ << l
Theory of small deformations
δ ≈ 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).
16 / 59
1. Continuity of material2. Homogenity and isotropy 3. Linear elasticity4. Small deformations theory
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
Basic principles of Elasticity and plasticity
17 / 59
1. Continuity of material2. Homogenity and isotropy3. Linear elasticity4. Small deformations theory5. 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).
(str. 4 učebnice)
Základní pojmy, výchozí předpoklady
Basic principles of Elasticity and plasticity
18 / 59
1. Continuity of material2. Homogenity and isotropy 3. Linear elasticity4. 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.
(str. 4 učebnice)
Základní pojmy, výchozí předpoklady
Basic principles of Elasticity and plasticity
19 / 59
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 easiercalculation of stresses in a member.
• the state of stress is influenced just in nearsurroundings of the load• farther from this load we have nearlyuniform distribution of stress
Used for:
replacement the surface load by the loadstatically equivalent but simpler for solution
Jean Claude Saint-Venant
(1797-1886)
20 / 59
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
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
21 / 59
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)
These theorems are valid when the basic principles are kept.
22 / 59
Department of Structural Mechanics
Faculty of Civil Engineering, VSB - Technical University Ostrava
Elasticity and Plasticity
Stress and Strain
23 / 59
External forces, stress
M
Fr
∆
Nr
∆
Vr
∆
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 distributedover a given section)
normal shear
24 / 59
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
25 / 59
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
26 / 59
Bending moment My , Mz ≠ 0
b
Rbz
a
Raz
F
M M
b
Rbz
a
Raz FMM
tension
compression
compression
-
+
Basic (simple) types of mechanical loading
1. Axial loading 3. Torsion 4. Shear2. Bending
tension
27 / 59
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
28 / 59
Shear force Vy , Vz ≠ 0
ba
VV
RbzRaz
F
VV-+
Basic (simple) types of mechanical stress
2. Bending 3. Torsion 4. Shear1. Axial loading
29 / 59
Type of the
loading
Internal force Stress
Axial Loading
(tension, compression)N σx - normal
Bending My, Mz σx- normal
Shear Vy, Vz τxy, τxz - shear
Torsion Mx τxy, τxz - shear
Basic (simple) types of mechanical stress
30 / 59
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.
31 / 59
Axial loading – tension, compression
Basic principles and condition of solution
The only one inner force in each cross-section is an axial force N.
0>N
0<N
0== zy VV 0== zy MM
… tension
… compression
ba
F
ba
Tension
compression
FRax
Rax
NN
NN
+
-
0=xM
32 / 59
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 ττγγ
.const=xσ … for x = const.
N N0== zy σσ
Daniel Bernoulli(1700 - 1782)
33 / 59
1. External load
Axis x = axis of a member
R
+N
axial force F → normal forces N → normal stress σx
(intensity of internal forces distributedover a given section)
[MPa](Tensile stress - positive sign
compressive stress – negative sign)
x
F
l
σx
Constant
a) Normal stress under axial loading
34 / 59
b) Strain under axial loading - deformation
Dimension changes:
l
lx
∆ε =
xzy υεεε −==
Axial strain(changes in length of a member– 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?
x
F
l ∆l - elongation
z
b
h
b´
h´y
(dimensionless quantity [-])
Deformations : elongation or contraction
35 / 59
2) Temperature changes
TTzTyTxT ∆αεεε ===
Tα - Coefficient of thermal expansion [ C-1]
lTl T ..∆α∆ =
If there is not defended the deformation of a member – doesn t
come up normal force and stress, later on indeterminate members
+∆T
εxT = ∆l/l = ∆b/b = ∆h/h = ∆d/d
l = l +
∆lb = b+∆b
h = h+∆h
a) stress
b) Strain (thermal strain)
ba
36 / 59
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
37 / 59
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).
�=
10
mP = 100 kN
+
N
RResults:
A = 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
38 / 59
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
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
Determine the total deformation of the rod in its length (see the picture).
Σ 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
39 / 59
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 3
(h´= 3,6027m, a´= 0,6005m, b´= 0,6005m)
0,60,6
h = 3,6m
x
40 / 59
Department of Structural Mechanics
Faculty of Civil Engineering, VSB - Technical University Ostrava
Elasticity and Plasticity
Design and assessment
of structures
41 / 59
Design and assessment of structures
� ultimate limit state – we compare carrying capacity and design inner force
� serviceability limit state – we compare deformation and limit given deformation
Limit 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).
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).
42 / 59
Characteristic and design load
Fk characteristic value of load
EU Czech republic
γG1,35 1,1 (1,35)
γQ1,50 1,50
γ.kd FF = 1≥γ
Coefficients of reliability –for load: permanent load γG , changeable load γQ
Fd design value of load
43 / 59
Design and assessment of tensile beams
� ultimate limit state
� serviceability limit state
EdRd NN ≥
maxall ll ∆≥∆
NRd carrying capacity
NEd inner normal force in design value (from Fd)
∆l – deformation in axial load = alongation or contraction
– load is given in characteristic values
∆lmax = ∆llim - limit given deformation
44 / 59
Strenght of material
Strenght of material: characteristic = fk and design fd = design
M
kd
ff
γ=
γM ... Coefficients of reliability for material
yk ff =
1≥γ
Strenght of material f = resistace of structure R (carrying
capacity) on the contrary of load get lower:
fy ... Stress at yield limit (from stress- strain diagram)fu ... Stress at ultimate limit
45 / 59
Design and reliability assessment of bar exposed to axial forces
Reliability assessment of design(Limit state of carrying capacity)
Design of carrying structure
Realization
Dimensioning
dreqEd fAN ,,
dRdEd fANN .=≤
Adjusted design
d
Edreq
f
NA =
M
kd
ff
γ=
1≤Rd
Ed
N
N fd = fyd
For steel:
(yield limit)
46 / 59
Rx,aF3 F2 F1
3213 FFFN −−−=
212 FFN −−=
11 FN −=
l3 l2 l1
x
A
Nx =σ
∑=∆ii
ii
AE
lNl
-Rx
,aTension and compression – basics of asssessment
Normal stress [Pa]
σ = const.
Deformation [m]
N
Cross sectional characteristic for tension and compression is area
Assessment of members in stress:
M
yk
yd
ff
γ=
A
Nf Ed
allowableyd =≥= maxσσ
Assessment of members in deformations :
∑=∆≥ii
iikallowable
AE
lNlδ
… Yield limit
yd
Edreq
f
NA =⇒
...=⇒ reqA
From the last lesson:
EdydrealRd NfAN ≥⋅=ULS:
SLS:
47 / 59
Make the design and assessment of the square section bar according to both limit
states. After your design of the cross-section calculate also normal stress in all parts of
the bar and draw its distribution. Determine the complete elongation of the bar.
F1=50kN fyk=235MPaF2=25kN ∆l2,max=0.035mm
l1=l2=l3=0.25m γG=1.35
E=210GPa γM=1.0
1- N-diagram 5- N2charakt=N2k 7- larger areq→areal(round up)
2- Nmax 8- Areal
3- NEd 6-
→areqII 9- both assessments:
4- AreqI→areq
I NRd>NEd
∆l2,max>∆l2,real
results.: NEd=67,5kN, areqI =16,9mm, areq
II =29,16mm, NRd=211,5kN, ∆l2 =0,033mm
AAAE
lNl k ==∆ 2
22
22 ,
Example – ULS + SLS
l1 l2 l3
F1 F2 F2 F1
Results:
48 / 59
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
a = 12 m,
b = 5 m,
Fk = 39,3 kN,
Fd=43,23kN
FN1
N2
γDetermination N – Two Equilibrium Conditions in the hinge a:
3846.0
ba
bsin
22=
+=γ 9231.0
ba
acos
22=
+=γ
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
b
c
a
Example 1 – ULS (ultimate limit state)
Rb=N1, Rc=N2)
Conditions of solution:
1) Hinges in nodes
2) Load in nodes
Then in rods just N forces
Rb
Rc
kNN
kNN
k
k
3,94
1,102
2
1
=
=
49 / 59
FN1
N2
σx = N/A
N2
N1
Diagram of N forces along the rods
Example 1 - ULS
Distribution of the stress in the section
- Normal stress is constant
Results:
Normal stress σx1 =134,9 Mpa, σx2 =124,5 Mpa, kNNkNfAN EdydRd 32,112 55,113 =≥=⋅=
50 / 59
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
Results:
Normal stress σx =237,78 Mpa, ∆lreal = 11,28mm ≤ ∆lall =10,00mm
a b
gl
ab
Ra
Rbx=0
Rbz
aRN =
AE
lNl =∆
1- Stress
2- Deformation
á
Example 2 - SLS
A
Nx =σreactions
51 / 59
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
1- N force from F
2- normal stress from N
3- elongation of the rod (from N + from temperature)
Solving:
Example 5
l1
l
l2F
∆t
Results: N=30kN, σx = 95,5MPa (tension), ∆lN =0,682mm, ∆lT =0,27mm
52 / 59
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.
cb d
b = 1,7 m
c = 1,1 m
d = 0,6 m
P1 = 20 kN
P2 = 10 kN
P3 = 20 kN
P1 P2 P3
y
σx = N/A
Example 6 - SLS
Results:
∆l = 1,376 mm < δlim = 5 mm
σ1=117,19MPa, σ2=39,06MPa, σ3=78,13MPa – all of them are tensile
53 / 59
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
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 ≥⋅=
54 / 59
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
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 ≥⋅=
55 / 59
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 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 ≥⋅=
56 / 59
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
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 ≥⋅=
57 / 59
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
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 ≥⋅=
58 / 59
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
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
1) Determine reactions, construct the N forces
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
59 / 59
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 ofthe II.order. Stress, state of the stresses of a member.
2. Relations between stresses and internal forces in a member, diferencialequilibrium conditions. The main types of stresses – simple and combined. Saint - Venant princip of local effect.
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 idealelastic-plastic material, ductility.
5. 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