ELASTICITY AND PLASTICITY I Ing. Lenka Lausová ... - …fast10.vsb.cz/lausova/ER_01_2013.pdf ·...

ELASTICITY AND PLASTICITY I

Ing. Lenka Lausová

LPH 407/1

tel. 59 732 1326

[email protected]

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


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


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.


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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)


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


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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)


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).


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ε

σ

Ideally elastic-plastic material

Stress-strain diagrams

concrete

steel

fe … Elasticity limitfy ... Yield limitf ... Ultimate limit


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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)


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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


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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


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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


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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)


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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:


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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)


1. Continuity of material2. Homogenity and isotropy 3. Linear elasticity4. Small deformations theory 5. Static loading


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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)


1. Continuity of material2. Homogenity and isotropy 3. Linear elasticity4. Small deformations theory 5. Static loading 6. No initial state of stress


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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)


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


Stress and Strain

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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


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


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


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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

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+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

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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).

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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)


Design and assessment

of structures

43 / 28



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).

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� 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:

49 / 28

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


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 ≥⋅=


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


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


2) Determine Areq and minimal diameter of the circle cross section dreq

3) Determine the real dreal (round up dreq)


5) Make the assessment EdydrealRd NfAN ≥⋅=

EdydrealRd NfAN ≥⋅=


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


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 ≥⋅=



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

58 / 28


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


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 ≥⋅=



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


59 / 28


2) Determine Areq and minimal side of the square cross section areq




Results:

NEd=-400kN Areq=2,0.10-3m2=2,0.103mm2 areq=20mm areal=20mm

Areal=2,0.103mm2



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



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

ELASTICITY AND PLASTICITY I Ing. Lenka Lausová ... - …fast10.vsb.cz/lausova/ER_01_2013.pdf ·...

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