elasticity and plasticity i ing. lenka lausová, ph.d. lph

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ELASTICITY AND PLASTICITY I

Ing. Lenka Lausová, Ph.D.

LPH 407/1

tel. 59 732 1326

[email protected]

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

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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


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


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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 elasticity4. Small deformation5. Static loading6. No initial state of stress (str. 4 učebnice)



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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)



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


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


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


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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


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


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)


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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)


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)



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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)



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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)



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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 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)

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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

.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.

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Stress and Strain

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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

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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

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Torsion moment Mx ≠ 0

+y

+z+x

1

2 3

F1

F2

F3

nv = 6


2. Bending 3. Torsion 4. Shear1. Axial loading

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Shear force Vy , Vz ≠ 0

ba

VV

RbzRaz

F

VV-+


2. Bending 3. Torsion 4. Shear1. Axial loading

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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


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




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


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


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 ≥⋅=

55 / 59


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


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 ≥⋅=


56 / 59


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


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 ≥⋅=


57 / 59


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


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



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



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

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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

elasticity and plasticity i ing. lenka lausová, ph.d. lph

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