Basic principles of Elasticity and plasticity Design and...

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Department of Structural Mechanics

Faculty of Civil Engineering, VSB - Technical University Ostrava

Elasticity and Plasticity

Basic principles of Elasticity and plasticity

Design and assessment of structures

Stress and Deformation of Bars in Tension

(Simple Pressure)

• Statically determinated examples

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Elasticity and plasticity in building engineering – studying thestrenght of material, theoretical basement for the theory ofstructures (important for steel, concret, timber structuresdesign) - to be able design safe structures (to resistmechanical load, 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, 6) no initial state of stress

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1. Continuity of material: A solid is a continuum, it has got its volume withoutany 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. 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.

Isotropy means that material has got characteristicsundepended 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 talkabout Hookes law = this is physical linearity.

4. Small deformation5. Static loading6. No initial state of stress

(str. 4 učebnice)



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Plasticity (on the contrary from linearity): This is anability of material to deform without any rupture by

non-returnable (malleable - tvárný) way. Afterremoving the load there are staying permanent

deformations.

(It is used at composit steel-concret structures).

ε

σ

Ideally elastic-plastic material


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1. Continuity of material2. Homogenity and isotropy3. Linear elasticity

4. Small deformations theory: Changes of a shape of a (solid) structure are smallwith aspect to the its size (dimensions). Than we can

use a lot of mathematical simplifications, which usuallylead 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 I-st orderTheory 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).

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1. Continuity of material2. Homogenity and isotropy3. Linear elasticity4. Small deformations theory

5. Static loading:It means gradually growing of load (not dynamiceffects)

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 (from the production).

(str. 4 učebnice)



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1. Continuity of material2. Homogenity and isotropy3. Linear elasticity4. Small deformations theory5. Static loading6. No initial state of stress

All these assumptions enable to use principal ofsuperposition which is based on linearity of all

mathematic relationship.

(str. 4 učebnice)



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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).

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Characteristic and design load

Fk characteristic value of load

EU Czechrepublic

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

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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

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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

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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)

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Rx,aF3 F2 F1

3213 FFFN −−−=

212 FFN −−=

11 FN −=

l3 l2 l1

x

A

Nx =σ

∑=∆ii

ii

AE

lNl

-Rx,a

Tension and compression – basics of asssessment

Normal stress [Pa]

σ = const.

Deformation [m]

N

Cross sectional characteristic fortension 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:

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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

=

=

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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 =≥=⋅=

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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

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α

∑ = 0Fix ∑ = 0Fiz ∑ = 0Mib

α= cos/RR axa

( )( ) axd

dax RllF

llqlRRaz ⇒=+−

+−+ 0.

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2

21

l1 = 4m

l = 3m

l2 = 2m

a

b

F

q

Ra

Raz

Rax N

Rbx

RbzN Rax/Ra = cosα

Raz/Ra = sinα

Results: NEd=129,9kN,

Make the assessment after ULS of the rod from material steel of fy = 275 MPa, γM =1,5 of circle

cross section (see the picture). d= 28mm, Fk = 12kN (γq =1,5), q=8,7kN/m(γg=1,3).

Example 3 - ULS

aRN =

a) reactions

b) N force

EdydRd NfAN ≥⋅=

….the rod is satisfying

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l1 = 4m

l = 3m

l2 = 2m

a

b

F

q

σx = N/A

N

Example 3 - ULS

Diagram of N forces along the rod

Behavior of the stress in the section

- Normal stress is constant

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Determine the maximum value from the stresses of this bar. Square cross section of the side a.

Determine the elongation of the part 2 (middle part). Calculate in given characteristic values.

a=30mm

F1k=50kN

F2k=25kN

l1=l2=l3=0.25m

E=210GPa

Solving:

1. diagram of N forces 4. N2

2. Nmax

5.

3. stress σmax

AA,AE

lNl 2

22

222 ==∆

l1 l2 l3

F1 F2 F2 F1

Example 4

Results: Nmax,k = 50kN, σmax= 55,55MPa (tension), N2k =25kN, ∆l2 =0,033mm

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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

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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

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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 assessmentEdydrealRd NfAN ≥⋅=

EdydRd NfAN ≥⋅=

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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 ≥⋅=

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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 ≥⋅=


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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 ≥⋅=


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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



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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


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 andcombined. 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. Changes Temperature Deformations.

6. Axial Stress – tension, compression

7. Deformations of a member in tension or compression.

Basic principles of Elasticity and plasticity Design and...

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