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1/26 M.Chrzanowski: Strength of Materials SM2-01: Summary of SM1 SUMMARY OF SM1

# 1 /26 M.Chrzanowski: Strength of Materials SM2-01: Summary of SM1 SUMMARY OF SM1

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1/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

SUMMARY OF SM1

2/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

MwII

SwII

SwI

MwI

P3

Pi

O

n

OP1

Pn

P2

n

{wI}{wII}

{wII} ≡ {ZI}{wI} ≡ {ZII}

SwII ≡ SzI

MwII ≡ MzI

SwI ≡ - SwII

MwI ≡ - MwII

SwI ≡ SzII

MwI ≡ MzII

SzII

MzII

MzI

SzI

Internal forces

{wII} ≡ {ZI}{wI} ≡ {ZII}

3/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

Components of internal forces resultants Swx , Swy , Swz and Mwx , Mwy , Mwz are called cross-sectional forces

In 3D vectors of internal forces resultants have three components each

Sw{ Swx , Swy , Swz } Mw{ Mwx , Mwy , Mwz }

x

y

z

Swz

Sny

Swx

SwMwz

MwxMwy

Mw

Cross-sectional forces

4/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

In 2D number of cross-sectional forces is reduced, because loading and bars axes are in the same plane (x, z):

Sw{ Sx , 0, Sz }

Mw{ 0, My , 0 }

x

y

z

P q

.M

M

Sx

Sz

My

We will use following notations and names for these components:

Sx=N - axial forces

Sz=Q - shear force

My = M - bending moment

Cross-sectional forces

5/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

Special cases of internal forces reductions are called:

TENSION – when internal forces reduce to the sum vector only, which is parallel to the bar axis

SHEAR – when internal forces reduce to the sum vector only, which is perpendicular to the bar axis

BENDING – when internal forces reduce to the moment vector only, which is perpendicular to the bar axis

TORSION – when internal forces reduce to the moment vector only, which is parallel to the bar axis

M

Ms

Q

N

Cross-sectional forces in 2D bars

6/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

n

AP1

Pn

I

{wI}

{ZI}

Internal forces – state of stress

Neighbourhood of point A

The sum of all internal forces acting on ΔA

ΔA – area of point A neighbourhood

Δw

pAA A

w

0lim Stress vector at A

nA

7/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

Ar

Stress vector is a measure of internal forces intensity and depends on the chosen point and cross section nr

A

wAAA

p ,lim0

x2

x1

x3

n1

n3

n2

p2

p3

p1

σ11 σ12

σ13

σ21

σ22

σ23

σ31

σ32

σ33

p1[σ11 , σ12 , σ13 ]

p2[σ21 , σ22 , σ23 ]

p3[σ31 , σ32 , σ33 ]

Point A image

σ11 , σ12 , σ13

σ21 , σ22 , σ23

σ31 , σ32 , σ33σ31 , σ32 , σ33

Tσ(σij)

Stress vectors:

Stress matrix

i,j = 1,2,3Components ij of matrix T are called stresses.

Stress measure is [N/m2] i.e. [Pa]

Internal forces – state of stress

8/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

11~

x2

x1

x3

n2

n1

n3

ΔA1

ΔA3

ΔA2

31~

21~

ν(νi )

1~

ΔAν

)~(~ip

X1= 0A

10~~~~

1331221111 AAAA

0~~~~1331221111

11 AAii AA …, …

0lim

A

3312211111

jjii

jiij

jj 11

jiji

If we assume:

then:

Internal forces – state of stress

9/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

x2

x1

x3

n1

n3

n2

p2

p3

p1

σ11 σ12

σ13

σ21

σ22

σ23

σ31

σ32

σ33

With given stress matrix in the chosen point and given loading one can find 3 perpendicular planes such that stress vectors (principal stresses) have only normal components (no shear components).The coordinate system defined by the directions of principal stresses is called system of principal axis.

)3( )2(

)1(

3

2

1

2

1

3

3

2

1

00

00

00

332331

232221

131211

Internal forces – state of stress

10/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

It can be proved that are extreme values of normal stresses (stresses on a main diagonal of stress matrix) . Customary, these values are ordered as follows

321 ,,

321

12

3

Surface of an ellipsoid with semi-axis (equatorial radii) equal to the values of principal stresses represents all possible stress vectors in the chosen point and under given loading.

Internal forces – state of stress

11/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

232

2,2,0 32 p

232

0,2,2 21 p

221

2,0,2 31 p

231

231

max

1

221

Mohr circles – represent 3D state of stress in a given point –

on the plane of normal and shear stresses

2

232

232

221

231

221

231

30

12/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

0

j

iji x

P

On the body surface stress vector has to be balanced by the traction vector

q

p

jijiiq

Stress on the body surface

Coordinates of vector normal to the surface

jijiq

This equation states statics boundary conditions to comply with the solution of the equation:

This equation (Navier equation) reflects internal equilibrium and has to be fulfilled in any point of the body (structure).

Stress distribution

13/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

0

j

iji x

P

We have to deal with the set of 3 linear partial differential equations.

Navier equation

0

0

0

3

33

2

32

1

313

3

23

2

22

1

212

3

13

2

12

1

111

xxxP

xxxP

xxxP

There are 6 unknown functions which have to fulfil static boundary conditions (SBC):

jijiq

We need more equations to determine all 6 functions of stress distribution. To attain it we have to consider deformation of the body.

Stress distribution

14/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

A

A’

'rur

'' BAAB

Displacement vector

AA’

'r

u

r

B

B’

rru

'

0'', BAAB

iuu

Au Bu

321 ,, xxxuu ii

State of deformation - strains

15/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

i

j

j

iij x

u

x

u

2

1 Small strains

Shear strains

when ij

Normal strains

when i=j

322

3

3

223 2

1

x

u

x

u

2

222 x

u

3

333 x

u

1

1

1

1

1

111 2

1

x

u

x

u

x

u

211

2

2

112 2

1

x

u

x

u

311

3

3

113 2

1

x

u

x

u

333231

232221

131211

T

Strain matrix – symmetrical by definition of

angular strains

1

j

i

x

u

State of deformation - strains

16/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

x1

x2

x3

State of deformation - strains

17/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

333231

232221

131211

TEigenvalues of strain matrix are normal strains on the planes where there are no shear strains.

Principal strains can be found by solving the secular equation:

3

2

1

00

00

00

T

0322

13 III

where I1, I2, I3 are invariants of strain matrix

When transfer of strain matrix is made to the new co-ordinate system then matrix transformation rule holds:

kljlikij ,

In the co-ordinates defined by principal directions of strain matrix it takes the diagonal form.

321 - principal strains

State of deformation - strains

18/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

Strain versus stressMaterial

deformability propertiesGeneral property of majority of solids is

elasticity (instantaneous shape memory)

P

u

Linear elasticity

Deformation versus internal forces

CEIIINOSSSTTUVUT TENSIO SIC VIS

ukP

which reads: „as much the extension as the force is”

where k is a constant dependent on a material and body shape

Constitutive equation - Hooke law

19/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

Physical quantities (measurable)

Mathematical quantities (non-measurable)

Dynamics Kinematics

P

u

ij

ijij f

Hooke, 1678

Navier, 1822ij

f - linear function of all strain matrix components defining all stress component matrix

To make Hooke’s law independent of a body shape one has to use state variables characterizing internal forces and deformations in a material point i.e. stress and strain.

Constitutive equation - Hooke law

20/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

The coefficients of this equation do depend only on the material considered, but not on the body shape.

As Hooke equation is a set of 9 linear algebraic equations then the number of coefficient in this set is 81 and can be represented as a matrix of 34=81 components:

klijklij C

Summation over kl indices reflects linear character of this constitutive equation.

Constitutive equation - Hooke law

21/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

ijkkijij G 2 ,G

Summation obeys ! Kronecker delta

Lamé constants [Pa]

)(2 3322111111 G

)(2 3322112222 G

)(2 3322113333 G

1212 2 G

2323 2 G

3131 2 G

Normal stress and normal strain dependences

Shear stress and shear strain dependences

This equation consists of two groups:

Constitutive equation - Hooke law

22/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

E

Summation obeys ! Kronecker’s delta

Poisson modulus [0]

E1212 1

Young modulus [Pa]

E)( 33112222

E)( 33221111

E)( 22113333

E2323 1 E3131 1

ijkkijij vE

11

3131 2 G

12EG

G23131

Normal stress and normal strain dependences

Shear stress and shear strain dependences

Constitutive equation - Hooke law

23/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

m

m

m

m

m

m

K

00

00

00

3

00

00

00

ijmijm K 3

KAA 3 Volume change law

Distortion law

ijmmij KG 332

ijmijkkijijmij KG 32

ijmmij GG 3322

ijmijijmij GGG 222

GDD 2

323 GK

Constitutive equation - Hooke law

24/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

i

j

j

iij x

u

x

u

2

1...

uSiu ...

uSj

i

x

u

0

j

iji x

P

jijiq

ijkkijij G 2

NE

CE

HE

SBC

KBC

The set of NE+CE+ HE equations consists of 15 linear differential-algebraic equations – and is always the same for any static problem (except of material constants in HE).

Individual problems are different only due to different boundary conditions,

which define body shape i , loading qi and displacements ui on the body

surface (supports). Here is where name Boundary Value Problem of Elasticity comes from.

,

BVP–Boundary Value Problem of Linear Elasticity

25/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

TENSION – when internal forces reduce to the sum vector only, which is parallel to the bar axis

SHEAR – when internal forces reduce to the sum vector only, which is perpendicular to the bar axis

BENDING – when internal forces reduce to the moment vector only, which is perpendicular to the bar axis

TORSION – when internal forces reduce to the moment vector only, which is parallel to the bar axis

M

Ms

Q

N

Remarks on the solutions of special cases

De SAINT-VENANT HYPOTHESIS

When solving BVP any change in boundary conditions - especially these of static - require that the separate solution of a complex set of BVP differential equations has to be searched for.

An important observation which simplifies this problem was made by de Saint-Venant . He noticed that for points of a body far away from the points where loading is applied the solution does not depend on the details of different loadings – provided that the resultants of loading are the same.

It is worthy to mention that the solution of BVP is very general one – though in many cases analytically not possible to be found.

26/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

TENSION – when internal forces reduce to the sum vector only, which is parallel to the bar axis

SHEAR – when internal forces reduce to the sum vector only, which is perpendicular to the bar axis

BENDING – when internal forces reduce to the moment vector only, which is perpendicular to the bar axis

TORSION – when internal forces reduce to the moment vector only, which is parallel to the bar axis

M

Ms

Q

N

Special cases of cross-sectional forces reduction

BERNOULLI HYPOTHESIS

If we will pay attention to the chosen bar cross-section we may judge the solution fulfilling BVP only on the basis of another experimental observation.

However, this solution is limited only to the stresses and strains; the Kinematic Boundary Conditions have to be considered for the displacement evaluation.

The Bernoulli hypothesis allows for much simpler – though more limited solutions - when compared to BVP ones.

27/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

stop

28/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

Force = mass * acceleration 1N = 1 kg (1 m/s2) If acceleration = 9,81 m/s2

1N = 0,102 kg* (9,81 m/s2); 1 kg yields 10 N, 1 ton yields 10 kN

Stress = force/area 1 Pa = 1N/(1 m2) If mass = 102 g = 0,102 kg (=mass of 1 square meter of standard writing paper) Pressure = 1 Pa If mass = 1kg (= mass of 1 liter of water = 0,001 m3 of water) Pressure = 10 P If mass = 1 t = 1000kg (= mass of a car) Pressure = 10 000 P = 10 kP

29/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

1. Hipoteza dSV:

W punktach ciała dostatecznie odległych od miejsca przyłożenia obciazenia rozkład naprężeń i odkształceń nie zależy od sposobu przyłożenia tego obciążenia – pod warunkiem, że są one statycznie równoważne

Pozwala ona na uogólnienie rozwiązania ZB TS na różne przypadki obciążenia (różne Statyczne Warunki Brzegowe), w wyniku czego otrzymujemy komplet rozwiązania w postaci naprężeń, odkształceń i przemieszczeń.

Rozwiązanie zadania brzegowego TS z uwzględnieniem hipotezy dSV uwalnia nas od posługiwania się pojęciem przypadków wytrzymałościowych

De Saint-Venant and Bernoulli hypotheses

30/26M.Chrzanowski: Strength of Materials

SM2-01: Summary of SM1

2. Hipoteza B:

Przekroje poprzeczne pręta płaskie przed przyłożeniem obciążenia pozostają płaskie także po jego przyłożeniu

Pozwala ona (szczególnych przypadków redukcji sił przekrojowych) na założenie - dla poszczególnych przypadków wytrzymałościowych - przemieszczeń przekroju poprzecznego pręta i na tej podstawie wyznaczenie odkształceń a następnie naprężeń. Wyznaczenie przemieszczeń wymaga uwzględnienia kinematycznych warunków brzegowych.

Hipoteza B pozwala na określenie naprężeń i odkształceń w dowolnym przekroju pręta bez potrzeby uciekania się każdorazowo do rozwiązywania skomplikowanego zadania brzegowego TS.

Hipotezę B stosuje się do podstawowych przypadków wytrzymałościowych: rozciąganie skręcanie a także ścinanie – przy odpowiednich ograniczeniach co do geometrii pręta (pręt pryzmatyczny, o długości znacznie większej niż wymiary poprzeczne)

De Saint-Venant and Bernoulli hypotheses

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