1 Unsymmetrical and/or inhomogeneous cross sections | CIE3109 CIE3109 Structural Mechanics 4 Hans...

1Unsymmetrical and/or inhomogeneous cross sections

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CIE3109Structural Mechanics 4

Hans Welleman

Module : Unsymmetrical and/or inhomogeneous cross section


| CIE3109

CIE3109 : Structural Mechanics 4

• 1-2 Inhomogeneous and/or unsymmetrical cross sections• Introduction

• General theory for extension and bending, beam theory

• Unsymmetrical cross sections

• Example curvature and loading

• Example normalstress distribution

• Deformations

• 3 Inhomogeneous cross sections • Refinement of the theory

• Examples

• 4-5 Stresses and the core of the cross section• Normal stress in unsymmetrical cross section and the core

• Shear stresses in unsymmetrical cross sections

• Shear centre

Lectures


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QuestionWhat is the stress and strain distribution in a crosssection due to extension and bending in case of aunsymmetrical and/or non-homogeneous cross section’ ?

z

zz

y

y

y

(a) (b) (c)

concrete

steel

E1

E1

E2

unsymmetrical inhomogeneous both


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BENDING and AXIAL LOADING

RELATION BETWEEN EXTERNAL LOADS AND DISPLACEMENTS

• Structural level• Cross sectional level

ux

uz

uy xz

y

z

x

y

Fz

FxFy

F

global definitions

sectional definitions

Vz

Vy


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fibre in cross section x at position y,z

ASSUMPTIONS

• FIBRE MODEL

• SMALL ROTATIONSOF THE CROSS SECTION

• UNIAXIAL STRESS SITUATION IN THE FIBRES

• LINEAR ELASTIC MATERIAL BEHAVIOUR

Hooke’s law : E


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SOLUTION PATH

• Determine the strain distribution due to the displacements of the cross section?

• Determine the stress distribution caused by these strains?

• Determine the resulting forces in the cross section?

Load Sectional forces Stress Strain Deformations Displacements(F en q) (N, V, M) (fibre) (fibre) (,) (u,)

Equilibrium Static Constitutive Compatibility Kinematischerelations relations relations relaties relaties

Constitutive relation (cross section)


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KINEMATIC RELATIONS

Relation between deformations and displacements

z

x-as

z-as

y

uz

ux

y

z

x

uzy d

d

y

x-as

y-as

zuy

ux

z y

x

u yz d

d


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FIBRE MODEL

• Horizontal displacementz

x-as

z-as

y

uz

ux

y

z

x y

x z

( , , )

( , , )

u x y z u z

or

u x y z u z u

z

y

y

y u

y

x-as

y-as

zuy

z y

ux

x

uzy d

d

x

u yz d

d


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RELATIVE DISPLACEMENT-STRAIN

x y z( , )y z u y u z u

x z y

x y z

( , , )

( , , )

u x y z u y z

or

u x y z u y u z u

( , , )( , )

u x y zy z

x

y z( , )y z y z with 2

2

2

2

d

d;

d

d;

d

d

x

u

x

u

x

u zz

yy

x


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STRAIN DISTRIBUTION

z

y

x-axis

z-axis

x-axis

y-axis

strain

strain

z

y

y z( , )y z y z

Conclusion:

Strain distribution is fully described with three deformation parameters


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STRAIN FIELD

Three parameters• strain in fibre which coincides with

the x-axis• slope of strain diagram in y-

direction• slope of strain diagram in z-

direction

y z( , )y z y z


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k

k

CURVATURE

• First order tensor• Curvature in x-y-plane• Curvature in x-z-plane• Plane of curvature k

2z

2y

zk

y

tan


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NEUTRAL AXIS

kk is perpendicular to neutral axis

0),( zy zyzy

neutral axis


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FROM STRAIN TO STRESS

• Constitutive relation : link between deformations and stresses

y z

( , ) ( , ) ( , )

( , ) ( , )

y z E y z y z

y z E y z y z


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STRESS AND SECTIONAL FORCES• Static relations : link between stresses

and sectional forces

y

z

( , )

( , )

( , )

A

A

A

N y z dA

M y y z dA

M z y z dA


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Elaborate …

AA

AA

AA

zdAzyzyEdAzyzM

ydAzyzyEdAzyyM

dAzyzyEdAzyN

zyz

zyy

zy

),(),(

),(),(

),(),(

zzzyyzz2

zyz

zyzyyyyz2

yy

zzyyzy

),(),(),(

),(),(),(

),(),(),(

EIEIESdAzzyEyzdAzyEzdAzyEM

EIEIESyzdAzyEdAyzyEydAzyEM

ESESEAzdAzyEydAzyEdAzyEN

A AA

A AA

A AA

approach with “double-letter” symbols


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CONSTITUTIVE RELATION

z

y

zzzyz

yzyyy

zy

z

y

κ

κ

ε

EIEIES

EIEIES

ESESEA

M

M

N

on cross sectional level

INDEPENDENT OF THE ORIGIN OF THE COORDINATE SYSTEM

SPECIAL LOCATION OF THE ORIGIN OF THE COORDINATE SYSTEM TO UNCOUPLE BENDING AND AXIAL LOADING


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NORMAL FORCE CENTRE

• Bending and axial loading are uncoupled:

z

y

zzzy

yzyy

z

y

0

0

00

κ

κ

ε

EIEI

EIEI

EA

M

M

N Axial loading

Bending

y z 0ES ES definition NC : if N = 0 then zero strain at the NC and the n.a. runs through the NC


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LOCATION NC y

z

NC NCy

NCz

y

z

dAzyE ),(

NCNC zzzyyy

NCzNC

z

NCyNC

y

),(),(

),(

),(),(

),(

zEAESdAzyEzzdAzyE

dAzzyEES

yEAESdAzyEyydAzyE

dAyzyEES

AA

A

AA

A

y

NC

zNC

ESy

EA

ESz

EA


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RESULT

• Bending and axial loading are uncoupled

• Moment is first order tensor

2z

2y MMM

y

zmtan

M

M


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MOMENT

M

m

x-axis

z-axis

y-axis

Mz

My

m

MOMENT

M

TENSION

COMPRESSION

N.C.

SUMMARY of formulas

TENSION

COMPRESSION

CURVATURE

k

n.a.

x-axis

n.a.

z-axis

y-axis

z

y

k

N.C.

z

y

zzyz

yzyy

z

y

zy

EIEI

EIEI

EA

M

M

N

zyzyEzy

zyzy

),(),(),(

),(


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PLANE OF LOADING SAME AS PLANE OF CURVATURE ?

y y

z z

yy yz y y

zy zz z z

M

M

M

EI EI

EI EI

0 eigenvalue problemyy yz

zy zz

EI EI

EI EI


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PRINCIPAL DIRECTIONS

y yy y

z zz z

2 21 1yy zz 2 2

12

0

0

, ( )

tan 2 ,( )

yy zz yy zz yz

yz

y zyy zz

M EI

M EI

EI EI EI EI EI EI

EI

EI EI


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ROTATE TO PRINCIPAL DIRECTIONS

y yy y y yy y

z zz z z zz z

0

0

M EI M EI

M EI M EI

direction (1)

direction (2)

z

y

m k

m k

yy zz

1) 0 principal direction

2) / 2 other principal direction

3) EI EI

z zz z zzm k

y yy y yy

tan( ) tan( )M EI EI

M EI EI

m k ??


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EXAMPLE 1

Determine the position of the plane loading m-m for the specified position of the neutral line which makes an angle of 30o degrees with the y-axis. The rectangular cross section is homogeneous.


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EXAMPLE 2

A triangular cross section is loaded by pure bending only. The cross section is homogeneous and the Youngs modulus is E. The normal stress in point A and C is equal to 10 N/mm2.

a) Calculate the magnitude and direction of the resulting bending moment

b) Compute the stress in point B


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F1 = HELP

tan

)(tan3

361

2

3361

3481

3361

bhI

bhhbI

bhI

yz

yy

zz


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Equillibrium conditions

2

2

2

2

d d0

d d

d d d0 and 0

d d d

d d d0 and 0

d d d

x x

y y yy y y

z z zz z z

N Nq q

x x

V M Mq V q

x x x

V M Mq V q

x x x


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Differential equations (structural level)

d'

dx

x

uu

x

2

2

d"

dy

y y

uu

x

2

2

d"

dz

z z

uu

x

Kinematics:

Constitutive relations:

z

y

zzzy

yzyy

z

y

0

0

00

κ

κ

ε

EIEI

EIEI

EA

M

M

N

d

d x

Nq

x

2

2

d

dy

y

Mq

x

2

2

d

dz

z

Mq

x

Equilibrium relations:

subst

ituteaxial

loading

bending

'''' ''''

'''' ''''

"x x

yy y yz z y

yz y zz z z

EAu q

EI u EI u q

EI u EI u q


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Result

2

2

"

"''

"''

xx

zz y yz zy

yy zz yz

yy z yz yz

yy zz yz

qu

EAEI q EI q

uEI EI EI

EI q EI qu

EI EI EI

2

2

"''

"''

zz y yz zyy y yy

yy zz yz

yy z yz yzz z zz

yy zz yz

EI q EI qEI u EI

EI EI EI

EI q EI qEI u EI

EI EI EI

"''

"''

yy y y

zz z z

EI u Q

EI u Q

2

2

zz yy y yz yy zy

yy zz yz

yz zz y yy zz zz

yy zz yz

EI EI q EI EI qQ

EI EI EI

EI EI q EI EI qQ

EI EI EI

modify the load in all forget-me-not’s

OR solve differential equation


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EXAMPLE (see lecture notes)41

, 9

4 2

2 4i jEI Ea

2 2 4

2 2 4

3"''

2

3"''

zz y yz z yz z zy

yy zz yz yy zz yz

yy z yz y yy z zz

yy zz yz yy zz yz

EI q EI q EI q qu

EI EI EI EI EI EI Ea

EI q EI q EI q qu

EI EI EI EI EI EI Ea

42 3

1 2 3 4 4

42 3

1 2 3 4 4

16

8

zy

zz

q xu C C x C x C x

Ea

q xu D D x D x D x

Ea

0 : ( 0; 0; 0; 0)

: ( 0; 0; 0; 0)

y z y z

y z y z

x u u

x l u u M M


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Result


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SUMMARY displacements

• solve differential equations• Use “pseudo” load for y - and z -direction and use this load

in standard engineering equations• Use curvature distribution in combination with moment

area theoremes to obtain displacements and rotations (see notes)

• Use principal directions, decompose load in (1) and (2) direction, and compute displacements in (1) and (2) direction with standard engineering equations. Finally transformate the displacements back to y - and z -direction

Original y-z coordinate system

Principal 1-2 coordinate system

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