Bending Induced Stresses

7/29/2019 Bending Induced Stresses

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SHIPPING AND MARINE TECHNOLOGYDIVISION OF MARINE DESIGN

Professor Jonas Ringsberg

p. 1

Lecture 6

CONTENTS:

REPETION: bending-induced stresses Introduction to torsion St Venant torsion theory (see compendium PART A: pp. 48 74)

Definition of simple case study Kinematic relations Constitutive relations Derivation of St Venant torsion moment, shear stress, shear stress flow, etc.

Summary of St Venant torsion derivation



p. 2

Learning objectives

To become familiar with torsion theory applied on ship structures Understand the difference between St Venant and Vlasov torsion theories Know how to calculate St Venant torsion shear stresses for an arbitrary

cross-section

Understand what the shear stress flow and distribution looks like duringpure St Venant torsion loading


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p. 3

Repetition of

bending-induced stresses



p. 4

General requirements - validity of beam theory

Plane sections of the beam must be plane after deformation Transverse sections must maintain the shape of the section after

deformation

Small deformations.

These requirements ensure that the distance between the neutral axis and

any longitudinal fiber of the beam is maintained during deformation

These requirements can be contained even if we add shear deformation to

the problem


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p. 5

General requirements - validity of beam theory

Beam theory works well also for non-prismatic beams as long as the twomain requirements of beam theory are fulfilled

The ship deviates from the perfect prismatic beam in many ways,however, due to the shape of the load and load effect the maximum

bending moment and the maximum shear forces of the hull will be located

between the non-prismatic ends

Therefore, the accuracy of the hull girder concept is very good.

Early efforts to compare full-scale measurements on ships with the beamtheory gave very good agreement



p. 6

Comparison of full-scale

measurements and

calculations of bending beamstresses and shear stresses of

a single skin tanker.

Ideal theory vs. measurements


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p. 7

Example of stress distributions

Bending shear and normal stress

distributions in a transverse

section.



p. 8

Plane sections remain plane after bending Sections keep the shape after bending Small deformations

This means that the distance to the neutral axis is the same

before and after bending for any fiber of the beam

Requirements, bending beam theoryNaviers theory


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p. 9

Bending normal stress



p. 10

Neutral surface/axis

Example: bending with My 0 and Mz = 0 of a straight bar with rectangularcross-section


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p. 11

The strain in a point (y, z) in the cross-section atxis:

Hookes law, , gives:

zy yz ++= 0

E=

dAyzyEdAyM

dAzzyEdAzM

dAzyEdAN

A

yz

A

xz

A

yz

A

xy

A

yz

A

x

++==

++==

++==

)(

)(

)(

0

0

0

Pure bending in two directionsNaviers theory



p. 12

The integrals are:

yz

A

y

A

z

A

AA

A

IdAyzIdAzIdAy

dAzdAy

AdA

===

==

=

,,

0,0

22



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p. 13

Kinematic relation bending deformation



p. 14

Normal strain and curvature can be derived as:

The stress in a point (y, z) in the cross-section atxis:

)()(220

yzzy

yzyzyz

yzzy

yzzzyy

IIIE

IMIM

IIIE

IMIM

EA

N

+=

+==

)(

)()(...

)(

2

0

yzzy

yzyzyzzy

yz

III

zIyIMyIzIM

A

N

zyEE

+==

++==



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p. 15

Bending shear stress



p. 16

Shear force

The figure shows bending of a homogeneous and two separate beams Assumption made in engineering beam theory:

Plane sections should remain plane after bending.


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p. 17

Sheared beam lamina of a homogenous beam

Equilibrium in thex-direction:

bD

dxDdAdAdxx

AA

=

=

+ 0

b is the length of the line in the

graph that separates areaA from

the full transverse section.

D is called shear flow (unit: N/m)

b



p. 18

bb

dx

dx

Sheared beam lamina of a homogenous beam


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p. 19


For thin-walled cross-sections holds:

The general expression for normal stress gives:

where:

=

A

dAx

bb

D

1

+

=

)(

)()(

2yzzy

yzyzyzzy

III

zIyIMyIzIM

A

N

xx

yzzy VMx

VMx

=

=



p. 20


Integration gives:

where and are called static moments

If the y- andz-axis are principal axes, we have:

)(

)()(

2yzzy

yzyyzyyzzzyz

IIIb

ISISVISISV

b

D

+=

==A

z

A

y dAySdAzS

y

yz

z

zy

bI

SV

bI

SV

b

D +=


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p. 21

The integrals are approximately equivalent with the sums




p. 22

bh

V

bbh

zhb

V

bhI

zh

bzh

AeS

xz

y

y

2

3

12

42

12

222

1

max

3

22

3

=

=

=

+==b

Static moment of a rectangular cross-section


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p. 23

T-shaped cross-section

D, shear flow

1

2

max

V



p. 24

The shear centre (SC)

Double symmetric:SC is in the COG

Open/single symmetric:SC is outside the structure

Crossing plates:SC is in the intersection between

the plates


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p. 25

Definition of bending shear stress:

We will make use of a coordinate system and

In this coordinate system, with the centroid Cas the origin (the neutral axispasses the centre of gravity):

the shear forces are called V and V, the static moments are called S and S and the moment area of inertia are called I and I.

y

yz

z

zy

bI

SV

bI

SV

b

D +=




p. 26

This equation is equivalent with Eq. (3.24) in the compendium.



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p. 27




p. 28

If we require that the cross-section will not rotate for the loads that weapply the loads must pass through the shear centre

I.e. bending without twisting!

To determine the shear centre, we make use of the fact that the shear

forces V and V must be statically equivalent to the shear stresses acting

on the beam cross section

This requirement determines the line of action for each of the shear forcesV and V, which both pass though the shear center when there is no twist



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p. 29

If the shear stresses xs are to be statically equivalent with the shearforces, their moment with regard to thex-axis or any longitudinal axis mustbe equal

This gives the relation:

where 0 and 0 are the coordinates of the shear centre with regard to a

provisional centre and h(s) is the moment arm for the shear flow




p. 30

Insert Eq. (3.24) [ ] into :

This equation must hold for all values of the shear forces and it is onlypossible if and only if:

y

yz

z

zy

bI

SV

bI

SV

b

D +=



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p. 31

And again, finally,

Bending shear and normal stress

distributions in a transverse

section.



p. 32

Torsion-induced stresses


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p. 33

Introduction to torsion

A catamaran in a sea-state which

gives rise to torsion structural response.



p. 34


An example of the structural response

of a container vessel on the North

Atlantic trade.


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p. 35


A YouTube movie which shows a good example of the

structural response of a container vessel in harsh weather

(http://www.youtube.com/watch?feature=player_detailpage&v=qEkErF51Uxg)



p. 36


Circular shaft in pure torsion Noncircular shaft in pure torsion


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p. 37


Warping displacement Out-of-plane deformation during torsion loading

Circular cross-section:no warping

Rectangular cross-section:very little warping

Thin-walled and open I-shaped cross-section:large amount of warping



p. 38

Cross-sections with and without warping All but two cross-sections below show warping displacements behavior. However, most of the cross-sections (a, c and d) will produce very little warping.



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p. 39


The analysis and understanding of the loading caseis very important for following stress and strain

analysis, see the example

Case study: A concentrated load, P, is acting on one of the flanges.

The structural response to this load, P, must bedivided into the following load and stress/strain

analyses:

Axial load, N Bending moment, My Bending moment, Mz Bimoment, B

y

xz



p. 40

Pure axial loading Normal stress, A

Pure bending condition

Normal stress, B Shear stress, B

Pure St Venant torsion Shear stress, SV

Vlasov torsion Normal stress, W Shear stress, W

TOT = A + B + W

TOT = B + SV + W

Superposition of stress

components for various

types of loading conditions.



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p. 41

Simple case study: Circular tube with constant thin wall thickness, h. Subjected to a torque, Tx, which is constant over the length, L.

Of symmetric reasons, we will have constant shear stresses along thecircumferential balancing the torque

St Venant torsion theory



p. 42

St Venant torsion theory

An arbitrary element Cof the tube wall,originally oriented along the

generatrise, is transformed by shear

deformation to a rhomb Cwith the

shear angle

In the figure we have that r = L ( = Greek letter gamma) Hookes generalized law gives = G

Thus,NOTE!

This is one of the most perfect

structures to be used for laboratory

determination of the shear modulus.


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p. 43

Geometric description of deformation in a continuum

Kinematic relation



p. 44

Equations of the geometric description of

deformation in a continuum:

In the limit as x and y

approaches zero we get:

Kinematic relation


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p. 45

The constitutive relations are concerned with material dependence I.e. relationships between stresses and strains.

If a material is elastic and isotropic, Hookes law can be applied Isotropic: the material is assumed to have similar properties in all its directions.

Constitutive relations



p. 46

The graph from slide 13 with

bending beam coordinates.

Derivation


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p. 47

The cross-section will not change itsshape, i.e. all strains in the yz-planeare zero

Thus, z = y= yz = 0

The displacements of the cross-section can be described as a rigidbody rotation of angle (x) around acentre of twist (VC)

Derivation



p. 48

Derivation

If the centre of twist hascoordinates y0 and z0 we get

expressions relating displacements

in y- and z-direction to the rotation

(x) according to the figure

)()(),,(

)()(),,(

0

0

yyxzyxw

zzxzyxv

=

=

Eq. (4.2)


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p. 49

Derivation

We will have use for the displacement in thex-direction:u = u(x,y,z), Eq. (4.3)

As we are dealing with torsion only (not bending), this function describesthe warping displacements

We will assume here that warping is not restrained anywhere in the beam

This means that:x= 0, Eq. (4.4)

In view of Eq. (4.1) and (4.4) it follows from Eq. (2.9) that:y= z = yz = x= 0, Eq. (4.5)



p. 50

Derivation

Shear strains are given by Eq. (2.8) and we introduce Eq. (4.2) into that:

and the shear stresses are:


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p. 51

Derivation

When there are no body forces(= forces from gravity or other inertia

forces), the equilibrium from Eq. (2.6)

gives:



p. 52

Derivation

There are no restraints (free warping), and therefore, there are no strainsin the longitudinal direction:

In combination with Eq. (4.7) we get:


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p. 53

Derivation

The equilibrium conditions of Eqs (4.8b) and (4.8c) are only satisfied if:



p. 54

Derivation

The displacement function u= f(y,z) is the basic unknown Insertion of Eq. (4.7) into Eq. (4.8a) gives us:

Introduce a stress function, (y,z), with the requirements that it is afunction ofyand z and twice differentiable and:

NOTE!

The stress function is used here as a

help to continue


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p. 55

Derivation

Derivate once with regard to dyand dy, respectively:

Combine Eqs (4.14) and (4.7):



p. 56

Derivation

Derivate!

Subtract!


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p. 57

Derivation

Locate an arbitrary point, P, on theboundary of a solid cross-section

Identify a tangential stress withcomponents in the direction of the

coordinate system



p. 58

Derivation

Then,

The stress function, , must be zero along the boundary In all previous relations, this function only appeared as derivatives

Any constant would do = 0.


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p. 59

Derivation



p. 60

Derivation

By definition,

Combine with Eq. (4.14):


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p. 61

Derivation

Note that in:

Hence, Eq. (4.18) can be rewritten:



p. 62

Derivation


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p. 63

Using Eq. (4.19):

we get that:

Since = 0 everywhere on the boundary, the value of the line integral onthe previous slide is zero and:

Derivation



p. 64

Summary of derivation

The formulation of St Venant torsion theory is now complete and we canstart to use it:

However, the solutions to the stress function with actual geometricboundaries are rather complex

This is why handbooks are full of pre-calculated torsional constants.


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p. 65

Summary of derivation

Thin-walled open section:

(4.28)



p. 66

Torsional constants and shear stress flow

Documents

Bending Induced Stresses