Dictaat UK v9

8/13/2019 Dictaat UK v9

1/99

STRUCTURAL MECHANICS 4CT3109

MODULE : NON-SYMMETRICAL ANDINHOMOGENEOUS CROSSSECTIONS

COEN HARTSUIJKER

HANS WELLEMAN

Civil EngineeringTU-Delft

March 2011


2/99

STRUCTURAL MECHANICS 4 Non-symmetrical and inhomogeneous cross sections

Ir C. Hartsuijker & Ir J.W. Welleman March 2011 ii

TABLE of CONTENTS

1. NON-SYMMETRICAL AND INHOMOGENEOUS CROSS SECTIONS ........................................... 1

1.1 SKETCH OF THE PROBLEM AND REQUIRED ASSUMPTIONS..................................................... ................. 11.2 HOMOGENEOUS CROSS SECTIONS ........................................................... .............................................. 4

1.2.1 Kinematic relations ....................................................... ........................................................... ....... 41.2.1.1 Curvature..... ..................... ...................... ..................... ..................... ...................... ..................... .......... 61.2.1.2 Neutral axis ................... ..................... ..................... ...................... ..................... ..................... .............. 7

1.2.2 Constitutive relations for homogeneous non-symmetrical cross sections....................................... 81.2.2.1 Moments............. ...................... ..................... ..................... ..................... ...................... ..................... . 101.2.2.2 Properties of the constitutive relation for bending... ...................... ..................... ..................... ............ 11

1.2.3 Equilibrium conditions............................................................ ...................................................... 131.2.4 Differential Equations................................................... ........................................................... ..... 141.2.5 Example 1 : Homogeneous non-symmetrical cross section ..................................................... ..... 15

1.2.6 Normal stresses in the y-z-coordinate system ...................................................... ......................... 191.2.7 Normal stresses in the principal coordinate system....................................................... ............... 201.2.8 Example 2 : Stresses in non-symmetrical cross sections......... ...................................................... 21

1.3 EXTENSION OF THE THEORY FOR INHOMOGENEOUS CROSS SECTIONS ................................................. 271.3.1 Position of the NC for inhomogeneous cross sections ................................................... ............... 301.3.2 Example 3 : Normal centre versus centroid ......................................................... ......................... 311.3.3 Example 4 : Stresses in inhomogeneous cross sections ........................................................... ..... 32

1.4 FORCE POINT OF THE CROSS SECTION...................................................... ............................................ 351.5 CORE OR KERN OF A CROSS SECTION ...................................................... ............................................ 37

1.5.1 Example 5 : Core of a non-symmetrical cross section ................................................... ............... 411.6 TEMPERATURE INFLUENCES*........................................................ ...................................................... 43

1.6.1 Example 6 : Static determinate structure under temperature load ............................................... 461.6.2 Example 7 : Static indeterminate structure under temperature load ............................................ 50

1.7 SHEAR STRESS DISTRIBUTION IN ARBITRARY CROSS SECTIONS ...................................................... ..... 541.7.1 Shear stress equations for principal coordinate systems ......................................................... ..... 55

1.7.1.1 Example 8 : Shear stresses in a composite cross section .................. ...................... ..................... ........ 571.7.2 General shear stress formula ............................................................ ............................................ 58

1.7.2.1 Example 9 : Shear stresses in a non-symmetrical cross section........ ...................... ..................... ........ 591.7.2.2 Example 10 : Shear force in a non homogeneous cross section................... ..................... ................... 65

1.7.3 Shear force center for thin walled non-symmeyrical cross sections ............................................. 671.7.3.1 Example 11 : Shear force center for thin walled cross sections........ ...................... ..................... ........ 68

APPENDIX A ........................................................ ............................................................ .................................. 74

APPENDIX B ........................................................ ............................................................ .................................. 76

2. ASSIGNMENTS .................................................... ........................................................... ......................... 78

2.1 CROSS SECTIONAL PROPERTIES ..................................................... ...................................................... 782.2 NORMAL STRESSES IN CASE OF BENDING .......................................................... .................................. 812.3 NORMAL STRESSES DUE TO BENDING AND EXTENSION ........................................................ ............... 862.4 INHOMOGENEOUS CROSS SECTIONS LOADED IN EXTENSION ........................................................... ..... 882.5 INHOMOGENEOUS CROSS SECTIONS LOADED IN BENDING .................................................... ............... 892.6 CORE .................................................... ............................................................ .................................. 922.7 SHEAR STRESSES DUE TO BENDING ......................................................... ............................................ 95


3/99


Ir C. Hartsuijker & Ir J.W. Welleman March 2011 iii

STUDY GUIDE

These lecture notes are part of the course CT3109 Structural Mechanics 4. Both theory and

examples are presented for self-study. Additional study material is available via the internet.

Sheets used and additional comments made in the lectures are available via BlackBoard or

also via internet:

http://home.hetnet.nl/~t-wmn/index.html

In theses notes reference is made to the first year lecture notes of CT1031 Structural

Mechanics 1 and CT1041 Structural Mechanics 2 and CT2031 Structural Mechanics 3 which

are covered with three books:

o Engineering Mechanics, volume 1 : Equilibrium, C. Hartsuijker and J.W. Wellemano Engineering Mechanics, volume 2 : Stresses, strains and displacements C. Hartsuijker and J.W.

Wellemano Toegepaste Mechanica , deel 3 : Statisch onbepaalde constructies en bezwijkanalyse, C.

Hartsuijker en J.W. Welleman (in Dutch)

These books will be referred to as MECH-1, MECH-2 and MECH-3.

With respect to the previous edition of these notes only minor changes have been made. The

general procedure to find the kernel of the cross section is now supported with fig 11b and fig

11c on page 40.

The answers of the assignments can be found on BlackBoard or via the above mentioned web

site. If needed additional information can be obtained from the Student Assistants ofStructural Mechanics.

Although these notes have been prepared with the utmost precision faults can not be

excluded. I will appreciate any comments made and invite students to read the material

carefully and make suggestions for improvement. Any reported faults will be printed on the

internet site to inform all students.

The lecturer,

Hans Welleman

pdf-edition, March [email protected]


4/99


Ir C. Hartsuijker & Ir J.W. Welleman March 2011 1

1. NON-SYMMETRICAL AND INHOMOGENEOUS CROSS SECTIONS

In this module the in MECH-2 introduced fiber model for beams will beextended and used on beams with a non-symmetrical and/or inhomogeneouscross section. With the presented theory a straight forward and easy methodis obtained to calculate the stresses and strains in cross sections made out ofdifferent materials (inhomogeneous) and or with no axis of symmetry (non-symmetrical).

1.1 Sketch of the problem and required assumptions

The cross sections used so far, always contained at least one axis of symmetry and the cross

section itself was always made out of one single material (homogeneous cross section). With

the fiber model as introduced in MECH-2 the cross section is modeled as a collection of

initially straight fibers which are parallel to the beam axis denoted asx-axis. The fibers are

kept together by infinite rigid cross sections which are by definition perpendicular to the beamaxis. In figure 1 this model is shown together with the coordinate system used. The origin of

the coordinate system of the cross section is the Normal Centre NC. A detailed description of

this model can be found in chapter 4 of MECH-21.

Figure 1 :Fiber model and a cross section with one axis of symmetry.

If a cross section is loaded with one single bending moment and a normal force, all fibers at

the tensile side will elongate and fibers at the compressive side will shorten. Due to the

assumption of the infinite rigidity of the cross section the plane cross sections will remain

plain. This is known as the hypothesis of Bernouilli. In figure 2a all sectional forces are

shown and in figure 2b the resulting strains in the fibers are shown. If a linear relation

between strains and stresses is assumed (Hookes law) the resulting normal stresses due to

combined bending and normal force can be presented as is shown in figure 2c .

(a) (b) (c)

Figure 2 :Bending and extension in a homogeneous cross section with one axis of symmetry.

1C. Hartsuijker and J.W. Welleman, Engineering Mechanics, Volume 2, ISBN 9039505942

N

fiber

fiber

beam axis

axis of symmetrycross sections


5/99



Based on this model, formulas for calculating stresses for combined bending and extension,

have been found in MECH-2. For non-symmetrical and or inhomogeneous cross sections the

found formulas can not be used. Examples of these situations are given in figure 3.

Figure 3 :Examples of non-symmetrical and/or inhomogeneous cross sections.

Example (a) shows a non-symmetrical cross section. In (b) the cross section is

inhomogeneous with one axis of symmetry. In (c) the cross section is inhomogeneous and

non-symmetrical. Apart from the shape and material of the cross section also the loading of

the cross section is important. In general a cross section can be loaded with three forces (two

shear forces and one normal force) and three moments (two bending moments and one

trosional moment). Thefiber modelonly describes the strains and stresses due to bending and

extension. The influence of the shear forces and torsional moments are therefore excluded.

The definitions used for the normal force, bending moments and the displacements as

introduced in section 1.3.2 of MECH-1 are shown in figure 4, see also section 1.2.2.

Figure 4 :Sectional forces and displacements.

The shear forces will not cause any strains in the fiber model. However with a simple model

as introduced in MECH-2 we can obtain the shear stresses due to shear forces. At the end of

these lecture notes a special chapter deals with the subject of shear stresses and the shearcentre SC.

For this chapter the central question to answer is :

How to describe the strains and stresses in non-symmetrical and/or inhomogeneous cross

section due to the combined loading of bending and extension ?

In order to answer this question in a structured way we can split the question in a number of

sub questions:

How can we find the strains due to the displacements of the cross section ?

How can we find the (normal) stresses due to the strains in the fibers ?

How can we find the (cross) sectional forces which belong to these (normal) stresses ?

z

zz

y

yy

(a) (b) (c)

concrete

steel

E1

E1

E2

ux

uzuy x

z

y

Fz

FxFy

F

z

xy


6/99



With this approach we follow the standard modeling technique in Structural Mechanics:

Loads Stresses Strains Displacements

( F, q) (N, M, V) ( , ) ( u, )

equilibrium constitutive kinematic

equations relations relations

Figure 5 :Basic modeling equations in Structural Mechanics.

This approach has already been introduced in MECH-2. The introduced assumptions also

holds for non-symmetrical and inhomogeneous cross sections:

1. Plane sections remain plane even after loading and deformations and the cross sectionsremain parallel to the beam axis which coincides with the direction of the fibers (

hypotheses of Bernoulli ). The cross sections are of infinite rigidity we therefore speaknormally of rigid cross sections.

2. Cross sectional rotations remain small,


7/99



1.2 Homogeneous cross sections

The modeling steps of figure 5 will be elaborated this section. We will start with

homogeneous cross sections which can be non symmetrical.

1.2.1 Kinematic relations

The kinematic relations relate the displacements of a cross section to the fiber strains in the

cross section. A cross section loaded in combined bending and extension may exhibits three

translations and three rotations :

zyxzyx ,,en,, uuu

We will use the earlier introduced definitions of the displacements, see figure 4.

If the displacement of the cross section can be described with these six degrees of freedom

then we can also describe the displacement u in the x-direction of any fiber in the crosssection. Suppose point P(x,y,z) is in the cross section at a distancexof the beams origin, see

figure 6.

Figure 6 :Point P(x,y,z) in a cross section at distancexfrom the origin.

The cross section moves with uxinx-direction and rotates with yalong they-axis and with

zalong thez-axis. The displacement uin the direction of the fiber at P can be written withthe assumption of small rotations (see section 15.3.2 from MECH-l) as:

yzx),,( zyuzyxu +=

In figure 7 this is clarified with some sketches of the displaced cross section (dashed) which is

subsequently rotated along they- andz-axis. Both the top and side view show the influence of

the rotations upon the displacement u.

Figure 7 :Displacement in the direction of the fibers due to the rotations yen z.

x

y

z

ux

x

y

x

z

top view side view

P yzy

zyz

fiber

cross section


8/99



Due to the assumed small rotations the influences in the displacement can be superposed. The

displacement quantities ux, yen zbelong to the cross section which contains P and aretherefore cross sectional related quantities. These displacements are thus only functions ofx.

In other words: the displacements of an arbitrary fiber in a cross section at a distance xcan be

described with three displacements quantities.

With the displacement of a point P also the straining of a fiber through P can be obtained. The

relative displacement which is known as the engineering strain of a fiber can be found with:

yzx

yzx

0

),(

),,(),,(lim),(

+=

+=

=

=

zyuzy

dx

dz

dx

dy

dx

du

x

zyxu

x

zyxuzy

x (a)

The rotations yen z can be expressed in the displacement quantities uyen uz. See figure 7.

y

y

z

zz

y

d

dd

d

ux

u

u

x

u

=+=

==

The change in sign is due tot the definitions of the rotations ( check this yourself !).

The strain according to (a) in the fiber through P can be written as:

zyx),( uzuyuzy = (b)

The strain is equal to the strain of the fiber atx-axis added with the strain due to bending

(curvature) along they- andz-axis of the beam axis. Expression (b) can be rewritten with the

introduction of the following three cross sectional deformationquantities:

yzz

zyy

x

==

==

=

u

u

u

(c)

These relations are known as the kinematic relations

and relate the cross sectional deformation quantities

to the cross sectional displacement quantities.

With the kinematic relations (c) the strain in a fiber

according to (b) can be written as:

zy),( zyzy ++=

In figure 8 is the strain distribution over the cross

section is visualised. From the assumption that plane

cross sections remain plane follows a strain

distribution which represents a plane. The slopes of

this plane iny- enz-direction is denoted with yand

z. From this also follows that yen z must beconstant and are therefore cross sectional quantities.

Figure 8 :Cross sectional strain distribution

neutral line


9/99



From figure 8 also follows that positive curvatures causes for positive values ofyandz

positive strains. This is in complete agreement with the definition of positive curvatures as

introduced in MECH-2.

1.2.1.1 Curvature

The curvature of a beam can be build out of a curvature in thex-y-plane and a curvature in the

x-z-plane. These curvatures are denoted with respectively yand z. These are the components

of the vector . In figure 9 this is visualised. The prove that a curvature behaves like a firstorder tensor is given in appendix A.

Figure 9 :Curvature as vector.

Being a vector means that the curvature has a magnitude and a direction. By definition a

positive curvature is visualised by an arrow which points from the concave side to the convex

side of the curved beam (see figure 9). In figure 10 the curvature is shown in the plane of the

cross section. The plane of the curvature is denoted with the letter k.

Figure 10 :Curvature as a vector in the cross sectional plane.

The magnitude of the curvature is :

2

z

2

y +=

The angle between the curvature and they-axis is defined as:

y

zktan

=

k

k

convex side

concave side


10/99



1.2.1.2 Neutral axis

With the expression of the strain distribution over the cross section, we can also find an

expression for fibers with zero strain. The fibers in the cross section with zero strains form a

line which is called the neutral line or neutral axis. In order to keep in line with the Dutch

edition of these notes we will use neutral line which is abbreviated as nl. With the zero strain

definition the expression for the neutral line becomes:

0),( zy =++= zyzy

In order to draw the neutral line in the cross section, two handy points are needed. These

points are the points of intersection with the coordinate axis. The neutral line crosses the

coordinate axis in:

Point of intersection with they-axis (z = 0) :y

=y

Point of intersection with thez-axis (y= 0):z

=z

In figure 11 the neutral axis is drawn in the cross sectional coordinate system.

Figuur 11 :Position of the neutral line in the cross section.

This figure also shows that the beams plane of curvature kis perpendicular to the neutral

axis. The arrow representing the curvature directs from the concave (smallest strain) to the

convex (largest strain) side. In this case from the compressive zonein to the tensile zone.

Assignment:

Give the proof for the observation that the plane of curvature is perpendicular tot the neutral

line.

k

neutral line nl.


11/99



1.2.2 Constitutive relations for homogeneous non-symmetrical cross sections

The fiber model used so far, assumes a linear elastic stress-strain relation. We will restrict

ourselves to this simple model, using Hookes law:

=E

In a cross section the strain and stress in a certain point is denoted with respect to the chosen

coordinate system:

),();,( zyzy ==

In case of a homogeneous cross section all fibers will have the same Youngs modulusE. The

stress in a certain point can easily be found from the computed strains with:

),(),( zyEzy =

In combined loaded sections (bending and extension), fibers will lengthen or shorten. Thedeformation behaviour of a particular cross section can be described with the earlier

introduced three cross sectional deformation quantities:

zy en,

The strain in any fiber (y,z) at the cross section is now known with:

zy),( zyzy ++=

Using Hookes law for the stress strain relation we can find the expression for the stress in

any particular (fiber) point of the cross section:

zy),( zyEzy ++=

The relations

between the stresses

in fibers and the

sectional forces can

be found in an

identical way as

introduced in section

4.3.2 of MECH-2.Figure 12 shows the

definitions used.

Figure 12 : Equivalent sectional force belonging to

normal stresses acting on an infinitesimalsmall area Aof the cross section.

NOTE:

The definitions used here are so called formal

definitions.Myis the bending moment in the

xy-plane andMzis the bending moment in the

xz-plane In engineering notation however

these moments are denoted asMzandMy,

bending about thezoryaxis. It is important

to check these definitions used in MECH-2.


12/99



The resulting normal forceNdue to the normal stresses becomes:

( )

++=++==

A AAAA

zdAydAdAEdAzyEdAzyN zyzy),(

The bending momentMyandMzbecome:

( )

( )

++=++==

++=++==

A AAAA

A AAAA

dAzyzdAzdAEzdAzyEdAzyzM

yzdAdAyydAEydAzyEdAzyyM

2

zyzyz

z

2

yzyy

),(

),(

In TM-CH-2 the following cross sectional quantities were defined to simplify the above given

expressions:

zz

2

zyyz

yy

2

y

IdAz

IIyzdASzdA

IdAySydAAdA

A

A

z

A

AAA

=

===

===

Ais the cross sectional area, Sis the first order area moment (static moment) andIis the

second moment of area (moment of inertia). Examples of these quantities can be found in

chapter 4 of MECH-2.

The relation between the sectional forces (N,MyandMz) and the sectional deformations (,

yen z) can be rewritten as:

zzzyzyzz

zyzyyyyy

zzyy

EIEIESM

EIEIESM

ESESEAN

++=

++=

++=

In matrix presentation this relation becomes:

=

z

y

zzzyz

yzyyy

zy

z

y

EIEIES

EIEIES

ESESEA

M

M

N

From this relation we can conclude that if from a fiber which coincides with thex-axis, the

strain and both curvatures are known, the cross sectional forces can be computed.

The matrix shown is called the stiffness matrix relating the generalised stresses (sectional

forcesN,MyandMz) to the generalised deformations (sectional deformations , yen z) andis also called the cross sectional constitutive relation. The derivation of it is based on the

linear elastic constitutive relation of a single fiber (Hookes law).

From the above we can conclude that all sectional properties of a beam can be assigned to a

single fiber which coincides with thex-axis. This is also why we are allowed to represent

beams according to the beam theory as single line elements in frame models.

As to now we worked with an arbitrary chosen position of they-z-coordinate system. By

choosing however a special position of the coordinate system the above found expressions for

the sectional forces can be significantly simplified.

NOTE:

This matrix is symmetrical and all diagonal terms

are positive.

NOTE:

The definitions used here are so called formal

definition. In engineering practice often a single

sub index is used. Iyyis then referred to asIzand

Izzis referred to asIy. It is very important tocheck these definitions used in MECH-2.


13/99



It is common use to chose the origin of the coordinate system such that the static moments Sy

en Szbecome zero. The position of they-z-coordinate system has to chosen at the normal

force centreNC of the cross section. For a homogeneous cross section the normal force centre

coincides with the centre of gravity. For inhomogeneous cross sections this no longer holds

which will be illustrated later.

With the origin of they-z-coordinate system at the normal force centreNC the static moments

become zero thus simplifying the constitutive relation:

=

z

y

zzzy

yzyy

z

y

0

0

00

EIEI

EIEI

EA

M

M

N

From this relation we can see that a normal forceN, acting at the normal force centreNC,

only causes strainsand no curvatures. See also section 2.4 from MECH-2. From the above

shown matrix representation we can also conclude that there is no interaction betweenextension and bending if the origin of the coordinate system is chosen at the normal centre

NC. The bending part of the equations is fully uncoupled from the extension part. The system

of equations can therefore also be written as:

=

=

z

y

zzzy

yzyy

z

y

EIEI

EIEI

M

M

EAN

In the bending part however we do see a coupling between bending in thexy-andxz-plane.

Compare the above shown relation with the earlier found relation in section 4.3.2 of MECH-

2. The coupling is caused by the non diagonal termEIyzwhich is non zero in case of a non-

symmetrical cross section.

1.2.2.1 Moments

In figure 13a a cross section is shown which is loaded in combined (double) bending and

extension. The momentsMyandMzcan be replaced by a resulting momentMwhich is shown

in figure 13b. The resulting momentMacts in a plane constructed by the beam axis (x-axis)

and the line m.

Figure 13 :Sectional forces.

From the above follows thatMyandMzare the components of a vector. In appendix A theproof is given that a moment is also a first order tensor. The components ofMcan also be

presented with straight arrows in they-z-plane as shown in figure 14.

extension

bending


14/99



The sign convention used here is that for a positive

moment the moment resultantMalways points from

the compressive zone to the tensile zone.

The magnitude of the resultant momentMis :

2

z

2

y MMM +=

The angle between the momentMand they-axis is

defined by:

y

zmtan

M

M=

The vector presentation with single arrow is different from the normal used angular vector

presentation with a double arrow. The momentMcan of course also be represented with the

bent moment arrowas shown in figure 15.

Figure 15 :Two possible presentations for a bending momentMin a cross section.

1.2.2.2 Properties of the constitutive relation for bending

If we only consider the constitutive relation for bending we will use the 22 system of

equations. A few remarks with respect to this system of equations can be made.

=

z

y

zzzy

yzyy

z

y

EIEI

EIEI

M

M

Both the momentMand the curvature are first order tensors. The stiffness matrix whichrelates two first order tensors is therefore a second order tensor and is referred to as the

bending stiffness tensor:

zzzy

yzyy

EIEI

EIEI

The bending stiffness tensoris a symmetrical matrix since: yzzy EIEI =

All known tensor transformation rules for second order tensor can be applied to the bendingstiffness tensorlike Mohrs circle and the transformation rules for coordinate system rotations

and the formulae for the principal values and principal directions.

Figure 14 : Bending moment as

vector in the -z- lane.

compression

tension

compression

tension


15/99



The general relation between the momentM and the curvature is shown in figure 16.

In this figure also the neutral line nhas been

drawn. The line of action kof the curvature isperpendicular to nas mentioned earlier. The

beam curves in a plane which is built by thex-

axis and k. This plane is also referred to as the

plane of curvature. The bending momentMacts

with the normal force in a plane built by thex-

axis and m. The sectional forces therefore act in

thex-m-plane which is therefore also referred to

as the loading plane.

In generalM and will not have the same line ofaction. This results in aplane of curvaturewhich

does not coincide with the loading plane. Bothmoment and curvature only act in the same plane

if the following relation holds:

=

z

y

z

y

M

M

If we substitute this into the constitutive relation we find:

0z

y

zzzy

yzyy

z

y

z

y

zzzy

yzyy

z

y=

=

=

EIEI

EIEI

EIEI

EIEI

M

M

We recognise the eigenvalue problem as was described

in the lecture note parts : Introduction into Continuum

Mechanics. If we apply the second order tensor theory to

this eigenvalue problem we can easily understand that

both first order tensorsMand only coincide if theplane of action coincides with one of the principal

directions.

If we rotate they-z-coordinate system to the principal

coordinate system zy as shown in figure 17, the

constitutive relation in this coordinate system becomes:

=

z

y

zz

yy

z

y

0

0

EI

EI

M

M

As can be seen from the above system both non diagonal terms are zero which is of course by

definition the case for a principal tensor. From this relation we can now also see that both

bending moments yM and zM -are fully uncoupled.

As a check we can look into the relation betweenplane of curvatureand the loading planefor

the principal coordinate system. We therefore check the angles andm k .

Figure 17 : Principal directions for

The bending stiffness

Figure 16 : Presentation of curvature and

moment in they-z-plane.


16/99



These can be found with:

)tan()tan( kyy

zz

yyy

zzz

y

zm

EI

EI

EI

EI

M

M===

These direction in deed only coincides if:

a) ;0== km M and act along the y -axis (a principal axis )

b) ;2 == km M and act along the z -axis (the other principal axis )

c) ;zzyy EIEI = km =

In the last case all directions are principal directions and Mohrs circle is represented by a

single dot (check this your self) !

1.2.3 Equilibrium conditionsAfter the kinematic and

constitutive relations the

equilibrium conditions rest to be

investigated (see figure 5). The

equilibrium conditions have

been formulated earlier in

section 11.2 of MECH-1 and in

4.3.3 of MECH-2. In case of a

non-symmetrical cross sections

it is important to describe both

the loading in thex-z-plane andthex-y-plane. Moments and

shear forces can be depicted

from figure 18. Shear forces

have components in both they-

andz-direction.

The equilibrium equations can be derived as described in MECH-1 and MECH-2 with:

z

z

z

z

z

z

y

y

y

y

y

y

x

qx

MV

x

Mq

x

V

qx

MV

x

Mq

x

V

qx

N

===+

===+

=+

2

2

2

2

d

d0

d

den0

d

d

d

d0

d

den0

d

d

0d

d

This results in three equilibrium conditions to relate the loads to the sectional forces in case of

(combined) bending and extension:

2 2

2 2

d0; ( )

d

d d; ( )

d d

x

y zy z

Nq extension

x

M Mq q bending

x x

+ =

= =

y

z

x

Vz

VyMz

My

qz

qy

N

N+dN

Vz+dVz

Vy+dVy

Mz+dMz

My+dMy

dx

Figure 18 :Equilibrium of a beam segment.

NOTE:Pay attention to the way the shear

force components are reduced

from the equations.


17/99



1.2.4 Differential Equations

With the found kinematic, constitutive- and equilibrium relationsit is possible to describe the

behaviour of a prismatic bar with a unsymmetrical and/or inhomogeneous cross section in the

displacements x y z, andu u u of the bar axis.

Kinematics:

d'

d

xx

uu

x = = ;

2

2

d"

d

y

y y

uu

x = = ;

2

2

d"

d

zz z

uu

x = =

Constitutive relations:

=

z

y

zzzy

yzyy

z

y

0

0

00

EIEI

EIEI

EA

M

M

N

Equilibrium relations:

d

dx

Nq

x=

2

2

d

d

y

y

Mq

x=

2

2

d

d

zz

Mq

x=

After substitution of these relations, three differential equations occur expressed in thedisplacements of the bar axis in the directions of the coordinate system:

"x xEAu q = extension

'''' ''''

'''' ''''

yy y yz z y

yz y zz z z

EI u EI u q

EI u EI u q

+ =

+ = (double) bending

Extension is uncoupled from the two differential equations for bending. The latter two

equations for bending are coupled. However we can rewrite them as two uncoupled

equations2. Thus resulting in three ordinary differential equations to describe the behaviour of

the bar axis:

2

2

"

"''

"''

xx

zz y yz z

y

yy zz yz

yy z yz y

z

yy zz yz

qu

EA

EI q EI qu

EI EI EI

EI q EI qu

EI EI EI

=

=

=

If the boundary conditions are specified we can find the displacement field of a bar for a

certain field in the usual way. Check for yourself the resulting differential equations if the

coordinate system coincides with the principal coordinate system of the cross section.

2Within the boundary conditions a coupling however may occur, see also APPENDIXB.


18/99



1.2.5 Example 1 : Homogeneous non-symmetrical cross section

With an example the described theory will be illustrated. In figure 19 two cross sections are

drawn with given neutral lines nl. The position of the neutral line depends on the load

condition of the cross section. Given the position of the neutral line the question is to find for

both cross sections the loading plane m.

Figure 19 :Cross sections with given neutral lines.

Cross section of figure 19a:

The relation between the neutral line and the plane of curvature is given as:

1. Neutral line : 0),( zy =++= zyzy

2. Plane of curvature is perpendicular to the neutral linenl.

The neutral line goes through the normal centre NC, the strain in a fiber which coincides

with thex-axis is therefore zero. From this we can conclude that also the normal forceNiszero. The first condition thus becomes:

zy0 zy +=

The slope of the neutral line is also given with a magnitude of 30 degrees. From this follows:

===

=

33tan

120

y

zk

o

k

The actual curvature is shown in figure 20.

The components of the bending momentMcan befound with:

=

z

y

zzzy

yzyy

z

y

EIEI

EIEI

M

M

For this square cross section holds:

0;)2( yz4

344

121

zzyy ==== IaaII

After substituting these values we find:

=

3

1

10

014

3

4

z

y

EaM

M

z-as

y-as

2a

2a

30o

(a)

z-as

y-as2a

2a

nl

nl

nlnl

(b)

z-as

y-as

2a

30o

Figure 20 :Curvature and neutral line

nl

nl

k

k

k

2a


19/99



The position of the loading plane mcan be determined with :

3)(

3tan

4

34

4

34

y

zm =

==

Ea

Ea

M

M

From this result we can conclude that the plane of curvature kcoincides with the loading

plane m. This is in agreement with the earlier made remarks in section 1.4.2 in which we

found that mand kcoincides when the coordinate system coincides with the principal

directions of the cross section or if the cross section has equal principal values which is the

case for this cross section.

Cross section of figure 19b:

We follow the same approach with the given

neutral line as in the previous example. The plane

of curvature kis found with:

0tan

90

===

=

y

zk

o

k

The actual curvature is shown in figure 21.The components of the bending momentMcan be

found with:

=

z

y

zzzy

yzyy

z

y

EIEI

EIEI

M

M

The moment of inertia of this triangular cross section can be found with the given formulae of

MECH-2:

4

92

43

yz

4

944

3613

361

zzyy

2

)2(

36

1

tan36

)2(

aabh

I

aabhII

===

====

Substituting these values results in:

=

=

420

4224 4

914

91

z

y EaEaMM

The loading plane mthus becomes:

22

4tan

4

91

4

91

y

zm =

==

Ea

Ea

M

M

In figure 22 this result is shown. In this example

the plane of curvature kdoes not coincide with the

loading plane m. Be aware that the angular difference between kand mis not due to torsion

but simply due to double bendingin case of a non-symmetrical cross section.

z-as

y-as

2a

2a

nlnl

k

k

k

Figure 21 :Curvature and neutral line.

z-as

y-as

2a

2a

nlnl

k

k

k

Figure 22:Curvature and loading

m

m

m


20/99



Cross section (b) from the previous example with a Youngs modulusEis used in the

clamped beam as is shown in the figure below. The structure is statically indeterminate,

deformation due to extension is neglected.

Figure 19-2 :Clamped beam with cross section (b)

The beam is loaded with a constant distributed load qz. The deflection of the beam is askedfor as well as the moment distribution in both thexy-andxz-plane. The bending stiffness

tensor of the cross section was earlier found as:

41, 9

4 2

2 4i j

EI Ea

=

Since no normal force acts in the cross section the deformation can be described with the two

following differential equations in which we take only into account the load qz.:

2 2 4

2 2 4

3

"'' 2

3"''

zz y yz z yz z z

y

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 q

u EI EI EI EI EI EI Ea

EI q EI q EI q qu

EI EI EI EI EI EI Ea

= = =

= = =

The general solution for the displacement field in both they- andz-direction becomes:4

2 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

= + + +

= + + + +

The eight boundary conditions can be described as:

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

= = = = =

= = = = =

From these the following integration constants can be obtained, see also the MAPLEinput on

the next page:

2

1 2 3 44 4

2

1 2 3 44 4

3 50 0

32 32

3 50 0

16 16

z z

z z

q l q lC C C C

Ea Ea

q l q lD D D D

Ea Ea

= = = =

= = = =

10 m x-as

qz= 8 kN/m

z-as

y-as 2a

2a

cross section

E=100 GPa

a = 0,1m


21/99



The MAPLEinput sheet is given below.

The moment distribution and the displacement field is shown below. Due to the load in z-

direction and the imposed boundary conditions the moment distribution in thexy-plane

becomes zero. The maximum moment at the clamped end is indeed 0,125

q

l

2

= 100 kNm.Due to the unsymmetrical cross section the member will deflect in both thexy-plane and the

xz-plane.

Figure 19-3 :Results forMand ufor a cross section of type (b)

Remark:

Although the used differential equations seem to be uncoupled a coupling may exist in the

boundary conditions. In particular the dynamic boundary conditions contain a coupling:

'' ''; '

'' ''; '

y yy y yz z yy y yz z y y

z yz y zz z yz y zz z z z

M EI EI EI u EI u V M

M EI EI EI u EI u V M

= + = =

= + = =

Special care should be given to the specified boundary conditions, see APPENDIXB.

> restart;

> EIyy:=(4/9)*E*a^4; EIyz:=(1/2)*EIyy; EIzz:=EIyy;

> uy:=C1+C2*x+C3*x^2+C4*x^3-qz*x^4/(16*E*a^4); phiy:=diff(uy,x): kappay:=-diff(phiy,x):

> uz:=D1+D2*x+D3*x^2+D4*x^3+qz*x^4/(8*E*a^4); phiz:=-diff(uz,x): kappaz:=diff(phiz,x):

>My:=EIyy*kappay+EIyz*kappaz: Mz:=EIyz*kappay+EIzz*kappaz:

> x:=0; eq1:=uy=0; eq2:=uz=0; eq3:=phiy=0; eq4:=phiz=0;

> x:=L; eq5:=uy=0; eq6:=uz=0; eq7:=My=0; eq8:=Mz=0;

> sol:=solve({eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8},{C1,C2,C3,C4,D1,D2,D3,D4}); assign(sol);

> x:='x':

> qz:=8;L:=10; E:=100e6; a:=0.1;

>plot([-My,-Mz],x=0..L,title="Moment My and Mz",legend=["My","Mz"]);

>plot([uy,uz],x=0..L,title="Displacements uy and uz",legend=["uy","uz"]);

M[kNm]

u

[m]

MomentMyandMz

Deflection uyand uz


22/99



1.2.6 Normal stresses in the y-z-coordinate system

If for a cross section the sectional forcesN,MyandMzare known, the sectional deformations

can be found with the constitutive relations:

( )

=

=

=

=

z

y

yyzy

yzzz

2

yzzzyyz

y

z

y

zzzy

yzyy

z

y 1

M

M

EIEI

EIEI

EIEIEI

EA

N

EIEI

EIEI

M

M

EAN

The stress in a fiber can be found with the earlier found relation:

zy),(),( zyEzyEzy ++==

In general it is of little use to elaborate this relation. For two cases however it is illustrative to

simplify the found relation.

Situation 1 : The y- and z-axis coincides with the principal axis of the cross section

If they-z-coordinate system coincides with the principal axis of the cross section the three

sectional forces are fully uncoupled. The sectional deformation quantities can be found as:

zz

zzzzz

yy

y

yyyyy

EI

MEIM

EI

MEIM

EA

NEAN

z ==

==

==

The stress distribution follows from:

zz

z

yy

y

zy

),(

),(),(

I

zM

I

yM

A

Nzy

zyEzyEzy

++=

++==

Situation 2 : The y- and z-coordinate system is thus positioned that one of the curvatures is

zero.In this situation, one of the curvature components is zero e.g. they-component. We then find:

y yz y z

z zz

with: 0 and

N EA

M EI

M EI

=

= = =

=

The stress distribution follows from:

( )zz

z0),(),(I

zM

A

NzyEzyEzy +=++==

Note that the componentMydoes not show up in the above relation for the determination ofthe stress although the curvature yiszero and the momentMyis notzero!


23/99



1.2.7 Normal stresses in the principal coordinate system

In the previous section the stresses in the cross section were found based on the definedy-z-

coordinate system. The disadvantage of this method is the coupling between the bending

components. The advantage is the straight forward method. According to the authors this

method is to be preferred. However in the engineering practice a different method is being

used. In most cases the method of the principal coordinate system is chosen. They-z-

coordinate system is rotated to the principal zy - coordinate system which coincides with

the principal directions of the cross section. In section 1.2.2.2 we proved that the constitutive

relations for bending are uncoupled if the principal coordinate system is used since the non

diagonal terms zyEI are zero by definition when using the principal directions as coordinate

system.

=

z

y

zz

yy

z

y

0

0

EI

EI

M

M

Using this approach leads in fact tot the previous outlined situation 1. The formula to compute

the stress from a cross section loaded in combined bending and extension becomes:

zz

z

yy

y),(

I

zM

I

yM

A

Nzy ++=

The advantage of a very simple and easy to memorise formula is however small. Additional

work has to be done since all quantities used have to be referred to the rotated coordinate

system:

a) First of all the principal directions of the cross section have to be determined. We canuse Mohrs circle or the transformation formulas for this.b) Then the bending moment components iny- andz- direction have to be decomposed

into the principal directions.

c) In order to find the stresses in e.g. the outer fibers all the distances of this fibers haveto be transformed into the principal coordinate system.

d) All deformations and related displacements found are with respect to the principaldirections. In order to find the displacements in the originaly-zcoordinate system the

results have to be transformed from the principal directions back to the original

coordinate system.

To illustrate the difference between both methods an example will be shown in which bothmethods are used.

NOTE:All quantities with respect to the

principal coordinate system.


24/99



1.2.8 Example 2 : Stresses in non-symmetrical cross sections

In figure 23a a simply supported beam is shown loaded with a concentrated loads at mid span.

We assume a zero normal force and also assume that the concentrated loads are applied in

such a way that no torsion occurs. In the last section of these lecture notes we will show that

this can be achieved if the load is applied in the shear force centre. A introduction in to this

topic can also be found in section 5.5 of MECH-2. We also assume sufficient rigidity of the

cross section to avoid local instability like plate buckling.

Figure 23 :Structure loaded in bending

For the structure loaded in bending the normal stress distribution at mid span is asked for. We

also want to know the displacements iny- andz-direction at mid span. In this example we will

use both previous outlined methods. We start with using they-zcoordinate system.

Method 1 : Computation in the y-z-coordinate systemThe load causes at mid span bending moments in both thex-yandx-zplane:

Nmm101012515002700025,0

Nmm1033751500900025,0

3

z41

z

3

y41

y

===

===

lFM

lFM

These components will act in the cross sectional NC. The position of the NC and the moments

of inertia can be obtained as is described in chapter 3 of MECH-2:

75 10 5 150 10 7551,67 mm

75 10 150 10

75 10 37,5 150 10 515,83 mm

75 10 150 10

upper side

right side

NC

NC

+ = =

+

+ = =

+

These distances are shown in figure 23b.

The moment of inertia with respect to they-z-coordinate system which has been chosen by

definition with its origin at the normal centre NC can be found as:

3 2 3 21 1yy 12 12

yz

3 2 3 21 1zz 12 12

10 75 75 10 (37,5 15,83) 150 10 150 10 (15,83 5)

75 10 (37,5 15,83) (5 51,67) 150 10 (5 15,83) (75 51,67)

75 10 75 10 ( 51,67 5) 10 150 150 10 (75 51,67)

I

I

I

= + + +

= +

= + + + +

750 mm

750 mm

27 kN

9 kN

75 mm

150mm

Fz=27 kN

Fy=9 kN

y-as

z-as

NC

x-as

15,83 mm

51,6

7

mm

(a) Loaded structure

(b) Cross section

t= 10 mm

E=210000 N/mm2


25/99



From this follows:

44

zz

44

yz

44

yy mm109,526;mm1075,113;mm102,89 === III

The components of the curvature in the y-z coordinate system follow from the constitutive

relation:

( )

=

=

z

y

yyzy

yzzz

2

yzzzyyz

y

z

y

zzzy

yzyy

z

y 1

M

M

EIEI

EIEI

EIEIEIEIEI

EIEI

M

M

Using the quantities found:

( )

=

3

3

2

4

z

y

1010125

103375

2,8975,113

75,1139,526

75,1139,5262,89

101

E

results in the following components of the curvature:

1/mm1099,171/mm1095,40 66 == zy

The stress in any point of the cross section can be found with:

zy),(),( zyEzyEzy ++==

Since there acts no normal force at the cross section, the stain in the fiber which coincideswith thex-axis (beam axis trough the NC) is also zero. This results in:

zyzy += 778,3600,8),(

The neutral line (points of zero strain) follows from:

0778,3600,8 =+ zy

For a number of key-points (A,B,C,D,E) the

stress can be computed. The results can be

found in the table below. The key-points

and their positions can be found in figure

24.

point y mm z mm N/mm2

A 59,17 -51,67 313,6B -15,83 -51,67 -331,4

C 59,17 -41,67 351,4

D -5,83 98,33 321,4

E -15,83 98,33 235,4

In figure 24 the stress distribution and the

neutral line are also shown. Most important

are the points C and B with the largest

(perpendicular) distance towards the neutral

axis. These point therefore have the largest

compressive and tensile strain and thus the

largest stresses.

With the known moment distribution it is possible to calculate the structural deformation.

75

150

y-as

z-as

NC

15,83

51,6

7

t= 10

A B

E

59,17

98,3

3length in mm

nl

nl

C

D

+

-

351,4 N/mm2

331,4 N/mm2

Figure 24 :Stress distribution.


26/99



Up to now we are used to calculate the displacements using the engineeringforget-me-not

formulae. However in case of non-symmetrical cross sections we hit a problem using this

formulae.

The practical method using the engineering formulae is based on uncoupled situations. In the uncoupled

situation the curvature can be found directly from the moment distribution:

EI

M=

This curvature is also known as the reduced moment distribution. In figure 25 the calculation of the

maximum displacement based on the curvature of a simple cantilever beam is used to refresh yourmemory.

Figure 25 :Basis forforget-me-notformulae

In case of a non-symmetrical cross section the reduced moment distributionMz/EIzzis not the curvature

component z. The correct curvature components should be obtained from the constitutive relation:

( )

=

z

y

yyzy

yzzz

2

yzzzyyz

y 1

M

M

EIEI

EIEI

EIEIEI

Only in case the coordinate system coincides with the principal directions of the cross section, we can

use the engineering formulae. For all other cases we should use the complete constitutive relation andcalculate the components of the curvature distribution.

Figure 26 shows the curvature distribution of the non-symmetrical beam. With the paradigma

of the curvature plane as described in section 8.4 in MECH-2, the displacements in they- en

z-direction can be found.

Figure 26 :Curvature and displacements in they-z-coordinate system.

With the boundary condition of zero displacements uyen uzat B, the rotations yen zat Acan be found. Subsequently the displacements at C can be found with the standard procedure:

mm37,30067,00

mm68,70154,00

61

421

)A()C()A(21

3)A(

61

221

)A()C()A(21

1)A(

====

====

llull

llull

yzyy

zyzz

F

l

zz

z

z EI

M=

EI

Fl

l32

= lw 32

EI

Fllw

EI

Fl

EI

Fll

3

23

32

2

21

==

==

x-axis

)A(z

x-as

l

y-as

1

1/mm1095,40 6=y

x-as

l

z-as

3

1/mm1099,17 6=z

A AB B

x-as

l21

y-as1/mm1095,40 6=y

A B2

)A(z

)A(y

x-as

l21

z-as1/mm1099,17 6=y

A B4

)A(y

l61

l61

engineering formula


27/99



Method 2: Computation in the principal coordinate system

The stresses in the key-points A to E can also be computed with the method based on the

coordinate system which coincides with the principal directions of the non-symmetrical cross

section. This method requires of course the principal directions which can be found using the

standard second order tensor formulae for the principal values and directions or by using

Mohrs graphical method. Both ways will be illustrated .

The moments of inertia in they-z-coordinate system were already found with method 1:

44

zz

44

yz

44

yy mm109,526;mm1075,113;mm102,89 === III

Using the second order tensor transformation rules the principal moments of inertia can be

found:

( )

44

2

44

1

2

yz

2

zzyyzzyy

2,1

mm1042,61;mm1067,554

2

4

2

==

+

+=

II

IIIIII

The principal directions can be found with:

yz o o

1,2

yy zz

2tan 2 13,73 ; 283,73

I

I I = =

The same result can be found by using Mohrs graphical method which is shown in figure 27.

(see the lecture notes : Introduction in to Continuum Mechanics).

Figure 27 :Mohrs circle for moments of inertia

The directions found can be presented within the drawing of the cross section which is shownin figure 28.

zz

yy

I

I

zyI

yzI

2I 1I

44 mm1025

44 mm1025

m

r

);( yzyy II

);( zyzz II

z

y

R.C.

(1)

(2)

o73,13

o73,283


28/99



Figure 28 :Principal directions of the cross section.

The stresses in the key-points can be found with the earlier determined stress formula with

respect to the principal 1-2-coordinate system. The principal axis are denoted with (1) and (2):

2

22

1

11)2,1(I

eM

I

eM+=

The position of the key-point in the principal coordinate system requires some calculus. The

perpendicular distances from the key-point to the coordinate axis can be found with the first

order tensor transformation rules. The distances in the 1-2-coordinate system are denoted with

the excentricities e1and e2. The calculation for point A is shown in figure 28.

The sectional moment also has to be transformed to the principal 1-2-coordinate system. With

the same first order tensor transformation rules the components of the sectional moment in the

1-2-direction become:

Nmm1017,5682)73,13sin()73,13cos(

Nmm1033,9034)73,13cos()73,13sin(

3

2

3

1

=+=

=+=

zy

zy

MMM

MMM

All the data for the key-points are summarised in the table below. With the above stress

formula the stress in the particular key-point can be obtained. Check the data and pay special

attention to the signs of e1and e2.

point y mm z mm e1 mm e2 mm N/mm2

A 59,17 -51,67 -64,24 45,21 313,6

B -15,83 -51,67 -46,43 -27,64 -331,4

C 59,17 -41,67 -54,53 47,59 351,4

D -5,83 98,33 96,90 17,68 321,4

E -15,83 98,33 99,28 7,97 235,4

The results match with those found with method 1.

In the principal directions, bending in the x-y-andx-z-plane are uncoupled. Therefore the

displacements in the 1-2-directions can be found with the engineeringforget-me-notformulae.

The concentrated loads which are used in the engineering formulae for the displacements alsohave to be transformed to the principal coordinate system using the first order tensor

transformation rules:

75

150

y-as

z-as

NC

15,83

51,6

7

t= 10

A B

E

length in mm

(1)

C

D

(2)

64,24

45,21

=

=

24,64

21,45

67,51

17,59

)73,13cos()73,13sin(

)73,13sin()73,13cos(

1

2

1

2

A

A

e

e

e

e


29/99



N15152,46)73,13sin()73,13cos(

N24091,55)73,13cos()73,13sin(

2

1

=+=

=+=

zy

zy

FFF

FFF

The displacements u1and u2in the principal directions can now be found with the engineering

formulae:

mm26012,81042,61101,248

150046,15152

48

mm4542,11067,554101,248

150055,24091

48

45

3

2

3

22

45

3

1

3

11

=

==

=

==

EI

lFu

EI

lFu

The displacements in the 1-2-directions have to be transformed to the originaly-z-directions

using the first order tensor transformation rules. This results in:

mm37,3)73,13sin()73,13cos(

mm68,7)73,13cos()73,13sin(

21)(

21)(

==

=+=

uuu

uuu

Cz

Cy

The total displacement of point C becomes:

mm4,837,368,722

=+=Cu

These results are in perfect agreement with those found with method 1. In figure 29 the

deformed structure is shown.

Figure 29 : Deformed structure

Concluding remarks

By using the method of the principal coordinate system more tensor transformations are

required than when using the original y-z-coordinate system. Also by using the method of the

principal directions the location of the neutral line can not easily be found. However the

equations used are quite easy to remember.

Since in todays engineering firms most of these calculations are done with help of computer

programs or spreadsheet calculations, the number of transformations is not very relevant any

more. The authors however advise the method in they-z-coordinate system since this method

automatically handles all possible situations and thus reducing the occurrence of erroneous

answers and overlooking limitations.

750 mm

750 mm

27 kN

9 kN

x-as

E=210000 N/mm27,68 mm

3,37 mm

y-as

z-as

Assignment

Make a sketch of the cross section

and show:

the plane of curvature

the loading plane

the neutral line

the principal directions


30/99



1.3 Extension of the theory for inhomogeneous cross sections

If the cross section is not made out of one single (homogeneous) material the cross section is

considered to be an inhomogeneous cross section. In the fiber model used the inhomogeneous

character can be implemented by using a function for the Youngs modulus or modulus of

elasticityE. The modulus of elasticity may vary between fibers based on material used foreach fiber or part of the cross section. The Youngs function used is denoted as:

),( zyE

Since the Youngs modulus only appears in the constitutive relations used in the model, only

this part of the model has to be extended.

The constitutive relation which relates the stresses tot the strains has to be modified slightly:

),(),(),( zyzyEzy =

Since the kinematic relation will not change the three cross sectional deformation quantitieszy ,, can still be used to describe the strain distribution of the cross section. This

strain distribution still appears to be a plane from which follows that cross sections remain

plane also after deformation.

For each fiber (y,z) of the cross section the strain can be determined with:

zy),( zyzy ++=

The stress in a specific fiber becomes in case of linear elasticity (Hookes law):

zy),(),( zyzyEzy ++=

The stress distribution will not be a linear distributed function. This will have consequences

for the evaluation of the integrals used to calculate the sectional forces.

The normal forceNcan be found with (see section 1.2.2):

( )

++=

++==

A AA

AA

zdAzyEydAzyEdAzyE

dAzyzyEdAzyN

),(),(),(

),(),(

zy

zy

The expressions for the components of the bending momentMyandMzyield:

( )

( )

++=

++==

++=

++==

A AA

AA

A AA

AA

dAzzyEyzdAzyEzdAzyE

zdAzyzyEdAzyzM

yzdAzyEdAyzyEydAzyE

ydAzyzyEdAzyyM

2

zy

zyz

z

2

y

zyy

),(),(),(

),(),(

),(),(),(

),(),(

As can be seen from these expressions, the Youngs functionE(y,z) remains under the

integral.

NOTE:

We will still consider a linear elastic stress-

strain relation for each fiber.


31/99



In order to obtain expressions which can be handled we will introduce new cross sectional

quantities which will be denoted with so-called double letter symbols:

zz

2

zyyz

yy

2

y

),(

),(),(

),(),(),(

EIdAzzyE

EIEIyzdAzyEESzdAzyE

EIdAyzyEESydAzyEEAdAzyE

A

A

z

A

AAA

=

===

===

The expressions found on the previous page can now be rewritten using these double letter

symbols. The cross sectional constitutive relation which relates the sectional forces (N,Myand

Mz) to the sectional deformations (, yand z) thus becomes:

zzzyzyzz

zyzyyyyy

zzyy

EIEIESM

EIEIESM

ESESEAN

++=

++=

++=

In matrix notation this constitutive relation becomes:

=

z

y

zzzyz

yzyyy

zy

z

y

EIEIES

EIEIES

ESESEA

M

M

N

When comparing this result with the earlier found result of section 1.2.2 we hardly observe

any difference. However the double letter symbolsrepresent the evaluation of an (extensive)

integral where as the symbols used in section 1.2.2 are the product of two quantities e.g.EArepresentsEtimesA. For inhomogeneous situations the double letter symbolEArepresents:

=A

dAzyEEA ),(

If we chose the origin of the coordinate system at the normal centre (NC) of the cross section

the coupling terms between extension and bending will vanish since these become zero due

tot the definition of the NC:

0),(

0),(

z

y

==

==

A

A

zdAzyEES

ydAzyEES

with respect to the coordinate system chosen at the NC

The constitutive relation can now be simplified to:

=

z

y

zzzy

yzyy

z

y

0

0

00

EIEI

EIEI

EA

M

M

N

basic formula (1)

Again we observe that bending and extension are uncoupled if we chose the origin of the

coordinate system at the NC.


32/99



In order to find the stresses in a cross section we first have to find with basic formula 1 the

cross sectional deformations. Subsequently we can find the strain of a specific fiber with:

zy),( zyzy ++= basic formula (2)

The stress in a specific fiber can be found with:

),(),(),( zyzyEzy = basic formula (3)

This strategy is summarised in the scheme of figure 30.

Figure 30 :Calculation scheme.

Since the modulus of elasticity may vary across the cross section, the stress distribution will

notbe congruous to the strain distribution. A single simplified expression for the stress can

therefore not be found for inhomogeneous situations.

- localise the NC - compute

EA,EIyy,EIzz,EIyz

- determine the cross sectional

forcesN,MyandMz

- calculate the cross sectional

deformations ),,( zy

- find the strain distribution

- find the stress distribution

basic formula 1

basis formula 2

basis formula 3


33/99



1.3.1 Position of the NC for inhomogeneous cross sections

The special location of the origin of the coordinate system for which the coupling terms (ESy

andESz) between extension and bending become zero is by definition called the normal force

centre NC. The result of this definition is that an axial forceNwhich acts at the NC only

causes straining and no bending. The coupling terms are also referred to as weighted static

momentsor weighted first order moments. To find the position of the coordinate system for

which the coupling terms become zero requires a tool.

Figure 31 :Weighted Static Momentof the cross section.

Assume a temporary zy -coordinate system from which the position is known and shown in

figure 31. The shift in origin of this coordinate system with respect to they-z-coordinate

system through the unknown NC is denoted with NCy and NCz . The temporary coordinate

system can be expressed in terms of they-z-coordinate system as:

NCNCzzzyyy +=+=

For the weighted static momentswith respect to the zy -coordinate system holds:

NCzNCz

NCyNCy

),(),(),(

),(),(),(

zEAESdAzyEzzdAzyEdAzzyEES

yEAESdAzyEyydAzyEdAyzyEES

AAA

AAA

+=+==

+=+==

By definition the weighted static momentswith respect to they-z-coordinate system are zero:

0;0 zy == ESES

From which the unknown position of the NC with respect to the known position of the zy -

coordinate system can be found:

EA

ESz

EA

ESy

zNC

y

NC

=

=

The temporary coordinate system is usually chosen along one edge of the cross section. With

an example this tool will be illustrated.

NCz

NCy

y

z

y

z

NC

dAzyE ),(


34/99



1.3.2 Example 3 : Normal centre versus centroid

A rectangular cross section is a composite of two materials 1 and 2. Both materials have the

same mass densities but have different Youngs moduli.

Figure 32 :Rectangular inhomogeneous cross section.

Consider a temporary zy -coordinate system at the upper right corner of the cross section.

The position of the normal force centre(NC) with respect to the chosen zy -coordinate

system can be found with the outlined method of the previous section using the double letter

symbolsfor inhomogeneous cross sections. We do not need integral calculus here since foreach material simple geometrical shapes can be recognized.

Vertical:

( ) ( )

( ) ( )

( )a

aE

aaE

aaEaaE

aaaEaaaE

EAEA

aEAaEA

EA

ESz z

65

2

33

21

21

221

1

21

221

1

NC 19

16

2

222=

+=

+

+=

+

+==

Horizontal:

( ) ( )

( ) ( )

( )a

aE

aaE

aaEaaE

aaaEaaaE

EAEA

aEAaEA

EA

ESy

y

2

1

2

33

21

21

21

221

1

21

21

221

1

NC 9

4

2

2=

+=

+

+=

+

+==

Since both materials have the same mass densities the centre of gravity of the cross section is

the centroid of the geometry which position with respect to the zy -coordinate system can

be found as:

NC63

ZW

21

ZW

1 zaz

ay

=

=

In case of inhomogeneous cross sections we can conclude that the normal force centre NC

does not necessarily coincides with the centroid of the cross section. This is a vital aspectwhen encountering inhomogeneous cross sections.

a

a

2a

Material 1 : EE =1

Material 2 : EE 42 =

z

y NC

yNC

zNC

y

z

(EA)1

(EA)2


35/99



1.3.3 Example 4 : Stresses in inhomogeneous cross sections

Consider a cantilever beam with a inhomogeneous cross section which consists of three parts

which are firmly glued together. The beam is loaded with a point load of 250 N at C. The

cross section is built out of two different materials as is shown in figure 33.

(a) : Loaded structure (b) : Cross section

Youngs moduli:E1= 6000 N/mm2E2= 12000 N/mm

2

Figure 33 :Inhomogeneous and non-symmetrical cross section.

From this loaded structure the normal stress distribution at the clamped support is requested.

For this a number of key points at the cross section are presented in figure 33 (b).

We will use the solution technique as given in the scheme of figure 30. For the cross sectional

quantities denoted with the so-called double letter symbolswe first need to find the location

of the normal centerNC.

The given cross section has rotational

symmetryaround the centroid of material 1.

The normal center NC of the total cross

section coincides therefore with this point.

In figure34 the sectional force distribution inthe structure is represented with the V- andM-

distribution. At the clamped support the

sectional forces are:

Nmm137500

N250

=

=

z

z

M

V

Figure 34 :Force distribution in beam AB.

250 N

0,55 m 0,55 m

x-axis

z-axis

A

BC

250 N

V-diagram

M-diagram

137500 Nmm

250 N

0,55 m 0,55 m

x-axis

z-axis

A

B

y-axis

E1

E2

10 mm

30 mm

20 mm

z-axisC

E2

10 mm

50 mm

50 mm

O P

Q S

T V

W

X

R

U


36/99



The sectional stiffness quantities represented with the double letter symbolscan be found

using the fact that the cross section is composed of simple geometrical elements from which

the local centroids are known. The distances inx- andy-direction between these local

centroids and the NC of the total cross section are given in figure 35.

Figure 35 :Distances between local centroids and the NC of the total cross section.

For the three rectangles of figure 35 the distances to the NC of the total cross section in they-

z-coordinate system are summarized:

Part Y z

Upper flange +15 mm -20 mm

Web 0 mm 0 mm

Lower flange -15 mm 20 mm

The double letter symbol quantities thus become:

( ) ( )

( ) ( ) 2923121

2

3

121

1zz

29

22yz

2923

121

2

3

121

1yy

Nmm1017,5201050105023020

Nmm1060,3)20)15(5010())20(155010(

Nmm1032,5155010501022030

=++=

=+=

=++=

EEEI

EEEI

EEEI

With basic formula 1 the curvatures can be obtained:

1-6

-16

9

mm1029,50

mm1003,34

17,56,3

6,332,510

137500

0

=

=

=

z

y

z

y

The strain distribution over the cross section is found with basic formula 2:

zy),( zyzy ++= basic formula (2)

web

upper and lower flange

upper flange lower flange

upper and lower flangeweb

y-as

E1

E2

10 mm

30 mm

20 mm

z-as

E2

10 mm

50 mm

50 mm

20 mm

20 mm

15 mm

NC

local NC of

the upper flange

local NC of

the lower flange

local NC of the

web (coincides

with the NC of the

total cross section)

NOTE :

The distance is taken from the NC of the

total cross section to the local NC of the

parts.


37/99



The stress at any point can be found by multiplying the strain of the considered fiber with its

Youngs moudulus:

),(),(),( zyzyEzy = basic formula (3)

Using a spread sheet like EXCELspeeds up the calculus for each key point as can be seenfrom the next table.

Units N, mm

Punt y [mm] z [mm] E-modulus [N/mm2] Strain [*10-3] Stress [N/mm2]

O 40 -25 12000 -0,10 -1,25

P -10 -25 12000 1,60 19,17

Q 40 -15 12000 -0,61 -7,28Material2

S -10 -15 12000 1,09 13,14

R 10 -15 6000 0,41 2,48

S -10 -15 6000 1,09 6,57

T 10 15 6000 -1,09 -6,57Material1

U -10 15 6000 -0,41 -2,48

T 10 15 12000 -1,09 -13,14

V -40 15 12000 0,61 7,28

W 10 25 12000 -1,60 -19,17Material2

X -40 25 12000 0,10 1,25

For the four points R,S,T and U two stress results are possible. Depending on the material

considered either the stress in material 1 or material 2 is found by multiplying the strain at

these points with the corresponding Youngs modulus of material 1 or material 2. The result

will be a jump in the stress distribution at these points. For point S this is shown in bold in the

table above. The stress results can also be presented graphically. To draw the stressdistribution it is important to know the position of the neutral line nl. The expression for the

neutral linebecomes in this case:

029,5003,34

01029,501003,340),( 66

=

==

zy

zyzy

Since the normal force is zero the strain at the NC is also zero. This results in a neutral line

which goes through the NC which is the origin of the coordinate system. In the graph on the

next page the neutral line is drawn, perpendicular to this line the plane of curvature kis also

shown.

As is usual, the stress distribution is drawn perpendicular to the neutral line. To avoid any

ambiguity it is advised to draw the stress distribution outside of the cross section in a separate

drawing.


38/99



In figure 36 the stresses in material 1 and 2 are presented respectively with the green and red

graphs.

Figure 36 :Neutral line, curvature and stresses in material 1 en 2.

1.4 Force point of the cross section

The expression for the strain which is used up to now is:

zyzy zy),( ++=

This expression depends on the loading. For zero normal forceN, the normal strain (at theNC) is also zero and the neutral line passes through the NC as was shown in the previous

example. For a non-zero normal forceNthe neutral line will not pass through the NC!

The expression of the strain can also be expressed in terms of the sectional forcesN,

MyandMz. We then need to substitute the constitutive relation for the cross section into the

original expression for the strain distribution. We use the expression found in section 1.2.5:

zyz2

yzzzyy

yy

y2

yzzzyy

yz

z

zyz2

yzzzyy

yz

y2

yzzzyy

zzy

MMM

EIEIEI

EIM

EIEIEI

EI

MMMEIEIEI

EIM

EIEIEI

EI

EA

N

zzyz

yzyy

+=

+

=

=

=

=

y-axis

E1

E2

10 mm

30 mm

20 mm

z-axis

E2

10 mm

50 mm

NC

n.l

n.l

k

k

19,17 N/mm2

19,17 N/mm

Material 2

6,57 N/mm

+

+-

-

Material 1


39/99



The strain distribution can thus be found with:

( ) ( ) zMMyMMEA

Nzy +++= zzzyyzzyzyyy),(

If the normal forceNis non-zero the moments in thexy-andyz-plane can also be expressed inthe eccentric applied normal forceN:

zz

yy

eNM

eNM

=

=

The three sectional forcesN,MyandMzare thus replaced with one single eccentric normal

forceNwhich acts in a point at location (ey, ez) towards the NC. This point is referred to as

theforce point. The expression for the strain can now be elaborated as:

( ) ( )

( ) ( )[ ]zeEAeEAyeEAeEAEA

Nzy

zeNeNyeNeNEA

Nzy

+++=

+++=

zzzyyzzyzyyy

zzzyyzzyzyyy

1),(

),(

This last expression shows the strain at a point (y,z) for a normal forceNacting at (ey, ez).

As an experiment of mind we can think of a forceNacting in (y,z) and observing the strain in (ey, ez). It appears to result in exactly the same strain. The experiment is shown in figure 36.

Figure 37 :Maxwells reciprocal theorem

In words we can summarise this phenomenon as:

The strain in P due to a forceNin Q is equal to the strain in Q due to a forceNin P.

This is also known as Maxwells reciprocal theoremand is general applicable to linear

elastic systems for which the superposition theorem holds. We will make use of this theorem

in the next sections.

Special case for force point:

If theyz-coordinate system coincides with the principal directions of the cross section the

expression for the strain can be simplified. In case of principal directions holds:

zz

zz

yy

yyyz

1;

1;0

EIEI

===

P

Q

P

Q

N

N


40/99



The strain can be expressed as:

+

+=

zz

z

yy

y1),(

EI

zeEA

EI

yeEA

EA

Nzy (symmetrical cross section)

After using the definition of the inertia radius i:

EA

EIi

EA

EIi zz

2

z

yy2

y ; == (symmetrical cross section)

We can further simplify the expression for the strain:

+

+=

2

z

z

2

y

y1),(

i

ze

i

ye

EA

Nzy (symmetrical cross section)

This latter result will be used in the next section on the core or kern of a cross section.

1.5 Core or Kern of a cross section

When the neutral line is inside the cross section both tensile and compressive zones will occur

on either side of the neutral line. Some materials however can hardly sustain tensile stresses

e.g. brick walls and unreinforced concrete. In case of these materials, cross sections should be

loaded in such a way that only compression occurs. The neutral line should then be outside

the cross section or just at its boundary. With this requirement the area can be determined in

which the force point should be positioned in order to prevent sign changes in the stresses.

This area is called the coreor kernof the cross section.

In section 4.9 of MECH-2 the core was introduced for a

rectangular cross section with dimensions bhas shown

in figure 38. The core appeared to be a diamond with

corner points at a distance to the NC iny-andz-

direction of respectively b61 and h

61 , see figure 38.

In this section a general method will be outlined to find

the core of inhomogeneous and or non-symmetrical

cross sections.

In the following we will make use of two important

properties which follow from the definition of the core:

o The neutral line never crosses the cross section.o Cross sections with straight edges have a

polygon as core.

The fir

Dictaat UK v9

Documents

Transcript of Dictaat UK v9