MODULE : NON-SYMMETRICAL AND INHOMOGENEOUS...

STRUCTURAL MECHANICS 4 CIE3109

MODULE : NON-SYMMETRICAL AND INHOMOGENEOUS CROSS SECTIONS

COEN HARTSUIJKER HANS WELLEMAN Civil Engineering TU-Delft

October 2017

STRUCTURAL MECHANICS 4 Non-symmetrical and inhomogeneous cross sections

Ir C. Hartsuijker & Ir J.W. Welleman October 2017 ii

TABLE of CONTENTS

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

1.1 SKETCH OF THE PROBLEM AND REQUIRED ASSUMPTIONS .......................................................................... 1

1.2 HOMOGENEOUS CROSS SECTIONS .............................................................................................................. 4

1.2.1 Kinematic relations ......................................................................................................................... 4

1.2.1.1 Curvature .................................................................................................................................................... 6

1.2.1.2 Neutral axis ................................................................................................................................................. 7

1.2.2 Constitutive relations for homogeneous non-symmetrical cross sections ....................................... 8

1.2.2.1 Moments ................................................................................................................................................... 10

1.2.2.2 Properties of the constitutive relation for bending .................................................................................... 11

1.2.3 Equilibrium conditions .................................................................................................................. 13

1.2.4 Differential Equations ................................................................................................................... 14

1.2.5 Example 1 : Homogeneous non-symmetrical cross section .......................................................... 15

1.2.6 Normal stresses in the y-z-coordinate system ............................................................................... 19

1.2.7 Normal stresses in the principal coordinate system ...................................................................... 20

1.2.8 Example 2 : Stresses in non-symmetrical cross sections ............................................................... 21

1.2.9 Concluding remarks ...................................................................................................................... 27

1.3 EXTENSION OF THE THEORY FOR INHOMOGENEOUS CROSS SECTIONS ...................................................... 28

1.3.1 Position of the NC for inhomogeneous cross sections .................................................................. 31

1.3.2 Example 3 : Normal centre versus centroid .................................................................................. 32

1.3.3 Example 4 : Stresses in inhomogeneous cross sections ................................................................ 33

1.4 FORCE POINT OF THE CROSS SECTION ...................................................................................................... 36

1.5 CORE OR KERN OF A CROSS SECTION ....................................................................................................... 38

1.5.1 Example 5 : Core of a non-symmetrical cross section .................................................................. 42

1.6 TEMPERATURE INFLUENCES* .................................................................................................................. 44

1.6.1 Example 6 : Static determinate structure under temperature load ............................................... 47

1.6.2 Example 7 : Static indeterminate structure under temperature load ............................................ 51

1.7 SHEAR STRESS DISTRIBUTION IN ARBITRARY CROSS SECTIONS ................................................................ 55

1.7.1 Shear stress equations for principal coordinate systems .............................................................. 56

1.7.1.1 Example 8 : Shear stresses in a composite cross section .......................................................................... 58

1.7.2 General shear stress formula ........................................................................................................ 59

1.7.2.1 Example 9 : Shear stresses in a non-symmetrical cross section ................................................................ 60

1.7.2.2 Example 10 : Shear force in a non homogeneous cross section ................................................................ 66

1.7.3 Shear force center for thin walled non-symmeyrical cross sections ............................................. 68

1.7.3.1 Example 11 : Shear force center for thin walled cross sections ................................................................ 69

APPENDIX A ...................................................................................................................................................... 75

APPENDIX B ...................................................................................................................................................... 77

2. ASSIGNMENTS ........................................................................................................................................ 79

2.1 CROSS SECTIONAL PROPERTIES ................................................................................................................ 79

2.2 NORMAL STRESSES IN CASE OF BENDING ................................................................................................. 82

2.3 NORMAL STRESSES DUE TO BENDING AND EXTENSION ............................................................................ 87

2.4 INHOMOGENEOUS CROSS SECTIONS LOADED IN EXTENSION..................................................................... 89

2.5 INHOMOGENEOUS CROSS SECTIONS LOADED IN BENDING ........................................................................ 90

2.6 CORE ....................................................................................................................................................... 93

2.7 SHEAR STRESSES DUE TO BENDING .......................................................................................................... 96


Ir C. Hartsuijker & Ir J.W. Welleman October 2017 iii

STUDY GUIDE

These lecture notes are part of the course CIE3109 Structural Mechanics 4 (CM4). 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://icozct.tudelft.nl/TUD_CT/

In these notes reference is made to the first year lecture notes of Structural Mechanics 1 and

Structural Mechanics 2 and Structural Mechanics 3 which are covered with three books:

o Engineering Mechanics, volume 1 : Equilibrium, C. Hartsuijker and J.W. Welleman

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

Welleman

o 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.

The answers to the assignments can be found via the above mentioned web site. If needed

additional information can be obtained from the Student Assistants of Structural 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 nr 11-3, October 2017

[email protected]


Ir C. Hartsuijker & Ir J.W. Welleman October 2017 1

1. NON-SYMMETRICAL AND INHOMOGENEOUS CROSS SECTIONS

In this module the in MECH-2 introduced fiber model for beams will be extended and used on beams with a non-symmetrical and/or inhomogeneous cross section. With the presented theory a straight forward and easy method is obtained to calculate the stresses and strains in cross sections made out of different 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 as x-axis. The fibers are

kept together by infinite rigid cross sections which are by definition perpendicular to the beam

axis. 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 Bernoulli. 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 (Hooke’s 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.

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

N

fiber

fiber

beam axis

axis of symmetry cross sections



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). The fiber model only 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 shear

centre 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

z z

y

y

y

(a) (b) (c)

concrete

steel

E1

E1

E2

ux

uz uy x

z

y

Fz

Fx Fy

F

ϕz

ϕx ϕy



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

Loads Stresses Strains Displacements

( F, q ) ( N, M, V) ( ε, κ ) ( u, ϕ )

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 sections

remain 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 speak

normally of rigid cross sections.

2. Cross sectional rotations remain small, ϕ << 1

3. The fibers represent a line stress situation (follows from the chosen fiber model )

4. The fibers show a linear elastic behaviour: σ = Eε ( Hooke’s law )

In the presented figures we use a x-y-z coordinate system. The x-axis always coincides with

the direction of the beam axis. The position of a cross section is given with its x-coordinate. In

the cross section itself the position of a fiber is given with the y- en z-coordinate. The x-axis

always coincides with the direction of the fibers. By definition the origin of the coordinate

system of the cross section is known as the Normal force Centre NC. The x-axis therefore

always crosses the NC, see figure 1. This special choice of the origin of the coordinate system

of the cross section is a priori unknown. However with some simple calculus the position of

the NC can be obtained as will be shown in the examples to come.

All quantities which vary over the cross section will be presented as functions of y and z . As

an example we mention stresses and strains which will vary over the cross section:

),(

),(

zy

zy

σ

ε

In the following sections we will distinct between homogeneous and inhomogeneous cross

sections. In the derivation of the formulae we will build the model step by step and start with

homogenous cross sections in which all fibers will have the same Young’s modulus (modulus

of elasticity) E. In case of an inhomogeneous cross section Young’s modulus can vary over

the cross section and a function of y and z will be used:

),( zyE

Most quantities will also vary with respect to the x–coordinate. Since however most equations

holds for a specific cross section the x-coordinate is omitted.

In the following sections the fiber model for bending and extension will be presented first for

homogeneous cross sections after which the model will be extended for inhomogeneous

situations. With the obtained knowledge we can then find the force point of the cross section

and the kernel of a section. As mentioned earlier the module is closed with the calculation of

shear stresses due to shear forces in non-symmetrical and/or inhomogeneous cross sections.



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 exhibit three

translations and three rotations :

zyxzyx ,,en,, ϕϕϕuuu

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 cross

section. Suppose point P(x,y,z) is in the cross section at a distance x of the beams origin, see

figure 6.

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

The cross section moves with ux in x-direction and rotates with ϕy along the y-axis and with

ϕz along the z-axis. The displacement u in the direction of the fiber at P can be written with

the assumption of small rotations (see section 15.3.2 from MECH-l) as:

yzx),,( ϕϕ zyuzyxu +−=

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

subsequently rotated along the y- and z-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 ϕy en ϕz.

x

y

z

ux

x

y

x

z

top view side view

P yϕz y

zϕy z

fiber

cross section



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

displacement quantities ux , ϕy en ϕz belong to the cross section which contains P and are

therefore cross sectional related quantities. These displacements are thus only functions of x.

In other words: the displacements of an arbitrary fiber in a cross section at a distance x can 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),(

ϕϕε

ϕϕε

′+′−′=

+−=∂

∂=

∆

∆=

→∆

zyuzy

dx

dz

dx

dy

dx

du

x

zyxu

x

zyxuzy

x (a)

The rotations ϕy en ϕz can be expressed in the displacement quantities uy en uz. See figure 7.

y

y

z

zz

y

d

d

d

d

ux

u

ux

u

′=+=

′−=−=

ϕ

ϕ

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

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

zyx),( uzuyuzy ′′−′′−′=ε (b)

The strain of the beam axis is equal to the strain of the fiber through the x-axis added with the

strain due to bending (curvature) along the y- and z-axis. Expression (b) can be rewritten with

the introduction of the following three cross sectional deformation quantities:

yzz

zyy

x

ϕκ

ϕκ

ε

′=′′−=

′−=′′−=

′=

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),( κκεε 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 in y- en z-direction is denoted with κy and

κz . From this also follows that κy en κz must be

constant and are therefore cross sectional quantities.

Figure 8 : Cross sectional strain distribution

neutral line



From figure 8 also follows that positive curvatures causes for positive values of y and z

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 the x-y-plane and a curvature in the

x-z-plane. These curvatures are denoted with respectively κy and κz. These are the components

of the vector κ. In figure 9 this is visualised. The prove that a curvature behaves like a first

order 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 κ is :

2

z

2

y κκκ +=

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

y

zktan

κ

κα =

k

k

convex side

concave side



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 =++= κκεε 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 the y-axis (z = 0) : yκ

ε−=y

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

ε−=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 beam’s plane of curvature k is 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 zone in to the tensile zone.

Assignment:

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

line.

k

neutral line nl.



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 Hooke’s law:

εσ ×= E

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

coordinate system:

),();,( zyzy εεσσ ==

In case of a homogeneous cross section all fibers will have the same Young’s modulus E. The

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

),(),( zyEzy εσ ×=

In combined loaded sections (bending and extension), fibers will lengthen or shorten. The

deformation behaviour of a particular cross section can be described with the earlier

introduced three cross sectional deformation quantities:

zy en, κκε

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


Using Hooke’s law for the stress strain relation we can find the expression for the stress in

any particular (fiber) point of the cross section:

( )zy),( κκεσ 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 infinitesimal

small area ∆A of the cross section.

NOTE:

The definitions used here are so called formal

definitions. My is the bending moment in the

xy-plane and Mz is the bending moment in the

xz-plane In engineering notation however

these moments are denoted as Mz and My ,

bending about the z or y axis. It is important

to check these definitions used in MECH-2.



The resulting normal force N due to the normal stresses becomes:

( )

++=++×== ∫ ∫∫∫∫

A AAAA

zdAydAdAEdAzyEdAzyN zyzy),( κκεκκεσ

The bending moment My and Mz become:

( )

( )

++=++×==

++=++×==

∫ ∫∫∫∫

∫ ∫∫∫∫

A AAAA

A AAAA

dAzyzdAzdAEzdAzyEdAzyzM

yzdAdAyydAEydAzyEdAzyyM

2

zyzyz

z

2

yzyy

),(

),(

κκεκκεσ

κκεκκεσ

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

=

===

===

∫

∫∫

∫∫∫

A is the cross sectional area, S is the first order area moment (static moment) and I is 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, My and Mz ) and the sectional deformations (ε,

κy en κz) can be rewritten as:

zzzyzyzz

zyzyyyyy

zzyy

κκε

κκε

κκε

EIEIESM

EIEIESM

ESESEAN

++=

++=

++=

In matrix presentation this relation becomes:

=

z

y

zzzyz

yzyyy

zy

z

y

κ

κ

ε

EIEIES

EIEIES

ESESEA

M

M

N

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

forces N, My and Mz ) to the generalised deformations (sectional deformations ε, κy en κz) and

is also called the cross sectional constitutive relation. The derivation of it is based on the

linear elastic constitutive relation of a single fiber (Hooke’s law).

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

single fiber which coincides with the x-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 the y-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. Iyy is then referred to as Iz and

Izz is referred to as Iy. It is very important to

check these definitions used in MECH-2.



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

en Sz become zero. The position of the y-z-coordinate system has to chosen at the normal

force centre NC 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 the y-z-coordinate system at the normal force centre NC the static moments

become zero thus simplifying the constitutive relation:

=

z

y

zzzy

yzyy

z

y

0

0

00

κ

κ

ε

EIEI

EIEI

EA

M

M

N

From this relation we can see that a normal force N, acting at the normal force centre NC,

only causes strains and 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 between

extension 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

κ

κ

ε

EIEI

EIEI

M

M

EAN

In the bending part however we do see a coupling between bending in the xy- and xz-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 term EIyz which 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 moments My and Mz can be replaced by a resulting moment M which is shown

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

and the line m.

Figure 13 : Sectional forces.

From the above follows that My and Mz are the components of a vector. In appendix A the

proof is given that a moment is also a first order tensor. The components of M can also be

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

extension

bending



The sign convention used here is that for a positive

moment the moment resultant M always points from

the compressive zone to the tensile zone.

The magnitude of the resultant moment M is :

2

z

2

y MMM +=

The angle between the moment M and the y-axis is

defined by:

y

zmtan

M

M=α

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

presentation with a double arrow. The moment M can of course also be represented with the

bent moment arrow as shown in figure 15.

Figure 15 : Two possible presentations for a bending moment M in 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 2×2 system of

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

=

z

y

zzzy

yzyy

z

y

κ

κ

EIEI

EIEI

M

M

Both the moment M and the curvature κ are first order tensors. The stiffness matrix which

relates 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 tensor is a symmetrical matrix since: yzzy EIEI =

All known tensor transformation rules for second order tensor can be applied to the bending

stiffness tensor like Mohr’s 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 y-z-plane.

compression

tension

compression

tension



The general relation between the moment M and the curvature κ is shown in figure 16.

In this figure also the neutral line n has been

drawn. The line of action k of the curvature κ is

perpendicular to n as mentioned earlier. The

beam curves in a plane which is built by the x-

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

plane of curvature. The bending moment M acts

with the normal force in a plane built by the x-

axis and m. The sectional forces therefore act in

the x-m-plane which is therefore also referred to

as the loading plane.

By definition M and κ will not have the same line

of action. This results in a plane of curvature

which does not coincide with the loading plane.

Both moment and curvature only act in the same

plane if the following relation holds:

=

z

y

z

y

κ

κλ

M

M

If we substitute this into the constitutive relation we find:

0z

y

zzzy

yzyy

z

y

z

y

zzzy

yzyy

z

y=

−

−⇒

=

=

κ

κ

λ

λ

κ

κλ

κ

κ

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 tensors M and κ only coincide if the

plane of action coincides with one of the principal

directions.

If we rotate the y-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

κ

κ

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 between plane of curvature and the loading plane for

the principal coordinate system. We therefore check the angles andm kα α .

Figure 17 : Principal directions for

The bending stiffness

Figure 16 : Presentation of curvature and

moment in the y-z-plane.



These can be found with:

)tan()tan( k

yy

zz

yyy

zzz

y

zm α

κ

κα

EI

EI

EI

EI

M

M===

These direction in deed only coincides if:

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

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

c) ;zzyy EIEI = km αα =

In the last case all directions are principal directions and Mohr’s circle is represented by a

single dot (check this your self) !

1.2.3 Equilibrium conditions

After 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 the x-z-plane and

the x-y-plane. Moments and

shear forces can be depicted

from figure 18. Shear forces

have components in both the y-

and z-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

Vy

Mz

My

qz

qy

N

N+dN

Vz+d Vz

Vy+d Vy

Mz+d Mz

My+d My

dx

Figure 18 : Equilibrium of a beam segment.

NOTE: Pay attention to the way the shear

force components are reduced

from the equations.



1.2.4 Differential Equations

With the found kinematic, constitutive- and equilibrium relations it 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

xε = = ;

2

2

d"

d

y

y y

uu

xκ = − = − ;

2

2

d"

d

zz z

uu

xκ = − = −

Constitutive relations:

=

z

y

zzzy

yzyy

z

y

0

0

00

κ

κ

ε

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 the

displacements 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

* *

2

"

"'' "'' ;

"'' "'' ;

x x

yy zz y yy yz z yy zz y yy yz z

yy y yy y y y

yy zz yz yy zz yz

yz zz y yy zz z yz zz y yy zz z

zz z zz z z z

yy zz yz yy

EAu q

EI EI q EI EI q EI EI q EI EI qEI u EI u q q

EI EI EI EI EI EI

EI EI q EI EI q EI EI q EI EI qEI u EI u q q

EI EI EI EI E

= −

− −= ⇔ = =

− −

− + − += ⇔ = =

− 2

zz yzI EI−

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

certain field in the usual way. The right hand side of the second and third ODE is marked with

a *. Apart from the right hand side, these two ODE are similar to the ODE for bending in case

of symmetrical cross sections. All forget-me-nots can therefore be used if the loading is

modified according to the expression marked with the *. This will be demonstrated later in an

example.

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



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 =++= κκεε zyzy

2. Plane of curvature is perpendicular to the neutral line nl.

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

with the x-axis is therefore zero. From this we can conclude that also the normal force N is

zero. The first condition thus becomes:

zy0 κκ zy +=

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

κ

κ

κ

κα

α

−=−==

=

33tan

120

y

zk

o

k

The actual curvature κ is shown in figure 20.

The components of the bending moment M can be

found with:

=

z

y

zzzy

yzyy

z

y

κ

κ

EIEI

EIEI

M

M

For this square cross section holds:

0;)2( yz

4

344

121

zzyy ==== IaaII

After substituting these values we find:

−×

=

3

1

10

014

34

z

y κEaM

M

z

y

2a

2a

30o

(a)

z

y 2a

2a

nl

nl

nl nl

(b)

z

y

2a

30o

Figure 20 : Curvature and neutral line

nl

nl

k

k

κ αk

2a



The position of the loading plane m can be determined with :

3)(

3tan

4

34

4

34

y

zm −=

−×

×==

κ

κα

Ea

Ea

M

M

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

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

found that m and k coincides 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 k is found with:

0tan

90

κ

κ

κα

α

=±∞==

=

y

zk

o

k

The actual curvature κ is shown in figure 21.

The components of the bending moment M can be

found with:

=

z

y

zzzy

yzyy

z

y

κ

κ

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

===

====

α

Substituting these values results in:

=

×

=

κ

κ

κ 4

20

42

244

914

91

z

yEaEa

M

M

The loading plane m thus becomes:

22

4tan

4

91

4

91

y

zm =

×

×==

κ

κα

Ea

Ea

M

M

In figure 22 this result is shown. In this example

the plane of curvature k does not coincide with the

loading plane m. Be aware that the angular difference between k and m is not due to torsion

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

z

y

2a

2a

nl nl

k

k

αk

Figure 21 : Curvature and neutral line.

z

y

2a

2a

nl nl

k

k

αk

Figure 22: Curvature and loading

m

m

αm



Cross section (b) from the previous example with a Young’s modulus E is 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 asked

for as well as the moment distribution in both the xy- and xz-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 zyy y y y

yy zz yz yy zz yz

yz y yy z yy z zzz z z z

yy zz yz yy zz yz

EI q EI q EI q qEI u q or u

EI EI EI EI EI EI Ea

EI q EI q EI q qEI u q or u

EI EI EI EI EI EI Ea

− − −= = = =

− −

− += = = =

− −

The general solution for the displacement field in both the y- and z-direction becomes:

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

= + + + −

= + + + +

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

ϕ ϕ= = = = =

= = = = =

From these the following integration constants can be obtained, see also the MAPLE input 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

qz= 8 kN/m

z

y 2a

2a

cross section

E=100 GPa

a = 0,1 m

y



The MAPLE input 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 the xy-plane

becomes zero. The maximum moment at the clamped end is indeed 0,125×q×l2

= 100 kNm.

Due to the unsymmetrical cross section the member will deflect in both the xy-plane and the

xz-plane.

Figure 19-3 : Results for M and u for 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:

Special care should be given to the

specified boundary conditions, see

APPENDIX B.

> 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(uz,x): kappay:=-diff(phiz,x):

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

Moment My and Mz

Deflection uy and uz

'' ''; '

'' ''; '

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

κ κ

κ κ

= + = − − =

= + = − − =



1.2.6 Normal stresses in the y-z-coordinate system

If for a cross section the sectional forces N, My and Mz are 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

κ

κ

ε

κ

κ

ε

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

( )zy),(),( κκεεσ 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 the y-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 ==

=⇒=

==

κκ

κκ

εε

The stress distribution follows from:

( )

zz

z

yy

y

zy

),(

),(),(

I

zM

I

yM

A

Nzy

zyEzyEzy

++=

++×=×=

σ

κκεεσ

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. the y-component. We then find:

y yz y z

z zz

with: 0 and

N EA

M EI

M EI

ε

κ κ κ κ

κ

=

= = =

=

The stress distribution follows from:

( )zz

z0),(),(I

zM

A

NzyEzyEzy +=+×+×=×= κεεσ

Note that the component My does not show up in the above relation for the determination of

the stress although the curvature κy is zero and the moment My is not zero!



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 defined y-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. The y-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

κ

κ

EI

EI

M

M

Using this approach leads in fact to 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 ++=σ

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 can

use Mohr’s circle or the transformation formulas for this.

b) Then the bending moment components in y- and z- 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 have

to be transformed into the principal coordinate system.

d) All deformations and related displacements found are with respect to the principal

directions. In order to find the displacements in the original y-z coordinate 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 both

methods are used.

NOTE: All quantities with respect to the

principal coordinate system.



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 in y- and z-direction at mid span. In this example we will

use both previous outlined methods. We start with using the y-z coordinate system.

Method 1 : Computation in the y-z-coordinate system

The load causes at mid span bending moments in both the x-y and x-z plane:

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 the y-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

150

mm

Fz=27 kN

Fy=9 kN

y

z

NC

x

15,83 mm

51,6

7

mm

(a) Loaded structure

(b) Cross section

t = 10 mm

E=210000 N/mm2

y



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

κ

κ

κ

κ

Using the quantities found:

( )

×

×

−−××=

−

3

3

2

4

z

y

1010125

103375

2,8975,113

75,1139,526

75,1139,5262,89

101

Eκ

κ

results in the following components of the curvature:

1/mm1099,171/mm1095,40 66 −− ×=×= zy κκ

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


Since there acts no normal force at the cross section, the stain ε in the fiber which coincides

with the x-axis (beam axis trough the NC) is also zero. This results in:

zyzy ×+×= 778,3600,8),(σ

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 σ N/mm2

A 59,17 -51,67 313,6

B -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

15

0

y

z

NC

15,83

51

,67

t = 10

A B

E

59,17

98

,33

length in mm

nl

nl

C

D

+

-

351,4 N/mm2

331,4 N/mm2

Figure 24 : Stress distribution.



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

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

formulae. Several methods to obtain the displacements will be demonstrated. First the

modified forget-me-nots will be used, then the displacement based on the curvature

distribution will be used and the last option is to solve the whole problem in the principal

directions.

Displacements in the original coordinate system with modified forget-me-nots

Since the boundary conditions are fully uncoupled the modified forget-me-nots can be used.

The load in y- and z-direction are re-written as:

*

2

*

2

20461,30936 N

53087,27357 N

yy zz y yy yz z

y

yy zz yz

yz zz y yy zz z

z

yy zz yz

EI EI F EI EI FF

EI EI EI

EI EI F EI EI FF

EI EI EI

−= =

−

− += =

−

At mid span the displacements can be found using the standard forget-me-nots by using the

modified loads:

* 3

* 3

7,68 mm48

3,37 mm48

y

y

yy

zz

zz

F lu

EI

F lu

EI

= =

= =

Displacements in the original coordinate system based on the curvature distribution:

The curvature distribution is known. With the paradigma of the curvature plane the deflection

can be obtained. In the following intermezzo a brush up of this method can be found.

In figure 25 a sketch of the curvature distribution is shown. All deformation is concentrated at the bend

and denoted by theta.

Figure 25 : Basis for forget-me-not formulae

In case of a non-symmetrical cross section the reduced moment distribution Mz/EIzz is not the curvature

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

( )

−

−

−=

z

y

yyzy

yzzz

2

yzzzyyz

y 1

M

M

EIEI

EIEI

EIEIEIκ

κ

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 and

calculate the components of the curvature distribution.

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

F

l

zz

z

zEI

M=κ

EI

Fl−

θ

l32

θ×= lw32

EI

Fllw

EI

Fl

EI

Fll

3

23

32

2

21

=×=

=××=

θ

θ

x-axis

engineering formula



With the paradigma of the curvature plane as described in section 8.4 in MECH-2, the

displacements in the y- en z-direction can be found.

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

With the boundary condition of zero displacements uy en uz at B, the rotations ϕy en ϕz at A

can 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

θϕϕθϕ

θϕϕθϕ

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

Mohr’s graphical method. Both ways will be illustrated .

The moments of inertia in the y-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 Iα α= ⇔ =

−

)A(zϕ

x

l

y

θ1

1/mm1095,40 6−×=yκ

x

l

z

θ3

1/mm1099,17 6−×=zκ

A A B B

x

l21

y 1/mm1095,40 6−×=yκ

A B θ2

)A(zϕ

)A(yϕ

x

l21

z 1/mm1099,17 6−×=yκ

A B θ4

)A(yϕ

l61 l

61



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

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

Figure 27 : Mohr’s circle for moments of inertia

The directions found can be presented within the drawing of the cross section which is shown

in figure 28.

Figure 28 : Principal directions of the cross section.

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

75

15

0

y

z

NC

15,83

51

,67

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



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+=σ

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 e1 and 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 e1 and e2.

point y mm z mm e1 mm e2 mm σ 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- and x-z-plane are uncoupled. Therefore the

displacements in the 1-2-directions can be found with the engineering forget-me-not formulae.

The concentrated loads which are used in the engineering formulae for the displacements also

have to be transformed to the principal coordinate system using the first order tensor

transformation rules:

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

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

2

1

=+=

=+−=

zy

zy

FFF

FFF

The displacements u1 and u2 in the principal directions can now be found with the engineering

formulae:

mm 26012,81042,61101,248

150046,15152

48

mm 4542,11067,554101,248

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 original y-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:

mm4,837,368,7 22 =+=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

1.2.9 Concluding remarks

Using nonsymmetrical cross sections will in general result in a plane of curvature which

differs from the plane of loading. Only in three specific cases both planes coincide:

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

b) ;2παα == km M and κ act along then z -as (other principal axis)

c) ;zzyy EIEI = km αα =

Due to this effect the standard engineering formula (forget me knots) to obtain the

displacements should not be applied. In this chapter three possible ways to find the correct

displacements haven been presented based up on:

1. Differential equation, displacements directly in the original coordinate system of the

cross section

2. Modified forget-me-nots, displacements directly in the original coordinate system of

the cross section

3. The paradigma of the curvature plane planes of curvature in y- and z-x plane,

displacements directly found in the coordinate system of the cross section

4. Using the uncoupled deformation in the two principal directions, displacements are

computed in the principal directions based upon the forget-me-nots

By using this latter method more tensor transformations are required than when using the

original y-z-coordinate system. First the principal directions have to be found, second the

loads have to be transformed in to the principal directions in order to find the displacements in

these principal directions. Finally these displacements have to be transformed back in to the

original coordinate system of the cross section. Also by using the method of the principal

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

for the displacements are quite easy to remember.

In today’s engineering firms most of these calculations are done with help of computer

programs or spreadsheet calculations, thorough knowledge of these matters is lacking.

However to really understand the observed phenomena, good knowledge of this theory is

essential although the outlined procedure will in most cases not be used. Method 2 is the

favorite method for hand calculations.

750 mm

750 mm

27 kN

9 kN

x

E=210000 N/mm2

7,68 mm

3,37 mm

y

z

Assignment

Make a sketch of the cross section

and show:

• the plane of curvature

• the loading plane

• the neutral line

• the principal directions



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 Young’s modulus or modulus of

elasticity E. The modulus of elasticity may vary between fibers based on material used for

each fiber or part of the cross section. The Young’s function used is denoted as:

),( zyE

Since the Young’s 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 to the strains has to be modified slightly:

),(),(),( zyzyEzy εσ ×=

Since the kinematic relation will not change the three cross sectional deformation quantities

zy ,, κκε 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:


The stress in a specific fiber becomes in case of linear elasticity (Hooke’s law):

( )zy),(),( κκεσ 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 force N can be found with (see section 1.2.2):

( )

∫ ∫∫

∫∫

++=

⇔++×==

A AA

AA

zdAzyEydAzyEdAzyE

dAzyzyEdAzyN

),(),(),(

),(),(

zy

zy

κκε

κκεσ

The expressions for the components of the bending moment My and Mz yield:

( )

( )

∫ ∫∫

∫∫

∫ ∫∫

∫∫

++=

⇔++×==

++=

⇔++×==

A AA

AA

A AA

AA

dAzzyEyzdAzyEzdAzyE

zdAzyzyEdAzyzM

yzdAzyEdAyzyEydAzyE

ydAzyzyEdAzyyM

2

zy

zyz

z

2

y

zyy

),(),(),(

),(),(

),(),(),(

),(),(

κκε

κκεσ

κκε

κκεσ

As can be seen from these expressions, the Young’s function E(y,z) remains under the

integral.

NOTE:

We will still consider a linear elastic stress-

strain relation for each fiber.



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, My and

Mz ) to the sectional deformations (ε, κy and κz) thus becomes:

zzzyzyzz

zyzyyyyy

zzyy

κκε

κκε

κκε

EIEIESM

EIEIESM

ESESEAN

++=

++=

++=

In matrix notation this constitutive relation becomes:

=

z

y

zzzyz

yzyyy

zy

z

y

κ

κ

ε

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 symbols represent the evaluation of an (extensive)

integral where as the symbols used in section 1.2.2 are the product of two quantities e.g. EA

represents E times A. For inhomogeneous situations the double letter symbol EA represents:

∫=A

dAzyEEA ),(

If we choose 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 to

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

κ

κ

ε

EIEI

EIEI

EA

M

M

N

basic formula (1)

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

coordinate system at the NC.



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),( κκεε zyzy ++= basic formula (2)

The stress in a specific fiber can be found with:

),(),(),( zyzyEzy εσ ×= 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

not be 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

forces N, My and Mz

- calculate the cross sectional

deformations ),,( zy κκε

- find the strain distribution

- find the stress distribution

basic formula 1

basis formula 2

basis formula 3



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

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

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

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

moments or 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 Moment of 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 the y-z-coordinate

system through the unknown NC is denoted with NC

y and NCz . The temporary coordinate

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

NCNCzzzyyy +=+=

For the weighted static moments with respect to the zy − -coordinate system holds:

NCzNC

z

NCyNCy

),(),(),(

),(),(),(

zEAESdAzyEzzdAzyEdAzzyEES

yEAESdAzyEyydAzyEdAyzyEES

AAA

AAA

×+=+×=×=

×+=+×=×=

∫∫∫

∫∫∫

By definition the weighted static moments with respect to the y-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 ),(



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 Young’s 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

symbols for inhomogeneous cross sections. We do not need integral calculus here since for

each 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

21

2

33

21

21

21

221

1

21

21

221

1

NC9

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:

NC6

3ZW

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 aspect

when 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



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

Young’s moduli: E1= 6000 N/mm2 E2 = 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 symbols we first need to find the location

of the normal center NC.

The given cross section has rotational

symmetry around 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 in

the structure is represented with the V- and M-

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

z

A

B C

250 N

V-diagram

M-diagram

137500 Nmm

250 N

0,55 m 0,55 m

x

z

A

B

y

E1

E2

10 mm

30 mm

20 mm

z C

E2

10 mm

50 mm

50 mm

O P

Q S

T V

W

X

R

U



The sectional stiffness quantities represented with the double letter symbols can be found

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

the local centroids are known. The distances in x- and y-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 the y-

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:

( ) ( )

( ) ( ) 2923

121

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

κ

κ

κ

κ

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

zy),( κκεε zyzy ++= basic formula (2)

web

yf

upper and lower flange

upper flange lower flange

upper and lower flange web

y

E1

E2

10 mm

30 mm

20 mm

z

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.



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

Young’s moudulus:

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

Using a spread sheet like EXCEL speeds up the calculus for each key point as can be seen

from the next table.

Units N, mm

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

-3] Stress [N/mm

2]

Mat

eria

l 2 O 40 -25 12000 -0,10 -1,25

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

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

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

Mat

eria

l 1 R 10 -15 6000 0,41 2,48

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

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

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

Mat

eria

l 2 T 10 15 12000 -1,09 -13,14

V -40 15 12000 0,61 7,28

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

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 Young’s 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 stress

distribution it is important to know the position of the neutral line nl. The expression for the

neutral line becomes in this case:

029,5003,34

01029,501003,340),( 66

=−−

=×−×−⇒= −−

zy

zyzyε

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 k is 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.



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.

For this structure the displacement can also be obtained by using the modified forget-me-nots

as presented earlier, this is left to the reader.

1.4 Force point of the cross section

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

zyzy zy),( κκεε ++=

This expression depends on the loading. For zero normal force N , the normal strain ε (at the

NC) is also zero and the neutral line passes through the NC as was shown in the previous

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

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

My and Mz. 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

MMMEIEIEI

EIM

EIEIEI

EI

MMMEIEIEI

EIM

EIEIEI

EI

EA

N

zzyz

yzyy

χχκ

χχκ

ε

+−=−

+−

−=

−=−

−−

=

=

y

E1

E2

10 mm

30 mm

20 mm

z

E2

10 mm

50 mm

NC

n.l

n.l

k

k

19,17 N/mm2

19,17 N/mm2

Material 2 6,57 N/mm

2

+

+ -

-

Material 1



The strain distribution can thus be found with:

( ) ( ) zMMyMMEA

Nzy ×+−+×−+= zzzyyzzyzyyy),( χχχχε

If the normal force N is non-zero the moments in the xy- and yz-plane can also be expressed in

the eccentric applied normal force N:

zz

yy

eNM

eNM

×=

×=

The three sectional forces N, My and Mz are thus replaced with one single eccentric normal

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

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

( ) ( )

( ) ( )[ ]zeEAeEAyeEAeEAEA

Nzy

zeNeNyeNeNEA

Nzy

××+×−+××−×+=

⇔××+×−+××−×+=

zzzyyzzyzyyy

zzzyyzzyzyyy

1),(

),(

χχχχε

χχχχε

This last expression shows the strain ε at a point (y,z) for a normal force N acting at (ey, ez).

As an experiment of mind we can think of a force N acting in (y,z) and observing the strain ε

in (ey, ez). It appears to result in exactly the same strain. The experiment is shown in figure 36.

Figure 37 : Maxwell’s reciprocal theorem

In words we can summarise this phenomenon as:

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

This is also known as Maxwell’s reciprocal theorem and 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 the yz-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=== χχχ

P

Q

P

Q

ε

ε

N

N



The strain can be expressed as:

××+

××+=

zz

z

yy

y1),(

EI

zeEA

EI

yeEA

EA

Nzyε (symmetrical cross section)

After using the definition of the inertia radius i :

EA

EIi

EA

EIi zz2

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

Nzyε (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 core or kern of the cross section.

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

rectangular cross section with dimensions b×h as shown

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

corner points at a distance to the NC in y- and z-

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 first property follows from the requirement that the stresses in the cross section do not

change in sign. Therefore the cross section is in tension or in compression which requires a

neutral line which is outside the cross section.

y h/3

b/3

z

h

b

Figure 38 : Core of a rectangular

cross section



As an example in figure 39 a cross section is shown. Valid positions of the neutral line are

AB, AH, HF, FE, ED and DB.

Figure 39 : Valid positions for the neutral line outside the cross section.

Based on these valid positions of the neutral line we can investigate six boundary positions of

the neutral line which will result in six force points corresponding to each position of the

neutral line, tangent to the cross sectional boundary. The force points form the core of the

cross section as will be explained in detail in an example.

The second property mentioned on the

previous page states that for cross sections

with straight edges the core also has straight

boundaries and is therefore a polygon. To

explain this we will look, as an example, to

the simple rectangular cross section given in

figure 40. From the two boundary lines 1-1

and 2-2 the according force points (core

points) are denoted as 1 and 2. The quest now

is to determine the boundary of the core

between these points 1 and 2. Force points at

the boundary of the core between 1 and 2

represent neutral lines which are tangent to the

cross section and intersect both the position of

the neutral line 1-1 and 2-2 which is at B. All

these force points between 1 and 2 will form a

straight line. We will explain this with the

Maxwell’s reciprocal theorem.

If the force point coincides with B as shown in figure 40, the neutral line can be found with

the expression for the strain in case of a symmetrical cross section, see page 38. The used

expressions for the inertia radius yield:

2

121

3

121

zz2

z

2

121

3

121

yy2

y ; hEA

Ebh

EA

EIib

EA

hEb

EA

EIi ======

With B as force point the eccentricities are:

hebe21

z21

y ; −==

A H

F G

E D

B

C

B

1 1

2

2

1

2

Figure 40 : Neutral line through corner.

z

y h

b

NOTE: A position of the neutral line which

coincide with BC can not be valid

since the neutral line then crosses

the cross section.



The expression for the neutral line becomes:

y z

2 2

y z

1 12 2

2 21 112 12

( , ) 1 0 (neutral line)

1 0

6 61 0

e yN e zy z

EA i i

b hy z

b h

y z

b h

ε × ×

= + + = ⇔

+ × − × = ⇔

+ − =

We found in words:

For a force in B all points on a line between core point 1 and core point 2 the stresses

(and strains) are zero.

Applying Maxwell’s reciprocal theorem will result in:

For all force points on a line between core point 1 and core point 2 the stress (and

strain) at B is zero.

This proves that the boundary of the core for cross sections with straight edges is a straight

line. This also holds for non-symmetrical and or inhomogeneous cross sections since the

neutral line is a straight line. The kinematic relations will not change in case of a non-

symmetrical or inhomogeneous cross section.

If the cross section is not made out of straight edges the determination of the core becomes

quite laborious. This is shown in figure 41.

Figure 41 : Arbitrary shaped cross sections.

The core of the cross section from figure 41 (a) will consist of a polygon with four sides since

the possible positions of neutral lines which are tangent to the cross section are the four

dashed lines. For the cross section given in figure 41 (b) this no longer holds since the cross

sectional edge from A to B is curved. Every neutral line which is tangent to a point on this

edge will result in a (unique) force point which is at the boundary of the core. This results in a

curved boundary of the core between the core points 1 and 2. Core point 1 is the force points

which belongs to a neutral line which coincide with edge 1-1 and core point 2 belongs to a

neutral line which coincides with 2-2.

(a) (b)

A

B

1

1

2 2

2

1

NOTE:

For y=0 this line crosses core point 1, for z=0

the neutral line crosses core point 2.



For arbitrary shaped cross sections we can not use the expressions given on page 38 since

these were limited to symmetrical cross sections. We therefore will use the general definition

of the neutral line as introduced in figure 11b, for cross sections with non zero normal forces:

y z

y z

( , ) 0

1 0 with: 0

y z y z

y z

ε ε κ κ

κ κε

ε ε

= + × + × = ⇔

+ × + × = ≠

To find the core points we assume a position of the

neutral line, tangent to the cross section. Suppose this

neutral line can be described with the following two

points, (y1,0) and (0,z1) as shown in figure 11c. These are

the points of intersection of the neutral line with the

coordinate axis. From the expression above we can relate

these points to the three cross sectional deformations:

z

1

y

1

κ

ε

κ

ε

−=

−=

z

y

The position of the force point which belongs to the

assumed neutral line can be found by using the cross

sectional constitutive relation:

=

=

z

y

z

y

z

y

zzzy

yzyy

z

y

:en

0

0

00

e

eN

M

M

κ

κ

ε

EIEI

EIEI

EA

M

M

N

Combining these, results in:

=

=

εκ

εκ

ε z

y

zzzy

yzyy

z

y

z

y 11

EIEI

EIEI

EAM

M

EAe

e

With the assumed position of the neutral line we now

have found the corresponding force point which is a

point at the boundary of the core:

−=

1

1

zzzy

yzyy

z

y

1

11

z

y

EIEI

EIEI

EAe

e basic formula (4)

Finding the core has reduced to a straight forward procedure in which a assumed position of

the neutral line (tangent to the cross section) is expressed by the two points of intersection

with the coordinate system from which with basic formula nr 4 the position of the core point

can be found. We will illustrate this procedure with an example.

z

y

z1

y1

n.a.

κz

κy

κ

y

z

y

z1

y1

n.l.

y ey

ez

force point

Figure 11b : Position of the neutral axis

in the cross section

Figure 11c : Position of the force point

which belongs to a neutral

axis which is tangent to the

cross section



1.5.1 Example 5 : Core of a non-symmetrical cross section

Find the core of the non-symmetrical cross section as

shown in figure 42.

Analysis:

If the neutral line coincides with the edge of the cross

section the corresponding force point is a corner point on

the boundary of the core. Five lines can be drawn tangent

to the cross section which will result in a core which is

five sided polygon. We therefore have to determine these

five corner points of the core.

Calculation fase:

We need the following quantities of the cross section for

our calculations:

o position of the NC ( y-z-coordinate system )

o zzyzyy ;;; EIEIEIEA

The position of the NC can be found with the earlier

explained method. With respect to the upper side of the cross section the NC position is:

( )

mm231120200400

601202002004002

2

NC =−×

×−××=z

With respect to the right side of the cross section the NC position is:

( )

mm108120200400

601201002004002

2

NC =−×

×−××=z

The position of the NC is shown in figure 42. With respect to this origin of the coordinate

system used we find all other cross sectional quantities:

46224

12123

121

zz

462

yz

46224

12123

121

yy

mm102,70517112012031200400400200

mm103,98)171()48(120318200400

mm103,221481201208200400200400

×=×−×−××+××=

×−=−×−×−×××=

×=×−×−××+××=

I

I

I

In order to find all core corner points we will tabulate the calculation and use a systematic

numbering of tangent lines to the cross section and their corresponding force points. This is

shown in figure 43. All calculus can be done with a spreadsheet in EXCEL. For each assumed

position of the neutral line the points of intersection (y1,0) and (0,z1) with the coordinate

system are first determined. The position of the force point can then be found with basis

formula nr 4:

−=

1

1

zzzy

yzyy

z

y

1

11

z

y

EIEI

EIEI

EAe

e

The results can be found in the table and the graph of figure 43.

80 mm 120 mm

120

mm

2

80

mm

400

mm

200 mm

108 mm

231

mm

NC y

z

Figure 42 : Cross section



If the normal force acts within the found core area, no change of sign in the stresses will occur

irrespectively of the magnitude of the normal force !

A specially for prestressed concrete beams this can be important. If the prestressing tendons

are within the core of the cross section no tensile stresses will occur. We mention again that

this is independent of the magnitude of the prestressing force!

For prestressed prefab girders it is important to avoid cracking due to dead load. For a

simply supported girder a save position of the tendons will be the lower core point of the core.

This results in a fast design of the prestressed girder. Concrete however can sustain some

tension which allows for a lower position of the tendon. For precise concrete calculations we

refer to the text books on reinforced and prestressed concrete.

80 mm 120 mm

120

mm

2

80

mm

108 mm

231

mm

NC y

z

Figure 43 : Core

1 1

2 2

3

3

4

4

5

5

12 mm

Table : Calculation results

line y1 z1 core point ey ez 1-1 ∞ -231 1 -6,5 46,5

2-2 ∞ 169 2 8,9 -63,6

3-3 92 ∞ 3 -36,6 16,3

4-4 -108 ∞ 4 31,2 -13,9

5-5 -219 -219 5 8,6 42,2

1

2

3

4

5



1.6 Temperature influences*

Constrained or restricted deformations due to temperature influences may lead to considerable

stresses in structures. These stresses can even be larger than stresses due to normal loading

conditions. Temperature loads can therefore not to be neglected. In section 4.12 of MECH-2

the influence of a linear temperature gradient over the depth of a cross section is applied to a

homogeneous and symmetrical cross section. We will consider, in this section, a non

homogeneous and or unsymmetrical cross section. The temperature distribution over the

depth of the cross section can be an arbitrary function of y and z.

Using the earlier introduced constitutive relations presented with the “double letter symbols”

will be used to generalise the introduced temperature influences as introduced in MECH-2.

The modulus of elasticity E as well as the coefficient of thermal expansion α and the

temperature function T are functions depending on y and z.

),(

),(

),(

zyTT

zy

zyEE

=

=

=

αα

In the following the dependency of y and z will be omitted to simplify the expressions. The

assumptions as presented in section 1.2 will also hold in this section.

The extension of the fiber model with temperature influences results in a fiber strain which is

the combination of a strain due to stresses and a strain due to temperature influences. In order

to clearly distinct between these two components we use a upper index T or α :

Tε : strain due to temperature influences

σε : strain due to stresses

The strain in a fiber due to a temperature influence, denoted with the temperature distribution

function T(y,z) , can be expressed as:

),(),(T zyTzy αε =

The strain as a result of the stress in a fiber which results from the constitutive relation, e.g.

Hooke’s law, can be expressed as:

),(

),(),(σ

zyE

zyzy

σε =

The total strain definition thus becomes:

),(

),(),(),(),(),( σT

zyE

zyzyTzyzyzy

σαεεε +=+=

The cross sectional strain distribution can be found with the earlier derived kinematic relation:

zyzy zy),( κκεε ++=



When we combine these latter two expressions we obtain:

{ }),(),(),( zy zyTzyzyEzy ακκεσ −++=

With this expression the cross sectional forces N, My and Mz can be determined as described

in section 1.4 using the well known “double letter symbols” :

Normal force:

{ }

dAzyTzyEESESEA

dAzyTzyzyEdAzyN

A

zzyy

AA

),(),(

),(),(),( zy

∫

∫∫

−++=

−++==

ακκε

ακκεσ

Bending moment:

dAzyTzyEzEIEIES

dAzyzM

dAzyTzyEyEIEIES

dAzyyM

A

A

A

A

),(),(

),(

),(),(

),(

zzzyzyz

z

zyzyyyy

y

∫

∫

∫

∫

−++=

=

−++=

=

ακκε

σ

ακκε

σ

With the special location of the coordinate system, chosen at the normal force center NC of

the cross section, the expressions for the sectional forces can be simplified to:

−

=

∫

∫

∫

A

A

A

AzyTzyEz

AzyTzyEy

AzyTzyE

EIEI

EIEI

EA

M

M

N

d),(),(

d),(),(

d),(),(

0

0

00

z

y

zzzy

yzyy

z

y

α

α

α

κ

κ

ε

basic formula (5)

From this expression the temperature influence becomes clear.

If a beam element is not subjected to sectional forces (unloaded) and can deform freely or

unconstrained, the deformations which may occur are only the result of a temperature

influence. The sectional deformations due to only temperature influences denoted with a

superscript T, can be found using basic formula (5):

−

=

∫

∫

∫

A

A

A

AzyTzyEz

AzyTzyEy

AzyTzyE

EIEI

EIEI

EA

d),(),(

d),(),(

d),(),(

0

0

00

0

0

0

T

z

T

y

T

zzzy

yzyy

α

α

α

κ

κ

ε



This system of equations can be solved with the earlier derived expression for the inverse

stiffness tensor:

1

yz 2zyy yz yy zz yz

zy zzzy yy

1

0 0

0 with: detdet det

0

det det

z

EAEA

EIEIEI EI EI EI EI

EI EIEI EI

−

= − = − −

The cross sectional deformations due to temperature influences can be found as:

+−=

−=

=

∫∫

∫∫

∫

AA

AA

A

AyzTzyEzEIAyzTzyEyEI

AyzTzyEzEIAyzTzyEyEI

AzyTzyEEA

d)(),(d)(),(det

1

d)(),(d)(),(det

1

d),(),(1

yyyz

T

z

yzzz

T

y

T

αακ

αακ

αε

basic formula (6)

With this expression, basic formula (5) can be simplified to:

−

−

−

=

T

zz

T

yy

T

zzzy

yzyy

z

y

0

0

00

κκ

κκ

εε

EIEI

EIEI

EA

M

M

N

For the special case in which the coordinate system coincides with the cross sectional

principal directions, the expressions can again be simplified since the bending in the x-y and

the x-z plane are then uncoupled. The constitutive relation then becomes:

( )

( )

( ) zz

T

z

T

zzzzz

yy

T

y

T

yyyyy

TT

/d),(),(

/d),(),(

/d),(),(

EIAzyTzyEzEIM

EIAzyTzyEyEIM

EAAzyTzyEEAN

A

A

A

∫

∫

∫

=−=

=−=

=−=

ακκκ

ακκκ

αεεε

With respect to the outlined sectional method to obtain stresses and strains, only the

constitutive relations have to be modified when introducing temperature influences. However

on a structural level also some problems will arise. We will address these problems in the

forthcoming sections. To do so, it is necessary to distinct between static determinate and static

indeterminate or hyper static structures.

Static determinate structures:

If a structural element can deform unconstrained as is the case with static determinate

structures, the displacements and rotations due to the temperature influences can be

determined directly from the obtained sectional temperature deformations T

z

T

y

T en, κκε . The

temperature influence does not effect the cross sectional forces, it only results in additional

deformations which may occur freely. The sectional forces N, My and Mz are therefore only

depending upon the loading and can be found directly from the equilibrium conditions.

(only valid for the

principal coordinate

system)



Static indeterminate structures:

In case of constrained deformations as is the case with static indeterminate structures, the

situations becomes more complex. The force distribution can not be determined based upon

equilibrium conditions only and the restricted deformation due to the temperature influences

will effect the force distribution. Using the compatibility and deformation conditions

combined with the constrained deformation due to temperature influences, the force

distribution in the structure can be determined. With the sectional forces (N, My en Mz ) the

sectional deformation quantities zy en, κκε can be found:

−

−

−

=

T

zz

T

yy

T

zzzy

yzyy

z

y

0

0

00

κκ

κκ

εε

EIEI

EIEI

EA

M

M

N

Subsequently the sectional stress distribution can be found with:

{ }{ }),(),(

),(),(),(

),(),(),(

zy

T

zyTzyzyE

zyzyzyE

zyzyEzy

ακκε

εε

εσ σ

−++=

−=

=

This approach will be illustrated with two examples. The first example contains a static

determinate structure, in the second example a statically indeterminate structure will be

presented.

1.6.1 Example 6 : Static determinate structure under temperature load

A non homogeneous prismatic cantilever beam with a T-shaped cross section is subjected to a

temperature gradient as is shown in figure 44.

Figure 44 : Cantilever beam subjected to a temperature gradient.

Both materials behave linear elastic. The web of the T-section has a modulus of elasticity E

and the flange has a modulus of elasticity of 0,75E. The coefficient of thermal expansion of

the flange is α. The temperature distribution over the depth of the flange is linear as can be

observed from figure 44. The temperature gradient is constant over the beam of the section.

The web of the section remains under constant temperature conditions.

Questions:

a) Find the force distribution in the structure,

b) Find the stress distribution in a cross section at A,

c) Draw a sketch of the deformed beam,

d) Determine the maximum vertical displacement of this cantilever beam.

E43

E

T

6a

2a

6a 6a 4a

l

A B x

z



Answers:

The structure is static determinate and can deform unconstrained. The force distribution can

be obtained directly from the equilibrium conditions and is independent of the temperature

load.

a) Since there is no load all sectional forces are zero.

b) To find the stress distribution we first have to find the sectional deformation quantities

due to the temperature load. We therefore need the position of the normal center and the

stiffness quantities denoted with the “double letter symbols”.

Temperature function:

The temperature function for the flange can be expressed as::

12

( , ) (1 ) with : 3z

T y z T a z aa

= − + − ≤ ≤ −

Normal center NC:

The location of the normal center NC must be on

the vertical axis of symmetry. Therefore only the

vertical position of the NC has to be found:

aaaEaaE

aaEaaaEa

EA

EAaEAaz

lijfflens

346162

)46(5)162(

)(5)(

43

43

NC

=××+××

×××+×××=

⇔×+×

=

Since this cross section has an axe of symmetry

which coincides with one of the axes of the

coordinate system, the coordinate system

coincides with the principal axis of the cross section. The coordinate system and the

location of the NC is shown in figure 45.

Axial stiffness:

The axial stiffness of the cross section can be found as:

2

43 4846162)()( EaaaEaaEEAEAEA lijfflens =××+××=+=

Bending stiffness:

Fibers at a distance z of the NC will all elongate with the same amount over the beam of

the cross section. This is of course an assumption which is reasonable for a acceptable

width of the flange. Due to symmetry the beam will not curve in the x-y-plane, 0T

y =κ .

Over the depth of the beam the fibers will elongate different. Therefore the beam will

curve in the x-z-plane. Thus we only need to determine the flexural stiffness EIzz :

{ }

{ } 423

121

23

121

43

zz

272)2(64)6(4

)2(216)2(16

EaaaaaaE

aaaaaEEI

=××+×××

+××+×××=

With basic formula (6) we can now find the sectional deformation quantities:

Figure 45 : NC of the section



a

TEIadz

a

zTEz

TEAadza

zTE

a

a

a

a

αακ

κ

ααε

687

zz

3

21

43T

z

T

y

41

3

21

43T

/16)1(

0

/16)1(

−=×

+−×××=

=

=×

+−××=

∫

∫

−

−

−

−

The strain distribution for the cross section can be expressed as:

zzy T

z

T),( κεε +=

And the stress distribution due to the straining becomes:

{ }{ }

{ }),(),(

),(),(),(

),(),(),(

),(),(),(

T

z

T

T

σ

zyTzzyE

zyTzyzyE

zyzyzyE

zyzyEzy

ακε

αε

εε

εσ

−+×=

−×=

−×=

=×=

If we elaborate this expression for the flange we find:

+=

++×−×=

a

zTE

a

zT

a

zTTEzy

27281

272153

21

687

41

43 1),( αααασ

The expression for the web of the cross section yields:

−=

×−×=a

zTE

a

zTTEzy

27228

27268

687

41),( ααασ

The total strain and stress distribution for the cross section is shown in figure 46.

Figure 46 : Strain and stress distribution over the beams depth.

values values

fla

nge

w

eb

z TEασ ×

flens -3a -90/272

flens -a 72/272

lijf -a 96/272

lijf +5a -72/272



It is remarkable that the stress distribution is not congruent to the strain distribution. Even

more noticeable is the double root in the stress distribution. One of the zero stress points

coincides with the position of the neutral axis or neutral line n.l. but the other zero stress

point does not! The fact that the we observe a double change of sign in the stress

distribution is a direct result of the elongation of the fibers due to the temperature gradient

and the horizontal equilibrium condition of the cross section. The elongation of fibers

must be such that the resulting stress distribution forms an equilibrium system (ΣN=0;

ΣM=0). The beam will deform unconstrained without sectional forces but with non-zero

normal stresses!

This example shows that the neutral line can not be regarded as the only position at the

cross section were zero normal stresses occur in case of temperature influences. The

neutral line only divides the cross section in to a part with positive and negative straining.

The use of the neutral line in case of temperature related problems is therefore only

limited to strains.

c) As a result of the temperature load and the unconstrained deformation of the beam, the

strain distribution will be constant along the beams axis. Therefore all cross sections will

exhibit the same stress and strain distribution as shown in figure 46. In figure 47 this is

also visualised with the stress and strain distribution as functions of x. Also a sketch is

given of the beam deformation in the x-z-plane.

Figure 47 : Stress and strain distribution along the beam axis and the beam deformation.

The elongation of the beam can be found with the strain:

Tllu x αε41T ==

The vertical displacement of B can be found with the curvature:

a

lTlu

a

lTl

z

2

1367

21

687T

z

αθ

ακθ

=×=

==



1.6.2 Example 7 : Static indeterminate structure under temperature load

The same cross section will be used in a static indeterminate structure. The temperature load

is also the same as in the previous example. Both the structure and the loaded cross section

are shown in the figure below.

Figure 48 : Fully clamped beam with temperature load.

The structure is clamped at A and roller supported at B.

Questions:

a) Find the force distribution in the structure,

b) Find the stress distribution in a cross section at A,

c) Draw a sketch of the deformed beam,

d) Determine the maximum vertical displacement of this clamped beam.

Answers:

The structure is a static indeterminate structure. The deformation due to the temperature load

is therefore constrained. To find the force distribution we also have to take the deformation

behaviour into account. Applying the force method, we first choose a so called Basic Static

(determinate) System and denote one of the reaction forces as the unknown static

indeterminate since the structure is static indeterminate to the first degree. We then formulate

the deformation or compatibility condition for the associated degree of freedom of the

unknown static indeterminate. See for this method MECH-3. Since we are dealing with a

temperature load we also have to take in to account the effect of the temperature load on the

deformation.

a) As Basic Static System we choose the cantilever

structure as shown in figure 49. The unknown

static indeterminate is the vertical support

reaction at B. The associated deformation

condition becomes:

0)B(z =u

The result of the previous example can be used here since the unrestricted deformation of

the cantilever beam subjected to a temperature load was already solved. The vertical

support reaction at B becomes thus:

l

aTEB

EI

lB

a

lTu

3

V

zz

3

V

2

1367

z 4203

)B( αα =⇒=−=

The moment and shear distribution can be found using the equilibrium conditions. The

result is shown in figure 50.

l

A B x

z BV

Figure 49 : Basic Static System

E43

E

T

6a

2a

6a 6a 4a

l

A B x

z



Figure 50 : Force distribution

b) At A the bending moment is:

3

z 42 TaEM α=

Since the y-z-coordinate system coincides with the principal coordinate system the

curvature in the x-z plane can be obtained directly from the moment Mz with:

a

T

Ea

TaE

EI

Mz

272

42

272

424

3

zz

z αακ ===

The associated stress distribution becomes:

a

TzzyEzzyEzyzyEzy

272

42),(),(),(),(),( z

ακεσ ×=×=×=

For a number of key points the stresses can be computed and the stress and strain

distribution over the depth of the beam at A can be shown as is seen in figure 51.

Figure 51 : Strain and stress distribution due to BV

z TEασ × Flange -3a -94,5/272

Flange -a -31,5/272

Web -a -42/272

Web +5a +210/272

values

values

fla

nge

web



The total strain and stress distribution can be found by superposition. The actual

distribution is the sum of distribution found in figure 51 and the distribution due to the

free or unrestricted deformation as found in the previous example. The result is shown in

figure 52.

Figure 52 : Total strain and stress distribution at A.

c) The curvature is also a summation of the earlier found curvature of the unconstrained

deformation due to the temperature load and the curvature due to the bending moment

distribution.

Figure 53 : Total curvature and deformed beam.

The displacements can be found based on the total curvature. The maximum vertical

displacement will occur at a location were the rotation ϕ is zero. From the curvature

diagram of the figure above we can observe that at C the rotation will be zero as can also

be seen from the following expression for the rotation:

values

values

fla

nge

web



C

C A z

A

0 0 0dxϕ ϕ κ= + = + =∫

The maximum displacement occurs at:

lx32

C =

The vertical displacement can be found using the rotations shown between parentheses in

figure 53 in which θ is:

1 1 12 3 6

2 225 1 4 4 14

z 9 9 54 54 272

and

(C)272

l k kl

Tl Tlu l l kl

a a

θ

α αθ θ

= × × =

= − + = − = − × ≈ −

Remark:

For both examples the sectional forces N and M at B are zero. The stress distribution at B

should therefore be the stress distribution found for the unrestricted (free) deformation

due to the temperature load of example 6. However in reality this stress distribution is not

very realistic since at the free edge of the beam the normal stresses should all be zero.

Over a small distance from the free edge at B the presented theory will therefore predict a

wrong stress and strain distribution. According to Saint Venant’s1 principle this will only

be the case for a distance equal to the depth of the beam and we can make use of this

theory without any hesitation but with this remark in mind.

1 Named after Barré the Saint Venant (1797-1886), French civil engineer who contributed to the development of

the theory of elasticity.



1.7 Shear stress distribution in arbitrary cross sections

Apart from normal stresses also shear stresses occur in cross sections. In chapter 5 of MECH-

2 a systematic method has been outlined for homogeneous cross sections with at least one

axis of symmetry. In these notes we will extend this approach to inhomogeneous and or non-

symmetrical cross sections. At the end of this section we will also pay attention to the location

of the shear centre (SC) of a cross section.

In case of non constant bending moments over a segment of a beam, shear forces will occur.

A shear forces in a cross section is the resultant force of the acting shear stresses in the cross

section. From basic equilibrium conditions we also know that shear stresses will occur on

perpendicular planes. For a horizontal cross section this will result in longitudinal shear

stresses in the horizontal plane parallel to the beam axis. In figure 54 a beam segment ∆x is

shown. In both cross sections the sectional forces Vz and Mz are shown. We assume only

positive sectional forces. Due to the acting loads qz and qy (not shown) the sectional forces

will vary. The change in magnitude is denoted with the ∆ symbol. An important assumption at

this moment is to assume a constant normal force over the segment.

Figure 54 : Shear forces in longitudinal direction.

In order to obtain the longitudinal force which act on the longitudinal cut (greyed) segment,

we introduce the sliding area or shearing area )a(A . In figure 55 the free body diagram of this

shearing segment is visualised. On both cross sections the resulting normal forces (a)R and (a) (a)R R+ ∆ are shown.

Figure 55 : Free body diagram of shearing segment with length ∆x.

x∆

)(aR

)()( aa RR ∆+

)(a

xs

x

z

y

)(aA

NC

y

z z

x

z

)(aR

),( zyσ ),( zyσ

)()( aa RR ∆+

x∆

NC

section (stress distribution

on left side)

(stress distribution

on right side)

zq

zMzz MM ∆+

zz VV ∆+

NN

zV

)(a

xs



Let us assume a shear flow (a)

xs acting on the longitudinal cut in the longitudinal direction as

shown in figure 55. This shear flow is a force per unit length. From the horizontal

equilibrium of the shearing segment with length ∆x we find the following expression for the

shear flow:

x

Rs

RRxsR

∆

∆−=

=∆++∆×+−(a)

(a)

x

(a)(a)(a)

x

(a) 0

Taking the limit for 0→∆x , the shear flow or shear force per unit of length becomes:

x

Rs

d

d (a)(a)

x −= (1)

The resulting normal force (a)R on the shearing segment follows directly from the normal

stress distribution on the shearing area )a(A :

∫=)(

d),((a)

aA

AzyR σ (2)

The normal stress distribution follows from the earlier found expression:

{ }zyzyEzy zy),(),( κκεσ ++×= (3)

The sectional deformation quantities zy en, κκε follow from the cross sectional constitutive

relation as introduced in the previous sections. We will illustrate this approach with a simple

example for which the coordinate system coincides with the principal coordinate system of

the cross section. For general situations we will derive an alternative method later.

1.7.1 Shear stress equations for principal coordinate systems

If the cross sectional y-z-coordinate system coincides with the principal coordinate system we

can use the following uncoupled constitutive relations:

zz

zz

yy

y

y

EI

M

EI

M

EA

N

=

=

=

κ

κ

ε

(4)

After combining expression (3) with expression (4) we find:

×+

×+×=

zz

z

yy

y),(),(

EI

zM

EI

yM

EA

NzyEzyσ (5)

If we substitute this result in (2) and use this in (1) we find:

∫ ∫

++×−=−=−=

)( )(

ddx

d

d

d

d

d1),(d

d

),(d

d

d z

zz

y

yy

(a)(a)

xa a

A A

AM

EI

z

x

M

EI

y

x

N

EAzyEA

x

zy

x

Rs

σ



With the welle known expressions for the shear force we can simplify this equation:

x

MV

x

MV

y

d

d

d

d

zz

y

=

=

Since we assumed a constant normal force for the beam segment we can assume:

0d

d=

x

N

For prismatic beams the stiffness quantities denoted with the “double letter symbols”

yy zz, andEA EI EI are constant and can be put in font of the integrals. The result of all this

work is a closed formulation for the shear flow per unit of length:

×+×−= ∫∫

)()(

d),(d),(zz

z

yy

y(a)

xaa

AA

AzzyEEI

VAyzyE

EI

Vs

With the ”double letter symbols” we can simplify this to a neat basic formula:

zz

(a)

zz

yy

(a)

yy(a)

xEI

ESV

EI

ESVs −−= basic formula (7)

This expression is almost identical to the expression found in MECH-2. New element here is

the extension to two directions, y and z.

From the shear flow the shear stress can be found by dividing the shear flow with the actual

width of the longitudinal cut of the shearing segment. This is shown in figure 56 while using

the in MECH-2 introduced local m-axis for an unambiguous definition of the shear stress.

Figure 56 : Shear stresses in a beam at distance z from the NC.

The shear stresses in the cross section are the same as those in the longitudinal cut of the

shearing segment, see for the proof section 5.3.1 of MECH-2:

mxxm σσ =

mxσ

(a)

(a)

xmx

b

s=σ

Remark:

We assume a constant shear stress

over the width (a)b of a section. In

reality this is not the case. However

for engineering purposes (relative slim

cross sections) this is acceptable.

x∆

)(a

xs

x

z

y

)(aA

NC

xmσ

m-as

(a)b

z



1.7.1.1 Example 8 : Shear stresses in a composite cross section

The concrete-steel composite beam is loaded in bending. The beam has one axis of symmetry.

In a cross section the shear force Vz is 40 kN as is shown in figure 57. The y-z-coordinate

system is chosen in the NC of the entire cross section.

Figure 57 : Concrete-steel composite beam.

The cross sectional properties of the steel sections are known. The concrete flange is without

slip firmly attached to the steel section. Therefore we can assume both parts to act together as

one rigid cross section. The known quantities are given as:

Cross area steel section steelA = 2mm32000

Second moment of area of steel zz,steelI = 46 mm10432 ×

Modulus of elasticity of steel steelE = 25 N/mm101,2 ×

Modulus of elasticity of concrete concreteE = 2N/mm14000

Shear force zV = kN40

Compute the longitudinal shear flow in the interface between steel and concrete.

Solution:

The shear flow can be found with basic formula (7). To use this formula we need to find the

cross sectional quantities (double letter symbols) first. We start therefore with the location of

the normal center NC of the entire cross section. Due to symmetry we only have to find the

vertical location of the NC. With respect to the shown zy − -coordinate system we find:

concrete steelNC

concrete steel

(200 2000) 100 (32000) 650400 mm

(200 2000) (32000)

E Ez

E E

× × × + × ×= =

× × + ×

Since the shear force in the y-direction is zero we only have to consider quantities in the z-

direction in basic formula (7). For the bending stiffness in z-direction we find EIzz :

3 2 6 3 21zz concrete steel12

12 2

zz

2000 200 (200 2000) (300) 432 10 32 10 (250)

1033,4 10 Nmm

EI E E

EI

= × × + × × + × + × ×

= ×

If we choose the concrete flange as shearing segment, the shear flow per unit of length in the

interface between the concrete flange and the steel section becomes according to basic

formula (7):

[ ]3(a)

concrete(a) zx 12

40 10 (2000 200) ( 300)65 N/mm

1033,4 10

z

zz

EV ESzs

EI

× × × × × −= − = − =

×

y

zz,

200 mm

800 mm NC

-350 mm

450 mm

2000 mm

Vz=40 kN

y y

z



1.7.2 General shear stress formula

If the principal coordinate system of the cross section does not coincide with the chosen y-z-

coordinate system we can not use the basic formula (7) since this formula was based on the

uncoupled bending terms of the constitutive relation which is only valid in the principal

coordinate system. For an arbitrary cross section with its y-z-coordinate system chosen

through the NC of the cross section we will derive a general shear stress formula. We start

with the result of the horizontal equilibrium of the shearing segment as shown in figure 54.

The shear flow per unit of length according to equation (1) is:

x

Rs

d

d (a)(a)

x −= (1)

The normal stress distribution on the shearing area A(a)

is linear and is the result of a

component due to the normal force N and a component due to the resulting moment M in the

cross section. The resultant of these stresses R(a)

can therefore also be split in to a component (a)

NR due to N and a component (a)

MR due to M:

McNcRRR 21

(a)

M

(a)

N

(a) +=+= with: 2

z

2

y MMM +=

Check for yourself that the constants c1 and c2 can be found as:

( ) ( )

1 2;a a

N MR Rc c

N M= =

Differentiating R(a)

and applying the chain rule we find for the shear flow (1):

VM

R

x

M

M

Rs

aaa

xd

d

d

d

d

d )()()( −=−= with: 2

z

2

y VVV +=

In this expression V is the resulting shear force in the cross section. Substituting the linear

relation for R(a)

with respect to N and M we can rewrite this expression as:

(a)

(a) 1 2 Mx 2

d( )

d

c N c M Rs V V c V

M M

+= − = − = − basic formula (8)

In words this formula says that the shear flow per unit of length is equal to the resultant shear

force V multiplied with a scaling factor. This scaling factor is the resultant of the normal

stresses on the shearing segment due to bending only divided by the resultant moment M in

the cross section.

Note that the scaling factor is independent of the magnitude of M. Important in the

application of this method is however that both the shear force V and the moment M

act in the same loading plane since we made use of the equilibrium condition:

x

MV

d

d=

If in a cross section only a shear force is known we can apply a dummy moment to calculate

the scaling factor c2. This will be illustrated in one of the examples.



1.7.2.1 Example 9 : Shear stresses in a non-symmetrical cross section

A cantilever beam is loaded with a horizontal force F as shown in figure 58(a). The steel

section can be regarded as thin walled and is shown in figure 58(b). Thin walled means that

the thickness t is relative small with respect to the length scale a.

Figure 58 : Cantilever beam.

Structural data:

dimensions l = mm2000

a = mm150

t = mm12

Modulus of elasticity steelE = 25 N/mm101,2 ×

Load F = kN85,19

Find the shear stress distribution for the entire cross section at A.

Solution:

The y-z-coordinate system at the NC does not coincide with the principal coordinate system of

this cross section. We can therefore not use the basic formula (7) for the shear flow but have

to use the general applicable basic formula (8). For this we need the normal stress distribution

over the cross section due to bending only. In this case the structure is only loaded in bending

since the normal force N is zero.

a) Normal stress distribution due to bending only

From point symmetry the location of the NC can be found and is shown in figure 58(b). The

sectional quantities with respect to the NC can be computed as:

2103

zyyz

2103

32

zz

2103

38

yy

Nmm105,850

Nmm10567

Nmm102268

×−=−==

×==

×==

tEaEIEI

tEaEI

tEaEI

The sectional forces at the clamped edge can be found based on equilibrium:

0

Nmm106,39

0

z

6

y

=

×−=×−=

=

M

lFM

N

(a) structure

x

l

z F

A

B

y

F y

z

a

t

a a

NC

(b) cross section

a



With the sectional constitutive relations we can find the three sectional deformation quantities

needed to describe the strain and stress distribution of the cross section:

( )

−

−

−=

−

−=×

−⇒

=

z

y

yyzy

yzzz

2

yzzzyyz

y

z

y126

z

y

zzzy

yzyy

z

y

1

67,5505,8

505,868,221010

0

7,39

M

M

EIEI

EIEI

EIEIEI

EIEI

EIEI

M

M

κ

κ

κ

κ

κ

κ

This results in:

1/mm100,61/mm100,4 66 −− ×−=×−= zy κκ

The stress at an arbitrary point of the cross section can be described with:


With zero normal force the strain ε in the fiber which coincides with the beam axis is also

zero. Thus the stresses can be described with:

5 6 2( , ) 2,1 10 ( 4 6 ) 10 0, 21 ( 4 6 ) N/mm with and in mmy z y z y z y zσ −= × × − − × = × − −

The neutral line follows from:

064 =−− zy

In figure 59 the stress distribution is shown in two ways. The first way uses the standard

method in which the stresses are plotted out side the cross section towards a vertical axis

which is perpendicular to the neutral line. The second way is to plot the stresses directly in the

section perpendicular to the thin walled section itself. This is of course only possible since the

stresses in thin walled sections (tickness t << a) are constant over the thickness of the

material.

Figure 59 : Normal stress distribution at A.

2

3

n.l.

n.l.

k

63

126

126

126

+

+

+

-

-

-

-

+

NC

P

Q

R

S

point y z ),( zyσ

N/mm2

P +150 -150 +63

Q +150 0 -126

R -150 0 +126

S -150 +150 -63

126

126

126

63



b) Shear stress distribution

Based on the normal stress distribution we can find the shear stress distribution with basic

formula (8):

VM

Rs

(a)

M(a)

x −=

The shear force V and moment M are the resulting sectional forces:

Nmm107,39

N19850

6

y

2

z

2

y

y

2

z

2

y

×−==+=

==+=

MMMM

VVVV

Substituting this in basic formula (8) results in:

(a) (a)

(a) (a)M Mx M6

19850N/mm with in N

39,7 10 2000

R Rs R

×= − = +

− ×

To make a sketch of the stress distribution we have to know the shear stress in a number of

key points. We will use the earlier found properties of shear stress distributions as described

in section 5.4.2 of MECH-2 ( Dutch edition, page 301 ) :

1 For constant normal stress distributions the shear stress distribution is linear.

2 For linear normal stress distributions the shear stress distribution is parabolic.

3 The shear stress has its extreme value at the points of intersection of the neutral line (due to

bending only) with the cross section.

4 The direction of the shear flow follows mostly from the direction of the acting shear force.

5 Zero shear stresses occur at the edges of the flanges.

6 At the connections of flanges and webs the sum of all shear forces must be zero. The total shear

flow in flux must be equal to the out flux.

Due to symmetry we only have to find the stresses on the part SR-NC of the total cross

section. The key points on this part are NC,R, C and S in which C is an extra point at the point

of intersection of the neutral line with the cross section SR, see figure 60.

Point S:

This is an free edge thus 0(a)

x =s .

Point C:

Shearing segment is CS, the resultant force of

the normal stresses due to bending only is :

N1890012506321(a)

M −=×××−=R

The shear flow per unit of length and the

associated shear stress at C thus becomes:

2

(a)

x

xm

(a)

M(a)

x

N/mm79,0

N/mm45,92000

−==

−==

t

s

Rs

σ

+

R

S

126

126

63

C

+

-

normal stresses

in N/mm2

NC

Figure 60 : Resultant of the normal

stresses on SC.

S 63

C

-

m n.l.

n.l.

50

50

50

150



+

R

S

126

126

63

C

+

-

normaalspanningen

in N/mm2

NC

Figure 61 : Resultant of normal

stresses on SR.

S 63

C

-

m

n.l.

n.l.

R 126

+

m

Figure 62 : Resultant of normal

stresses on SR-NC.

Point R

The shearing segment is now SCR. The resultant

force of the normal stresses on this segment due

to the bending moment M is:

N567001210012612506321

21(a)

M =×××+×××−=R


associated shear stress at R is:

2

(a)

x

xm

(a)

M(a)

x

N/mm36,2

N/mm35,282000

==

==

t

s

Rs

σ

Point NC

The shearing segment is now SR-NC. The

resultant force of the normal stresses on this

segment due to the bending moment M is:

N17010012150126

12100126125063

21

21

21(a)

M

+=×××

+×××+×××−=R


associated shear stress at the NC is:

2

(a)

x

xm

(a)

M(a)

x

N/mm09,7

N/mm05,852000

==

==

t

s

Rs

σ

In all calculations a temporary, outward normal, m-axis is used to define the positive direction

of the shear stresses. The direction of the shear stress found in NC complies with the expected

direction since the (horizontal) shear force acts in the same direction as the obtained shear

stress.

With the previous summarised properties of shear stress distribution we can draw the shear

stress distribution of the entire cross section. In figure 63 this distribution is shown. The

direction of the shear stress is shown with the bold arrows. Notice the three points in which

the shear stress reaches its (local) extreme values at the points of intersection of the cross

section with the neutral line.



Figure 63 : Shear stress distribution.

Since the normal stress distribution is linear for all parts of the cross section, the shear stress

distribution will be parabolic on all segments. For segment SR the distribution can be found

rather quickly since the parabolic distribution reaches its extreme value at C. At D therefore

the shear stress must again be zero!

c) Exact shear stress distribution

The exact expression for the shear stress distribution can also be obtained by using basic

formula (8). As an example we will derive this expression for segment SR.

To describe the shear stress distribution for segment SR we use a shearing segment SW with a

local (outward) normal m-axis as shown in figure 64.

The linear stress distribution on SR can be written

as a function of m with:

mm150

18963)( +−=σ

This normal stress distribution is due to bending

only. This is an important requirement when using

basic formula (8).

The resulting normal force due to bending on the shearing area SW is:

( ) mmmtmR

t

××+−=×= ∫ 12)(63d)(21

0

(a)

M σσ

Combining these two expressings results in:

2

2

1(a)

M25

18975612

150

1896363 mmmmR +−=××

+−+−=

2

3

n.l.

n.l.

NC

P

Q

R

S

2,36

2,36

0,79

7,09

4,73

2,36

0,79

z

y

50

50

50 V = 19850 N

Shear stress distribution in N/mm2

C

D

50

50

50

Figure 64 : Normal stress on SR.

S 63

C

-

m-as

n.l.

σ

σ(m)

m

W



With basic formula (8) the shear flow per unit of length and the associated shear stress on the

segment SR as function of m can be found as:

122000

25

189756

)(

2000

25

189756

2000

2)(

xm

2(a)

M(a)

x

×

+−==

+−==

mm

t

sm

mmR

s

a

xσ

This result shows a parabolic shear stress distribution as was already shown in figure 63.

The shear stress has its extreme value when the first derivative to m becomes zero. This

results in:

mm50024000

225

189756

0d

)(d

=⇒=+

⇔=

m

m-

m

mxmσ

This result is in perfect agreement with the earlier found result. The point were the shear

stress is extreme coincides with the point of intersection with the neutral line. For a few

values of m the shear stress on segment SR is shown in the following table.

m (mm)

xmσ

(N/mm2)

0 0

50 -0,79

100 0

150 2,36

Finding the exact expression for the distribution is of course not a practical way to determine

the shear stress distribution in a cross section. It does however clearly show the earlier

mentioned properties for shear stress distribution and gives additonal qualitative information.

Note

Since the shear force distribution is constant for the entire beam, the shear stress distribution

for all cross sections will be the same. If however the shear stress distribution would have

been asked for in the cross section at B we would have a problem since the bending moment

M at B is zero. The normal stress distribution due to bending also will be zero and the shear

flow with basic formula (8) could not be found. The solution for this problem is to assume a

dummy moment acting in the loading plane of the shear force at B. The resulting normal stress

distribution due to this bending moment can then be used to obtain the scaling factor c2 in

basic formula (8). This scaling factor is independent of the magnitude of M :

(a)

M2

Rc

M=

By doing so the same shear stress distribution will be found. Check this yourself!



1.7.2.2 Example 10 : Shear force in a non homogeneous cross section

In example 4 the stress distribution was found for the non homogeneous cross section. From

this structure we want to find the shear flow in the glued interface RS over the beam section

AC. The structure and the cross section with its properties are shown in figure 65.

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

Properties : E1= 6000 N/mm2 E2 = 12000 N/mm

2

Figure 65 : Example 10

The y- and z-axis of the coordinate system does not coincide with the principal coordinate axis

which means that the shear flow can only be found by using the general method of basic

formula (8). For this we need the normal stress distribution due to bending only which we

already determined inexample 4. We refer to example 4 for these results.

To determine the shear flow in the glued interface we consider the segment OPQS as shearing

segment. The normal stresses at the four corners of this flange were already found in example

4 and are shown in figure 66:

Point Stress [N/mm2]

O -1,25

P 19,17

Q -7,28

S 13,14

Figure 66 : Shearing segment OPQS

Since the normal stress distribution over OPQS is linear, the resulting normal force on the

shearing segment can be found by multiplying the average normal stress with the cross

sectional shearing area:

( )

N5,297210504

14,1328,717,1925,1(a)

M =××+−+−

=R

250 N

0,55 m 0,55 m

x

z

A

B

y E1

E2

10 mm

30 mm

20 mm

z C

E2

10 mm

50 mm

50 mm

O P

Q S

T V

W

X

R

U

10 mm E2

50 mm

O P

S R Q

20 mm m



The outward normal of the shearing interface RS is denoted with the m-axis as can be seen

from figure 66. De shear flow per unit of length can be found with basic formula (8):

VM

Rs

(a)

M(a)

x −=

The shear force V and the moment M in the cross section were already found in example 4:

N137500

N250

−=

=

M

V

The shear flow per unit of length in the direction of the beam axis and in the longitudinal cut

(glued interface) of the shearing area RS becomes:

N/mm4,5250137500

5,2972(a)

x =×−

−=s

Assuming a uniform shear stress distribution in the glued interface we can find the shear

stress in the glued interface with:

2

(a)

(a)

x

mx N/mm27,020

4,5===

b

sσ

The interface should be designed in such a way that it can resist this shear stress. If so the

interface can be seen as a rigid interface without slip. The composite cross section then acts as

a rigid cross section.



1.7.3 Shear force center for thin walled non-symmeyrical cross sections

In case of thin walled cross sections we assume constant shear stresses over the thickness of

the material. The resultant force of these shear stresses is of course equal to the acting shear

force in the cross section. However its line of action will in most cases not act in the normal

force center NC but in the so-called shear force center SC, see also section 5.5 of MECH-2 .

The shear force center is defined as:

The shear force center SC is a point in the plane of the cross section were the shear force

must be applied in order to cause shear stresses without torsion.

In case of rotation symmetrical cross section the shear force center SC conincides with the

normal force center NC. For arbitrary shaped cross sections however this will not be the case.

If the cross section has an axis of symmetry, the shear force center SC will be located on this

axis. For a number of standard sections this has been illustrated in MECH-2 (Dutch edition,

page 331).

In this section we will show how to find the location of the shear force center SC in case of

arbitrary shaped thin walled cross sections without an axis of symmetry. The procedure which

has to be followed is an extension of the outlined method of the previous sections. The

neccesary steps to find the shear force center are:

Step 1 : Find for an arbitrary shear force the shear stress distribution in the cross section.

Determine the line of action of the resulting shear force and denote this force with its

line of action as R1 .

Step 2 : Find for a second shear force which is different with respect to its direction to

the one from step 1, the shear stress distribution on the cross section.

Determine the line of action of the resulting shear force and denote this force with its

line of action as R2 .

Step 3 : The shear force center is the point of intersection of the two lines R1 and R2 .

Since we try to find the point of intersection of two lines of action it is essential to choose two

shear forces which are not parallel to each other.

The term arbitrary also means that we can choose a clever or intelligent special set of shear

forces since we are free in our assumption of the directions and the magnitude of the set of

shear forces as long as these two shear forces are not parellel to each other.

If we choose two unity shear forces which acts in the direction of the coordinate system we

can simplify the calculations as will be illustrated in the following example.



1.7.3.1 Example 11 : Shear force center for thin walled cross sections

The thinwalled cross section PQRS of figure 67 has no axis of symmetry. Find the location of

the shear force center SC.

Properties : a = 10 mm; t = 6 mm; E = 2,0×105 N/mm

2

Figure 67 : Example 11, thin walled non-symmetrical cross section.

The shear stress distribution associated to an assumed shear force will be determined using a

dummy moment as described earlier. We can therefore use both unity shear forces and unity

bending moments acting in the cross section. To find the stress distribution we need the

secional quantities and the location of the normal center. We therfore start with the location of

the normal center NC.

Cross sectional quantities

With help of a temporary coordinate zy − -system as shown in figure 67, we can find the

location of the NC as:

mm33,533

16

864

884624

mm33,233

7

864

083664

NC

NC

==++

×+×+×=

==++

×+×+×=

a

atatat

aataataatz

a

atatat

ataataaty

With the NC as origin of the y-z-coordinate system we can find the sectional quantities, the

so-called double letter symbols, as:

3 2

yy

3 2

zz

3 2

yz zy

118 708000 Nmm

160 960000 Nmm

104 624000 Nmm

EI Ea t E

EI Ea t E

EI EI Ea t E

= =

= =

= = − = −

Vy

y

z 4a

t

6a

8a

Vz

SC

NC

y

z

P

Q R

S



The shear stress distribution can only be found using the general method described in section

1.7.2 since the section is non symmetrical. In this method we need a normal stress distribution

due to bending only. We will use a dummy moment in the same loading plane as the acting

shear force. The magnitude of this moment is of no importance. We therefore choose a unity

moment as dummy.

To find the location of the shear force center we have to compute two shear stress

distributions. The first distribution is the result of a unity shear force Vy and the associated

(dummy) unity moment My . The second distribution is the result of a unity shear force Vz

with the associated (dummy) unity moment Mz. All calculations can be made using MAPLE.

Load case 1 : Nmm0,0

0,1;N

0,0

0,1

=

=

z

y

z

y

M

M

V

V

Load case 2 : Nmm0,1

0,0;N

0,1

0,0

=

=

z

y

z

y

M

M

V

V

With this “intelligent” choice we can simplify basic formula (8) to:

b

R

b

s

RVM

Rs

(a)

M

(a)

x

xm

(a)

M

(a)

M(a)

x

−==

−=−=

σ

( modified formula 8 )

With this expression we can easily find the shear flow and shear stress distribution due to a

unity shear force.

Normal stress distribution for load case 1

In the same manner as presented in example 9 the normal stress distribution can be found

starting with the sectional constitutive relation:

( )

−

−

−=

−

−×=

⇒

=

0

0,11

960624

624708100,2

0

0,1

0

0,1

yyzy

yzzz

2

yzzzyyz

y

z

y8

z

y

zzzy

yzyy

EIEI

EIEI

EIEIEI

EIEI

EIEI

κ

κ

κ

κ

κ

κ

The components of the curvature results from the above expression as:

1/mm10107474,01/mm10165344,0 1010 −− ×=×= zy κκ

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


The strain ε in the fiber through the normal center NC is zero since the normal force is zero.



The expression for the stress thus becomes:

5 5 2( , ) 0,330688 10 0, 214947 10 N/mm with and in mmy z y z y zσ − −= × + ×

The stresses in the key point P, Q,R and S can be found as:

point y z ),( zyσ

N/mm2

P +36,67 -53,33 0,066e-4

Q +36,67 -13,33 0,926e-4

R -23,33 -13,33 -1,058e-4

S -23,33 +66,67 0,661e-4

The resulting normal stress distribution is shown in figure 68.

Figure 68 : Normal stress distribution of load case 1 in N/mm2 .

To find the shear stress distribution we only need the determine shear stresses at a few points

using the following considerations:

2. the shear stress is zero at P and S

3. the shear stress is extreme for zero normal stresses ( at A and B)

4. the shear stress distribution is parabolic for linear normal stress distributions

To draw an accurate shear stress distribution we only need the shear stresses in four points; A,

B, Q and R. With the modified basic formula (8) we find the following shear flows:

N/mm10105,6)10058,1623,49(

N/mm10524,9)10058,1632(

N/mm10683,19)10926,0628(

N/mm10905,1110)926,0066,0(640

34

21

Rx(a)

Bx(a)

34

21

Ax(a)

Rx(a)

34

21

Qx(a)

Ax(a)

34

21(a)

M

(a)

MQx

(a)

−−−−

−−−−

−−−−

−−−

×=×−×××−+=

×−=×−×××−+=

×−=××××−+=

×−=×+×××−=−=×−=

ss

ss

ss

RVM

Rs

60 mm

normal stresses

×10-4

N/mm2

0,066

0,926

0,926

1,05

1,05

0,66

nl

A

B

11,9

66

40 m

m

80 m

m

28,0 mm

49,2

3 m

m

+ -

+

+

-



The shear flow distribution is shown in figure 69. The arrows show the direction in which the

shear flow actually works.

Figure 69 : Shear flow distribution of load case 1.

For each segment of the cross section the resultant force can be found by integrating the shear

flow function and multiply this result with the constant thickness of the segment. In order to

find the expressions for the shear flow distributions for each of the segments DQ, QR and RS

only the shear flow in Q and R are needed.

For segment PQ we choose the variable x from P towards Q and find for the resulting shear

force:

N1693,04

)(4

0

PQ21

P

PQ −=

×

−+××−= ∫

a

dxa

xxtR

σσσ

For segment QR we choose the variable x from Q to R and find for the resulting shear force:

N0,16

)(6

0

QR21

QQ-x(a)QR −=

×

−+××−= ∫

a

dxa

xxtsR

σσσ

For segment RS we choose the variable x from R to R and find for the resulting shear force:

N1693,08

)(8

0

RS21

RR-x(a)RS =

×

−+××−= ∫

a

dxa

xxtsR

σσσ

The sum of the horizontal and vertical components of the shear force are indeed equal to the

applied shear forces Vy (1,0) en Vz (0,0).

The direction of the shear flow in segment QR can be found directly from the acting shear

force or with help of a local outward normal m-axis. Draw a small sketch to check this

yourself!

60 mm

shear flow

×10-3

N/mm

11,9

11,9

9,5

4

nl

A

B

11,9

66

40 m

m

80 m

m

28 mm

49,2

3 m

m

19,7

9,5

4

P

Q

R

S

6,1



The resulting forces on each segment are shown in figure 70. Based on equilibrium the

resulting force and its line of action can be determined. This is also shown in figure 70. The

resulting force for load case 1 is of course equal to the unity shear force Vy .

Figure 70 : Line of action of the resulting shear force for load case 1.

The shear force center, we are looking for, must lie somewhere on this line of action. To find

the exact location we need a second line of action. At the point of intersection of these two

lines we find the location of the shear force center. The whole procedure has therefore to be

repeated for load case 2. We will only show the brief results since this is a repetition of steps.

Normal stress distribution for load case 2

With the (dummy) unity moment Mz we can find the sectional defomations with the sectional

constitutive relation:

( )

−

−

−=

0,1

01

yyzy

yzzz

2

yzzzyyz

y

EIEI

EIEI

EIEIEIκ

κ

The components of the curvature become:

1/mm10121941,01/mm10107474,0 1010 −− ×=×= zy κκ

The expression for the normal stress distribution will become:

mminenmetN/mm10243882,010214947,0),( 255 zyzyzy −− ×+×=σ

The stresses at key points P, Q,R and S can be found as:

point y z ),( zyσ

N/mm2

P +36,67 -53,33 -0,513e-4

Q +36,67 -13,33 0,463e-4

R -23,33 -13,33 -0,827e-4

S -23,33 +66,67 1,124e-4

60 mm

0,1693 N

40 m

m

80 m

m

P

Q R

S

0,1693 N

1,0 N

1,0 N

line of action of the resultant

of all shear stresses in the

cross section. e=10,16 mm



To find the shear flow distribution for the segments PQ, QR and RS we need the shear flow at

the points Q and R:

N/mm10143,7)10)827,0463,0(660(

N/mm10595,010)463,0513,0(640

34

21

Qx(a)

Rx(a)

34

21(a)

M

(a)

MQx

(a)

−−−−

−−−

×=×−×××−+=

×−=×+−×××−=−=×−=

ss

RVM

Rs

The resulting forces per segment can be found after integration:

N090,04

)(4

0

PQ21

P

PQ =

×

−+××−= ∫

a

dxa

xxtR

σσσ

N06

)(6

0

QR21

QQ-x(a)QR =

×

−+××−= ∫

a

dxa

xxtsR

σσσ

N910,08

)(8

0

RS21

RR-x(a)RS =

×

−+××−= ∫

a

dxa

xxtsR

σσσ

The total force must equal the applied shear force Vz (1,0). The results are shown in figure

71(a).

Figure 71 : Line of action for load case 2 and the shear force center SC.

In figure 70(b) the shear force center is located at the point of intersection of the two line of

actions of the applied shear forces for load case 1 and 2. The normal force center is also

shown. It is clear that for non symmetrical cross sections these two centers do not coincide.

Conclusion

• If only a normal force acts on a cross section through the normal force center NC, the

cross section is loaded in extension only and no bending will occur.

• If only a shear force acts on a cross section through the shear force center, the cross

section is loaded in shear only and no torsion will occur.

60 mm

0,91 N

40 m

m

80 m

m

P

Q R

S

0,09 N

0 N

1,0 N

line of action of the resultant

of all shear stresses in the

cross section.

e=54,60 mm

54,60 mm

(a) : Line of action of load case 2. (b) : Normal force center NC and

shear force center SC

10,16 mm

NC

y

z

SC



APPENDIX A

The proof that the curvature can be considered as a first order tensor will be given below. In

order to do so we have to examine the effect of a rotation of the coorinate system on the

expression of the strain field in a cross section.

y z( , )y z y zε ε κ κ= + +

For a given point P, see the figure below, we can find the relation between its position in the

original y-z-coordinate system and the rotated y z− -coordinate system as:

Figure 72 : Coordinate transformation due to a rotation.

cos sin

sin cos

y y z

z y z

α α

α α

= +

= − + and the inverse :

cos sin

sin cos

y y z

z y z

α α

α α

= −

= +

These latter relations are the tensor transformation rules for first order tensors. The strain in

the y-z-coordinate system can thus be expressed in terms of the rotated coordinate system

with:

( ) ( )

( ) ( )

y z

y z

y z y z

y z

( , )

cos sin sin cos

cos sin sin cos

y z y z

y z y z

y z

y z

ε ε κ κ

ε α α κ α α κ

ε κ α κ α κ α κ α

ε κ κ

= + +

= + − + +

= + + + − +

= + +

From this latter expression follows that the transformation of the curvature complies with the

transformation rule of a first order tensor from which we can deduce that the curvature is a

first order tensor.

More or less in the same way we can proof that also the moment vector M is a first order

tensor which will be shown on the next page.



To do so we investigate the effect of a rotation on the expressions of the sectional forces,

normal force N and moment M. The previous found relations also hold in the rotated

coordinate system, thus:

( , )d

( , )d

( , )d

y

z

N y z A

M y y z A

M z y z A

σ

σ

σ

=

=

=

∫

∫

∫

In het original system we find:

( , )d

( , )d

( , )d

y

y

N y z A

M y y z A

M z y z A

σ

σ

σ

=

=

=

∫

∫

∫

By using the inverse relation for the coordinate transformation from the previous page,

cos sin

sin cos

y y z

z y z

α α

α α

= −

= +

we find:

( ) ( )

( ) ( )

( , )d

( cos sin ) ( , )d

( , )d cos ( , )d sin

cos sin

( sin cos ) ( , )d

( , )d sin ( , )d cos

sin cos

y

y z

z

y z

N y z A

M y z y z A

y y z A z y z A

M M

M y z y z A

y y z A z y z A

M M

σ

α α σ

σ α σ α

α α

α α σ

σ α σ α

α α

=

= −

= −

= −

= +

= +

= +

∫

∫

∫ ∫

∫

∫ ∫

The components of the moment M transform like:

cos sin

sin cos

y y z

z y z

M M M

M M M

α α

α α

= −

= +

This transformation is exactly the same as the transformation rule for a first order tensor.

Both the curvature and the moment vector can be regarded as first order tensors..



APPENDIX B

The moment distribution in the xy- and the xz-plane of a statically indeterminate element

exhibits a coupling if the boundary conditions in both planes are not the same. As an example

we consider the same structure as in figure 19-2 and will vary the boundary conditions. The

beam is loaded with a constant distributed load qy and qz and the two differential equations to

be solved are:

2

2

"''

"''

zz y yz z

y

yy zz yz

yz y yy z

z

yy zz yz

EI q EI qu

EI EI EI

EI q EI qu

EI EI EI

−=

−

− +=

−

The general solution of these ODE are:

4 3 2

1 23 42

4 3 2

1 23 42

( )

24( ) 6 4

( )

24( ) 6 4

zz y yz z

y

yy zz yz

yz y yy z

z

yy zz yz

EI q EI q x C x C xu C x C

EI EI EI

EI q EI q x D x D xu D x D

EI EI EI

−= + + + +

−

− += + + + +

−

Situation 1:

The beam is fully clamped at the left end at x = 0 and simply supported at the right end x = l.

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

ϕ ϕ= = = = =

= = = = =

Solving the ODE we find for the moment distributions:

2 218

2 218

( ) (4 5 )

( ) (4 5 )

y y

z z

M x q x xl l

M x q x xl l

= − − +

= − − +

In this case we find the normal moment distribution which we could also have obtained with a

simple analysis in one plane. The moment at the clamped edge is:

218

218

(0)

(0)

y y

z z

M q l

M q l

= −

= −

Situation 2:

The beam is only fully clamped in the xz-plane at the left end at x = 0. In the xy-plane the

beam is at this location simply supported. At the right end x = l the beam is in both the xy- and

the xz-plane simply supported. 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 M

x l u u M M

ϕ= = = = =

= = = = =



Solving the ODE we find for the moment distributions:

12

2 2 2

( ) ( )

1( ) (4 5 )

8

y y

z yy z yy z yy z yz y yz y

yy

M x q x x l

M x EI q x EI q xl EI q l EI q xl EI q lEI

= − −

= − − + + −

From this result the coupling is quite evident. The moment in the xz-plane at the clamped

edge is reduced due to the coupling in the constitutive relation and becomes:

2 2

218

( )(0) 0; (0)

8 8

yy z yz y yz y

y z z

yy yy

EI q EI q l EI q lM M q l

EI EI

− + ×= = = − +

The MAPLE code shown below can be used to verify these results. The boundary conditions

for situation 1 and 2 can be activated by removing or adding the # character at the start of the

line. ( All text after the # character is regarded as comment .)

> restart;

Double bending for unsymmetrical and or inhomogeneous cross sections,

one side clamped beam and simply supported at the other end,

loaded with a constant distributed load qy and gz :

> DV1:=diff(uy(x),x$4)=(EIzz*qy-EIyz*qz)/(EIyy*EIzz-EIyz^2);

> DV2:=diff(uz(x),x$4)=(EIyy*qz-EIyz*qy)/(EIyy*EIzz-EIyz^2);

General solution of the two differential equations:

> sol:=(dsolve({DV1,DV2},{uy(x),uz(x)})); assign(sol);

> uy:=uy(x); uz:=uz(x);

related quantaties:

> phiz:=diff(uy,x): kappay:=diff(-phiz,x):

> phiy:=-diff(uz,x): kappaz:=diff(phiy,x):

> My:=EIyy*kappay+EIyz*kappaz: Vy:=diff(My,x):

> Mz:=EIyz*kappay+EIzz*kappaz: Vz:=diff(Mz,x):

8 boundary conditions, clamped end at x=0 and a simply supported end at x=L:

> #x:=0: eq1:=uy=0: eq2:=phiy=0: eq3:=uz=0: eq4:=phiz=0: # situation1

> x:=0: eq1:=uy=0: eq2:=My=0: eq3:=uz=0: eq4:=phiy=0: # situation2

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

> x:='x':

solve the integration conastants:

> sol:=solve({eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8},

{_C1,_C2,_C3,_C4,_C5,_C6,_C7,_C8}): assign(sol):

Moment and Shear distribution in the beam:

> simplify(My); simplify(Vy):

> simplify(Mz); simplify(Vz):

> x:=0: simplify(My); simplify(Mz);

Example from lecture notes:

> x:='x': L:=10; qy:=0; qz:=8;

> Emod:=100e6: a:=0.1:

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

joke: very unsuspected behaviour, remove the # to play with this !!

> #EIzz:=10000; EIyy:=20000; EIyz:=4000;

plot: I plot graphs with a minus to obtain visually positive downwards graphs

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

plot: I plot graphs with minus to obtain visually positive moments downwards

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

> x:=0: evalf(My); evalf(Vy); evalf(Mz); evalf(Vz);

> x:=L/2: evalf(uy); evalf(uz); evalf(My); evalf(Mz);

coupling



2. ASSIGNMENTS

2.1 Cross sectional properties

All specified problems are related to the determination of the cross sectional quantities of

homogeneous non-symmetrical cross sections.

Problem 1

Question:

a) Centre of gravity Z.

b) The second moments of area for the given

coordinate system.

c) The second moments of area for a coordinate

system which is rotated by 45o .

d) The second moments of area for a coordinate

system with its origin at C.

Problem 2

Question:

a) Centre of gravity

b) , , yy zz yzI I I

c) Principal directions

d) Principal second moments of area (moments of inertia)



Problem 3

Given: 3' ' 1/12z zI bh= ×

Question:

a) Check with Steiner’s parallel theorem.

b) Compute the magnitude of '' ''z zI

Problem 4

Question:


b) The second moments of area : yzzzyy ,, III

d) The principal directions

e) Principal second moments of area (moments of inertia)

Problem 5

Question:


b) , , yy zz yzI I I

c) Pricipal directions

d) Principal second moments of

area (moments of inertia)

e) The coordinate system for

which the magnitude of ' 'y zI

is at largest.



Problem 6

Question:

a) Centre of gravity.

b) Quantities: , , yy zz yzI I I .

c) Draw Mohr’s circle and find the principal

directions and the principal values of the

second area moments 1 2 en I I

in mm, all walls

thickness t = 10



2.2 Normal stresses in case of bending

All problems are related to non symmetrical cross sections loaded in pure bending only.

Problem 1

Question:

Find the normal stress distribution for the cross section at the clamped edge of the cantilever

beam. Note: assume the thin-walled theory since t << a.

Problem 2

A cross section is loaded with a bending moment M of

100 Nm. The force line is shown in cross section. The load

configuration is shown in the figure at the left where F and l are

unknown.

Question:

a) The normal stress distribution of the cross section.

b) The location of the neutral line.

c) The direction of flexure.

force line



Problem 3

The cross section is loaded in bending in such a way that the neutral axis coincides with the y-

axis.

Question:

a) Find directly from the stress diagram the force line m.

b) Find the quantities , , yy zz yzI I I and the principal coordinate system.

c) Find the under a) force line m again but now from the neutral line

towards the principal coordinate system.

Problem 4

The location of the neutral line is known and shown in the figure at the

right.

Question:

a) Find the location of the force line in the cross section to the right.

b) Is it possible to find the force line from the stress distribution in a

simple way ?

Problem 5

A cross section of a beam loaded in pure bending is shown in the figure

to the right. In A and C we know that the same normal stresses occur

with a magnitude of +10 MPa.

Question:

a) Find the magnitude of the bending moment.

b) Find the stress in point B.

neutral line



Problem 6

A cantilever beam with a I-section type of HE 160A is loaded by two forces and y zF F as

shown in the figure.

Properties:

4 4yy

4 4

max

HE 160A

612 10 mm

1673 10 mm

100 MPa

160 mm; 152 mm; 3000 mm

zz

I

I

b h l

σ

= ×

= ×

=

= = =

Question:

Find the maximum values for and y zF F for which holds: zy FF = .

Problem 7

A beam with a cross section as shown in the figure is

loaded in pure bending. The largest compressive

stress is –50 MPa and the largest tensile stress is +70

MPa.

Question:

a) Find the magnitude and the direction of the

moment in this cross section.



Problem 8

A thin walled section is loaded with:

2

2

80

52

y

z

M a t

M a t

= + σ

= + σ

Question:

a) Find the stresses at the corner points A, B, C

and D.

b) Find the location of the neutral line.

c) Determine the second moment of area yyI in

the direction perpendicular to the neutral line.

d) Check the stress at A which results when using the stress formula.

Problem 9

Beam AB with a L-section as shown in the figure is clamped at A with the vertical (web) of

the section. At B the beam is loaded with a vertical force F = 500 N. The force acts at the

normal force centre NC of the thin walled cross section.

Question:

a) Draw the normal stress distribution for a cross section at A. Put the values and the signs in

the graph.

b) Find the location of the shear force center S.C. and the torsional moment in this section.

c) Calculate the displacement of the normal center NC at B.

d) Find the maximum shear stress and its location in a cross section.

NC

Properties:

GPa200

mm105,13

mm105,22

44

yz

44

zzyy

=

×−=

×==

E

I

II



NC

NC

Problem 10

A cantilever beam with a L-section 100 / 100 / 10 is clamped at the vertical plane A-D. The

normal force center NC at the end B is attached to a tension rod BC in such a way that the

cross section at B can only move in vertical direction. The deformation of the tension rod BC

is negligible. The vertical load of 1000 N is applied at the normal force centre. Use a Young’s

modulus E = 210 GPa. The cross section can NOT be regarded as thin walled.

Question:

a) Find the normal force in the tension rod in such a way that the cantilever beam AB only

shows a vertical displacement. HINT : Use the pseudo-load.

b) Show that for each cross section over AB the neutral line is a horizontal line.

c) Draw the normal stress distribution due to the bending moment at the clamped support,

put the values and signs in the graph.

d) Find the maximum shear stress in this cross section due to the total applied shear force.

For this computation you may regard the cross section as thin walled.

e) Find the torsional moment in beam AB and show its direction.

NOTE:

The influence of the torsional moment on the shear stresses is not asked for but you can

add these to the shear stress found due to shear only.



2.3 Normal stresses due to bending and extension

The following assignments all deal with unsymmetrical and or inhomogeneous cross sections loaded

in bending and/or extension.

Problem 1

A column is loaded by an excentric

load as shown in the figure.

Question:

a) Determine the stress distribution of

the normal stress in cross sesction

C-C for F = 10 kN. The point of

application of this force is point A

as shown in the top view.

b) The same question but now for a

force which acts in point B as

shown in the top view.

Problem 2

Answer the same questions for a thin

walled column with a cross section as

shown in the figure to the right.

Top view

cross section C-C

dimensions in mm

dimensions in mm

t= 10 mm

dimensions in mm.

a) b)



Problem 3

For both cross section the normal stress distribution is geven. Each cross section is loaded by an

excentric normal force.

Question:

a) Find the normal force N

b) Find the location of the force point in the cross section.

Cross section 1 Cross section 2



2.4 Inhomogeneous cross sections loaded in extension

Problem 1

A concrete column of 4 meters in length is reinforced

with a steel HEA-section. The column must transfer

a compressive force of 160 kN.

200 GPa 14 GPast bE E= =

questions:

a) The distribution of normal forces over the cross-

section

b) The shortening of the column

c) Determine the maximal compressive force on the

column when

concrete steel10 MPa and 160 MPaσ = σ =

Problem 2

The shown concrete cross-section is reinforced with 8

steel bars 25∅ .

210 GPa 14 GPas bE E= =

The allowed stress in the concrete and steel is 10 MPa

and 160 MPa respectively.

questions:

a) the maximal compressive force on the column

b) The point of application of this compressive force

in the case where no bending occurs in the column.

Problem 3

A column of length 3 ml = with a square cross-section is

built up of four squares of 40 × 40 mm2 of different materials.

1 330 GPa 100 GPaE E= =

2 460 GPa 200 GPaE E= =

questions:

a) Determine the elastic center of gravity for this

inhomogenuous cross-section (Normal centre).

b) If for all the materials the maximal compressive stress is

100 MPa, what is the maximal compressive force F on the

column that acts on the elastic center of gravity.



2.5 Inhomogeneous cross sections loaded in bending

problem 1

A wooden beam 6 meters long is reinforced with steel

strips in various ways but in such a manner that wood and

steel have a perfect bond. They work as one.

15 GPa 210 GPa

7 MPa 140 MPa

h st

h st

E E= =

σ = σ =

questions:

Determine the maximal load F if:

a) The wooden beam is not reinforced

If the beam is reinforced with:

b) Two plates of 120 × 10 mm on the sides,

c) One strip of 60 × 10 mm at the top,

d) Two strips of 60 × 10 mm, one at the top and one at

the bottom

Problem 2

A cantilever beam is built up from 3 strips of 60 × 10

mm2, connected in such a way that they do not move

relative to each other.

1

2 3

500 N 60 GPa

80 GPa 100 GPa

F E

E E

= =

= =

question:

Determine the stress diagram of the cross-section at the

clamped end.

maten in mm



Problem 3

A bi-metal beam, built up of strips 2 × 20 mm2, of 120 mm long undergoes a temperature decrease of

200 K. 6 -1

1 1

6 -12 2

200 GPa 12 10 K

67 GPa 15 10 K

E

E

−

−

= α = ×

= α = ×

question:

The stress diagram over the cross-section.

Problem 4

A concrete column is reinforced on the short

side with: 2 2

1 21000 mm resp. 500 mmA A= = .

In the reinforcement with 1A there is a tensile

stress of 20 MPa and in 2A a compressive stress

of 60 MPa. In the tensile area we assume the

concrete is cracked.

15st

b

E

E=

questions:

a) Determine the force point ( )e

b) Determine the magnitude of the load

Problem 5

A beam of 200 × 400 mm2 of a material with a modulus of elasticity

of 14 GPa is reinforced with steel rebar ( 210 GPa)steelE = at a

distance of 50 mm from the bottom. The area of the rebar is 4800

mm2.

question:

The distribution of the normal stresses for pure bending in the

x,z-plane, such that the reinforcement is loaded in tension.



Problem 6

If the beam from the previous exercise is made from reinforced concrete, the low tensile strength of

the concrete must be taken into account. To do this most of the time a fully cracked cross-section is

assumed in the tensile zone, which means no strength remains there in the concrete. Determine in that

case the distribution of the normal stresses.



2.6 Core

Problem 1

Determine the core of the shown cross-section:

a) analytically

b) grafically

Determine furthermore:

c) The stresses occuring in corners A, B and C due to a

bending moment MyM =

d) The force line when the neutral axis coincides with the y-axis

Problem 2

Determine the core for these three cross-sections

Problem 3

a) Determine for these cross-sections the core.

b) Determine the stress distribution if the cross-section is loaded with a compressive force in A.

2

4a

3 3



Problem 4

The given cross-section transfers a compressive force of

a 100kN.

questions:

a) Determine the location of the force point.

b) Draw the stress distribution.

Problem 5

The drawn profile is the cross-section of a simply

supported concrete beam with a length l loaded with a

variable uniformly distributed load.

Indicate how pretension cables have to be placed in

this cross-section such that the maximal bending

moment is maximized. l = 8 m.

Problem 6

The drawn cross-section transfers a bending

moment 150 kNmzM = .

questions:

a) Determine the core of the cross-section

b) Draw the normal stress distribution with

values and signs.

c) For what direction of the force line is the

deflection direction vertical?



Problem 7

Two beams:

a) Determine the bending stresses occurring in the middle cross-section of two simply supported

beams loaded with a uniformly distributed load of 2kN/m. The length of the beams is 2 meters and

the cross-sections are as the shown profiles in a and b above.

b) How must the beams be rotated around their axis such that the occurring bending stresses are

minimized?

c) How must the beams be rotated around their axis such that the direction of deflection is vertical?

Problem 8

A hollow thin-walled column has a cross-section as

indicated.

questions:

a) Determine the moments of inertia for the given

coordinate system through the center of gravity.

b) Draw the core of this cross-section.

c) If it is given that the neutral line due to a normal

force of 2 kN coincides with AE, sketch the stress

distribution and determine the location of the force

point.

300

300

300



2.7 Shear stresses due to bending

Problem 1

Given: a beam with a profile as drawn has to transfer a vertical shear

force Q downwards.

question:

Determine the shear stress distribution.

What is the line of application of Q such that no torsion occurs?

Problem 2

Answer the same questions as the previous problem for this cross-

section.

Problem 3

The drawn profile has to transfer a vertical shear force

Q without torsion. ( , 2 / 3t h b h<< = × )

questions:

a) Determine the line of application of Q

b) Determine the distribution and direction of shear

stresses in the cross-section.



Problem 4

A cantilever beam with a thin-walled cross-section is loaded by two forces F1 and F2, as

indicated in the figure.

Given:

F F=1

2 2F F=

Questions:

Determine the shear stress distribution in the cross-section A .

t

z

2a

z

a

a y

F2

F1

A

E

l y

Cross-section profile: B

C

D

A

MODULE : NON-SYMMETRICAL AND INHOMOGENEOUS...

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