Lecture Notes

DEN-102: Handout No. 1

1. Equilibrium of Forces & Moments

Force

There are many kinds of force, such as gravity, gas pressure, atmospheric pressure and wind pressure, magnetic and electromagnetic attractions, nuclear. Four fundamental forces are recognised in physics relating to the interaction of particles (i) gravitational, (ii) electromagnetic, (iii) strong forces bonding atomic nuclei, (iv) weak forces bonding atomic nuclei.

Definition: A force is an action on a body that maintains or alters its position or distorts it or changes its velocity if it is moving.

The unit of force is the NEWTON - symbol N in the SI system. Multiples of this basic unit are commonly used.

Kilonewton kN = 103N Meganewton MN = 106N Giganewton GN = 109N

Newton’s laws

Newton's laws concern the action and movement of solid bodies at rest or in motion. The subject of bodies at rest is termed STATICS, while that of bodies in movement is called DYNAMICS. The first law states that - if a body is at rest or moving at constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. The second law is a quantitative description of the changes that a force can produce in the motion of a body. Simply stated it is onacceleratiMassForce ⋅= In the SI system, the unit of mass is the kilogram (kg), while the unit of acceleration is 1m/s2. Hence 1N = 1kgm/s2 The third law states - "Reaction is always equal and opposite to action". Thus if a book is pushing down on a table with its own


gravitational force (mass x local specific force due to gravity), there is an equal and opposite force from the table pushing upwards on the book - the reaction. In turn the gravitational force of the book and table on the floor is subject to a reaction from the floor on the table.

Stress Analysis/Mechanics of Materials

The subject of stress analysis is mainly concerned with statics, ie bodies and structures at rest under the action of several forces. The structure is simply a series of 'bodies' joined together. Examples of structures are:

- rods in tension or compression - beams in bending - space frames for buildings - an aircraft in flight under the action of many forces - a crank-shaft and piston in an engine - a composite material with fibres and a matrix.

Stress analysis deals with the effects of applied loads which are external to a body or structure. The external forces result in internal reacting forces together with deformations and displacements.

A system must be defined which is an identifiable collection of matter on which the attention is focused and analysed. A system may consist of a single component (e.g. a rod) or a number of components connected together in a specified manner. It is sometimes useful to draw a boundary around the system.


Figure 1.1 The applied load and the reactions are forces external to the system. Internal forces are found in the cable and the beam. To find the internal forces, it becomes necessary to consider part of the whole system called a subsystem. A diagram showing forces on the subsystem is referred to as a free body diagram (FBD).

Figure 1.2

Conditions of Static Equilibrium

A body is said to be in equilibrium when the combined action of the forces and moments acting on that body cause no bodily movement (i.e. dynamic movement). The forces and moments can, and do, cause distortion of the body.


1.1.1 Coplanar, concurrent forces

Consider first forces F1, F2, ... acting through a common point (i.e. concurrent forces) and in the same plane (coplanar).

Figure 1.3 For equilibrium we have

∑=

=n

i

iFF1

where F is the resultant (the vector sum) of all the forces. It follows that:

∑∑

=

=

iyy

ixx

FF

FF

where suffixes x and y denote components in these directions.

Example 1.1 - Resolution of forces at a point:

Fig 1.4 shows three forces acting at a point.


If the system is in equilibrium, the three conditions for equilibrium apply. Since all forces act through the same point, no moments operate. Hence only resolution of forces in two orthogonal directions can be used to determine the values of each force. Resolving in the y direction

)90cos()30cos()45cos( 321321 FFFFFFFF yyyyy ++=⇒++=

Resolving in the x direction

)90sin()30sin()45sin( 321321 FFFFFFFF xxxxx ++=⇒++=

Three unknown forces exist in these equations, F1, F2 and F3, but

there are only two equations. Hence the value of one of the forces must be defined before the equations can be used to find all force values absolutely. Set F1 = 10kN. Then

∴F2 = 8.16kN

∴F3 = 11.15kN

1.1.2 Coplanar, non-concurrent forces

The non-concurrent forces may, or may not, be parallel to each other.

x

y

F1

F3 F2

45o

30o


Moments

Figure 1.5 A moment is the product of a force acting on lever arm. Consider a force, F, acting orthogonally at the end of a bracket (or rod) of length lo. Under the action of the force, the rod will tend to rotate

unless it is constrained. If the rod is held, the action is a moment, M, equal to

)90sin(⋅⋅=⇒×= FlMFlM oo

Clearly there must be a reaction, preventing the rotation occurring, the restraint can be replaced by reaction in the opposite sense to the action (Newton's 3rd law). Parallel Coplanar Forces

Figure 1.6 The resultant force as well as the moment of all the forces about any arbitrary point (e.g. A) must be zero for equilibrium. Thus and

F M

lo

X1

X2

X3

X4

F1

F3

F2

F4

A


where MA denotes the moment of the forces about an axis through

A, perpendicular to the plane. Example 1.2 Consider the beam shown in Fig 1.5 subjected to a force, F, at distance, a, from one support of the beam and b from the other. Find out how this beam is in equilibrium.

Figure 1.7 First replace the supports by equivalent restraints, in this case vertical reactions, R1 and R2.

Figure 1.8 (a) Resolving vertically, we obtain

21 RRF +=

(b) Resolving horizontally does not add further information - there are no horizontal forces. (c) Take moments about any point in the system. To eliminate one of the unknown reactions, we take the point through which one of these act.

0)(2 =+⋅+⋅−= baRaFM A

a b

F

R1 R2 A

B

a b F


Note that the two forces act in opposite directions. R2 acts

anticlockwise, taken as positive, while F acts clockwise, taken as negative, about A. Non-parallel Coplanar Forces

Figure 1.9 In this case, the resultants in two directions as well as moments about an arbitrary point must be zero, i.e.,

a4

a1

a3

a2

F1

F2

F3 F4


Types of Structural and Solid Body Components

'Tie' prevents parts A and B from moving apart when a pull is applied.

'Strut' prevents parts A and B moving towards each other when compressive force is applied. 'Column' supports mass of a structure 'Cable' - flexible member under tension 'Beam' – typically a horizontal member carrying vertical loading and supported from below Cantilever – a beam with one end built in and the other free Shaft – Transmits a torque T

W

F F A B

F F B A


1.1.3 Typical X-Sections of Components

Angle

I-Section

Circular (solid)

Channel

T-section

Circular (hollow)


1.1.4 Types of support and connections

(a) Built in support Equivalent force system

There will be reactions Rx and/or Ry and a "fixing moment", M, at

the built-in end to keep beam in equilibrium. There is no linear or angular displacement at the support. (b) Pinned joint Equivalent force system

Rx

Ry

M

Ry

Rx

weld


There will be no moments but reactions Rx and Ry can exist. No

linear displacement will occur but angular displacement possible (rotation of link) (c) Roller support Equivalent force system

There will be no moment and only a vertical reaction, Ry. Both

horizontal and angular displacement can occur. (d) Sliding support Equivalent force system

There is a reaction, Rx and there can be a moment. Vertical sliding

motion is possible.

Rx

Ry


2. Stress and Strain

The concept of stress

Stress is defined as the transmission of force across a boundary. Consider the solid body shown in Figure 2.1 separated by a plane

into two parts A and B. Each part exerts a

force on the other, and over a small area δA

at a point P. Let this force on part A be δF in an arbitrary direction.

The intensity of this force is δF/δA and the stress at P is defined as

A

F

AP δ

δσ

δ 0lim

→= (2.1)

Figure 2.1 The units of stress are those of (Force/Area). Hence, typical values are N/mm2 or MN/m2. The force action has magnitude and direction and in addition, the plane through P has an orientation. These three quantities must

be specified for the stress σ to be completely known.

Average stresses

2.1.1 Tensile stress.

When a tensile force is applied to a member, it gives rise to tensile stresses. Figure 2.2 shows a rod of uniform cross-section, area Ao, subjected to an axial

tensile force F. Figure 2.2

A

B

PδF

δAA P

FF

Area Ao


The average stress throughout the rod is tensile and is given by

oA

F=σ (2.2)

Tensile stresses are assigned as positive (+ve). Tensile stresses tend to stretch or extend the length of the member. Consider the tapering rod of length l, of area A0 at one end and A1 at the other, shown in Figure 2.3. The force is uniform throughout the rod. The area at any distance x from the smaller end is equal to: Figure 2.3

( )oo AAl

xAA −+= 1 (2.3)

The average stress at any value of x is tensile and is equal to

( )oo

av

AAl

xA

F

A

F

−+==

1

σ (2.4)

Hence the stress varies along the length of the rod, from a high value of F/A0 at the smaller end to the lowest F/A1 at the wide end. We may also look at the stresses generated in a bar in terms of the forces transmitted along the bar in conjunction with equilibrium. Figure 2.4

Area AoArea A1

xl

F F

pF FP Q

q


Consider a straight tie bar of uniform X-section loaded by an axial tensile force, F (Figure 2.4). Figure 2.5 Imagine that the bar is 'cut' into two parts P and Q at some section pq at right angles to the direction of F - that you are standing in the middle holding the 2 parts together (Figure 2.5). We can then consider a free body consisting of part P only, drawing the FBD (Figure 2.6). Equilibrium requires that the force F on its LHS be balanced by a force F to the right. This force can only be the force applied by part Q on part P at the cross-section pq. We assume that the force F is distributed uniformly over the cross- section An. Figure 2.6

Since the stress acts normal to the area, it is called the normal stress. In the example considered, the force F is tensile, i.e.

tending to stretch the bar. The normal stress, σ, is thus along the outward normal to the surface, i.e. it is a tensile stress.

2.1.2 Compressive stress

When a compressive force is applied to a member, it gives rise to compressive stresses. The average stress throughout the rod shown in Figure 2.7 is compressive, equal to Figure 2.7

oA

F−=σ (2.5)

Compressive stresses are assigned as negative. Compressive stresses tend to compress or shorten the length of the member.

pF P

q

F F

Area Ao

pF FP Q

q

p

q


Note that the average stress depends only on the total area of the section at any position. Hence the shape of the section is irrelevant. The shape of the section can be circular, square, rectangular, hollow (tubular) or irregular.

2.1.3 Shear Stress

Materials can also be subjected to shear force. A shear force tends to slide (or shear) one part relative to another. Consider the uniform cross-section square bar, area Ao in a shear

block shown in Figure 2.8. The left hand side of the shear block is moving Figure 2.8 upwards under a force F, while the right hand side is moving down under an equal opposing force F. The material in the block is sheared along the dotted line and is subject to a shear force across the section (Figure 2.9). The average shear stress tending to cut or shear the bar is

oA

F=τ (2.6) Figure 2.9

Now suppose that the bar considered in Figure 2.5 is 'cut' along 'mn' where mn is not normal to F as shown in Figure 2.10. Figure 2.10 Again consider part P as a free body (Figure 2.11). Let A be the cross sectional area of section mn, larger than the 'normal' cross sectional area, An. In this case,

part Q exerts a net force F onto P axially. Figure 2.11

Area Ao

F

F

Area Ao

F

F

ΘP QF Fm

n

ΘPFm

n

F

F

n

s


This may be resolved into

(a) a force, Fn = F sinΘ normal to area A

(b) a force, Fs = F cosΘ along the surface As above the normal force Fn divided by A is the normal stress

denoted here by σ.

A

Fn=σ (2.7)

The force Fs along the surface is called a shear force and Fs/A is

called the shear stress, denoted by τ:

A

Fs=τ (2.8)

The shear stress τ will cause shear (angular) deformation. In many situations, combinations of tensile, compressive and shear stress can act simultaneously, e.g. stresses in a simple rivetted joint.

2.1.4 Shear stress in a rivet in single shear

Consider the forces acting on the rivet shown in Figure 2.12. The top plate, which is being pulled to the left by force F, presses against the top half of the rivet and exerts a force F to the left. Figure 2.12 Similarly, the lower plate exerts a force F on the lower half of the rivet (Figure 2.13). These two forces tend to shear the rivet across its cross-section mn. This is resisted by shear force FS (= F ) acting over the cross-section of the rivet at mn. Figure 2.13

tF

F

Rivet (dia. d)

F

F

m nσb

σbFs


Thus, there are shear stresses τ acting across the section mn which

has an area of πd2/4. If we assume the stresses to be uniform over this area, the shear stress is given by

( )4/2d

F

πτ = (2.9)

In addition to the shear stress, a contact stress called bearing stress, is also induced. Assuming that the bearing stress is uniformly distributed, it is defined as the force divided by the projected area of the appropriate part of the rivet. Considering the top half, the projected area is a rectangle of dimensions t x d. Thus

d t

F=σb x

(2.10)

The bearing stresses are direct stresses which tend to crush the rivet, and also cause permanent deformation or enlargement of the rivet hole.


2.1.5 Rivet in double shear

Consider a rivetted joint shown in Figure 2.14. In this case, too, there will be shear stresses and bearing stresses. If the thickness of the three plates is the same, the bearing stress on the central third projected area is twice that on the top and bottom third projected areas. Figure 2.14 As shown in Figure 2.14, the force tending to shear the rivet is F/2, i.e. half that for a rivet in single shear. Therefore, the shear stress in the rivet is given by

)4/(

)2/(2

d

F

πτ = (2.11)

Stress concentrations

The previous sections assumed that the applied force is uniformly distributed across the loaded section, hence the use of the phrase average stress. In situations where the cross-section changes rapidly, local stresses can by higher than the average stress. Consider a wide strip, subjected to an axial tensile force, F, containing an elliptical hole of major diameter a and minor diameter b as shown in Figure 2.15. Figure 2.15

F/2

F/2

F

F/2

F/2

Fσb

σb

2σbF/2

F/2

(a) (b)(c)

F

F

F

F

3σav

σavx

w

b

a


Let the width of the strip be w and its thickness be t as shown in Fig 3.15. The average tensile stress in the strip, well away from the hole is

t w

F=σav x

(2.12)

The stress at the tip of the ellipse is increased by a factor

( ))/(21 ba+

For a circular hole (a = b), the stress at the horizontal diameter of a circular hole is increased by [1+2(a/a)]= 3 over the average stress in the strip.

The stress at point x is 3σav and decreases rapidly as the distance

from the hole increases, reducing to σav well away from the hole,

shown in Figure 2.15. Since stress can lead to the failure of materials, failure is likely to initiate from the positions of stress concentration. This has serious implications in many engineering situations e.g. rivetted joints as considered previously.

Non-uniform force / stress distribution

We have assumed that the normal and shear forces are uniformly distributed over the surface, A. In general this need not be the case and the force per unit area, over equal elements of area, will vary with the position of the element.

Consider an element of area δA in the whole area A as shown in Figure 2.16. Let δFn and δFs be the

normal and shear forces on this area. The local normal and shear stresses will be

dA

dF

A

F

dA

dF

A

F

Ss

A

nn

A

==

==

→

→

δδ

τ

δδ

σ

δ

δ

0

0

lim

lim

(2.13)

Figure 2.16

δFn

δFs

δA


It is clear that the total forces Fn and Fs are obtained from

dAdFn ⋅= σ

∴ ∫ ⋅=A

n dAF σ (2.14)

Similarly, ∫ ⋅=A

S dAF τ where the integrals are taken over the whole

area A.

Complementary Shear stress

A block of material is subject to shear stress,

τ, separated by distance y as shown in Figure 2.17. The resultant shear forces constitute a couple and would lead to rotation of the block. For equilibrium, additional shear stress must act around the block in a complementary manner so as to prevent the block from rotating, i.e they should provide an equal and opposite couple. This implies that each shear stress acting on a particular plane must have its Figure 2.17 complement acting on a plane orthogonal to it. Consider a small element

of sides ∆x, ∆y and ∆z in rectangular cartesian co-ordinate system (x,y,z) as shown in Figure 2.18. Let the shear stress on its top

surface be τxy where x

indicates the direction in

which τ acts and y indicates that the normal to the surface is in the y-direction. The force on the element in positive x-direction will be: Figure 2.18

zxxy ∆⋅∆⋅τ

y

τ

τ

τxy

τyxτyx

τxy

y

xo

∆y

∆x

A B

D C


For the element to be in equilibrium, there must be a force of the same magnitude but in the -ve x-direction. Thus shear stress on

lower face must be τxy in the negative x-direction.

The forces (τxy ∆x ∆y) in opposite directions on AB and CD are in

linear equilibrium but they constitute a clockwise couple:

zyxM xy ∆∆∆=τr

(2.15)

For equilibrium there must be an anticlockwise couple of the same magnitude. This can only be provided by the shear stresses acting

on the left and right faces AD and BC. Let τyx be the shear stress on

these faces [subscript y denotes the direction of the stress and x the direction of the normal to the surface]. The forces on AD and

BC are zyxxy ∆∆∆τ and the anticlockwise couple provided by them is

xzyM yx ∆∆∆=τs

(2.16)

For equilibrium, clockwise M = anticlockwise M. Therefore, from Eq. 3.15 and 3.16,

yxxy ττ = (2.17)

τxy and τyx are called complementary shear stresses. Henceforth

we shall use only one symbol τxy for shear stresses on all four faces.

In Figure 2.18, we assumed that τxy on the top face CD acts in the

+ve x-direction. This shear stress is taken as positive. If we had assumed that the shear stress on the top face is in the -ve x-direction then all the signs of other stresses would be reversed. To make the sign convention clearer, we first define the +ve face of an element as one for which the outward normal is in the +ve direction of the axis as Figure 2.19 shown in Figure 2.19.

y

xo

Outward Normal

Positive Faces

Negative Faces


Thus a shear stress is +ve if it acts in the +ve direction of the axes on the +ve face and in the -ve direction of the axes on the -ve face. Thus, stresses shown above are +ve. We have omitted consideration of the normal and shear stresses in the z-direction. However, everything will also work in this direction.

The concept of Strain

Under the action of the applied stress, the material deforms. Such deformation is measured in terms of strain. Two types of strain exist, matching the two types of stress. Normal stresses (tensile or compressive) are associated with normal strains. Shear stresses give rise to shear strains.

2.1.6 Normal strain

If a prism of uniform material is subject to a uniform normal stress, it will elongate by an amount e as shown in Figure 2.20. Figure 2.20

The normal strain ε is defined as the extension (or compression) per unit length, i.e.

l

e=ε (2.18)

This is a dimensionless quantity. Tensile strains (i.e. those arising from tensile stresses) are considered positive, since the sign of e is positive. Compressive strains are assigned negative values.

2.1.7 Shear strain

A plane element (as the dotted square in Figure 2.21a) if subjected to equal shears on all four faces will deform as shown in a parallelogram. The angle subtended by any face w.r.t. either the

vertical or the horizontal plane changes by an amount φ . φ is defined as the shear strain, and like normal strain, is a dimensionless quantity. Usually in engineering applications, an axis parallel to one of the

L e


deformed sides is taken as shown in Figure

2.21b, and γ = 2φ is then taken as “engineering” shear strain. Note that

both γ and φ are angles measured in radians. Figure 2.21

2.1.8 Volumetric strain

Volumetric strain is defined as the change in the volume divided by the original volume.

o

VV

V∆=ε (2.19)

Figure 2.22 Consider a rectangular parallelopiped whose sides before loading are lx, ly, and lz in x, y and z directions respectively (Figure 2.22). Therefore, the initial volume V = lx ly lz .

Suppose the element is loaded so that εx, εy, and εz are the strains in the x, y and z directions respectively.

x

xx

l

l∆=ε

y

y

yl

l∆=ε

z

zz

l

l∆=ε

Then, the new lengths are ll ∆+ or

φ φ

(a) (b)

y

x

z∆lz

∆lx

∆ly


)1( xxl ε+ )1( yyl ε+ )1( zzl ε+ (2.20)

Therefore, the increase in volume is

( )[ ] ( )[ ] ( )[ ] zyxzzyyxxoldnew llllllVVV −+⋅+⋅+=−=∆ εεε 111 (2.21)

We are dealing with small strains, i.e. ε of the order of 0.01 or less.

Therefore, ε2 and ε3 terms are negligble, and

∴ )()( zyxozyxzyx VlllV εεεεεε ++=++=∆ (2.22)

The volumetric strain εv is then given by

∴ zyx

o

VV

Vεεεε ++=

∆= (2.23)

Stress-strain relationships

The relationship between the strain as shown in Figure 2.20 and the stress applied to produce this strain can exhibit one of the four characteristic behaviours illustrated in Figure 2.23a to 2.23d.

σ

ε

x

ε ε εx

σ σ σ

(a) (b) (c) (d)

Figure 2.23 In each of these, the slope of the curve relating stress to strain is the vital factor. This slope is symbolised by E. E is defined as the

ratio of incremental change in stress (dσ) to incremental change in

strain (dε).


• In Figure 2.23a, E = ∞ , i.e. there is no strain and the material is said to be rigid.

• In Figure 2.23b, E = constant, and the stress-strain relationship is linear.

• In Figure 2.23c, E = f(σ).

• In Figure 2.23d, E = 0: there is no definite strain for any value of stress and the material is said to be plastic.

The behaviour on unloading can follow either the full or the dotted curves shown in Figure 2.23b and 2.23c. Whenever the material returns to its original dimension along the loading curve, it is said to be elastic, but if it behaves as indicated by dotted lines, i.e. if it departs from the original loading curve, it is said to be inelastic. Materials which fracture when the strains are small are known as brittle whilst materials which undergo considerable deformation before failure are said to be ductile. Real materials generally have complex stress-strain relationships, but for purposes of analysis the behaviour is simplified into either:

• that exemplified by the single curve of Figure 2.24a, or

• behaviour as rigid material up to some yield stress σy, after which it is plastic (Fig.2.24b), or

• a combination of linear elastic and plastic behaviour (Fig. 2.24c).

ε

σ

A

ε

σ

ε

σ

O(a) (b) (c)

Figure 2.24


A uniform rod of length L and cross-sectional area A (Figure 2.25) which is subjected to a uniform

state of stress σ, i.e. a

force F=σA., and with a value of E which is constant for the stress range used, has Figure 2.25

A

FStress

L

eStrain

=

=

Hence, Ae

FL

Strain

StressE == (2.24)

This constant value of E is called Young’s modulus or Modulus of Elasticity. At any point A in curve of Figure 2.24a, the stress-strain relationship can be defined in two ways. One is the tangent

(dσ/dε), i.e. the slope of the curve at A, and this value is known as

the tangent modulus. The other is the slope σ/ε of the line OA, and this is known as the secant modulus. For linear elastic material, these two modulii are the same. For shear strain as in Figure 2.21c, similar behaviour exists but the

relationship between the shear stress τ and the engineering shear

strain γ is known as the shear modulus or modulus of rigidity and is denoted by G. It is given by

γτ

=G (2.25)

Poisson's Ratio

The element of Figure 2.26, which initially has the shape indicated by the full lines, will deform under a tension as shown to the shape indicated by the dotted lines.

The longitudinal strain εx is accompanied by a lateral strain

εy and the ratio ex/ey is known as the Poisson’s ratio and is

generally denoted by ν. Figure 2.26

σ

F

A

L e

} F

σx σx

εy

εx


If, in addition to the force producing stress σx, there is a force on

the perpendicular plane producing stress σy then the strains εx and

εy due to combined stresses are given by

( )yxxE

νσσε −=1

(2.26)

( )xyyE

νσσε −=1

(2.27)

Poisson’s ratio not only gives the relationship between the longitudinal and lateral strains, but also enables us to formulate the relationship between the modulus of elasticity and modulus of rigidity. Consider an element ABCD of unit length sides as shown

in Figure 2.27 which is subjected to shear stresses τ on each face.

This produces a direct tension and compression of magnitude τ on the two diagonals. The strain on the diagonal BD is therefore equal to

( )ντ

νττ +=+ 1)(1

EE (2.28)

If the effect of the shear is to cause the deformation shown by the dotted lines in Figure 2.27 with B moving to B’, then BB’ is the shear strain and is

approximately equal to τ/G. The increase in length of the

diagonal DB is BB’/√2, i.e.

τ/G√2, and since the original

length was √2, the strain is

τ/2G. But this strain

is also equal to (τ/E)(1+ν). Figure 2.27

A B B’

CDτ

τ τ

τ

σn=τ σn=τ

BB'

2


Hence,

GE 2

11=

+ν (2.29)

or

)1(2 ν+

=E

G (2.30)

Generalised Hooke’s Law

For a linear elastic material, the stress is directly proportional to the strain with Young’s modulus E as the constant of proportionality:

εσ ⋅= E or E

σε = (2.31)

The above relationship between stress and strain is deceptively simple and is an incomplete statement of the state of stress in the material. It considers only one stress and one strain (in the direction of the applied stress). Let us index this direction as x. Hence,

E

xx

σε = (2.32)

Due to the Poisson’s Ratio effect, there must also be strains in the two orthogonal directions to direction x, which will be called direction y and direction z, and are known as secondary strains (Figure 2.28). These strains are given by

xxyE

σν

νεε −=−= (2.33)

and

xxzE

σν

νεε −=−= (2.34)

Figure 2.28

y

z

x


Hence, the uniaxial stress, σx, applied in direction x gives rise to three mutually perpendicular strains or a component of strain in the three perpendicular directions.

E

xx

σε =

E

xy

νσε

−=

E

xz

νσε

−= (2.35)

Alternatively the single stress could be applied in direction y as σy

(where σy ≠ σx). The strains due to σy would be:

E

y

x

νσε

−= in x-direction

E

y

y

σε = in y-direction (2.36)

E

y

z

νσε

−= in z-direction

Similarly, the single stress could have been applied in direction z as

σz (where σz ≠ σy ≠ σz). The strains due to σz would be:

E

zx

νσε

−= in x-direction

E

zy

νσε

−= in y-direction (2.37)

E

zz

σε = in z-direction

In the most general case, the three components of stress (σx, σy, σz),

could be simultaneously applied to a block of material. The resulting strains in the three directions would then be the sum of the separate components. By simply superimposing (or adding) equations 2.35, 2.36 and 2.37, the general linear elastic stress-strain relationships are derived:

EEE

zyxx

νσνσσε −−=

EEE

zyzy

νσσνσε −+

−= (2.38)

EEE

zyxz

σνσνσε +−−=


The equations are usually rearranged and written in the form:

( )[ ]zyxxE

σσνσε +−=1

( )[ ]zxyyE

σσνσε +−=1

(2.39)

( )[ ]yxzzE

σσνσε +−=1

These are termed the Linear elastic constitutive equations or Generalised Hooke’s Law. These equations form the basis for analysis of many mechanics of materials problems, where the deformation is (or can be assumed to be) linear elastic. They are subject to the following constraints: (a) Linear-elastic deformation (b) Small scale deformation (c) Time independent deformation

Thermal stresses and strains

Most engineering materials expand when their temperature is raised. If this expansion is restrained, then stresses may be induced. Consider a rod of length l1 at temperature T1 being held between rigid supports (Figure 2.29). If the temperature is raised to T2, then the rod, if its expansion was not restrained, would expand to a length l2. In restraining the expansion, the supports exert a compressive load on the rod. Figure 2.29

2.1.9 Thermal strain

We wrote the Generalised Hooke's Law in the form

( )[ ]zyxxE

σσνσε +−=1

(2.40)

with similar equations for the linear

strains, εy and εz, in the y and z directions. Figure 2.30

l1

l2

At tem peratu re T 1

At tem peratu re T 2

l 1

l


Consider now a piece of material with length lo in the x-direction at temperature To (Figure 2.30). If the temperature is raised to T and the rod is free to expand, the increase in length will be

( )oo TTlll −=−= αδ 1 (2.41)

where α is the coefficient of linear expansion of the rod material.

The quantity (δl/lo) is clearly a linear strain arising from temperature changes. It is called the thermal strain and denoted

by εt. It is given by

( )o

o

t TTl

l−== α

δε (2.42)

It is clear that if the material expands equally in all directions (α is the same for x, y, and z directions as the material is isotropic) then this strain will be equal in all directions. Thus when temperature changes occur, the total strain in any direction must also include the thermal strain

( )[ ] ( ) txozyxx TTE

εεασσνσε σ +=−++−= )(1

(2.43)

where (εx)σ is the strain resulting from the stresses σx, σy, and σz.

Similar equations will hold for εy and εz . Note that εt is the same

for all directions.

2.1.10 Thermal stress

Let us return to the simple problem we started with and find the stress arising from the restrained expansion (Figure 2.31). In the x-direction there is no net strain,

i.e. εx = 0. Also σy = σz = 0 since there is no restraint in these directions nor is there any imposed load. Thus

[ ] )(01

0 12 TTE

x −+−= ασ ∴ )( 12 TTEx −−= ασ

Since T2 > T2 it follows σx < 0, i.e. the

thermally induced stress is compressive as we had intuitively expected. Figure 2.31

l1At temperature T1

At temperature T2

l1


2.1.11 Practical Example

A simple example is that of tie-bars in old houses. The walls of old houses tend to buckle outwards with the lapse of time under the weight of the roof or movement of the foundations. To stop the house collapsing, tie-bars are inserted to pull the side walls together. The ends of such bars are passed through ‘X’ or ‘S’ plates to spread the load on the walls. Rather than tensioning the bars by turning the nuts on the tie-bars, it was customary to light a fire under the tie-bar, thus causing it to expand. Then the nuts could be easily nipped up. As the tie-bar then cooled, it contracted, pulling the walls together. Clearly the tie-bar had to be heated to the right amount, otherwise problems could occur.


3. Beams - flexural loaded members

Figure 3.1 Beam supports may be simple (e.g. roller supports) or built in. The stresses (or internal forces) in the beam material will depend on: (a) magnitude and distribution of loading (b) nature of the supports (c) size and shape of the beam corss section The strains (proportional to the changes in length and angle) depend on (a) stresses

(b) properties of the beam material (E, ν) First we can consider the internal forces and moments reulting from a specific loading case.

The concepts of 'shear force' and 'bending moment'

Consider e.g. the loading on a skate board. This is essentially a simply supported beam with a load which is (nearly) concentrated at a point.

Figure 3.2

α β

Subsystem β Subsystem α

Cut

RA

z

L

a b RB

W

Concentrated or point-load Distributed load

non-uniform

Beam

Reaction Reaction


3.1.1 Equilibrium

For equilibrium, resolve y-wise & take moments about A or B for the whole system Fy = O W -RA - RB = O

∴ RA + RB = W (3.1)

MB = O Wb - RAL = O

∴ RA = Wb

L (3.2)

Therefore RB = Wa

L (3.3)

Next, to find how the load is sustained internally by the beam material consider subsystems A and B and draw free body diagrams for them

Figure 3.3 Consider the equilibrium of subsystem A. For vertical equilibrium there must be a force F = RA acting

downwards. Further RA has a moment, RA z (clockwise) about an axis at the cut.

The subsystem will rotate unless there is a moment , M= RA z

(anticlockwise) about the cut. The only way the balancing force and moment can be applied, is if subsystem B applies them on bubsystem A at the cut. By Newton's 3rd law - A must apply an equial and opposite force and moment

on β.

M M F

F z

RA

α β

W


F is called the 'shear force'. It prevents A from shearing off relative to B.

Figure 3.4 M is called the 'bending moment'. It is associated with the tendency of the beam to bend.

Figure 3.5 Before considering how F and M vary along the length of a beam, we need to adopt some coordinate axes and a sign convention for +ve F and M. Axes: we use a right handed co-ordinate system with axes ox, oy, oz

Figure 3.6 i.e. if a screw with a right handed thread is turned in the sense oy to oz (anticlockwise in figure), the screw would move in the +ve x-direction.

z

y

x

O

M M

RA

α β

RA

α

β α

F

F RA


We shall take the beam to lie along the z-axis

Figure 3.7 For a horizontal beam we shall take +ve direction of load to be vertically downwards, i.e. in the +ve y direction.

3.1.2 "+ve and -ve faces" of a part of a beam

Figure 3.8 Right handed face (i.e. one with larger value of z) is the +ve face or the +ve face is one for which the outward normal is in the +ve direction.

3.1.3 +ve Shear Force

The shear force is +ve if it acts in the +ve y-direction on the +ve face and in the -ve y-direction on the -ve face.

Figure 3.9

-ve face

z

y

x

z z+δz

-ve face +ve face

outward movement from face

z z

y

x

z

y

x

O

W +

+ve face


3.1.4 +ve Bending Moment

The B.M. is +ve if it tends to "sag" the beam i.e. to make it concave on upper face.

Figure 3.10 Note: Some books use the opposite convention for +ve SF and +ve BM. Whatever convention you want to use, stick to it.

Determination of Shear Force and Bending Moments

Obtain expressions for F and M at the section shown

Figure 3.11 Consider subsystem consisting of beam from one end (RH or LH end) and draw a f.b.d. for it. Let F and M be the shear force & B.M. at the section. Show them in the +ve sense in the f.b.d.

Figure 3.12

R1

W1 W2

F M

z

z1

z2

R1 R2

W1 W2 section

+

Sagging moment

-

Hogging moment


For vertical equilibrium F + W1 + W2 - R1 = 0

F = R1 - W1 - W2 (1)

For moment equilibrium, take moments about axis shown M + W1 z1 + W2 z2 - R1 z = 0

∴ M = R1 z - W1 z1 - W2 z2 (2)

From (1) F = sum of all upward forces on subsystem i.e. all loads/reactons to the left of the section. M = sum of all clockwise moments about axis due to forces to left of the section. Shear Force (SF) and Bending Moment (BM) diagrams - are graphical displays of the variation of F & M along length of beam.


Example 1 Simply supported beam with concentrated loads. e.g. Painter (mg = 700 N) on a wooden plank supported on tressles. Dimensions as shown. Draw the shear force and bending moment diagrams

A BC

R1 R20.25 m 1 m 2 m 0.25 m

y

z0x

1 2 3 4


Example 2 A cantilever beam with concentrated load

A B

C

z

L

y

z0x

1 2

Mc

Rca


The method of considering subsystems and drawings f.b.d. to get F

and M seems cumbersome but in initial stages it helps to get

insight into internal forces/moments. After practice, it is perfectly

accpetable to imagine a 'cut' anywhere along the beam and to find F

as the total upward force to the left of the cut. Similarly for M.

Some General Relations among Loading Intensity, Shear Force and Bending Moment

At any point z, load/length = w (in general, varying with z).

Consider a short length δz of beam which carries point loads and a distributed load. Over this length, load/length = w

∴ load = w δz Let F, M = SF and BM on LH (-ve) face

and F + δF, M + δM = SF & B M on R H (+ve) face

load

beam

δz

F

F+δF Wδz

M+δM M

z δz

z

y x


For vertical equilibrium -F + F + δF + wδz ≈ 0

∴ δF ≈ - wδz

δF

δz = - w

Αs δz → 0, δF/δz → dF/dz so that

dF

dz = - w (3.3.1)

Note that w may vary along beam length i.e. w = w (z). Between positions z1 and z2

F2 - F1 = ⌡⌠

F1

F2 dF = -

⌡⌠

z1

z2 wdz = (3.3.2)

= minus area under graph of w v z from z1 to z2

If w = 0, then dF/dz = 0 i.e. F = constant If there is a point load, e.g.

then we have no variation of w with z, so for equilibrium W + F2 - F1 = 0

∴ F2 - F1 = -W (3.3.3)

For moment equilibrium (moment about horizontal axis through O)

(M + δM) - M - Fδ z + (w δz) δz2 = 0

W

F1

δz z

z1

Z2

F2

Wδz


δM = Fδz - (w δz2 δz)

∴ δM

δz = F -

w2 δz

As δz → 0, δM

δz →

dM

dz and

w

2 δz = 0. Thus

∴ dM

dz = F (3.3.4)

Between z1 and z2

M2 - M1 = ⌡⌠

M1

M2 dM =

⌡⌠

z1

z2 Fdz = B

= area under graph of F v z from z1 to z2

If a couple Mo is applied at a point along beam

then for equilibrium M2 + Mo - M1 = O

∴ M2 - M1 = - Mo

Mo M2

M1


4. Deformations and stresses in pure bending

Figure 4.1 Initially straight beam will bend due to application of M. The lower part of beam is stretched (in tension) and uppermost part is compressed. There is also shear deformation due to shear force F. We can neglect this as it is small for long slender beams.

Assumptions (refer to figure)

(a) Plane transverse sections remain plane after bending, i.e. PQ,

AC etc remain plane while PAB is bent to P'A'B' (b) Beam bends in circular arc in vertical plane only; transverse

sections are perpendicular to circular arcs having common centre of curvature. But radius of curvature & centre of curvature will vary from point to point along beam.

(c) Beam is subject to pure couples (M) resulting in tension &

compression in the z-direction; there are no shear stresses. (d) Radius of curvature, R >> transverse 'depth' of beam (e) E is the same in tension and compression (f) Beam x-section has vertical axis of symmetry 0y - simplifies

presentation.

F F

M M

F=M=0


Neutral Surface

Consider initially parallel sections AC and BD, a distance δz apart. After bending:

CD stretches to C'D' i.e. εz +ve

AB contracts to A'B' i.e.εz -ve

Between AB and CD there is some layer FG which neither stretches

nor contracts, i.e. for which εz = 0

∴ straight length FG = curved length F'G' = δz Layer FG is called the NEUTRAL SURFACE and axis 0x on the x-section is called the NEUTRAL AXIS. Take z-axis to be in the neutral surface and through the line of symmetry of the cross section.

Figure 4.2

M

Neutral surface

B’

G’

K’

A’

F’

H’

C’ D’

Q’

P’

δθ R R+y

y

M

A P B

G F

H K

C D

x

x

δz

x

+y

z

Q

Neutral surface

O

C D

A B

OZ x

+y

Area εA

Neutral axis

Section XX

-y


Bending equation - σσσσz

y = ER

Consider layer HK at distance y from the neutral axis. This stretches to H'K' in the bent position. The longitudinal (i.e. z-wise) tensile strain in HK is

εz = H'K' - HK

HK (=

increase in length

initial length) (4.1)

Now HK = FG = F'G' = δz = Rδθ (4.2) where R = radius of curvature of neutral surface

δθ = angle (in radians) subtended by F'G' at the centre of curvature

Also H'K' = (R + y) δθ (4.3)

∴ εz = (R+y) δθ - Rδθ

Rδθ = y

R (4.4)

i.e. εz varies linearly from 0 on the neutral axis to a maximum at

the bottom (C'D') and top (A'B'). Since y is +ve at the bottom, εz is

+ve there. And εz is -ve at top.

From Hooke's Law

εz = 1

E [σz - ν (σx + σy)] =

σz

E

since in this case σx and σy are zero. Substituting εz = σz/E in (4.4)

gives

σz

E =

y

R i.e.

σz

y =

E

R

(46.5)

i.e. σz = E

R y and varies linearly with y in same way as εz.


Bending equation - MIx

= ER

Figure 4.3

The force, δFz on an element of area, δA

δFz = σz δA = E

R y δA

and its resultant is obtained by integrating δFz over whole the area

A, i.e.

Fz = ⌡⌠

A

dFz = ⌡⌠

A

E

R y d A =

E

R ⌡⌠

A

y d A (4.6)

Consider the equilibrium of portion P'Q'D'B'. Since loading on P'Q' is just a pure couple M (clockwise) the resultant force on P'Q' = 0. Thus, for equilibrium we must have Fz = 0. From 6.6 it follows that

Fz = E

R ⌡⌠

A

y d A = O therefore ⌡⌠

A

y d A = O (4.7)

But

⌡⌠

A

y d A = "first moment" of area about the neutral axis and for

this to be 0, the neutral axis must pass through the "centroid" of area A, i.e. point O in Figure is the centroid of area.

C D

A B

OZ x

+y

Area dΑ

Axis of symmetry

Neutral axis OZ

dF2=σ2dΑ

y

O

σz


The neutral axis passes through the "centroid" of area Consider moment equilibrium of P'Q'D'B'. Since there is a clockwise couple M at P'Q', there must be a counter-clockwise couple M at

B'D'. This arises from the total moment of forces δFz about neutral

axis xx.

moment of δFz about xx = δM = δFz y = y δFz

total moment M =

⌡⌠

A

y dFz and since δFz = (E/R) y δA

∴ M = ⌡⌠

A

y (E/R)y d A = E

R ⌡⌠

A

y2 d A (4.8)

But

x

A

IdAy =∫2

(4.9)

"second moment" of area about axis xx (analogous to "second moment of mass" or "moment of inertia" used in dynamics) Subscript x on I denotes that distance y in y2 dA is measured from axis xx. Substitute 6.9 in 6.8

M = E

R Ix

M

Ix =

E

R (4.10)

From 6.5 and 6.10

MIx

= σσσσz

y = ER (4.11)

Eqn. 11.11 is called the "bending formula".

σz is called the axial stress due to bending moment M, i.e. the

"bending stress".

Since σz = (M/Ix)y then for a given M &y, σz depends on Ix.

Ix depends on geometry of beam x-section. The first step is to find

the neutral axis (which passes through the centroid)


Co-ordinates of centroid of area

Figure 4.4 Consider an arbitrary x-section of area A, with neutral axes 0x, 0y through 0. Let OX, OY be axes parallel to 0x, 0y.

Coordinates of area δA are (x,y) or (X,Y) depending on axes chosen.

Coordinates of centroid 0 are (X--

, Y-- ).

By definition, the first moment of area A about axes 0x or 0y is zero.

The first moment of δA about OX is YδA

∴ total moment about axis OX is φx =

⌡⌠

A

Y d A (4.12)

But since area A can be taken to be concentrated at centroid 0, we have also

φx = AY_ (4.13)

From 4.12 and 4.13

AY_ =

⌡⌠

A

Y d A

so that

dA

y-axis Y-axis

X-axis

x-axis

X xXX +=

Y

yYY +=

O

o Centre of area


Y_ =

1

A ⌡⌠

A

Y d A (4.14)

Similarly considering first moment about axis OY

X--

=1

A ⌡⌠

A

X d A (4.15)

Equations 6.14, 6.15 give coordinates (X--

, Y-- ) of the centroid

Second moment of area

The second moment of area about axis OX is defined as IX =

⌡⌠

A

Y2 d A (4.16)

subscript X on I denotes that distance Y is measured from the X axis. Similarly about 0x Ix =

⌡⌠

A

y2 d A (4.17)

Also IY =

⌡⌠

A

X2 d A (4.18)

Iy = ⌡⌠A x2 d A (4.19)


Theorem of parallel axes

Referring to figure in section 4.5

Y = Y_ + y.

Thus IX = ⌡⌠

A

Y2 dA = ⌡⌠

A

( Y_ + y) 2 dA

= ⌡⌠

A

Y_

2 dA +

⌡⌠

A

2Y_ y dA +

⌡⌠

A

y2 dA

= Y_

2 ⌡⌠

A

dA + 2Y_ ⌡⌠

A

y dA + ⌡⌠

A

y2 dA

⌡⌠

A

d A = total area A , ⌡⌠

A

y d A = 0 since this is the first moment

about centroid which must be 0 and ⌡⌠

A

y2 d A = Ix by definition.

Thus Ix = A Y_

2 + Ix = Ix + AY

_

2 (4.20)

Similarly Iy = A X_

2 + Iy = Iy + A X

_ 2 (4.21)

This is the theorem of parallel axes. Notation in books is often I = Ic + AH2 i.e. Ic replaces Ix (or Iy) as the

value of I about axis through centroid and H replaces distance Polar second moment of area (see notes on Torsion)

Figure 6.5 By definition polar second moment about axis OZ is

dA

y-axis Y-axis

X-axis

x-axis

X

Y

OZ

oz

x

y

R

r

axes OZ, oz are⊥ lar to figure

R = distance of δA from OZ

r = distance of δA from oz


JZ =

⌡⌠

A

R2 dA = ⌡⌠

A

(X2 + Y2) dA = ⌡⌠ X2 d A +

⌡⌠ Y2 dA

JZ = Ix + Iy (4.22)

This is theorm of perpendicular axes. Similarly Jz =

⌡⌠

A

r2 dA = ⌡⌠

A

(x2 + y2) dA = ⌡⌠

A

x2 A + ⌡⌠

A

y2 dA

Jz = Ix + Iy (4.23)

Second moments of area of some typical x-sections

4.1.1 Rectangle

Centroid o is at intersection of diagonals

Figure 4.6 Consider

Ix = ⌡⌠

A

y2 d A= ⌡⌠

-D/2

D/2 y2 Bdy =

BD3

12 (4.24)

By inspection

Iy = DB3

12 (4.25)

Consider axis OX parallel to ox. By parallel axes theorem

dA=Bdy

x

X B

D

y Y

o

2/DY =


IX = Ix + A Y_

2 and Y

_ = D/2 ; A = BD

= BD3

12 + BD (D

2 ) 2 =

BD3

3 (4.26)

But IY = Iy + A X_

2 = Iy since X

_ = o

4.1.2 Rectangular 'box' section

Figure 4.7 Box section can be thought of as rectangle (1) minus rectangle (2). Thus Ix = Ix1 - Ix2

= BD3

12 -

bd3

12 =

1

12 (BD3 - bd3) (4.27)

For a 'square' box section (B = D ), b = d )

Ix = 1

12 (D4 - d4) (4.28)

B

b

x

y

d D

y

x d

b


4.1.3 Circle

Consider axis oz, ⊥ lar to plane of circle. The polar 2nd moment of area is:

Figure 6.8 Jz =

⌡⌠

A

r2 d A = ⌡⌠

o

D/2 r2 2πr dr

= πD4

32 (4.29)

(see notes on Torsion)

by ⊥ lar axes theorem Jz = Ix + Iy and by symmetry, Ix = Iy.

Thus 2Ix = 2Ix = Jz ∴ Ix = Iy = 1

2 JZ =

π64

D4 (4.30)

4.1.4 Annular x-section

Figure 6.9

x

y

D1 D2

As for box section Ix= Iy = Ix1 - Ix2 =

π64

( D24 - D14)

R

r

dr

x

y

D


4.1.5 Symmetrical I sections

By symmetry centroid is at intersection of diaganols AC and BD.

Figure 6.10 To calculate Ix, note:

∴ Ix = Ix1 - Ix2 - Ix3

but Ix3 = Ix2

∴ Ix = Ix1 - 2Ix2

all dimensions in mm

∴ Ix = 100x2003

12 - 2

(50-2.5) x 1803

12 mm4 = 20.50 x 106 mm4

To calculate Iy, note:

Iy = Iy1 + Iy2 + Iy3 and Iy2 = Iy3

Iy = Iy1 + 2Iy2 = 180x53

12 + 2

10x1003

12 mm4 = 1.67 x 106 mm4

5

200

10

100

10

x

y

1

2 3


5. Deflection of Beams

Bending formula: M

Ix =

E

R =

σz

y (5.1)

A beam subjected to loads will also be deflected

Figure 5.1

Let v = deflection of the N.A. at distance z along the N.A. before bending.

v is +ve downwards (in the +ve y direction ). Note: v varies with z ⇒ v = v(z)

Both v and its rate of change with z (= dv/dz) are very small, i.e

v << z (5.2)

dv

dz << 1 (5.3)

Consider two neighbouring points A and B on the N.A. and distance

δz apart. v = deflection at A

v + δv = deflection at B

φΑ = φ = angle between tangent at A and z-axis

φB = φ + δφ = angle between tangent at A and z-axis

δz

Ο

y

x

NA after

loading A B

v v+δv

NA before

loading


Note that with the beam bent as shown, φB is actually less than φA,

i.e. δφ < o

δs = arc length AB = R δθ (δθ in radians)

Figure 5.2

Right angled ∆ABC (see Figure .2) AB2 = BC2 + CA2

(δs)2 = (δz)2 + (δv)2

= (δz)2 [1 + (δv

δz )2 ]

≈ (δz)2 since δv

δz << 1

∴ δs = δz ≈ Rδθ

dθdz

= 1

R (5.4)

From Figure 5.2

δθ = φA - φB = φ - (φ + δφ) = - δφ (5.5)

Therefore (5.4) becomes

dθdz

= - dφdz

= 1

R (5.6)

δφ

φΒ φΑ v+δv

v

A B

D

R R

E

δθ

B

A

C δz

z

δs

v+δv v

O


From Figure 5.2

tan φ ≈ δv

δz →

dv

dz as δz → o (5.7)

Since φ is small, tan φ = φ (φ in radians)

φ = tan φ = dv

dz (5.8)

∴ dφdz

= d2v

dz2 (5.9)

Replacing (5.9) in (5.6):

d2v

dz2 = - 1

R (5.10)

From (5.1)

M

E Ix =

1

R (5.11)

Therefore from (5.10) and (5.11)

d2vdz2 = - M

EIx

(5.12)

Equation (5.12) is the differential equation which, when solved,

gives the deflection curve of the beam, i.e. v = v(z). Since it is 2nd

order, 2 boundary conditions are required to determine the

constants of integration. The equation is approximate since we have

considered small v, φ & dv/dz. It can be shown that, more exactly:

d2v

dz2

[1 + (dv

dz)2]3/2

= - M

E Ix (5.13)


Example 1: Cantilever with end loads

For equlibrium: Ro = W and Mo = WL

For moment equilibrium M + Wz = O ∴ M = Wz (A)

d2v

dz2 = - M

EIx = -

1

EIx (-Wz) =

W

EIx z (B)

Integrate twice with respect to z

dv

dz =

W

EIx

z2

2 + A (C)

v = W

EIx

z3

6 + Az + B (D)

Need 2 boundary conditions at z = L, v = O (E) z = L, dv/dz = O (F) Since beam is "built in" at B Using (F) in (C)

0 = W

EIx

L2

2 + A ∴ A = -

W

EIx

L2

2

L

W Mo

Ro


Using (E) in (D)

0 = W

EIx

L3

6 + AL + B

B = W

EIx [-

L3

6 +

L2

2 L ] ∴ B =

WL3

3EIx

Thus (D) becomes :

v = W

EIx [z

3

6 -

L2z

2 +

1

3 L3]

Maximum v occurs at z = O, vmax = WL3/3EIx


Example 3 Simply supported beam with u.d.l.

RA = RB = wL/2

At section, M = wL

2 z -

wz2

2 =

w

2 [Lz - z2] (1)

∴ d2v

dz2 = - M

EIx = -

w

2EIx [Lz - z2] (2)

∴ dv

dz = -

w

2EIx [

Lz2

2 -

z3

6 ] + A (3)

∴ v = - w

2EIx [

Lz3

6 -

z4

24 ] + Az + B (4)

At z = 0, v = 0; also z = L, v = 0, gives B = 0 and A = - wL3/24EIx.

Thus (4) becomes:

v = - w

12EIx [Lz3 -

z4

2 -

L3z

2 ] (5)

vmax occurs at z = L

2 and vmax =

5

384

wL4

EIx (6)

L

z


Example 4 Simply suppoprted beam with concentrated load

Load W acts at distance z from Taking moments about B:

RA = W (L-a)

L (1)

At section (1), i.e. o < z < a

M = W (L-a)z

L (2)

∴ d2v

dz2 = - M

EIx = -

W

EIx

(L-a) z

L (3)

∴ dv

dz = -

W

EIx

L-a

L

z2

2 + A1 (4)

∴ v = - W

EIx

L-a

L

z3

6 + A1z + B1 (5)

Eqns (2) to (5) form "equation set 1" which apply to points from A to C (i.e. o < z < a) only. We are only free to impose boundary conditions in this range. At section (2) between C and B (a < z < L) :

M = W (L-a)

L z - W (z-a) (6)

∴ d2v

dz2 = - M

EIx =

W

EIx [-

(L-a)

L z + (z-a)] (7)

∴ dv

dz =

W

EIx {-

(L-a)

L

z2

2 +

(z-a)

2

2} + A2 (8)

∴ v = W

EIx { -

(L-a)

L

z3

6 +

(z-a)3

6 } + A2z + B2 (9)

L

a

A B

RA RB

C

W 1 2


"set 2" equations apply from C to B (i.e. a < z < L) only We have 4 unknown constants A1, B1, A2, B2 so we need 4

boundary conditions. Two are obvious:

(a) z = o, v = o using eqn (5) → not (9)

(b) z = L, v = o using eqn (9) → not (5) The remaining two boundary conditions are obtained by noting that at point C (i.e. z = a), both sets of eqns are valid. Thus slope (dv/dz) and deflection (v) from both sets must be equal. Thus

(c)

dv

dz z=a

using eqn (4) =

dν

dz z=a

using eqn (8)

(d) (v)z=a using eqn (5) = (v)z=a using eqn (9)

Using (c) : - W

EIx

L-a

L

a2

2 + A1 = -

W

EIx

L-a

L

a2

2 + A2

∴ A1 = A2 = A (10)

Using (d) -W

EIx

L-a

L

a3

6 + Aa + B1 = -

W

EIx

L-a

L

a3

6 + Aa + B2

∴ B1 = B2 = B (11)

Then using (a) in (5) gives B=0 and using (b) in (9) [we cannot use (b) in (5)]

0 = W

EIx { -

(L-a)

L

L3

6 +

(L-a)3

6 } + A L

∴ A = W

EIx

I

L {

(L-a)

L L3

6 -

(L-a)3

6 } =

W

6EIx

a(L-a) (2L-a)

L

subst for A1 = A and A2 = A in (5) and (9) respectively

(5) ⇒ v = - W

6EIx

L-a

L z3 +

W

6EIx

a(L-a) (2L-a)

L z


= W

6EIx

L-a

L [a(2L-a) z-z3] for 0 ≤ z ≤ a (12)

similarly (9) becomes

v = W

6EIx

L-a

L [ a(2L-a) z - z3 + (z-a)3

L

L-a ] for a ≤ z ≤ L

(13) In particular, at point C (z=a) deflection is obtained from (12) or (13) as

vc = W

6EIx

L-a

L [ a2(2L-a - a3] =

W

6EIx

2a2(L-a)2

L (14)

If loading is at mid-point (a = L/2) (BEAM EXPERIMENT)

vc = W

6EIx

2 (L

2)2 (L -

L

2)3

L =

WL3

43EIx (15)

for rectangular x-section beam of breadth b and depth d, as in the experiment,

Ix = bd3/12. Thus

vc = WL3

48E x

12

bd3 =

L3

4E

W

bd3 (16)

∴ graph of vc v/s W/bd3 should give straight line whose slope =

L3

4E

from which E can be calculated as

E = L3/(4 * slope)


6. Two-dimensional pin-jointed frameworks

A framework of members joined at their ends is called a truss. Trusses are commonly used for bridges, roof support structures, towers, antennae, etc. The structural members in trusses include I-beams, channels, angles and tubular members. In this course, we shall consider a particular type of truss, where all of the members lie in one plane: such frameworks are called two-dimensional or plane trusses. Plane trusses are often used in pairs, e.g. one on each side of a bridge with cross-member connections that provide stability to the assembled three-dimensional structure also transfer the applied loads from roadways to trusses. Similarly, a roof truss may be composed of several plane trusses. Peep into the attic of your house to confirm this !!

The members of a truss may be joined at their ends by welding, rivets or bolts and gusset plates, pins, adhesives, etc.

Figure 6.2 Frictionless point joint Rivetted joint

Figure 6.1


In this course, we shall consider one type of connection only, by means of frictionless pins, i.e. we shall deal with pin-jointed trusses. There are two reasons for this. Firstly, we shall see that this restriction simplifies greatly the analysis of a truss. Secondly, it may turn out that for any truss with slender members, any type of joint may be modelled (i.e. idealised) as a pin-joint for the purpose of analysis: this gives reasonably good predictions of the bar forces in the real structure, providing the centrelines of adjoining members are concurrent at each joint.

Triangulated trusses

Three bars joined by pins at their ends forms a rigid framwork (Figure 6.3a) whereas four bars form a non-rigid frame or mechanism (Figure 6.3b). The addition of one extra bar to the truss of Figure 6.3b transforms it into a triangulated truss, which consists of a series of triangles. Triangulated trusses are rigid.

The trusses in Figures 6.3a and 6.3c are both statically determinate because their bar forces can be determined solely from considerations of equilbrium. It is a characteristic of statically determinate trusses that the bars can be assembled without forcing, even though they may not all be exactly of the correct length. It follows that changes of length of the members of statically determinate trusses due to such causes as change in temperature or humidity (especially in case of wooden trusses) do not produce additional stresses in the structure.

Figure 6.3

(a) (b) (c)


The truss shown in Figure 6.4, obtained by adding a bar to the truss of Figure 6.3c, is statically indeterminate, i.e. its bar forces can not be determined by equilibrium conditions alone. Analysis of statically indeterminate trusses is out of scope of this course.

The easiest way of designing a truss which is both rigid and statically determinate is to form a series of triangles and then to arrange for suitable supports. The design of more general, non-triangulated trusses which are both rigid and statically determinate requires a proper matrix analysis which again is not the part of this course.

External loads and internal forces for a truss

A truss consists of slender elements and, for a general loading, there will be the following three internal forces at any section of any member:

• axial force T;

• bending moment M; and

• shear force S Figure 6.5 shows a typical straight member AB of a truss and the internal forces T,M, and S at a particular section. It is assumed that member AB is rigidly connected to other members of the truss, and hence all three internal forces can be applied at its ends.

Figure 6.4


In this course, we shall consider only pin-jointed trusses subject to point loads applied only at the joints. As these loads are usually large in comparison to the self-weight of the members, we shall also neglect the self-weight of the members. Let us consider a typical member of such a truss, as shown in Figure 6.6: this member is subjected to only one force applied at each pin end. It follows from considerations of statical equilibrium that the two forces at either end must be equal in magnitude, opposite in direction and collinear. Thus, each member is in a state of pure tension or pure compression, i.e. it is an axial force member.

In particular note that the bending moment M is zero at all sections of this member. Bending moment could arise only if the pins were not smooth, or if the member was loaded in between the joints, for example, by a uniformly distributed weight. From a design perspective, a pin-jointed truss carries the applied loads from the

T

B

S=0

T

M=0

A

A

T

Figure 6.6

TB

SB

MB SA MA

TA

P

W

SB

Figure 6.5

TB

SA MA

TA


joints to the foundation by setting up tension and compression in its members.

Calculation of bar forces in a pin-jointed truss

Two alternative methods can be used to calculate the axial forces in the bars: the method of joints and the method of section.

6.1.1 Method of Joints

The method of joints provides a systematic way of calculating all the bar forces in a structure by considering the equilibrium of forces at each joint. As the forces are concurrent, we can resolve them in two directions to obtain two independent equilibrium equations. Thus, we can find two unknown bar forces (or reactions) at each joint. Example 6.1 Two bar truss: find the axial forces in the bar

Free-body diagram for joint A Triangle of forces

X

Y

TAC

TAB

45o

W

450

WTAC −=

W

WTAB 2=

B

A C

L

45o

Figure 6.7


∑ =−⋅⇒= 0)45sin(0 0WTF ABy ∴ WTAB 2=

∑ =⋅+⇒= 0)45cos(0 0

ABACx TTF ∴ WTAC −=

We can check this result by drawing a triangle of forces.

Example 6.2 Four-bar truss: find the axial forces in the bar

FBD for joint A

WT

WT

AC

AB

−=

= 2

FBD for joint B Triangle of forces

∑ =−−⇒= 02

2

20

WTF BC

y ∴ )(2 ncompressioWTBC −=

02

2

20 =+−−⇒=∑

WTTF BC

BDx ∴ )(2 tensionWTBD =

We can check this result again by drawing a triangle of forces.

X

Y

TBC

TBD

45o

45o

WTBD 2=

WTAB 2=

WTAB 2=

45o

WTBC 2−=

B

A C

45o

Figure 6.8

45o

D

L

W


6.1.2 Method of Section

If we are required to know the force in only one or two members of a truss and the members are remote from supports, the method of joints can involve a lot of work. In such cases, the method of section can save much time. It involves following three steps:

1. Make an imaginary cut through the structure to create a free body. The cut must pass through the member of interest. The cut often divides the structure into two sections.

2. Draw a separate FBD for a section of the structure, imagining it

to be completely removed from the remainder of the structure but subject to axial forces (of unknown magnitude) acting on the cut member.

3. Resolve and/or take moments for the section as necessary. In

general, up to three independent equilibrium equations can be obtained from which up to three unknown bar forces can be determined.

In step 3, often the bar force of interest can be obtained directly by means of a single equilibrium equation, either by taking moments about a specially chosen point, or by resolving forces in a suitable direction.

Example 6.3 14 bar tower with diagonals at 45°°°°: find the axial forces in bars DF and EG

C

D

F

H

L W

A

B

E

G

J

L W

A

B

E

C

D

F

2L

TOF

TEF

TEG

Figure 6.9


The method of joints would involve analysing 5 joints (A-E) in succession. Having cut the structure through the section DEFG, we note that the upper section is subject to 3 unknown forces which can be found from 3 equilibrium equations. We can obtaine TDF directly from one of these equations, by taking moments about E (where TEF and TEG intersect):

∑ =⋅−⋅⇒= 020)( LTLWEM DF ∴ WTDF 2=

Similarly for bar EG

∑ =⋅−⋅⇒= 030)( LTLWFM EG ∴ WTEG 3−=

How would you find TEF? (See Examples Paper).

Final remarks

If all joints of a truss are connected to more than two bars (where each foundation reaction counts as one bar), it is necessary to start by determining the support reaction by considering the overall equilibrium of the structure (see Example 2.4). The calculation of bar forces can be simplified by noting the following special cases for unloaded joints:

Figure 6.10

T2=T1

T3=0

T1

3 members of which 2 collinear

T2=0

T1=0

2 members

T3=T1

T1

T2

T4=T1

2 pairs of collinear members


Example 6.4 Roof truss, vertical load W at joint E: find the axial force in DE.

Whether we use the method of joints or the method of section, we must start by finding reactions at A and B: we could not start at joint E with the method of joint because there are three unknown forces there and we only have two equilibrium conditions (in x and y directions). Considering the equilibrium of the whole structure:

∑ =⋅−⋅⇒= 040 1)( LWLRM BA ∴ 125.0 WRB =

Considering the equilibrium of the right hand section:

∑ =⋅+⋅⇒= 0220)( LTLRM DEBC ∴ BDE RT 2−= ∴ 1354.0 WTDE −=

If instead we wanted TAC then:

175.00 WTM ACE =⇒=∑

or for TCE:

1707.00 WTM CEA −=⇒=∑

For W = 1, the full set of tensions turn out to be:

Figure 6.11

HA=0

RB

B

W1

RA

L L L L

D

F

E

A

C

L L

D

F

E

C

TOE

TCE

TAC

RB


Figure 6.12

0.25

1

0.75

-0.354

-0.354

0.25

0.75

-1.061

-0.354


7. Torsion

A common engineering mode of deformation is that of torsion- where a solid/tubular member is subjected to a torque about its long axis; this results in twitsting

7.1 Stress-strain considerations for a structural member in torsion

Torsion refers to twisting of a member when loaded by couples that produce rotation about its longitudinal axis.

Figure 7.1 In the case shown, the torque (or twisting moment or twisting couple) is force x perpendicular distance between the forces F

torque dFT ⋅=

For equilibrium of the member, there must be a torque T = Fd acting in the direction opposite to the torque due to the applied force F. In the figure above this acts the fixed (left hand) end.

d

T F

F


7.2 Torsion of thin walled cylindrical tube

AIM: to establish a relationship between τ and θ Consider a thin-walled tube of mean radius r and wall thickness t (<<r), subjected to a torque at each end, shown in Fig 4.2. Assume that 'lower' end of tube is fixed. Due to application of torque, a straight line AB which is parallel to

the tube axis twists through a small angle, φ, to AB'. Similarly CD twists to CD'. We assume AB' and CD' are straight lines. An element PQRS deforms to P'Q'R'S'. The shear strain in the element is equal to the decrease in angle QPS, which is initially 90o.

The decrease is seen to be angle φ. In fact the shear strain for any

other elements (e.g. ABDC) is also φ.

∴ shear strain = φ (7.1)

From BOB' length of arc BB' = rθ (7.2)

where θ = angle BOB' (in radians) is the "angle of twist" over the whole length L of the tube.

Figure 7.2

T T

P

P’

S’

S

Q

Q’

R

R’

B

B’

D

D’

dFs

dA

θ

r t

dα

L

φ A

C

θ

O

B’

B

L

A φ


From BAB' length of arc BB'= Lφ (7.3)

where φ = BAB' (in radians) is the shear strain Equating 7.2 and 7.3

Lφ = rθ

∴ φ = rθL

(7.4)

The Shear modulus, G, for a material is given by Hooke's Law stating that shear sress is proportional to shear strain in the linear elastic region, just as normal stress and strain are related by Young's modulus

τφ = G

∴ τr =

GθL

(7.5)

or τ = GθL

r (7.6)

Since we are considering a thin tube, the variation of τ across the

tube thickness is negligible, i.e. τ is constant.

AIM: to establish a relationship between T and θ

Consider a small element of area subtending an angle dα at the centre of the tube, see Figure 7.2

The shear force on element due to τ is

dFs = τ dA = τ (r dα) t (7.7)

Torque about axis due to dFs is

dT = dFs r = τ dΑ r (7.8)

T = ⌡⌠A

τ r dA

as τ = GθL

r, T = ⌡⌠ GθL

r2 dA

G,θ and L are constants and

⌡⌠ r2 dA = J, an expression called the polar second moment of

area

Therefore τr =

T

J (7.9)


and from 7.5 τr =

GθL

∴ τr =

GθL

= T

J (7.10)

7.3 Torsion of hollow circular shaft (Thick walled tube)

Consider hollow shaft of inner diameter D1, outer diameter D2 and

length L subjected to torque T, shown in fig.4.3.

Figure 7.3 We assume (a) shaft is straight and of uniform x-section (b) torque is constant at all x-sections perpendicular to z-axis (c) plane sections remain plane after twisting (d) radial lines remain radial The hollow shaft can be thought of as being built from thin

tubes of thickness δr and mean radius r. From thin tube theory we have

τr =

GθL

T T

L

A

C r

r2

r1

dr

r

τ


τ = GθL

r (7.11)

Figure 7.4

Since G, θ, L are constant, τ = constant x r i.e. τ increases linearly with r.

At inner radius τ = GθL

D1

2 and at outer radius τ =

GθL

D2

2

Torque on elemental ring of radius r and thickness δr is

dT = τ dΑ r = τ 2πr dr r (7.12)

= GθL

r 2π r2 dr = GθL

2πr3 dr

total torque, T = ⌡⌠

D1/2

D2/2

GθL

2πr3 dr

∴ T = GθL

⌡⌠

D1/2

D2/2

r2 (2πr dr) (7.13)

Now 2πrdr = dA = area of elemental ring

∴ ⌡⌠

r2 2πr dr =

⌡⌠

r2 dA = 2nd moment of area about

axis = J

∴ T = GθL

J or GθL

= T

J

r

τ


As Gθ/L = τ/r

τr =

GθL

= T

J (7.14)

To calculate the polar second moment of area, described in equation 7.13

J = ⌡⌠

D1/2

D2/2 r2 (2πr)dr = 2π

⌡⌠ r3dr

= 2π [r4

4 ] =

π32

(D24 - D1

4) (7.15)

7.4 Torsion of solid circular shaft

This is just a special case of hollow shaft with D1 = 0 (no hole) and

D2 = D. Equation 7.14 applies with 7.15 changing to J = πD4/32

7.5 Concentric shafts

Take two shafts of different materials, 1 and 2

Figure 7.7

1

2

L

r1 r2

r3


Using equilibrium T = T1 + T2

As we have considered before, from the geometry θ = θ1 = θ2

the angle of twist is the same if there is no slip between the shafts i.e. the ends are rigidly connected

Using T

J =

GθL

θ1 = θ2 = T1 L1

G1 J1 =

T2 L2

G2 J2

As L1 = L2 T1

G1 J1 =

T2

G2 J2

When both shafts are made of the same material

T1

J1 =

T2

J2


Appendix


8- Mohr’s Stress Circle

As derived in Handout No. 7, the direct and shear stresses on a

plane inclined at an angle θ w.r.t. the x-axis are given by

)2sin()2cos()(2

1)(

2

1θτθσσσσσ θ xyyxyxx +−++= (8.1)

)2sin()2cos()(2

1)(

2

1θτθσσσσσ θ xyyxyxy −−++= (8.2)

)2cos()(2

1θτσσσ θ xyyxx ++−= (8.3)

These are the parametric equations of a circle. If we choose a set of

rectangular axes and plot a point M of abscissa σxθ and ordinate τθ for any value of the parameter θ, all the points will lie on

a circle. Let us try to plot this circle. For clarity, let us use the following notation:

σσσ =+ )(2

1yx ayx =− )(

2

1σσ bxy =τ

Then, equation 8.1 for σxθ and equation 8.2 for τθ become

)2sin()2cos( θθσσ θ bax +=− (8.4)

)2cos()2sin( θθτθ ba +−= (8.5)

Squaring equations 8.4 and 8.5 and adding them, we obtain

( ))2cos()2sin(2)2cos()2sin(2

)2(cos)2((sin)2(sin)2((cos 22222222

θθθθ

θθθθτσσ θ

abab

baxy

−+

++++=+−

∴ ( ) 2222bax +=+− θθ τσσ (8.6)

We used equation 8.1 for σxθ to obtain equation 8.6. We could use

equation 8.2 for σyθ and would still obtain equation 8.6. This is due to the simple fact that when squaring the terms, their sign does not

matter. Therefore, for further clarity, let us replace σxθ by σθ in equation 8.6. It is also worth noting that

( ) 22

2

22

2

1Rba xyyx =+

+=+ τσσ (8.7)


Therefore, equation 8.6 becomes

( ) 222R=+− θθ τσσ (8.8)

Equation 8.8 is an equation showing the variation between σθ and

τθ, i.e. it describes a curve which relates τθ (plotted on vertical axis,

for example) to σθ (plotted on horizontal axis). It is an equation of a

circle whose centre has coordinates (σ, 0) and whose radius is R as shown in Figure 8.1. This circle is called Mohr’s stress circle.

Figure 8.1 Mohr’s Stress Circle Any point on Mohr’s stress circle (e.g., point P in Figure 8.1) represents values of direct and shear stresses on a plane inclined at

an angle θ w.r.t. the x-axis. Note that the angle of inclination the

radius of Mohr’s stress circle passing through point P w.r.t. the σθ

axis is 2θ. It is due to the fact that the parametric equations of

Mohr’s stress circle are in terms of 2θ and not θ. The four points A, B, C, and D on Mohr’s stress circle (as shown in Figure 8.1) have

special significance. Points A and B lie on the σθ axis, i.e. there are no shear stresses at points A and B. Therefore, they must represent principal stresses. Point A represents minimum or minor principal stress and point B represents maximum or major

B

aσ

+

E

R

θ2

C )0,(σ

D

A

+

-

-

θτ


principal stress. Similarly, points D and E represent maximum shear stress.

8.1. Construction of Mohr’s stress circle

8.1.1. Sign convention for direct and shear stresses

Before we study the method of construction of Mohr’s stress circle, let us fix the sign convention for direct and shear stresses. In the case of direct stresses, the tensile stress is taken positive and the compressive stress is taken negative. In the case of shear stress, if the action of shear stress would result in an anticlockwise rotation of the element, it is taken as positive. On the other hand, if the action of shear stress results in a clockwise rotation, it is taken as negative. This is clearly illustrated in Figure 8.2. You can use a reverse sign convention if you want to. It will not make any difference on the construction of Mohr’s stress circle. But you must be consistent in the use of the adopted sign convention when you present the results obtained from Mohr’s stress circle. Figure 8.2

8.2. Method of construction

We are given a plane stress system shown in Figure 8.3 which consists of two-dimensional stress

components σx, σy, τxy and τyx ( = τxy) on any pair of orthogonal reference planes aligned in the x and y directions. We are required to find out the direct and shear stresses

on a plane inclined at an angle θ w.r.t. the x-axis using Mohr’s stress circle. The construction of Mohr’s stress circle involves following steps: (Refer to Figure 8.4). Figure 8.3 1. Draw the direct stress (tensile +ve) axis and the shear stress

(anticlockwise +ve) axis and fix the scale of these axes. Make

σx

τxy


sure that you take the same scale for both the axes otherwise your Mohr’s stress circle will turn out to be an ellipse !

2. Plot point X which has coordinates (σx, τxy) and point Y which has

coordinates (σy, -τxy) as shown in Figure 8.4. At θ = 0 or 90°, σxθ =

σx, σyθ = σy and τθ = τxy. Thus, (σx, τxy) acting on the face normal to

x-axis and (σx, -τxy) acting on face normal to y-axis are two points on Mohr’s stress circle. The shear stress for the face normal to y-axis is negative because it applies a clockwise couple to the

element. Since the planes on which the stresses (σx, τxy) and (σx, -

τxy) act are orthogonal (θ = 90°), they will be plotted diametrically opposite on Mohr’s stress circle. This is because Mohr’s stress

circle is parametric in terms of 2θ which will be equal to 180° for the two orthogonal planes. In other words, the centre of Mohr’s stress circle will lie on the line passing through points X and Y.

Figure 8.4 3. Join points X and Y by a line. This line will intersect the direct

stress axis at point C which is the centre of Mohr’s stress circle. The radius of Mohr’s stress circle will then be either CX or CY.

The coordinates of the centre will be (σ, 0) where σ is the

arithmetic mean of the two direct stresses σx and σy. Draw Mohr’s stress circle with centre at C and radius as CX (or CY).

4. The principal stresses are defined by points B and D where the

shear stresses are zero.


5. The maximum shear stresses are defined by points G and H. These two points lie on either end of the vertical diameter of Mohr’s stress circle. Note that points B and D representing the principal stresses are on either end of the horizontal diameter of Mohr’s stress circle. Hence, the angle between the planes of

maximum shear and the principal planes should be 90°/2 = 45° (as proved in Handout No. 7).

8.3. Pole on Mohr’s stress circle

By now, it may have become apparent to you that this business of

denoting the stresses on Mohr’s stress circle in terms of θ2 is not very convenient. It would be better if we could somehow draw on the Mohr’s stress circle the planes on which these stresses act. This can be achieved if we define a point known as the pole of Mohr’s circle. The essential characteristic of the pole is that a line passing through the pole and a point representing direct and shear stresses on Mohr’s stress circle will be parallel to the plane on which the represented stresses act. There is only one pole for a Mohr’s stress circle and it can be located as follows: 1- After constructing the Mohr’s stress circle using the method given above, draw a line passing trhough point X which is parallel to the vertical plane on which stresses ),( xyx τσ act, as shown in

Figure 8.5.


2- Similarly, draw a line passing through point Y which is parallel to the horizontal to the horizontal plane on which stresses ),( xyx τσ −

act. These two lines intersect each other on Mohr’s stress circle at a point P. This point is the pole of Mohr’s stress circle since any line passing through P and point representing a stress state (e.g. point X or Y) is parallel to the plane on which the stresses act. 3- It becomes fairly simple now to obtain the directions of principal plane or planes of maximum shear stress. All you have to do is to draw a line passing through the pole and the point representing the principal stress or maximum shear stress as shown in Figure 8.5. 4- Similarly, the direction of planes of maximum shear stress can be obtained by drawing a line passing through the pole and one of the points of maximum shear which are located on either end of the vertical diameter of Mohr’s stress circle. (Figure 8.6).

Figure 8.5 Pole of Mohr’s stress circle and principal planes


5- Using the reverse procedure, the stresses on any plane inclined at an angle θ w.r.t. the horizontal plane (or x-axis) can be obtained. In this case, a line parallel to the plane in question is drawn passing through the pole of Mohr’s stress circle. This line will cut Mohr’s stress circle at a point S as shown in Figure 8.7. The direct and shear stresses acting on this inclined plane are the co-ordinates of point S. If a line perpendicular to the inclined plane is drawn passing through the pole, it will cut Mohr’s stress circle at point T as shown in Figure 8.7. It can be seen from Figure 8.7 that point T is diametrically opposite to point S and that its coordinates will give the direct and shear stresses on a plane which is orthogonal to the inclined plane.

Figure 8.6 Planes of maximum shear stress


In each of the three cases, you can either measure the angles and the stresses off the diagram or use the equations given in Handout No.7 to compute them. However, the beauty of Mohr’s stress circle is that all the angles and stresses can be obtained graphically and therefore, there is not much point in using the equations once you have succesfully drawn Mohr’s stress circle unless great accuracy is required.

Figure 8.7 Stresses on any inclined plane


9. Multi-dimensional Stress States Multi-dimensional Stress States

AUTOMOTIVE Stress analysis of front suspension Source: www.ansys.com

AEROSPACE Stress analysis of fuselage impact Source: ONERA


MEDICAL Stress analysis of hip endoprostheses

Source: www.endolab.de

Maximum Principle Stresses (Example)

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