Calculating stress and strain on Loading Systems.

MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : COMPLEX LOADING SYSTEMS

LESSON 1 : STRESS AND STRAIN

MP - 1 - 1

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

It is essential that any product, machine or structure is safe and stable under the

loads exerted on it during any foreseeable use. In this module a range of

mechanical principles will be covered, which underpin the design and

operation of mechanical engineering systems.

In order to study mechanical principles, we should first know mechanical

properties of engineering materials, which involves the determination of

stresses in loading-carrying members and the deflection or deformation of

those members. In general terms, this requires the study of stress and strain.

The content presented in this first lesson of the module will be enable you to

determine the magnitude of both stress and strain created in members

subjected to uniaxial, biaxial or triaxial forces, either tensile or compressive.

Under these loadings, engineering materials such as metals, polymers and

ceramics show their different aspects of mechanical properties described by

modulus of elasticity, shear modulus and bulk modulus.

Other important mechanical properties and the tensile test will also be

presented to give you a further understanding of the mechanics of engineering

materials.

In this lesson, the basic concepts in the strength of materials will be presented.

You will need them for the further study in this module.

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YOUR AIMS ________________________________________________________________________________________

After completing this lesson, you should be able to:

• define stress and strain in a body under the action of different forces

• define modulus of elasticity, shear modulus and bulk modulus

• determine volumetric strain and change in volume

• calculate material change within elastic limit under tensile or

compressive stresses.

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________________________________________________________________________________________

STRESS AND STRAIN ________________________________________________________________________________________

STRESS

When an external force acts on a material, which infers that an equivalent force

acts in the opposite direction to maintain equilibrium, internal forces are set

up. These internal forces are known as stresses. Thus, stress is the internal

resistance offered to an externally applied load by a unit area of the material

from which a member is made.

Stress provides a measure of force F scaled to the area A over which the force

operates:

There are three fundamental types of stress: tensile stress, compressive stress

and shear stress. Conditions giving rise to these stresses are illustrated in

FIGURE 1.

FIG. 1 The three simple stresses: A is in tension, B is in compression, C and

D are in shear

Stresses that result in extension, or tensile stresses, are defined as positive, and

compressive stresses as negative in sign.

P

A

P

P

B

P P

P

C

P

P 2

P 2

D

Stress force area

Paσ = = ( )F A

.......................... 1( )

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STRAIN

If a solid object is subjected to forces that tend to stretch, shear or compress the

object, its shape changes. If the object returns to its original shape when the

forces are removed, it is said to be elastic. Most objects are elastic for forces

up to a certain maximum, called the elastic limit. If the forces exceed the

elastic limit, the object does not return to its original shape but is permanently

deformed.

FIGURE 2 shows a solid bar of length l0 subjected to a stretching or tensile

force F acting equally to the right and to the left. The bar is in equilibrium, but

the forces acting on it tend to increase its length. The fractional change in

length of the bar ∆l is called the strain:

FIG. 2 Elongation of a bar in tension due to the acting of tensile force F

l 0

l

F (σ)F (σ)

strain deformation in length

original leng ε =

tth = = ( )∆l

l

l l

l0

0

0

2 –

..........

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Example 1

An 8 kg lamp is supported by two rods AB and BC, as shown in the figure

below. If AB has a diameter of 10 mm and BC has a diameter of 8 mm,

determine the average normal stress in each rod.

Solution

From the question, the weight of the lamp is:

The diameters of the rods are:

To determine the normal or uniaxial forces acting on the rods, we must first

analyse the force balance on the lamp. The free-body diagram of the lamp is

shown overleaf.

d

d

AB 3

BC 3

mm 10 10 m

mm 8 10 m

= = ×

= = ×

10

8

–

–

W mg= = × =8 9 81 78 48. . N

B

4 3

5

60°

A C

5

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In the x direction, we have

In the y direction, we have

Thus,

F F W

F F

F

BA BC

BC BA

BA

sin 60 sin° + =

=

=

θ –

. – .

.

0

0 8 0 5 0

1 6FF

F F

F F

BC

BA BC

BC BC

0 866 0 6 78 45

1 3856 0 6 7

. . .

. .

+ =

+ = 88 45

1 9856 78 45

39 5

6

.

. .

.

F

F

F

BC

BC

BA

N

and

=

=

= 33 2. N

F FBC BAcos cos 60θ – ° = 0

sin

cos

θ

θ

=

=

3 5

4 5

B

4 3

5

60°

y

F BC

F BA

W

x θ

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By Newton's third law of action, equal but opposite reaction, these forces

subject the rods to tension throughout their length. Applying equation (1) we

have:

σ BC BC

BC

BC

BC 2

= =

= × ×( )

=

F

A

F

d π

π

4

39 5

4 8 10

785 82

3 2

.

–

88

4

63 2

4 10

Pa 0.786 MPa

AB BA

AB

BA

AB 2

=

= =

= ×

σ F

A

F

d π

π .

××( )

= =

10

804 687

3 2–

Pa 0.805 MPa

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Example 2

A cylindrical casting of radius 220 mm and

length 840 mm is made of steel having a

density of 7850 kg m–3, as shown in the

figure (right). Determine the average

compressive stress acting at the base.

Solution

From the question

The force acting at the base is due to the weight of the casting. In the vertical

direction, the force balance is

Remember, m = ρV, where ρ is the densisty of the casting and V is the volume of the casting.

Then P mg Vg A= = = × × ×

= ×

ρ ρ 840 10 9 81

7850

3– .

00 152 840 10 9 81

9832

3. .–× × ×

= N

P W

P W mg

– =

∴ = =

0

A r= = ×( ) =

=

π π2 3 2220 10 0 152

7850

–

–

. m

kg m

2

3ρ

W

P

220 mm

84 0

m m

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Then, the average compressive stress acting at the base is:

The negative sign means that it is a compressive stress.

σ = =

=

=

– – .

–

–

P

A

9832 0 152

64 684 Pa

64.7 kPa

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________________________________________________________________________________________

ELASTICITY ________________________________________________________________________________________

Mechanical properties are almost always quoted when a new design is being

considered even if the component's strength or other mechanical behaviour is

not the prime attribute, e.g. an electrical insulator. This is because of the way

mechanical properties affect forming characteristics and because even an

electrical unit needs to have sufficient strength to last for its service period.

Elasticity is a very important mechanical property of a solid material.

In a material showing elastic behaviour, stress is directly proportional to strain.

This can apply to three types of loading system:

• uniaxial or tensile loading

• biaxial or shear loading

• triaxial or hydrostatic loading.

UNIAXIAL OR TENSILE LOADING

A measure of the stiffness of a material is found by calculating the ratio of the

normal stress on an element to the corresponding strain on the stress element,

as shown in FIGURE 2. This ratio is called the modulus of elasticity or

Young's modulus, E. This is

Equation (3) is also referred to as one-dimensional Hooke's law.

E = σ ε

........................................... 3( )

= = ( )

=

where normal stress Pa

Y

σ F A

E ooung's modulus of elasticity Pa

normal

( )

=ε strain, defined in equation 2( )

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A material with a higher value of E means it will deform less under a given

stress than one with a lower value of E. Some typical values of E for different

materials are listed in TABLE 1.

TABLE 1 Typical values of E of some engineering materials

Example 3

A wire 1.5 m long has a cross-sectional area of 2.4 mm2. It is hung vertically

and streches 0.32 mm when a 10 kg block is attached to it. Find the stress, the

strain and Young's modulus for the wire. From TABLE 1 determine the

material that the wire is probably made from.

Material Modulus of elasticity (GPa)E

Stainlesss steel 193

Copper and its al

– 203

lloys

Titanium alloy

C

110 131

110 120

–

–

ooncrete

Wood

22 30–

Ceramic

10 14–

Rubber

300 400

7 10 4 –

––× 44 10 3× –

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Solution

From the question

The force from the block is:

The normal stress

The strain

Using equation (3):

we have

According to TABLE 1, the material of the wire is probably stainless steel.

E

E

=

= × ×

= × =

σ ε

Pa 192 GP

4 09 10 2 13 10

192 10

7

4

9

. . –

aa

σ

ε

ε

= = ×

= ×

=

= ×

F

A

l

l

98 1 2 4 10

4 09 10

0 32 10

6 7

0

. .

.

.

– Pa

∆

–– –

. .

3 4

1 5 2 13 10= ×

F mg= = × =10 9 81 98 1. . N

l

A

l

0

6

1 5

2 4 2 4 10

0 32

=

= = ×

= = ×

.

. .

.

–

m

mm m

mm 0.32 10

2 2

∆ ––3 m

kgm = 10

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BIAXIAL OR SHEAR LOADING

Under a biaxial or shear loading, shear stress is applied to a material as shown

in FIGURE 3. The ratio of the shear stress τ to the shear strain γ is called the modulus of elasticity in shear, or shear modulus, or the modulus of rigidity G,

which is

where G = shear modulus (Pa)

τ = shear stress (Pa) γ = shear strain, defined as the angle change (measured in radians).

In practical problems, only very small values of γ are encountered. Thus, we can have γ = tan α ≈ α.

FIG. 3 Body acted on by shear stress

Shear modulus is a property of the material, which describes the material's

response to shearing strains. Some typical values are shown in TABLE 2

overleaf. By comparing TABLES 1 and 2, you can find that, in general, the

shear modulus G is less than the elastic modulus E for a given material.

τ

τ

α

γ = tan α

G = ( )τ γ

................................ 4

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TABLE 2 Typical values of G of some engineering materials

TRIAXIAL OR HYDROSTATIC LOADING

When a triaxial or hydrostatic loading acts on a material, there will be a three-

dimensional change within the material. The bulk elastic properties of a

material determine how much it will compress under a given amount of

external pressure p. FIGURE 4 shows a material deformation under the action

of an external pressure p in 3-dimensions.

FIG. 4 Deformation of a material in 3-dimensions under a

hydrostatic loading p

p

VV 0

Material Shear Modulus (GPa)G

Stainless steell 75 83

Copper and its alloys 37 39

–

–

TTitanium alloy 39 44

Wood

–

0.8 12

Ceramic

–

20 200

Rubber

–

. ––2 0 10 4× 11 0 10 3. –×

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The ratio of the change in pressure ∆p to the fractional volume compression is called the bulk modulus of the material K.

where ∆p = pressure change (Pa)

K = bulk modulus, which gives the change in volume of a solid

substance as the pressure on it is changed (Pa)

V0 = original volume (m 3)

∆V = V – V0 = Volume change due to the pressure p (m3)

= volume strain or bulk strain.

A representative value for the bulk modulus for steel is 160 GPa and that for

water is 2.2 GPa. That means water is about 80 times more compressible than

steel. For gases, bulk modulus varies with pressure and temperature, for

example, at standard atmospheric pressure (101 325 Pa) and 20 °C,

Kair = 151 767 Pa. The reciprocal of K is called the compressibility of the

substance. The amount of compression of solids and liquids is seen to be very

small. Therefore, essentially, K measures a substance's resistance to uniform

compression.

In all of these stress situations the same types of relationships apply for tension

and compression.

∆V V0

K p V

V

= ( )– .............................∆∆ 0

5

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Example 4

A liquid compressed in a cylinder has a volume of 1 litre at 1 MPa and a

volume of 995 cm3 at 2 MPa. What is its bulk modulus?

Solution

From the question, the pressure change is

The volume change is

Then, using equation (5), we have

K p V

V

= = × ×

×

= ×

=

– – – –

–

∆ ∆

0

6

6

3

6

1 10 5 10 1 10

200 10 Pa

200 MMPa

∆V

V

= × × = ×

= = ×

995 10 1 10 5 10

1

6 3 6

0

– – –

–

– – m

litre 1 10

3

33 3m

∆ p = = = ×2 1 1– MPa 1 10 Pa6

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________________________________________________________________________________________

STRENGTHS IN TENSION ________________________________________________________________________________________

We can obtain a range of mechanical properties of materials from the results of

stress–strain curves carried out on tensile test pieces which are appropriate for

the particular material.

For example, for metals:

for polymers:

A typical stress-strain curve of a metal is sketched in FIGURE 5.

FIG. 5 Typical Stress–Strain Curve for Metals

Notice how the yield stress σy, and ultimate tensile strength, σu, are defined from the curve.

Stress

Yield stress σy

Fracture stress σf

Strain ε

Ultimate tensile strength σu

Slope, E = σ ε

Yield point

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Sometimes if the yield point is not readily identifiable the proof stress, PS, is

used, as shown in FIGURE 6.

FIG. 6 Determination of the yield point

For a range of steels a discontinuous yield may be observed (FIGURE 7).

FIG. 7 Discontinuous yield stress observed in steel

Strain

Stress

x% strain

PS

Strain

Stress

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For polymers the forms of stress–strain curves are often quite different from

those for metals, as shown in FIGURE 8.

FIG. 8 Typical Stress–Strain Curves for Polymers

The properties of polymers are very dependent on temperature and structures.

The curves in FIGURE 8 can be explained on the basis of temperature. There

is a temperature known as the glass transition temperature, Tg, below which

extensive deformation by the mechanism of interatomic bond rotation is very

difficult. The type of stress–strain curve displayed by a material depends on

the value of its glass transition temperature relative to the temperature, T, at

which the curve is produced.

Brittle, <<

Limited plasticity,

Cold drawing,

Viscous flow, >>

T T

T T

T T

T T

g

g

g

g

≈

≈

0 8.

Stress σ

σy

Strain ε~1%

σy

Brittle

Limited plasticity

Cold drawing

Viscous flow

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We can define stress during testing as follows.

However, during tensile testing the cross-sectional area of a test piece

decreases and it may be more appropriate (and accurate!) to define true stress,

σT, as

Almost every engineering design involves stresses limited to the elastic region,

e.g. it would not be appropriate for the girders of a bridge to deform plastically

as traffic passed over.

True stress, force,

instantaneous cross σ T =

P

--sectional area of test piece, A

P

P

Original area Ao

Engineering stress, force,

original cros σ = P

ss-sectional area of test piece, Ao

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Distinguish between true stress and engineering stress. From your understanding of

these two parameters, sketch typical stress–strain curves for both types.

________________________________________________________________________________________

The answer is given on the next page.

Strain

Stress

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Answer to In-text Question on page 21.

'Area' refers to cross-sectional area. As specimens are tested, cross-sectional

area changes, decreasing for tensile tests. Hence for a given load the effective

stress increases.

Strain

Stress

True

Engineering

Engineering stress force

original area =

True stress force

actual area =

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________________________________________________________________________________________

OTHER MECHANICAL PROPERTIES ________________________________________________________________________________________

PLASTICITY

A material is perfectly plastic if no strain disappears when it is relieved of

stress. Plasticity is a mechanical property of a material that enables the

material to undergo permanent deformation without rupture. Under very high

stresses, a solid will 'flow' in much the same way as a liquid flows. Metallic

materials have this property so that they can be made into different shapes by

the use of pressing, forging, drawing and extrusion processes.

DUCTILITY AND BRITTLENESS

Ductility is the amount of deformation that a component will sustain without

fracture. It is normally defined as the increase in length at fracture,

∆l or lf – l0 divided by the original length, l0

Alternatively, if a component is being significantly reduced in cross-sectional

area ductility can be expressed as

where ∆A is the final cross-sectional area, Af, minus the original cross- sectional area, A0.

ductility = ∆A A0

i.e. ductility = ∆l l0

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It is necessary to know about ductilities so that the best forming process can be

chosen for a particular material–component combination. For example, it

would not be a good idea to try to shape a brittle material into a wire or column

by a mechanical working process such as forging! Similarly, there would be

problems over using a very brittle material for components such as automotive

exhaust pipes which experience vibrations and impacts from small projectiles

thrown up by wheels.

By contrast, brittleness is another property of a material that makes it break

easily without bending. Have you ever dropped something and seen it shatter?

Glass does this because of its brittleness.

MALLEABILITY

Malleability is the capability of a material to be deformed in all directions

without cracking, i.e. in the conditions likely to be encountered under the

compressive forces of forging. When a material can be beaten or rolled into

thin sheet, it is said to be malleable, e.g. gold beaten into gold leaf.

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NOTES ________________________________________________________________________________________

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SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. The bar of rectangular cross-section shown in the figure below has a constant

width of 35 mm and a thickness of 10 mm. Determine the maximum average

normal stress in the bar when it is subjected to the loadings shown.

2. A 500 kg mass is hung from a 3 m steel wire with a cross-sectional area

of 0.15 cm2. How much does the wire stretch? (Take Esteel = 200 GPa.)

3. The following observations were taken for the load and extension during

a test of a bar of length 250 mm and diameter 22.6 mm.

Load (kN) 0 30 60 90 100 105

Extension (mm) 0 0.094 0.19 0.284 0.317 0.333

Load (kN) 110 115 118 119 120

Extension (mm) 0.356 0.419 0.53 0.89 1.75

Using the graph at the top of page 27, estimate

(a) Young's modulus

(b) the elastic limit

(c) the yield stress.

A 4 kN

35 mm

4 kN

9 kN

9 kN

10 mm B C D

12 kN 22 kN

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4. A pressure vessel of internal volume 2.0 m3 is filled with water at

atmospheric pressure. What is the reduction in volume of the water, bulk

modulus 2.1 GPa, when the gauge pressure in the vessel is 2 MPa?

120

100

80

60

40

20

0 0.25 0.5 0.75 1.0 1.25 1.5 1.75

Extension mm

B

A Yield point

F or

ce k

N

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. From the question, the cross-sectional area is:

The loadings in different sections are all constant yet have different

magnitudes, as shown below.

The normal force diagram representing these results is illustrated below.

A

30

20

10

B C D

P(kN)

A

4 kN

4 kN

9 kN

9 kN

B

C

AB 12 kN

22 kN

P AB

= 12 kN

B

12 kN P

BC = 12 + 9 + 9 =30 kNBC

CD

C D

P CD

= 22 + 4 + 4 =30 kN

P CD

A = × = = ×35 10 350 350 10 6mm m2 2–

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Therefore, by inspection, the largest loading is in section BC.

Then the maximum average normal stress

2. From the question

Using equations (1), (2) and (3)

we have E

F

A l

l

Fl

A l = = =σ

ε ∆ ∆ 0

0

σ ε σ ε

= = =F A

l

l E,

∆

0

and

A

E

= = ×

= = ×

0 15 0 15 10

200

4. . –cm m

GPa 200 10 P

2 2

steel 9 aa

m

N

l

F mg

0

3

3

500 9 81 4 905 10

=

= = × = ×. .

σ max max

A

Pa

85.7 MP

= = × ×

= ×

=

P 30 10 350 10

85 7 10

3

6

6

–

.

aa

P PBC max kN= = 30

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Then, the increase in length of the steel wire is:

3.

(a) Young's modulus

Stress is proportional to strain up to a load of 105 kN, with a

corresponding extension of 0.333 mm, which is the elastic limit of

the material.

Stress at this point load area

=

= × ×

105 10 401 10

3

––

.

6

82 62 10= ×

=

Pa

Strain extension

original lengtth

= × ×

=

0 333 10 250 10

0 00133

3

3

.

.

–

–

Area of bar 4

mm m2 2= × = = ×π 22 6 401 401 102 6. –

∆l Fl

AE =

= × × × × ×

=

0

3

4 9

4 905 10 3 0 15 10 200 10

4 905

. .

.

–

××

=

10 3– m

4.905 mm

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(b)

(c) Yield-point stress

The yield occurs at 110 kN = 110 × 103 N

Yield-point stress 110 10 401 10

3

6 = ×

×

= ×

–

.2 74 108 PPa

Elastic-limit stress 2.62 10 Pa, as above.8= ×

Young's modulus, stress strain

E =

= ×2 62 10 0 00

8. . 1133

197 109= ×

=

Pa

197 GPa

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4. From the question:

Using equation (5)

we have that the volume reduction is:

∆p is the increase of the pressure, which is ∆p = p – p0. If we use gauge pressure to calculate, the initial gauge pressure p0 = 0. Remember

that absolute pressure = gauge pressure + atmospheric pressure.

Therefore

then

(Negative sign means the volume is being reduced.)

∆

∆

p

V

= × = ×

= × × ×

= ×

2 10 0 2 10

2 10 2 2 1 10

1 9 1

6 6

6

9

–

– .

– .

Pa

00 3– m3

∆ ∆

V pV

K = – 0

K p V

V

= – ∆∆

0

V

p

K

p

0

0

2 0

1

2 1

2

=

( ) =

= = ×

= = ×

.

.

m

atm

GPa 2.1 10 Pa

MPa 2 10 Pa

3

abs

9

gauge 6

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

In engineering, the internal resultant loadings in a body and deformation of the

body are specified using the concepts of normal and shear stresses and strains.

In this lesson, we have given definitions to them and shown how they can be

determined for various types of problem due to different loadings. Also, some

basic mechanical properties of materials have been presented. Among these,

modulus of elasticity, shear modulus and bulk modulus are particularly

important both for our further study and in engineering generally.

We can now carry out investigations on complex loading systems in the next

lesson, in which the loadings are in two or three dimensions.

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Teesside University Open Learning (Engineering)

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