Calculating stress and strain on Loading Systems.

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MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : COMPLEX LOADING SYSTEMS

LESSON 2 : POISSON'S RATIO AND LOADINGS

MP - 1 - 2

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

A more complete understanding of deformation of a member subjected to

normal stress is frequently required in engineering design or operation, which

involves two or three-dimensional strains. In this lesson, we are going to

consider these subjects in a little more detail by determining the behaviour of

materials under complex loading conditions.

FIGURE 1 shows that one-dimensional stress can lead to three-dimensional

strain. The calculation of these strains uses a ratio first defined by Poisson.

Therefore, we start our study with the introduction of Poisson's ratio, then further

discussions on two and three-dimensional loading systems will be carried out.

FIG. 1 One-dimensional stress leading to three-dimensional strain

Lateral contraction in y-direction

σ x

Axial extension in x-direction

σ x

Lateral contraction in z-direction

Lateral compression

σ x

Axial extension

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________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

After completing this lesson, you should be able to:

• define Poisson's ratio

• understand the relationships between modulus of elasticity, shear

modulus, bulk modulus and Poisson's ratio in an elastic material

• determine strains in the x, y and z-directions under two and three-

dimensional loading

• calculate material change in dimensions.

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________________________________________________________________________________________

POISSON'S RATIO ________________________________________________________________________________________

When a deformable body is subjected to an axial tensile force, not only does it

elongate but it also contracts laterally. For example, if a rubber band is

stretched, you will find that the thickness and width of the band are decreased.

On the contrary, when a rubber block is compressed due to a force acting in

one direction, its sides get bigger.

FIGURE 2 shows deformation of an object with a square cross-section under

tensile loading. When the load F is applied to it, the length of the object

changes by an amount ∆l = l – l0 and the width by ∆h = h – h0. Strains in the longitudinal or axial direction (εl) and the in the lateral or transverse direction (εt) can be presented, respectively, as

FIG. 2 Deformation of an object under tensile loading

l 0

h 0 h

ε l = = ∆l l

l l

– .......................................

– .....

( )

= =and tε ∆h h

h h

h ................................ 2( )

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In the early 1800s, the French scientist S. D. Poisson realized that the

deformations ∆l and ∆h are proportional and the ratio of these strains εl and εt is a constant for a given solid material within its elastic limit. This constant is

referred to as Poisson's ratio ν and it has a numerical value that is unique for a homogeneous and isotropic material.

Thus, Poisson's ratio ν is defined as

The negative sign means that a longitudinal elongation (positive strain) always

accompanies a lateral contraction (negative strain), and vice versa.

It should be noted that:

• Poisson's ratio is a measure of the simultaneous change in elongation and

in cross-sectional area within the elastic range. During a tensile test,

when a material has the reduction in cross-sectional area and the

increment in length, as shown in FIGURE 2, it is said to have a positive

Poisson's ratio.

• The lateral strain is the same in all lateral directions and is caused only by

the axial or longitudinal force; i.e. no force or stress acts in a lateral

direction in order to strain the material in this direction.

• Virtually all common materials become narrower in cross-section when

they are stretched. The reason why, in the continuum view, is that most

materials resist a change in volume as determined by the bulk modulus K

more than they resist a change in shape, as determined by the shear

modulus G. For example, as mentioned in the last lesson, Ksteel = 160 GPa,

but Gsteel = 75 – 83 GPa.

Poisson's ratio transverse strain

longitu ν = –

ddinal strain t

= – ........................ ε ε

..... 3( )

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Poisson's ratio is seen to be dimensionless. For a perfectly incompressible

material, the Poisson's ratio would be exactly 0.5 and rubber is close to 0.5. In

particular, an ideal material having no lateral movement when it is stretched or

compressed will have ν = 0 and cork is close to 0. Most practical engineering materials have ν between 0 and 0.5. Typical values of Poisson's ratio for common engineering materials are listed in TABLE 1.

TABLE 1 Typical values of ν for some engineering materials

By special design, some materials or structures, mostly polymer foams, can

have a negative Poisson's ratio; if these auxetic materials are stretched in one

direction, they become thicker in perpendicular directions.

Material Poisson's ratio ν Stainless steel 0.27 0.3

Aluminium

–

Cast iron

0 33.

Copper and its alloys

0 2 0 3. – .

Titanium alloy

0 33 0 36. – .

Marble

0 3 0 36. – .

Concrete (non-reinforced)

0 2 0 3

0 2 0

. – .

. – .33

0 29 0Wood . – ..

. –

0 22Ceramics 00 26

0 4

.Rubber 55 0 5

– .

Nylon ..4

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Example 1

A metal box with E = 200 GPa and ν = 0.32 has the dimensions shown in the diagram below. If an axial force F = 80 kN is applied to the bar,

determine the change in its length and the change in the dimensions of its

cross-section under this load. Assume the material behaves elastically.

Solution

From the question:

The normal stress in the bar is:

σ = = ×

× × ×

= ×

80 10 50 10 100 10

16 10

3 3

– –

= = ×

200

1 5

GPa 200 10 Pa

0.32

mm 50 1

00 m

mm 1 0 10 m

kN 80 10 N

–

–x

0 100 0

= = ×

1.5 m

50 mm

100 mm

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The strain in the z-direction is:

Then, the deformation in the z-direction or axial elongation is:

Using equation (3) , we have that the contraction strains in

both the x and y-directions are:

Therefore the changes in the dimensions of the cross-section are

∆ x xx= = × × ×

= × =

ε 0 5 3

2 56 10 100 10

2 56 10

– .

– . –

– –

– m 2.556 m

∆ y yy= = × × ×

= ×

ε 0 5 3

2 56 10 50 10

1 28 10

– .

– –

– == – .1 m28 µ

ε ε νεx y z= =

= × ×

= ×

–

– .

–

0 32 8 10

2 56 10

ν ε ε

ε ε

= =– –x z

∆z zz= = × ×

= × =

ε 0 5

8 10 1 5

1 2 10

–

. m 120 mµ

ε σ

z z

z E = = =

× ×

= ×

∆

16 10 200 10

8 10–

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Poisson's ratio is another way of stating the elasticity of a material. For most

applications it is not necessary to measure every modulus of elasticity as they

can all be calculated from E, the modulus in tension (or compression), using

the following relationships:

Hence G and K can be calculated with reasonable accuracy by incorporating

values of ν and E.

E G

G K =

+ ( ) 3

1 3/ ..........................................

....................

2 1

( )

= +( )G

ν ........................... 5

( )

=and K E

11 2– ....................................

ν( ) ......... 6( )

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Example 2

A specimen of titanium alloy is tested in torsion and the shear stress-strain

diagram is shown below.

Determine the shear modulus G, the proportional limit and the ultimate shear

stress. Also, determine the maximum distance d that the top of a block of this

material, shown in the figure below, could be displaced horizontally if the

material behaves elastically when acted upon by a shear force F. What is the

magnitude of F necessary to cause this displacement?

F d

75 mm

100 mm

50 mm

τ γ

y y

yield stress

and strain respectively

γγ u ultimate shear stress and strain respectiv

eely

600

500

400

300

200

100

0 γ y = 0.008

γ (rad)

F o rc

e kN

γ u = 0.54 0.73

τ (MPa)

τ u = 503

τ y = 360

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Solution

Shear modulus G represents the slope of the straight line portion OA of the

τ – γ curve. The coordinates of point A are (0.008 rad, 360 MPa). Thus

Remember, , then the proportional limit or the yield stress is

τy = 360 × 106 Pa = 360 MPa since the graph ceases to be linear at point A.

From our study in Lesson 1, we know that the ultimate shear stress τu is the value of τ at point B which represents the maximum shear stress. By inspection of the graph again, τu = 503 MPa = 503 × 106 Pa.

Since the maximum elastic shear strain is 0.008 rad, a very small angle γ will be displaced horizontally, which can be determined by

Then, d = × ×

= × =

0 008 50 10

4 10

. –

– m 0.4 mm

tan 0.008 rad mm m

( ) ≈ = = ×

0 008 50 50 10 3

. –

d d

G y

= τ γ

G = ×

= × =

360 10 0 008

4 5 10

. Pa 45 GPa

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The corresponding average shear stress in the block is τy = 360 MPa = 360 × 106 Pa. So the shear force F needed to cause the displacement is given by:

Then F A= = × × ×( ) × ×

= × =

τ 100 10 75 10 360 10

2 7 10

3 3 6

– –

. N 2.77 MN

τ = F

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________________________________________________________________________________________

POISSON'S RATIO AND TWO-DIMENSIONAL LOADING ________________________________________________________________________________________

In a two-dimensional stress system, all the stresses lie within one plane, such

as the plane of this paper, the x – y plane. Now, we consider a flat plate, shown

in FIGURE 3, which is subjected to two-dimensional loading.

FIG. 3 Two-dimensional loading system

From our previous study, within the elastic limit of a material its deformation

follows Hooke’s law:

Also,

In this particular case, εt = εxx, and εl = εyy.

Poisson's ratio ν ε ε

= – .....................t l

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

E = σ ε

σ x

σ y

x x

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In the xx direction, there are two strains:

• strain due to the stress σx along xx direction is

• strain due to the stress σy along xx direction is

Note that minus sign for compressive strain according our convention.

Therefore, the combined strain in xx direction is

Using a similar method, we have the combined strain in yy direction is

After solving the simultaneous equations (9) and (10), we have

σ ν

ε εx x y E

v= +( ) 1 2–

.....................................

– .............

1 2

( )

= +( )σ ν ε εy y x E

v ........................ 12( )

ε σ νσ

y y x

E E = – ................................................. 10( )

ε σ νσ

x x y

E E = – ................................................. 9( )

– .................. νσ y

E compressive strain( ) ............... 8( )

σ x E

tensile strain( ) ........................................... 7( )

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Example 3

If the plate loaded as shown in the figure below is made from a steel with

ν = 0.3 and E = 205 GPa, determine the changes in dimension in both the x and y directions.

3 m m

y2 0 0 m

12 kN

9 kN

10 0 m

9 kN

12 kN

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Solution

From the question:

In the x direction the plate is subject to a load Fx acting over an area Ax. Thus,

the stress in the x direction

Similarly, the stress in the y direction

σ y y

A = =

× × × ×

= × =

9 10 3 10 100 10

30 10

3 3

– –

Pa 30 MPa

σ x x

A = =

× × × ×

= × =

12 10 3 10 200 10

20 10

3 3

– –

Pa 20 MPaa

ν =

= = ×

0 3

205

100 100 100 3

–

GPa 205 10 Pa

mm m

== = ×

= = ×

200

3 3 100 3

mm 200 10 m

mm m

3–

–z

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Using equation (9), the total strain in the x direction is

Since , then dimension change in the x direction is:

Similarly, the total strain in the y direction is:

Dimension change in the y direction is:

∆y yy=

= × × ×

= × =

ε 0

4 3

1 17 10 200 10

2 34 10

– –

– m 23.4 µmm

ε σ νσ σ νσ

y y x y x

E E E = =

= × × ×

– –

– .30 10 0 3 20 10 205 10

6 6

99 41 17 10= ×. –

∆ x xx=

= × × ×

= × =

ε 0

5 3

5 37 10 100 10

5 37 10

– –

– m 5.37 µmm

ε x x

x =

∆

ε σ νσ σ νσ

x x y x y

E E E = =

= × × ×

– –

– .20 10 0 3 30 10 205 10

6 6

55 37 10= ×. –

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________________________________________________________________________________________

POISSON'S RATIO AND THREE-DIMENSIONAL LOADING ________________________________________________________________________________________

STRAINS IN THE x, y AND z DIRECTIONS

Based on our previous study, we can now extend the method used in the

analysis of two-dimensional loading to three-dimensional loading. In the x

direction, there are three strains:

• strain due to the stress σx along xx direction is

• strain due to the stress σy along xx direction is

• strain due to the stress σz along xx direction is

Then, the combined strain in x direction is

ε σ νσ νσ

σ νσ νσx x y z

x y zE E E E = = ( )– – – – ..........1 16(( )

– .................. νσ z

E compressive strain( ) ............ 15( )

– .................. νσ y

E compressive strain( ) ........... 14( )

σ x E

tensile strain( ) ........................................... 13( )

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Similarly, the strains in the y and z directions in terms of the mutually

perpendicular stresses are represented by

ε σ νσ νσ

σ νσ νσy y x z

y x zE E E E = = ( )– – – – ............1 ..

– – – – ......

( )

= = ( )ε σ νσ νσ σ νσ νσz z x y z x yE E E E ........ 18( )

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PRINCIPAL STRAINS IN TERMS OF STRESS

In general cases, stresses on a point in three-dimensions have 9 components

(only 6 of them are independent, for example, τxy = τyx) as shown in FIGURE 4. When the state of stress is represented by the principal stresses, no

shear stress will act on the element. Thus, the principal stresses represent the

maximum and minimum normal stress at the point.

FIG. 4 General state of stress

σ y

σ x

σ z

τ yz

τ xz

τ yz

τ xy

τ yxτxz

τ xy

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When shear stress is in absence, the stresses σx, σy and σz on the faces of an object, shown in FIGURE 5, are in fact principal stresses. So the principal

strain in a given direction may be obtained from the principal stresses. For

principal stresses and strains the suffices x, y and z are replaced by 1, 2 and 3

respectively. Therefore, we may rewrite the above equations in terms of

principal stresses and strains as

FIG. 5 Principal stresses

σ x

(σ 1 )

σ z (σ

3 )

σ y

(σ 2 )

σ x

(σ 1 )

σ z (σ

3 )

σ y

(σ 2 )

ε σ νσ νσ1 1 2 3 1

= ( ) E

– – ...................................

– – .............

1 2 2 1 3

( )

= ( )ε σ νσ νσ E

......................

– –

1 3 3 1 2

( )

= ( )ε σ νσ νσ E

................................... 21( )

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VOLUMETRIC STRAIN

In the last lesson, the definition of volumetric strain or bulk strain was given as

Now, it can be shown that the volumetric strain equals the sum of the linear

strains in the x, y and z directions or

After substituting equations (16) to (18) into equation (22), we have

Rearranging the above equation, we get

Now we consider a special case: a circular cross-sectional bar subjected to a

uniaxial tensile stress σx only, shown in FIGURE 6.

ε σ σ σ νV x y zE = + +( ) ( ) ( )1 1 2 24– ..............

ε σ νσ νσ σ νσ νσV x y z y x zE E = ( ) + ( )1 1– – – –

+ 1 E z

σ ν– σσ νσx y–( )

ε ε ε εV x y z= + + ..................................... 23( )

εV V

V =

∆

................................................... 22( )

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FIG. 6 A circular cross-sectional bar subjected to a uniaxial stress

In this case, the only stress in the bar is a tensile stress along the axis of the

bar, then the stresses in the y and z directions are zero. Therefore, using

equation (24), we have

We also know that , then, the general equation for calculating the

volumetric strain of the circular cross-sectional bar is:

Example 4

A bar of circular cross-section is subject to a tensile load of 100 kN, which is

within the elastic range. The bar is made from a steel with E = 210 GPa and

Poisson's ratio = 0.3, it has a diameter of 25 mm and is 1.5 m long.

Determine the extension of the bar, the decrease in diameter and the increase in

volume of the bar.

ε σ

νV V

V E = = ( )∆

1 2– ................................. 25( )

εV V

V =

∆

ε σ

σ σ σ σ

x y z

E = ( )

= = =

1 2

–

where as

σ x

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Solution

From the question:

Since the loaded bar is in the elastic range, then we can find the axial strain

from:

ε σ

σ π

= =

= ×

× ×

d where

100 10

4 25 10

––

3 2

203 7 10

203 7 10 210 10

( )

= ×

∴ = ×

== ×9 7 10 4. –

= = ×

100

210

kN 100 10 N

GPa 210 10 Pa

0.3

mmm 25 10 m

3= ×

–

.l0 1 5

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To find the decrease in diameter, we should first find the transverse strain εt from Poisson's ratio by

Since decrease in diameter = transverse strain × original diameter then

The change in volume is given by equation (25)

ε σ

σ ν

V V

V E

V V

d l

= = ( )

∴ = ( )

∆

2 0

1 2

–

π ×× ×

× ×( )

= ×

203 7 10

210 10 1 2 0 3

2 86 10

286

. – .

. – m 3

mmm 3

∆d d= = × × ×

= × =

ε t

m 7.2

– .

– . –

– –

–

2 91 10 25 10

7 275 10

4 3

6 775 mµ

ε νεt = = × ×

= ×

– – . .

– .

–

0 3 9 7 10

2 91 10

Deformation in length ∆l l= = × ×

ε 0 49 7 10 1 5

. .

–

4455 10 3×

– m

1.455 mm

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

...................................................................................................................................................

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. An aluminium specimen shown in the figure below has a diameter of

d0 = 25 mm and a gauge length of l0 = 250 mm. If a force of 165 kN

elongates the gauge length by 1.20 mm, determine the modulus of

elasticity. Also, determine by how much the force causes the diameter of

the specimen to contract. Take Gal = 26 GPa and σy = 440 MPa.

2. Assume that the pressure vessel, shown in the figure below, contains a

pressurised liquid which subjects the vessel to volumetric (hydrostatic)

stress. Show, stating all assumptions, that the elastic contants E and K,

and Poisson's ratio are related by the expression:

Pressure

l y

K E

= ( )3 1 2– ν

l o

d o

165 kN165 kN

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3. An aluminium alloy has a modulus of elasticity of 80 GPa and a modulus

of rigidity of 33 GPa. Determine the value of Poisson's ratio and the bulk

modulus.

4. A rod 30 mm in diameter and 0.5 m in length is subjected to an axial

tensile load of 90 kN. If the rod extends in length by 1.1 mm and there is

a decrease in diameter of 0.02 mm, determine the value of Poisson's ratio

and the values of E, G and K.

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. From the question:

The average normal stress in the specimen is:

and the average normal strain or longitudinal strain is:

εl l

l = =

× ×

= ×

∆

1 2 10 250 10

4 8 10

–

σ = = ×

×( )

= × =

165 10

4 25 10

336 1 10

3 2

π –

. Pa 336.1 MPPa

= = ×

= =

165

250

1 2

kN 165 10 N

mm 250 10 m

3–

.∆ 11 2

440

. –×

= = ×

10 m

GPa 26 10 Pa

MPa 440 10

yσ PPa

mm 25 10 m3d0 25= = × –

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Since σ < σy, then the material behaves elastically. Thus, the modulus of elasticity is:

Using equation (5) we have:

Using equation (3), , we can determine the transverse strain εt.

(The negative sign means that there is contraction on the cross-section of

the specimen.)

Therefore, the deformation of the diameter is

A contraction of 41.5 µm.

∆d dt= = × × ×

= × =

ε – .

– . –

– –

–

1 66 10 25 10

4 15 10

3 3

5 m 41.5 µµm

ε νεt l= = × ×

= ×

– – . .

– .

–

0 346 4 8 10

1 66 10

ν ε ε

= – t l

G E

= +( )

∴ =

= ×

× × =

2 1

70 10 2 26 10

1 0 9

–

– ..346

E = = ×

= × =

σ ε

336 1 10 4 8 10

70 10

. . –

Pa 70 GPa

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2. In order to find this relationship we need to remember one very important

fact, which is

Now, from the question, we have that the stresses in the x, y and z

directions are equal for a hydrostatic loading, thus:

Assumption: all surfaces of the pressure vessel are subject to equal

pressure (ignore the weight of the liquid contained in the vessel).

Therefore, the material has the same mechanical properties.

From equations (16) and (17) we have

Then

After rearrangement

σ νσ σ νσ

σ σ ν σ σ

x y y x

– –

= ( )or

ε σ νσ νσ

σ νσ νσ

= ( )

x y z

y x z

– –

ε ε σ νσ νσ

= = ( )

x x y z

y y x z

– –

ε ε ε ε

ε ε

x y z

= = =

=and 3

volumetric strain the sum of the strains in= the , and directionsx y z

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Since ν ≠ 0, the only possibility for the above equation to be true is that σx = σy.

Similarly, solving equations (16) and (18), and (17) and (18), we have the

results as σx = σz and σy = σz.

Therefore, σz = σy = σz = σ.

Actually, σ is the stress caused by the pressure in vessel or σ = p.

From equation (24), the volumetric strain is

Since the bulk modulus

then

Substituting equation (B) into (A), we have

3 1 2

σ ν

E K

K E

–

( ) =

= ( )

K p

V V

= =

pressure volumetric strain ε

σ ε

ε σ

............................................. ..... B( )

ε σ σ σ ν

σ ν

V x y zE

= + +( ) ( )

= ( )

1 1 2

3 1 2

–

– ................................... A( )

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3. From the question:

Using equation (5), , the Poisson's ratio can be found from

Using equation (6), , the bulk modulus can be determined

K = ×

× ×( )

= ×

80 10 3 1 2 0 212

4 63 10

– .

. Pa

46.3 GPa

K E

= ( )3 1 2– ν

ν = = ×

× ×

G2 1

80 10 2 33 10

0 212

9 – –

G E

= +( )2 1 ν

= = ×

33 3

GPa 80 10 Pa

GPa 3 10 Pa

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From the question:

Thus

The Poisson's ratio is

ν ε ε

= = ×

× =– –

– . .

. –

– t

6 67 10 2 2 10

0 303 4

εl l

l = =

= ×

∆

1 1 10 0 5

2 2 10

. .

–

– ((elongational strain)

εt d

d = =

× ×

∆

30 02 10 30

– . –

110

6 67 10

–

–– .= × (contraction straiin)

= = ×

0 5

1 1

kN 90 10 N

mm 30 10 m

3–

.∆ mmm 1.1 10 m

mm 0 10 m

= ×

= = ×

–

–– . – .∆d 0 02 02

0.5 m

90 kN 30 mm

90 kN

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Assuming the bar is in the elastic limit under the loading of F, then

. We have

Applying equation (5), , then

Applying equation (6), , we have

K = ×

× ×( )

= × =

5 79 10 3 1 2 0 303

4 90 10

. – .

. Pa 49 GPa

K E

= ( )3 1 2– ν

G = ×

× +( )

= × =

5 79 10 2 1 0 303

2 22 10

. .

. Pa 22.2 GPa

G E

= +( )2 1 ν

A F

Al l l = = =

= ×

× ×( ) × ×

σ ε ε ε

3 2

90 10

4 30 10 2 2 1

π – . 00

5 79 10

–

.= ×

57.9 GPa

ε σ

σl E F

A = = and

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

In this lesson, the equations describing the relationships between stress and

strain in two and three-dimensional loading systems have been derived based

on the definitions of Poisson's ratio and modulus of elasticity, shear modulus

and bulk modulus. The examples given in this lesson show that these theories

play an important role in engineering design and operation.

In the first topic of this module, we have analysed the internal reactions of a

material within elastic limit under different loading systems. The relationships

between stress and strain have been discussed based on the definitions of

Poisson's ratio, modulus of elasticity, shear modulus and bulk modulus. A

very important test, the tensile test, has been introduced. The results, found

from pulling on a specimen of known size, are plotted as normal stress on the

vertical axis and normal strain on the horizontal axis. The above investigations

will lead us to further study of mechanical principles.

Teesside University Open Learning (Engineering)

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