EDEXCEL NATIONAL CERTIFICATE/DIPLOMA

ADVANCED MECHANICAL PRINCIPLES AND APPLICATIONS

UNIT 18
NQF LEVEL 3

OUTCOME 1
BE ABLE TO DETERMINE THE EFFECTS OF UNIAXIAL AND COMPLEX
LOADING ON ENGINEERING COMPONENTS

Uniaxial loading: expressions for longitudinal and transverse strain; application of Poisson’s
ratio; determination of dimensional changes in plain struts and ties

Complex loading: expressions e.g. strain in x and y directions due to 2D loading, strain in x, y
and z directions due to 3D loading; changes e.g. dimensional in rectangular plates, dimensional
and volume for cubic elements

INTRODUCTION

An important part of your study of stress and strain involves calculating the dimensional changes
that occur not only in the direction of the applied stress but in other directions as well. It is a well
known fact that when an elastic material is stretched, it gets thinner and when it is squashed, it gets
thicker. This is the new theme that must be studied.

The first four sections of this tutorial revise basic stress and strain and may be skipped if you think
you are already proficient at calculating them. The new material starts at section 5.

You should judge your progress by completing the self assessment exercises.

© D.J.Dunn www.freestudy.co.uk 1
1. DIRECT STATIC FORCES

STATIC FORCES will deform a body in one or more of the following manners.

a. It may stretch the body, in which

case it is called a TENSILE FORCE.
A rod or rope used in a frame to take a
tensile load is called a TIE.

b. It may squeeze the body in which

case it is called a COMPRESSIVE
FORCE. A part of a structure in
compression may be a COLUMN and in
a frame is a STRUT.

2. DIRECT STRESS 

When a force is applied to an elastic body, the body deforms. The way in which the body deforms
depends upon the type of force applied to it.

A compression force makes the body shorter. A tensile force makes the body longer.

Tensile and compressive forces are called DIRECT FORCES.

Stress is the force per unit area upon which it acts.

Stress =  = Force/Area N/m2 or Pascals.

Note the cross section may be any shape. The symbol  is called SIGMA

NOTE ON UNITS The fundamental unit of stress is 1 N/m2 and this is called a Pascal. This
is a small quantity in most fields of engineering so we use the multiples kPa, MPa and GPa.

Areas may be calculated in mm2 and units of stress in N/mm2 are quite acceptable. Since 1
N/mm2 converts to 1 000 000 N/m2 then it follows that the N/mm2 is the same as a MPa

© D.J.Dunn www.freestudy.co.uk 2
3. DIRECT STRAIN 

In each case, a force F produces a deformation x. In engineering we usually change this force into
stress and the deformation into strain and we define these as follows.

Strain is the deformation per unit of the original length

Strain =  = x/L

The symbol  is called EPSILON

Strain has no units since it is a ratio of length to length. Most engineering materials do not stretch
very much before they become damaged so strain values are very small figures. It is quite normal to
change small numbers in to the exponent for of 10 -6. Engineers use the abbreviation  (micro
strain) to denote this multiple.

For example a strain of 0.000068 could be written as 68 x 10-6 but engineers would write 68 .

Note that when conducting a British Standard tensile test the symbols for original area are S o
and for Length is Lo.

WORKED EXAMPLE No. 1

A metal wire is 2.5 mm diameter and 2 m long. A force of 12 N is applied to it and it stretches
0.3 mm. Assume the material is elastic. Determine the following.
i. The stress in the wire .
ii. The strain in the wire .
SOLUTION
πd 2 π x 2.52 F 12
A   4.909 mm 2 σ   2.44 N/mm 2
4 4 A 4.909
Answer (i) is hence 2.44 MPa
x 0.3 mm
   0.00015 or 150 
L 2000

SELF ASSESSMENT EXERCISE No. 1

1. A steel bar is 10 mm diameter and 2 m long. It is stretched with a force of 20 kN and extends
by 0.2 mm. Calculate the stress and strain.
(Answers 254.6 MPa and 100 )

2. A rod is 0.5 m long and 5 mm diameter. It is stretched 0.06 mm by a force of 3 kN. Calculate
the stress and strain.
(Answers 152.8 MPa and 120)

© D.J.Dunn www.freestudy.co.uk 3
4. MODULUS OF ELASTICITY E

Elastic materials always spring back into shape when

released. They also obey HOOKE'S LAW. This is the law of
a spring which states that deformation is directly
proportional to the force. F/x = stiffness = k N/m

The stiffness is different for different materials and

different sizes of the material. We may eliminate the size by
using stress and strain instead of force and deformation as
follows. If F and x refer to direct stress and strain then
F σA FL 
F = A x = L hence  and 
x εL Ax 

The stiffness is now in terms of stress and strain only and this constant is called the MODULUS of
ELASTICITY and it has a symbol E.
FL 
E 
Ax 

A graph of stress against strain will be a straight line with a gradient of E. The units of E are the
same as the units of stress.

WORKED EXAMPLE No. 2

A Steel column is 3 m long and 0.4 m diameter. It carries a load of 50 MN. Given that the
modulus of elasticity is 200 GPa, calculate the compressive stress and strain and determine
how much the column is compressed.

SOLUTION

πd 2 π x 0.4 2 F 50 x10 6
A   0.126 m 2 σ   397.9 x10 6 Pa
4 4 A 0.126
6
σ σ 397.9 x10
E so ε    0.001989
ε E 200 x10 9
x
ε so x  ε L  0.001989 x 3000 mm  5.97 mm
L

© D.J.Dunn www.freestudy.co.uk 4
 WORKED EXAMPLE No. 3

A metal bar which is part of a frame is 50 mm diameter and 300 mm long. It has a tensile force
acting on it of 40 kN which tends to stretch it. The modulus of elasticity is 205 GPa. Calculate
the stress and strain in the bar and the amount it stretches.

SOLUTION

F = 40 x 103 N.
A = D2/4 =  2/4 = 1963 mm2
 = F/A = (40 x10 )/(1963 x 10-6) = 20.37 x 106 N/m2 = 20.37 MPa
3
E = ζ/ε = 205 x 109 N/m2
ε = ζ/E = (20.37 x 106)/(205 x 109) = 99.4 x 10-6 or 99.4 
ε = L/L
L=  x L = 99.4 x 10-6 x 300 mm = 0.0298 mm

SELF ASSESSMENT EXERCISE No. 2

1. A bar is 500 mm long and is stretched to 500.45 mm with a force of 15 kN. The bar is 10 mm
diameter.

Calculate the stress and strain.

Assuming that the material has remained within the elastic limit, determine the modulus of
elasticity.

If the fail stress is 250 MPa calculate the safety factor.

(Answers 191 MPa, 900με , 212.2 GPa and 1.31.)

2. The maximum safe stress in a steel bar is 300 MPa and the modulus of elasticity is 205 GPa.
The bar is 80 mm diameter and 240 mm long. If a factor of safety of 2 is to be used, determine
the following.
i. The maximum allowable stress. (150 MPa)
ii. The maximum tensile force allowable. (754 kN)
iii. The corresponding strain at this force. (731.7 με)
iv. The change in length. (0.176 mm)

3. A circular metal column is to support a load of 500 Tonne and it must not compress more than
0.1 mm. The modulus of elasticity is 210 GPa. The column is 2 m long. Calculate the
following.
i. The cross sectional area(0.467 m2)
ii. The diameter. (0.771 m)

Note 1 Tonne is 1000 kg.

© D.J.Dunn www.freestudy.co.uk 5
5. POISSON'S RATIO

Consider a bar or tie that is stretched with a force F

as shown. The bar will get longer in the direction it
is stretched but it will also get thinner as shown.

Let's reduce this to two dimensions x and y.

The stress in the x direction is x and there is no stress

in the y direction. When it is stretched in the x
direction, it causes the material to get thinner in all the
other directions at right angles to it. This means that a
negative strain is produced in the y direction. This is
called the lateral strain. For elastic materials it is found
that the lateral strain is always directly proportional to the applied such that
εy
 ν
εx
 (Nu) is an elastic constant called Poisson’s' ratio.

The strain produced in the y direction is y = - x

If stress is applied in the y direction then the resulting strain in the x direction would similarly be
x = - y

NOTE that a force in the x direction acts on a plane in the y direction and many text books take ζ x
and εx to mean the stress and strain on the x plane. Here we take it to mean the stress and strain in
the x direction.
NOTE that we do not have to confine ourselves to the x and y directions and that the formula
works for any two stresses at 90o to each other. In general we use 1 and 2 with corresponding
strains 1 and 2. In the following examples we use L and D to mean in the direction of the length
and any diameter.

WORKED EXAMPLE No. 4

A bar is 500 mm long and is stretched to 500.45 mm. The bar is 10 mm diameter.
Given that ν = 0.23, calculate the new diameter.

SOLUTION
Δ 0.45
εL  L   900 με
L 500
ε D   νσL  0.23 x 900  -207 με

ΔD = εD x D = -207 x 10-6 x 10 = -2.07 x 10-3 mm

The new diameter is 10 - 0.00207 = 9.99793 mm

© D.J.Dunn www.freestudy.co.uk 6
 WORKED EXAMPLE No. 5

A tie bar 2 m long and 10 mm diameter is stretched with a stress of 2 MPa. Given the elastic
constants are E = 205 GPa and  = 0.27, calculate the strains in both the longitudinal and
diametral directions. Calculate the change in length and change in diameter.

SOLUTION
σL 2 x 106
εL    9.756 με
E 205x10 9
ε D   νσD  0.27 x 9.756  -2.634 με
ΔL  ε LL  9.756 x 106 x 2000  0.0195 mm (length change)
ΔD  ε DD  2.634 x 106 x 10  26.34 x 106 mm (change in diameter)

WORKED EXAMPLE No. 6

A metal bar 0.5 m long and 0.2 m diameter is compressed by an axial load of 800 kN. Given the
elastic constants are E = 200 GPa and  = 0.25, calculate the stresses and strains in both the
longitudinal and diametral directions. Calculate the change in length and change in diameter.

SOLUTION

A = π D2/4 = 31.42 x 10-3 m2

ζL = -F/A = 800 x 103/31.42 x 10-3 = -25.462 MPa (Compressive)

σ - 25.462 x 106
εL  L   127.3 με
E 200 x 109
ε D   νσD  0.25 x (-127.3)  31.826 με

ΔL  ε LL  127.3 x 106 x 500  63.65 x 10-3 mm (length change)

ΔD  ε DD  31.826 x 106 x 200  6.365 x 103 mm (change in diameter)

SELF ASSESSMENT EXERCISE No. 3

1. A tie is 1 m long and 5 mm diameter is stretched by 0.3 mm. Given that ν = 0.25, calculate the
new diameter. (4.99962 mm)

2. A metal bar 0.2 m long and 0.1 m diameter is compressed by an axial load of 1 MN. Given the
elastic constants are E = 200 GPa and  = 0.25, calculate the stresses and strains in both the
longitudinal and diametral directions. Calculate the change in length and change in diameter.
(-127.307 x 10-3 mm and 15.913 x -3 mm)

© D.J.Dunn www.freestudy.co.uk 7
6. STRESS IN TWO MUTUALLY PERPENDICULAR DIRECTIONS.

Now consider that the material has an applied stress in

both the x and y directions.

The resulting strain in any one direction is the sum of

the direct strain and the lateral strain.
Hence
σ σ σy
ε x  x  νσy  x  ν
E E E
1
 
ε x  σ x  νσy ............(A)
E

Similarly ε y 
1
E
 
σ y  νσx ............(B)

The modulus E must be the same in both directions and such a material is not only elastic but
ISOTROPIC.

WORKED EXAMPLE No. 7

A material has stresses of -2 MPa in the x direction and 3 MPa in the y direction. Given the
elastic constants E = 205 GPa and = 0.27, calculate the strains in both direction.

SOLUTION
1

ε x  σ x  νσy 
E
1
205x10 9
 
- 2 x 106  0.27 x 3 x 106  13.7 

1

ε y  σ y  νσx 
E
1
205x10 9
 
3 x 106  0.27 x (-2 x 106 )  17.3

WORKED EXAMPLE No. 8

A material has stresses of 2 MPa in the x direction and 3 MPa in the y direction. Given the
elastic constants E = 205 GPa and  = 0.27, calculate the strains in both direction.

SOLUTION

εx 
1
E

σ x  νσy   1
205x10 9
 
2 x 106  0.27 x 3 x 106  5.8 με

1

ε y  σ y  νσx 
E
 1
205x10 9
 
3 x 106  0.27 x 2 x 106  12 με

© D.J.Dunn www.freestudy.co.uk 8
 WORKED EXAMPLE No. 9

A thin flat plate is 200 mm x 100 mm forms part of a structure and when in service a stress of
100 MPa is produced in the long direction and 150 MPa in the short direction. Given the elastic
constants E = 205 GPa and  = 0.25, calculate the strains in both direction. Calculate the change
in dimensions ands area of the plate.

SOLUTION

εL 
1
E 205x10
 
σ L  νσS   1 9 100 x 106  0.25 x 150 x 106  304.9 με

εS  σS  νσL  
1
E
1
205x10 9
 
150 x 106  0.25 x 100 x 106  609.8 με

Change in length = εL x 200 = 60.976 x 10-3 mm

Change in side = εS x 100 = 60.976 x 10-3 mm
Change in area = 60.976 x 10-3 x 60.976 x 10-3 = 3.718 x 10-3 mm2

SELF ASSESSMENT EXERCISE No. 4

1. Solve the strains in both directions for the case below.

E = 180 GPa  =0.3 1= - 3 MPa 2= 5 MPa

(Answers 32.78  and -25 )

2. A large cylindrical pressure vessel is constructed from thin plate. When pressurised a tensile
stress of 60 MPa is produced in the longitudinal direction and 120 MPa in the direction around
the circumference. Calculate the strain in these directions.
Part of the shell is a plate 100 mm x 100 mm and this may be treated as a flat plate. Calculate
the change in dimensions and area.

Take E = 205 GPa and ν = 0.25

(146.3 με and 512.2 με )

(14.364 x 10-3 mm and 51.22 x10-3 mm)
(6.585 mm2)

© D.J.Dunn www.freestudy.co.uk 9
7. THREE DIMENSIONAL STRESS AND STRAIN

Equations A and B were derived for a 2 dimensional system. Suppose a material to be stressed in 3
mutually perpendicular directions x, y and z. The strain in any one of these directions is the direct
strain plus the lateral strain from the other two directions. It follows that:
1
  1
  
ε x  σ x  νσy  νσz  σ x  ν σ y  σ z ..............(C)
E E
   
ε y  σ y  νσx  νσz  σ y  νσ x  σ z  ..............(D)
E
1
E
1

1
  1
  
ε z  σ z  νσx  νσy  σ z  ν σ x  σ y ..............(E)
E E
If the stress is the same in all directions, then the strains are the same.

WORKED EXAMPLE No. 10

A solid cube of metal has sides of 200 mm. It is compressed by a pressure of 80MPa on all its
faces. Determine the change in length of each side and the reduction of volume.
E is 71 GPa and Poisson's ratio  is 0.34.

SOLUTION

εx 
1
E
 
σ x  ν σ y  σz   1
71 x 109
  
80 x 106  0.34 80 x 106  80 x 106  360.6 με

The strain will be the same in the y and z directions

Change in length = δ = εx x L = 360.6 x 10-6 x 200 = 72.113 x 10-3 mm
The change in volume in each direction is δ x L2 = 72.113 x 10-3 x 2002 = 2 885 mm3
The total change in volume is 3 x 2 885 = 8654 mm3
Note this is not quite true because we should also subtract the corner common to all three which
is (72.113 x 10-3)3 = 0.000 375 mm3and is negligible.

SELF ASSESSMENT EXERCISE No. 5

1. A cube is stressed in 3 mutually perpendicular direction x, y and z. The stresses in these

directions are

x = 50 kPa y = 80 kPa z = -100 kPa

Determine the strain in each direction.

 is 0.34 and E is 71 GPa.
(Answer 800 x 10-9, 1.366 x 10-6 and -2.031 x 10-6)

2. A cube of metal of side 30 mm is placed inside a pressure vessel and the pressure is raised to
20 MPa. Given that E = 205 GPa and ν = 0.25 determine the change in volume of the cube.
(Answer 3.951 mm3)

3. Repeat question 2 fore a solid sphere 30 mm diameter. (-2.069 mm3)

© D.J.Dunn www.freestudy.co.uk 10