This section provides a very useful introduction to the concepts of direct and shear stress and strain as encountered in mechanical engineering structures. The standard notation used is explained in some detail. Young’s modulus is introduced and explained with worked examples of standard types of simple calculations. Factors of safety are examined and their importance is explored with once again some sample calculations included. Some simple examples of basic shear stress are also given.

1. Introduction to Stress and Strain
HNC in Engineering- Mechanical Science
Edexcel Unit: Engineering Science (NQF L4)
Author: Leicester College
Date created:
Date revised: 2009
Abstract
This section provides a very useful introduction to the concepts of direct and shear stress and strain as
encountered in mechanical engineering structures. The standard notation used is explained in some detail. Young’s
modulus is introduced and explained with worked examples of standard types of simple calculations. Factors of
safety are examined and their importance is explored with once again some sample calculations included. Some
simple examples of basic shear stress are also given.
Contents
Introduction to Stress and Strain....................................................................................................................1
Direct loading – stress and strain...................................................................................................................2
Factors of Safety............................................................................................................................................5
Factors of Safety in SHEAR mode................................................................................................................6
Credits............................................................................................................................................................7
These files support the Edexcel HN unit – Engineering Science (mechanical)
Unit Key words
outcome
Stress 1.1 Stress, strain, statics, young’s modulus
introduction
BM, shear 1.1 Shear force, bending moment, stress
force
diagrams
Selecting 1.2 Beams, columns, struts, slenderness ratio
beams
Torsion 1.3 Torsion, stiffness, twisting
introduction
Dynamics 2.1/2.2 Linear motion, angular motion, energy, kinetic, potential, rotation
introduction
For further information regarding unit outcomes go to Edexcel.org.uk/ HN/ Engineering / Specifications
© Leicester College 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

2. Introduction to stress and strain
Direct loading – stress and strain
Stress
When a material has a force exerted on it the material is said to be ‘stressed’.
If a rod is stretched the force is TENSILE
If a rod is squashed the force is ‘COMPRESSIVE’
Force (N)
Stress = Force / Area
=F/A
Units N/m2 or Pascal (Pa)
We give stress the letter σ (sigma)
So that σ = F/A
Extension due to load
Strain
If a material has a stress exerted on it the material will either lengthen or shorten. This is known
as STRAIN
Strain is the ratio of the extension due to the stress divided by its original length;
Strain = extension / original length
= x / l (we give strain the letter ε (Epsilon)
So ε = x/l
Example 1
A tie bar has a cross sectional area of 125mm2 and is subjected to a tensile load of 10kN.
Determine the stress.
F = 10kN and area = 125mm2
σ = F/A
= 10000/ 125 = 80 N/mm2
But 1m2 = 1000 000 mm2
so 80N/mm2 = 80MN/m2 OR MPa
(M = mega 106)
Example 2
Page 2 of 7

3. Introduction to stress and strain
A piece of steel 12 mm in diameter is compressed by a load of 15 kN. Determine the induced
stress in MPa.
F = 15kN and area = π r2 = π x 6 mm2
σ = F/A = 15000 / 113.1
= 132.63 N/mm2 or MPA
Strain example 1
A tie bar of original length 30mm is extended by 0.01mm when a tensile load is applied.
Determine the strain.
Original length = 30mm, Extension = 0.01mm
Strain, ε = x/l
= 0.01 / 30 = 3.3333 x 10-4
Note the units – There are none – it is a ratio !
Strain example 2
A steel column of 1.5 m length is compressed by 0.04mm when a compressive load is applied.
Calculate the strain in the column.
Original length = 1.5 m , Extension = 0.04mm
Strain, ε = x/l
= 0.04 / 1500 = 2.667 x 10-5
On strain calculations note that we need to be consistent with the units m/m OR
mm/mm
Modulus of Elasticity OR Young’s Modulus
Engineering materials possess a property known as ELASICITY. If a piece of material is
strained and the forces producing the strain are removed the material will regain its original
dimensions. IT IS ELASTIC.
This situation only happens up to a certain point in Engineering materials. This point is known
as;
The ELASTIC LIMIT
or LIMIT of PROPORTIONALITY
We can now link together stress and strain by using the Modulus of Elasticity or E;
Page 3 of 7

4. Introduction to stress and strain
E = Stress / Strain
= σ /ε (In the elastic range only)
The units of E are the same as stress (Pa).
But it is usually a very large number, typically GPa (Giga – 109)
Table 1 – Typical values of E for Engineering Materials
Material Approx value of E (GPa)
Rubber 0.007
Steel 210
Diamond 1200
Wood 14
Aluminium 72
Typical Plastic 1.4
Combined Example
A steel test specimen, 5 mm in diameter and 25mm long has a tensile load of 4.5kN exerted on
it. It is observed to stretch by 0.05mm under this load and revert back to its original length when
unstrained.. determine the value of E for this material.
In this example;
F = 4.5kN, Original l = 25mm, Diameter = 5 mm and Extension = 0.005mm.
Theory;
σ = F/A : ε = x/l : E = σ /ε
σ = F/A = F / π r2 = 4500 / (π x 2.52)
= 229.18 N/mm2 (MPa)
ε = x/l = 0.05 / 25
= 2 x 10 -3
And;
E = σ /ε = 229.18 x 106 / 2 x 10 -3
= 114.6 x 109 N/m2 (GPa)
Further Example
A steel specimen 6mm in diameter and originally 30mm long has a 8kN load applied to it. Under
this load it is observed to extend by 0.025mm. determine E for this material.
Page 4 of 7

5. Introduction to stress and strain
Factors of Safety
When designing a component or structure that will be under some form of stress a Factor of
Safety has to be considered to make certain that the working stresses keep within safe limits.
For a brittle material the factor of safety (FOS) is defined in terms of the Ultimate tensile
strength;
Factor of Safety = Ultimate tensile strength
Maximum working stress
For ductile materials the factor of safety is more usually defined in terms of the YIELD stress.
Yield stress is the value of the stress when the material goes from elastic to plastic.
Factor of Safety = Yield stress
Maximum working stress
Determining the factor of safety
The size of the factor of safety chosen depends on a range of conditions relating to the function
f the component or structure when in service some of them are listed below;
• Possible overloads
• Defects of workmanship
• Possible defects in material
• Deterioration due to wear, corrosion etc.
• The amount of damage that might occur if there is failure
• The possibility of the loads being applied suddenly or repeatedly
•
Worked examples
1. A cable used on a crane is made from a material with an ultimate tensile stress of 600
MPa. Determine the maximum safe working stress if a factor of safety of 4 is used.
Maximum working stress = Ultimate tensile strength
Factor of Safety
= 600 / 4 = 150 MPa
2. What is the factor of safety of a column made of a material with a UTS of 500 MPa if the
maximum working stress should be 200 MPa
Factors of Safety = UTS
Max working stress
= 500/200 = 2.5
Page 5 of 7

6. Introduction to stress and strain
Factors of Safety in SHEAR mode
Factors of safety can be used in shear mode but instead of using the UTS the material USS
(Ultimate Shear Stress) is used;
Factor of Safety = USS
Max working shear stress
Example
Determine the factor of safety used for a shear pin if the USS for the material is 300 MPa and
the maximum working stress should not exceed 100 MPa.
Factor of Safety = USS
Max working shear stress
= 300 / 100 = 3
Page 6 of 7

7. Introduction to stress and strain
Credits
This resource was created Leicester College and released as an open educational resource
through the Open Engineering Resources project of the Higher Education Academy
Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE
and part of the JISC/HE Academy UKOER programme.
© 2009 Leicester College
This work is licensed under a Creative Commons Attribution 2.0 License.
The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England &
Wales Licence. All reproductions must comply with the terms of that licence.
The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided
that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher.
The Leicester College name and logo is owned by the College and should not be produced without the express permission of the College.
Page 7 of 7