edexcel hnd unit- engineering science (nqf l4)

Edexcel HN Unit: Engineering Science (NQF L4)

Author: Leicester CollegeDate created:Date revised: 2009

AbstractThe aim of this unit is to investigate a number of major scientific principles that underpin thedesign and operation of engineering systems. It is a broad-based unit, covering both mechanicaland electrical principles. It is intended to give an overview that will provide the basis for furtherstudy in specialist areas of engineering.

Summary of learning outcomesTo achieve this unit a learner must:

1. Analyse static engineering systems2. Analyse dynamic engineering systems3. Apply DC and AC theory4. Investigate information and energy control systems.

ContentsSection 1. Stress..............................................................................................................................2Section 2. Strain...............................................................................................................................3Section 3. Modulus of Elasticity OR Young’s Modulus.....................................................................5Section 4. Combined Example.........................................................................................................6Section 5. Factor of Safety...............................................................................................................7Credits..............................................................................................................................................9

In addition to the resources found below there are supporting documents which should be used in combination with this resource.

These files support the Edexcel HN unit – Engineering Science (NQF L4)Unit outcome

Key words

1.1 Stress, strain, statics, young’s modulus

1.1 Shear force, bending moment, stress

1.2 Beams, columns, struts, slenderness ratio

1.3 Torsion, stiffness, twisting

2.1/2.2 Linear motion, angular motion, energy, kinetic, potential, rotation

For further information regarding unit outcomes go to http://www.edexcel.com/quals/hn/engineering/mechanical/Pages/default.aspx

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Edexcel HN Unit- Engineering Science (NQF L4)

Section 1. StressWhen 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 (sigma) So that = F/A

Extension due to load

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Section 2. Strain

If a material has a stress exerted on it the material will either lengthen or shorten. This is known as STRAINStrain 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 (Epsilon)

So = 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

= F/A

= 10000/ 125 = 80 N/mm2

But 1m2 = 1000 000 mm2

so 80N/mm2 = 80MN/m2 OR MPa(M = mega 106)

Example 2

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

= 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, = x/l = 0.01 / 30 = 3.3333 x 10-4

Note the units – There are none – it is a ratio !

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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, = 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

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Section 3. 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;

E = Stress / Strain

= / (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 0.007Steel 210Diamond 1200Wood 14Aluminium 72Typical Plastic 1.4

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Section 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, Dia = 5 mm and Extemsion = 0.005mm.

Theory;

= F/A : = x/l : E = /

= F/A = F / π r2 = 4500 / (π x 2.52)

= 229.18 N/mm2 (MPa)

= x/l = 0.05 / 25= 2 x 10 -3

And;

E = / = 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.

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Section 5. Factor 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;

Factors 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.

Factors of Safety = Yield strength 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 Factors 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

Factor 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;

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Factors 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.

Factors of Safety = USS Max working shear stress

= 300 / 100 = 3

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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.

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