Introduction to Stress

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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.

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Introduction to Stress

  1. 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. 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. 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. 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. 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. 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. 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

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