Selecting Columns And Beams

Selecting Columns And Beams

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  Selecting Columns andBeams HNC In Engineering – Mechanical Science Edexcel HN Unit: Engineering Science (NQF L4) Author: Leicester College Date created: Date revised: 2009 Abstract: Selecting the correct column in order to support a given load without over engineering the situation takes in many factors. This section takes a step by step approach to one method of selection. The key terminology and calculations are explored by use of a simple example. Extracts of column tables are used to extract key variables. The tables used are extracted from standard construction engineering data sheets. © Leicester College 2009. This work is licensed under a Creative Commons Attribution 2.0 License .
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  Contents Overview BeamsZ – Modulus of Section Columns Effective Length Selecting Columns and Beams Selecting Columns Selecting Columns – Example Credits These files support the Edexcel HN unit – Design for Manufacture (NQF L4) For further information regarding unit outcomes go to Edexcel.org.uk/ HN/ Engineering / Specifications File Name Unit Outcome Key Words Stress introduction 1.1 Stress, strain, statics, young’s modulus BM, shear force diagrams 1.1 Shear force, bending moment, stress Selecting beams 1.2 Beams, columns, struts, slenderness ratio Torsion introduction 1.3 Torsion, stiffness, twisting Dynamics introduction 2.1/2.2 Linear motion, angular motion, energy, kinetic, potential, rotation
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  When selecting thecorrect size beam for a particular situation the following must be taken into account; If the beam selected is to thin it will buckle under the applied loading If the beam is to thick then there is a cost implication – we are spending more than we need to. It therefore follows that selecting the right section size for a particular application is important. Beams
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  Z – Modulusof Section Max Bending Moment = And Z (section Modulus) = I / y max Different sections of beam have differing formulae for I Such that for a square section I = bd 3 12 And for a simply supported beam with a UDL over its full length M = wL 2 / 8 I  max y max
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  Columns Columns areloaded by DIRECT COMPRESSIVE STRESS The two basic modes of failure are kneeing and buckling The critical buckling stress depends on the material concerned and how slender the column is. A measure of slenderness used is known as the SLENDERNESS RATIO SR = Effective length / Radius of gyration The effective L depends on how the beam is supported
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  Effective Length Effectivelength – length between points at which the member bows out For a pin ended member L e = L Fixed ended member L e = 0.7 L When one end is fixed L e = 0.85 L (The other being pinned) When one end is fixed and the other is free L e = 2 L (such as a cantilever)
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  Selecting Columns andBeams Beams When selecting the correct size beam for a particular situation the following must be taken into account; If the beam selected is to thin it will buckle under the applied loading If the beam is to thick then there is a cost implication – we are spending more than we need to. It therefore follows that selecting the right section size for a particular application is important.
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  Selecting Columns Inorder to select the correct column for a particular application Determine the effective length of the column required Select a trial section Using the radius of gyration value for this trial section calculate the slenderness ratio. If the slenderness ratio is greater than 180, try a larger cross section trial section. Using the slenderness ratio obtain the compressive strength from tables. (from the Y-Y axis)
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  Selecting Columns -Example A column, pin ended, length 5m has an axial load of 1500kN. The steel has a yield stress of about 265MPa Effective L = 1 x 5 = 5m Choose a trial section – 203 x 203 / 86 From table 1 - Rad. Of Gyration = 5.32 SR = 5 / 0.0532(Needs to be in metre) = 94
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  Extract from Table2 Axis of Buckling – Y – Y Yield (MPa) SR = 94 Comp Strength = 200MPa (Approx) SRatio 265 275 340 25 258 267 328 50 221 228 275 75 187 192 221 100 138 141 153 150 74 74 77
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  From Table 2– Compressive strength is approx 150MPa The actual stress (  ) = Load / CSA = 1500 x 10 3 / 110.1 x 10 -4 = 132 MPa This value is 18MPa below the required stress value for this section We should now repeat the process until we have a stress value JUST below the 150 MPA value TRY IT YOURSELF
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  This resource wascreated 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.