Selecting Columns and BeamsHNC In Engineering Mechanical Science Edexcel HN Unit: Engineering Science (NQF L4)Author: Leicester CollegeDate 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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ContentsOverviewBeamsZ Modulus of SectionColumnsEffective LengthSelecting Columns and BeamsSelecting ColumnsSelecting Columns ExampleCredits

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 NameUnit OutcomeKey WordsStress introduction1.1Stress, strain, statics, youngs modulusBM, shear force diagrams1.1Shear force, bending moment, stressSelecting beams1.2Beams, columns, struts, slenderness ratioTorsion introduction1.3Torsion, stiffness, twistingDynamics introduction2.1/2.2Linear motion, angular motion, energy, kinetic, potential, rotation

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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 loadingIf 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 Modulus of SectionMax Bending Moment =

And Z (section Modulus) = I / y max

Different sections of beam have differing formulae for ISuch that for a square section I = bd312And for a simply supported beam with a UDL over its full length M = wL2 / 8 I smax y max

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ColumnsColumns are loaded by DIRECT COMPRESSIVE STRESS

The two basic modes of failure are kneeing and bucklingThe critical buckling stress depends on the material concerned and how slender the column is. A measure of slenderness used is known as the SLENDERNESS RATIOSR = Effective length / Radius of gyrationThe effective L depends on how the beam is supported

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Effective LengthEffective length length between points at which the member bows out

For a pin ended member Le = LFixed ended member Le = 0.7 LWhen one end is fixed Le = 0.85 L

(The other being pinned)When one end is fixed and the other is free Le = 2 L

(such as a cantilever)

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Selecting Columns and BeamsBeamsWhen 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 loadingIf 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 ColumnsIn order to select the correct column for a particular application Determine the effective length of the column requiredSelect a trial sectionUsing 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 - ExampleA 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 = 5.32SR = 5 / 0.0532(Needs to be in metre) = 94

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Serial size (mm)

Mass per unit length (kg/m)

Area of section (cm2)

Min Rad of Gyration (cm)

305 x 305

283

360.4

8.25

198

252.3

8.02

137

174.6

7.82

97

123.3

7.68

254 x 254

167

212.4

6.79

107

136.6

6.57

89

114.0

6.52

73

92.9

6.46

203 x 203

86

110.1

5.32

71

91.1

5.28

60

75.8

5.19

52

66.4

5.16

46

58.8

5.11

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Extract from Table 2Axis of Buckling Y Y

Yield (MPa)

SR = 94CompStrength = 200MPa(Approx)

SRatio265275340252582673285022122827575187192221100138141153150747477

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From Table 2 Compressive strength is approx 150MPaThe actual stress (s)= Load / CSA= 1500 x 103 / 110.1 x 10-4= 132 MPaThis 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 valueTRY IT YOURSELF

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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.

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