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Understanding Area Centroid and Moments: Application in Engineering Calculations

This guide explores the concept of the centroid of an area, which represents the point where the total area is considered to be concentrated. It's crucial for analyzing bending stresses in beams, as the position of the centroid influences the distribution of compression and tension. The guide covers the first moment of area, second moment of area, section modulus and methods to calculate these properties for various shapes using symmetry, differential calculus, and established tables. Examples illustrate its application in structural engineering.

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Understanding Area Centroid and Moments: Application in Engineering Calculations

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  1. Properties of Area Centroid 1st Moment of area 2nd Moment of area Section Modulus

  2. CENTROID OF AREAS • Centroid of an area is the point at which the total area may be considered to be situated for calculation purposes. • Corresponds to the centre of gravity of a lamina of the same shape as the area • Often possible to deduce the centroid by SYMMETRY of the area. • Need to know position of the centroid of a section as bending occurs with compression above and tension below this axis. • Distance from centroid to axis of rotation (x or y) is 1st moment of area /total area

  3. 1st Moment of Area B Moment of Force = F x d F d A Likewise; First Moment of Area about the line CD = A x d D AreaA d C G = centroid of area

  4. CENTROID OF AREAS y Total Area A x G x y y Elemental area a x

  5. 1st moment of Area - Example 30 60 30 Find centroid of the composite beam section shown 50 7 15 35 20 Dia 65

  6. 1st moment of Area – Example (Ans)

  7. 2nd Moment of Area • A property of area used in many engineering calculations (e.g. stress in beams) • Elemental Elemental area a Second Moment of Area about the line CD = I D x C

  8. Standard Results for I • Using differential calculus we can formulate standard solutions, eg: • Rectangle about its base • Rectangle about its centre • For more complicated shapes can use compound areas and parallel axes theorem • Or, easier, use tables from steel joist manufacturers b b d d

  9. Example / Exercise • Loaded Timber beam has max BM of 5 kNm, find stress in the section. 5 kNm 100 compression 300 tension Section Stress block BMD Hence I = bd3 / 12 = 100 x 3003 mm4 12 = 102 x 33 x 1003 12 = 27 x 102 x 106 12 = 2.25 x 108 mm4 Hence f = 5 x 103 x 103Nmm x 150 mm 2.25 x 108 mm4 = 750 x 106 225 x 106 = 3.33 N/mm2

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