PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by

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1 PROPERTES OF RES Centroid The concept of the centroid is prol lred fmilir to ou For plne shpe with n ovious geometric centre, (rectngle, circle) the centroid is t the centre f n re hs n is of smmetr, the centroid will e somewhere on tht is For shpes tht re mde up of numer of elements with known centroids, we cn find the _ R R position of the centroid of the whole such shpes process kin to tking moments For emple, consider shpe consisting of two circulr res touching There is n is of smmetr the is so the centroid will e somewhere on this is To find the position of the centroid on this, tke moments of res out the is The re of ech circle is ssumed to ct t its centroid (its centre) The moment of the whole shpe lnces the moments from the individul circulr res: π R R + πr (R + R ) ( πr + πr ) From this, we get R + RR + R R + R n generl, nd for n irregulr shpe, the definition of the centroid t position (, ) is given d () d where is the re of the shpe The integrls men integrte over the whole re of the shpe To see wht this mens, look t simple emple with n ovious nswer squre of side First find Wht we re in effect doing is tking moments of re out the is We d therefore divide the re into strips tht re t constnt distnce from this is, ie, prllel to it These strips cn e thought of s hving re d From the digrm, it is cler tht d d We now replce d in the first of equtions () with d d This gives us n integrl which we cn hndle:

2 d where the integrl now hs upper nd lower limits defined the rnge of Since, this ecomes d which is the ovious result We cn do similr eercise for the co-ordinte We re now tking moments out the is, so our choice of strip for the re d is now prllel to the is Now, d d in the second of equtions () The integrl ecomes d d d gin giving the epected result Centroid of tringle n generl the lengths of the strips of mteril d re not constnt Tke the cse of right-ngled tringle of height nd se Suppose we wnt to find Then we need verticl strips s shown, with height vring with To determine the height, we need the eqution for the tringle s olique side This is given d d The height h is just given the vlue, h We proceed s efore to determine the re d Thus, d hd d Putting this in the first of the equtions () gives d d Putting in the limits gives

3 This is well-known result Emple Show tht Second moment of re This is quntit tht will e required lter for use in finding em deflections nd stresses n stndrd ending es, the plne of interest is lelled the - plne Second moment of re is defined out prticulr is There re stndrd formule for common shpes (rectngle, circle) which re out the smmetr es Second moment of re out n is lelled the -is is defined s: d () where is perpendiculr the distnce from the is Tke the emple of rectngle, with the is n is of smmetr (which must lws pss through the centroid) We re tking moments out the is, so, s with the centroid, we choose strips prllel to this is Hence, d d nd eqution () ecomes d d d d / d / d shpe ntegrting, d/ d/ d d 8 8 d where the limits on the integrl re defined the rnge of vlues over the () This is the stndrd result for the rectngulr section Note tht, for the out verticl is through the centroid, we cn simpl swp for d nd get d

4 These re the stndrd results for this section nother useful stndrd result is for solid circulr section out dimeter, which is given 4 πr 4 for circle rdius R Note tht the units for second moment of re re length 4, ie mm 4 or m 4 Prllel is theorem Sometimes we hve the stndrd result for n re, which is for the second moment out n is of smmetr, nd we need to find the second moment out different is prllel to the originl is The prllel is theorem provides convenient w of doing this For n originl is, nd prllel is perpendiculr distnce from, the originl second moment is relted to the second moment out the epression: + (4) llustrtion Find the second moment of re of the rectngulr section out the is shown d d Solution Use eqution (), with the distnce from the is The integrl ecomes + d/ d d/ On integrtion, we hve + d/ ) d/ which on multipling out ecomes [( + d / ) ( d / ] [ + d / + d / 4 + d / 8 ( d / + d / 4 d /8)] d [ d /8 + d] + d Note tht we could get the sme result using the stndrd result of eqution () together with eqution (4), given tht d 4

5 dding second moments Suppose we hve shpe consisting of two seprte res We cn get the second moment for the whole shpe dding together the second moments for the two seprte res, provided these two second moments re out the sme is Emple Find the second moment of re for the shpe illustrted out the is R Outline solution Use stndrd formul to get for squre out its centre Use Prllel is Theorem to get for squre out the is squre Use stndrd formul to get for circle out its dimeter Use Prllel is Theorem to get for circle out the is circle The required result is squre + circle Hollow sections The dditive nture of second moments mkes possile net w of evluting second moments of hollow sections Suppose we hve to find the second moment of the hollow circulr section illustrted, out dimeter Let the second moment of this section e hollow, nd the second moments of solid circles of rdius r nd R respectivel r nd R Since we re tking moments out the sme is, we m write r R hollow + r R Therefore, hollow R - r Using the stndrd result, it follows tht for this 5

6 section 4 4 πr πr π 4 4 hollow (R r ) The two prolems given in the summr file for this chpter mke use of this technique 6

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