Consider a cross section with a general shape such as shown in Figure B.2.1 with the x axis normal to the cross section. Figure B.2.1.

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1 ppendix B rea Properties of Cross Sections B.1 Introduction The area, the centroid of area, and the area moments of inertia of the cross sections are needed in slender bar calculations for stress and deflection. To simplif the problem we place the x axis so that it coincides with the loci of centroids of all cross sections of the bar. In our examples the cross sections lie in the plane. Furthermore, for beam bending analsis in these chapters we orient the and axes so that the are principal axes of inertia of the cross section area. This simplifies the equations for stress and displacement. Just what this means is explained in the following sections. B. Centroids of Cross Sections Consider a cross section with a general shape such as shown in Figure B..1 with the x axis normal to the cross section. Figure B..1 The x axis is a centroidal axis if d = 0 d = 0 (B..1) nalsis of Structures: n Introduction Including Numerical Methods, First Edition. Joe G. Eisle and nthon M. Waas. 011 John Wile & Sons, Ltd. Published 011 b John Wile & Sons, Ltd.

2 60 ppendix B: rea Properties of Cross Sections If the cross sectional is smmetrical the centroid is easil found since it will alwas lie on the axis of smmetr. For sections with double smmetr, that is, smmetr about both the and axes, such as those sections in Figure B.., the location is obvious. (a) (b) (c) Figure B.. For sections with smmetr about just one axis we know the centroid lies on that axis but we must locate just where on that axis. For the sections in Figure B..3 we use the formulae d d ȳ = d as appropriate. = (B..) d (a) (b) (c) Figure B..3 When an area can be divided into sub areas with simple geometr so that the centroid of the sub area is easil identified the process of finding the centroid of the original area is simplified to ȳ = s s s = s s s (B..3)

3 ppendix B: rea Properties of Cross Sections 603 where s and s represent the distances from base axes to the centroids of the sub areas and s represents the areas of the sub areas. n example will help. Example B..1 Consider the cross section in to which some dimensions have be added. 0 We place the axis on the axis of smmetr where we know the centroid lies and the axis convenientl at the right edge. The cross section is divided into three rectangular areas for which their centroids are known. To find = = = (a) For slender bar analsis the axis is moved to the new location. Centroids of some common shapes are given in the last section of this appendix. B.3 rea Moments and Product of Inertia The area moments of inertia are I = d I = d (B.3.1) and the area product of inertia is I = d (B.3.)

4 604 ppendix B: rea Properties of Cross Sections We are interested primaril in values referred to centroidal axes. In man cases I = 0. This occurs when either the x or the x axes plane is a plane of smmetr or the axes are oriented so that I = 0. Then the axes are called principal axes of inertia. For sections with double smmetr the integration is often straight forward. n example will help. Example B.3.1 Consider first the rectangular cross section with double smmetr as shown in. h b Given a height h and a width b we have h I = bd = 1 b h 1 bh3 I = hd = 1 b h b 1 b3 h I = dd = 0 b h (a) Moments of inertia of tpical double and single smmetr sections are given in the last section of this appendix. For sections made up of subsections with known moments of inertia about the centroids of the sub sections there is a transfer process. It is known as the parallel axis theorem. Let the c c axes be centroid axes for an area whose moments and product of inertia are known. We wish to find the moments of inertia of this area with respect to a set of axes. Let ȳ c and c be the distances from the axes to the c c axes as shown in Figure B.3.1. I = ( c + ȳ c ) d = c d + ȳ c c d+ c d (B.3.3) Since c d = 0 d = (B.3.4)

5 ppendix B: rea Properties of Cross Sections 605 c c c Figure B.3.1 this becomes I = I c c + ȳ c (B.3.5) Likewise I = I c c + c (B.3.6) and I = I c c + ȳ c c (B.3.7) Example B.3. Find the area moments of inertia with respect to centroidal axes for the cross section in Example B..1. Units are millimeters

6 606 ppendix B: rea Properties of Cross Sections Transferring the three sections from top to bottom: I = 1 1 (0)3 + 0 (40) (60) (0)3 + 0 (40) = 6,893,333 mm 4 (a) I = ()3 + 0 (9.3) (0) (30.77) ()3 + 0 (9.3) (b) = 4,850,56 mm 4 From smmetr the product of intertia I = 0. For a section with no smmetr the process requires also finding the product of inertia. We show an example next. Example B.3.3 Consider the following section. Find the centroid and the area moments and product of inertial with respect to the centroidal axes. 0 c c c c 50 First find the centroid: c = = = 3.6 mm ȳ c = = = 59.5 mm (a)

7 ppendix B: rea Properties of Cross Sections 607 Now the moments of inertia: I = 1 1 (0)3 + 0 (30.48) (60) (9.5) (0) (49.5) = 4,879,048 mm 4 I = ()3 + 0 (17.38) (b) (0) (.6) (50) (7.6) = 3,191,190 mm 4 I = = 1,695,38 mm 4 Since in the main text we do all analsis with respect to principle axes, that is, axes for which I = 0, we must reorient the axes to appl those methods. Consider the rotated axes in Figure B.3.. θ Figure B.3. point with the coordinates and with respect to the axes have the coordinates and with respect to the axes. The transformation equations are = cos θ + sin θ = cos θ sin θ (B.3.8) From this we obtain I = ( ) d= ( cos θ + sin θ) d = I cos θ + I sin θ + I sin θ cos θ (B.3.9) = I + I I I cos θ + I sin θ

8 608 ppendix B: rea Properties of Cross Sections I I = ( ) d = ( cos θ sin θ) d = I cos θ + I sin θ I sin θ cos θ (B.3.10) = I + I + I I cos θ I sin θ = d = ( cos θ + sin θ)( cos θ sin θ) d = I sin θ cos θ I sin θ cos θ + I sin θ cos θ (B.3.11) = I I sin θ + I cos θ Thereisavalueofθ for which I x = 0. I ma be found b setting I I sin θ + I cos θ = 0 tan θ = I I I (B.3.1) It ma be noted that when I x = 0thenI is either a maximum or a minimum and I is a corresponding minimum or a maximum. These values are I max min = I + I ± ( I I ) + I (B.3.13) Example B.3.4 Find the rotation angle of the axes to obtain principal axes of inertia and the resulting values for the cross section in Example B.3.. c c c c The angle of rotation is tan θ = I 1,695,38 = I I 4,879,048 3,191,190 = θ =.56 (a)

9 ppendix B: rea Properties of Cross Sections 609 The principal moments of inertia are I max = 5,98,805 mm 4 I min =,141,433 mm 4 (b) B.4 Properties of Common Cross Sections c h = bh I = bh 3 /1 I = hb 3 /1 I = 0 b c = πr c R I = I = πr 4 /4 J = πr 4 / I = 0 c = bh/ h I = bh 3 /36 c h/3 I = hb 3 /36 I = b h /7 b b/3