Chapter 2 GEOMETRIC ASPECT OF THE STATE OF SOLICITATION

Transcription

1 Capter GEOMETRC SPECT OF THE STTE OF SOLCTTON. THE DEFORMTON ROUND PONT.. Te relative displacement Due to te influence of external forces, temperature variation, magnetic and electric fields, te construction bodies are deformed and distorted (modifing teir dimensions and sape). Te geometric aspect studies tis deformation as a geometrical penomenon, produced b te relative displacement of te points from te studied bod. To observe te modifications of geometrical nature produced b external causes, we consider a bod into an ortogonal rectangular sstem of axis, wit te origin in O (Fig..). Due to te external forces tat subject te bod, point O reaces te position O. Te vector OO is called relative displacement, and: OO= ' d Fig.. Projecting d on te axis Ox, O and O, we obtain te components of te relative displacements: on Ox axis it is noted wit u, on O axis it is noted wit v and on O axis it is noted wit w.

2 From figure., we ma write: u v w cos α=, cos β=, cos γ= (.) d d d Squaring and ten adding tese relations, we get: u + v + w cos α + cos β + cos γ = = (.) d Te relative displacement is: d + = u + v w (.) Function te ratio between te relative displacements and te initial dimensions of te bod (for bars, te cross section dimensions), te problems from Structural Mecanics are studied wit: a. Te st order Teor: if te relative displacements are ver small wit respect to te initial dimensions of te construction bod. n tis categor are placed most of te problems from Mecanics of Materials. n tis situation, te mode of supporting and loading of te bod is not influenced b te structure deformation, being defined for te initial undeformed position of te structure. b. Te nd order Teor: if te relative displacements are comparable wit te initial dimensions of te construction bod. Te displacement of a point wit respect to its initial position, called relative displacement, defined above, represents te exterior aspect of te deformation of a bod. Te interior aspect is given b te modification of te volume and te sape of te bod, wic is a complex deformation, but it can be illustrated b two simple deformations (strains): - te elongation (linear strain) - te sliding (angular strain).. Te elongation Te elongation is a linear deformation wic will be analed on an axis Ox (Fig...a). segment from a deformable bod is in te initial position B. fter deformation, tis segment reaces te position B, due to te potesis of te continuit of displacements inside a bod. Te initial segment B as te lengt l (Fig...a), wic becomes l+ l for te segment B. Tis increased lengt of te segment B makes evident te lengtening (elongation) l of te initial segment B. Terefore te elongation represents te quantit wit wic te initial lengt of a segment is modified. B convention l is positive if it is a lengtening, respectivel negative if it is a sortening.

3 s te elongation l depends on te initial lengt of te segment, it isn t an adequate measure to caracterie te linear deformation. Tat s w it is introduced a new notion called specific elongation, wic represents te total lengtening of te unit lengt (tat s w it is named also unit elongation): ε x l = l (.4) Fig.. We observe tat for te unit lengt l =, te specific elongation ε x is even l: (ε x ) l= = l (.5) Te specific elongation ε x is a non-dimensional notion and te above relation is valid onl if l is uniforml distributed along te entire lengt l, so onl if ε x = const. Hence, te total lengtening isn t produced b identical lengtening of eac unit of lengt. Tat s w we consider again a differential segment MN of lengt dx on te same axis Ox (Fig...b). Te displacement of point M is u, but te one of point N will var wit te differential measure du, so te displacement of N will be u+du. Te total elongation of te differential lengt dx will be: dx = du (.6)

4 Considering tat tis total elongation is uniforml distributed on te differential lengt dx, te specific elongation will be now: dx = dx x = du dx ε (.7) n te relation written above it was considered tat u is function onl of x. n realit it is function of all tree coordinates u = u(x,,) and te differential du as te general expression: u u u du = dx + d + d (.8) x But as te segment MN is on Ox axis, te single coordinate tat varies is x and te oter derivatives, wit respect to O and O axis, are 0. So, du will be: u du = dx (.9) x nd ε x is: u dx du ε x u x = = ε x = (.0) dx dx x Similarl, te specific elongations along O and O axis ma be written: v ε = (.) and w ε = (.) Relations (.0) (.) are te first tree Cauc s relations representing te differential relations between te specific elongations and te components of te relative displacement... Te sliding (sear strain) Cause b te fact tat te sliding is an angular deformation it will be analed in a plan from a deformable bod. We consider a differential rectangle in a plan xo, a corner of te rectangle being even te origin O (Fig...a). dmitting tat point O is fixed, point will reac a position perpendicular to Ox axis, = dw, respectivel point B will reac a position perpendicular to O axis B, BB = du. Tis deformation doesn t cange te rectangle area, onl its sape, transforming it into a parallelogram. s a result of tis deformation te initial straigt angle xo (Fig...a,b) is modified, b te rotation of Ox axis wit te angle α x and respectivel O axis wit α x.

5 a. b. Fig.. ssuming te plan is made from strips parallel to x and axis (Fig..b), we observe tat te above rotations are produced in fact b relative translations of te strips, called slidings. Hence, te angles α x and α x are a measure of tese slidings. Teir sum is te specific sliding γ x : γ x = α x + α x (.) Te specific sliding can be defined as te modification of te initial straigt angle. Te specific sliding (sear strain) is positive if decreases te initial straigt angle (it becomes an acute angle), respectivel negative if it increases tis (becomes an obtuse angle). Taking into account te potesis of te small deformations, te angles α x and α x ma be written: w dx dw x w α x tgα x = = = (.4) dx dx x and u d du u α x tgα x = = = (.5) d d Replacing in (.) we obtain: x w u = + x γ (.6) Similarl, te specific slidings in xo plan and O plan are: and γ γ x u v = + x v w = + (.7) (.8)

6 Relations (.6) (.8) are te oter tree Cauc s relations representing te differential relations between te specific sliding and te components of te relative displacement. ugustin-louis Cauc ( ). GEOMETRCL CHRCTERSTCS OF THE CROSS SECTON To use te notion of cross section in all calculations made in Mecanics of Materials, certain important caracteristics sould be known. Tese geometrical caracteristics ma be grouped, approximatel, in two categories:. Caracteristics tat define te relative position of te sstem of reference O eiter it as an arbitrar origin, or identical to te cross section centroid G or te sear center C. n tis categor tere are: te area, te centroid G, te sear center C, but also te cross section sape wic is ver important in te calculation of te construction members at certain solicitations.. Caracteristics connected to te potesis of te movement of te cross section, wic ave also a mecanical interpretation. n tis group te following caracteristics are included: te first moment of area (static moment): axial and sectorial, te second moments of area (moments of inertia): axial and sectorial, te radius of gration (radius of inertia).

7 .. Te area. Te first moment of te area. Te centroid We consider a certain cross section in an ortogonal sstem of axis O, wit te origin in O (Fig..4). Te cross section of area is made from infinite differential areas d. a. Te area is: = d (.9) Terefore, te area is te infinite sum of all te elementar areas d, on te entire area. t is alwas measured in [lengt] units: cm, m, mm. b. Te first moment of te area (static moment): S = d S = d (.0) Te static moment of an area wit respect to an axis, is te infinite sum of all products between tat area and te distance between tat area centroid and te respective axis (Fig..4). Te static moment can be positive or negative and it is alwas measured in [lengt] units: cm, m, mm. Fig..4 c. Te centroid (center of gravit) position is given b its coordinates G and G (Fig..4), were wit G we note te centroid.

8 We write Varignon s teorem, wic sas tat te moment of te entire area is equal to te sum of all moments of te elementar areas: = d (.) G Pierre Varignon (654-7) Respectivel = d (.) G From (.) and (.) te centroid coordinates are: and G G = = d S = d S = (.) (.4) Tese coordinates are alwas measured in [lengt] units: cm, m, mm. Observing relations (.) and (.4) we ma conclude tat if te point O is identicall to te centroid G, = = 0 and implicit G = G = 0. Tis means tat in tis case te static moments are null S = S = 0, and tese axes: and are called central axis.!!! f wit respect to an axis te static moment is null, tat axis is a central axis. f te cross section is made up of man sections aving te area i and te centroid position G i known (Fig..5), te integrals from (.9), (.0), (.) and (.4) are transformed into finite sum:

9 n = i (.5) i= S S n = i i (.6) i= n = i i (.7) i= G n i i i= = (.8) Fig..5 G n i i i= = (.9).. Te second moment of te area (area moment of inertia) and oter geometrical caracteristics Te area moment of inertia represents te cross section inertia to its movement of rotation around an axis included in its plane. Te second moment of area is a measure of resistance to bending of a loaded section. n engineering contexts, te area moment of inertia is often called simpl "te" moment of inertia even toug it is not equivalent to te usual moment of inertia (wic as dimensions of mass times lengt squared and caracteries te angular acceleration undergone b a solids wen subjected to a torque). We ma define: a. Te axial moment of inertia of an area (Fig..6) wit respect to an axis comprised in its plane represents te infinite sum of all products between te elementar area d and te square of te distance between tis area and tat axis. Wit respect to te central axis G and G (Fig..6) te moments of inertia are: and = = d d (.0) (.)

10 ! Te axial moments of inertia and Fig..6 are alwas positive and te are measured in [lengt] 4 units: cm 4, m 4, mm 4. b. Te centrifugal moment of inertia (te product moment of area) of an area (Fig..6) wit respect to a sstem of axis comprised in its plane represents te infinite sum of all products between te elementar area d and te distances between tis area and tat rectangular sstem of axis. Wit respect to te central sstem of axis G (Fig..6) te centrifugal moment of inertia is: = d (.) Te centrifugal moment of inertia reflects troug value and signs te cross section repartition in te fourt quadrants of te sstem of axis (Fig..6). t is also measured in [lengt] 4 units: cm 4, m 4, mm 4.! Wile and are alwas positive, te centrifugal moment of inertia ma be positive, negative or even ero. Proprieties of te centrifugal moment of inertia: - f te cross section as at least an axis of smmetr (Fig..7.a), te centrifugal moment of inertia is equal to ero. ssuming te cross section from figure.7.a can be described as a pair of elementar areas, smmetricall disposed, te centrifugal moment of inertia is:

11 [ ( ) d+ ( )( ) d] = d = = 0 a. b. Fig..7 - f te sstem of axis G is rotated about G wit 90 0 in te clock-wise direction (Fig..7.b) te centrifugal moment of inertia canges its sign. From figure.7.b we ma write te relations between te initial sstem G and te rotated one 90G 90 : 90= ; 90 = Replacing in te centrifugal moment of inertia: = ( ) d = 90 90d = c. Te polar moment of inertia t measure also te cross section inertia, but te rotation is made around an axis perpendicular to te cross section plane, wic intersect te cross section in te centroid G. From figure.6 we ma write: ρ = + (.) Te polar moment of inertia, written similar to te axial moment of inertia, will be wit (.): ( ) p = ρ d = + d = d + d Result: = + (.4) p

12 From (.4) we ma conclude tat te sum of te axial moments of inertia wit respect to rectangular axis wit te same origin in G (Fig..6) is an invariant to te rotation of te sstem of axis. d. Te radius of gration (radius of inertia) f we want to increase te moment of inertia of a cross section wit respect to an axis, ten we move it awa from tis, witout increasing te cross section area. s te material consumption is directl proportional to te cross section area, te above solution is as economic as te ratio / is bigger. Te caracteristic connected to te moment of inertia of a cross section wit respect to an axis is te radius of gration. t is defined (Fig..8) as te distance from an inertia axis to a fictitious point Q, in wic if te entire cross section area is concentrated, te moment of inertia (punctual) wit respect to tat axis is equal to te real moment of inertia. Fig..8 ccording to tis definition we ma write te equalit: d = i (.5) = From (.5) te radius of gration wit respect to G axis is: i = (.6) Similarl wit respect to te oter axis G : i = (.7) Te radius of gration is alwas measured in [lengt] units: cm, m, mm. e. Te strengt modulus Te ratio between te axial moment of inertia and te distance to te fartest point of te cross section from tat axis is called strengt modulus. n figure.6 we note te distances from G axis till te extreme points: superior s and inferior i, and te strengt modulus wit respect to G axis in tese points are:

13 W = and s s W = (.8) i i Te strengt modulus can be positive or negative and it is alwas measured in [lengt] units: cm, m, mm... Moments of inertia for some simple cross sections a. Rectangle We consider te rectangular section from figure.9, for wic we want to calculate troug direct integration te moments of inertia wit respect to O and O axis, tangent to te rectangle sides, and ten wit respect to te central axis G and G. We consider an elementar area d, a rectangular strip parallel to G axis, of widt b and eigt d, so tat d = b d. Fig..9 Wit respect to O and O, d = b d (Fig..9) and te area, te static moments and te centroid coordinates will be: = d = b d = b 0 b S = d = b d = 0 b b b S = d= d = G S b = = and 0 G S = = Te axial moments of inertia wit respect to O and O are: = d b d = = 0 b

14 and respectivel: b = Te centrifugal moment of inertia: b = d b d = = 0 b 4 Wit respect to te central sstem of axis G, d = b d (Fig..9) and te same caracteristics are now: Similarl: = d = b d = b = b S = d b d b 0 = = = S = 0 We found a result tat was obvious, because G and G are even te central axis, and we ave seen in paragrap.. tat wit respect to tese axiss = S = 0. mplicit: G= G= 0 Te axial moments of inertia wit respect to G and G are: b b = d = b d = = Similarl: b = (.9) b = (.40) Te centrifugal moment of inertia: = 0 (.4) Relation (.4) results from te first propert of te centrifugal moment of inertia, as te rectangle as smmetr axes.

15 b. Triangle Let s consider te triangle from figure.0 of widt b and eigt. We want to calculate te moments of inertia wit respect to O and O axes, wit te origin O in te triangle vertex, ten wit respect to O and O axes and finall wit respect to te central axis of te triangle, G and G. Fig..0 We consider an elementar area d as a rectangular strip parallel to G axis, wit te widt b and eigt d, so tat d = b d. For For we ma write b = and d = b d. Wit tese: b b b = d b d d = = = b b : b = ( ) and ( ) For : d d =. Terefore: ( ) b b = d = d = 0 b b = and b d = d. Hence: b b = d = d = 6 0 Similarl we ma write te moments of inertia, and. We ma keep in mind te moments of inertia of te triangle, wit respect to te central axes:

16 b = 6 = b 6 b 7 = (.4) c. Circle R 0 0 Fig.. π R π D = d = ρ dρ dϕ = π = π R = 4 Te elementar area d (Fig..) is comprised between two radius situated to an angle dφ and two concentric circles of radius ρ and ρ+d ρ: d = ρ d ρ dϕ Wit te limits of integration: 0 ρ R and 0 ϕ π and integrating on te entire circle surface, we get: Te distances from te elementar area centroid to G axis, respectivel G axis are: = ρ cosϕ = ρ sinϕ Te axial moments of inertia wit respect to G and G are: 4 π R cos ϕ R sin ϕ π π R π D ρ sin ϕ ρ ρ ϕ 0 4 ϕ ϕ = d = d d = d = = = So, te moments of inertia of te circle, wit respect to te central axes, are: 4 π D = = and 0 64 = (.4)..4 Te variation of te moments of inertia wit te translation of axes Let s consider a plane cross section of area wit an ortogonal sstem of axis O. Knowing te values of te moments of inertia and, we want to determine te moments of inertia and wit respect to O and O axis, parallel to te first axis (Fig..).

17 = + b = + a Fig.. Wit respect to O axis, te moment of inertia is: ( ) = = + = + + = + + d a d d ad a d a S a Similarl, wit respect to O axis, te moment of inertia is: = + b S + b Te centrifugal moment of inertia is: ( )( ) = + a + b d = d + a d + b d + ab d = + a S + b S + ab n te above relations regarding, and, S and S are static moments wit respect to O and O axis. f tese axes are even te central axes of te cross section (O G and O G ), tese static moments are ero: S = S = 0 Te relations become: a = + (.44) b = + (.45) ab = + (.46) Relations (.44), (.45) and (.46) represent te formulas of te moments of inertia wit respect to translated axes, called Steiner s formulas. Eac first term from relations represent te moment of inertia wit respect to te centroid axes G and

18 G, wile te second term is te translation term equal wit square distance between te translated axes multiplied b te corresponding area. Jakob Steiner (796-86) nversel, if we ave te moments of inertia wit respect to a sstem of axes O, te moments of inertia wit respect to te centroid axes G are: = a = b = ab Let s ceck tese formulas for te rectangular section. From...a paragrap, we got (Fig..9): b =, b =, b = 4 Replacing in te relation written above (see Fig..9): b b = b = b b b = b = b b = b = 0 4 We found te relations (.9), (.40) and (.4).

19 ..5 Te variation of te moments of inertia wit te rotation of axes Let s assume tat for a cross section, wit respect to te central sstem of axes G, te moments of inertia are known and we intent to compute te moments of inertia wit respect to an ortogonal sstem of axes α G α (Fig..), rotated wit an angle α (α > 0 for clockwise sense). From Fig...b: GC=, CD=, GB= α, DB= α G=, C= tgα, cosα D = CD C = tgα B = D sin α = ( tgα ) sinα a. b. Fig.. GB = G + B = + sinα tgα sinα = ( sinα ) + sinα = cosα + sinα cosα cosα cosα Finall te coordinates in te rotated sstem of axes, are: = cosα + sinα (.47) α DB = D cos α = ( tgα ) cosα sinα cosα α = + (.48)

20 Te moments of inertia wit respect to te rotated sstem of axes α G α are: ( ) = α d = sinα + cosα d = sinα d sinα cosα d + cosα d α Replacing (.0), (.) and (.) we ave: = cosα + sinα sin α α n a similar manner we find te oter central moment of inertia and te centrifugal moment of inertia: = sinα + cosα + sin α α α = sin α + cos α α sinα cosα + = + cos α sin α α cos + cos α Replacing = and =, te moments of inertia are: α α α α + = cos α + sin α = sin α + cos α dding te first two relations we get: + = + α α (.49) (.50) (.5) n conclusion, te sum of te moments of inertia wit respect to ortogonal axes aving te same origin is an invariant...6 Principal moments of inertia. Principal axis of inertia Te moments of inertia and are continuous and periodical functions of α. α α For tese functions, we can find a value of α wic correspond to an extreme value of te moments of inertia. We make te first derivative of and we equalie it 0: d α ( )sin α cos α 0 dα = = We get, from tis equation: But also: tgα = d α = ( sin α + cos α ) = = 0 α α dα α (.5)

21 We get tat = 0, wat means tat axes α and α are conjugated axes, so wit α α respect to tese axes te central moments of inertia and ave extreme values. We sall name tese axes principal axes of inertia (noted wit G and G) and te corresponding moments of inertia are te principal moments of inertia (noted wit and ). Te principal axes orientation is given b te equation: te solutions: βi α 0 nπ Te first two solutions are: were: = + ( n= 0,,,...) n = 0 β = α β = α 0 0 π n = β = α 0 + π β = α 0 + α 0 = arctg n conclusion, te principal axes of inertia are ortogonal axes. We agree tat π β We write anoter condition of extreme, for α=β : < 0 : + 4 < 0 But: = + 4 < 0 + < 0 < 0 tgα =, wic as Tis final relation tells us tat te centrifugal moment of inertia and te angle must alwas ave different signs. lwas G is te strong axis of inertia, so: > Wit relations (.49), (.50) and (.5) te final relations for te principal moments of inertia and are:

22 , = ± + 4 (.5) Te principal axes of inertia are tose to wic te centrifugal moment of inertia is ero. f a cross section as at least one axis of smmetr, te central axes of inertia are identicall to te principal axes of inertia. pplication For te quarter-circle in te figure bellow (Fig..4) calculate te principal moments of inertia and te direction of te principal axes. Fig..4 n te cosen reference sstem O te differential element of area d written in polar coordinates is: = Te area is: = = = = 4 Te distances from te centroid of te differential element to te axes O, respectivel O are: = =

23 Te second moments of area (static moments) about O axis, respectivel O axis, are: = = = = = = = Te coordinates of te centroid: = = = 4 4 = = = = 4 4 n order to calculate te central moments of inertia and and te centrifugal moment of inertia we calculate first te moments of inertia about te reference axes,, and : = = = = = = = 6 = 6 = = = = Te central moments of inertia and are calculated b translating: 4 = = 4 = = = = = = = Te principal moments of inertia calculated wit formula (.5) are: = + and =

24 Te rotation of te principal axes: = = Te solutions are: = and = Te first solution is te correct one, due to te condition: < 0 Te final angle of rotation: =