ENGINEERING MECHANICS

Transcription

1 Egieerig Mechaics CHAPTER ENGINEERING MECHANICS. INTRODUCTION Egieerig mechaics is he sciece ha cosiders he moio of bodies uder he acio of forces ad he effecs of forces o ha moio. Mechaics icludes saics ad damics. Saics deals wih he special case of a bod a res or a bod ha moves wih cosa veloci. A bod a res or movig wih cosa veloci is said o be i equilibrium. This is someimes also called as saic equilibrium. Whe he bod moves wih fiie veloci or acceleraio, he priciples of saics are o loger applicable. The mechaics of such a ssem is called damics. Whe he bod has o roaioal moio, i is called as paricle. Usuall, each poi o a rigid-bod is alwas a a cosa disace from a oher poi i he bod. Damics is furher divided io kiemaics ad kieics. Kiemaics defies he relaioships amog displaceme, veloci, ad acceleraio of a movig bod. Kieics defies he relaioship bewee he forces ha ac o a bod ad he moio of he bod. Aalsis of he bodies ca eiher be coduced i plae or i hree-dimesios. A rigid-bod i space has si degree of freedom. This chaper provides a brief review of various opics i Mechaics.. NEWTONIAN MECHANICS Collecivel, he sud of saics ad damics is called classical mechaics. Classical mechaics reas he moio of bodies of ordiar size ha move a speeds ha are small compared o he speed of ligh. Newo (64-77) formulaed he law of uiversal graviaio ad he mahemaics of calculus. Newo iroduced he coceps of force ad mass. Newo formulaed he hree laws of moio ha are he basis of egieerig applicaios of mechaics. The classical mechaics is ofe called Newoia mechaics..3 NEWTON S LAWS OF MOTION Newo s laws are fudameall applicable direcl o a paricle ha is a bod which ma be reaed as havig poi mass. Newo s laws ma be saed i erms of a paricle parameers as follows:

2 Solvig Egieerig Mechaics Problems wih MATLAB Firs Law: I he absece of applied forces, a paricle origiall a res or movig wih cosa speed i a sraigh lie will remai a res or coiue o move wih cosa speed i a sraigh lie. Secod Law: If a paricle is subjeced o a force, he paricle will accelerae. The acceleraio of he paricle will be i he direcio of he force, ad he magiude of he acceleraio will be proporioal o he force ad iversel proporioal o he mass of he paricle. Newo s secod law ca be epressed i he form: a = F...(.) m where F = force m = mass a = acceleraio Newo s secod law is he basis for he sud of kieics of a paricle. Third Law: For ever acio, here is a equal ad opposie reacio, or he muual forces eered b wo paricles o each oher are alwas equal ad opposiel direced..4 RESULTANTS OF COPLANAR FORCE SYSTEMS I saics, i is ofe ecessar o fid he sae of equilibrium or he resula of several forces acig i plae. I coplaar ssem, he resula R for a cocurre forces is give b ad ( ) ( ) R = Σ F + ΣF a ΣF θ = ΣF...(.) where ΣF, ΣF = algebraic sums of he ad compoes of he forces of he ssem respecivel. θ = he agle ha he resula R makes wih he -ais. For a parallel ssem, he resula R is give b R=ΣF ad Ra =ΣM O where ΣF = algebraic sum of he forces of he ssem O = a mome ceer i he plae a = perpedicular disace from he mome ceer O o he resula R Ra = mome of R wih respec o O ΣM O = algebraic sum of he momes of he forces of he ssem wih respec o O...(.3)

3 Egieerig Mechaics 3 The resula R for a o-cocurre ad o-parallel ssem is give b ( ) ( ) R = Σ F + ΣF Σ ad θ = a F ΣF...(.4) where ΣF, ΣF = algebraic sums of he ad compoes of he forces of he ssem respecivel θ = he agle ha he resula R makes wih he -ais. The acio of he lie of he resula force is give b Ra = ΣM o...(.5) where O = a mome ceer i he plae a = perpedicular disace from he mome ceer O o he resula R Ra = mome of R wih respec o O ΣM O = algebraic sum of he momes of he forces of he ssem wih respec o O.5 RESULTANTS OF NON-COPLANAR FORCE SYSTEMS I o-coplaar force ssem, he forces are o i oe plae. The resula R for a cocurre ssem is give b wih he direcio cosies ( ) ( ) ( ) z R = Σ F + Σ F + ΣF...(.6) ΣF ΣF ΣF cos θ =, cosθ = ad cos θ z = R R R where ΣF, ΣF, ΣF z = algebraic sums of he, ad z compoes of he forces of he ssem respecivel. For a parallel ssem, he resula R is give b R = ΣF R = ΣM z...(.7) Rz = ΣM where ΣF = algebraic sum of he forces of he ssem = perpedicular disace from he z plae o he resula z = perpedicular disace from he plae o he resula ΣM, ΣM z = algebraic sum of he momes of he forces of he ssem abou he ad z aes respecivel z

4 4 Solvig Egieerig Mechaics Problems wih MATLAB If ΣF = 0, he resula couple C whe eiss is give b C = ( Σ M ) + ( ΣM ) z ad φ = a Σ M z Σ M...(.8) where φ is he agle ha he vecor represeig he resula couple makes wih he -ais. The magiude of he resula R of he o-cocurre ssem a he origi (, ad z aes is placed wih heir origi a he base poi) is give b R = z ( Σ F ) + ( Σ F ) + ( ΣF )...(.9) wih he direcio cosies ΣF ΣF ΣFz cosθ =,cosθ = ad cosθ z =...(.0) R R R The magiude of he resulig couple C is give b C = z ( Σ M ) + ( Σ M ) + ( ΣM )...(.) wih he direcio cosies ΣM ΣM ΣM z cos φ =,cosφ = ad cosφ z =...(.) C C C where ΣM, ΣM, ΣM z = algebraic sums of he momes of he forces of he ssem abou, ad z-aes respecivel. φ, φ, φ z = agles which he vecor represeig he couple C makes wih he, ad z-aes respecivel..6 EQUILIBRIUM OF COPLANAR FORCE SYSTEMS The ecessar ad sufficie codiios for he equilibrium of a coplaar force ssem are: R = ΣF = 0 ad C = ΣM = 0...(.3) where ΣF = vecor sum of all forces of he ssem ΣM = vecor sum of he momes of all he forces of he ssem. For cocurre ssem, a of he followig ses of equaios esures equilibrium (he resula is zero). The cocurrec is assumed a he origi. Se : ΣF = 0 ΣF = 0

5 Egieerig Mechaics 5 Se : ΣF = 0 ΣM A = 0...(.4) (A ma be chose a place i he plae ecep o he -ais) Se 3: ΣM A = 0 ΣM B = 0 (A ad B ma be chose a place i he plae ecep A, B ad he origi do o lie o he same sraigh lie) where ΣF, ΣF = algebraic sum of he ad compoes of he forces of he ssem respecivel. ΣM A, ΣM B = algebraic sum of he momes of he forces of he ssem abou A ad B respecivel. For parallel ssem, a of he followig ses of equaios esures equilibrium (he resula is eiher a force or a couple). Se : ΣF =0 ΣM A =0...(.5) Se : ΣM A =0 ΣM B =0 (A ad B ma chose a place i he plae provided i he lie joiig A ad B is o parallel o he forces of he ssem) where ΣF = algebraic sum of he forces of he ssem parallel o he acio lies of he forces ΣM A, ΣM B = algebraic sum of he momes of he forces of he ssem abou A ad B respecivel. For o-cocurre ad o-parallel ssem, a of he followig ses of equaios esures equilibrium (he resula is eiher a force or a couple). Se : ΣF =0 ΣF =0 ΣM A =0...(.6) Se : ΣF =0 ΣM A =0 ΣM B =0...(.7) (Provided ha he lie joiig A ad B is o perpedicular o he -ais) Se 3: ΣM A =0 ΣM B =0 ΣM C =0...(.8) (Provided ha A, B ad C do o lie o he same sraigh lie) where ΣF, ΣF, ΣF z = algebraic sum of he, ad z compoes of he forces respecivel ΣM A, ΣM B, ΣM C = algebraic sum of he momes of he forces of he ssem abou a hree pois A, B ad C i he plae respecivel.

6 6 Solvig Egieerig Mechaics Problems wih MATLAB.7 EQUILIBRIUM OF NON-COPLANAR FORCE SYSTEM The ecessar ad sufficie codiios ha R ad C be zero vecors are give b R = ΣF = 0 ad C = ΣM = 0...(.9) where ΣF = vecor sum of all he forces of he ssem ΣM = vecor sum of he momes of all he forces of he ssem relaive o a poi. For a cocurre ad o-coplaar ssem, he followig se of equaios mus be saisfied: ΣF = 0 ΣF = 0...(.0) ΣF z = 0 where ΣF, ΣF, ΣF z = algebraic sums of he, ad z compoes of he forces of he ssem respecivel. For a parallel o-coplaar ssem, he se of equaios o be saisfied for equilibrium are: ΣF =0 ΣM = 0...(.) ΣM z =0 where ΣF = algebraic sum of he forces of he ssem alog he -ais which is seleced parallel o he ssem. ΣM, ΣM z = algebraic sums of he momes of he forces of he ssem abou he ad z aes respecivel. The ecessar ad sufficie codiios required for equilibrium for a o-cocurre o-coplaar ssem are give b ΣF =0 ΣF =0 ΣF z =0 ΣM =0 ΣM =0 ΣM z = 0...(.) where ΣF, ΣF, ΣF z = algebraic sums of he, ad z compoes of he forces of he ssem respecivel. ΣM, ΣM, ΣM z = algebraic sums of he compoes of he forces of he ssem abou he, ad z aes respecivel..8 TRUSSES A russ is a ssem of sleder members ha are pied ogeher. The members are free o roae a he pied jois ad carr forces ol. All he eeral forces ac a he jois. Trusses are eamples of coplaar force ssems i equilibrium. Trusses are assumed o be rigid members all locaed i oe plae. The weighs of he russ members are egleced. Forces are rasmied

7 Egieerig Mechaics 7 from oe member o aoher hrough pi jois. These members are called wo-force members. A woforce member is i equilibrium uder he effec of wo resula forces oe a each ed. The wo-force members will be eiher i esio or compressio. There are wo mehods available for he aalsis of russes. For a sable plaar russ, he followig codiio applies: = m + p...(.3) where = umber of jois m = umber of members p = umber of ukow eeral forces I a sable hree-dimesioal russ, he codiio ha holds is 3 = m + p...(.4) Mehod of Jois I his mehod, a free-bod diagram of a pi i he russ is draw. Maimum of wo ukow forces ac o ha pi. Proceed from oe pi o aoher uil all ukows have bee obaied. Mehod of Secios A free-bod diagram of a secio of he russ is draw. The forces i he members cu ac as eeral forces. The ssem is a o-cocurre ad o-parallel oe. I a oe secio o more ha hree ukow forces are o be foud..9 ANALYSIS OF BEAMS A beam is a log member ha is subjeced o eeral laeral forces, F ad eeral laeral momes, T. These cause ieral laeral forces V called shear forces ad ieral laeral momes M kow as bedig momes. The shear forces ad he bedig momes iduce laeral deformaios, also called bedig. Shear ad Mome Diagrams Shear ad mome diagrams show he variaio of V ad M across a beam. Summaio of he forces ad momes of forces are used o plo V ad M direcl for he beams. The slope of he shear diagram a a secio alog he beam is he egaive of he load per ui legh a ha poi. The chage i shear bewee wo secios of a beam carrig a disribued load equals he egaive area of he load diagram bewee he wo secios. The slope of he mome diagram a a secio alog he beam is he value of he shear a ha secio. The chage i he mome bewee wo secios of a beam equals he area of he shear diagram bewee he wo secios..0 FRICTION The ageial force ha opposes he slidig of oe bod relaive o he oher is he saic fricio bewee he wo bodies. Kieic fricio is he ageial force bewee wo bodies afer moio begis. The bod begis o slide ol whe he applied force eceeds he saic fricioal force a he floor. Referrig o Fig.., he followig erms are defied: Limiig fricio F is he maimum value of saic fricio ha occurs whe moio is impedig. Coefficie of saic fricio is he raio of he limiig fricio F o he ormal force N, or µ= F' N

8 8 Solvig Egieerig Mechaics Problems wih MATLAB Coefficie of kieic fricio is he raio of he kieic fricio o he ormal force. Agle of repose, α is he agle o which a iclied plae ma be raised before a objec resig o i will move uder he acio of he force of gravi ad he reacio of he plae. I Fig.., R is he resula of F ad N ad α = φ. W F R φ N Fig.. Slidig fricio Bel Fricio: Whe a bel passes over a pulle, he esios i he bel o he wo sides of he pulle will differ. Whe slip is abou o occur, he esios T ad T are give b T = T e µβ...(.5) where T = larger esio T = smaller esio µ = coefficie of fricio β = agle of wrap (radia) e =.78 (base of aural logarihms) Screw fricioal force occurs whe a u is screwed over a bol.. FIRST MOMENTS AND CENTROIDS The ceroidal posiio vecor r of a geomer composed of areas d, d, d 3,..., d locaed a pois P, P, P 3,, P represeed b posiio vecors r, r, r 3,, r respecivel is defied as where r = i i i= i= r d d i d i = area of ih eleme r i = posiio vecor of he ih eleme d i = oal area of all elemes i= r d = firs mome of area of all he elemes relaive o seleced poi O. i i i=

9 Egieerig Mechaics 9 The ceroid ca be wrie i erms of, ad z coordiaes as = = i i i= i= i= d d i i i= i d d i z = i i i= i= zd d where d i = magiude of area of he ih eleme i z,, = coordiaes of ceroid of he assemblage i, i, z i = coordiaes of P i a which d i is coceraed. The ceroid of a coiuous quai of mass m is give b rdm r = dm or dm = = dm Q z m dm = = dm zdm z = = dm Q z m Q m where Q, Q z, Q z = firs momes wih respec o, z ad z plaes respecivel.. VIRTUAL WORK A virual displaceme δs of a paricle is defied as a ifiiesimal chage i he posiio of he paricle cosise wih he cosrais imposed o he paricle. Virual work δu doe b a force, F is defied as F δs, where F is he magiude of he compoe of he force alog he virual displaceme δs. Similarl, virual work δu doe b a couple of mome M is defied as Mδθ, where δθ is he virual agular displaceme.

10 0 Solvig Egieerig Mechaics Problems wih MATLAB Equilibrium: The ecessar ad sufficie codiio for he equilibrium of a paricle is zero virual work doe b all he forces acig o he paricle durig a virual displaceme δs. The ecessar ad sufficie codiio for he equilibrium of a rigid-bod is zero virual work doe b all he eeral forces acig o he bod durig a virual displaceme cosise wih he cosrais imposed o he bod. The equilibrium of a ssem of rigid bodies eiss if he poeial eerg V has a saioar value. Sable equilibrium occurs if he poeial eerg V is a miimum. Usable equilibrium occurs if he poeial eerg V is a maimum. Neural equilibrium eiss if a ssem remais i a posiio i which i is placed. If is he variable ad V he poeial eerg of he ssem, he we have dv > 0 d dv d dv = d < 0 0 (sable equilibrium) (usable equilibrium) (eural equilibrium).3 KINEMATICS OF A PARTICLE Kiemaics is he sud of moio wihou cosiderig he forces or oher facors ha ifluece he moio. Reciliear moio: The geeral epressio for displaceme(s), veloci(v) ad acceleraio(a) are derived from he followig hree differeial relaios: a= dv/d v= ds/d ads = vdv For moio of a poi alog a sraigh lie, he followig formulas are valid for cosa acceleraio a = k: v = v 0 + k v = v 0 + ks s = v 0 + s = ( v+ v ) 0 where v 0 = iiial veloci v = fial veloci k = cosa acceleraio = ime s = displaceme. k...(.5a)

11 Egieerig Mechaics Curviliear moio: I a plae, curviliear moio is moio alog a plae curve (pah). The veloci ad acceleraio of a poi o such a curve ca be epressed eiher as: (a) (b) (c) recagular compoes ageial ad ormal compoes radial ad rasverse compoes (a) Recagular compoes The posiio vecor r(), he veloci vecor v(), ad he acceleraio vecor a() are give b r() = i () + () j+ zk () v() = v () i+ v () j+ v () k = () i+ () j+ z () k z a() = a () i+ a () j+ a () k = () i+ () j+ z() k...(.6) z where he over-do represes ime differeiaio. The recagular compoes of veloci ad acceleraio are give b v () = (), v() = (), v() = z () z a () = (), a () = (), a () = z()...(.7) The compoes of posiio are give b () = () = z() = ( ) + v ( s) ds ( ) + v ( s) ds z z ( ) + v( sds )...(.8) The compoes of veloci are give b Also v () = v () = v( ) + a( s) ds v( ) + a( s) ds z v z () = vz( ) + az( s) ds...(.9) v () = v( ) + a( ) d...(.30)

12 Solvig Egieerig Mechaics Problems wih MATLAB (b) Tageial ad ormal compoes Referrig o Fig.., he veloci vecor v ca be wrie as v = v where v = magiude of he veloci vecor = ageial ui vecor direced alog he veloci vecor....(.3) Also ad =θ = θ...(.3) () () v () Fig.. Veloci vecors where = he ormal ui vecor defied o be perpedicular o he ageial ui vecor. I Fig..3, he radius of curvaure ρ of he pah a ime is show which is obaied b he iersecio of he lies eedig from () ad ( + ) where is ime icreme. The agle θ chages a icremeal amou θ ad he poi moves a icremeal amou s. θ ρ ( + ) ( + ) () θ θ( + ) () θ() () Fig..3 Radius of curvaure of pah Now ρ θ = s or ρθ = v...(.33) where θ = d d θ v = ds d

13 Egieerig Mechaics 3 Differeiaig Eq. (.3) wih respec o ime gives a = a + a...(.34) where a =v v a = ρ...(.35) (c) Radial ad rasverse compoes The polar form of a posiio vecor is r = r r...(.36) where r = r is he magiude of r r = ui vecor i he direcio of r r = he radial ui vecor θ = he circumfereial ui vecor The circumfereial ui vecor is perpedicular o r. Hece, r ad θ are relaed o i ad j i Fig..4 as r () = cos θ()i + si θ()j θ () = si θ()i + cos θ()j...(.37) θ r r θ Fig..4 Radial ad rasverse compoes Differeiaig Eq. (.37) wih respec o ime gives r = θ si θ i+θ cos θ j =θ ( si θ i+ cos θ j) =θ θ...(.38) ad θ = θ r...(.39) Hece, he derivaives of he ui vecors are give b =θ r θ θ = θ r...(.40) Similarl, from Eqs. (.36) ad (.40), we ge v = v r r + v θ θ...(.4) where v r =r v θ =rθ

14 4 Solvig Egieerig Mechaics Problems wih MATLAB Differeiaig Eq. (.4) wih respec o ime gives a = arr + a θ θ...(.4) where a r = r rθ a θ = rθ+ r θ...(.43).4 D ALEMBERT S PRINCIPLE Kieics of paricles begis wih Newo s secod law, which relaes forces, acceleraios ad ime. The secod form of wriig Newo s law is called D Alember s priciple which gives he codiio for damic equilibrium. I saes ha ΣF ma = 0...(.44) where ΣF = vecor sum of all he forces acig o he paricle m = mass of he paricle a = acceleraio of he paricle. Hece, a imagiar force (also kow as ieria force) which is colliear wih ΣF bu opposie i sese ad of magiude ma would cause i o be i equilibrium if applied o he paricle. All he equaios of equilibrium are he applicable. Kieics of paricles ca also be aemped wih workeerg priciple ad impulse-momeum priciple, jus as a rigid-bod. Whe velociies ad displacemes are give isead of acceleraio as i sprig problems, he problem should be aemped wih workeerg priciple. O he oher had whe veloci ad ime are give as i recoil or impac problems, he equaios of moio are formulaed from impulse-momeum relaios..5 KINEMATICS OF A RIGID BODY IN PLANE MOTION A rigid-bod is cofigured b specified dimesios ad roaioal moio should be cosidered i addiio o raslaio. I plae moio of a rigid bod, ever poi i he bod remais a a cosa disace from a fied plae. As show i Fig..5, B is a arbirar poi i he bod ad --z is a oroaig referece frame. Y A B ρ φ z r B r A X Z Fig..5 Represeaio of plae moio of rigid-bod

15 Egieerig Mechaics 5 The posiio vecor r A of a poi A (fied or movig) i he lamia is give b r A = r B + ρ...(.45) where r B = posiio vecor of B ρ = vecor BA Differeiaig Eq. (.45) gives v A = r = r + ρ = r +ρωe...(.46) A B B φ where r B = liear veloci of B relaive o he fied aes X, Y ad Z ω = agular veloci (magiude) of ρ abou a lie parallel o Z-ais e φ = ui vecor perpedicular o ρ (i he direcio of icreasig φ) The acceleraio a A is give b..... a A = v = r = r ρω e + ραe...(.47) A A B ρ φ where r B = acceleraio of B relaive o fied aes X, Y ad Z e ρ = ui vecor alog ρ direced from B oward A e φ = ui vecor perpedicular o ρ (i he direcio of icreasig φ) α = agular acceleraio (magiude) of ρ abou a lie parallel o he Z-ais. Equaios (.46) ad (.47) ca also be wrie i vecor form as follows: v A = v B + ω ρ or v A = v B + v A/B...(.48) ad a A = a B + ω (ω ρ) + α ρ a A = a B + a A/B...(.49) Here is a vecor (cross) produc. also ω = φk = ωk α = φ k = ω k = αk...(.50) ραe ρ = ω ρ ραe φ = α ρ ρω e ρ = ω (ω ρ) i which v A/B = relaive veloci of A as i roaes aroud B. a A/B = relaive acceleraio of A as i roaes aroud B. Thus i plae moio he erm ω (ω ρ) becomes ρω. Aoher mehod of compuig velociies ad acceleraios i rigid bodies is o draw he vecor diagrams. These are draw based o he cocep ha relaive veloci is alwas perpedicular o he lie joiig he wo pois ad he wo compoes of relaive acceleraio are perpedicular o each oher. Someimes a roaig frame of referece is used o represe he moio of a bod raslaig wih veloci v, ad acceleraio a, o aoher roaig bod as i case of a crak ad sloed lever likage. Here he acceleraio epressio becomes v A = v B + (ω ρ) + v ad a A = a B + ω (ω ρ) + α ρ+ ω v AB + a Here he erm ω v AB is called coriolis compoe of acceleraio.

16 6 Solvig Egieerig Mechaics Problems wih MATLAB.6 MOMENTS OF INERTIA The aial mome of ieria or he secod mome of ieria I, of a eleme of area abou a ais i is plae (Fig..6) is give b di = da di = da...(.5) The polar mome of ieria J, of a eleme abou a ais perpedicular o is plae is he produc of he area of he eleme ad square of is disace from he ais. Referrig o Fig..6, he polar mome of ieria is dj = ρ da = ( + )da = di + di...(.5) The produc of ieria of a eleme of area i Fig..6 is give b di = da...(.53) da ρ Fig..6 Defiiio of produc of ieria The aial mome of ieria of a area is he sum of he aial momes of ieria of is elemes I = I = da da...(.54) The radius of graio of a area wih respec o a ais is give b k = I/A...(.55) The polar mome of ieria of a area is he sum of he polar momes of ieria of is elemes J = ρ da...(.56) The produc of ieria of a area is he sum of he producs of ieria of is elemes I = da...(.57) The parallel-ais heorem saes ha he aial or polar mome of ieria of a area abou a ais equals he aial or polar mome of ieria of he area abou a parallel ais hrough he ceroid of he area plus

17 Egieerig Mechaics 7 he produc of he area ad he square of he disace bewee he wo parallel aes. Referrig o Fig..7, we have I = I + Αm I = I + Α I z = J + Ar I = I + Am...(.58) where A =area I = produc of ieria of he area wih respec o ad aes I = produc of ieria abou wo parallel ceroid aes ad m, = coordiaes of G relaive o he (, ) aes hrough O or he coordiaes of O relaive o he (, ) aes hrough G., = a aes hrough O, = coplaar parallel aes hrough he ceroid G. r G m O Fig..7 Parallel-ais heorem The aial or polar mome of ieria or produc of ieria of a composie area is he sum of he aial or polar momes of ieria, or producs of ieria, of he compoe areas of he whole area. Mass mome of ieria of a rigid-bod is defied as: I = r dm, where r is perpedicular disace from he z-ais o he arbirar eleme dm. Thus he value of I differs for each ais abou which i is compued. Usuall i plaar kieics, he ais which is geerall chose for aalsis passes hrough bod s mass ceer G ad is alwas perpedicular o he plae of moio. I s uis are kgm. Someimes I is give as radius of graio abou ceer G. The radius of graio k of a bod wih respec o a ais is give b k = Im /...(.59) Cosider Fig..8, he aial mome of ieria of a mass dm is give b ( ) ( ) ( ) I = I = + z dm ad I = I = dm I = I = + z dm ad I = I = z dm z z I = I = + dm ad I = I = z dm...(.60) zz z z z where dm = mass of eleme I, I, I z = aial momes of ieria wih respec o he, ad z aes respecivel.

18 8 Solvig Egieerig Mechaics Problems wih MATLAB + z dm + + z z Fig..8 Polar mome of ieria of composie area I hree dimesios, a bod has si compoes of ieria for a specified,, z aes. Three of hese are momes of ieria abou each of he aes I, I, I z ad hree are producs of ieria each defied from wo orhogoal plaes I, I z, I z. If eiher oe or boh of he plaes are plaes of smmer, he he produc of ieria wih respec o hese plaes will be zero. Such aes are pricipal aes of ieria. The produc of ieria of a mass is give b I = dm The parallel ais heorem saes ha he mome of ieria of a bod abou a ais is equal o he mome of ieria I abou a parallel ais hrough he ceer of gravi of he bod plus he produc of he mass of he bod ad he square of he disace bewee he wo parallel aes. For eample, mome of ieria of sleder rod abou oe ed ca be compued i erms of mome of ieria abou mass l ceer G accordig o he followig relaio: I C = IG + m. A summar of momes of ieria of some geomerical shapes are give i Appedi-I. z.7 DYNAMICS OF A RIGID BODY IN PLANE MOTION Whe he moio is specified i erms of acceleraio ad forces, Newo s secod law or D Alember s priciple ca be direcl used. For a rigid-bod i plae moio, here are hree equaios describig damic equilibrium: Two force relaios for raslaio alog ad direcio ad oe mome equaio abou a poi usuall he ceer of mass G. I vecor form he equaios of plae moio are give b ΣF = m a...(.6) ( ) Σ M = I α k+ mr a = I α+ ma ma k...(.6) O O G / O O O O O where ΣF = resula of he eeral forces acig o he bod ΣM O = resula of he eeral momes acig o he bod

19 Egieerig Mechaics 9 m = mass of he bod a = acceleraio of he mass ceer of he bod a o = acceleraio of referece poi O α = agular acceleraio of he bod I o = mome of ieria of he bod relaive o he referece poi O, = coordiaes of he mass ceer relaive o he referece poi O r G/o = posiio vecor of he mass ceer relaive o he referece poi O a o, a o = magiude of he compoes of he acceleraio of he referece poi O alog he ad aes. The scalar equaios of he plae moio are give b ΣF = ma ΣF = ma Σ M = I α...(.6a) where ΣF, ΣF = algebraic sums of he magiudes of he compoes of he eeral forces alog he ad aes respecivel. m = mass of he bod a, a = compoes of he liear acceleraio of he mass ceer i ad direcios respecivel. ΣM = algebraic sum of he momes of he eeral forces abou he mass ceer. I = mome of ieria of he bod abou he mass ceer. α = magiude of he agular acceleraio of he bod. The scalar equaios for raslaio of a rigid bod are give b ΣF = ma ΣF = ma...(.63) ad ΣM =0 where ΣF, ΣF = algebraic sums of he compoes of he eeral forces i he ad direcios respecivel. m = mass of he bod a, a = acceleraio compoes i he ad direcios respecivel. ΣM = sum of he momes of he eeral forces abou he mass ceer of he bod. The scalar equaios of moio of a rigid bod uder he acio of a ubalaced force ssem for a bod wih a plae of smmer ad roaes abou a fied ais perpedicular o he plae are give b ΣF = mrω ΣF = mr α...(.64) ΣM = α o I o where ΣF = algebraic sum of he compoes of all eeral forces (which are he applied forces F, F, F 3, ec., he graviaioal force o he bod, ad he reacio R of he ais o he bod)

20 0 Solvig Egieerig Mechaics Problems wih MATLAB alog he ais, which is he lie draw bewee he ceer of roaio O ad he mass ceer G; oe ha he posiive sese is from G oward O because a = r ω has ha sese ΣF = algebraic sum of he compoes of he eeral forces alog he ais, which is perpedicular o he ais a O; oe ha he posiive sese alog his ais agrees wih ha of a = rα ΣM o = algebraic sum of he momes of he eeral forces abou he ais of roaio hrough O; oe ha posiive sese agrees wih he assumed sese of he agular acceleraio α m = mass of he bod G = ceer of mass of he bod r = disace from he ceer of roaio O o he mass ceer G I o = mome of ieria of he bod abou he ais of roaio ω = agular speed of he bod α = magiude of he agular acceleraio of he bod This pe of roaio is called o-ceroidal roaio. Eample: a lever oscillaig abou a poi of suspesio. Whe G ad O coicide (roaio abou a fied ais hrough G) ad F = 0, he equaios of moio are give b ΣF = 0, ΣF = 0, Σ M = I α...(.65) where ΣF = algebraic sum of he compoes of he eeral forces alog a ais chose as he -ais. ΣF = algebraic sum of he compoes of he eeral forces alog he -ais ΣM = algebraic sum of he momes of he eeral forces abou he ais of roaio hrough he mass ceer G (ais of smmer) I = mome of ieria of he bod abou he ais of roaio hrough he mass ceer G α = magiude of he agular acceleraio of he bod This pe of roaio is called ceroidal roaio..8 WORK AND ENERGY The work doe o move paricle or a bod from poi o poi b he resula force F acig o he paricle is give b U = F. dr...(.66) The kieic eerg T of a paricle wih mass m ad movig wih speed v is defied as mv. I coeced rigid bodies he oal kieic eerg of all he bodies a a cofiguraio is compued.

21 Egieerig Mechaics The work doe o a paricle b he resula force as i moves from poi o poi is equal o he chage i kieic eerg. U = T T...(.67) where T,T = iiial ad fial kieic eerg respecivel a pois ad. The kieic eerg T of a rigid-bod i raslaio is T = mv The kieic eerg T of a rigid-bod i roaio is T = I o ω where I O = mass mome of ieria of he bod abou he ais of roaio ω = agular speed The kieic eerg T of a bod i plae moio is give b...(.68) T = ω I where ω = agular speed...(.69) I = mome of ieria abou a ais hrough he mass ceer parallel o he z-ais. The chage i poeial eerg ma be defied as he egaive of he work doe b he coservaive force acig o he bod i brigig i from he daum o a fial posiio. The selecio of he daum is arbirar. The priciple of work ad eerg saes ha he work doe b he eeral forces acig o a rigid bod durig a displaceme is equal o he chage i kieic eerg of he bod durig he same displaceme. The sum of he work doe b he o-coservaive eeral forces such as fricio ad he work doe b he ieral forces acig o a ssem of paricles is equal o he chage i he oal (kieic ad poeial) eerg of he ssem of he paricles over he ime ierval of he acio. or E = T + V...(.70) where E = oal eerg (kieic ad poeial) of coservaive ssem T = kieic eerg V = poeial eerg The law of coservaio of eerg saes ha if a paricle (or bod) is aced upo b a coservaio force ssem, he sum of he kieic eerg ad poeial eerg is a cosa. Thus for o-coservaive ssem, priciple of work ad eerg is used ad for coservaive ssem, priciple of coservaio of eerg ca be emploed..9 IMPULSE AND MOMENTUM I paricles, here is ol liear momeum sice i has o agular moio. The liear momeum of he ih paricle is defied as L i = m i v i...(.7) where m i = mass of he ih paricle v i = ih paricle s veloci.

22 Solvig Egieerig Mechaics Problems wih MATLAB The liear momeum of he ssem of paricles is he sum of he liear momeum of he paricles. or L = L i...(.7) i= The liear impulse acig o he ssem impared over a ime ierval is give b G = F d...(.73) The he priciple of impulse ad momeum ca be wrie as: G = L L...(.74) where L p = liear momeum of he ssem a a sae p. Equaio (.74) saes ha he liear impulse acig o he ssem is equal o he chage i he ssem s liear momeum over he ime ierval. For a rigid bod, here is agular momeum ad agular impulse i addiio o liear couerpars. Agular impulse is creaed b a mome of a force while he agular momeum is due o ieria of he bod ad agular veloci of roaio. Agular momeum someimes called as he mome of he liear momeum is defied as: H o = r L...(.75) where H O is he agular momeum abou he poi O. The agular momeum of a ssem of rigid bodies abou a poi A is give b H = H = r L = r mv...(.76) A Ai i/ A i i/ A i i i= i= i= where A is a poi. The resula eeral mome abou he poi A is give b M A = MAi = ri /A F i...(.77) i= i= Differeiaig Eq. (.76) wih respec o ime ields H A = v A L + M A...(.78) If he poi A is fied he v A = 0 ad if he poi A is he mass ceer C he v A L = v C mv C = 0. Thus, wheher he poi A represes a fied poi O or he mass ceer C: H A = M A...(.79) Equaio (.79) saes ha he ime derivaive of he agular momeum abou a fied poi O (or he mass ceer O) is equal o he resula eeral mome abou he fied poi O (or he mass ceer C). Iegraig Eq. (.79) ields N A = d d ( ) ( ) M = H = H H = H H A A A A A A...(.80) where N A = M d agular impulse impared over he ime ierval A Hece N A = H A H A...(.8)

23 Egieerig Mechaics 3 Equaio (.8) saes ha he agular impulse acig o he ssem abou he fied poi O (or he mass ceer G) is equal o he chage i he ssem s agular momeum abou he poi O (or he mass ceer G) over he ime ierval. This is illusraed i Fig..9, where iiial ad fial velociies are ieres are ake a mass ceer G. G I G ω M d G + = G I G ω m(v G ) m(v G ) F d mg( ) F d Fig..9 Priciple of Impulse-momeum of rigid bod Coservaio of liear momeum i a give direcio occurs if he sum of he eeral forces i ha direcio is zero. For eamples i case of wo balls (paricles) collidig head-o wih each oher eiher cerall or obliquel, he liear momeum is coserved alog he lie of collisio. Mahemaicall, i is wrie as L p = 0. Coservaio of agular momeum abou a ais occurs if he sum of he momes of he eeral forces abou ha ais is zero. Mahemaicall i is wrie as: H p = 0. This occurs i cases where he roaig or oscillaig bod suffers differe speeds whe he poi of suspesio chages. As a eample, a diver jumpig from a heigh io waer maiais he cosa agular momeum durig his moio..0 THREE-DIMENSIONAL MECHANICS I hree dimesioal moio, he agular veloci ad acceleraio vecor has compoes i more ha oe ais, ulike i plae moio where for eample ω = ω ˆk a sigle compoe parallel o z-ais. Similar o wo dimesioal moio he moio of wo pois A ad B o a bod, or a series of coeced bodies ca be relaed usig relaive moio aalsis wih roaig ad raslaig aes a B. If a bod udergoes geeral moio, he he moio of a poi A i he bod ca be relaed o he moio of aoher poi B usig a relaive moio aalsis alog wih raslaig aes a B: v A = v B + ω r AB a A = a B + α r AB + ω (ω r AB ) There are hree scalar equaios of raslaioal moio for a rigid-bod ha moves i hree dimesios. F = m(a G ) F = m(a G ) F z = m(a Gz ) The hree scalar equaios of roaioal moio deped upo he locaio of he,, z referece. M = I ω (I I z ) ωω z M = I ω (I z I ) ωω z M z = I z ω z (I I ) ωω

24 4 Solvig Egieerig Mechaics Problems wih MATLAB Mos ofe, hese aes are orieed so ha he are he pricipal aes of ieria. If he aes are fied i ad move wih he roaio ω of he bod, he he equaios are referred o as Euler equaios of moio. The agular moio of a groscope is bes described usig he chages i moio of he hree Euler agles. These agular veloci compoes are he precisio φ, uaio θ ad spi ψ. If ψ = 0 ad φ ad θ are cosa he he moio is referred o as sead precisio. The agular veloci of he bod is specified ol i erms of Euler agle θ as: ω = ω i + ω j + ω z k = θ i + φsi θ j + ( φ cos θ+ψ ) k The spi veloci is give b Ω = Ω i + Ω j + Ω z k = θ i + φ si θ j + φ cos θ k I is spi of a gro roor ha is resposible for holdig he roor from fallig dowward ad isead causig i o precess abou a verical ais. This pheomeo is called he groscopic effec.