FE Exam Review 2013 Statics. Bi Brittany Coats, PhD Mechanical Engineering Dept.

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1 FEExamReview2013 Statics Bi BrittanyCoats,PhD MechanicalEngineeringDept.

2 Scalars haveamagnitude Vectors haveamagnitude&direction

3 VectorPrinciples 2. Vectorddition B R GraphicallyidentifyRusingtheparallelogram law B R OR R B B

4 VectorPrinciples 3. VectorSubtraction R B OR R ( B ) Sameasvectoraddition,but1ismultipliedto oneofthevectors B B R R

5 SolvingfortheResultant GraphicalSolution: lawofsines: lawofcosines: B sin C 2 2 C sin sin B C B 2 2B cos

6 RectangularComponents Fx =Fcos Fy =Fsin Fx &Fy arestillvectorsw/amagnitude&direction.

7 Cartesian Coordinates CartesianCoordinates Breaksdownthemagnitudeanddirectionofthevector j F F i F y x ˆ ˆ 2 2 F x F y F y x y F 1 t x y F 1 tan Bemindfulofpositive&negativeiandjdirections

8 CartesianVector 3D iˆ x y ˆ j k ˆ z 2 x 2 y 2 z

9 Cartesian Vector 3D CartesianVector 3D unit vector u unitvector,u u = = k j i z y x ˆ ˆ ˆ u = = k j i z y x z y x cos cos cos k j i u ˆ cos ˆ cos ˆ cos k j i u cos cos cos

10 Position Vectors PositionVectors z B(x B,y B,z B ) k z j y i x r ˆ ˆ ˆ (x y z ) r k z j y i x r kˆ ˆ ˆ r B y (x,y,z ) r r B k z j y i x r B B B B x y k z z j y y i x x r B B B B ˆ ˆ ˆ

Defineforcevector frompositionvector Definethevector,F,usingtheprovided magnitudeoftheforceandtheposition f d h ii vectortodeterminethedirection.

11 Defineforcevector frompositionvector Definethevector,F,usingtheprovided magnitudeoftheforceandtheposition f d h ii vectortodeterminethedirection. Magnitude=F(given) Direction=sameaspositionvectorfromtoB Step1:Findtheunitvectorfromthe positionvector u F u r r r Step2:Multiplyunitvectorbyforce magnitude F Fu r

12 Canbeusedto: DotProduct Findanglebetweentwovectors bt t t Findaforceperpendicular( )orparallel ( )toaline B B cos B

13 B B cos DotProduct i j =0 j k =0 k i =0 k j ii=1 jj=1 kk=1 i iˆ x y ˆ j kˆ z B B iˆ x B y ˆ j B z kˆ B ( iˆ ˆj k ˆ) ( B iˆ B ˆj B k ˆ) x y z x y z B B x x y B y z B z

14 DotProductpplications 1. nglebetweentwovectors: B B cos 2. Componentofavector toalineof to a (projectionofavectoronaline) cos u a a 3. Componentofvector toaline a sin 2 2 a a a

15 Moments Momentarm M 0 =Fd

16 z MomentDirection Userighthandrule x y +M:thumbpointsin directionofpositiveaxis(3d) ORthumbpointsoutofthe out of the page(2d). y x M:thumbpointsindirection ofnegativeaxis(3d)or thumbpointsintothepage (2D).

17 CalculatingMoments M o =rxf o

18 CrossProduct lwaysresultsinavector. Resultingvectorisalwaysperpendicular tothetwooriginatingvectors. x B ( B sin ) u B B

19 CrossProduct i x j = k j x k = i k x i = j k j i x i =0 j x j =0 k x k =0 i j x i = k k x j = i i x k = j x B ( B sin ) u B

20 ResultantMoments O d 1 F 1 F 3 d 3 d 2 F 2 M F d F d F d O

21 ResultantMoments O F 1 F 3 r r 3 1 r 2 F 2 M O ( r1xf 1 ) ( r2xf 2 ) ( r3xf 3 )

22 Momentaboutxis M a Fd a M u ˆ ( r x F ) a a

23 EquivalentCouple 2d F d 2F FF M=F(2d)=2Fd= 2Fd 2F M=(2F)d=2Fd= 2Fd Magnitude&directionofeachcouplearesame

24 EquivalentForceMomentSystems F 1 d F 1 O M=F 1 d 1 F 1 F F 1 M F d O 1

25 DistributedLoads F R x R F w x dx d r ( ) x R L L L xw ( x ) dx w( x) dx F R xd d

26 Equilibrium: objectatrestwillstayatrest t t t t t OR objectwithaconstantvelocity willstayatthatconstantvelocity t th t t t l it F=ma F=00 0

27 Equilibrium Drawfreebodydiagram(FBD) Sumupalltheforcesandsetequalto zero zero F=F = x i +F F y j+f F z k =0 Sumupallthemomentsandsetequalto zero zero M=Mx i+m y j+m z k =0 M x =0 M =0 y M z =0 F x =0 F =0 y F z =0

28 Freebodydiagrams diagrams Findallforcesactingat acting at T D T B T C

29 Typesofforces:weight forces: particle W=mg Wmg W=0

30 Typesofforces:cables/ropes forces: cables/ropes Cbl Cablesarealwaysintensionl i

31 Typesofforces: cablesinfrictionlesspulleys l ll T W T Cbl Cablesarealwaysintensionl i

32 Typesofforces:normalforces forces: forces W W normalforce normalforce

33 Typesofforces:springs forces: springs F=ks k=springconstant s=length final length initial +s: s:

34 Reactionforces Reactionforcesresisttranslationorrotation R x roller R y pin R y R x M R y fixedsupport R rolleron angle

35 Trick#1 TwoForce ForceMember Ifonlytwoforcesactonarigidbody,theymustbe equalandoppositeinmagnitude,andactalongtheand in and act along the samelineofaction

36 Trick#2 ThreeForce ForceMember Ifthreeforcesactonarigidbody,theymustbeeither concurrentorparallelor

37 Example: Equilibrium

38 Example: FreeBodyDiagrams

39 Trusses Loadsareappliedtothejoints Membersareattachedwithsmoothpins Weightofmembersisconsideredsmall is small comparedtoloadsbeingapplied Structuremakesthemrigidt th iid Eachmembermustbe undercompressionor tension

40 SolveforReactionForces x y C y DrawFBDofentiretruss

41 SolveforInternalForces UsingMethodofJoints B 500N F B F BC F B x F C x F BC C y y C y F C C y Createequilibriumequationsforeachjoint

42 MethodofSectionsof

43 ZeroForce ForceMembers Rule#1:Ifthereareonly2membersactingonajoint andnoexternalforces(appliedorreaction),thenthoseno or reaction), then those membershavezeroforce.

44 ZeroForce ForceMembers Rule#2:Ifthereareonly3membersactingonajoint(and noappliedorreactionforces)andtwomembersare collinear,thenthe3 rd memberhaszeroforce.

45 SolvingTrussProblems Seeifyoucansolvefortheforcesby inspection Zeroforcemembers Symmetry Easy joints Solveforreactionforces Usemethodofjointsiftheunknownsare nearaknownforce Usemethodofsectionsiftheunknowns arenotnearaknownforce.

46 Frames&Machines F B F B F B

47 InternalForces B

48 InternalForces

49 InternalForces(3D)

50 Load,Shear,MomentRelationships 10N 15N 11N 14N Shearchangesbythe magnitudeanddirectionof concentratedload. 11N 11N*m 1N 1N*m 14N 14N*m Momentchangeisequalto theareaundertheshear curve. Slopeofshearisequalto integralofload,slopeof momentisequaltointegral ofshear.

51 Load,Shear,MomentRelationships 10N 9N/m 11N 2N*m 17N dditionofamoment makesthemomentdiagram jumpbythemagnitude 11N 1N 1N 11N*m 9N Shearisintegralof 12N*m 11N N*m distributedload,momentis 10N*m

52 Friction forcethatopposesmotionorpotential motionbetweentwocontactingsurfaces f N

53 KineticFriction Ifappliedload,P,surpassesthelimitingstatic frictionalforce,frictionisreducedtothe kineticfrictionalforce. f k = k N ngleofkineticfriction: 1 k =tan 1 ( k )

54 Frictionvs.ppliedLoad

55 FrictionProblems Equilibrium:Whatisthefrictionalforce? CheckthatF a N andf c C N C

56 FrictionProblems Impendingmotionatallcontactpoints:What isthesmallestanglethataladdercanbe placedalongawall?

57 FrictionProblems Impendingmotionatasinglecontactpoint: Whatistheminimumappliedloadthat neededtocausemotionanywhere?

58 Wedges Transformanappliedforceintoamuchlarger forcesatapproximatelyarightangletothe appliedforce.

59 FlatBeltFriction Friction

60 FlatBeltFriction Friction T2 1 T e

61 CenterofMass Centerofrea CenterofaLine: L L L L

62 IdentifyingCentroids Line Pickyourdifferentialelement IdentifyarelationshipofdL todx anddy dl dx 2 dy 2 Identifyarelationshipofdx tody dx 2 ydy Plugintoourequationsfor xandycentroid

63 IdentifyingCentroids rea Pickyourdifferentialelement Identifyarelationshipofd todx ordy d xdy Plugintoourequations forxandycentroid

64 Centroids compositebodies

65 MomentsofInertia(I) of y 2 I 2 d x I x 2 d y J r 2 o d J o I x I y

66 Parallelxis xistheorem I x I d 2 x' y I y I d 2 y' x J C I I x' y' J O J C d 2

67 RadiusofGyration k x I x k y I y k o J o

68 MomentofInertiaof composite 1) Computemomentofinertiafor eachsegment. 2) ddupmomentofinertias BUT theymustbeaboutthe sameaxisinordertoaddthemup! axis in add them

69 MomentsofInertiaof Example2: Determinethemomentofinertiaofthecrosssectionalarea ofthechannelwithrespecttothexaxis

Calculating Truss Forces. Method of Joints

Calculating Truss Forces. Method of Joints Calculating Truss Forces Method of Joints Forces Compression body being squeezed Tension body being stretched Truss truss is composed of slender members joined together at their end points. They are usually