FE Exam Review 2013Statics

B i C PhDBrittany Coats, PhDMechanical Engineering Dept.

brittany.coats@utah.edu

Scalars have a magnitude

Vectors have a magnitude & direction

Vector PrinciplesVector Principles

2 Vector Addition2. Vector Addition

RBA

Graphically identify R using the parallelogramlaw B

R ORA

RAA

R OR

B

R

BB

Vector PrinciplesVector Principles

3 Vector Subtraction3. Vector Subtraction

ABR ABR )(OR

Same as vector addition, but 1 is multiplied to

)(

one of the vectorsB

RB

AR

Solving for the ResultantSolving for the Resultant

Graphical Solution:Graphical Solution:law of sines:

CBAsinsinsin

law of cosines:

CBA

cos2222 ABBAC

B

AC

A

Rectangular ComponentsRectangular Components

Fx = Fcos

Fy = Fsin

Fx & Fy are still vectors w/ a magnitude & direction.

Cartesian CoordinatesCartesian CoordinatesBreaks down the magnitude and direction of the vector

jFiFF yxˆˆ

22yx FFF yx

yF1tx

y

F1tan

Be mindful of positive & negative i and j directions

Cartesian Vector �– 3DCartesian Vector 3D

jAiAA ˆˆ kA ˆjAiAA yx kAz

222zyx AAAA

Cartesian Vector �– 3DCartesian Vector 3D

unit vector uAunit vector, uA

Au = = kAj

AiA zyx ˆˆˆ

AuA= = k

Aj

Ai

Azyx

AA

AA

AA zyx coscoscos

AAA

kjiu ˆcosˆcosˆcos kjiuA coscoscos

Position VectorsPosition Vectorsz

B (xB, yB, zB)kzjyixrA ˆˆˆ

A (x y z ) r

kzjyixr AAAA

k̂ˆˆrAB

y

A (xA, yA, zA)

rA

rBkzjyixr BBBB

x

y

kzzjyyixxr ABABABABˆˆˆ

Define force vectorffrom position vector

Define the vector, F, using the providedi d f h f d h i imagnitude of the force and the position

vector to determine the direction.

Magnitude = F (given)g (g )Direction = same as position vector from A to B

Step 1: Find the unit vector from the

rruu rF

position vector

rStep 2: Multiply unit vector by forcemagnitude

ruFF

Dot ProductDot Product

Fi d l b t t tCan be used to:

�• Find angle between two vectors�• Find a force perpendicular ( ) or parallel(||) to a line

cosBABA cosBABA

AA

B

Dot ProductcosBABA

Dot Productki j = 0 i i = 1

j

jj k = 0k i = 0

j j = 1k k = 1

i

kAjAiAA ˆˆˆ kBjBiBB ˆˆˆkAjAiAA zyx kBjBiBB zyx

ˆˆ )ˆˆˆ()ˆˆˆ( kBjBiBkAjAiABA zyxzyx

zzyyxx BABABABA

Dot Product Applications1. Angle between two vectors:

Dot Product Applications

2. Component of a vector || to a linecosBABA

2. Component of a vector || to a line(projection of a vector on a line)

uAAA cos3. Component of vector to a line

aa uAAA cos||

sinAA a

22AAA 2||aa AAA

A||

a

MomentsMoments

Moment arm

M = FdM0= Fd

Moment DirectionMoment Directionz Use right hand rule

+M: thumb points indirection of positive axis (3D)OR thumb points out of the

y

x

OR thumb points out of thepage (2D).

x

M: thumb points in directiony

of negative axis (3D) ORthumb points into the page(2D)

x

(2D).

Calculating MomentsCalculating MomentsMo = r x Fo

Cross ProductCross Product�• Always results in a vector.�• Resulting vector is always perpendicularto the two originating vectors.

ABuBABxA )sin( AB

A

BB

Cross ProductCross Productki x j = k i x i = 0

j

jj x k = ik x i = j

j x j = 0k x k = 0

ij x i = kkk x j = ii x k = j

ABuBABxA )sin( AB)(

Resultant MomentsResultant Moments

F F

O

F1 F3

d

d3

O d1

F2d2

332211 dFdFdFMO 332211O

Resultant MomentsResultant Moments

F F

O

F1 F3r1

r3O

F2

r2

)()()( 332211 FxrFxrFxrMO )()()(O

Moment about AxisMoment about Axisaa FdM

)(ˆ FM )(ˆ FxruM aa

Equivalent CoupleEquivalent Couple2F

Fd2d

2FFF

M=F(2d) = 2Fd M=(2F)d = 2FdM=F(2d) = 2Fd M=(2F)d = 2Fd

Magnitude & direction of each couple are same

Equivalent Force Moment SystemsEquivalent Force Moment Systems

FFd

F1F1

M=F1d

O

1

FFF1

dFM

FF1

dFMO 1

Distributed LoadsDistributed Loads

AdAdF )(xR

FR

AdAdxxwFALr )(

xdAdxxxw )(

R

A

A

L

LR dA

xdA

dxxw

dxxxwx

)(

)(

FR

AL

Equilibrium:Equilibrium:

bj t t t ill t t tobject at rest will stay at restOR

object with a constant velocityill t t th t t t l itwill stay at that constant velocity

F0

F = ma

F 0F = 0

EquilibriumEquilibrium�• Draw free body diagram (FBD)�• Sum up all the forces and set equal tozero F = F i + F j+ F k = 0zero

�• Sum up all the moments and set equal tozero

F = Fxi + Fyj+ Fzk = 0

zeroM = Mxi+ Myj+ Mzk = 0

Mx = 0My = 0

Fx = 0Fy = 0y

Mz = 0yFz = 0

Free body diagramsFree body diagrams

Find all forces acting at AFind all forces acting at A

TAD

AA

TACTAB

Types of forces: weightTypes of forces: weight

W=mgparticleW=0W mg W=0

Types of forces: cables/ropesTypes of forces: cables/ropes

C bl l i t iCables are always in tension

Types of forces:bl f l llcables in frictionless pulleys

C bl l i t i

TT W

Cables are always in tension

Types of forces: normal forcesTypes of forces: normal forces

WW

WW

l fnormal force

normal force

Types of forces: springsTypes of forces: springs

F=ksF=ksk= spring constant

s=lengthfinal �– lengthinitial

+s:+s:

s:

Reaction forcesReaction forcesReaction forces resist translation or rotation

Rx

Ryroller pinRy

RyRx

M

roller onangle

fixed supportRyangle

R

Trick #1 �– Two Force MemberTrick #1 Two Force MemberIf only two forces act on a rigid body, they must beequal and opposite in magnitude and act along theequal and opposite in magnitude, and act along the

same line of action

Trick #2 �– Three Force MemberTrick #2 Three Force MemberIf three forces act on a rigid body, they must be either

concurrent or parallelconcurrent or parallel

EquilibriumEquilibriumExample:

Free Body DiagramsFree Body DiagramsExample:

TrussesTrusses�• Loads are applied to the joints�• Members are attached with smooth pins�• Weight of members is considered small�• Weight of members is considered smallcompared to loads being appliedSt t k th i id�• Structure makes them rigid

�• Each member must beunder compression ortension

Solve for Reaction ForcesSolve for Reaction Forces

Axx

Ay Cy

Draw FBD of entire truss

Solve for Internal ForcesSolve for Internal ForcesUsing Method of Joints

B 500 N

FBA

FBA FBC A

BA

FACAx

Ax

Ay

C

FBC

x

Ay Cy

C

FAC

Cy

Create equilibrium equations for each joint

Method of SectionsMethod of Sections

Zero Force MembersZero Force Members

Rule #1: If there are only 2 members acting on a jointand no external forces (applied or reaction), then thoseand no external forces (applied or reaction), then those

members have zero force.

Zero Force MembersZero Force Members

Rule #2: If there are only 3 members acting on a joint (andno applied or reaction forces) and two members arepp )

collinear, then the 3rd member has zero force.

Solving Truss ProblemsSolving Truss Problems�• See if you can solve for the forces byinspection�– Zero force members�– Symmetry�– Easy �‘joints�’

�• Solve for reaction forces�• Use method of joints if the unknowns arenear a known force

�• Use method of sections if the unknownsare not near a known force.

Frames & MachinesFrames & Machines

F

FAB

FAB

FAB

Internal ForcesInternal Forces

BB

Internal ForcesInternal Forces

Internal Forces (3D)Internal Forces (3D)

Load, Shear, Moment Relationships

Shear changes by the15 N10 N

g ymagnitude and direction ofconcentrated load.14 N11 N

11 N

14 N1 N Moment change is equal tothe area under the shearthe area under the shearcurve.

1 N*m

Slope of shear is equal tointegral of load, slope of

11 N*m 14 N*m

moment is equal to integralof shear.

Load, Shear, Moment Relationships

Addition of a moment9 N/m10 N

makes the moment diagramjump by the magnitude17 N11 N

2 N*m

11 N1 N 1 N

9 N Shear is integral ofdistributed load, moment isi l f h

12 N*m11 N*m integral of shear.11 N*m

10 N*m

11 N m

FrictionFriction�• A force that opposes motion or potentialmotion between two contacting surfaces

f

N

Kinetic FrictionKinetic Friction�• If applied load, P, surpasses the limiting staticfrictional force, friction is reduced to thekinetic frictional force.

fk = kN

1( )Angle of kinetic friction:

k = tan 1( k)

Friction vs. Applied LoadFriction vs. Applied Load

Friction ProblemsFriction Problems�• Equilibrium: What is the frictional force?

Check that Fa ANA and Fc CNC

Friction ProblemsFriction Problems�• Impending motion at all contact points: Whatis the smallest angle that a ladder can beplaced along a wall?

Friction ProblemsFriction Problems�• Impending motion at a single contact point:What is the minimum applied load thatneeded to cause motion anywhere?

WedgesWedges�• Transform an applied force into a much largerforces at approximately a right angle to theapplied force.

Flat Belt FrictionFlat Belt Friction

Flat Belt FrictionFlat Belt Friction

eTT eTT 12

Center of Mass

Center of AreaA A

AA

C t f LiCenter of a Line:L L

LL

Identifying Centroids LineIdentifying Centroids Line

�• Pick your differential element

�• Identify a relationship of dL to dx and dy

y

22 dydxdL

�• Identify a relationship of dxto dy

ydydx 2�• Plug into our equations forx and y centroid

Identifying Centroids AreaIdentifying Centroids Area�• Pick your differential element�• Identify a relationship of dA to dx or dy

xdydA xdydA

�• Plug into our equationsg qfor x and y centroid

Centroids �– composite bodiesCentroids composite bodies

Moments of Inertia (I)Moments of Inertia (I)

2dAyIx2

dAxI y2

dArJo2

o

IIJ yxo IIJ

Parallel Axis TheoremParallel Axis Theorem

22' yxx AdII

2' xyy AdII

'' yxC IIJ y

2AdJJ CO AdJJ CO

Radius of GyrationRadius of Gyration

AIk x

xAI

k yyA Ay

Jk o

Ako

Moment of Inertia compositeMoment of Inertia composite1) Compute moment of inertia for

each segment.2) Add up moment of inertias�…) p

BUT�…they must be about thesame axis in order to add them up!same axis in order to add them up!

Moments of InertiaMoments of Inertia

Example 2:Example 2:

Determine the moment of inertia of the cross sectional areaof the channel with respect to the x axis