FE am w 2013 ics · ames & Machines F F AB AB F AB. ernal s B. ernal s. ernal s (3D) Load, , t s...
Transcript of FE am w 2013 ics · ames & Machines F F AB AB F AB. ernal s B. ernal s. ernal s (3D) Load, , t s...
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FE Exam Review 2013Statics
B i C PhDBrittany Coats, PhDMechanical Engineering Dept.
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Scalars have a magnitude
Vectors have a magnitude & direction
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Vector PrinciplesVector Principles
2 Vector Addition2. Vector Addition
RBA
Graphically identify R using the parallelogramlaw B
R ORA
RAA
R OR
B
R
BB
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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
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Solving for the ResultantSolving for the Resultant
Graphical Solution:Graphical Solution:law of sines:
CBAsinsinsin
law of cosines:
CBA
cos2222 ABBAC
B
AC
A
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Rectangular ComponentsRectangular Components
Fx = Fcos
Fy = Fsin
Fx & Fy are still vectors w/ a magnitude & direction.
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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
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Cartesian Vector �– 3DCartesian Vector 3D
jAiAA ˆˆ kA ˆjAiAA yx kAz
222zyx AAAA
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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
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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
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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
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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
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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
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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
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MomentsMoments
Moment arm
M = FdM0= Fd
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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).
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Calculating MomentsCalculating MomentsMo = r x Fo
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Cross ProductCross Product�• Always results in a vector.�• Resulting vector is always perpendicularto the two originating vectors.
ABuBABxA )sin( AB
A
BB
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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)(
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Resultant MomentsResultant Moments
F F
O
F1 F3
d
d3
O d1
F2d2
332211 dFdFdFMO 332211O
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Resultant MomentsResultant Moments
F F
O
F1 F3r1
r3O
F2
r2
)()()( 332211 FxrFxrFxrMO )()()(O
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Moment about AxisMoment about Axisaa FdM
)(ˆ FM )(ˆ FxruM aa
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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
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Equivalent Force Moment SystemsEquivalent Force Moment Systems
FFd
F1F1
M=F1d
O
1
FFF1
dFM
FF1
dFMO 1
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Distributed LoadsDistributed Loads
AdAdF )(xR
FR
AdAdxxwFALr )(
xdAdxxxw )(
R
A
A
L
LR dA
xdA
dxxw
dxxxwx
)(
)(
FR
AL
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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
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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
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Free body diagramsFree body diagrams
Find all forces acting at AFind all forces acting at A
TAD
AA
TACTAB
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Types of forces: weightTypes of forces: weight
W=mgparticleW=0W mg W=0
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Types of forces: cables/ropesTypes of forces: cables/ropes
C bl l i t iCables are always in tension
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Types of forces:bl f l llcables in frictionless pulleys
C bl l i t i
TT W
Cables are always in tension
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Types of forces: normal forcesTypes of forces: normal forces
WW
WW
l fnormal force
normal force
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Types of forces: springsTypes of forces: springs
F=ksF=ksk= spring constant
s=lengthfinal �– lengthinitial
+s:+s:
s:
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Reaction forcesReaction forcesReaction forces resist translation or rotation
Rx
Ryroller pinRy
RyRx
M
roller onangle
fixed supportRyangle
R
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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
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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
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EquilibriumEquilibriumExample:
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Free Body DiagramsFree Body DiagramsExample:
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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
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Solve for Reaction ForcesSolve for Reaction Forces
Axx
Ay Cy
Draw FBD of entire truss
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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
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Method of SectionsMethod of Sections
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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.
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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.
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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.
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Frames & MachinesFrames & Machines
F
FAB
FAB
FAB
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Internal ForcesInternal Forces
BB
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Internal ForcesInternal Forces
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Internal Forces (3D)Internal Forces (3D)
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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.
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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
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FrictionFriction�• A force that opposes motion or potentialmotion between two contacting surfaces
f
N
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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)
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Friction vs. Applied LoadFriction vs. Applied Load
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Friction ProblemsFriction Problems�• Equilibrium: What is the frictional force?
Check that Fa ANA and Fc CNC
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Friction ProblemsFriction Problems�• Impending motion at all contact points: Whatis the smallest angle that a ladder can beplaced along a wall?
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Friction ProblemsFriction Problems�• Impending motion at a single contact point:What is the minimum applied load thatneeded to cause motion anywhere?
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WedgesWedges�• Transform an applied force into a much largerforces at approximately a right angle to theapplied force.
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Flat Belt FrictionFlat Belt Friction
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Flat Belt FrictionFlat Belt Friction
eTT eTT 12
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Center of Mass
Center of AreaA A
AA
C t f LiCenter of a Line:L L
LL
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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
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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
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Centroids �– composite bodiesCentroids composite bodies
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Moments of Inertia (I)Moments of Inertia (I)
2dAyIx2
dAxI y2
dArJo2
o
IIJ yxo IIJ
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Parallel Axis TheoremParallel Axis Theorem
22' yxx AdII
2' xyy AdII
'' yxC IIJ y
2AdJJ CO AdJJ CO
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Radius of GyrationRadius of Gyration
AIk x
xAI
k yyA Ay
Jk o
Ako
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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!
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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