fe statics 2013
TRANSCRIPT
FE Statics Review
Sanford Meek Department of Mechanical Engineering
Kenn 226 (801)581-8562 [email protected]
Statics
forces = 0!
moments = 0!
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Vectors
Scalars have magnitude only Vectors have magnitude and direction
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Vectors
Vector addition and subtraction !A+!B =!R
!R!!B =!A
A
-B
R
A
B
R
B
A
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Vectors
Resultants
sin!A
=sin"B
=sin#C C
A
B
Law of sines
Law of cosines
C2 = A2 +B2 ! 2ABcos!
α
β
γ
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Vectors
Cartesian coordinates
The only system that is generally used in statics
x, y, z axes i, j, k unit vectors
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Vectors
2 2 2x y zA mag A A A= = + +A
!
ˆ ˆ ˆ ˆyx zAA A
A A A A! "! " ! "= = + +# $# $ # $
% & % &% &
AA i j k!
ˆ ˆ ˆx y zA A A= + +A i j k
!
ij
k A
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Direction cosines
ˆ ˆ ˆ ˆyx zAA A
A A A A! "! " ! "= = + +# $# $ # $
% & % &% &
AA i j k!
The components of the UNIT vector,
are the cosines of the angles made by the vector with the coordinate axes.
cos
cos
cos
x
y
z
AAAAAA
!
"
#
=
=
=
2 2 2cos cos cos 1! " #+ + =8
Position Vectors
!rA = xAi + yA j+ zAk!rB = xB i + yB j+ zBk
!rAB = xB ! x( ) i + yB ! yA( ) j+ zB ! zA( ) k
Final point (head of arrow) – initial point (tail of arrow)
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Force Vectors Simply the force magnitude times the unit position vector
ˆ ˆ ABAB
AB
F F Fr
= = =rF F r!!
ˆ( 5')ˆ( 3'cos20 )ˆ(3'sin 20 6')
AB = !
+ ! °
+ ° +
r i
j
k
!
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Dot Product Use: find the component of a vector along a certain direction – component of force along a particular axis
cos
x x y y z z
AB
A B A B A B
!=
= + +
A BA B
! !i! !i
A!
!
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Dot Product Use: find the component of a vector along a certain direction – component of force along a particular axis The result is a scalar.
cos
x x y y z z
AB
A B A B A B
!=
= + +
A BA B
! !i! !i
A!
!
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Dot Product
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What part of A is aligned with the line? What part is perpendicular?
cos ˆa aA !=A u!
ˆ au!A
!=!A"!Aa
ˆcos aA ! = A u!i
Projected component (scalar):
=perpendicular part
Moment (Torque)
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Only the perpendicular part of force gives torque
Cross Product
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IEEE logo
Remember Right Hand Rule
Cross Product
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In terms of i, j, k
i x j = k j x k = i k x i = j
j x i = -k k x j = -i i x k = -j
i x i = 0 j x j = 0 k x k = 0
i ⇒j ⇒k ⇒i positive
i ⇐ j ⇐ k ⇐ i negative
Moment of a Force
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posiBon vector from O to any point on the line of acBon of F
Primary use of the cross product in statics
Moment About an Axis
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Triple scalar product
In this application, axis “a” is the y-axis, making unit axis vector ua simply base vector j.
Resultant Moments
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Resultant Moments
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Force Couples
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answers.com
Equal and opposite forces acting at a distance
No net force
Force Couples
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These all have the same net moment!
Q: Why is this a useful observaBon? A: It speeds up computaBons. Look for couples!
Force Couples
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d r!
M Fd= !
=
rM F! !!vector connecting any two
points on lines of actionperpendicular distance between fo
The force at th
r
e
ces
tip of
d
=
=
=
r
F r!
!
!
Equivalent Force-Moment Systems
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PRINCIPLE OF TRANSMISSIBILITY
P
F
P F
d
M=Fd F
For both of these, the equivalent load at P is the same…
You can slide any force along its line of action without affecting the equivalent load!
Equivalent Force-Moment Systems
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An equivalent shifted force always exists for a set of coplanar forces. The net force for coplanar forces is itself in the plane. The moment for coplanar forces is perpendicular to the plane.
For these, you can always find a location to place the net force such that the moment about “O” is unchanged.
( )R OM
dF
=
Distributed Loads
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Demand the moment to be the same:
Equilibrium
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forces = 0!
moments = 0!
No acceleration – at rest or constant velocity
Equilibrium
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Free Body Diagram
Define coordinate system Write all known and unknown forces and moments in terms of the coordinate system
Sum all forces to zero Sum all moments to zero
Support Reactions
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zero horizontal resistance. has vertical resistance. zero moment resistance.
has horizontal resistance. has vertical resistance. zero moment resistance.
has horizontal resistance. has vertical resistance. has moment resistance.
Two-Force Members
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A two-‐force member is a body or part that is subjected to exactly two forces, and no moments. The equilibrium equaBons then imply that those two forces must be equal in magnitude, opposite in direcBon, and be collinear (i.e., act along the same line of ac4on.)
Trusses
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Usually, one support is a roller to allow expansion and contraction from temperature changes
• A structure composed of two-force members – tension or compression
• Loads are applied only at the joints • Joints are frictionless pins • Member weight is insignificant
Trusses
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Method of Joints – FBD at each joint (pin joints)
Method of joints: apply force balance at pins. Since force balance has only two equaBons (ΣFx=0 and ΣFx=0), pick joints that have only two unknown forces!
Force balance gives: C T
T
As part of your checks, first guess tension “T” or compression “C.”
Trusses Method of Joints – FBD at each joint (pin joints)
Joint (pin) FBD has the applied loads and link loads. Link (spar or element) FBD has only axial load! Links must be two-‐force members. (thus equal and opposite forces)
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Trusses Method of sections – make a ‘cut’ in the structure
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Trusses Method of sections: Make cuts with no more than 3 unknowns.
Apply 3 statics equations (ΣFx=0, ΣFy=0, ΣM=0) to get the 3 unknowns!
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Trusses Steps for solving truss problems
• Find the reactions by looking at the truss as a single rigid body
• Use the method of joints if the unknown forces to be solved are near or at the joint with a known (applied) force.
• Use the method of sections if the unknown forces to be solved are not at or near the known (applied) forces.
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Frames Not necessarily two-force members – loads can be applied anywhere on themembers
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Frames Internal forces
They don’t appear in this FBD:
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Internal Forces
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Make a virtual cut in a member and analyze the internal forces
Internal Forces
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Sign conventions
Internal Forces
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Load, shear, and moment relationships
• Shear is the integral of the load
• Moment is the integral of the shear
• Beam slope is the integral of the moment
• Beam deflection is the integral of the slope
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Friction
configuration matters!
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Friction
F P=( )F Nµ!
N W=
To be in equilibrium, friction is however large it needs to be treated as a pin support at an unknown location
moment balance:
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Friction
Frictional limit: At some point, P will reach a peak value, Fs, large enough to start dragging the block.
s sF Nµ=
Impeding motion
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Friction
F = µsN
F = µkN
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Friction
sF Nµ!( only at impending motion)sF Nµ=
Methodology: assume that fricBon is a pin support located at an unknown posiBon. Find these three unknowns (horizontal and verBcal pin reacBons, as well as pin locaBon) by enforcing three equaBons of equilibrium. Validate “no slip” assumpBon by confirming that F < µsN.
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Friction To slip, both ends must slip, which implies that BOTH ends will be at impending slip. Seing FA= µANA and FB= µBNB, the FBD has THREE unknowns: NA, NB, and angle θ. Use the THREE equilibrium equaBons to solve for them! Answer:
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Wedges
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Screws Screws are wedges around a cylinder
How many threads?
Lead = Number of threads x Pitch
Pitch = 1/(number of threads per inch)
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Screws
Upward Motion Self locking
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Screws Downward Motion
Non-‐self-‐locking
FricBon angle is smaller than the thread (wedge) angle
The moment must resist the moBon
Self-‐locking
FricBon angle is bigger than the thread (wedge) angle
The moment must start the moBon
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Belts
Tension is different at each end because of fricBon! (Contrast with a pully)
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Belts
dT dT
µ !" =
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Belts Tension increases in the direction of impending motion
2 1( )2
1
T eT
µ ! !"= 2!1!
If fricBonless (µ=0), then T2= T1 (like pulley). With fricBon, however, T2> T1 (consistent with limit µ→∞).
Usually written as: where β is the angle of contact
Moment of Inertia
about an axis polar moment of inerBa
Moment of Inertia Parallel axis theorem
Use this if you know Moment of Inertia about the primed axes passing thru the centroid, but you seek it with respect to some different set of parallel axes.
!
d2 = dx2 + dy
2
Moment of Inertia Radius of Gyration
The equivalent moment of inertia of a body that is a point mass at a distance (the radius of gyration) from the axis.
!
Ix = Akx2
kx =IxA
!
Iy = Aky2
ky =IyA!
Io = Ako2
ko =IoA
Moment of Inertia
Moment of inertia for Composite Areas:
Add them together (subtract for holes).
Make sure that all the moments of inertia for all parts are found about the same axis.
Use tables and parallel axis theorem.
Centroids - Lines
Centroids - Areas
Centroids – Composite Bodies