Transcript AE2120 Review Material

COE2001 Review Material
• Basic equilibrium equations are from Physics I
– Reinforce fundamental understanding of force & moments
– Techniques for applying to mechanical problems
• Applications
–
–
–
–
–
Rigid body problems
Trusses
Multi-force members: axial, shear forces & bending moments
Machines
Problems with friction (more complex equilibrium conditions)
• Calculation of centroids
– 1D (wires), 2D (areas) and 3D (volumes)
– Distributed forces and their resultants
COE 2001 Statics
Summer 2005
1
Particle Equilibrium
 1D  1equation

2 D  2 equations
 3D  3 equations

F2
F1
0   F   ( Fxiˆ  Fy ˆj  Fz
F3
Finite Body Equilibrium
0  F
F2
F1
M1
0   M (at any point,all points )
2 D  2 equations

 3D  3 equations
F3
where
M 0  rOA  FA
M 0  rOA FA sin 
O
FA
rOA
A
d

• FA can act anywhere along
“line of action”
• +θ rotates rOA into FA
 FA d
and
M O  rOA & M O  FA
COE 2001 Statics
Summer 2005
2
Moments
1. Moment of a force about a line (e.g., a hinge) is the projection of M on to the line.
M  (M  uˆ ) uˆ  ((r  F )  uˆ ) uˆ
2. Only the component of F perpendicular to
û
contributes to the moment.
3. M due to F is equal to M due to all components of F.
M  r  ( FA  FB )  r  FA  r  FB  M A  M B
4. To compute M @ Q given M @ P and Fi use:
M
Q
 rQP   Fi   M P
5. Couple: moment developed by equal and opposite forces separated by d; direction is perpendicular to both F
and d. Since SFi=0 we see that SMQ=0+MP so a couple produces a moment that is constant
Equipollent Systems
Two force & moment systems are equipollent (“equal power”) if they have the same resultant at any point.
Notes:
1. Resultant is defined as SF and SM
2. For rigid bodies, equipollent systems produce the same results on the body
3. Must satisfy:
 F    F 
 M    M 
i
P
COE 2001 Statics
i
1
1
2
P
for any (and all) points, P, so we can pick any point for test and making equivalent
2
Summer 2005
3
FR
Simplest Resultant
Definition: an equipollent system
consisting of only a force and NO moment.
Exists for:
1.
2.
3.
4.
Q
Q
P
P
Concurrent forces (where MP=0)
Co-planar forces
Parallel forces
3D: Screwdriver (wrench) concept
?
Q
MR
P
Want M Q  0, but M P  M Q  rPQ  FR
so : M P  0  rPQ  FR
Must find rPQ as solution to above eqn
Distributed Loads
1.
2.
3.
4.
FR
Forces per unit volume: gravity, acceleration, electromagnetic…
Forces per unit area: pressures
Forces per unit length: line loads in 2D problems
Concentrated forces: defined in the limit as area → 0
Line Loads:
Simplest resultant :
q(x)
r0x
Mr0
x
dx
Ex:
Fr
x
b/2
q0
Fr
q0b
M ro  rox  Fr
x
General :
L
x
q0
b
x
xr 
b/3 ½q b
0
x
x
 x q( x) dx
0
L
 q( x) dx
0
b
COE 2001 Statics
Summer 2005
4
General Equilibrium
0  F
0   M (at any point,all points )
Procedure:
1. Idealize loads, constraints and define reactions
2. Determine solvability:
- determinate or indeterminate
- external vs internal redundancy
Notes:
1. 3D problem: 6 equations
2D problem: 3 equations
2. # unknowns = # equations → determinate (SOLVABLE)
# unknowns > # equations → indeterminate (many solutions)
# unknowns < # equations → mechanism
3. Free body diagram (FBD) to reveal and isolate forces &
moments
4. Equilibrium must apply to ANY and to ALL FBD’s
5. Constraints: create reactions
(pin, slot, normal force, rotating disk, rope & pulley, etc)
3. Create appropriate FBD’s
4. Solve equilibrium equations (Note: FBD’s
simply help you decouple and solve equilibrium
equations in the simplest way. This is not a
unique process and many ways are possible.
Use all of the above techniques for working
with forces and moments.)
Applications:
1. Bars: 1D prismatic members supporting only coaxial forces
through end pins
- forces & loads are applied only at ends
- no lateral (transverse) loads
2. Beams: 1D prismatic members carrying lateral loads
- loads applied in a single plane (planar bending)
- beam-column combines beam and bar behavior
COE 2001 Statics
Summer 2005
5
Trusses (application of bars)
A
FAB
B
B
FBD
C
D
Bar
Joint
+ Tension
Tension = away
-Compression
Compression = towards
This eqn. is a
necessary but not
sufficient condition
FBC
Q1
Solvability
Equations:
Unknowns:
Reactions:
Q2
2J (3D = 3J)
M (# members)
R
M+R=2J (=3J)
= determinate
> indeterminate (redundants exist)
< mechanism
Method of Joints
Method of Sections (use when only one or two member forces are needed)
1. Label all joints
1. Label all joints
2. Find external reactions
2. Cut to construct a FBD to expose bar
forces and eliminate other unknown
forces and reactions
3. Formulate a joint equilibrium equation
4. Solve for unknown member force
3. Solve for bar force (usually using 0=SM)
5. Repeat from #3 until done…
Tips
1. Force can act anywhere along its line of action
(pick location to avoid added moment calculations).
2. To avoid including a bar force, sum forces in
direction  to that bar force.
3. Consider using virtual points for 0=SM to eliminate
bar forces acting through that point.
COE 2001 Statics
Summer 2005
6
Centroids & Centers of Mass
Basic definition of centroid:
Q rOC   r dQ
where Q is either: (a) volume (V), (b) area (A), or (c) line (L)

and Q   dQ

Area:
y=f(x)
y=f(x)
y2
y
y=g(x)
A x   x dA   x( f ( x)  g ( x))dx
y
where A   dA   ( y2  y1 )dx
dy
x1
where A   dA   ( x2  x1 )dy
x2
y1
x
dx
y=f(x)
A y   y dA   y ( x2  x1 )dy
y=g(x)
x
Alternate method for computing y-bar:
y2
y
A y   y * dA   12 ( y2  y1 )dA   12 ( y2  y1 )( y2  y1 )dx   12 ( y2 2  y12 )dx
y=g(x)
y*
where A   dA   ( y2  y1 )dx
y1
x
dx
Composite Areas:
y
A1
x
A2
A3
x A   xi Ai  x1 A1  x2 A2  x3 A3
y A   yi Ai  y1 A1  y2 A2  y3 A3
y
x
xi , yi are locations of centroids of component areas
COE 2001 Statics
Summer 2005
7
Friction
W
Static friction: Fm=msN
P
Fm
F
Fs
Fk
Kinetic friction: Fm=mkN
P
N
Model:
• Friction force depends only on surfaces (m)
and normal force (N)
• Magnitude cannot exceed Fs before motion
and Fk during motion
• Direction always opposes motion (opposite to
relative velocity)
• Will lock up if force magnitude drops below Fk
Modes:
1. P=0: no friction force is developed; no motion
2. P=F<Fm: no motion but F is whatever is needed to maintain equilibrium (cannot use F=mN to compute F)
3. P=Fm=msN: motion is “impending” (use F=msN to compute friction force)
4. P>Fm: motion is present (problem must now include dynamic forces)
Angle of Friction
W
W
P
P
N
Fm
tan fs 
fs
R
Fm
 ms
N
Use:
• Replace Fm and N with R acting at angle fs from N
• Knowing ms allows calculation of fs
• Particle equilibrium problem: R acts on line of action
defined by fs
• Valid at instant of impending motion only
General Solution Approach:
Case A:
All forces are specified and all reactions & internal forces to maintain equilibrium are computed. If any forces
that must develop from friction are above Fm in magnitude, then motion occurs, otherwise we have a simple
static solution. Note that this means F<Fm=msN.
Case B:
Motion is impending and we find critical values for ms or for the geometry of the problem (angles, lengths)
Case C:
Determine valuse of a load necessary to cause motion. For this case, we must choose directions for friction
forces (opposing impending motion) and at least one friction force is at the limit value.
COE 2001 Statics
Summer 2005
8