ENGR-36_Lec-17_Machines_H13e

Report
Engineering 36
Chp 6:
Machines
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Introduction: MultiPiece Structures
• For the equilibrium of structures made of several
connected parts, the internal forces as well the
external forces are considered.
• In the interaction between connected parts, Newton’s
3rd Law states that the forces of action and reaction
between bodies in contact have the same magnitude,
same line of action, and opposite sense.
• Three categories of engineering structures are
considered:
• Frames: contain at least one multi-force
member, i.e., a member acted upon by
3 or more forces.
• Trusses: formed from two-force members, i.e.,
straight members with end point connections
• Machines: structures containing moving parts
designed to transmit and modify forces.
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Analysis of Machines
• Frames and Machines are structures with at least one
multiforce member. Frames are designed to support loads
and are usually stationary. Machines contain moving parts
and are designed to transmit & modify forces.
• A free body diagram of the complete frame is used to
determine the external forces acting on the frame.
• Internal forces are determined by dismembering the
frame and creating free-body diagrams for each
component.
• Forces on two force members have known lines of
action but unknown magnitude and sense.
• Forces on multiforce members have unknown
magnitude and line of action. They must be
represented with two unknown components.
• Forces between connected components are equal,
have the same line of action, and opposite sense.
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Pin Equilibrium
 Consider Actions of
Members AD & CF
on Pin-C
 Member CF
• Pulls Pin RIGHT
• Pushes Pin UP
 Member AD
• Pulls Pin LEFT
• Pushes Pin Down
 FBD
AD
y
AD x
Engineering-36: Engineering Mechanics - Statics
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CF x
CF y
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Pin Equilibrium
 Note that Pin-C is in
Equilibrium
AD
y
CF x
AD x
 The forces on the
MEMBERS caused
by the Pin are Equal
and Opposite as
Predicted by
Newton’s 3rd Law
CF y
• By ΣFx = ΣFy = 0
AD
x
  CF
x
AD
y
  CF
y
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Machines
• Machines are structures designed to transmit
and modify forces. Their main purpose is to
transform input forces into output forces.
• Given the magnitude of P, determine the
magnitude of Q.
• Create a free-body diagram of the complete
machine, including the reaction that the
wire exerts.
• The machine is a nonrigid structure. Use
one of the components as a free-body.
• Taking moments about A,

Engineering-36: Engineering Mechanics - Statics
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M
A
 0  aP  bQ
Q 
a
P
b
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Example: CenterPull Brake
 For the Center-Pull
Bicycle Brake
shown at right find
Normal Force
Exerted by the
BrakePad on the
Rim
• Neglect the brakeopening Spring
Force
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Example: CenterPull Brake
 There is a Single
cable connecting
Pts B, G, & C
• Thus TGB = TGC
 The FBD for the
Cable Connector
 The Forces at the
connector are
CONCURRENT →
Particle Applies
 Take the ΣFy = 0
 T BGCy  T BGCy  200 N  0
T BGCy  T BGC sin 
• Note:

6 . 727
sin  
Opp
Hyp
Engineering-36: Engineering Mechanics - Statics
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
5 cm
 0 . 7433
6.727cm
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Example: CenterPull Brake
 The ΣFy = 0 at G
 T BGCy  T BGCy  200 N  0
 The FBD for Brake
Arm BAE
 2 T BGCy   200 N  0
T BGCy   200 N  2  100 N
 Use Trig & Geometry
to find TBCG
T BGC  T BGCy
sin 
T BGC  100 N 
6 . 727
5
T BGC  134 . 5 N
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Example: CenterPull Brake
 Find Fnormal by
ΣMA = 0
• Note
T BGCx  T BGC cos 
T BGCx  134 . 5 N 
4 .5
6 . 272
T BGCx  90 N
 Use F∙d to find
Moments about A
 T BGCy  4 . 5 cm   T BGCx  2 cm   F normal 9 cm
F normal 
 450
 18 0  N  cm


9 cm
9
F normal  70 N (15.7 lbs)
Engineering-36: Engineering Mechanics - Statics
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630
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
N

Example: CenterPull Brake
70 N
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
WhiteBoard Work
Let’s Work
This Nice
Problem
Big Ol’ Bolt Cutter
 For the Cutters with Applied Hand Force, P,
Find the Force Applied to the Gripped Bolt
Engineering-36: Engineering Mechanics - Statics
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[email protected] • ENGR-36_Lec-17_Frames.pptx
Engineering-36: Engineering Mechanics - Statics
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[email protected] • ENGR-36_Lec-17_Frames.pptx
Pliers (4-pc Structure)
 AB is TWO-Force Member
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Double Toggle Machine
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Engineering 36
Appendix
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Friction-Filled Bearing
 Consider NonFrictionless Bearing
r
O
F
Engineering-36: Engineering Mechanics - Statics
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 From the Diagram
• r ≡ Brg Radius
• F ≡ Brg Support
Force as determined
by Static Analysis
 If the Brg has nonnegligible Friction
the Moment that
OPPOSES rotation
M O  r  F
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx
Friction-Filled Bearing
 The Rotation
Resisting Moment:
M O  r  F
 Where
• µ ≡ the Bearing’s
COEFFICIENT of
FRICTION
 µ Discussed in
Detail in Chp08
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-17_Frames.pptx

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