ENGR-36_Lec-12_2D

Report
Engineering 36
Chp 5: 2D Equil
Special Cases
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
1
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
2D Equil → Special Cases
 PARTICLE: Size & Shape
of the Object can be
neglected as long as all
applied Forces have a
Point of Concurrency
• Covered in Detail in Chp03
 TWO-FORCE MEMBER: A
Structural Element of
negligible Wt with only 2
Forces acting on it
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
2D Equil → Special Cases
 THREE-FORCE MEMBER:
A structural Element of
negligible Wt with only 3
Forces acting on it
• The forces must be either
concurrent or parallel.
– In the PARALLEL Case the
PoC is located at Infinity
– The NONparallel Case can
be Very Useful in Load
Analysis
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
2D Equil → Special Cases
 FRICTIONLESS PULLEY: For a
frictionless pulley in static equilibrium,
the tension in the cable is the same on
both sides of the pulley
• Discussed Briefly in Chp03
– Will Prove the T1 = T2 = T Behavior Today
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
2D Planar System Equilibrium
 In 2D systems it is assumed that
• The System Geometry resides
completely the XY Plane
• There is NO Tendency to
– Translate in the Z-Direction
– Rotate about the X or Y Axes
 These Conditions Simplify The
Equilibrium Equations
F
x
F
0
Engineering-36: Engineering Mechanics - Statics
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y
0
M
z
0
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
2D Planar System:
F
x
0
F
y
0
M
z
0
 No Z-Translation → NO Z-Directed Force:
F
x
F
0
y
0
F
0
z
 No X or Y Rotation → NO X or Y
Applied Moments
M
x
0
M
Engineering-36: Engineering Mechanics - Statics
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y
0
M
z
0
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Special Case: 2-Force Member
 A 2-Force Member/Element is a Body
with negligible Weight and Only two
applied Forces.
 Some Special Properties of 2-Frc Ele’s
• the LoA’s of the Two Forces MUST Cross
and thus Produce a PoC
– Treat as a PARTICLE
• The Crossed LoA’s Define a PLANE
– Treat as PLANAR System
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
2-Force Element Equilibrium
 Consider a L-Bracket plate subjected
to two forces F1 and F2
 For static equilibrium, the sum of
moments about Pt-A must be zero.
Thus the moment of F2 About Pt-A
must be zero. It follows that the line of
action of F2 must pass through Pt-A
 Similarly, the line of action of F1 must
pass through Pt-B for the sum of
moments about Pt-B to be zero.
 Requiring that the sum of forces in
any direction be zero leads to the
conclusion that F1 and F2 must have
equal magnitude but opposite sense.
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Special Case: 2-Force Element
 Mathematically
• Since the Two Forces Must be Concurrent
M
PoC

r
PoC  F
F 
all F's
0F  0
all F's
• Since the System is in Equilibrium ΣF’s =0.
F 0F
A
 FB

FB   F A
all F's
– Thus the two force are Equal and Opposite;
that is, the forces CANCEL
Engineering-36: Engineering Mechanics - Statics
9
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Special Case: 3-Force Member
 A 3-Force Element is a PLANAR Body
with negligible Weight with Exactly 3
applied Forces (No applied Moments).
 Claim: If a Planar 3-Force Element is
in Equilibrium, Then the LoA’s for the
3-Forces must be CONCURRENT
• If the Claim is TRUE, then the 3-Force
Element can be treated as a PARTICLE
Engineering-36: Engineering Mechanics - Statics
10
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
3-Force 2D Body Equilibrium
 Consider a Planar rigid body subjected
to forces acting at only 3 points.
 The lines of action of intersect F1 &
F2, at Pt-D. The moment of F1 and F2
about this point of intersection is zero.
 Since the rigid body is in equilibrium,
the sum of the moments of F1, F2, and
F3 about ANY Pivot-Pt must be zero. It
follows that the moment of F3 about D
must be zero as well and that the line
of action of F3 must pass through D.
 The lines of action of the three forces
must be Concurrent OR Parallel.
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
3-Force 2D Body: Parallel Forces

x
d1
F1
O
d3
d2
If 3 Parallel Forces Maintain a
Rigid Body in Static
Equilibrium, The following
Conditions MUST be Satisfied
•
For Translation Equilibrium
F
•
F2
0

F3  F1  F 2
For Rotation Equilibrium
M
F3
O
0
 r  F   0
d 1 F1  d 3 F3  d 2 F 2
Engineering-36: Engineering Mechanics - Statics
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x
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx

Special Case: 3-Force Element
 Mathematically for ||-Forces
• Since a Body in Equil. Has NO Net Moment
M
O

r
O F
F 
all F' s
d
m
Fm  0
all F' s, d' s
• Since the System is in Equilibrium ΣF’s =0.
F 0 F
A
 FB  FC
all F's
• In Summary: The dmFm products and,
3 Forces, Sum to Zero
Engineering-36: Engineering Mechanics - Statics
13
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Special Case: 3-Force Element
 A Graphical Summary
AB is 3F Member
(BC is 2F Member)
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Example  Pole Raising

Solution Plan
•
Create a free-body diagram
of the joist.
–
•

A man Raises a 10 kg
Joist, of Length 4 m, by
pulling on a rope.
Find the TENSION in
the rope and the
REACTION at A.

Engineering-36: Engineering Mechanics - Statics
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Note that the joist is a 3
force body acted upon by
the ROPE, its WEIGHT, and
the REACTION at A
The three forces must be
concurrent for static
equilibrium. Therefore, the
reaction R must pass
through the intersection of
the lines of action of the
weight and rope forces.
Determine the direction of
the reaction force R.
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Example  Pole Raising

Use LoA’s & Trigonometry to
Determine the direction of the
reaction force R

Create a free-body
diagram of the joist
AF  BF  AB cos 45    4 m  cos 45   2 . 828 m
CD  AE 
1
2
AF  1 . 414 m
BD  CD cot( 45   25  )  1 . 414 m  tan 20   0 . 515 m
70

CE  BF  BD   2 . 828  0 . 515  m  2.313 m
tan  
CE

AE
  58 . 6
Engineering-36: Engineering Mechanics - Statics
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2 . 313
 1 . 636
1 . 414

A LARGE, SCALED
Diagram is REALLY
Useful in this
Problem
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
70º Angle Analysis
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Example  Pole Raising


Draw the Force
Triangle to Scale
Use the Law of the Sines to
Find the Reaction Force R
T
sin 31 . 4


R

sin 110

98 . 1 N

sin 38.6
Solving find
T  81 . 9 N
R  147 . 8 N
 = 58 .6 °
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx

Special Case: Frictionless Pulley
 A FrictionLess Pulley is Typically used
to change the Direction of a Cable or
Rope in Tension
Pulley with
PERFECT Axel
(FrictionLess)
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Special Case: FrictionLess Pulley
 A Perfect Axel Generates NO
moment to Resist Turning.
 Consider the FBD for a
Perfect Pulley
• Since the LoA’s for FAx & FAy
Pass Thru the Axel-Axis Pt-A
they Generate No moment
about this point .
• T1 and T2 have Exactly the
SAME Lever arm, i.e., the
Radius, R, of the Pulley
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Special Case: FrictionLess Pulley
 Since the Pulley is in
Equilibrium ΣMA = 0
 Writing the Moment Eqn
M
A

rT  0

all T's
 R  T1 kˆ    R  T 2  kˆ   0
or
R kˆ T
1
 T2   0
 Thus for the
NO-Friction
Perfect Pulley
T1  T 2
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
FritionFilled Pulley
 Consider the case where we have a
pulley that is NOT Free Wheeling; i.e.,
the pulley
resists rotation
 Example: Automobile
alternator changes
thermal-mechanical
energy into
electrical energy
T1
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
T2
FrictionFilled Pulley
 In Alternator
Operation the
generation of
electricity produces a
resisting moment
that counters the
direction of spin; The
FBD in this case →
MAz
 The ΣMA = 0
 R  T1 kˆ    R  T 2  kˆ   M A  kˆ   0
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
FrictionFilled Pulley
 Thus a RESISTING
Moment causes a
DIFFERENCE
between the two
Tensions
R T1  T 2   M
A
MAz
0
OR
T1  T 2  M
A
R
 More on This when we Learn Chp08
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
FrictionLess Pulley; 3F Mem
 In the System at
Right Member ABC,
which is a FOURForce System, can
be reduced to a
3-Force System
using and
Equivalent
Resultant-Couple
System at the Pulley
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
FrictionLess Pulley; 3F Mem
 Recall that Forces Can be MOVED to a
new point on a Body as long as the
Rotation Tendency caused by the move
is accounted for by the Addition of a
COUPLE-Moment at the new Point
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
FrictionLess Pulley; 3F Mem
 Apply the Equivalent
Loading Method to a
FrictionLess Pulley
T2
T1
T1  T 2
Engineering-36: Engineering Mechanics - Statics
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 From the Previous
Discussion the
MOMENT about the
Axle (Pin) of a
Frictionless pulley
produced by the
Tensions is ZERO
 Thus Can Move the
T’s to the Pin with a
Couple of ZERO
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
FrictionLess Pulley; 3F Mem
 The Equivalent Systems by MA = 0
T
T
T
T
Engineering-36: Engineering Mechanics - Statics
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TR
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
FrictionLess Pulley; 3F Mem
 Moving the
FrictionLess
Pulley ForceResultant to the
Pin at Pt-A
produces the
FBD Shown
At Right
TR
B
C
• Now can Draw the
Force Triangle
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
FrictionLess Pulley - Important
 For a FrictionLess Pulley the Tension
Forces and be to the Pulley Axel (Pin)
WithOUT the Addition of a Couple
T
T
T
Engineering-36: Engineering Mechanics - Statics
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=
T
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Special Cases Summarized
 Particle:
F
2D
3D
F
x
0
x
F
0
F
y
y
0
0
F
z
0
 2-Force Element: F B   F A
 3-Force Planar F  F  F  0
A
B
C
Element:
 FrictionLess Pulley:
Engineering-36: Engineering Mechanics - Statics
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T1  T 2
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
WhiteBoard Work
Lets Work
These Nice
Problems
Engineering-36: Engineering Mechanics - Statics
32
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Engineering 36
Appendix
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
33
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Jib Problem
 The upper portion of the
crane boom consists of
the jib AB, which is
supported by the pin
at A, the guy line BC,
and the backstay CD, each cable being separately
attached to the mast at C. If the 5-kN load is
supported by the hoist line, which passes over the
pulley at B, determine the magnitude of the resultant
force the pin exerts on the jib at A for equilibrium, the
tension in the guy line BC, and the tension T in the
hoist line. Neglect the weight of the jib. The pulley at
B has a radius of 0.1 m.
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
Disk Problem
 The smooth disks
D and E have a
weight of 200 lb
and 100 lb,
respectively.
Determine the largest horizontal force P
that can be applied to the center of disk
E without causing the disk D to move up
the incline.
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx

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