Transcript ENGR-36_Lec-12_2D
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
2 Forces acting on it
Engineering-36: Engineering Mechanics - Statics
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
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3 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 both sides
of the pulley on • Discussed Briefly in Chp03 – Will Prove the T 1 = T 2 = T Behavior Today
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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
0
F y
0
M z
0
Engineering-36: Engineering Mechanics - Statics
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2D Planar System:
F x
0
F y
0
M z
0 No Z Translation → NO Z-Directed Force:
F x
0
F y
0
F z
0 No X or Y Rotation → NO X or Y Applied Moments
M x
0
M y
0
M z
0
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6 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
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2
-Force Element Equilibrium
Consider a L-Bracket plate subjected to two forces F 1 and F 2 For static equilibrium, the sum of moments about Pt-A must be zero. Thus the moment of F 2 About Pt-A must be zero. It follows that the line of action of F 2 must pass through Pt-A Similarly, the line of action of F 1 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 F 1 and F 2 must have equal magnitude but opposite sense.
Engineering-36: Engineering Mechanics - Statics
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Special Case: 2-Force Element
Mathematically • Since the Two Forces Must be Concurrent
M
PoC
all F's
r
PoC
F
F
all F's 0
F
0 • Since the System is in Equilibrium Σ
F
’s =0. all F's
F
0
F
A
F
B
F
B
F
A
– Thus the two force are Equal and Opposite; that is, the forces CANCEL
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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
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3
-Force 2D Body Equilibrium
Consider a Planar rigid body subjected to forces acting at only 3 points.
The lines of action of intersect F 1 & F 2 , at Pt-D. The moment of F 1 and F 2 about this point of intersection is zero.
Since the rigid body is in equilibrium, the sum of the moments of F 1 , F 2 , and F 3 about ANY Pivot-Pt must be zero. It follows that the moment of F 3 about D
must be zero
as well and that the line of action of F 3 must pass through D.
The lines of action of the three forces must be
Concurrent Engineering-36: Engineering Mechanics - Statics
OR
Parallel
.
11 Bruce Mayer, PE [email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
F
2
3
-Force 2D Body: Parallel Forces
F
1 d 1
O
d 2
x
d 3 If 3 Parallel Forces Maintain a Rigid Body in Static Equilibrium, The following Conditions MUST be Satisfied • For Translation Equilibrium
F x
0
F
3
F
1
F
2
F
3 • For Rotation Equilibrium
M
O
0
r
F
d
1
F
1
d
3
F
3
d
2
F
2 0
Engineering-36: Engineering Mechanics - Statics
12 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
all F' s
r
O
F
F
all F' s, d' s
d m F m
0 • Since the System is in Equilibrium Σ
F
’s =0. all F's
F
0
F
A
F
B
F
C
• In Summary: The d m F m 3 Forces, Sum to Zero products and,
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Special Case: 3-Force Element
A Graphical Summary AB is 3F Member (BC is 2F Member)
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Example
Pole Raising
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.
• • Solution Plan Create a free-body diagram of the joist. – 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.
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Example
Pole Raising
Create a free-body diagram of the joist Use LoA’s & Trigonometry to Determine the direction of the reaction force R
AF
BF
AB
cos 45 cos 45 2 .
828 m
CD BD CE
AE CD BF
1 2
AF
cot( 45
BD
1 .
414 m 25 ) 2 .
828 1 .
414 m 0 .
515 m tan 20 2.313
m 0 .
515 m 70 tan
CE AE
2 .
313 1 .
414 1 .
636 58 .
6 A LARGE, SCALED Diagram is REALLY Useful in this Problem
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70º Angle Analysis
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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
Engineering-36: Engineering Mechanics - Statics
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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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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 F Ax & F Ay Pass Thru the Axel-Axis Pt-A they Generate No moment about this point . • T 1 and T 2 have Exactly the SAME Lever arm, i.e., the Radius, R, of the Pulley
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Special Case: FrictionLess Pulley
Since the Pulley is in Equilibrium ΣM A = 0 Writing the Moment Eqn
M
A
r
T
0
R
T
1
k
ˆ
R k
ˆ
all T's or
R T
1
T
2
T
2
k
ˆ
0 0 Thus for the NO-Friction Perfect Pulley
T
1
T
2
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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
T
1
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T
2
FrictionFilled Pulley
In Alternator Operation the generation of electricity produces a resisting moment that counters the direction of spin; The FBD in this case → M Az The ΣM A
R
T
1
k
ˆ
= 0
R
T
2
A k
ˆ
0
Engineering-36: Engineering Mechanics - Statics
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FrictionFilled Pulley
Thus a RESISTING Moment causes a DIFFERENCE between the two
R
Tensions
T
1
T
2
M A
0
T
1
T
2
OR
M A R
M Az More on This when we Learn Chp08
Engineering-36: Engineering Mechanics - Statics
24 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 FOUR Force System, can be
reduced 3-Force
to a System using and
Equivalent Resultant-Couple
System at the Pulley
Engineering-36: Engineering Mechanics - Statics
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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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FrictionLess Pulley; 3F Mem
Apply the Equivalent Loading Method to a FrictionLess Pulley
T
1
T
2
T
1
T
2 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
Engineering-36: Engineering Mechanics - Statics
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FrictionLess Pulley; 3F Mem
The Equivalent Systems by
M
A = 0
T T T T Engineering-36: Engineering Mechanics - Statics
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T
R
FrictionLess Pulley; 3F Mem
Moving the FrictionLess Pulley Force Resultant to the Pin at Pt-A produces the FBD Shown At Right • Now can Draw the Force Triangle
B T
R
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C
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
Special Cases Summarized
Particle: 2D 3D
F x
0
F x
0
F y
0
F y
0
F z
2-Force Element:
F
B
F
A
0 3-Force Planar Element:
F
A
F
B
F
C
0 FrictionLess Pulley:
T
1
T
2
Engineering-36: Engineering Mechanics - Statics
31 Bruce Mayer, PE [email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx
WhiteBoard Work
Lets Work These Nice Problems
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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.
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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.
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35 Bruce Mayer, PE [email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx