[email protected] engr-36_lec-12_2d-equilibrium_specialcases_2n3_f_members.pptx 1 bruce...
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[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx1
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 36
Chp 5: 2D Equil
Special Cases
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx2
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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 LoadAnalysis
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
000 zyx MFF
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
2D Planar System: 000 zyx MFF
No Z-Translation → NO Z-Directed Force:
000 zyx FFF
No X or Y Rotation → NO X or Y Applied Moments
000 zyx MMM
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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.
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Special Case: 2-Force Element
Mathematically• Since the Two Forces Must be Concurrent
00 F'sall F'sall
FFrM FPoCPoC
• Since the System is in Equilibrium ΣF’s =0.
ABBA FFFFF 0 F'sall
– Thus the two force are Equal and Opposite; that is, the forces CANCEL
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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.
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
3-Force 2D Body: Parallel Forces
F1
F2
F3
d1d2
d3
O
If 3 Parallel Forces Maintain a Rigid Body in Static Equilibrium, The following Conditions MUST be Satisfied• For Translation Equilibrium
2130 FFFFx • For Rotation Equilibrium
223311
00
FdFdFd
O
FrM
x
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Special Case: 3-Force Element
Mathematically for ||-Forces• Since a Body in Equil. Has NO Net Moment
0sd' s,F' allsF' all
mmFOOFdFrM
• Since the System is in Equilibrium ΣF’s =0.
CBA FFFF 0 F'sall
• In Summary: The dmFm products and, 3 Forces, Sum to Zero
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Special Case: 3-Force Element
A Graphical Summary
AB is 3F Member(BC is 2F Member)
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example Pole Raising 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.
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.
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example Pole Raising Use LoA’s & Trigonometry to
Determine the direction of the reaction force R
Create a free-body diagram of the joist
636.1414.1
313.2tan
m 2.313m 515.0828.2
m 515.020tanm 414.1)2545cot(
m 414.1
m828.245cosm445cos
21
AE
CE
BDBFCE
CDBD
AFAECD
ABBFAF
6.58A LARGE, SCALED Diagram is REALLY
Useful in this Problem
70
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
70º Angle Analysis
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example Pole Raising Use the Law of the Sines to
Find the Reaction Force R Draw the Force
Triangle to Scale
38.6sin
N 1.98
110sin4.31sin RT
N 8.147
N9.81
R
T = 58.6° = 58.6°
Solving find
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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)
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx21
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Special Case: FrictionLess Pulley
Since the Pulley is in Equilibrium ΣMA = 0
Writing the Moment Eqn
0or
0
0
21
21
T's all
TTkR
kTRkTR
A
ˆ
ˆˆ
TrM
Thus for the NO-Friction Perfect Pulley 21 TT
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
1T
2T
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
FrictionFilled Pulley
In Alternator Operation the generation of electricity produces a resisting moment that counters the direction of spin; The FBD in this case →
The ΣMA = 0
MAz
021 kMkTRkTR Aˆˆˆ
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
FrictionFilled Pulley
Thus a RESISTING Moment causes a DIFFERENCE between the two Tensions
More on This when we Learn Chp08
MAz
RMTT
OR
MTTR
A
A
21
21 0
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
FrictionLess Pulley; 3F Mem In the System at
Right Member ABC, which is a FOUR-Force System, can be reduced to a 3-Force System using and Equivalent Resultant-Couple System at the Pulley
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
FrictionLess Pulley; 3F Mem Apply the Equivalent
Loading Method to a FrictionLess Pulley
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
1T
2T
21 TT
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
FrictionLess Pulley; 3F Mem
The Equivalent Systems by MA = 0
T
T
T
T
RT
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
RT
C
B
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
T=
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx31
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Special Cases Summarized
Particle:
AB FF
3-Force Planar Element:
0 CBA FFF
2-Force Element:
0003D
002D
zyx
yx
FFF
FF
21 TT FrictionLess Pulley:
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
WhiteBoard Work
Lets WorkThese NiceProblems
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 36
Appendix
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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.
[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx35
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Disk Problem
The smooth disksD 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.