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Chp 5: 2D Equil Special Cases

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Presentation on theme: "Chp 5: 2D Equil Special Cases"— Presentation transcript:

1 Chp 5: 2D Equil Special Cases
Engineering 36 Chp 5: 2D Equil Special Cases Bruce Mayer, PE Licensed Electrical & Mechanical Engineer

2 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 PARTICLES covered in ENGR-36_Lec-06_Particle-Equilibrium_H13e.pptx

3 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 Graphics => https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=st&chap_sec=05.5&page=theory

4 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 PARTICLES covered in ENGR-36_Lec-06_Particle-Equilibrium_H13e.pptx

5 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

6 2D Planar System: No Z-Translation → NO Z-Directed Force:
No X or Y Rotation → NO X or Y Applied Moments

7 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

8 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.

9 Special Case: 2-Force Element
Mathematically Since the Two Forces Must be Concurrent Since the System is in Equilibrium ΣF’s =0. Thus the two force are Equal and Opposite; that is, the forces CANCEL

10 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

11 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. If Parallel, then the resultant is still zero, but the three || forces have Lever arms to arbitrary point such that the Sum Of the Moments Must be zero The lines of action of the three forces must be Concurrent OR Parallel.

12 3-Force 2D Body: Parallel Forces
If 3 Parallel Forces Maintain a Rigid Body in Static Equilibrium, The following Conditions MUST be Satisfied For Translation Equilibrium x d3 d1 F1 O d2 For Rotation Equilibrium F2 F3

13 Special Case: 3-Force Element
Mathematically for ||-Forces Since a Body in Equil. Has NO Net Moment Since the System is in Equilibrium ΣF’s =0. In Summary: The dmFm products and, 3 Forces, Sum to Zero

14 Special Case: 3-Force Element
A Graphical Summary AB is 3F Member (BC is 2F Member)

15 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.

16 A LARGE, SCALED Diagram is REALLY Useful in this Problem
Example  Pole Raising Create a free-body diagram of the joist Use LoA’s & Trigonometry to Determine the direction of the reaction force R Tan70 = CD/BD A LARGE, SCALED Diagram is REALLY Useful in this Problem

17 70º Angle Analysis

18 Example  Pole Raising Draw the Force Triangle to Scale
Use the Law of the Sines to Find the Reaction Force R Solving find 38.6 =

19 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)

20 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

21 Special Case: FrictionLess Pulley
Since the Pulley is in Equilibrium ΣMA = 0 Writing the Moment Eqn Thus for the NO-Friction Perfect Pulley

22 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

23 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

24 FrictionFilled Pulley
Thus a RESISTING Moment causes a DIFFERENCE between the two Tensions MAz More on This when we Learn Chp08

25 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

26 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

27 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

28 FrictionLess Pulley; 3F Mem
The Equivalent Systems by MA = 0

29 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

30 FrictionLess Pulley - Important
For a FrictionLess Pulley the Tension Forces and be to the Pulley Axel (Pin) WithOUT the Addition of a Couple =

31 Special Cases Summarized
Particle: 2-Force Element: 3-Force Planar Element: FrictionLess Pulley:

32 Lets Work These Nice Problems
WhiteBoard Work Lets Work These Nice Problems H13e: P5-44, P5-23 => See ENGR36_H13_Tutorial_3-Force_Members_1207.pptx

33 Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical Engineer

34 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. H13e: P5-44 => See ENGR36_H13_Tutorial_3-Force_Members_1207.pptx

35 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. H13e: P5-23 => See ENGR36_H13_Tutorial_3-Force_Members_1207.pptx


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