Report

Engineering 36 Chp 5: 2D Equil Special Cases Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering-36: Engineering Mechanics - Statics 1 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 2 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 3 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 4 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 5 y 0 M z 0 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 6 y 0 M z 0 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 7 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 8 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 0F 0 all F's • Since the System is in Equilibrium ΣF’s =0. F 0F 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 BMayer@ChabotCollege.edu • 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 BMayer@ChabotCollege.edu • 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 11 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 12 x Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 BMayer@ChabotCollege.edu • 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 14 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 15 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 BMayer@ChabotCollege.edu • 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 16 2 . 313 1 . 636 1 . 414 A LARGE, SCALED Diagram is REALLY Useful in this Problem Bruce Mayer, PE BMayer@ChabotCollege.edu • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 70º Angle Analysis Engineering-36: Engineering Mechanics - Statics 17 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 18 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 19 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 20 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 rT 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 21 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 22 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 23 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 24 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 25 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 26 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 27 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 BMayer@ChabotCollege.edu • 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 28 TR Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 29 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 30 = T Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 31 T1 T 2 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 BMayer@ChabotCollege.edu • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering-36: Engineering Mechanics - Statics 33 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 34 Bruce Mayer, PE BMayer@ChabotCollege.edu • 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 35 Bruce Mayer, PE BMayer@ChabotCollege.edu • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx