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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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## Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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**Engineering 36**Chp5: FBDs2D/3D Systems Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu**Free Body Diagrams (FBDs)**• A free-body diagram is a sketch of an object of interest with all the surrounding objects stripped away to reveal all of the forces acting on the body • The purpose of a free-body force diagram is to assist with determination of the Net Force and/or Moment acting on a body SpaceDiagram Free BodyDiagram**Select an object or group of objects to focus on as the**"body“: i.e., the system. Sketch the body by itself, "free" of its surroundings Draw only those forces/moments that are acting directly on the body. Include both the magnitude and the direction of these forces. Do not include any forces that the body exerts on it surroundings, they do NOT act ON the body. However, there is always an equal reaction force acting on the body. For a compound body (e.g. Trusses, Machines) you do NOT need to include any INTERNAL forces acting between the body's SUBPARTS these internal forces come in action-reaction pairs which cancel out each other because of Newton's Third Law. Choose a coordinate system and sketch it on the free-body diagram. Often choose one of the axes to be parallel one or more forces it can sometimes simplify the equations to be solved. Constructing a free-body diagram**Structural Supports**• NonMoving Structures are typically Connected to Some Sort of Supporting Base • The connection between the Structure and Base are usually Called “Structural Supports” • The Force and/or moments exerted on the Structure Base are usually called “Structural Reactions” (RCNs for Short)**Structural Supports**• A Support that Prevents Linear Motion (sliding, translating) of the structure then exerts a Force on the structure • A Support that Prevents Rotating Motion (twisting, turning) of the structure then exerts a CoupleMoment on the structure**Recall SLIDING & FREE Vectors**• Forces are SLIDING Vectors; They can applied at ANY-POINT on the Vector Line of Action (LoA) • COUPLE-Moments are FREE Vectors; They can be applied at ANY Point, On or Off the Body**2D Support ReActions**• Cable can only Generate TENSION • WeightLess Link is 2-Force Element**2D Support ReActions**• Note that in BOTH these Cases the Support ReAction is NORMAL (Perpendicular) to the Supporting Surface • RCN can only PUSH, and NOT PULL**2D Support ReActions**• Note that in BOTH these Cases the Support ReAction is NORMAL (Perpendicular) to the Supporting Surface • RCN can PUSH or PULL**2D Support ReActions**• Only the Supports (9) & (10) Can Generate a Couple-MomentReAction**Center of Gravity**• If the Weight of the Rigid Body is Not Negligible, then the Entire Weight of the Body can be concentrated at a Single Point Called the Center of Gravity (CG) • Many times the CG location is Given • Can Calculate using Centroid Methods which will be covered later**2D Free-Body Diagram**• First Steps for Static Rigid-Body Equilibrium Analysis • Identification of All Forces & Moments Acting on the Body • Formulation of the Free-Body Diagram • Free Body Diagram Construction Process • See next slide**2D Free-Body Diagram cont**• Select the extent of the free-body and detach it from the ground and other bodies • Indicate for external loads: • Point of application • Magnitude & Direction Of External Forces • Including The Body Weight. • Indicate point of application and ASSUMED direction of UNKNOWN applied forces**2D Free-Body Diagram cont.2**• The Unknown Forces Typically Include REACTIONS through which the GROUND and OTHER BODIES oppose the possible motion of the rigid body • Include All dimensions Needed to Calculate the Moments of the Forces**Consider Rocker & Pin Supported Truss**Example: Truss Structure • Analyze Loading • Four External Force Loads as shown • Truss Weight, W • RCN at Pt-A by Rocker • Expect NORMAL to support Pad • RCN at Pt-B by Pin • Expect • in plane of Truss • Arbitrarily Directed • Draw the FBD for this Structure**Example: Truss – Draw FBD**• This Dwg is, in fact, a Full Free Body Diagram RBx RB RA RBy φ W**3D Support ReActions**• Same as 2D ReActions of this Type**3D Support ReActions**• Ball-n-Socket is the 3D analog to the 2D Smooth Pin or Hinge**3D Support ReActions**• This configuration is Commonly Known as a “Pillow Block Bearing”. • Type of support is (obviously) designed to allow the shaft to SPIN FREELY on its AXIS**3D Support ReActions**• The Sq-Shaft Bearing System does NOT Allow the shaft to spin completely freely, Thus the My**3D Support ReActions**• These supports are (obviously) designed to allow the Free Spin on the Pin Axis**3D Support ReActions**• This Type of support is commonly Known as a CANTILEVER. • Generates the Maximum Amount of Unknowns for 3D systems**ROUGH SURFACE ReActions**• Friction on a Rough Surface will Generate RCNs Parallel to the Supporting Surface • 2D • 3D**Given Bar supported by Hinge at Pt-A and rests on the Rough**x*y*z* Surface at Pt-B Example: Hinge & Rough-Surf • Analyze Rcn at Pt-A. By 5.2-(9) the Single Axially Constrained Hinge will • Provide Lateral (y & z) and Axial (x) Support • Resist twisting about the y and z axes**Example: Hinge & Rough-Surf**• PUSH (not PULL) Normal to the Surface • In this case the y* direction is normal to the supporting plane • Resist Sliding in any direction WITHIN the supporting plane • Analyze Rcn at Pt-B. Support Leg on a rough surface will**Example: cont.**• If the Weight of the Bar is negligible, then All Forces are accounted for and this is, in fact, the FBD FAx MAy FAy FAz FBz* MAz FBx* FBy***Symmetry City**• If We’re Lucky enough to have a Plane of Symmetry for BOTH Loading and Structural GEOMETRY then we can treat real world 3D problems as 2D • OtherWise we need to Operate in full 3D No Symmetry Can Treat as 2D Must Treat as 3D**Consider Leaning Utility Pole**Example: Utility Pole • Determine the Loads acting on the BASE of the Pole • Analyze Rcn at Base • This is a FIXED support which is often call a CantiLever • Cantilever supports resist both forces and moments in ALL 3 Spatial Directions**Draw in the BASE ReActions**Example: Pole • This Diagram is NOT a FBD as it does not account for these forces acting on the pole • Pole Weight • Cable Tension MAx FAx MAy FAy FAz MAz**Distributed Forces/Loads**• In Some Cases Forces are concentrated at Points; this is simplest case • Often times a Load cannot be identified with a single point; Instead the Load is Spread Out over a supporting surface • Such Forces are Called “Distributed” • Distributed Loads are indicated with a Load Profile**Distributed Force Profiles**• A uniformily Dist Load Has the same action at every point on it’s region of application. • It’s profile is “Flat” • NonUniform Loads are also common • They may be kinked, curved, or arbitrary**Distributed-Force Equivalent**• In Chp4 we discussed how to Replace a Distributed-Load with an Equivalent Point-Load placed at a Specific Location • Units for Distributed Forces • 2D → Force per Length (lb/ft, lb/in N/m) • 3D → Force Per Area (Pa, PSI, PSF)**Example: Hydraulic Cylinder**• The Hydraulic Cylinder Pumps Fluid in & out of the Cylinder Reservoir as Shown at Right • Draw The loads on the Piston Assy • Game Plan: • Isolate Piston Assy as Free Body • CareFully Account for all Pt-Force and PRESSURES acting on the Piston Assy**Example – Cont.**FOr FOr • Load-1 = 100 kg (220lb, 9.81kN) CounterWt WP • Load-2 = Weight of the Piston Rod WR • Load-3 = Weight of the Piston • Load-4 = Lateral Restraining Forces Exerted by the Cylinder Wall on the O-Ring 9.81kN**Pair**Example – Cont. FOr FOr • Load-5 = The Air Pressure on Top of the Piston WP Pfluid • Load-6 = The Hydraulic Fluid Pressure on the Bottom of the Cylinder WR 9.81kN**Pair**Example – Cont. FOr FOr • We can SIMPLIFY the analysis by making assessments about the relative significance of the loads WP Pfluid WR • The Weight of the Rod and Piston are likely negligible compared to the Counter Weight 9.81kN**Pair**Example – Cont. FOr FOr • Additional Symplifications Pfluid • The SideWall Forces on the O-Ring must cancel if the Cylinder is Balanced • The AIR pressure is negligible compared to the FLUID pressure 9.81kN**Example – Cont.**• Thus in the NonMoving Simplified System the Fluid Pressure balances the Counter Weight. • Mathematically Pfluid 9.81kN**WhiteBoard Work**None Today;Did byPowerPoint**Engineering 36**Appendix Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu