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Engineering 36 Chp 5: Mech Equilibrium Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Engineering-36: Vector Mechanics - Statics 1 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx ReCall Equilibrium Conditions Rigid Body in Static Equilibrium Characterized by • Balanced external forces and moments – Will impart no Tendency toward Translational or Rotational motion to the body The NECESSARY and SUFFICIENT condition for the static equilibrium of a body are that the RESULTANT Force and Couple from all external forces form a system equivalent to zero Engineering-36: Vector Mechanics - Statics 2 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Equilibrium cont. Rigid Body Equilibrium Mathematically F 0 and MO r F 0 Resolving into Rectangular Components the Resultant Forces & Moments Leads to an Equivalent Definition of Rigid Body Equilibrium F 0 F 0 F 0 M 0 M 0 M 0 x y x Engineering-36: Vector Mechanics - Statics 3 z y z Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx 2D Planar System 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 Engineering-36: Vector Mechanics - Statics 4 F y 0 M z 0 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx 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 Engineering-36: Vector Mechanics - Statics 5 M y 0 M z 0 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx 2D Planar System: F x 0 F y 0 M z 0 If r Lies in the XY Plane, then rz = 0. With Fz = 0 the rxF Determinant: iˆ ˆj r F rx Fx ry Fy kˆ 0 ry 0 0 Fy iˆ rx 0 Fx 0 ˆj rx Fy ry Fx kˆ 0 Mx 0 My 0 Mz 0 So in this case M due to rxF is confined to the Z-Direction: r F XY M z M z kˆ rx Fy ry Fx kˆ Engineering-36: Vector Mechanics - Statics 6 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Example Crane problem Solution Plan • • Create a free-body diagram for the crane Determine Rcns at B by solving the equation for the sum of the moments of all forces about Pin-A – A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. Engineering-36: Vector Mechanics - Statics 7 • Note there will be no Moment contribution from the unknown reactions at A. Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components. Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Example Crane problem Reaction Analysis • PIN at A – • X & Y Reactions ROCKER at B – NORMAL Reaction Only +X Rcn in This Case Solution Plan cont. • Check the values obtained for the reactions by verifying that the moments about B of all forces Sum to zero. Engineering-36: Vector Mechanics - Statics 8 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Example Crane problem M A 0 : B1.5m 9.81 kN2m Draw the Free Body Diagram (FBD) 23.5 kN6m 0 B 107.1 kN Determine Pt-B Rcn by solving the equation for the sum of the moments of all forces about Pt-A. Engineering-36: Vector Mechanics - Statics 9 Determine the reactions at Pt-A by solving the eqns for the sum of all horizontal & vertical forces Fx 0 : Ax B 0 Ax 107.1kN Fy 0 : Ay 9.81kN 23.5 kN 0 Ay 33.3 kN Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Example Coal Car Solution Plan • • A loaded coal car is at rest on an inclined track. The car has a gross weight of 5500 lb, for the car and its load as applied at G. The car is held in position by the cable. Determine the TENSION in the cable and the REACTION at each pair of wheels. Engineering-36: Vector Mechanics - Statics 10 • Create a free-body diagram for the car with the coord system ALIGNED WITH the CABLE Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle, in the LoA of the Pull Cable Determine the cable tension by solving the equation for the sum of force components parallel to the cable Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Example Coal Car Create Free Body Dia. Determine the reactions at the wheels M A 0 : 2320 lb 25in. 4980 lb 6in. R2 50in. 0 R2 1758 lb M B 0 : 2320 lb 25in. 4980 lb 6in. R1 50in. 0 R1 562 lb Wx 5500lb cos 25 4980lb Wy 5500lbsin 25 2320lb Engineering-36: Vector Mechanics - Statics 11 Determine cable tension Fx 0 : 4980 lb T 0 T 4980 lb Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Example Cable Braced Beam Solution Plan • • The frame supports part of the roof of a small building. The tension in the cable is 150 kN. Determine the reaction at the fixed end E. Engineering-36: Vector Mechanics - Statics 12 Reaction Analysis • Create a free-body diagram for the frame and cable. Solve 3 equilibrium equations for the reaction force components and Moment at E. The Support at E is a CANTILEVER; can Resist – Planar Forces – Planar Moment Also DF 4.52 62 7.5m Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Triangle Trig Review adj cos hyp Engineering-36: Vector Mechanics - Statics 13 opp sin hyp opp tan adj Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Example Cable Braced Beam Create a free-body diagram for the frame and cable Fx 0 : E x 4.5 150 kN 0 7.5 E x 90.0 kN Fy 0 : E y 420 kN sin 6 150 kN 0 7.5 E y 200 kN cos M E 0 : 20 kN7.2 m 20 kN5.4 m 20 kN3.6 m 20 kN1.8 m 6 150 kN 4.5 m M E 0 7.5 Solve 3 equilibrium eqns for the reaction force M E 180.0 kN m components and couple Engineering-36: Vector Mechanics - Statics 14 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Rigid Body Equilibrium in 3D SIX scalar equations are required to express the conditions for the equilibrium of a rigid body in the general THREE dimensional case Fx 0 Fy 0 Fz 0 Mx 0 My 0 Mz 0 These equations can be solved for NO MORE than 6 unknowns • The Unknowns generally represent REACTIONS at Supports or Connections. Engineering-36: Vector Mechanics - Statics 15 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Rigid Body Equilibrium in 3D The scalar equations are often conveniently obtained by applying the vector forms of the conditions for equilibrium F 0 MO r F 0 Solve the Above Eqns with the Usual Techniques; e.g., Determinant Operations • A Clever Choice for the Pivot Point, O, Can Eliminate from the Calculation as Many as 3 Unknowns (a PoC somewhere) Engineering-36: Vector Mechanics - Statics 16 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Example 3D Ball-n-Socket Solution Plan • • A sign of Uniform Density weighs 270 lb and is supported by a BALL-ANDSOCKET joint at A and by two Cables. Determine the tension in each cable and the reaction at A. Engineering-36: Vector Mechanics - Statics 17 Create a free-body diagram of the Sign Apply the conditions for static equilibrium to develop equations for the unknown reactions. Reaction Analysis • Ball-n-Socket at A can Resist Translation in 3D – • Can NOT Resist TWIST (moment) in Any Direction Cables Pull on Their Geometric Axis Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx 3D Ball-n-Socket Prob cont. Create a free-body diagram of the Sign The x-Axis rotation Means that the Sign is Only Partially Constrained. Mathematically M • Ax Anything But If the Sign is NOT Swinging (i.e., it’s hanging STILL) The Following Condition Must Exist M 18 0 With The Absence of the Mx Constraint This Problem Generates only 5 Unknowns Note: The Sign can SWING about the x-Axis • (Wind Accommodation?) Engineering-36: Vector Mechanics - Statics Ax MX = 0 by X-Axis P.O.A. for W, TEC & TBC Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx 3D Ball-n-Socket Prob cont.2 rAD rAB TBD TBD TBD rAD rAB TBD TBD TEC TEC State in ComponentForm the Two CableTension Vectors TEC TEC Engineering-36: Vector Mechanics - Statics 19 rBD rBD 8iˆ 4 ˆj 8kˆ 12 23 iˆ 13 ˆj 23 kˆ rAC rAE EC TEC rAC rAE EC 6iˆ 3 ˆj 2kˆ 7 6 iˆ 3 ˆj 2 kˆ 7 7 7 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx 3D Ball-n-Socket Prob cont.3 F A TBD TEC 270lbˆj 0 iˆ : Ax 23 TBD 76 TEC 0 ˆj : Ay 1 TBD 3 TEC 270 lb 0 3 7 kˆ : Az 23 TBD 72 TEC 0 M A rAB TBD rAE TEC 4 ftiˆ 270 lb ˆj 0 ˆj : 5.333TBD 1.714TEC 0 kˆ : 2.667TBD 2.571TEC 1080ft lb 0 Apply the conditions for static equilibrium Solving 5 Eqns in 5 Unkwns to develop TBD 101.3 lb TEC 315lb equations for the unknown reactions. A 338lbiˆ 101.2lb ˆj 22.5lbkˆ Engineering-36: Vector Mechanics - Statics 20 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx WhiteBoard Work Solve 3D Problem by Mechanics & MATLAB >> r = [5 -3 -7] r = 5 -3 -7 >> F = [-13 8 11] F = -13 8 11 >> rxF = cross(r,F) rxF = 23 36 1 Determine the tensions in the cables and the components of reaction acting on the smooth collar at G necessary to hold the 2000 N sign in equilibrium. The sign weight may concentrated at the center of gravity. Engineering-36: Vector Mechanics - Statics 21 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Engineering-36: Vector Mechanics - Statics 22 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Engineering-36: Vector Mechanics - Statics 23 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Engineering-36: Vector Mechanics - Statics 24 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected] Engineering-36: Vector Mechanics - Statics 25 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx F y 0 R T1 kˆ R T2 kˆ M A kˆ 0 Engineering-36: Vector Mechanics - Statics 26 OR RkˆT T M kˆ 1 2 Bruce Mayer, PE [email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx A