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NCSU MAE 206 - VECTOR MECHANICS FOR ENGINEERS

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Slide 1ContentsIntroductionFree-Body DiagramReactions at Supports and Connections for a Two-Dimensional StructureSlide 6Equilibrium of a Rigid Body in Two DimensionsStatically Indeterminate ReactionsSample Problem 4.1Slide 10Sample Problem 4.3Slide 12Sample Problem 4.4Slide 14Equilibrium of a Two-Force BodyEquilibrium of a Three-Force BodySample Problem 4.6Slide 18Slide 19Equilibrium of a Rigid Body in Three DimensionsReactions at Supports and Connections for a Three-Dimensional StructureSlide 22Sample Problem 4.8Slide 24Slide 25VECTOR MECHANICS FOR ENGINEERS: STATICSEighth EditionFerdinand P. BeerE. Russell Johnston, Jr.Lecture Notes:J. Walt OlerTexas Tech UniversityCHAPTER© 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4Equilibrium of Rigid Bodies© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 2ContentsIntroductionFree-Body DiagramReactions at Supports and Connections for a Two-Dimensional StructureEquilibrium of a Rigid Body in Two DimensionsStatically Indeterminate ReactionsSample Problem 4.1Sample Problem 4.3Sample Problem 4.4Equilibrium of a Two-Force BodyEquilibrium of a Three-Force BodySample Problem 4.6Equilibrium of a Rigid Body in Three DimensionsReactions at Supports and Connections for a Three-Dimensional StructureSample Problem 4.8© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 3Introduction•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,   00 FrMFO000000zyxzyxMMMFFF•Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the conditions for static equilibrium,•For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body.© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 4Free-Body DiagramFirst step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a f ree-body diagram.•Select the extent of the free-body and detach it from the ground and all other bodies.•Include the dimensions necessary to compute the moments of the forces.•Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body.•Indicate point of application, magnitude, and direction of external forces, including the rigid body weight.© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 5Reactions at Supports and Connections for a Two-Dimensional Structure•Reactions equivalent to a force with known line of action.© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 6Reactions at Supports and Connections for a Two-Dimensional Structure•Reactions equivalent to a force of unknown direction and magnitude.•Reactions equivalent to a force of unknown direction and magnitude and a couple.of unknown magnitude© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 7Equilibrium of a Rigid Body in Two Dimensions•For all forces and moments acting on a two-dimensional structure,OzyxzMMMMF  00•Equations of equilibrium become   000AyxMFFwhere A is any point in the plane of the structure.•The 3 equations can be solved for no more than 3 unknowns.•The 3 equations can not be augmented with additional equations, but they can be replaced   000BAxMMF© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 8Statically Indeterminate Reactions•More unknowns than equations•Fewer unknowns than equations, partially constrained•Equal number unknowns and equations but improperly constrained© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 9Sample Problem 4.1A 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.SOLUTION:•Create a free-body diagram for the crane.•Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no 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.•Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero.© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 10Sample Problem 4.1•Create the free-body diagram.•Check the values obtained.•Determine B by solving the equation for the sum of the moments of all forces about A.     0m6kN5.23m2kN81.9m5.1:0 BMAkN1.107B•Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces.0:0 BAFxxkN1.107xA0kN5.23kN81.9:0 yyAFkN 3.33yA© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighthEdition4 - 11Sample Problem 4.3A loading car is at rest on an inclined track. The gross weight of the car and its load is 5500 lb, and it is applied at at G. The cart is held in position by the cable. Determine the tension in the cable and the reaction at each pair of wheels.SOLUTION:•Create a free-body diagram for the car with the coordinate system aligned with the track.•Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle.•Determine the cable tension by solving the equation for the sum of force components parallel to the track.•Check the values obtained by verifying that the sum of force components perpendicular


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