东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A04 Equilibrium of Rigid Bodies

Equilibrium of Rigid Bodies

Equilibrium of Rigid Bodies

ContentsIntroduction（绪论）Free-BodyDiagram（受力简图）Reactions at Supports and Connections for a Two-DimensionalStructure（支撑与连接处的作用力与反作用力）Equilibrium of a Rigid Body in Two Dimensions（两维刚体的平衡条件）EquilibriumofaTwo-ForceBody（二力构件的平衡）·EquilibriumofaThree-ForceBody（三力构件的平衡）2

Contents • Introduction（绪论） • Free-Body Diagram（受力简图） • Reactions at Supports and Connections for a Two-Dimensional Structure（支撑与连接处的作用力与反作用力） • Equilibrium of a Rigid Body in Two Dimensions（两维刚体的 平衡条件） • Equilibrium of a Two-Force Body（二力构件的平衡） • Equilibrium of a Three-Force Body（三力构件的平衡） 2

Introduction. For a rigid body in static equilibrium, the external forces andmoments are balanced and will impart no translational or rotationalmotion to the body.. The necessary and sufficient condition for the static equilibrium of abody are that the resultant force and couple from all external forcesform a system equivalent to zero,ZF=0 ZMo=Z(r×F)=0: Resolving each force and moment into its rectangular componentsleads to 6 scalar equations which also express the conditions for staticequilibrium,ZF, =0ZFx=0ZF,=0ZM=0ZM=0ZM, =03

Introduction • 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,  F  0 MO  r  F  0                 0 0 0 0 0 0 x y z x y z M M M F F F • 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. 3

Free-Body DiagramFirst step in the static equilibrium analysis of arigidbody is identification of all forces acting on thebody with a free-body diagram.A2400 kgGKSelecttheextentofthefree-bodyanddetachit1.5m土from the ground and all other bodies.B. Indicate point of application, magnitude, and4mdirection of external forces, including the rigid+2m-bodyweight..Indicate point of application and assumeddirection of unknown applied forces. These23.5kNusually consist of reactions through which theground and other bodies oppose the possible1.5mmotion of the rigid body.B9.81kNB:Includethedimensions necessaryto compute4m←2m→the moments of theforces.4

Free-Body Diagram First step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-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. 4

Reactions at Supports and ConnectionsReactionsequivalenttoaforcewithknownlineofaction.ForcewithknownFrictionlessRockerRollersline of actionsurfaceShort cableShort linkForce withknownlineofaction900CollaronForce with knownFrictionlesspininslotfrictionless rodlineofaction5

Reactions at Supports and Connections • Reactions equivalent to a force with known line of action. 5

Reactions at Supports and Connections1Reactionsequivalenttoaforce of unknowndirectionand magnitude.Frictionless pinRough surfaceForce of unknownorhingedirectionReactions equivalent to aorforceofunknowndirectionandmagnitudeand a couple.ofunknownmagnitudeFixed supportForce and couple6

Reactions at Supports and Connections • 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 6

Equilibrium of a Rigid Body in Two DimensionsQ. For all forces and moments acting on a two-dimensional structure,DCF,=0 Mx=M,=0 M,=Mo.Equations of equilibrium becomeAZFx=0 ZF,=0 ZMA=0(a)where A is any point in the plane of thestructure.DThe3eguationscanbesolvedfornomorethan 3 unknowns.B. The 3 equations can not be augmented withadditional equations, but they can be replacedBZFx=0 ZMA=0 ZMB=0(b)7

Equilibrium of a Rigid Body in Two Dimensions • For all forces and moments acting on a two￾dimensional structure, Fz  0 M x  M y  0 M z  MO • Equations of equilibrium become  Fx  0  Fy  0 M A  0 where 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  Fx  0 M A  0 MB  0 7

Sample ProblemSOLUTION:: Create a free-body diagram for the craneA2400 kg: Determine B by solving the equation forG+1.5mthe sum of the moments of all forces土about A. Note there will be noBcontribution from the unknownreactions at A.4m←2m→: Determine the reactions at A byA fixed crane has a mass of 1000 kgsolving the equations for the sum ofand is used to lift a 2400 kg crate. Itall horizontal force components andis held in place by a pin at A and aall vertical force componentsrocker at B. The center of gravity of·Checkthevaluesobtainedforthethe crane is located at G.reactions by verifying that the sum ofDetermine the components of thethemomentsaboutBofallforcesisreactions at A and Bzero.8

Sample Problem 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. 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. 8

: Determine B by solving the equation for thesum of the moments of all forces about AZMA =0: +B(1.5m)-9.81kN(2m)23.5kN- 23.5 kN(6m)= 0B = +107.1kN9.81kNBB4m+2m+: Determine the reactions at A by solving theequations for the sum of all horizontal forcesCreatethefree-bodydiagramand all vertical forces.ZFx=0: Ax+B=0Ax =-107.1kNZF,=0: A,-9.81kN-23.5kN=0A, = +33.3kN: Check the values obtained9

• 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.     23.5 kN6m 0 0 : 1.5m 9.81kN 2m    M A   B  B  107.1kN • Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces.  Fx  0 : Ax  B  0 Ax  107.1kN  Fy  0 : Ay 9.81kN  23.5kN  0 Ay  33.3 kN 9

Sample ProblemSOLUTION:24in. Create a free-body diagram for the carwith the coordinate system alignedwiththetrackG25.inDetermine the reactions atthe wheels30in.by solving equations for the sum of25in.moments about points above eachaxle:Determine the cable tension byA loading car is at rest on an inclinedsolving the equation for the sum oftrack. The gross weight of the car andforce components parallel to the track.its load is 5500 lb, and it is applied atat G. The cart is held in position by: Check the values obtained by verifyingthe cable.that the sum of force componentsperpendicular to the track are zero.Determine the tension in the cable andthe reaction at each pair of wheels10

Sample Problem A 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 to the track are zero. 10