3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM

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1 3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM Consider rigid body fixed in the x, y and z reference and is either at rest or moves with reference at constant velocity Two types of forces that act on it, the resultant internal force and the resultant external force Resultant internal force f i is caused by interactions with adjacent particles 3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM Resultant external force F i represents the effects of gravitational, electrical, magnetic, or contact forces between the i th particle and adjacent bodies or particles not included within the body Particle in equilibrium, apply Newton s first law, F i + f i = CONDITIONS FOR RIGID-BODY EQUILIBRIUM When equation of equilibrium is applied to each of the other particles of the body, similar equations will result Adding all these equations vectorially, F i + f i = 0 Summation of internal forces = 0 since internal forces between particles in the body occur in equal but opposite collinear pairs (Newton s third law) 3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM Only sum of external forces will remain Let F i = F, F= 0 Consider moment of the forces acting on the i th particle about the arbitrary point O By the equilibrium equation and distributive law of vector cross product, r i X (F i + f i ) = r i X F i + r i X f i = 0 1

2 3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM Similar equations can be written for other particles of the body Adding all these equations vectorially, r i X F i + r i X f i = 0 Second term = 0 since internal forces occur in equal but opposite collinear pairs Resultant moment of each pair of forces about point O is zero Using notation M O = r i X F i, M O = CONDITIONS FOR RIGID-BODY EQUILIBRIUM Equations of Equilibrium for Rigid Body F= 0 M O = 0 A rigid body will remain in equilibrium provided the sum of all the external forces acting on the body = 0 and sum of moments of the external forces about a point = 0 For proof of the equation of equilibrium, - Assume body in equilibrium 3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM -Force system acting on the body satisfies the equations F= 0 and M O = 0 -Suppose additional force F is applied to the body F + F = 0 M O + M O = 0 where M O is the moment of F about O -Since F= 0 and M O = 0, we require F = 0 and M O -Additional force F is not required and equations F= 0 and M O = 0 are sufficient FBD is the best method to represent all the known and unknown forces in a system FBD is a sketch of the outlined shape of the body, which represents it being isolated from its surroundings Necessary to show all the forces and couple moments that the surroundings exert on the body so that these effects can be accounted for when equations of equilibrium are applied 2

3 Support Reactions If the support prevents the translation of a body in a given direction, then a force is developed on the body in that direction If rotation is prevented, a couple moment is exerted on the body Consider the three ways a horizontal member, beam is supported at the end - roller, cylinder -pin - fixed support 3

4 Support Reactions Roller or cylinder Prevent the beam from translating in the vertical direction Roller can only exerts a force on the beam in the vertical direction Support Reactions Pin The pin passes through a hold in the beam and two leaves that are fixed to the ground Prevents translation of the beam in any direction Φ The pin exerts a force Fon the beam in this direction Support Reactions Fixed Support This support prevents both translation and rotation of the beam A couple and moment must be developed on the beam at its point of connection Force is usually represented in x and y components Cable exerts a force on the bracket Type 1 connections Rocker support for this bridge girder allows horizontal movements so that the bridge is free to expand and contract due to temperature Type 5 connections 4

5 Concrete Girder rest on the ledge that is assumed to act as a smooth contacting surface Type 6 connections Floor beams of this building are welded together and thus form fixed connections Type 10 connections Utility building is pin supported at the top of the column Type 8 connections External and Internal Forces A rigid body is a composition of particles, both external and internal forces may act on it For FBD, internal forces act between particles which are contained within the boundary of the FBD, are not represented Particles outside this boundary exert external forces on the system and must be shown on FBD FBD for a system of connected bodies may be used for analysis Weight and Center of Gravity When a body is subjected to gravity, each particle has a specified weight For entire body, consider gravitational forces as a system of parallel forces acting on all particles within the boundary The system can be represented by a single resultant force, known as weight Wof the body Location of the force application is known as the center of gravity 5

6 Weight and Center of Gravity Center of gravity occurs at the geometric center or centroid for uniform body of homogenous material For non-homogenous bodies and usual shapes, the center of gravity will be given Idealized Models Needed to perform a correct force analysis of any object Careful selection of supports, material, behavior and dimensions for trusty results Complex cases may require developing several different models for analysis Idealized Models Consider a steel beam used to support the roof joists of a building For force analysis, reasonable to assume rigid body since small deflections occur when beam is loaded Bolted connection at A will allow for slight rotation when load is applied => use Pin Support at B offers no resistance to horizontal movement => use Roller Building code requirements used to specify the roof loading (calculations of the joist forces) Large roof loading forces account for extreme loading cases and for dynamic or vibration effects Weight is neglected when it is small compared to the load the beam supports 6

7 Idealized Models Consider lift boom, supported by pin at A and hydraulic cylinder at BC (treat as weightless link) Assume rigid material with density known For design loading P, idealized model is used for force analysis Average dimensions used to specify the location of the loads and supports Example 5.1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg. Free-Body Diagram Support at A is a fixed wall Three forces acting on the beam at A denoted as A x, A y, A z, drawn in an arbitrary direction Unknown magnitudes of these vectors Assume sense of these vectors For uniform beam, Weight, W = 100(9.81) = 981N acting through beam s center of gravity, 3m from A 7

8 Example 5.5 The free-body diagram of each object is drawn. Carefully study each solution and identify what each loading represents. 3.3 EQUATIONS OF EQUILIBRIUM For equilibrium of a rigid body in 2D, F x = 0; F y = 0; M O = 0 F x and F y represent the algebraic sums of the x and y components of all the forces acting on the body M O represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to x-y plane and passing through arbitrary point O, which may lie on or off the body 8

9 3.3 EQUATIONS OF EQUILIBRIUM Alternative Sets of Equilibrium Equations For coplanar equilibrium problems, F x = 0; F y = 0; M O = 0 can be used Two alternative sets of three independent equilibrium equations may also be used F a = 0; M A = 0; M B = 0 When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis 3.3 EQUATIONS OF EQUILIBRIUM Alternative Sets of Equilibrium Equations Consider FBD of an arbitrarily shaped body All the forces on FBD may be replaced by an equivalent resultant force F R = Facting at point A and a resultant moment M RA = M A If M A = 0 is satisfied, M RA = EQUATIONS OF EQUILIBRIUM Alternative Sets of Equilibrium Equations If F R satisfies F a = 0, there is no component along the a axis and its line of axis is perpendicular to the a axis If M B = 0 where B does not lies on the line of action of F R, F R = 0 Since F= 0 and M A = 0, the body is in equilibrium 3.3 EQUATIONS OF EQUILIBRIUM Alternative Sets of Equilibrium Equations A second set of alternative equations is M A = 0; M B = 0; M C = 0 Points A, B and C do not lie on the same line Consider FBD, if If M A = 0, M RA = 0 M A = 0 is satisfied if line of action of F R passes through point B M C = 0 where C does not lie on line AB F R = 0 and the body is in equilibrium 9

10 3.3 EQUATIONS OF EQUILIBRIUM Example The link is pin-connected at a and rest a smooth support at B. Compute the horizontal and vertical components of reactions at pin A 3.3 EQUATIONS OF EQUILIBRIUM FBD Reaction N B is perpendicular to the link at B Horizontal and vertical components of reaction are represented at A 3.3 EQUATIONS OF EQUILIBRIUM 3.3 EQUATIONS OF EQUILIBRIUM Equations of Equilibrium M = 0; 90N. m 60N(1m ) + N (0.75m) = 0 N + F A x B A = 200N x o 200sin 30 N = 0 Ax = 100N = 0; B + F o A 60N 200cos30 N = 0 y A = 233N y y = 0; 10

11 3.4 TWO- AND THREE-FORCE MEMBERS Simplify some equilibrium problems by recognizing embers that are subjected top only 2 or 3 forces Two-Force Members When a member is subject to no couple moments and forces are applied at only two points on a member, the member is called a two-force member 3.4 TWO- AND THREE-FORCE MEMBERS Two-Force Members Example Forces at A and B are summed to obtain their respective resultants F A and F B These two forces will maintain translational and force equilibrium provided F A is of equal magnitude and opposite direction to F B Line of action of both forces is known and passes through A and B 3.4 TWO- AND THREE-FORCE MEMBERS Two-Force Members Hence, only the force magnitude must be determined or stated Other examples of the two-force members held in equilibrium are shown in the figures to the right 3.4 TWO- AND THREE-FORCE MEMBERS Three-Force Members If a member is subjected to only three forces, it is necessary that the forces be either concurrent or parallel for the member to be in equilibrium To show the concurrency requirement, consider a body with any two of the three forces acting on it, to have line of actions that intersect at point O 11

12 3.4 TWO- AND THREE-FORCE MEMBERS Three-Force Members To satisfy moment equilibrium about O, the third force must also pass through O, which then makes the force concurrent If two of the three forces parallel, the point of currency O, is considered at infinity Third force must parallel to the other two forces to insect at this point 3.4 TWO- AND THREE-FORCE MEMBERS Bucket link AB on the back hoe is a typical example of a two-force member since it is pin connected at its end provided its weight is neglected, no other force acts on this member The hydraulic cylinder is pin connected at its ends, being a twoforce member 3.4 TWO- AND THREE-FORCE MEMBERS The boom ABD is subjected to the weight of the suspended motor at D, the forces of the hydraulic cylinder at B, and the force of the pin at A. If the boom s weight is neglected, it is a three-force member The dump bed of the truck operates by extending the hydraulic cylinder AB. If the weight of AB is neglected, it is a two-force member since it is pin-connected at its end points 3.4 TWO- AND THREE-FORCE MEMBERS Example 5.13 The lever ABC is pin-supported at A and connected to a short link BD. If the weight of the members are negligible, determine the force of the pin on the lever at A. 12

13 3.4 TWO- AND THREE-FORCE MEMBERS FBD Short link BD is a two-force member, so the resultant forces at pins D and B must be equal, opposite and collinear Magnitude of the force is unknown but line of action known as it passes through B and D Lever ABC is a three-force member 3.4 TWO- AND THREE-FORCE MEMBERS FBD For moment equilibrium, three nonparallel forces acting on it must be concurrent at O Force F on the lever at B is equal but opposite to the force F acting at B on the link Distance CO must be 0.5m since lines of action of Fand the 400N force are known 3.4 TWO- AND THREE-FORCE MEMBERS Equations of Equilibrium 3.5 EQUILIBRIUM IN THREE DIMENSIONS Solving, 0.7 tan 1 o θ = = F = 0; o o F cos60.3 F cos N = 0 A + F o o F sin 60.3 F sin 45 = 0 A F = 1.07kN A y x = 0; F = 1.32kN 13

14 3.5 EQUILIBRIUM IN THREE DIMENSIONS 3.5 EQUILIBRIUM IN THREE DIMENSIONS 3.5 EQUILIBRIUM IN THREE DIMENSIONS 3.5 EQUILIBRIUM IN THREE DIMENSIONS Ball and socket joint provides a connection for the housing of an earth grader to its frame Journal bearing supports the end of the shaft 14

15 3.5 EQUILIBRIUM IN THREE DIMENSIONS Thrust bearing is used to support the drive shaft on the machine Pin is used to support the end of the strut used on a tractor 3.5 EQUILIBRIUM IN THREE DIMENSIONS Example 5.14 Several examples of objects along with their associated free-body diagrams are shown. In all cases, the x, y and z axes are established and the unknown reaction components are indicated in the positive sense. The weight of the objects is neglected. 3.5 EQUILIBRIUM IN THREE DIMENSIONS 3.5 EQUILIBRIUM IN THREE DIMENSIONS 15

16 3.5 EQUILIBRIUM IN THREE DIMENSIONS 3.5 EQUILIBRIUM IN THREE DIMENSIONS 3.6 EQUATIONS OF EQUILIBRIUM Vector Equations of Equilibrium For two conditions for equilibrium of a rigid body in vector form, F= 0 M O = 0 where Fis the vector sum of all the external forces acting on the body and M O is the sum of the couple moments and the moments of all the forces about any point O located either on or off the body 3.6 EQUATIONS OF EQUILIBRIUM Scalar Equations of Equilibrium If all the applied external forces and couple moments are expressed in Cartesian vector form F= F x i+ F y j+ F z k= 0 M O = M x i+ M y j+ M z k= 0 i, jand kcomponents are independent from one another 16

17 3.6 EQUATIONS OF EQUILIBRIUM Scalar Equations of Equilibrium F x = 0, F y = 0, F z = 0 shows that the sum of the external force components acting in the x, y and z directions must be zero M x = 0, M y = 0, M z = 0 shows that the sum of the moment components about the x, y and z axes to be zero 3.7 CONSTRAINTS FOR A RIGID BODY To ensure the equilibrium of a rigid body, it is necessary to satisfy the equations equilibrium and have the body properly held or constrained by its supports Redundant Constraints More support than needed for equilibrium Statically indeterminate: more unknown loadings on the body than equations of equilibrium available for their solution 3.7 CONSTRAINTS FOR A RIGID BODY Redundant Constraints Example For the 2D and 3D problems, both are statically indeterminate because of additional supports reactions In 2D, there are 5 unknowns but 3 equilibrium equations can be drawn 3.7 CONSTRAINTS FOR A RIGID BODY Redundant Constraints Example In 3D, there are 8 unknowns but 6 equilibrium equations can be drawn Additional equations involving the physical properties of the body are needed to solve indeterminate problems 17