Engineering Mechanics: Statics

Engineering Mechanics: Statics
Chapter 5: Equilibrium of a Rigid Body Engineering Mechanics: Statics

Chapter Objectives To develop the equations of equilibrium for a rigid body. To introduce the concept of the free-body diagram for a rigid body. To show how to solve rigid-body equilibrium problems using the equations of equilibrium.

Chapter Outline Conditions for Rigid Equilibrium Free-Body Diagrams
Equations of Equilibrium Two and Three-Force Members Equilibrium in Three Dimensions Constraints for a Rigid Body

5.1 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 fi is caused by interactions with adjacent particles

Resultant external force Fi represents the effects of gravitational, electrical, magnetic, or contact forces between the ith particle and adjacent bodies or particles not included within the body Particle in equilibrium, apply Newton’s first law, Fi + fi = 0

When equation of equilibrium is applied to each of the other particles of the body, similar equations will result Adding all these equations vectorially, ∑Fi + ∑fi = 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)

Only sum of external forces will remain Let ∑Fi = ∑F, ∑F = 0 Consider moment of the forces acting on the ith particle about the arbitrary point O By the equilibrium equation and distributive law of vector cross product, ri X (Fi + fi) = ri X Fi + ri X fi = 0

Similar equations can be written for other particles of the body Adding all these equations vectorially, ∑ri X Fi + ∑ri X fi = 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 ∑MO = ∑ri X Fi, ∑MO = 0

Equations of Equilibrium for Rigid Body ∑F = 0 ∑MO = 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

- Force system acting on the body satisfies the equations ∑F = 0 and ∑MO = 0 - Suppose additional force F’ is applied to the body ∑F + F’ = 0 ∑MO + MO’= 0 where MO’is the moment of F’ about O - Since ∑F = 0 and ∑MO = 0, we require F’ = 0 and MO’ - Additional force F’ is not required and equations ∑F = 0 and ∑MO = 0 are sufficient

5.2 Free-Body Diagrams 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

5.2 Free-Body Diagrams

5.2 Free-Body Diagrams 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

5.2 Free-Body Diagrams 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

5.2 Free-Body Diagrams 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 F on the beam in this direction

5.2 Free-Body Diagrams 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

5.2 Free-Body Diagrams 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

5.2 Free-Body Diagrams Concrete Girder rest on the ledge that is assumed to act as a smooth contacting surface Type 6 connections Utility building is pin supported at the top of the column Type 8 connections

5.2 Free-Body Diagrams Floor beams of this building are welded together and thus form fixed connections Type 10 connections

5.2 Free-Body Diagrams 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

5.2 Free-Body Diagrams 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 W of the body Location of the force application is known as the center of gravity

5.2 Free-Body Diagrams 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

5.2 Free-Body Diagrams 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

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

5.2 Free-Body Diagrams 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

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

5.2 Free-Body Diagrams Procedure for Drawing a FBD
1. Draw Outlined Shape Imagine body to be isolated or cut free from its constraints Draw outline shape 2. Show All Forces and Couple Moments Identify all external forces and couple moments that act on the body

5.2 Free-Body Diagrams Procedure for Drawing a FBD Usually due to
- applied loadings - reactions occurring at the supports or at points of contact with other body - weight of the body To account for all the effects, trace over the boundary, noting each force and couple moment acting on it 3. Identify Each Loading and Give Dimensions Indicate dimensions for calculation of forces

5.2 Free-Body Diagrams Procedure for Drawing a FBD
Known forces and couple moments should be properly labeled with their magnitudes and directions Letters used to represent the magnitudes and direction angles of unknown forces and couple moments Establish x, y and coordinate system to identify unknowns

5.2 Free-Body Diagrams Example 5.1
Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg.

5.2 Free-Body Diagrams Solution Free-Body Diagram

5.2 Free-Body Diagrams Solution Support at A is a fixed wall
Three forces acting on the beam at A denoted as Ax, Ay, Az, 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

5.2 Free-Body Diagrams Example 5.2 Draw the free-body diagram of
the foot lever. The operator applies a vertical force to the pedal so that the spring is stretched 40mm and the force in the short link at B is 100N.

5.2 Free-Body Diagrams Solution Lever loosely bolted to frame at A
Rod at B pinned at its ends and acts as a short link For idealized model of the lever,

5.2 Free-Body Diagrams Solution Free-Body Diagram
Pin support at A exerts components Ax and Ay on the lever, each force with a known line of action but unknown magnitude

5.2 Free-Body Diagrams Solution
Link at B exerts a force 100N acting in the direction of the link Spring exerts a horizontal force on the lever Fs = ks = 5N/mm(40mm) = 200N Operator’s shoe exert vertical force F on the pedal Compute the moments using the dimensions on the FBD Compute the sense by the equilibrium equations

5.2 Free-Body Diagrams Example 5.3 Two smooth pipes, each
having a mass of 300kg, are supported by the forks of the tractor. Draw the free-body diagrams for each pipe and both pipes together.

5.2 Free-Body Diagrams Solution For idealized models,
of pipe A

For weight of pipe A, W = 300(9.81) = 2943N Assume all contacting surfaces are smooth, reactive forces T, F, R act in a direction normal to tangent at their surfaces of contact Free-Body Diagram at pipe B

*Note: R represent the force of A on B, is equal and opposite to R representing the force of B on A Contact force R is considered an internal force, not shown on FBD Free-Body Diagram of both pipes

5.2 Free-Body Diagrams Example 5.4 Draw the free-body diagram
of the unloaded platform that is suspended off the edge of the oil rig. The platform has a mass of 200kg.

Idealized model considered in 2D because by observation, loading and the dimensions are all symmetrical about a vertical plane passing through the center Connection at A assumed to be a pin and the cable supports the platform at B

Direction of the cable and average dimensions of the platform are listed and center of gravity has been determined Free-Body Diagram

5.2 Free-Body Diagrams Solution Platform’s weight = 200(9.81) = 1962N
Force components Ax and Ay along with the cable force T represent the reactions that both pins and cables exert on the platform Half of the cables magnitudes is developed at A and half developed at B

5.2 Free-Body Diagrams Example 5.5
The free-body diagram of each object is drawn. Carefully study each solution and identify what each loading represents.

5.2 Free-Body Diagrams Solution

5.3 Equations of Equilibrium
For equilibrium of a rigid body in 2D, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 ∑Fx and ∑Fy represent the algebraic sums of the x and y components of all the forces acting on the body ∑MO 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

Alternative Sets of Equilibrium Equations For coplanar equilibrium problems, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be used Two alternative sets of three independent equilibrium equations may also be used ∑Fa = 0; ∑MA = 0; ∑MB = 0 When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis

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 FR = ∑F acting at point A and a resultant moment MRA = ∑MA If ∑MA = 0 is satisfied, MRA = 0

Alternative Sets of Equilibrium Equations If FR satisfies ∑Fa = 0, there is no component along the a axis and its line of axis is perpendicular to the a axis If ∑MB = 0 where B does not lies on the line of action of FR, FR = 0 Since ∑F = 0 and ∑MA = 0, the body is in equilibrium

Alternative Sets of Equilibrium Equations A second set of alternative equations is ∑MA = 0; ∑MB = 0; ∑MC = 0 Points A, B and C do not lie on the same line Consider FBD, if If ∑MA = 0, MRA = 0 ∑MA = 0 is satisfied if line of action of FR passes through point B ∑MC = 0 where C does not lie on line AB FR = 0 and the body is in equilibrium

Procedure for Analysis Free-Body Diagram Establish the x, y, z coordinates axes in any suitable orientation Draw an outlined shape of the body Show all the forces and couple moments acting on the body Label all the loadings and specify their directions relative to the x, y axes

Procedure for Analysis Free-Body Diagram The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed Indicate the dimensions of the body necessary for computing the moments of forces

Procedure for Analysis Equations of Equilibrium Apply the moment equation of equilibrium ∑MO = 0 about a point O that lies on the intersection of the lines of action of the two unknown forces The moments of these unknowns are zero about O and a direct solution the third unknown can be obtained

Procedure for Analysis Equations of Equilibrium When applying the force equilibrium ∑Fx = 0 and ∑Fy = 0, orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components If the solution yields a negative result scalar, the sense is opposite to that was assumed on the FBD

Example 5.6 Determine the horizontal and vertical components of reaction for the beam loaded. Neglect the weight of the beam in the calculations.

Solution FBD 600N force is represented by its x and y components 200N force acts on the beam at B and is independent of the force components Bx and By, which represent the effect of the pin on the beam

Solution Equations of Equilibrium A direct solution of Ay can be obtained by applying ∑MB = 0 about point B Forces 200N, Bx and By all create zero moment about B

Solution

Solution Checking,

Example 5.7 The cord supports a force of 500N and wraps over the frictionless pulley. Determine the tension in the cord at C and the horizontal and vertical components at pin A.

Solution FBD of the cord and pulley Principle of action: equal but opposite reaction observed in the FBD Cord exerts an unknown load distribution p along part of the pulley’s surface Pulley exerts an equal but opposite effect on the cord

Solution FBD of the cord and pulley Easier to combine the FBD of the pulley and contracting portion of the cord so that the distributed load becomes internal to the system and is eliminated from the analysis

Solution Equations of Equilibrium Tension remains constant as cord passes over the pulley (true for any angle at which the cord is directed and for any radius of the pulley

Solution

Example 5.8 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

Solution FBD Reaction NB is perpendicular to the link at B Horizontal and vertical components of reaction are represented at A

Solution Equations of Equilibrium

Solution

Example 5.9 The box wrench is used to tighten the bolt at A. If the wrench does not turn when the load is applied to the handle, determine the torque or moment applied to the bolt and the force of the wrench on the bolt.

Solution FBD Bolt acts as a “fixed support” it exerts force components Ax and Ay and a torque MA on the wrench at A

Solution

Solution Point A was chosen for summing the moments as the lines of action of the unknown forces Ax and Ay pass through this point and these forces are not included in the moment summation MA must be included Couple moment MA is a free vector and represents the twisting resistance of the bolt on the wrench

Solution By Newton’s third law, the wrench exerts an equal but opposite moment or torque on the bolt For resultant force on the wrench, For directional sense, FA acts in the opposite direction on the bolt

Solution Checking,

Example 5.10 Placement of concrete from the truck is accomplished using the chute. Determine the force that the hydraulic cylinder and the truck frame exert on the chute to hold it in position. The chute and the wet concrete contained along its length have a uniform weight of 560N/m.

Solution Idealized model of the chute Assume chute is pin connected to the frame at A and the hydraulic cylinder BC acts as a short link

Solution FBD Since chute has a length of 4m, total supported weight is (560N/m)(4m) = 2240N, which is assumed to act at its midpoint, G The hydraulic cylinder exerts a horizontal force FBC on the chute Equations of Equilibrium A direct solution of FBC is obtained by the summation about the pin at A

Solution

Solution Checking,

Example 5.11 The uniform smooth rod is subjected to a force and couple moment. If the rod is supported at A by a smooth wall and at B and C either at the top or bottom by rollers, determine the reactions at these supports. Neglect the weight of the rod.

Solution FBD All the support reactions act normal to the surface of contact since the contracting surfaces are smooth Reactions at B and C are acting in the positive y’ direction Assume only the rollers located on the bottom of the rod are used for support

Solution Note that the line of action of the force component passes through point A and this force is not included in the moment equation Since By’ is negative scalar, the sense of By’ is opposite to shown in the FBD

Solution Top roller at B serves as the support rather than the bottom one

Example 5.12 The uniform truck ramp has a weight of 1600N ( ≈ 160kg ) and is pinned to the body of the truck at each end and held in position by two side cables. Determine the tension in the cables.

Solution Idealized model of the ramp Center of gravity located at the midpoint since the ramp is approximately uniform FBD of the Ramp

Solution Equations of Equilibrium By the principle of transmissibility, locate T at C

Solution Since there are two cables supporting the ramp, T’ = T/2 = N

5.4 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

Two-Force Members Example Forces at A and B are summed to obtain their respective resultants FA and FB These two forces will maintain translational and force equilibrium provided FA is of equal magnitude and opposite direction to FB Line of action of both forces is known and passes through A and B

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

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

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”

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 two-force member

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

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.

Solution 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

Solution FBD For moment equilibrium, three non-parallel 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 F and the 400N force are known

Solution Equations of Equilibrium Solving,

5.5 Equilibrium in Three Dimensions (FBD)
Consider types of reaction that can occur at the supports Support Reactions Important to recognize the symbols used to represent each of these supports and to clearly understand how the forces and couple moments are developed by each support As in 2D, a force that is developed by a support that restricts the translation of the attached member

Support Reactions A couple moment is developed when rotation of the attached member is prevented Example The ball and socket joint prevents any translation of connecting member; therefore, a force must act on the member at the point of connection This force have unknown magnitude Fx, Fy and Fz Magnitude of the force is given by

Support Reactions The force’s orientation is defined by the coordinate angles α, β and γ Since connecting member is allow to rotate freely about any axis, no couple moment is resisted by a ball and socket joint Single bearing supports, single pin and single hinge are shown to support both force and couple moment components

Support Reactions However, if these supports are used with other bearings, pins or hinges to hold a rigid body in equilibrium and the supports are properly aligned when connected to the body, the force reactions at these supports alone may be adequate for supporting the body Couple moments become redundant and not shown on the FBD

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

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

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.

Solution

Vector Equations of Equilibrium For two conditions for equilibrium of a rigid body in vector form, ∑F = 0 ∑MO = 0 where ∑F is the vector sum of all the external forces acting on the body and ∑MO 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

Scalar Equations of Equilibrium If all the applied external forces and couple moments are expressed in Cartesian vector form ∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0 ∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0 i, j and k components are independent from one another

Scalar Equations of Equilibrium ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 shows that the sum of the external force components acting in the x, y and z directions must be zero ∑Mx = 0, ∑My = 0, ∑Mz = 0 shows that the sum of the moment components about the x, y and z axes to be zero

5.7 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

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

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

Improper Constraints Instability of the body caused by the improper constraining by the supports In 3D, improper constraining occur when the support reactions all intersect a common axis In 2D, this axis is perpendicular to the plane of the forces and appear as a point When all reactive forces are concurrent at this point, the body is improperly constrained

Improper Constraints Example From FBD, summation of moments about the x axis will not be equal to zero, thus rotation occur In both cases, impossible to solve completely for the unknowns

Improper Constraints Instability of the body also can be caused by the parallel reactive forces Example Summation of forces along the x axis will not be equal to zero

Improper Constraints Instability of the body also can be caused when a body have fewer reactive forces than the equations of equilibrium that must be satisfied The body become partially constrained Example If O is a point not located on line AB, loading condition and equations of equilibrium are not satisfied

Improper Constraints Proper constraining requires - lines of action of the reactive forces do not insect points on a common axis - the reactive forces must not be all parallel to one another When the minimum number of reactive forces is needed to properly constrain the body, the problem is statically determinate and equations of equilibrium can be used for solving

Procedure for Analysis Free Body Diagram Draw an outlined shape of the body Show all the forces and couple moments acting on the body Establish the x, y, z axes at a convenient point and orient the axes so that they are parallel to as many external forces and moments as possible Label all the loadings and specify their directions relative to the x, y and z axes

Procedure for Analysis Free Body Diagram In general, show all the unknown components having a positive sense along the x, y and z axes if the sense cannot be determined Indicate the dimensions of the body necessary for computing the moments of forces

Procedure for Analysis Equations of Equilibrium If the x, y, z force and moment components seem easy to determine, then apply the six scalar equations of equilibrium,; otherwise, use the vector equations It is not necessary that the set of axes chosen for force summation coincide with the set of axes chosen for moment summation Any set of nonorthogonal axes may be chosen for this purpose

Procedure for Analysis Equations of Equilibrium Choose the direction of an axis for moment summation such that it insects the lines of action of as many unknown forces as possible In this way, the moments of forces passing through points on this axis and forces which are parallel to the axis will then be zero If the solution yields a negative scalar, the sense is opposite to that was assumed

Example 5.15 The homogenous plate has a mass of 100kg and is subjected to a force and couple moment along its edges. If it is supported in the horizontal plane by means of a roller at A, a ball and socket joint at N, and a cord at C, determine the components of reactions at the supports.

Solution FBD Five unknown reactions acting on the plate Each reaction assumed to act in a positive coordinate direction

Solution Equations of Equilibrium Moment of a force about an axis is equal to the product of the force magnitude and the perpendicular distance from line of action of the force to the axis Sense of moment determined from right-hand rule

Solution Components of force at B can be eliminated if x’, y’ and z’ axes are used

Solution Solving, Az = 790N Bz = -217N TC = 707N The negative sign indicates Bz acts downward The plate is partially constrained since the supports cannot prevent it from turning about the z axis if a force is applied in the x-y plane

Example 5.16 The windlass is supported by a thrust bearing at A and a smooth journal bearing at B, which are properly aligned on the shaft. Determine the magnitude of the vertical force P that must be applied to the handle to maintain equilibrium of the 100kg bucket. Also, calculate the reactions at the bearings.

Solution FBD Since the bearings at A and B are aligned correctly, only force reactions occur at these supports

Example 5.17 Determine the tension in cables BC and BD and the reactions at the ball and socket joint A for the mast.

Solution FBD Five unknown force magnitudes

Solution Solving,

Example 5.18 Rod AB is subjected to the 200N force. Determine the reactions at the ball and socket joint A and the tension in cables BD and BE.

Solution FBD

Solution Equations of Equilibrium Since rC = 1/2rB,

Solution Solving,

Example 5.19 The bent rod is supported at A by a journal bearing, at D by a ball and socket joint, and at B by means of cable BC. Using only one equilibrium equation, obtain a direct solution for the tension in cable BC. The bearing at A is capable of exerting force components only in the z and y directions since it is properly aligned.

Solution FBD Six unknown Three force components caused by ball and socket joint Two caused by bearing One caused by cable

Solution Equations of Equilibrium Direction of the axis is defined by the unit vector

Chapter Summary Free-Body Diagram
Draw a FBD, an outline of the body which shows all the forces and couple moments that act on the body A support will exert a force on the body in a particular direction if it prevents translation of the body in that direction A support will exert a couple moment on a body is it prevents rotation

Chapter Summary Free-Body Diagram
Angles are used to resolved forces and dimensions used to take moments of the forces Both must be shown on the FBD

Chapter Summary Two Dimensions
∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be applied when solving 2D problems For most direct solution, try summing the forces along an axis that will eliminate as many unknown forces as possible

Chapter Summary Three Dimensions
Use Cartesian vector analysis when applying equations of equilibrium Express each known and unknown force and couple moment shown on the FBD as a Cartesian vector Set the force summation equal to zero Take moments about point O that lies on the line of action of as many unknown components as possible

Chapter Summary Three Dimensions
From point O, direct position vectors to each force, then use the cross product to determine the moment of each force The six scalar equations of equilibrium are established by setting i, j and k components of these force and moment sums equal to zero

Chapter Review

Engineering Mechanics: Statics

Similar presentations

Presentation on theme: "Engineering Mechanics: Statics"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Engineering Mechanics: Statics

Similar presentations

Presentation on theme: "Engineering Mechanics: Statics"— Presentation transcript:

Similar presentations

About project

Feedback