ch 05111
TRANSCRIPT
Engineering Mechanics: StaticsEngineering Mechanics: Statics
Chapter 5: Equilibrium of a Rigid
Body
Chapter 5: Equilibrium of a Rigid
Body
Chapter ObjectivesChapter 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 OutlineChapter Outline
Conditions for Rigid EquilibriumFree-Body DiagramsEquations of EquilibriumTwo and Three-Force MembersEquilibrium in Three DimensionsEquations of EquilibriumConstraints for a Rigid Body
5.1 Conditions for Rigid-Body Equilibrium
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
5.1 Conditions for Rigid-Body Equilibrium
5.1 Conditions for Rigid-Body Equilibrium
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
5.1 Conditions for Rigid-Body Equilibrium
5.1 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,
∑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)
5.1 Conditions for Rigid-Body Equilibrium
5.1 Conditions for Rigid-Body Equilibrium
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
5.1 Conditions for Rigid-Body Equilibrium
5.1 Conditions for Rigid-Body Equilibrium
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
5.1 Conditions for Rigid-Body Equilibrium
5.1 Conditions for Rigid-Body Equilibrium
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
5.2 Free-Body Diagrams5.2 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 Diagrams5.2 Free-Body Diagrams
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
5.2 Free-Body Diagrams5.2 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 Diagrams5.2 Free-Body Diagrams
Support ReactionsRoller or cylinderPrevent 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 Diagrams5.2 Free-Body Diagrams
Support ReactionsPin 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 Diagrams5.2 Free-Body Diagrams
Support ReactionsFixed 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 Diagrams5.2 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 Diagrams5.2 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 Diagrams5.2 Free-Body Diagrams
Floor beams of this building are welded together and thus form fixed connections
Type 10 connections
5.2 Free-Body Diagrams5.2 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 Diagrams5.2 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 Diagrams5.2 Free-Body Diagrams
Weight and Center of GravityCenter 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 Diagrams5.2 Free-Body Diagrams
Idealized ModelsNeeded to perform a correct force
analysis of any objectCareful selection of supports, material,
behavior and dimensions for trusty results
Complex cases may require developing several different models for analysis
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
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
5.2 Free-Body Diagrams5.2 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
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
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
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
Procedure for Drawing a FBD1. 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 Diagrams5.2 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 Diagrams5.2 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 Diagrams5.2 Free-Body Diagrams
Example 5.1Draw the free-body diagram of the
uniform beam. The beam has a mass of 100kg.
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
SolutionFree-Body Diagram
5.2 Free-Body Diagrams5.2 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 Diagrams5.2 Free-Body Diagrams
Example 5.2Draw 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 Diagrams5.2 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 Diagrams5.2 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 Diagrams5.2 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 leverFs = 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 Diagrams5.2 Free-Body Diagrams
Example 5.3Two 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 Diagrams5.2 Free-Body Diagrams
SolutionFor idealized models,
Free-Body Diagram of pipe A
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
Solution 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
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
Solution*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 Diagrams5.2 Free-Body Diagrams
Example 5.4Draw 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.
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
Solution 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
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
SolutionDirection 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 Diagrams5.2 Free-Body Diagrams
SolutionPlatform’s weight = 200(9.81) =
1962NForce 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 Diagrams5.2 Free-Body Diagrams
Example 5.5The free-body diagram of each object
is drawn. Carefully study each solution
and identify what each loading represents.
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
SolutionView Free Body Diagram
5.2 Free-Body Diagrams5.2 Free-Body Diagrams
Solution
5.3 Equations of Equilibrium5.3 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Alternative Sets of Equilibrium EquationsConsider FBD of an arbitrarily shaped
bodyAll 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Procedure for AnalysisFree-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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Procedure for AnalysisFree-Body DiagramThe 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Procedure for AnalysisEquations 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Procedure for AnalysisEquations 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.6Determine the horizontal and vertical components of reaction for the beam
loaded. Neglect the weight of the beam in the calculations.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionFBD 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionEquations 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
NB
BN
M
x
x
B
424
045cos600
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
NB
BNNNN
F
NA
mAmNmNmN
M
y
y
y
y
y
B
405
020010045sin600319
;0
319
0)7()2.0)(45cos600()5)(45sin600()2(100
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionChecking,
NB
mBmN
mNmNmN
M
y
y
A
405
0)7()7)(200(
)5)(100()2.0)(45cos600()2)(45sin600(
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.7The 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.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionFBD 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionFBD 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionEquations 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
NT
mTmN
M A
500
0)2.0()2.0(500
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
NA
NNA
F
NAx
NA
F
y
y
y
x
x
933
030cos500500
;0
250
030sin500
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.8The link is pin-connected at a and rest a smooth support at B. Compute the
horizontal and vertical components of reactions at pin
A
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionFBDReaction NB is perpendicular to the
link at BHorizontal and vertical
components of reaction are represented at A
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionEquations of Equilibrium
NAx
NA
F
NN
mNmNmN
M
x
x
B
B
A
100
030sin200
;0
200
0)75.0()1(60.90
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
NA
NNA
F
y
y
y
233
030cos20060
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.9The 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.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionFBDBolt acts as a “fixed support” it exerts
force components Ax and Ay and a torque MA on the wrench at A
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionEquations of Equilibrium
NAx
NNA
F
mNM
mNmNM
M
x
x
A
A
A
00.5
060cos3013
552
;0
.6.32
0)7.0)(60sin30()3.0(13
1252
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
NmM
mNmNM
MCCW
NA
NNA
F
A
A
y
y
y
y
6.32
07.060sin303.013
1252
;0
0.74
060sin3013
1252
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
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
1.8600.5
0.74tan
1.740.7400.5
1
22
N
N
NFA
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution Checking,
0.8.51.6.32.2.19
0)7.0(0.74.6.32)4.0(1312
52
;0
mNmNmN
mNmNmN
MC
Example 5.10Placement 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.
5.3 Equations of Equilibrium
5.3 Equations of Equilibrium
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionFBD 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
NAx
NA
F
NF
mN
mNmF
M
x
x
BC
BC
A
7900
07900
;0
7900
0)0625.0(30sin2240
)2(30cos2240)5.0(
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
Checking,
0)0625.0(30sin2240)1(30cos2240
)30cos1(2240)5.0(7900
;0
2240
02240
;0
mNmN
mNmN
M
NA
NA
F
B
y
y
y
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.11The 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.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionFBD 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionEquations of Equilibrium
030cos30cos300
;0
030sin30sin
;0
0)8)(30cos300()6(.4000)2(
;0
''
''
''
yy
y
xyy
x
yy
A
BCN
F
ABC
F
mNmCmNmB
M
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
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
kNNC
kNNB
y
y
35.14.1346
10.1000
'
'
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution Top roller at B serves as the support
rather than the bottom one
NA
ANN
x
x
173
030sin0.100030sin4.1346
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.12The 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.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution Idealized model of the ramp Center of gravity located at the midpoint since
the ramp is approximately uniform
FBD of the Ramp
View Free Body Diagram
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionEquations of Equilibrium
By the principle of transmissibility, locate T at C
md
md
NT
N
TmT
M A
0154.120sin
2
10sin
5985
0)30cos5.1(1600
)30cos2(20sin)30sin2(20cos
;0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionSince there are two cables
supporting the ramp,
T’ = T/2 = 2992.5N
5.4 Two- and Three-Force Members
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
5.4 Two- and Three-Force Members
5.4 Two- and Three-Force Members
Two-Force MembersExample 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
5.4 Two- and Three-Force Members
5.4 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
5.4 Two- and Three-Force Members
5.4 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
5.4 Two- and Three-Force Members
5.4 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”
5.4 Two- and Three-Force Members
5.4 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 two-force member
5.4 Two- and Three-Force Members
5.4 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
5.4 Two- and Three-Force Members
5.4 Two- and Three-Force Members
Example 5.13The 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.
5.4 Two- and Three-Force Members
5.4 Two- and Three-Force Members
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
View Free Body Diagram
5.4 Two- and Three-Force Members
5.4 Two- and Three-Force Members
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
5.4 Two- and Three-Force Members
5.4 Two- and Three-Force Members
Solution Equations of
Equilibrium
Solving,
kNF
kNF
FF
F
NFF
F
A
A
y
A
x
32.1
07.1
045sin3.60sin
;0
040045cos3.60cos
;0
3.604.0
7.0tan 1
5.5 Equilibrium in Three Dimensions (FBD)
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
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
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 by222zyx FFFF
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
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
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
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
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (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
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
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
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
Example 5.14Several 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.
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
Solution
View Free Body Diagram
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
Solution
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
Solution
5.5 Equilibrium in Three Dimensions (FBD)
5.5 Equilibrium in Three Dimensions (FBD)
Solution
5.6 Equations of Equilibrium5.6 Equations of Equilibrium
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
5.6 Equations of Equilibrium5.6 Equations of Equilibrium
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 = 0i, j and k components are
independent from one another
5.6 Equations of Equilibrium5.6 Equations of Equilibrium
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
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
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Redundant ConstraintsExample 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
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Redundant ConstraintsExample 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
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
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 ConstraintsExample 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
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Improper Constraints Instability of the body also can be
caused by the parallel reactive forcesExampleSummation of
forces along the x axis will not be equal to zero
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
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 constrainedExample If O is a point not located on line AB,
loading condition and equations of equilibrium are not satisfied
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
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
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Procedure for AnalysisFree 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
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Procedure for AnalysisFree 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
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Procedure for AnalysisEquations 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
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Procedure for AnalysisEquations 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
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Example 5.15The 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.
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionFBDFive unknown reactions acting on the
plateEach reaction assumed to act in a
positive coordinate direction
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionEquations 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
0981300;0
0;0
0;0
NNTBAF
BF
BF
Czzz
yy
xx
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Solution
Components of force at B can be eliminated if x’, y’ and z’ axes are used
0)3(.200)5.1(981)5.1(300
;0
0)2()2(300)1(981;0
0.200)3()3()5.1(981)5.1(300
;0
0)2()1(981)2(;0
'
'
mTmNmNmN
M
mAmNmNM
mNmAmBmNmN
M
mBmNmTM
C
y
zx
zz
y
ZCx
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionSolving,
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 5.16The 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.
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionFBDSince the bearings at A and B are
aligned correctly, only force reactions occur at these supports
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionEquations of Equilibrium
0
0)8.0(
;0
3.424
0)4.0)(6.377()8.0()5.0(981
;0
6.377
0)30cos3.0()1.0(981
;0
y
y
z
z
z
y
x
A
mA
M
NA
mNmAmN
M
NP
mPmN
M
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionEquations of Equilibrium
NB
B
F
B
B
F
A
F
z
z
z
y
y
y
x
x
934
06.3779813.424
;0
0
00
;0
0
;0
Example 5.17Determine the tension in cables BC and
BD and the reactions at the ball and socket
joint A for the mast.
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionFBDFive unknown force
magnitudes
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionEquations of Equilibrium
0)9
6707.0()
9
61000()
9
3707.0(
0
;0
9
6
9
6
9
3
707.0707.0
}1000{
kTTAjTAiTTA
TTFF
F
kTjTiTr
rTT
kTiTT
kAjAiAF
NjF
DCzDyDCx
DCA
DDDBD
BDDD
CCC
zyxA
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionEquations of Equilibrium
0)96
96
93
707.0707.01000(6
0)(
;0
096
707.0;0
096
1000;0
093
707.0;0
kTjTiTkTiTjXk
TTFXr
M
TTAF
TAF
TTAF
DDDCC
DCB
A
DCzz
Dyy
DCxx
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Solution
Solving,
NA
NA
NA
NT
NT
TTM
TM
jTTiT
z
y
x
D
C
DCy
Dx
DCD
1500
0
0
1500
707
0224.4;0
060004;0
0)224.4()60004(
Example 5.18Rod 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.
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionFBD
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionEquations of Equilibrium
0)200()()(
0
;0
}200{
kAjTAiAA
FTTFF
F
NkF
iTT
iTT
kAjAiAF
zDyEx
DEEA
DD
EE
zyxA
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionEquations of Equilibrium
Since rC = 1/2rB,0)()221()200()115.0(
0)(
;0
0200;0
0;0
0;0
jTiTXkjikXkji
TTXrFXr
M
AF
TAF
TAF
DE
DEBC
A
zz
Dyy
Exx
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
Solution
Solving,
NA
NA
NA
NT
NT
TTM
TM
TM
kTTjTiT
z
y
x
E
D
EDz
Ey
Dx
EDED
200
100
50
50
100
02;0
01002;0
02002;0
0)2()1002()2002(
Example 5.19The 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.
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionFBD Six unknown Three force components
caused by ball and socket joint
Two caused by bearing One caused by cable
View Free Body Diagram
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionEquations of Equilibrium Direction of the axis is defined by the
unit vector
0)(
707.0707.0
21
21
rXFuM
ji
jirr
u
DA
DA
DA
5.7 Constraints for a Rigid Body
5.7 Constraints for a Rigid Body
SolutionEquations of Equilibrium
NT
kTiTji
kXj
kTjTiTXjji
XWrXTru
B
BB
BBB
EBB
572857.0
5.490
0]286.0)5.4908577.0).[(707.0707.0(
0)]981()5.0(
)7.0
6.0
7.0
3.0
7.0
2.0()1).[(707.0707.0(
0)(
Chapter SummaryChapter 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 SummaryChapter Summary
Free-Body DiagramAngles are used to resolved forces
and dimensions used to take moments of the forces
Both must be shown on the FBD
Chapter SummaryChapter Summary
Two Dimensions∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be
applied when solving 2D problemsFor most direct solution, try
summing the forces along an axis that will eliminate as many unknown forces as possible
Chapter SummaryChapter 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 SummaryChapter 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 ReviewChapter Review
Chapter ReviewChapter Review
Chapter ReviewChapter Review