Chap. 2 Force Vectors

Chap. 5 Equilibrium of a Rigid Body

5-2

Chapter Outline

Conditions for Rigid-Body EquilibriumFree-Body DiagramsEquations of EquilibriumTwo and Three-Force MembersConstraints and Statical Determinacy

5-3

Conditions for Rigid-Body Equilibrium

The equilibrium of a body is expressed as

Consider summing moments about some other point, such as point A, we require

( )

==

==

0

0

OOR

R

MM

FF

( ) =+= 0ORRA MFrM

5-4

FREE BODY DIAGRAMS (TWO DIMENSION)

(constraint)

5-5

(continued with ch5a.ppt)

5-7

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

5-8

Free-Body DiagramsSolutionFree-Body Diagram

5-9

Free-Body DiagramsExample 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.

5-10

Free-Body DiagramsSolution

For idealized models,

Free-Body Diagram of pipe A

5-11

p.216, 5-9. Draw the free-body diagram of the beam, which is pin- connected at A and rocker-supported at B.

5-12

5.3 Equations of EquilibriumFor equilibrium of a rigid body in 2D,

Fx = 0; Fy = 0; MO = 0Fx and Fy represent the algebraic sums of the x and y components of all the forces acting on the bodyMO 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

5-13

5.3 Equations of EquilibriumAlternative 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 = 0When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis

5-14

Alternative Sets of Equilibrium EquationsA second set of alternative equations is

MA = 0; MB = 0; MC = 0Points A, B and C do not lie on the same line

5.3 Equations of Equilibrium

5-15


Procedure for AnalysisFree-Body Diagram

Establish the x, y, z coordinates axes in any suitable orientationDraw an outlined shape of the bodyShow all the forces and couple momentsacting on the bodyLabel all the loadings and specify their directions relative to the x, y axes

5-16


Procedure for AnalysisFree-Body Diagram

The sense of a force or couple moment having an unknown magnitude but known line of action can be assumedIndicate the dimensions of the body necessary for computing the moments of forcesEquations of Equilibrium

5-17

5.3 Equations of EquilibriumExample 5.5Determine the horizontal and vertical components of reaction for the beam loaded. Neglect the weight of the beam in the calculations.

5-18

5.3 Equations of EquilibriumSolutionFBD

600N force is represented by its x and y components200N 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

5-19

5.3 Equations of EquilibriumSolutionEquations of Equilibrium

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

NBBN

F

x

x

x

424045cos600

;0

==

=+o

5-20


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

=

=+

=+

=

=+

=

o

oo

5-21


SolutionChecking,

NB

mBmNmNmNmN

M

y

y

A

405

0)7()7)(200()5)(100()2.0)(45cos600()2)(45sin600(

;0

=

=+

=oo

5-22

5.3 Equations of EquilibriumExample 5.6The 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.

5-23

5.3 Equations of EquilibriumSolutionFBD of the cord and pulley

Principle of action: equal but opposite reaction observed in the FBDCord exerts an unknown load distribution p along part of the pulleys surfacePulley exerts an equal but opposite effect on the cord

5-24


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

5-25

5.3 Equations of EquilibriumSolutionEquations 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

NTmTmN

M A

5000)2.0()2.0(500

;0

==+

=

5-26

Two- and Three-Force Members

F3 F1 F2

Slender rod (beams)

5-27

5.4 Two- and Three-Force MembersExample 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.

5-28

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 collinearMagnitude of the force is unknown but line of action known as it passes through B and DLever ABC is a three-force member

5-29


Solution FBD

For moment equilibrium, three non-parallel forces acting on it must be concurrent at OForce F on the lever at B is equal but opposite to the force F acting at B on the linkDistance CO must be 0.5m since lines of action of F and the 400N force are known

5-30


Solution Equations of

Equilibrium

Solving,kNFkNF

FF

FNFF

F

A

A

y

A

x

32.107.1

045sin3.60sin

;0040045cos3.60cos

;0

3.604.07.0tan 1

==

=

=+

=+

=+

=

=

oo

oo

o

5-31

p.227, Problem 5-20The train car has a weight of 120 kN and a center of gravity at G. It is suspended from its front and rear on the track by six tires located at A, B, and C. Determine the normal reactions on these tires if the track is assumed to be a smooth surface and an equal portion of the load is supported at both the front and rear tires.

5-32

120 kN

1.5 m

1.2 m

5-33

p.230, Problem 5-32The jib crane is supported by a pin at C and rod AB. If the load has a mass of 2 Mg with its center of mass located at G, determine the horizontal and vertical components of reaction at the pin C and the force developed in rod AB on the crane when x = 5 m.

5-35

p.235, 5-60. The uniform rod has a length l and weight W. It is supported at one end A by a smooth wall and the other end by a cord of length s which is attached to the wall as shown. Show that for equilibrium it is required that h = [(s2 - l2)>3]1>2.

5-38

Free Body Diagrams (Three Dimension)

5-41

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.

5-42

Solution

5-43

Equation of Equilibrium

Vector

Scalar

00=

=

OMF

00===

===

ZYX

ZYX

MMMFFF

5-44

Constraints for a Rigid Body

Redundant constraints

5-45


Improper Constraints-intersect points on a common axis

5-46


Improper Constraints-intersect points on a common axis2-D reaction force

5-47

Constraints for a Rigid BodyImproper Constraints-reactive force are all parallel

5-48


Partially Constrained

5-49

p.253, 5-67. Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these components have weights WA = 225 kN, WB = 40 kN, and WC = 30 kN, determine the normal reactions of the wheels D, E, and F on the ground.

5-50

0X = M

0= YM

RE

5-51

p.255, 5-73. Determine the force components acting on the ball-and-socket at A, the reaction at the roller B and the tension on the cord CD needed for equilibrium of the quarter circular plate.

5-52

p.257, 5-85. The circular plate has a weight W and center of gravity at its center. If it is supported by three vertical cords tied to its edge, determine the largest distance d from the center to where any vertical force P can be applied so as not to cause the force in any one of the cables to become zero.

5-53

d

d

(< r(1+W/P) )

5-54

p. 261, prob. 5-96 The symmetrical shelf is subjected to a uniform load of 4 kPa. Support is provided by a bolt (or pin) located at each end A and A and by the symmetrical brace arms, which bear against the smooth wall on both sides at B and B. Determine the force resisted by each bolt at the wall and the normal force at B for equilibrium.

5-55

Chap. 5 Equilibrium of a Rigid BodyChapter OutlineConditions for Rigid-Body Equilibrium 4 5 6Free-Body DiagramsFree-Body DiagramsFree-Body DiagramsFree-Body Diagrams 115.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of Equilibrium5.3 Equations of EquilibriumTwo- and Three-Force Members 5.4 Two- and Three-Force Members5.4 Two- and Three-Force Members5.4 Two- and Three-Force Members5.4 Two- and Three-Force Members 31 32 33 34p.235, 5-60. The uniform rod has a length l and weight W. It is supported at one end A by a smooth wall and the other end by a cord of length s which is attached to the wall as shown. Show that for equilibrium it is required thath = [(s2 - l2)>3]1>2. 36 37 38 39 40Example 5.14SolutionEquation of EquilibriumConstraints for a Rigid Body Constraints for a Rigid BodyConstraints for a Rigid BodyConstraints for a Rigid BodyConstraints for a Rigid Bodyp.253, 5-67. Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these components have weights WA = 225 kN, WB = 40 kN, and WC = 30 kN, determine the normal reactions of the wheels D, E, and F on the ground. 50p.255, 5-73. Determine the force components acting on the ball-and-socket at A, the reaction at the roller B and the tension on the cord CD needed for equilibrium of the quarter circular plate.p.257, 5-85. The circular plate has a weight W and center of gravity at its center. If it is supported by three vertical cords tied to its edge, determine the largest distance d from the center to where any vertical force P can be applied so as not to cause the force in any one of the cables to become zero. 53p. 261, prob. 5-96The symmetrical shelf is subjected to a uniform load of 4 kPa. Support is provided by a bolt (or pin) located at each end A and A and by the symmetrical brace arms, which bear against the smooth wall on both sides at B and B. Determine the force resisted by each bolt at the wall and the normal force at B for equilibrium. 55

Chap. 2 Force Vectors

Documents

Transcript of Chap. 2 Force Vectors