CHAP 5 Equilibrium of a Rigid body

5.1 Conditions for Rigid body equilibrium

Consider a rigid body which is at rest or moving with x y z reference at constant velocity

x

2F

3F

4F

ij

y

z

1F

rigid body

Free body diagram of ith particle of the body

ijf

iF

ir

x

y

z

i

j

External force (外力 ) : gravitational, electrical, magnetic or contact force

iF

Internal force (內力 )if

if

n

j 1

ijf ij

Force equilibrium equation for particle i

0ii fF (Newton first law)

0 ii fF

0if

0FFi

Force equilibrium equation for the whole body

(Newton’s 3rd law 作用力與反作用力 )

Moment of the forces action on the ith particle about pt. O

iiijo fFrM

0 iiii frFr

Moment equilibrium equation for the body

0

0

0

0

jjo

jj

jjjj

jjjjo

FrM

fr

frFr

fFrM

Equations of equilibrium for a rigid body are

0 F

0 oM

力平衡

力矩平衡

1. Free-body Diagram (1) F.B.D A sketch of the outlined shape of the body represents it as being

isolated or “free” from its surrounding , i.e ., a free body”.

5.2 Equilibrium in Two Dimensions

(2) Support Reactions

A .Type of support : see Table 5-1

B . General rules for support reaction:

If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction . Likewise, if rotation is prevented, a couple moment is exerted on the body.

(a) roller or cylinder support

AyF

ByF

xF

Fy

xF

Fy

(b) pin support

Examples:

(c) Fixed support

M

FAy

FAx

(3) External and Internal forces

A. Internal force

Not represented on the F.B.D. became their net effect on the body is zero.

B. External force

Must be shown on the F.B.D.

(a) “Applied” loadings

(b) Reaction forces 反作用力(c) Body weights 重力

(4) Weight and the center of gravity

The force resultant from the gravitational field is referred as the weight of the body, and the location of its point of application is the center of gravity G.

w

body

平行力 等效力系

(ch4)

w

P=G( 重心 )

2. Equations of Equilibrium for 2D rigid body

y

x Couple moment

1F

2F

3F

1M

2M

0M0F 0

0M

0F

0F

0

y

x

Here: Fx

Fy

0M

algebraic sum of x components of all force on the body.

algebraic sum of y components of all force on the body.

algebraic sum of couple moments and moments of allthe force components about an axis xy plane and⊥passing 0.

(1) Conditions of equilibrium

(2) Alternative equilibrium equation

0M

0M

0F

B

A

a

When the moment points A and B do not

lie on a line that is “perpendicular” to the axis a.

(A)

ARA

R

MM

FF

aaFF

MM

Ra

RAA

0

00 0/ RRBAB FoFrM

0M

0M

0M

C

B

A

a

a

A

B

C AF

AM

(B)

Points A, B and C do not lie on the same line

ARA

R

MM

FF

0

00

BR

RAA

MBAcollinearF

MM0/ RRCAC FoFrM

600N

200N

2m 3m 2m

100N

y

xBxA B

ByAy

3 unknown Ax, Bx, By

Equations of equilibrium

0.0245sin600545sin600721000

020010045sin6000

045cos6000

AyM

ByAyFy

BxFx

B

3 equations for 3 unknowns

(3) Example

BAR FFF 0

0Mo

5.3 Two-and Three-Force Members

1. Two-Force member

A member subject to no couple moments and forces applied at only two points on the member.

A

B

A

B

FA

FBEquations of Equilibrium

2. Three-Force member

0F 0F 0Mo 0Mo

A member subject to only three forces, which are either concurrent or parallel if the member is in equilibrium.

(1)Concurrent (2)parallel(3力交於O 點 ) (3力相交無限遠處 )

o

F1F2

F3

F1 F2

F3

5.4 Equilibrium in Three Dimensional Rigid Body

1. Free Body Diagrams

(1)F.B.D

Same as 2D equilibrium problems

(2)Support Reactions

A. Types of support:see Table 5-2

B. General rules for reaction

Same as two-dimensional case

Examples:

(a) Ball and Socket joint

No translation along any direction

Rotate freely about any axis

Fx

Fy

Fz

3 reaction forces

(b) single journal bearing

Rotate freely about its longitudinal axis

Translate along its longitudinal direction

two unkown forces

and couple momentsx

yMx

Mz

z

Fz

Fx (c) single pin

Only allow to rotate about a specific axis.

x

y

z

Mz

My

FxFy

Fz

Three unkown forces

and two couple moments

A. Vector equations of equilibrium

B. Scalar equations of equilibrium

0F 0Mo

0Fx

0Fy

0Fz

0Mx

0My

0Mz

2. Equations of Equilibrium

1. Redundant constraints

(1) Redundant constraints

Redundant supports are more than necessarily to hold a body in equilibrium.

2KN-m 500N

5.6 Constraints for a rigid body

Ex:

Equation of motion=3

5 unknown reactions >3 equation of motion

there are two support reactions which are redundant supports and more than necessarily.

(2) Statically indeterminate 靜不定 There are more unknown loadings on the body than equations of equilibrium available for the solution.

F.B.D of above example

x

y

A BC

2KN-m

Ay By Cy

MA

Ax

500N

unknown loadings AX,AY,MA,BY,CY;5

Equations of equilibrium ΣFX=0,ΣFY=0,ΣMA =0;3

5>3

Statically indeterminate structure

(3)Solutions for statically indeterminate structure

Additional equations are needed ,which are obtained from the deformation condition at the points of redundant support based on the mechanics of deformation, such as mechanics of materials.

Equations of Equilibrium for above example are

+ ΣFX=0 AX=0

+ΣFY=0 500-AY-BY-CY=0

+ΣMA=0 MA-2-DYBY-DCCY=0

Need two more equations to solve the five unknown forces.

(1) Reaction force = equations of equilibrium

If this kind of improper constraint occurs then system is instable

A. The lines of action of the reactive forces intersect points on a common axis (concurrent).

2.Improper Constraints

A

B

C

0.2m

100N

Ao

B

C100NF.B.D FC

FAFB

body will rotate about Z-axis or point O0

0

o

yx

M

FF

o

FC

A

B C

FB

FA

100N

F.B.D

Body will translate along x direction.

B. The reactive forces are all parallel

100NA

B C

0 xF

(2) Reaction forces < equations of equilibrium

If the body is partially constrained then it is in instable condition

100N

Stable?

F.B.D

0

0

0

o

y

x

M

F

F

FA FB

100N

o

Not in equilibrium