6161103 5.7 constraints for a rigid body

5.7 Constraints for a Rigid Body5.7 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 supportsor 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 ConstraintsExample� For the 2D and 3D problems, both are

statically indeterminate because of additional supports reactionssupports reactions

� In 2D, there are 5 unknowns but 3 equilibrium equations can be drawn


Redundant ConstraintsExample

� In 3D, there are 8 unknowns but 6 equilibrium equations can be drawn 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 � 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


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


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� The body become partially constrainedExample� 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 axisinsect 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 AnalysisFree Body Diagram� Draw an outlined shape of the body� Show all the forces and couple moments

acting on 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 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 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, 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 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 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 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 � Each reaction assumed to act in a positive coordinate direction


Solution

Equations of Equilibrium

0;0

0;0

==∑

==∑

BF

BF xx

� 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

=−−++=∑

==∑

NNTBAF

BF

Czzz

yy


Solution

0.200)3()3()5.1(981)5.1(300

;0

0)2()1(981)2(;0

=−−−+

=∑

=+−=∑

mNmAmBmNmN

M

mBmNmTM

y

ZCx

� 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

'

'

=+−−

=∑

=−+=∑

=−−−+

mTmNmNmN

M

mAmNmNM

mNmAmBmNmN

C

y

zx

zz


Solution

Solving,

Az = 790N Bz = -217N TC = 707N

� The negative sign indicates Bz acts downward� 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.


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 correctly, only force reactions occur at these supports


Solution


0)30cos3.0()1.0(981

;0

=−

=∑ x

mPmN

Mo

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

=

=−

=∑

=

=++−

=∑

=

=−

y

y

z

z

z

y

A

mA

M

NA

mNmAmN

M

NP

mPmN


Solution


A

Fx

0

;0

=

=∑

NB

B

F

B

B

F

A

z

z

z

y

y

y

x

934

06.3779813.424

;0

0

00

;0

0

=

=−+−

=∑

=

=+

=∑

=

Example 5.17

Determine the tension in cables BC and BD

and the reactions at the ball and socket joint

A for the mast.


A for the mast.


Solution

FBD

� Five unknown force magnitudesmagnitudes


Solution


}1000{

++=

−=

kAjAiAF

NjFrrrr

rr

0)9

6707.0()

9

61000()

9

3707.0(

0

;0

9

6

9

6

9

3

707.0707.0

=−−+++−+−+

=+++

=∑

−+−=

=

−=

++=

kTTAjTAiTTA

TTFF

F

kTjTiTr

rTT

kTiTT

kAjAiAF

DCzDyDCx

DCA

DDDBD

BDDD

CCC

zyxA

rrr

rrrr

r

rrrr

r

rrr


Solution


6

093

707.0;0 =−+=∑ TTAF DCxx

0)96

96

93

707.0707.01000(6

0)(

;0

096

707.0;0

096

1000;0

=−+−−+−

=++

=∑

=−−=∑

=++−=∑

kTjTiTkTiTjXk

TTFXr

M

TTAF

TAF

DDDCC

DCB

A

DCzz

Dyy

rrrrrrr

rrrr

r


Solution

TTM

TM

jTTiT

Dx

DCD

0224.4;0

060004;0

0)224.4()60004(

=−=∑

=+−=∑

=−++−rr

Solving,

NA

NA

NA

NT

NT

TTM

z

y

x

D

C

DCy

1500

0

0

1500

707

0224.4;0

=

=

=

=

=

=−=∑

Example 5.18

Rod AB is subjected to the 200N force.

Determine the reactions at the ball and

socket joint A and the


socket joint A and the

tension in cables BD

and BE.


Solution

FBD


Solution


=

++=

iTT

kAjAiAF zyxArr

rrrr

0)200()()(

0

;0

}200{

=−++++

=++++

=∑

−=

=

=

kAjTAiAA

FTTFF

F

NkF

iTT

iTT

zDyEx

DEEA

DD

EE

rrr

rrrrr

r

rr

rr


Solution


0;0

0;0

=+=∑

=+=∑

TAF

TAF

Dyy

Exx

Since rC = 1/2rB,0)()221()200()115.0(

0)(

;0

0200;0

0;0

=+−++−+

=++

=∑

=−=∑

=+=∑

jTiTXkjikXkji

TTXrFXr

M

AF

TAF

DE

DEBC

A

zz

Dyy

rrrrrrrrrr

rrrrr

r


Solution

TM

TM

kTTjTiT

Ey

Dx

EDED

01002;0

02002;0

0)2()1002()2002(

=+−=∑

=−=∑

=−++−+−rrr

Solving,

NA

NA

NA

NT

NT

TTM

z

y

x

E

D

EDz

200

100

50

50

100

02;0

=

−=

−=

=

=

=−=∑

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


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 � Three force components caused by ball and socket joint

� Two caused by bearing

� One caused by cable


Solution


� Direction of the axis is defined by the unit vectorvector

0)(

707.0707.0

21

21

=∑⋅=∑

−−=

−−==

rXFuM

ji

jirr

u

DA

DA

DA


Solution


XWrXTru EBB

6.03.02.0

0)( =+⋅

NT

kTiTji

kXj

kTjTiTXjji

B

BB

BBB

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(

==

=++−−−

=−−+

+−−−−

6161103 5.7 constraints for a rigid body

Business

rigid bodyprocedure

supportsredundant constraints

satisfiedthe body

whena body

equations equilibrium

equilibrium equations

analysisfree body diagramdraw

fewer reactive forces