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ME2560 – Statics Chapter VI. Equilibrium of a Rigid Body Fall 2018 1 ME 2560 STATICS Chapter VI. Equilibrium of a Rigid Body

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Page 1: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

1

ME 2560 STATICS

Chapter VI. Equilibrium of a Rigid Body

Page 2: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

2

1. Conditions for Rigid Body Equilibrium• A rigid body (RB) is form by an infinite number of particles. • Individual particles of the RB are subjected to internal and external forces.• Internal forces (fi) are caused by the interaction of the particle with the particles adjacent to it.• External forces (Fi) are those caused by external effects such as gravitational, electrical, magnetic, or contact forces.•The condition required for the particle to be in equilibrium is that

0 ii fF

Page 3: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

3

If the equilibrium equation is applied to each of the other particles in thebody:

• The summation of the internal forces will equal zero since the internalforces between particles within the body will occur in equal but oppositecollinear pairs (Newton’s third law).• Thus, only the summation of external forces remains in the equationfor equilibrium:

•For summation of moments acting on the ith particle about an arbitrarypoint, the equilibrium equation for a specific particle results as

0 ii fF

0 iF

0)( iii fFr

Page 4: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

4

Writing similar equations for all the particles forming the body, the summation of moments for all the particles is

The second term is zero since the internal forces occur in equal but opposite collinear pairs, therefore the resultant moment of each pair of forces about point O is zero. Thus,

Therefore, the equilibrium equations for a rigid body are

0 iiii frFr

0 oM

00 oMF

Page 5: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

5

2. Free–Body Diagrams (FBD)A FBD is a sketch of an element of a system where all the remainingcomponents of the system are removed and all of the forces acting onthe body of interest are shown.

•The correct computation of the forces and moments acting on a RBhighly depends on the correct presentation of the FBD.

•There are a number of different structural supports, which imposedifferent sort of forces on a rigid body.

•If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction.

•If a support prevents rotation, then a couple moment is exerted on the body.

Page 6: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

6

Three of the most common supports are:•Roller Support.•Pin Support.•Fixed Support

Page 7: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

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Three of the most common supports are:•Roller Support.•Pin Support.•Fixed Support

Page 8: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

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External and Internal ForcesSince a rigid body is formed by an infinite number of particles, bothexternal and internal loadings may act on it.However, when preparing a free–body diagram the internal forces arenot presented in the diagram since these forces always occur in oppositecollinear pairs having a net effect of zero on the body.

Weight and Center of Gravity

When a body is subjected to a gravitational field, each of its particleshas a specified weight. The sum of all the individual weights representsthe weight of the body W and the location where this weight W actscorresponds to the center of gravity.

Page 9: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

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3. Equations of Equilibrium

The conditions sufficient and necessary to ensure equilibrium in a rigidbody are:•Summation of forces equal to zero•Summation of moments equal to zero.

In a coplanar system, these equations are reduced to:

0

00

o

yx

M

FF

Page 10: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

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4. Two and Three Force MembersTwo–Force Members: When a member is subjected to no couplemoments and forces are applied only at two points on the member, theelement is called two–force member.

FB= –FA

Two force members frequently found.

Page 11: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

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Two Force Members

Page 12: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

12

Three–Force Members. If a member is subjected to only three forcesthen it is necessary that the forces be concurrent or parallel for themember to be in equilibrium.

Three force member.

F1

Page 13: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

13

5. Equilibrium in Three DimensionsThe first step when analyzing a three dimensional problem is to draw a FBD. This must include:•The proper reactions at the supports. •The reactive forces and couple moments acting on the body analyzed

As in the two dimensional case, a force is developed by a support that restricts the translation of the attached member, whereas a couple moment is developed when the rotation of the attached member is prevented.

Page 14: ME 2560 STATICS - Western Michigan Universityhomepages.wmich.edu/~s9montef/ME2560PresentationChVI.pdf · ME2560 –Statics Chapter VI. Equilibrium of a Rigid Body Spring 2018 11 5

ME2560 – Statics

Chapter VI. Equilibrium of a Rigid Body

Fall 2018

14

The conditions for equilibrium if a rigid body subjected to a threedimensional force system require that both the resultant force and theresultant couple moment acting on the body be equal to zero.

The two equations for equilibrium in a rigid body can be expressed invector form as:

These two equations can also be expressed as

Or

00 oMF

0ˆˆˆ

0ˆˆˆ

kMjMiMM

kFjFiFF

zyxo

zyx

000

000

zyx

zyx

MMM

FFF