chapter 04 part 1

7/28/2019 Chapter 04 Part 1

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Force System Resultants4

Engineering Mechanics:

Statics in SI Units, 12e

Copyright © 2010 Pearson Education South Asia Pte Ltd




Chapter Objectives

• Concept of moment of a force in two and three

dimensions

• Method for finding the moment of a force about a

specified axis.• Define the moment of a couple.

• Determine the resultants of non-concurrent force

systems

• Reduce a simple distributed loading to a resultant forcehaving a specified location




Chapter Outline

1. Moment of a Force – Scalar Formation

2. Cross Product

3. Moment of Force – Vector Formulation

4. Principle of Moments5. Moment of a Force about a Specified Axis

6. Moment of a Couple

7. Simplification of a Force and Couple System

8. Further Simplification of a Force and Couple System

9. Reduction of a Simple Distributed Loading




4.1 Moment of a Force – Scalar Formation

• Moment of a force about a point or axis – a measure

of the tendency of the force to cause a body to rotate

about the point or axis

• Torque –

tendency of rotation caused by Fx or simplemoment (Mo) z





Magnitude

• For magnitude of MO,

MO = Fd (Nm)

where d = perpendicular distancefrom O to its line of action of force

Direction• Direction using “right hand rule”





Resultant Moment

• Resultant moment, MRo = moments of all the forces

MRo = ∑Fd


http://slidepdf.com/reader/full/chapter-04-part-1 7/25Copyright © 2010 Pearson Education South Asia Pte Ltd

Example 4.1

For each case, determine the moment of the force about

point O.



Solution

Line of action is extended as a dashed line to establish

moment arm d.

Tendency to rotate is indicated and the orbit is shown as

a colored curl.

)(.200)2)(100()( CW mN mN M a o



Solution

)(.5.37)75.0)(50()( CW mN mN M b o

)(.229)30cos24)(40()( CW mN mmN M c o



Solution

)(.4.42)45sin1)(60()( CCW mN mN M d o

)(.0.21)14)(7()( CCW mkN mmkN M e o



4.2 Cross Product

• Cross product of two vectors A and B yields C, which

is written as

C = A X B

Magnitude• Magnitude of C is the product of

the magnitudes of A and B and

sine of angle θ

• For angle θ, 0° ≤ θ ≤ 180° C = AB sinθ



4.2 Cross Product

Direction

• Vector C has a direction that is perpendicular to the

plane containing A and B such that C is specified by

the right hand rule• Expressing vector C when

magnitude and direction are known

C = A X B = ( AB sinθ)uC




4.2 Cross Product

Laws of Operations

1. Commutative law is not valid

A X B ≠ B X A

Rather,

A X B = - B X A

• Cross product B X A yields a

vector opposite in direction to C

B X A = -C




4.2 Cross Product

Laws of Operations

2. Multiplication by a Scalar

a( A X B ) = (aA) X B = A X (aB) = ( A X B )a

3. Distributive Law

A X ( B + D ) = ( A X B ) + ( A X D )

• Proper order of the cross product must be maintained

since they are not commutative




4.2 Cross Product

Cartesian Vector Formulation

• A more compact determinant in the form a special

utility for a cross-product in Cartesian vector form

z y x

z y x

B B B

A A A

k ji

B X A




4.3 Moment of Force - Vector Formulation

• Moment of force F about point O can be expressed

using cross product

MO = r X F

Magnitude

• For magnitude of cross product,

M O = rF sinθ





Direction

• Direction and sense of MO are determined by right-

hand rule

*Note:- “curl” of the fingers indicates the sense of rotation

- Maintain proper order of r and F since cross product

is not commutative





Principle of Transmissibility

• For force F applied at any point along the line of action

of the force, moment about O is MO = r x F

• F has the properties of a sliding vector, thus

MO = r 1 X F = r 2 X F = r 3 X F





Cartesian Vector Formulation

• For force expressed in Cartesian form,

• With the determinant expended,

MO = (r y F z – r z F y )i

– (r x F z - r z F x ) j + (r x F y – y F x )k

z y x

z y xO

F F F

r r r

k ji

F X r M





Resultant Moment of a System of Forces

• Resultant moment of forces about point O can be

determined by vector addition

MRo = ∑(r x F)




Example 4.4

Two forces act on the rod. Determine the resultant

moment they create about the flange at O. Express the

result as a Cartesian vector.




Solution

Position vectors are directed from point O to each force

as shown.

These vectors are

The resultant moment about O is

m254

m5

k jir jr

OB

OA

mkN604030

304080

254

204060

050

k ji

k jik ji

F r F r F r M OBOAO




4.4 Principles of Moments

• Also known as Varignon’ s Theorem

“Moment of a force about a point is equal to the sum of

the moments of the forces’ components about the point”

• Since F = F1 + F2,

MO = r X F

= r X (F1 + F2)

= r X F1 + r X F2




Example 4.5

Determine the moment of the force about point O.




Solution

The moment arm d can be found from trigonometry,

Thus,

Since the force tends to rotate or orbit clockwise about

point O, the moment is directed into the page.

m898.275sin3 d

mkN5.14898.25 Fd M O

chapter 04 part 1

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