Ch.04 Force System Resultants

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04. Force System Resultants

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Engineering Mechanics Statics 4.01 Force System Resultants

Chapter Objectives

To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions

To provide a method for finding the moment of a force about an axis

To define the moment of a couple

To present methods for determining the resultants of non-concurrent force systems

To indicate how to reduce a simple distributed loading to a resultant force having a specified location



1. Moment of a Force Scalar Formulation

- The moment of a force about a point provides a measure of

the tendency for rotation (sometimes called a torque)


= = = = = 0



- Magnitude

=

- Direction

- Resultant Moment

+ ()= : ()= 11 22 + 33




- Example 4.1 For each case illustrated in the figure,

determine the moment of the force about point

Solution

= 100 2 = 200

= 50 0.75 = 37.5

= 40 (4 +2300) = 229




- Example 4.1 For each case illustrated in the figure,

determine the moment of the force about point

Solution

= 60 1450 = 42.4

= 7 (4 1) = 21.0



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2. Cross Product

The cross product of two vectors

=

- Magnitude

= , =

- Direction

Specified by the right-hand rule

curling the fingers of the right hand from vector (cross) to

vector , the thumb points in the direction of



2. Cross Product

- Laws of Operation

Commutative law

=

Multiplication by a scalar

= = =

Distributive law

+ = ( ) + ( )



2. Cross Product

- Cartesian Vector Formulation

=



3. Moment of a Force-Vector Formulation

- The moment of a force about point

=

- Magnitude

= = =

- Direction

Determined by the right-hand rule

as it applies to the cross product




- Principle of Transmissibility

= 1 = 2 = 3

- Cartesian Vector Formulation

= =




- Resultant Moment of a System of Forces

= ( )



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- Example 4.3 Determine the moment produced by the force about point . Express the result as a Cartesian vector

= 12

= 4 + 12

=

= 2 (4 + 12 ) (12 )

42 + 122 + (12)2

= 0.4588 + 1.376 1.376

=

= 0 0 12

0.4588 1.376 1.376

= 16.5 + 5.5



Solution =


- Example 4.4 Two forces act on the rod, determine the

resultant moment they create about the flange at . Express the result as a Cartesian vector

Solution

= 5

= 4 + 5 2

=

= 1 + 2

= 0 5 0

60 40 20

+ 4 5 280 40 30

= 30 40 + 60



4. Principle of Moments (Varignons Theorem)

- The moment of a force about a point is equal to the sum of the moments of the components of the force about the point

= = 1 + 2 = 1 + 2

For two-dimensional problems

= =




- Example 4.6 Determine the moment of the force about point

Solution 1

= 3 750

= 2.898

=

= 5 2.89

= 14.5

Solution 2

+ =

= 5450 3300

5450 3300

= 14.5

= 14.5




Solution 3

+ =

= 5750

= 14.5

= 14.5




- Example 4.6 Force acts at the end of the angle bracket, determine the moment of the force about point

Solution 1

+ = 400300 0.2 400300 0.4

= 98.6

= 98.6



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- Example 4.6 Force acts at the end of the angle bracket, determine the moment of the force about point

Solution 2

= 0.4 0.2

= 400300 400300 = 200 346.4

= =

0.4 0.2 0200.0 346.4 0

= 98.6



Fundamental Problems

- F4.1 Determine the moment of the force about point
















- F4.5 Determine the moment of the force about point . Neglect the thickness of the member



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5. Moment of a Force about a Specified Axis

- Scalar Analysis

= =

- Vector Analysis

= 0 = ( )




- The moment of about a axis

= ( + + )

+ ( )

= ( ) =

,, : ,, components of the unit vector of axis

,, : , , components of the position vector

extended from any point on the axis to any

point on the line of action of the force

,, : ,, components of the force vector



= ( )

=

=


- Example 4.7 Determine the resultant moment of the three

forces 1 60 , 2 50 , 3 40 about the ,, axes

Solution

= 1 1 + 2

2 + 3

3

= 60 2 + 50 2 + 0 = 220

and

= 0 50 3 40 2 = 230

= 0 + 0 40 2 = 80




- Example 4.8 Determine the moment produced by the force

, which tends to rotate the rod about the axis

Solution

=

where

=

=0.4 + 0.2

0.42 + 0.22

= 0.8944 + 0.4472

is directed from any point on the axis to any point on the line of action of

, such as: , , , , choose

= = 0.6

= 300




= 0.8944 + 0.4472

= 0.6

= 300

=

=0.8944 0.4472 0

0.6 0 00 0 300

= 80.5

Expressing as a Cartesian vector

=

= 80.50 0.8944 + 0.4472

= 72.0 + 36.0 ()



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- Example 4.9 Determine the magnitude of the moment of force

about segment of the pipe assembly

Solution

=

where

=

=0.3 + 0.4

0.32 + 0.42= 0.6 + 0.8

= = 0.5 + 0.5

=

= 3000.4 0.4 + 0.2

0.42 + (0.4)2+0.22

= 200 200 + 100 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



= 0.6 + 0.8

= 0.5 + 0.5

= 200 200 + 100

=

=0.6 0.8 00.5 0 0.5200 200 100

= 100

Expressing as a Cartesian vector

=

= 100 0.6 + 0.8

= 60.0 + 80.0 ()




- F4.13 Determine the magnitude of the moment of the force

= 300 200 + 150 () about the axis. Express the result as a Cartesian vector





= 300 200 + 150 () about the axis. Express the result as a Cartesian vector




- F4.15 Determine the magnitude of the moment of the 200 force about the axis





about the axis



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= 50 40 + 20 () about the axis. Express the result as a Catersian vector




- F4.18 Determine the magnitude of the moment of force about the , , axes. Use a scalar analysis



6. Moment of a Couple

- A couple: two parallel forces that have the same magnitude but

opposite in direction separated by a perpendicular distance

- A couple moment: the moment produced by a couple

= +

=

=




- Scalar Formulation

=

- Vector Formulation

=

- Equivalent Couples




- Resultant Couple Moment

=




- Example 4.10 Determine the resultant couple moment of the

three couples acting on the plate

Solution

+ = : = 11 + 22 33

= 200 4 + 450 3 300 5

= 950

= 950



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- Example 4.11 Determine the magnitude and direction of the

couple moment acting on the gear

Solution

+ = : = 600300 0.2 600300 0.2

= 43.9

+ = : = 600300 0.2 600300 0.2

= 43.9 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien


or


Note

The same result can also be obtained using

=

: the perpendicular distance between the lines of action of the couple forces




- Example 4.12 Determine the couple moment acting on the

pipe. Segment is directed 300 below the plane

Solution 1

= 25 + 25

= 8 25 + (6300 + 8 6300) 25

= 200 129.9 + 200

= 130 () HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



It is easier to take moments of the couple forces about a point

lying on the line of action of one of the forces, e.g., point A

In this case the moment of the force at is zero, so that

= 25

= (6300 6300) 25




Solution 2

Taking moments of the forces about either point or point

= = 25 6300 = 25 5.196 = 129.9

Applying the right-hand rule, acts in the direction, thus




- Example 4.13 Replace the two couples acting on the pipe

column by a resultant couple moment

Solution

1 = = 150 0.4 = 60 1 = 60

2 = = 0.3 1254

5 125

3

5 = 22.5 +30

= 1 + 2 = 60 + 22.5 + 30 ()



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- F4.19 Determine the resultant couple moment acting on the

beam





triangular plate




- F4.21 Determine the magnitude of so that the resultant couple moment acting on the beam is 1.5 clockwise




- F4.22 Determine the couple moment acting on the beam





pipe assembly




- F4.24 Determine the couple moment acting on the pipe

assembly and express the result as a Cartesian vector



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7. Simplification of a Force and Couple System

- System of Forces and Couple Moments

=

= +

If the force system lies in the plane and any couple moments are perpendicular to this plane, then the above

equations reduce to the following three scalar equations

= ,

= , = +




- Example 4.14 Replace the force and couple system by an

equivalent resultant force and couple

moment acting at point

Solution

Force summation

+ = 3300 +

3

55 = 5.598

+

= 3300 4

55 4

= 6.50 = 6.50

Moment summation

+ = 3300 0.23300 0.1

+3

550.1

4

550.540.2

= 2.46 = 2.46




Summation

+ = 3300 +

3

55 = 5.598

+

= 6.50 = 6.50

+ = 2.46

Using the Pythagorean theorem

= 2 +

2

= 5.5982 + 6.502

= 8.58

Its direction

= 1

=

16.50

5.598= 49.30




- Example 4.15 Replace the force and couple system acting on

the member by an equivalent resultant

force and couple moment acting at point

Solution

Force summation

+ =

3

5500 = 300 = 300

+

=4

5500 750 = 350 = 350

Moment summation

+ =4

5500 2.5

3

5500 1

750 1.25 + 200

= 37.5 = 37.5

and = 3002 + 3502 = 461, = 1 350

300= 49.40




- Example 4.16 The structural member is subjected to a couple

moment and forces 1 and 2. Replace this system by an equivalent resultant force and couple moment acting at its

base, point

Solution

1 = 800

2 = 300

= 300

=0.15 + 0.1

(0.15)2+0.12

= 249.6 + 166.4

= 4

5500 +

3

5500 = 400 + 300




Force summation

= = 1 + 2 = 800 249.6 + 166

= 250 + 166 800 ()

Moment summation

= +

= + 1 + 2

= 400 + 300 + 800

+

0.15 0.1 1249.6 166.4 0

= 166 650 + 300 ()



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- F4.25 Replace the loading system by an equivalent resultant





























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8. Further Simplification of a Force and Couple System

- Concurrent force system: the system in which the lines of

action of all the forces intersect at a common point

=




- Coplanar force system



=

=

=

=

= /


- Parallel force system



=

=

= = /





- Reduction to a wrench

= +



, = /


- Example 4.17 Replace the force and couple moment system

acting on the beam by an equivalent resultant force, and find where

its line of action intersects the beam, measured from point

Solution

Force summation

+ =

3

58 = 4.8 = 4.8

+

= 4 +4

58 = 2.4 = 2.4

and = 4.82 + 2.42 = 5.37, = 1

2.4

4.8= 26.60



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= 4.8

= 2.4

Moment summation

+ =

= 4 1.5 15 3

58 0. 5

4

58 4.5

= 2.25




- Example 4.19 The jib crane is

subjected to three coplanar forces.

Replace this loading by an equivalent

resultant force and specify where the

resultants line of action intersects the column and boom

Solution

Force summation

+ =

3

5250 175 = 325

+

= 4

5250 60 = 260

and = 3252 + 2602 = 416

= 1260

325= 38.70




Moment summation

+ = 325 260 0

= 175 5 60 3

+3

5250 11

4

5250 8

= 2.29

By the principle of transmissibility,

can be placed at a distance where it intersects

+ = 325 11 260

= 175 5 60 3

+3

5250 11

4

5250 8

= 10.9




- Example 4.18 The slab is subjected to four parallel forces.

Determine the magnitude and direction of a resultant force

equivalent to the given force system and locate its point of

application on the slab

Force summation

+ = 600+100400500

= 1400 = 1400

Moment summation

+ = 1400

= 100 5 400 10

= 2.50

+

= 1400

= 600 8 100 6

= 3 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien


Solution



force and specify where the resultants line of action intersects the beam measured from








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force and specify where the resultants line of action intersects the member measured from




- F4.35 Replace the loading system by an equivalent single

resultant force and specify the and coordinates of its line of action




- F4.36 Replace the loading system by an equivalent single

resultant force and specify the and coordinates of its line of action



9. Reduction of a Simple Distribute Loading

- Surface forces are caused by direct contact

- Concentrated forces are applied to a point

- Linear Distributed forces () (force/length) are idealizations

The resultant force is equivalent to the area under distributed loading curve and acts through centroid or geometric center of

this area

- Body force is developed when one body exerts a force on

another body without direct physical contact HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



- Uniform Loading Along a Single Axis

= /2

= = ()



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- Magnitude of Resultant Force

+ =

=

=

- Location of Resultant Force

= =

=

=




- Example 4.21 Determine the magnitude and location of the

equivalent resultant force acting on the shaft

Solution

+ =

= 6022

0

= 603

3 0

2

= 160

The location of measured from

=

= 602

2

0

6022

0

=

604

4 0

2

160

= 1.5




- Example 4.22 A distributed loading of Pa acts over the top

surface of the beam. Determine the magnitude and location of

the equivalent resultant force

Solution

= 800 0.2 = 160/

=1

29 1440 = 6480 = 6.48

= 9 1

39 = 6




- Example 4.23 The granular material exerts the distributed

loading on the beam. Determine the magnitude and location of

the equivalent resultant of this load

Solution

Divide the trapezoidal loading into a rectangular and triangular

loading

a trapezoidal loading = a rectangular + a triangular loading



+ =

+ =


The magnitude of the force represented by each of these

loadings is equal to its associated area

1 =1

29 50 = 225/, 1 =

1

39 = 3

2 = 9 50 = 450/, 2 =1

29 = 4.5



+ =


The two parallel forces 1 and 2 can be reduced to a single

resultant

+ = = 225 + 450 = 675

+ = 675 = 3 225 + 4.5 450 = 4



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- F4.37 Determine the resultant force and specify where it acts

on the beam measured from




























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Ch.04 Force System Resultants

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