Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3-1 EML 3004C Chapter 3: Force...

Namas ChandraIntroduction to Mechanical engineering

HibbelerChapter 3-1

EML 3004C

Chapter 3: Force System ResultantsChapter 3: Force System Resultants


HibbelerChapter 3-2

EML 3004C

Cross ProductCross Product

The Cross product of two vectors and A B

C A B

Magnitude:C=AB sin

Direction: C is perpendicular to both A and B


HibbelerChapter 3-3

EML 3004C

Laws of Operation for Cross ProductLaws of Operation for Cross Product

Commutative law is not valid

In fact

A B B A

A B B A

A B A B A B A B

Scalar Multiplication

Distributive Law

A B D A B A D


HibbelerChapter 3-4

EML 3004C

Cartesian Vector Formulation Cartesian Vector Formulation (sec 3.1)(sec 3.1)

Using general definition,

ˆ ˆ

Magnitude: ( )( )(sin )

ˆDirection: R.H. Rule

ˆˆ ˆ

i j

i j

k

i j k

0

0

0

i j k i k j i i

j k i j i k j j

k i j k j i k k


HibbelerChapter 3-5

EML 3004C

Cross Product of Two Vectors Cross Product of Two Vectors (sec 3.1)(sec 3.1)

x y z

x y z

A A i A j A k

B B i B j B k

( - )

- ( - )

( - )

x y z

x y z

y z z y

x z z x

x y y x

i j k

A B A A A

B B B

i A B A B

j A B A B

k A B A B

Let


HibbelerChapter 3-6

EML 3004C

Moment Systems Moment Systems (sec 3.2)(sec 3.2)

The moment of a force about an axis (sometimes represented as a point in a body) is the measure of the force’s tendency to rotate the body about the axis (or point). The magnitude of the moment is:

Direction R.H. RuleoM Fd


HibbelerChapter 3-7

EML 3004C

Moment Systems of System of Forces Moment Systems of System of Forces (sec 3.2)(sec 3.2)

0

1, 2 3

1, 2 3

Consider a system of Forces and

They are at and from point 0.

CCW R

F F F

d d d

M Fd It is customary to assume CCW as the positive direction.

Resultant Moment of four Forces:

50 (2 ) 60 (0) 20 (3sin 30 )

40 (4 3cos30 )

334 =334 Nm(CW)

oRM N m N N m

N m m

Nm


HibbelerChapter 3-8

EML 3004C

Moment of a Force-Vector Formulation Moment of a Force-Vector Formulation (sec 3.3)(sec 3.3)

0

The moment of a Force about a point O,

is the position vector of between O

and any point on the line of action of

F

M r F

r F

F

sin for any d,

:Note

r d


HibbelerChapter 3-9

EML 3004C

Moment of a Force-Vector Formulation Moment of a Force-Vector Formulation (sec 3.3)(sec 3.3)

Let

and

Then,

x y z

x y z

o x y z

x y z

F F i F j F k

r r i r j r k

i j k

M r F r r r

F F F

The axis of the moment is perpendicular to the plane that contains both F and r

The axis passes through point O


HibbelerChapter 3-10

EML 3004C

Moment of Force Systems-Vector FormulationMoment of Force Systems-Vector Formulation (sec 3.3)(sec 3.3)

Let a system of forces act upon a body. We like to compute the net moment of all the forces about the point O.

Net moment is the sum of moment

of each force with separate

oR

F r

M r F

The moment will have three

components in , and o

o

R

R x y z

M

x y z

M M i M j M k



EML 3004C

CCW ;

4 3 375 (11) (500) (5) (500) (0)

5 5 160 cos 30 (0) 160 sin 30 (0.5)

6165 lb ft 6.16 kip ft C CW

B

B

RB

RB

R

M M

M

M

Problem 3-10 (page 84, Section 3.1-3.3)Problem 3-10 (page 84, Section 3.1-3.3)

3.10 Determine the resultant moment about point B on the three forces acting on the beam.

Solution:



EML 3004C


3.20 The cable exerts a 140-N force on the telephone pole as shown. Determine the moment of this force at the base A of the pole. Solve the problem two ways, i.e., by using a position vector from A to C, then A to B.

Solution:

2 2 2

Position Vector:

r 6k m r 2i -3j m

Force Vector:

(2-0) i (-3-0) j (0-6) k F 140

(2 0) ( 3 0) (0-6)

40i - 60 j -12 0k N

AB AC



EML 3004C


Solution-Con’t

360i 240j N m

Moment about point :

M r FA

A

Use r r

i j k

0 0 6

40 -60 -120

0 (-120) - (-60)(6) i - 0(-120)-40(6) j 0(-60)-40(0) k

AB



EML 3004C


Solution-Con’t

Use r r

i j k

M 2 -3 0

40 -60 -120

( 3) (-120) - (-60)(0) i - 2(-120)-40(0) j 2(-60)-40(-3) k

360i 240j N m

AC

A



EML 3004C

Section 3.1-3.3 (In-class Exercise)Section 3.1-3.3 (In-class Exercise)

If the man B exerts a force P=30 lbs on his rope,

determine the magnitude of F the man at C must exert

to prevent pole from tipping.

Solution:

Net moment should be zero

45

Assume CCW +ve.

30(cos 45)(18) ( )(12

39

)

.8lb

F

F



EML 3004C

Section 3.1-3.3 (In-class Challenge Exercise)Section 3.1-3.3 (In-class Challenge Exercise)

The foot segment is subjected to the pull of the two

plantar flexor muscle. Determine the moment of

each force about the point of contact A on the ground.

Solution:

120cos30(4.5) 20sin 30(4)

118 (cw)AM

lb

230cos30(4.0) 30sin 70(3.5)

140 (cw)AM

lb

1 1

258 (cw)

A A AM M M

lb



EML 3004C

3.4 Principle of Moments 3.4 Principle of Moments (sec 3.4)(sec 3.4)

The moment of a force is equal to the sum of the moment of the force’s component about a point. (Varginon’s theorem 1654-1722)

0 1 2 1 2M r F r F F r F r F

ACable exerts F on pole with moment M .

F can be slided by the

Note A x y

principle of transmissibility

M F h F b F d



EML 3004C

3.5 Moment of a Force about a specified axis3.5 Moment of a Force about a specified axis

0When the moment of a force F is computed using M =r F,

the axis is perpendicular to r and F.

If we need the moment about other axis still through O, we can use either scalar or vector analysis.

Here we have F=20 N applied. Though the typical equation gives moment with respect to b-axis, we require it through y-axis.



EML 3004C

3.5 Moment of a Force about a specified axis-23.5 Moment of a Force about a specified axis-2

Step 1. Find M about using cross produc

(0.3 0.4 ) ( 20 )

8 6 Nm

t.

o AM r F i j k

i j

ˆStep 2. Find M about the given axis

ˆ ( 0.8 6 )

6 Nm

=j.

y o A

Au

M M u i j j



EML 3004C

3.5 Moment of a Force about a specified axis-33.5 Moment of a Force about a specified axis-3

The two steps in the previous analysis can be combined with the definition of a scalar triple product. Since dot product is commutative

ˆIf and then

ˆ

x y z

x y z

o a a o

a a

a a a x y z

x y z

a a a

x y z

x y z

M r F M u M

M u r F

i j k

u i u j u k r r r

F F F

u u u

r r r

F F F



EML 3004C

3.6 Moment of a Couple-13.6 Moment of a Couple-1

A couple is defined as two parallel forces with same magnitude and opposite direction. Net force is zero, but rotates in specified direction.

is the moment of the couple.Sum of the moment is same about any point.Moment about O,

Since does not depend on O, the momentis same at any point.

A B

B A

Couple moments

M r F r F

r r F r Fr



EML 3004C

3.6 Equivalent Couples -23.6 Equivalent Couples -2

Two couples are equivalent if they produce the same moment. The forces should be in the same or parallel planes for two couple to be equivalent.

Couple moments are free vectors. They can be added at any point P in the body.

1 2

1 2

There are two couples, with moments and +

FR

R

M MM M MM r



EML 3004C


3.39The bracket is acted upon by a 600-N force at A. Determine the moment of this force about the y axis.

Solution:

Force Vector:

F 600 (cos 60 i cos 60 j cos45 k)

300i 300j 424.26 k N

Position Vector:

r -0.1 i 0.15 k m



EML 3004C

3.39The bracket is acted upon by a 600-N force at A. Determine the moment of this force about the y axis.

Magnitude of the moment along axis:

ˆ j (r F)

0 1 0

-0.1 0 0.15

300 300 424.26

=0-1 (-0.1)(424.26)-(300)(0.15) 0

8

y

y

M

7.4 N m

In cartesian vector form

87

:

. 4 j N M m y

Solution-Con’t



EML 3004C


3.54 Two couples act on the frame. If d = 6 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 3-14) and (b) summing the moments of all the force components about point A

1

2

1 2

(a)

100cos30 (6)

519.6 lb ft CW

4 (150)(4) 480 lb ft CCW

5 519.6-480

39.6 lb ft CW

R

M

M

M M M

Solution:



EML 3004C

3.54 Two couples act on the frame. If d = 6 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 3-14) and (b) summing the moments of all the force components about point A

(b)

CCW ;

4 100cos30 (3) (150)(4) 100cos30 (9)

5 39.6 lb ft

39.6 lb CW ft

R B

R

R

M M

M

M

Solution-Con’t



EML 3004C


Solution:

a. Find the normal distance for each case first.

Two couples act on the frame. If d = 4ft, find the resultant couple moment by (a) direct method, and (b) resolving the x and y components (take moment about A).

4540cos30(4) 60 (4)

53.4 lb.ft (CW)cM

3 34 45 5 5 5

40cos30(2) 40cos30(6)

60 (3) 60 (7) 60 (7) 60 (7)

53.4 lb.ft (CW)

cM



EML 3004C


Solution:

The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its coordinate direction angles.

1

2

1 2

50 N.m

20cos 20sin 30 20cos 20cos30

20sin 20 N.m

= 9.397 16.276 6.840 N.m

9.397 16.276 6.840 50 N.m

= 9.397 16.276 56.840 N.m

R

M k

M i j

k

i j k

M M M

i j k k

i j k



EML 3004C

Section 3.4-3.6 (In-class Challenge Exercise-2)Section 3.4-3.6 (In-class Challenge Exercise-2)

Solution—contd.

2 2

1 9.37959.867

1 16.27659.867

1 56.84059.867

9.379 16.276 + 56.840 N.m

59.867 N.m=59.9 N.m

cos 99.0

cos 106.0

cos 18.3

RM



EML 3004C

3.7 Movement of a Force on a Rigid Body-13.7 Movement of a Force on a Rigid Body-1

A single force on a body can cause it to rotate (moment) and translate (force).

In the first example, the ruler causes a force F and in addition a moment M=Fd.

In the example, the ruler causes a force F and NO ADDITIONAL moment.



EML 3004C

3.7 Movement of a Force on a Rigid Body-23.7 Movement of a Force on a Rigid Body-2

Extend this idea to a general 3-D case. Now, the force can be moved

Force now causes the force at any point 0 and then a couple.



EML 3004C

3.8 Resultant of a Force and Couple System-13.8 Resultant of a Force and Couple System-1

0

By applying the same concepts we have

x

y

z

o

R

R x

R y

R z

R

F F

F F

F F

F F

M M



EML 3004C

3.9 Further Reduction on Force/Couples-13.9 Further Reduction on Force/Couples-1

If resultant force and moment is known then it is possible to

reduce them to a single force at P. d=

o

o

R R

R

R

F M

M

F



EML 3004C


Concurrent Force Systems

Only equivalent force

Coplanar Force Systems A single force at d from point 0



EML 3004C


Parallel Force Systems

Here we have parallel forces and moments that are perpendicular.

Resultant moment (see b):

A single force =

oR C

R

M M r F

F F



EML 3004C


3.103 The weights of the various components of the truck shown. Replace this system of forces by an equivalent resultant force and couple moment acting at point A.

Force Summation:

;

3500 5500 1750

=10750lb 110.75lb

R y

R

F F

F

Moment Summation:

CCW ;

3500(20) 5500(6) 1750(2)

=99500lb ft 99.5 kip ft

RA A

RA

M M

M

Solution:



EML 3004C


3.93 The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x,y) on the slab.

;

30 50 40 20 140kN

M ;

-140 50(3) 30(11) 40(13)

M M ;

140 50(4) 20(

140kN

7.

10) 40(10)

14m

5.

R x

R

R xx

R yy

y

x

F F

F

M

y

x

71m

Solution:



EML 3004C

3.69 The gear is subjected to the two forces shown. Replace these forces by an equivalent resultant force and couple moment acting at point O.


Force Summation:

;

40 (2.25) 3sin 60 -0.40295 kN

41

;

9 (2.25) 3sin 60 -1.0061 kN

41

Rx x

Rx

Ry y

Ry

F F

F

F F

F

��

Solution:



EML 3004C

3.69 The gear is subjected to the two forces shown. Replace these forces by an equivalent resultant force and couple moment acting at point O.

Solution-Con’t2 2

2 2

-1 -1

1.08 kN

68.2

( 0.40295) ( 1.0061)

1.0061 tan tan

0.40295

Moment Summation:

CCW ;

Rx RyR

Ry

Rx

Ro o

F F F

F

F

M M

0

0

3sin 60 (0.175cos 20 )

403cos60 (0.175sin 20 )+ (2.25)(0.17

0.901 kN

5)41

m CCW

RoM



EML 3004C


Replace the force system by an equivalent force and couple

moment at the point A.

Solution:

1 2 3

300 100 400 100

100 50 500

400 300 650 N

R RF F F F F F

i j

k

i j k



EML 3004C

Section 3.7-3.9 (In-class Exercise..2)Section 3.7-3.9 (In-class Exercise..2)

Solution (contd).

1 2 3

0 0 12 0 0 12

300

31

400 100 100 100 5

00 4800 N.m

0

0 1 12

0 0 500

R A

AB AB AB

M M

r F r F r F

i j k i j k

j

i j

i

k



EML 3004C


The weights of the various components of the truck are shown. Replace this system by an equivalent resultant force and specify its location from point A.

Solution: Equivalent force

1750 5500 3500

107 10.75Kip50 s

R y

y

F F

F

lb

Location of force

10750( ) 3500(20) 5500(6) 1750(

9

2

2 f

)

. 6 t

AR ACCW M M

d

d



EML 3004C

Chapter 3: Force System Resultants…Chapter 3: Force System Resultants…concludesconcludes

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Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 3-1 EML 3004C Chapter 3: Force...

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