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Three Dimensional Force Systems - Dokuz Eylül · PDF fileThe same principles can be enlarged to three dimensional force systems. The resultant of forces acting on a body ... The resultant

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Three Dimensional Force Systems

The same principles can be enlarged to three dimensional force

systems. The resultant of forces acting on a body

can be obtained by moving them to a desired point. In this way, the

given force system will be converted to

1) Three dimensional, concurrent forces comprising the same

magnitudes and directions as the original forces,

2) Three dimensional couples.

nFFFF

...,,,, 321

By calculating the resultants of these forces and couples, a single

resultant force and a single couple can be obtained.

The resultant force,

222

321 ...

zyx

zzyyxx

n

FFFR

FRFRFR

FFFFFR

The resultant couple moment,

The selection of point O is arbitrary, but the magnitude and direction

of 𝑀 will depend on this point; whereas, the magnitude and direction

of 𝑅 are the same no matter which point is selected.

CFrM

CC

As a special case, if the resultant couple 𝑀 is perpendicular to the resultant

force 𝑅, these two vectors can further be simplified to obtain a single resultant

force 𝑅. The force 𝑅 can be slided a distance d to form a moment - 𝑀 , which

is equal in magnitude and opposite in direction 𝑀, so that they will cancel each

other out. The distance d will be equal to 𝒅 = 𝑴

𝑹.

1. The special-purpose milling cutter is subjected to the force of 1200 N and a

couple of 240 N.m as shown. Replace the given force system with an equivalent

force-couple system at O.

2. Represent the resultant of the force system acting on the pipe

assembly by a single force and a couple at A .

kjiFR

100180120 --

SOLUTION:

j

A

A

jk

kiikM

CFrM

40

25.016050

1003.012025.0

-

kjM A

50100

Equivalent force-couple system at A

A

y

x

z

kjiR

100180120 --

kjM A

50100

Sum of the moments about point A

Resultant force

3. The pulley wheels are subjected to the loads shown.

Determine the equivalent force-couple system at point O.

kjFR

5001000

jkijki

kjikjiM

CFrM

o

o

60018.02.140018.02.1

30024.06.020024.06.0

-

-

kjiM o

120030012 -

Equivalent force-couple system at point O

O

x

y

z

kjR

5001000

kjiM o

120030012 -

SOLUTION:

Sum of the moments about point O

Resultant force

4. Two forces and a couple (applied to portion AB) are applied to the car’ s exhaust system as

shown. Replace the given force system with an equivalent force-couple system at D.

25 N.m

115 N

100 N

kjR

kjjFR

FF

5039.28

30sin10030cos100115

21

--

--

SOLUTION:

Resultant force

25 N.m

115 N

100 N

CFrM D

kjiM

kjiM

D

D

868.179825.8

207.43568.4115243.94475.85

--

----

kjjkikikFr

505.86225.03.079.046.031.036.022 ---

jikikFr

11569.046.031.036.011 --

ikFr

3.947.4311

kjiFr

568.4124745.8522 ---

y

z

C

kjkj

C

2015375.0

3.0225.025 -

-

Equivalent force-couple system at D

--

--

kjiM

kjR

D

868.179825.8

5039.28

Sum of the moments about point D

B

A

When the resultant couple vector 𝑀 is parallel to the resultant force 𝑅, the resultant is called

a “wrench”.

The wrench is the simplest form in which the resultant of a general force system may be

expressed. By definition, a wrench is positive if the couple and force vectors point in the

same direction, and negative if they point in opposite directions.

A common example of a wrench is found with the

application of a screw driver.

MR

//RR

RR//R

MMMM

nnMMM

-

2

1

Equivalent force-couple

system at point O

M is resolved into components M1 along the

direction of R and M2 normal to R.

Positive wrench

R

Md 2

All force systems can be reduced to a wrench acting at a particular line of action.

M

R

MR

5. For the position shown, the crankshaft of a small two-cylinder compressor is subjected

to the 400 N and 800 N forces exerted by the connecting rods and the 200 N·m couple.

Replace this loading system by a force-couple system at point A. Show that 𝑅 is not

perpendicular to 𝑀𝐴. Then replace this force-couple system by a wrench. Determine the

magnitude M of the moment of the wrench, the magnitude of the force of the wrench

and the coordinates of the point in the x-z plane through which the line of action of the

wrench passes.

6. The force-couple system acting at O is equivalent to the wrench acting at A.

If and , determine . NkjiR

7001400600 mNM //R 1200

oM

kjikji

R

RnR

418.0835.0358.07001400600

7001400600

222

(Unit vector of resultant force)

kjikjinMM RRR

6.50110026.429418.0835.0358.01200////

kjikjijiRrM R

300021001400700140060023 -

kjiMMM RRO

6.350110986.1829// -

SOLUTION:

7. The threading dye is screwed onto the end of the fixed pipe which is bent

through an angle of 20°. Replace the two forces by an equivalent force at O and a

couple . Find and calculate the magnitude M of the moment which tends to screw

the pipe into the fixed block about its angled axis through O.

Nm.k.i.kiMMM

k.i.kcosisinn

kiM

ji.k.kcos.isin.

ji.k.kcos.isin.M

jjjFR

oOC

OC

o

o

68859403408517

9403402020

8517

150250202015020150

200250202015020150

50200150

---

--

-

--

--

--

y

x

500 N

1700 N

3400 N

3

4

qtan q

30 cm

50 cm

34 cm

50 cm

50 cmz

800 N.m

q

tan q15

8

8. The pulleys and the gear are subjected to the loads shown. For these forces,

determine the equivalent force-couple system at point A.

15

8

jiR

jiijiFR

jijiF

iF

jijiF

35201040

800150050027202040

800150017

81700

17

151700

500

272020405

43400

5

33400

3

2

1

--

----

--

-

----

kjiCFrM

kC

kjijikjFr

kijFr

kjiFr

jikjijijikFr

CFrM

A

A

6.11391770960

800

75075040080015005.05.0

1505003.0

6.48910201360

272020405.016.03.02720204017

8

17

1534.05.0

33

22

11

11

--

-

--

-

--

-----

-

9. a) Reduce the general three-dimensional force system to a force-couple system at O.

b) Replace the force-couple system obtained by a wrench and determine the coordinates of the

point in the yz plane through which the line of the wrench passes.

Dimensions in meters.

a) Reduce equivalent force-couple system at O.

kj2iM,kj48iR

kC

kj8ikj4i

C

kj4ikikj3iBDBEN

N

NnnCC

kj6ikj2i

C

kj2ikij3iEFEDNN

NnnCC

k48ikj48ijFr

CFrM

kj48ikji

F

o

3

2

22

1

11

o

10186246

65

3624

18412

181244

18124645

918

326

3621

362

1922468

246

481

04800154

222

2

2

222

222

11

111

222

---

---

---

------

--

--

-

---

(Equivalent force-couple system at O)

b) Reduction to a wrench.

my,m.z

146zj

18448z24yi

kjii48zj6zi24yky

Mkj48ikzjyMRr

kjikjikj2iMMM

kjiM,kj48iR

kjikjinMM

Nmkjikj2inMM

kji54

kj48in

MRMR

RR

R//R

R//

R//R//

R//

oo

ooo

o

Roo

Roo

R

oo

3332

18141846

246

1814184816210186

8162246

81629

4

9

8

9

118

189

4

9

8

9

110186

9

4

9

8

9

1246

0972

--

-

------

--

-

-

---

--

(This force-couple system can be reduced a wrench – Positive Wrench)

(Wrench Resultants)

(Coordinates of the point in the yz plane which the line of action of the wrench passes)

C3 = 80 Nm (in plane ABCD)qy > 90o

F1 = 30 NF2 = 75 NF3 = 40 NC1 = 60 NmC2 = 100 Nm (in yz plane)

O (0, 0, 0) mA (12, 0, 0) mB (in xz plane)C (12, 8, 0) mE (6, 10, -3) mG (10, 4, 4) m

10. Replace the system comprising two forces, two couples and a positive wrench with an

equivalent force-couple acting at point O. Then, reduce the system further into a wrench and

determine the coordinates of point P, of which the line of action of the wrench cuts through

the yz plane.

A

B

C

O

D

E

G

37°

X Y

Z

45°

60°qy

53°

30°

6 m

4 m

1C

1F

2C

3F

3C

2F

k1ji.kcos60jcos6icos45F

cos6cos45cos oyy

222

5152121030

6010

1

qq

Force:

C3 = 80 Nm (in plane ABCD)qy > 90o

F1 = 30 NF2 = 75 NF3 = 40 NC1 = 60 NmC2 = 100 Nm (in yz plane)

A

B

C

O

D

E

G

37°

X Y

Z

45°

60°qy

53°

30°

6 m

4 m

1C

1F

2C

3F

3C

2F

Force:

C3 = 80 Nm (in plane ABCD)qy > 90o

F1 = 30 NF2 = 75 NF3 = 40 NC1 = 60 NmC2 = 100 Nm (in yz plane)

A

B

C

O

D

E

G

37°

X Y

Z

45°

60°qy

53°

30°

6 m

4 m

1C

1F

2C

3F

3C

2F

k.j.i.

k)(j)(iF

884688461535

884

8008181275

2222 -

--

O (0, 0, 0) mA (12, 0, 0) mB (in xz plane)C (12, 8, 0) mE (6, 10, -3) mG (10, 4, 4) m

Force:

C3 = 80 Nm (in plane ABCD)qy > 90o

F1 = 30 NF2 = 75 NF3 = 40 NC1 = 60 NmC2 = 100 Nm (in yz plane)

A

B

C

O

D

E

G

37°

X Y

Z

45°

60°qy

53°

30°

6 m

4 m

1C

1F

2C

3F

3C

2F

O (0, 0, 0) mA (12, 0, 0) mB (in xz plane)C (12, 8, 0) mE (6, 10, -3) mG (10, 4, 4) m

k.j.i.

...

k.j.i.

R

Rn

k.j.i.FR

k.jikcosjsincosicoscosF

.

222R

7806300150

52968877951

52968877951

52968877951

643416123040536040536040

04124

3

-

-

-

kj.2i.k.jikji6Fr

k.j562.56i.k.j.i.ji12Fr

k.j.k1ji.k4ji10Fr

CFrMo

248443439464341612310

84843043758846884615358

1665166551521214

33

22

11

---

--

-

Moment:

C3 = 80 Nm (in plane ABCD)qy > 90o

F1 = 30 NF2 = 75 NF3 = 40 NC1 = 60 NmC2 = 100 Nm (in yz plane)

A

B

C

O

D

E

G

37°

X Y

Z

45°

60°qy

53°

30°

6 m

4 m

1C

1F

2C

3F

3C

2F

O (0, 0, 0) mA (12, 0, 0) mB (in xz plane)C (12, 8, 0) mE (6, 10, -3) mG (10, 4, 4) m

A

B

C

O

D

E

G

37°

X Y

Z

45°

60°qy

53°

30°

6 m

4 m

1C

1F

2C

3F

3C

2F

k3i

2432

k24iC

k24ijki6BDBA

iC

kji.4kcosjcosicos45C

223

2

1

26432

80

3248

100

3030422606060

40

--

--

---

C3 = 80 Nm (in plane ABCD)qy > 90o

F1 = 30 NF2 = 75 NF3 = 40 NC1 = 60 NmC2 = 100 Nm (in yz plane)

O (0, 0, 0) mA (12, 0, 0) mB (in xz plane)C (12, 8, 0) mE (6, 10, -3) mG (10, 4, 4) m

kj.i.M

k3iikji.4

kj.2i.k.j562.56i.k.j.M

o

o

8675684186847

2641003030422

2484434394848430437516651665

-

--

----

Moment:

k86756.841i86.847M

k52.9677.88i95.1

o

-

-

j

jR

Equivalent force-couple system at point O

k86756.841i86.847M

k52.9677.88i95.1

o

-

-

j

jR

Equivalent force-couple system at point O

Reduction to a wrench in yz plane

k.j.ik.j.i..nMM

Nm.k.j.i.kj.i47.86nMM

R////

Ro//

0210402842780630015035133

351337806300150867568418

--

--

Positive wrench

k02.10402.48i2M

k52.9677.88i95.1

//

-

-

j

jR

O

xy

z

k02.10402.48i2M //

- j

k52.9677.88i95.1

- jR

oM

M

Positive wrench

The coordinates of point P, of which the line of action of the wrench cuts through the yz plane:

myyk

mzzj

jizjziyky

jj

j

jj

Rr

27.39198.76295.1

66.47458.92595.1

k98.76258.259i49.8688.7745.152.9695.1

k98.76258.259i49.868k52.968.77i95.1kzjy

k98.76258.259i49.868M

k02.10402.48i2k867841.56i47.868MMM

M

//o

--

---

--

-

----

Oxy

z

k02.10402.48i2M //

- j

k52.9677.88i95.1

- jR

Positive wrench

P(0;391.27;474.66)r