PDF Ch3B Couple static

7/29/2019 PDF Ch3B Couple static

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MOMENT OF A COUPLE

Lessons Objectives:Students will be able to

a) define a couple, and

b) determine the moment of a couple.


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A couple is defined as apair of forces having samemagnitudebut opposite directions, parallel lines of action, separated by a

perpendicular distance d.

Since couple forces are equal but opposite, they produce a zeroresultant force (i.e. F = 0).

Effect of couple: produces a rotation or tendency of rotation in aspecified direction.

Moment produced by couple is called a couple moment and has amagnitude M = Fd.

Moment Of A Couple


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Moment of A Couple

Acouple moment M is produced by two parallel (ie. non-collinear)forces that are equal in magnitude but opposite in direction.

A

d2

(F = F)

d

d1

F

F

Magnitude: M = FdF = magnitude of one of the forcesd = perpendicular distance between forces F and F

Fx = 0; Fy = 0MA 0

MA

= Fd1

Fd2

= F(d1 d2)= Fd

+


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Concept ofcouple moment is better illustrated by the examplebelow:

100N

100NA

B C

D

1 m

3 m

2 m

Observe that MA = MB = MC = MD Moment of couple can be determined

about any point. Does not need any

specific point. Only depend on M = Fd

Moment of a couple is a free vector, whichmeans you can place the moment M any

where on the rigid body.

M= F d

+ve CCW : MA = 100(4) = 400 N m

+ve CCW : MB = 100(3) 100(1) = 400 N m

+ve CCW : MC = 100(4) = 400 N m

+ve CCW : MD = 100(6) +100(2) = 400 N m


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If two couples produce a moment with the same magnitudeand same direction, then these two couples are equivalent.

For example, each couple moment below has a magnitude ofM = Fd= 30N(0.4m) = 40N(0.3m) = 12 N.m and rotates

clockwise.


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300 N

300 N

50 mm

100 mm

50 mm

150 mm

200 N

200 N150

mm

120 N 120 N

250 mm

30 N.m

50 mm

100 mm

50 mm

150 mm

Figure 3.9

30 N.m

50 mm

100 mm

50 mm

150 mm


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F1d1

F1

=

F1

F1

A

B

C

D

P

P

F1

F1

A

B

C

D= =

Q

Q

F2

F2


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Given: If F = 2000 N, determine theresultant couple moment M.

Soln: First, resolve each force into x-ycomponents. Next:

(a) Find moment of each couple

and sum them up, OR

(b) Sum the moments of all theforce components about anarbitrary point (point A, or B or C).

Answer: M = 260 N.m (CW)

Example 3.3

C


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Example 3.3

+ve CCW :

M1500 = 1500cos30(0.4) 1500sin30(0.4) = 819.6 N m

M2000 = 20004

5

(0.2)+ 2000

3

5

(0.2) = 560 N m

MR = M1500 +M2000

= 819.6 + 560 = 259.6 N m clockwise

Resolve 1500-N and 2000-N force couples into x-y components.

Determine the couple moments produced by 1500-N couple and 2000-Ncouple, respectively.

Sum these couple moments to obtain resultant couple moment.

Solution: METHOD 1


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Example 3.3

+

ve CCW : MR=

MA= 1500 cos 30(0.6) 1500 sin30(0.4)

+ 20004

5

(0.2) + 2000

3

5

(0.6)

20003

5

(0.4) + 20004

5

(0)

+1500 cos 30(0.2) 1500 sin30(0)

= 779.4 300 + 320 + 720 480 + 0 + 259.8 0

= 259.6 N m clockwise

Resolve 1500-N and 2000-N force couples into x-y components. Treat each force component as an individual force. Find moment of each

force component about point A.

Sum the moments of all force components about point A.

Solution: METHOD 2


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Example 3.3

Solution: METHOD 2

+ve CCW : MR= MB

= 1500 cos 30(0)1500 sin30(0.4)

+ 20004

5

(0.2)+ 2000

3

5

(0)

+ 20003

5

(0.2)+ 2000

4

5

(0)

1500 sin30(0.4)+1500 cos30(0)

= 300 + 0 + 320 + 0 + 240 + 0 519.6 0

= 259.6 N m clockwise

Resolve 1500-N and 2000-N force couples into x-y components. Treat each force component as an individual force. Find moment of each

force component about point B.

Sum the moments of all force components about point B.


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Given: Two force couples act on thestructure. The resultant coupleis zero.

Find: Magnitudes of P and F and

distance d.

1) Use definition of a couple moment to determine P and Fi.e. M = Fd

2) Resolve the 300-N force into x-y components.3) Determine the net (resultant) moment.

4) Equate resultant moment to zero and solve for d.

Ans: P = 500N, F = 300 N, d = 3.96 m

Example 3.4

Method:


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Definition: A couple has twoparallel forces of samemagnitude but oppositedirections. Hence:

P = 500 N and F = 300 N

Solve above equation for d.d= (1000 + 60 sin 30)/(300 cos 30) = 3.96 m

Given M = 0:

+ M = (500N)(2m) + (300N cos 30)(d) (300N sin 30)(0.2m) = 0

Example 3.4


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Practice Problem 1

Answer: M = 80.15 N.m clockwise

Two couples act on the frame. Ifd= 1.2m, determine the resultant

couple moment.

First, resolve 200-N and 300-N force couplesinto x-y components.

To find resultant couple moment M:

Find moment of each force couple and sumthem up. (Method 1)

Sum the moments of all force componentsabout a point (A, B or C) (Method 2)

Method:

C


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Practice Problem 2

Two couples act on the frame. If the resultant couple moment is

zero, determine the distance dbetween the 200-N couple forces.

Answer: d = 1.663m


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Determine the required magnitude of force F if the resultant

couple moment on the frame is 200 Nm, clockwise.

Practice Problem 3

Answer: 2213 N


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EQUIVALENT FORCE-COUPLE SYSTEM

Lessons Objectives:

Students will be able to:1) Determine the effect of moving a force.

2) Find an equivalent force-couple system for a system offorces and couples.


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B

A

A

B

A= =d M = Fd

F

F F

Equivalent force-

couple system

3.10 Resolution of A Given Force Into A Force Acting At AGiven Point And A Couple (continue)

That means ..Any force F acting on a rigid body may be moved to any point A,provided that a couple M is added at the new point. Moment ofthis couple must equal to the moment ofF (in its originalposition) about point A, MA= Fd

Force F originally

acts at B


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B

Sample Prob 3.4

50 N

A

50 mm

100 mm

C

30

30 mm

A 50-N force is applied to a plate as shown.

Replace this force by:

(a) An equivalent force-couple systemacting at point A.

(b) An equivalent system consisting of a

150-N force at B and another force at A.


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Sample Prob 3.4

(a) Force-couple system at A.

i) We move 50-N force to the designated location i.e. point A;direction must be the same as the original force.

ii) Next find the couple moment MA; magnitude must be equal tomoment of 50-N force (in its original position) about point A.

MA= Fxy + Fyx= (50 N sin 30)(0.050m) (50 N cos 30)(0.100m)

MA= 3.08 Nm = 3.08 Nm CWAnswer: F = 50N MA= 3.08 Nm CW60 3.08 Nm

50 N

B

30

A

This is an important step because it

determines the direction of the couple,whether clockwise or counter clockwise.


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150 N

50 N

Sample Prob 3.4

(b) 150-N force at B and a force at A.

i) Assume clockwise couple MA= 3.08 Nm consists of two150-N parallel forces acting at B and A, respectively.

MA= 3.08 Nm = (150N)(d)

d= 0.02053 m

d= 0.02053 m = (0.030m)(cos )

= cos-1 (0.02053/0.03) = cos-1 (0.6844) = +46.8

ii) At point A, we have two forces acting: 150-N and 50-N.Find resultant of these two forces.

FAX = 50N sin 30 150N cos 46.8 = 77.67 N

FAY= 50N cos 30 150N sin 46.8 = 152.65 N

FA = 171.3 N

= tan-1(152.65/77.67) = 63.0

FB = 150 N = 46.8

B

150 N

A

30

30 mm

d

FA = 171.3 N = 63.0 Answer:


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150 N

50 N

Sample Prob 3.4

150-N force at B and a force at A

(alternative answer)

Know that:

= cos-1 (0.02053/0.03) = cos-1 (0.6844) = +46.8

ii) At point A, we have two forces acting 150-N and 50-N.Find resultant of these two forces.

FAX = 50N sin 30 150N cos 46.8 = 77.67 N

FAY= 50N cos 30 + 150N sin 46.8 = 66.05 N

FA = 101.96N;

= tan-1 (66.05/77.67) = 40.37

FB = 150 N = 46.8

B150 N

A

30

FB = 150 N = 46.8 Answer: FA = 101.96 N = 40.37


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3.08 Nm

Sample Prob 3.4

150 N

B101.96 N

A

46.8

40.37

150 N

B

171.3 N

A

46.8

63

50 N

B

30

A

Summary: The equivalent systems

= =


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125 NC

400 mm

500 mm500 mm

BA

Example 3.5

(a) FB = 125 N ; MB = 50 N-m CCW

(b) FA = 200N 30 ; FB = 314.5N 18.54


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866 N

500 N

C B

1000 N

300

A

100 mm200 mm

Example 3.6

(a) FC = 1000N 30 and MC = (500N)(0.1m) = 50 N.m CW

(b) M = 50Nm = Fd ; d = 0.2 m and F = 250 N.

FA= 250N FCx = -866N, FCy = -750N, FC = 1145.6N, = 40.89


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A 500-N force is applied at C to the bracket as shown below. Replacethe 500-N force by:a) An equivalent force-couple system at Bb) An equivalent system formed by two parallel forces at A and B.

C

B

500N

30

A

50mm

100mm100mm

Example 3.7


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EQUIVALENT SYSTEM:

FORCE-COUPLE SYSTEM

When several forces and couple momentsact on a body, they can be combined intoa single force and couple moment havingthe same external effect.

Move each force and its associated couplemoment to a common point O.

Then, add all forces and couple momentstogether.

Find one resultant force-couple momentpair.


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3.11 Reduction of a system of forces into one singleforce-couple system (point A)

F2

F1

F3

A

F2

F3

A

F1

M2M3

M1=

(a) (b)

= A =

R

M

d = M/R

A

R

( c ) (d)


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=

(F1)x

(F1)y(F2)x

(F2)y

(F3)x

(F3)y

A

Ry

RxA

M

(a) (b)

Rx = FxRy = FyMA = M


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=

R

dx= M/Ry

A B Rx

Ry

dy= M/Rx

R

A

Rx

Ry

C


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Given:A system of forces is as shown.

Find: (a) equivalent resultant force andcouple moment acting at A;

(b) equivalent single force

location along the beam AB.

Plan:

1) Sum all the (Rx) and (Ry) components of the forces to find FR.

2) Sum all the moments resulting from moving each force to A.3) Shift the FR to a distance dsuch that d= MRA/FRy

Example 1


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FR = (42.52 + 50.312)1/2 = 65.9 N

= tan-1 (50.31/42.5) = 49.8

+ Fx = Rx = 25N + 35N sin 30 = +42.5 N

+Fy = Ry =

20N

35N cos 30 =

50.31 N

+ MRA = 35N cos 30 (0.2 m) 20N (0.6m)

+ 25N (0.3m)

=10.56 N.m

Location of equivalent resultant force FRon beam AB at adistance d measured from the left end (point A):

d = MRA/FRy = 10.56 N.m/50.31 N = 0.21 m.

FR

Example 1


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Example 2

+ Fx = Rx = 175 N sin 30

+125 N = 212.5 N

+ Fy = Ry = 175 N cos 30

+100 N = 251.6 N

R= (212.5)2 + (251.6)2 = 329.3 N

= tan1251.6

212.5

= 49.8

Location where line of action intersects beam AB :

+ MR= MA = 251.6 (d) = 154.8 N m

= 175 cos 30(0.6) +100 (1.8) 125 (0.9)

d = 0.630 m

SOLUTION


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100 mm

50 mm

50 mm200 mm

60o

60o

60oO

500 N

1000 N

FA

B

Example 3.8

(a) F = 750N(b) M = Fd: d = 0.206m, F = 728.16 N

C


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30

4 m 4 m

6 m

AB

C

D

E

6 kN

9 kN

3 kN

Example 3.9

tan-1 (12/8.196) = = 55.67Fx = 8.196 kN , Fy = 12 kN, R = 14.53 kN

(From test 1 Feb 2002)


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B

Knowing that the line of action of 600-N force passes through point C,

replace the 600-N force by:a) An equivalent force-couple system at Db) An equivalent system formed by two parallel forces at B and D.

120mm

A

C

600N

D

160mm 80mm

Beer, 5eProblem 3.35

FD = 600.0 N 36.9

MD = 28.8 Nm (CCW)

FB = 200 N 36.9

FD = 400 N 36.9


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B

Replace the 600-N force by:

a) An equivalent force-couple system at Cb) An equivalent system consisting of a 120-N force at B and another

force at C

120mm

A

C

600N

30

D

160mm 80mm

Beer, 5e

Problem 3.37

FD = 600.0 N 30

MD = 14.35 Nm (CW)

FB = 120 N 48.4

FD = 588 N 41.5


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PDF Ch3B Couple static