Project “The development of the didactic potential of Cracow University of Technology in the range of modern

construction” is co-financed by the European Union within the confines of the European Social Fund

and realized under surveillance of Ministry of Science and Higher Education.

1

Exercise 3.

I. For a given force system find the sum vector,

the total moment about points B, O and E.
Verify the correctness of your computations.

II. Find an equivalent force couple system at point B.
III. Determine the simplest equivalent force system.

x

z

y

F

1

F

F

3

2

O

a

a

2a

A

C

E

D

B

P

F

2

1



,

P

F

2

2



,

P

F

6

3



I.

The components of the forces.



 



0

2

0

2

0

2

0

2

1

1

P

a

a

P

BA

BA

F

F















 

P

P

a

a

P

F













0

2

1

0

1

2

2











P

P

P

a

a

P

F









2

6

1

2

1

6

3

The sum vector



 

 

 



P

P

P

P

P

P

P

F

F

F

S

2

0

0

2

0

0

2

0

3

2

1

























The total moment about point B

B

B

B

B

M

M

M

M

3

2

1







Moments

B

M

1

and

B

M

3

vanish, because the lines of actions of forces

1

F and

3

F

pass through point B.

Hence

The total moment of the force system about point B equals





0

0

Pa

M

B





.













0

0

0

0

0

2

Pa

a

P

P

M

M

B

B















The total moment about point O

O

O

O

O

M

M

M

M

3

2

1



















Pa

a

a

P

M

O

2

0

0

0

2

0

2

0

1

























0

2

0

0

2

3

Pa

Pa

a

P

P

P

M

O























Pa

Pa

a

a

a

P

P

M

O

2

0

2

2

0

2

















Project “The development of the didactic potential of Cracow University of Technology in the range of modern

construction” is co-financed by the European Union within the confines of the European Social Fund

and realized under surveillance of Ministry of Science and Higher Education.

2





0

4

Pa

Pa

M

O





The correctness of above computations can be verified by changing the moment center.

BO

S

M

M

B

O









 

 

 



0

4

0

2

a

Pa

M



2

0

0

0

0

P

a

a

P

Pa

O

















The total moment about point E

E

E

E

E

M

M

M

M

3

2

1

















Hence





0

0

Pa

M

E





.

Verification:

BE

S

M

M

B

E







II. An equivalent force coupe system at point B

An equivalent force couple system comprises force





P

S

b

2

0

0







applied to point B, and a couple

with a moment





0

0

Pa

M

B





.

Finding a coupe of a given moment (one of infinitely many)

The couple must satisfy two conditions

1.

0







B

B

M

F

M

F



The remaining components

are arbitrary (but can’t simultaneously equal zero).

z

x

F

F ,

For example

,

, hence

P

F

x



0



z

F





0

0

P

F



i





0

0

P

F







.

2.

BG

F

M

B







 

 



)

2

(

0

2

0

0

a

y

P

Pz

z

a

y

a

x

P

BG

F

























0

0

2

0

0

1

a

E



0

2

0

Pa

P

M







0

2 E



M







0

2

0

2

2

3

Pa

Pa

a

a

P

P

P

M

E











 

 

 



0

0

0

0

2

0

0

0

0

Pa

a

P

Pa

M

E













































F

F

F

y

x























z

y

x

G

F

F

a

a

B

F

F

F

z

z

y

x

,

0

2









0

0







F

F

F

F



0

0











B

F

Pa

Pa

F

M



y

y

z

y

x

Project “The development of the didactic potential of Cracow University of Technology in the range of modern

construction” is co-financed by the European Union within the confines of the European Social Fund

and realized under surveillance of Ministry of Science and Higher Education.

3

so



 

2

(

0

0

0

a

y

P

Pz

Pa



)









a

z

Pa

Pz











a

y

a

y

P

2

0

)

2

(









The above set of equations defines a line. We choose one point of that line e.g.

ne of the couples resulting in a mome

II. The simplest equivalent force system.

Determining the parameter of a system





a

a

G

2

0



O

nt consists of:





0

0

Pa

M

B

















































 

a

a

G

P

F

a

a

B

P

F

2

0

0

0

,

0

2

0

0

I



 



0

0

0

2

0

0







Pa

P

K



Conclusion: Because





M

S

B



0

0







S

, the system of fo

reduced to a resultant force

k

rces can be



S

W

2

0

0









P applied at an unknown point





z

y

x

H

,

,

,

, about which the total moment of the force

system equals zero.

0









BH

S

M

M

B

H



 

 



0

2

2

0

0

0

0















z

a

y

a

x

P

Pa





0

0

)

(

2

)

2

(

2











Pa

a

x

P

a

y

P





2



























2

0

)

(

2

0

)

2

(

2

a

x

a

y

a

x

P

a

y

P

- the central axis of a system.

Setting

2

x



, we select a point on the central axis

a









a

a

H

2

2

.

The central axis contains point H an



 a

d is parallel to the sum vector. Hence, the parametric equation of the

central axis takes the form:























a

z

a

y

a

x

2

2

Project “The development of the didactic potential of Cracow University of Technology in the range of modern

construction” is co-financed by the European Union within the confines of the European Social Fund

and realized under surveillance of Ministry of Science and Higher Education.

4

y

a

z

x

F

1

F

2

a

F

3

2a

A

E

B

C

D

H

O

2

a

x



a

y

2



- central axis