background image

 

 

 

 

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

 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

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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

,

               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

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

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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

 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

 

 

 

 

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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