B06 exercise03 DJ

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

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

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.

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

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.

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

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.

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



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