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 4.3

10 kN

16 kNm

2

2 kN

q

2

=2 kN/m

q

1

=4 kN/m

O

2m

2m

6m

4m

=2 kN/m2

=2 kN/m2

45°

Given the structure shown above

I.

Check if the force system acting on it can be reduced to the resultant force.

II.

If yes, determine whether the central axis passes through point O.

I

The necessary and sufficient condition for a reduction of a planar force system to a resultant force states
that the sum vector must be different from zero.

F

3

=10 kN

M

1

=16 kNm

2

2 kN

q

2

=2 kN/m

q

1

=4 kN/m

O

2m

2m

4m

=2 kN/m2

=2 kN/m2

45°

W

4

W

3

W

2

W

1

3m

3m

F

2

F

1

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

 Replacing the distributed loads by their resultants.

]

[

6

]

[

6

])

/

[

2

]

/

[

4

(

2

1

1

kN

m

m

kN

m

kN

W











]

[

12

]

[

6

]

/

[

2

2

kN

m

m

kN

W







]

[

24

]

[

12

]

/

[

2

2

2

3

kN

m

m

kN

W







]

[

32

]

[

16

]

/

[

2

2

2

4

kN

m

m

kN

W







 Resolving the oblique forces into components

]

[

2

2

)

45

cos(

1

kN

F







,

]

[

2

2

)

45

sin(

2

kN

F







,

]

[

2

2

1

kN

F

F





 Determining the sum vector

)

,

,

(

O

S

S

S

y

x



4

3

1

W

F

F

S

x







]

[

44 kN

S

x



3

2

1

2

W

W

W

F

S

y









]

[

40 kN

S

y





)

],

[

40

],

[

44

(

O

kN

kN

S





0

0







k

S

 Answer

The system of forces can be reduced to the resultant force

II.

If the central axis of the system passes through point O, than the total moment of the system about this point

must be equal to zero.

0



O

M

.

 Determining the total moment about point O

)

,

0

,

0

(

Oz

O

M

M



]

[

0

]

[

16

]

[

2

]

[

32

]

[

3

]

[

24

]

[

3

]

[

12

]

[

4

]

[

6

]

[

4

]

[

10

]

[

6

]

[

2

kNm

kNm

m

kN

m

kN

m

kN

m

kN

m

kN

m

kN

M

Oz

































)

0

,

0

,

0

(



O

M

 Answer

Point O lies on the central axis of the system.