1
Exercise 4_1: Reduce a force system depicted below at a point belonging to the central axis.
Solution:
1. Finding the coordinates of the forces
(- 4,5,3)m
F1 = 2 50kN = (- 8,10,6)kN
50m
(- 4,2,-2)m
F2 = 4 6kN = (- 8,4,-4)kN
2 6m
F3 = (16,0,0)kN
F4 = (0,-14,0)kN
2. Calculating the sum vector.
S = (0,0,2)kN
3. Computing the total moment about a chosen point (point L).
(- 8,4,-4)kN (16,0,0)kN
M = F2 EL + F3 KL = + = (12,40,16)kNm + (0,0,80)kNm = (12,40,96)kNm
L
(- 4,0,3)m (0,5,0)m
4. Determining the parameter of the system.
2 2
k = S o M = (0,0,2)o (12,40,96)kN m = 192kN m ą 0
L
Parameter of the system is different from zero hence at the points belonging to the central axis the system is
reduced to the wrench.
5. Finding the moment of the wrench
2
k 192kN m
M = S = (0,0,2)kN = (0,0,96)kNm
P
2 2
4kN
S
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
6. Determining the equation of the central axis
M = M + S LP
P L
P = (x, y, z), L = (0,5,5)
(0,0,2)kN
(0,0,96)kNm = (12,40,96)kNm +
(x, y - 5, z - 5)m
0 = 12 - 2y +10
0 = 40 + 2x
96 = 96 + 0
The equation of the central axis.
y = 11
x = -20
7. Choosing an arbitrary point belonging to the central axis.
P(- 20,11,0)
8. Answer
At the point P belonging to the central axis the system is reduced to the wrench comprising one force, equal
to the sum, applied at this point and a couple with the moment M , parallel to the sum vector.
P
ć F = S = (0,0,2)kN
and a couple M = (0,0,96)kNm .
P
P(- 20,11,0)m
Ł ł
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.
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