The equivalent force-couple system at O is the force and couple experienced when at point O due to the applied force at point A
The equivalent force couple system at O due to force F are;
Force, F = (8.65·i - 4.6·j) KN
Couple, M₀ ≈ 40.9 k kN·m
The reason the above values are correct is as follows:
The known values for the cantilever are;
The height of the beam = 0.65 m
The magnitude of the applied force, F = 9.8 kN
The length of the beam = 4.9 m
The angle away from the vertical the force is applied = 26°
The required parameter:
The equivalent force-couple system at the centroid of the beam cross-section of the cantilever
Solution:
The equivalent force-couple system is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;
The equivalent force [tex]\overset \longrightarrow F[/tex] = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j
Which gives;
The equivalent force [tex]\overset \longrightarrow F[/tex] ≈ (8.65·i - 4.6·j) KN
The couple acting at point O due to the force F is given as follows;
The clockwise moment = 9.8 kN × cos(28°) × 4.9
The anticlockwise moment = 9.8 kN × sin(28°) 0.65/2
The sum of the moments = Anticlockwise moment - Clockwise moments
∴ The sum of the moments, ∑M, gives the moment acting at point O as follows;
M₀ = 9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9 ≈ 40.9 kN·m
The couple acting at O, due to F, M₀ ≈ 40.9 kN·m
The equivalent force couple system acting at point O due the force, F, is as follows
F = (8.65·i - 4.6·j) KN
M₀ ≈ 40.9 k kN·m
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