Question

The cantilevered W530 x 150 beam shown is subjected to a 9.8-kN force F applied by means of a welded plate at A. Determine the equivalent force-couple system at the centroid of the beam cross-section at the cantilever O.

Answer 1

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 $\overset \longrightarrow F$ = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j

Which gives;

The equivalent force $\overset \longrightarrow F$ ≈ (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

Learn more about equivalent force systems here:

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