Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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.

The Cantilevered W530 X 150 Beam Shown Is Subjected To A 98kN Force F Applied By Means Of A Welded Plate At A Determine The Equivalent Forcecouple System At The class=

Sagot :

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

Learn more about equivalent force systems here:

https://brainly.com/question/12209585