Uniformly distributed around the circumference Loads (Statics and Mec)

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SUMMARY

The discussion centers on the interpretation of a 30 kN force uniformly distributed around the circumference of a circular plate. Participants clarify that while the force could be expressed in terms of kN/m, it is more practical to represent it simply as 30 kN in the free body diagram (FBD). This representation indicates that the load is evenly distributed, eliminating concerns about eccentric axial loads on the shafts. The total load remains 30 kN, and the specific circumference is not necessary for this analysis.

PREREQUISITES
  • Understanding of statics and mechanics of materials
  • Familiarity with free body diagrams (FBD)
  • Knowledge of load distribution concepts
  • Basic principles of circular plate mechanics
NEXT STEPS
  • Study the principles of load distribution in circular plates
  • Learn how to create and interpret free body diagrams (FBD)
  • Research eccentric loading effects on shafts
  • Explore the mechanics of materials under uniform loading conditions
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Engineering students, mechanical engineers, and anyone involved in structural analysis or design of circular plates and load distribution systems.

Willjeezy
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Homework Statement


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My question is this, they say that at point B the 30kN force is uniformly distributed

If a force is uniformly distributed shouldn't the units be 30kN/mIf so, wouldn't that imply the point force would act in the center of the disk and the force would be be 30kN multiplied by the area of that disc?

How come in the FBD they just drew it as 30kN? Hoping someone can clear this up for me.
 
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The total load is 30 kN. You don't know the exact circumference of the circular plate, so 30 kN/m would not mean anything useful.

IMO, they are saying the load is distributed evenly around the circumference of the circular plate to suggest there are no eccentric axial loads placed on the shafts.
 

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