Static Equilibrium Reaction Question about Fixed Point

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Homework Help Overview

The discussion revolves around a static equilibrium problem involving forces and moments acting on a beam. Participants are analyzing the forces, including a distributed load and a couple, to determine the reactions at fixed points A and B.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up equations for the sum of vertical forces and moments to find the reactions at points A and B. Some participants question the presence of an unknown force at point B and suggest considering the torque at point A. Others seek clarification on specific values used in the calculations, such as the 0.75 kN force, and whether the beam remains in equilibrium with the calculated reaction at A.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach and calculations. Some guidance has been offered regarding the need to consider both forces and moments for equilibrium, and there is an exploration of the implications of the assumptions made in the problem setup.

Contextual Notes

Participants note the absence of information regarding the beam's weight and the nature of the applied moment, suggesting that these factors are crucial for a complete analysis. There is also mention of the need to clarify the role of the 0.75 kN value in the context of the problem.

Grandpa04
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Homework Statement
Determine the reaction at fixed point A
Relevant Equations
ΣFx = 0
ΣFy = 0
ΣM = 0
Screenshot 2024-02-26 at 12.40.43 PM.png

For this problem I set ΣFy = -3kN - 0.75kN - 3kN + Ay + By (with By representing the right side vertical force, and Ay representing the left side force). For ΣM, I used the rectangular distributed load of 0.5 kN at 3m from point A as the pivot:
ΣM = -3m*Ay - (0.5kN*4m) - 3kN*6m - 5kN*m + By*6m.
I solved the system of the two equations and found Ay to be 1.722 kN. I am unsure if I used the correct/best method for the problem and if the answer is correct.
 
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I see no indication of an unknown force at B. Instead, I suggest there is a torque at A.
 
Could you explain the 0.75kN value?
If you represent the calculated value for Ay in the diagram, would the beam remain in equilibrium of forces and moments?
 
Grandpa04 said:
I am unsure if I used the correct/best method for the problem and if the answer is correct.
Hi @Grandpa04. Your answer isn't correct. In addition to the other replies, this might help.

If you have posted the complete question, then we assume that the beam’s weight is negligible - as there is no mention of the weight.

The given externally applied forces are the distributed load and the 3kN force.

The given externally applied moment (torque) is the one shown as 5 kN.m (CW.). Note that this is 'pure' moment - e.g. produced by a couple. (You may have to read-up on this if you are not familiar with it. E.g. look-up 'moment of a couple'.)

For the beam to be in equilibrium, the reaction at A must balance the given externally applied forces and moment.

So the reaction at A consists of two parts: a force (##F_A##) and a moment (##M_A##). You are required to find both of these.

There are no forces in the x-direction (as you correctly imply in your working).

You need to set up 1 equation for the sum of the y-forces; solving this gives you ##F_A##.

You need to set up 1 equation for the sum of the moments about A; solving this gives you ##M_A##.

Have another go and post your working and answer.
 

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