Reaction forces in simple beam involving applied moments

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Discussion Overview

The discussion revolves around calculating reaction forces in a simple beam subjected to an applied moment of 50 kN·m. Participants explore the implications of this moment on the equilibrium equations used to determine the reaction forces at supports A and B.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant questions whether the applied moment should be treated as a force and how it affects the calculations for reaction forces.
  • Another participant clarifies that an applied moment is considered a pair of equal and opposite forces and does not count as a linear force.
  • A participant presents calculated reaction forces of A = 4.3 kN and B = 11.7 kN, based on their working.
  • Another participant expresses uncertainty about the correctness of the working and the calculations presented.
  • A later reply emphasizes the importance of maintaining significant digits in calculations and correct unit formatting, while also correcting a notation error regarding the reaction force label.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the calculations, as participants express uncertainty and seek clarification on the working steps. Multiple viewpoints on the treatment of the applied moment exist, indicating a lack of agreement.

Contextual Notes

Participants have not fully resolved the assumptions regarding the applied moment's role in the calculations, and there are unresolved steps in the mathematical reasoning presented.

nick.falconer
Messages
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Homework Statement



IMAG0195.jpg


Homework Equations





The Attempt at a Solution



The problem I'm having is that I'm not sure how to work out the reaction forces because of the applied moment of 50kN.m. I know that the formulas are the sum of forces in a single direction equals 0 and the sum of moments about a point equal 0. But does the applied moment count as a force? And as a moment do I multiply it by the distance from the point I'm taking?
 
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welcome to pf!

hi nick! welcome to pf! :wink:
nick.falconer said:
… does the applied moment count as a force? And as a moment do I multiply it by the distance from the point I'm taking?

an applied moment counts as a pair of equal and opposite (but not in-line) forces …

so no, it doesn't "count" as a linear force, and its moment is the same about any point :smile:
 
So if my working is correct is the reaction force at A=4.3kN and B=11.7kN
 
nick.falconer said:
So if my working is correct is the reaction force at A=4.3kN and B=11.7kN

if your working is correct, then your answer will be correct also :confused:

(what working? :redface:)
 
Oops left out the working.

Taking down as positive and and clockwise as positive

sum of forces = 0 = -RA + 4x4 - RB

sum of moments about A = 0 = 16x2 + 50 - RBx7

So RB = 11.7kN and RA = 4.3kN
 
nick.falconer: Your answers in post 3 were correct, except generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three (or four) significant digits. RB in your post should instead be RD. Also, always leave a space between a numeric value and its following unit symbol. E.g., 4.29 kN, not 4.29kN. See the international standard for writing units[/color] (ISO 31-0[/color]).
 

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