Statics: moments around beam subjected to 3 forces

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Discussion Overview

The discussion centers around determining the reactions at the supports of a beam subjected to two vertical forces and one horizontal force. Participants explore the application of static equilibrium equations and the implications of different support types on reaction forces.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equations of static equilibrium, attempting to calculate the reactions at supports A and C, but expresses uncertainty about the vertical reactions.
  • Another participant clarifies that point A is a roller and questions whether point C is pinned, noting a mistake in the moment arm length used in calculations.
  • Further discussion reveals that while point A cannot resist a moment, both supports can exert forces on the beam.
  • One participant calculates the vertical reaction at C as 15 kN but questions the implication of a negative reaction at A for force balance.
  • Another participant identifies an arithmetic mistake in the moment calculations, prompting a correction of the moment value from 15 * 12.
  • Concerns are raised about the equilibrium of vertical reactions, suggesting that both cannot be positive due to the nature of the applied forces.
  • Another participant emphasizes the need to correct arithmetic errors in the moment equation and clarifies the directional capabilities of the reactions at the supports.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the reactions at the supports, particularly regarding the implications of the roller and pinned connections. There is no consensus on the correct values for the reactions or the interpretation of equilibrium conditions.

Contextual Notes

Participants note limitations in their calculations, including potential arithmetic errors and assumptions about the nature of the supports. The discussion reflects ongoing uncertainty regarding the correct approach to balancing forces and moments.

Who May Find This Useful

This discussion may be useful for students studying statics, particularly those interested in understanding the implications of different types of supports on reaction forces in equilibrium problems.

mh1985
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Homework Statement


NDCw2.jpg

The beam is subjected to two oppositve vertical forces and one horizontal force as shown. Determine the reactions at the supports A & C

Homework Equations



\sumforces in plane = 0
moment = force x perpendicular distance

The Attempt at a Solution



\sumY = 15 + RyC + RyA - 15 = 0
\sumx = 20 + RxC = 0
\sumMoments around A = (-15 * 3) + (15 *12)+(RyC * 9)

Reaction at C in x = 20 kN
Not sure about reactions in Y at either, tried to calculate this from the moments but neither of these types of joint provide any reaction force to moments?

Thanks in advance
 
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Point A is a roller. Is point C pinned?

BTW, the moment arm from A to C is 12 m according to the figure, not 15 m as shown in the calculation.
 
yep point C is pinned, and thanks for spotting that mistake! I still can't see how to calculate the moment reaction though because A is a roller so has no reaction to the moments?

Following on from that I get RyC = 135/9 = 15 kN... but that would mean the RyA would have to be -15kN?
 
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The roller at A and the pin at C can both exert force on the beam, even though not being able to resist a moment.

Initially, static reactions at A and C are assumed to exist. Then, by applying the equations of static equilibrium, the magnitudes and directions of the reactions can be determined. Even though neither point A or point C can resist a moment, nevertheless, the sum of the moments about either point due to the applied loads and reactions must sum to zero, just like the forces do.
 
Moments around A = (-15 * 3) + (15 *12)+(RyC * 9) = 0

= -45 + 135 + 9 Ry at C = 0
= 90 + 9 Ry at C = 0

90/9 = Ry C = 10

But that must mean that Ry at A is negative in order for the forces in the y direction to balance?
 
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You have an arithmetic mistake in your moment calculations. What is 15 * 12?
 
SteamKing said:
You have an arithmetic mistake in your moment calculations. What is 15 * 12?

oops that's what I get for working on it so late...180
 
Regardless of that error, this still doesn't make sense to me though, there are 2 vertical reactions...and they cannot be equal to 0 because the moments would not be in equilibirum.

However since the two known vertical forces are equal and opposite, both of the vertical reactions cannot be positive (upwards) or else the beam would not be in equilibrium? But how could the support provide a downwards force?
 
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You must correct the arithmetic in your moment equation and calculate the reaction at C.
The roller at A can only offer an upward reaction, but the pinned connection at C can have a reaction oriented in any direction.
 

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