Sign convention of fixed end moment at cantilever beam

AI Thread Summary
The discussion centers on the confusion regarding the sign convention for the fixed end moment at a cantilever beam, specifically at point A. The fixed end moment is stated to be 10 kNm clockwise, but there is a discrepancy in the moment diagram, which shows it as anticlockwise. Participants clarify that for equilibrium, the internal moment at A must counteract the applied loading, resulting in a clockwise moment. The confusion arises from the unconventional representation of the moment in the diagram, which does not align with standard sign conventions. Ultimately, understanding the relationship between internal and reaction moments is crucial for resolving the discrepancies in the moment calculations.
fonseh
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Homework Statement


I don't understand why the fixed end moment at AE is 10KNm ( clockwise) .

Homework Equations

The Attempt at a Solution


Forger about the sign convention here , i have the resultant moment = 10KNm , which ic clockwise , why in the 2nd picture ( SFD and BMD part) , the moment is anticlockwise ?

The moment at AE= 5(2) - 10(2) = -10(anticlcockwise) , why it's clockwise as shown in the diagram in the 2nd picture?
 

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fonseh said:

Homework Statement


I don't understand why the fixed end moment at AE is 10KNm ( clockwise) .

Homework Equations

The Attempt at a Solution


Forger about the sign convention here , i have the resultant moment = 10KNm , which ic clockwise , why in the 2nd picture ( SFD and BMD part) , the moment is anticlockwise ?

The moment at AE= 5(2) - 10(2) = -10(anticlcockwise) , why it's clockwise as shown in the diagram in the 2nd picture?
You must sum moments equal to 0 for equilibrium, so when drawing a free body diagram of AE and summing moments about the cut section at A, the applied loading is 10 kNm anticlockwise, so the moment in the beam at A in that diagram must be 10 kNm clockwise.

In terms of normal sign convention, clockwise moments acting on a left hand section are negative (hogging moments, beam bending convex downwards ), but unfortunately, the author drew the moment diagram unconventionally; it should me mirrored for conventional signage.
 
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PhanthomJay said:
You must sum moments equal to 0 for equilibrium, so when drawing a free body diagram of AE and summing moments about the cut section at A, the applied loading is 10 kNm anticlockwise, so the moment in the beam at A in that diagram must be 10 kNm clockwise.

In terms of normal sign convention, clockwise moments acting on a left hand section are negative (hogging moments, beam bending convex downwards ), but unfortunately, the author drew the moment diagram unconventionally; it should me mirrored for conventional signage.
Do you mean that to produce sum of moment = 0 , the moment should counter The moment at AE= 5(2) - 10(2) = -10(anticlcockwise) , so , the moment is clockwise ?
 
PhanthomJay said:
clockwise moments acting on a left hand section are negative
The moment is clockwise , so they it's neagtive -10 as stated by the author , right ? anything wrong with it ?
 
fonseh said:
The moment is clockwise , so they it's neagtive -10 as stated by the author , right ? anything wrong with it ?
I'm not sure what the author is stating. The internal moment at A acting on Section AE is clockwise. The picture is confusing I think. Clockwise moments acting on left hand sections are negative, by convention. But the author goes against convention when drawing the moment diagram, and shows the moment as positive. You can argue that being unconventional is ok, but it is wrong as far as I am concerned, because it does not follow the calculus of dM/dx = V
 
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PhanthomJay said:
I'm not sure what the author is stating. The internal moment at A acting on Section AE is clockwise.
Can you explain why the internal moment is clockwise of 10 ?

Should i explain in this way ? Since internal moment = reaction moment , so at left of A , the moment resultant = 10Nm anticlockwise , reaction moment has to be 10NM clockwise ??
 
fonseh said:
Can you explain why the internal moment is clockwise of 10 ?

Should i explain in this way ? Since internal moment = reaction moment , so at left of A , the moment resultant = 10Nm anticlockwise , reaction moment has to be 10NM clockwise ??
Since internal moment = reaction moment , so at left of A , the moment resultant = 10Nm anticlockwise , reaction moment has to be 10NM clockwise, in order to satisfy the equilibrium requirement that the sum of all moments about any point must equal 0.[/color]
 
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