# Fixed end moment in beam

1. Apr 27, 2017

### fonseh

1. The problem statement, all variables and given/known data
I dont understand the fixed end moment AE and also fixed end moment EA ? shouldnt they be the same ?
2. Relevant equations

3. The attempt at a solution
For the fixed moment AE / fixed moment EA , shouldnt it = -10(2) +5(2) = -10 , for fixed end moment , we always assume clockwise as positive and anticlockwise as negative , right ?

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2. Apr 27, 2017

### scottdave

If you use the right-hand rule, then clockwise will be negative (Thumb pointing away from you), while counter-clockwise will be positive (thumb toward you).

3. Apr 27, 2017

### fonseh

So , in this example , the auhthor chose that the anticlockwise moment as positive ?
So , the author's working of Fixed moment AE 10(2) -5(2) = 10 kNm , the 10kNm is anticlockwise moment ?

then , why shouldn't the fixed moment EA = -10kNm ?

4. Apr 27, 2017

### fonseh

@PhanthomJay stated that we can choose either clockwise or anticlockwise as positive ?

But , when we draw the bending moment diagram , we must always keep clockwise as positive ...So , that's why the -10kNm ( antoiclockwise moment ) has negative value in the bending moment diagram ?

5. Apr 27, 2017

### scottdave

So as long as your convention (right hand/ left hand) is consistent throughout, then you should be OK. The 5 kN pointing up from the left side of the pivot does create clockwise. The 10 kN/m pointing down on the left side creates anticlockwise. This creates a larger value moment so the anticlockwise wins.

6. Apr 28, 2017

Bump

7. May 2, 2017

### PhanthomJay

There are lots of ways to go wrong with the minus sign. I urge you to use clockwise moments as plus when determining unknown reactions. And to assume the unknown moment reactions as clockwise .When you solve for it , and if it comes out minus, then it is actually anticlockwise.
In your example, I would write the equation as 5(2) - (10)(2)(1) + MAE = 0. Solve MAE = +10, clockwise.