Integration method (Navier) calculating deflection

[Engineer_s]

Homework Statement

A beam is given with a constant load. Calculate the deflection at the end of the beam. Use the integration method or method of navier with delta functions.

Homework Equations

See equations in my attached file.

The Attempt at a Solution

The red load you see on the drawing is to cancel out the upper force. Can anyone help me or say what I am doing wrong? Thanks in advance.

 

[haruspex]

Not sure what your sign convention is for moments. Standard for me is anticlockwise positive. That would make M₁ positive and A₁x negative. You have the two terms both positive.
I don't understand the presence of the fourth term, q(a-x) etc. You should only include moments coming from one side of the point x.

 

[Engineer_s]

Not sure what your sign convention is for moments. Standard for me is anticlockwise positive. That would make M₁ positive and A₁x negative. You have the two terms both positive.
I don't understand the presence of the fourth term, q(a-x) etc. You should only include moments coming from one side of the point x.

Thanks for your reply. The convention is added beside the drawing. Positive is anticlockwise for me.
M1 is negative but if you cut the beam at let say an distance x from the wall you'll have a moment equal to M1 but in the opposite direction to cancel M1 out and to reach an equilibrium?

q(x-a) is added to cancel out the q*x*x/2 because this last load will continue till the end of the beam but this is not the case. Therefore I had to add the same force in the opposite direction at (x-a) to cancel it out. (this represents the red part on the drawing.)

I hope my explanation is clear. Thanks in advance to help me with my problem.

Best regards

 

[haruspex]

Thanks for your reply. The convention is added beside the drawing. Positive is anticlockwise for me.
M1 is negative but if you cut the beam at let say an distance x from the wall you'll have a moment equal to M1 but in the opposite direction to cancel M1 out and to reach an equilibrium?

q(x-a) is added to cancel out the q*x*x/2 because this last load will continue till the end of the beam but this is not the case. Therefore I had to add the same force in the opposite direction at (x-a) to cancel it out. (this represents the red part on the drawing.)

I hope my explanation is clear. Thanks in advance to help me with my problem.

Best regards

ok, I understand about the (x-a) terms. I should have noticed the δ(x-a).

But you do have a sign error.
I am still confused by your sign convention because you drew the arrow circling clockwise, implying clockwise is positive. You also wrote all of the moments in the M(x) expression as clockwise positive, except the first one. So I am going to assume you are actually taking clockwise positive.
With that comvention, the moment expression is M₁+A₁x etc., where M₁ = -qa²/2 and A₁ is qa.
Whatever your convention the those first two terms should have opposite sign.

 

[Engineer_s]

Yes, you are right. I made a mistake with my conventions. Look at my new PDF, that is the method I used. Normally the moment expression is now correct.
Do you agree with my method/calculation now or not? Thanks for your help.

Comment: I used a yellow remarker but it is hardly visible.

 

[Engineer_s]

The file.

 

[Engineer_s]

ok, I understand about the (x-a) terms. I should have noticed the δ(x-a).

But you do have a sign error.
I am still confused by your sign convention because you drew the arrow circling clockwise, implying clockwise is positive. You also wrote all of the moments in the M(x) expression as clockwise positive, except the first one. So I am going to assume you are actually taking clockwise positive.
With that comvention, the moment expression is M₁+A₁x etc., where M₁ = -qa²/2 and A₁ is qa.
Whatever your convention the those first two terms should have opposite sign.

See the replies above. Sorry I forgot to quote you the first time.

 

[Engineer_s]

Topic can be closed. Problem solved. Problem was my moment was positieve and it had to be negative. Thanks for your help.