Beam deflection integration help

AI Thread Summary
The discussion revolves around understanding the right-hand side (RHS) of the equations EIy' and EIy'' in beam deflection problems. Participants are questioning whether the transition from EIy'' to EIy' involves integration with respect to x, and they explore the implications of singularity functions in this context. There is debate over the correct formulation of the equations, with one participant suggesting an alternative expression for EIy'' that includes terms not present in the original author's equation. The importance of adhering to the author's intended mathematical framework is emphasized, particularly regarding singularity functions and their integration rules. The conversation highlights the complexities of beam deflection analysis and the need for clarity in mathematical interpretations.
chetzread
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Homework Statement


can someone explain about the RHS of EIy' and EIy'' ?
how to get the RHS of EIy" from RHS of EIy' ??
It's not integration of dx , am i right?
BT25I3i.jpg

Homework Equations

The Attempt at a Solution


if it's integration of dx, it should look like this , right?[/B]
EIy' = 0.25P(x^2) - 0.5P(x^2) +0.5PLx ??
 
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chetzread said:

Homework Statement


can someone explain about the RHS of EIy' and EIy'' ?
how to get the RHS of EIy" from RHS of EIy' ??
It's not integration of dx , am i right?

Why don't you think it's integration to go from EIy" to EIy'? What else would it be?
BT25I3i.jpg

Homework Equations

The Attempt at a Solution


if it's integration of dx, it should look like this , right?[/B]
EIy' = 0.25P(x^2) - 0.5P(x^2) +0.5PLx ??[/QUOTE]

Not necessarily.
 
SteamKing said:
Why don't you think it's integration to go from EIy" to EIy'? What else would it be?
BT25I3i.jpg

Homework Equations

The Attempt at a Solution


if it's integration of dx, it should look like this , right?[/B]
EIy' = 0.25P(x^2) - 0.5P(x^2) +0.5PLx ??

Not necessarily.[/QUOTE]for the EIy" , why shouldn't it = 0.5Px - P(x-0.5L) - 0.5P( L-x) ??
 
chetzread said:
Not necessarily.
for the EIy" , why shouldn't it = 0.5Px - P(x-0.5L) - 0.5P( L-x) ??[/QUOTE]
It's hard to make out from the image provided, but it appears you start with:

##EIy" = \frac{1}{2} Px - P<x-L>##

Now, the expression P<x-L> usually represents some kind of singularity function, and you don't split up the expression inside the <>.
These expressions usually have some special integration rules which must be followed.
 
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does the author left out something ? as we can see from the diagram , there are 2 P/2 force at 2 different ends...
 
Last edited:
chetzread said:
It's hard to make out from the image provided, but it appears you start with:

##EIy" = \frac{1}{2} Px - P<x-L>##

Now, the expression P<x-L> usually represents some kind of singularity function, and you don't split up the expression inside the <>.
These expressions usually have some special integration rules which must be followed.
##EIy" = \frac{1}{2} Px - P<x-L>## this is the author's idea , my idea is = 0.5Px - P(x-0.5L) - 0.5P( L-x)
Which is correct ? the author ? or me ?[/QUOTE]
Obviously, the author is the expert on the interpretation of his own text.

You are not allowed to make up your own mathematics if it does not follow what the author intends.
 
SteamKing said:
##EIy" = \frac{1}{2} Px - P<x-L>## this is the author's idea , my idea is = 0.5Px - P(x-0.5L) - 0.5P( L-x)

why the author ignore the moment 0.5P( L-x) ?
 
for this question, why did C2=0?
in singularity function, <x-a> = 0 only if x<a , and <x-a> =1 if x > a ,
but, in this question, x is drawn beyond 0.5: (a) , am i right?
So, there's also a possibility of <x-a> =1 , am i right?
 

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