# Beam deflection

1. Jul 2, 2016

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. The attempt at a solution
if it's integration of dx, it should look like this , right?

EIy' = 0.25P(x^2) - 0.5P(x^2) +0.5PLx ??

2. Jul 2, 2016

### SteamKing

Staff Emeritus
Why don't you think it's integration to go from EIy" to EIy'? What else would it be?

2. Relevant equations

3. The attempt at a solution
if it's integration of dx, it should look like this , right?

EIy' = 0.25P(x^2) - 0.5P(x^2) +0.5PLx ??[/QUOTE]

Not necessarily.

3. Jul 2, 2016

Not necessarily.[/QUOTE]

for the EIy" , why shouldnt it = 0.5Px - P(x-0.5L) - 0.5P( L-x) ??

4. Jul 2, 2016

### SteamKing

Staff Emeritus

for the EIy" , why shouldnt 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.

5. Jul 2, 2016

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: Jul 2, 2016
6. Jul 2, 2016

### SteamKing

Staff Emeritus
$EIy" = \frac{1}{2} Px - P<x-L>$ 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.

7. Jul 2, 2016

why the author ignore the moment 0.5P( L-x) ?

8. Jul 12, 2016