• Support PF! Buy your school textbooks, materials and every day products Here!

Beam deflection integration help

  • Thread starter chetzread
  • Start date
  • #1
801
1

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 ??
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666

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.
 
  • #3
801
1
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 shouldnt it = 0.5Px - P(x-0.5L) - 0.5P( L-x) ??
 
  • #4
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
Not necessarily.

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
801
1
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:
  • #6
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
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.
 
  • #7
801
1
##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) ?
 
  • #8
801
1
for this question, why did C2=0?
in singularity fucntion, <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?
 

Related Threads on Beam deflection integration help

Replies
3
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
6
Views
642
Replies
4
Views
19K
Replies
10
Views
3K
  • Last Post
Replies
23
Views
8K
  • Last Post
Replies
1
Views
568
  • Last Post
Replies
6
Views
630
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
24
Views
2K
Top