Beam Deflection: Solving for Unknowns at a Given Point

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The discussion centers on calculating beam deflection using singularity functions. The primary confusion arises around the definition of the singularity function <x-a>, particularly why it equals zero when x is less than a, which is clarified by the function's definition. Participants debate the correct application of these functions in determining maximum deflection points, emphasizing the importance of understanding the beam's support locations. The conversation highlights that singularity functions simplify deflection calculations by avoiding complex piecewise analysis of varying loads. Ultimately, recognizing the behavior of singularity functions is crucial for accurately solving beam deflection problems.
chetzread
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Homework Statement


determine deflection of beam
m6vCPT1.jpg

RoiNzgx.jpg

Homework Equations

The Attempt at a Solution


at x=0, y=0 , i found that C2 = LP(a^2)/6 (circled part) but the author gt 0, which is correct?
 
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chetzread said:

Homework Statement


determine deflection of beam
m6vCPT1.jpg

RoiNzgx.jpg

Homework Equations

The Attempt at a Solution


at x=0, y=0 , i found that C2 = LP(a^2)/6 (circled part) but the author gt 0, which is correct?

The author is correct.

##EIy = -\frac{b}{6a}Px^3 + \frac{L}{6a}P<x-a>^3+C_1x+C_2##

When x = 0, y = 0, and the expression <x-a> = 0, due to the definition of the singularity function <x-a>, and C2 = 0 as described in the text.
 
SteamKing said:
The author is correct.

##EIy = -\frac{b}{6a}Px^3 + \frac{L}{6a}P<x-a>^3+C_1x+C_2##

When x = 0, y = 0, and the expression <x-a> = 0, due to the definition of the singularity function <x-a>, and C2 = 0 as described in the text.
why <x-a> = 0 ?
 
chetzread said:
why <x-a> = 0 ?
The way a singularity function is defined:

<x-a> = 0, if x - a < 0
<x-a> = 1, if x - a > 0

Some other properties are found in this article:

https://en.wikipedia.org/wiki/Singularity_function

You should check your text for the precise definition.
 
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SteamKing said:
The way a singularity function is defined:

<x-a> = 0, if x - a < 0
<x-a> = 1, if x - a > 0

Some other properties are found in this article:

https://en.wikipedia.org/wiki/Singularity_function

You should check your text for the precise definition.
my notes doesn't hv explanation on this
 
SteamKing said:
The way a singularity function is defined:

<x-a> = 0, if x - a < 0
<x-a> = 1, if x - a > 0

Some other properties are found in this article:

https://en.wikipedia.org/wiki/Singularity_function

You should check your text for the precise definition.
since x - a > 0, so <x-a> = 1 , am i right ? why the author take <x-a> = 0 ?
 
chetzread said:
since x - a > 0, so <x-a> = 1 , am i right ? why the author take <x-a> = 0 ?
When x = 0, <0 - a> = -a; -a < 0; <0-a> = 0, according to the definition of the singularity function
 
chetzread said:
my notes doesn't hv explanation on this
What about your textbook? Do you have a textbook for this course? What does it say?
 
SteamKing said:
What about your textbook? Do you have a textbook for this course? What does it say?
ok , i have another part of question here . in the working , the maximum deflection occur at x= (1/sqrt rt 3) ,
why the author ignore the LP[(x-a)^3]/ (6a) at x = (1/sqrt rt 3) ?
by ignoring the at LP[(x-a)^3]/ (6a) at x= (1/sqrt rt 3) , the author assume x-a <0 ?
but , the value of a is unknown , how to know that x-a<0 ?

WceUCUV.jpg

Pa51WwX.png
 
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  • #10
chetzread said:
ok , i have another part of question here . in the working , the maximum deflection occur at x= (1/sqrt rt 3) ,
why the author ignore the LP[(x-a)^3]/ (6a) at x = (1/sqrt rt 3) ?
by ignoring the at LP[(x-a)^3]/ (6a) at x= (1/sqrt rt 3) , the author assume x-a <0 ?
but , the value of a is unknown , how to know that x-a<0 ?

WceUCUV.jpg

Pa51WwX.png

I think the author is clumsily trying to say that the singularity function <x-a> = 0 where the deflection is at a maximum. Instead of saying "do not exist", it would probably be better to say that "<x-a> vanishes" at that location.

Anyhow, once that fact is recognized, then the location x of the maximum deflection can be solved for as illustrated.
 
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  • #11
SteamKing said:
I think the author is clumsily trying to say that the singularity function <x-a> = 0 where the deflection is at a maximum. Instead of saying "do not exist", it would probably be better to say that "<x-a> vanishes" at that location.

Anyhow, once that fact is recognized, then the location x of the maximum deflection can be solved for as illustrated.
how could that be ? the value of a is unknown
 
  • #12
chetzread said:
how could that be ? the value of a is unknown
No, it isn't. a is the distance between R1 and R2. Look at the beam diagram.
 
  • #13
SteamKing said:
No, it isn't. a is the distance between R1 and R2. Look at the beam diagram.
if it' so , x > a , so x-a > 0 , so <x-a> should be =1 , right ?
 
  • #14
chetzread said:
if it' so , x > a , so x-a > 0 , so <x-a> should be =1 , right ?
It's not a which is unknown, it is the value of x at which the deflection is a maximum.
 
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  • #15
SteamKing said:
I think the author is clumsily trying to say that the singularity function <x-a> = 0 where the deflection is at a maximum. Instead of saying "do not exist", it would probably be better to say that "<x-a> vanishes" at that location.

Anyhow, once that fact is recognized, then the location x of the maximum deflection can be solved for as illustrated.
why <x-a> = 0 when deflection is at a maximum ??
 
  • #16
chetzread said:
why <x-a> = 0 when deflection is at a maximum ??
Specifically, the problem is looking for the location of the maximum deflection between the supports.

The quantity <x-a> is going to be zero because x is located between between the supports and x = a is the location of the right support. That's just the definition of the singularity function in this case. It helps to check the beam diagram for this problem.
 
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  • #17
SteamKing said:
Specifically, the problem is looking for the location of the maximum deflection between the supports.

The quantity <x-a> is going to be zero because x is located between between the supports and x = a is the location of the right support. That's just the definition of the singularity function in this case. It helps to check the beam diagram for this problem.
x is located between supports ? but , in the diagram , x span from R1 and beyond R2 ?
What do you mean ?
 
  • #18
chetzread said:
x is located between supports ? but , in the diagram , x span from R1 and beyond R2 ?
What do you mean ?
x is the length coordinate of the beam with origin at the left support, i.e. x = 0 there.

Sure, x can be used to locate stuff along the entire length of the beam (even beyond the location of the right support), but this particular question asks the student to find the location of maximum deflection between the supports, i.e. find max. δ(x) such that 0 ≤ x ≤ a. Again, refer to the diagram.
 
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  • #19
SteamKing said:
x is the length coordinate of the beam with origin at the left support, i.e. x = 0 there.

Sure, x can be used to locate stuff along the entire length of the beam (even beyond the location of the right support), but this particular question asks the student to find the location of maximum deflection between the supports, i.e. find max. δ(x) such that 0 ≤ x ≤ a. Again, refer to the diagram.
Can you explain what is the purpose of having singularity function in beam problem?
 
  • #20
chetzread said:
Can you explain what is the purpose of having singularity function in beam problem?
Their use can simplify the deflection calculations, especially when the double integration method is used to calculate deflections.
 
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  • #21
SteamKing said:
Their use can simplify the deflection calculations, especially when the double integration method is used to calculate deflections.
if they are not used,then the calculation will become complicated?
I'm still blurred. Taking the case from post#1 as an example, why singularity function said that <x-a> =0 if x<a ?
 
  • #22
chetzread said:
if they are not used,then the calculation will become complicated?

Yes. The beam must be split into many different pieces, depending on how it is loaded.
I'm still blurred. Taking the case from post#1 as an example, why singularity function said that <x-a> =0 if x<a ?
Because that is how the singularity function is defined.

If x < a, then the singularity function <x-a> = 0
If x > a, then the singularity function <x-a> = (x - a)
 
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  • #23
SteamKing said:
Yes. The beam must be split into many different pieces, depending on how it is loaded.

Because that is how the singularity function is defined.

If x < a, then the singularity function <x-a> = 0
If x > a, then the singularity function <x-a> = (x - a)
what do you mean by
The beam must be split into many different pieces ?
 
  • #24
chetzread said:
what do you mean by
The beam must be split into many different pieces ?
Just that. If you have many different loads, then the shear force curve and the bending moment curve will be affected by all these different loads and their location along the beam. If you have a single distributed load running the entire length of the beam, that is the simplest loading to analyze. If you have a mix of point loads and distributed loads, or just several point loads, the analysis becomes more complicated. It's hard to describe without having an example problem to use as an illustration.
 
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  • #25
SteamKing said:
Just that. If you have many different loads, then the shear force curve and the bending moment curve will be affected by all these different loads and their location along the beam. If you have a single distributed load running the entire length of the beam, that is the simplest loading to analyze. If you have a mix of point loads and distributed loads, or just several point loads, the analysis becomes more complicated. It's hard to describe without having an example problem to use as an illustration.
can you explain using the example in post #1?
 

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