A problem with limits ( i think)

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Homework Help Overview

The problem involves determining the maximum deflection of a beam as a parameter approaches zero, using the equation y=M/P(sec(μL/2)-1) with μ defined as √(P/EI). The context is mathematical analysis, particularly focusing on limits and derivatives.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss substituting variables and rearranging the equation to isolate constants. There are attempts to apply derivatives and limits, with some questioning the correctness of the original equation and the implications of using P versus P^2.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the equations involved. Some have offered alternative formulations, while others express confusion about the changes made to the original equation. There is no clear consensus on the correct approach or the validity of the limits being discussed.

Contextual Notes

Participants note potential discrepancies in the original problem statement and the limit provided. There is uncertainty regarding the application of L'Hôpital's rule and the handling of the limit as P approaches zero.

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Homework Statement


The maximum deflection of a beam is given by the equation
y=M/P(sec(μL/2)-1)
μ=√(P/EI) Where EI Is a constant.
Show that as P→0 y→ML^2/8EI

This is a mathCAD problem by the way, but I'm very novice at it so i want to try formulise some sort of solution on paper before i attempt mathCAD

Homework Equations


8991dfbd9db5990224ae803c727464a7.png



The Attempt at a Solution


I'm really stuck, firstly i would replace μ with √(P/EI)
∴y = M/P(sec(√(P/EI)*L)/2)-1)
I think the next step is to try get the EI out the function as it is a constant and doesn't need to be differntiated.
The derivivative of secx is tanxsecx which is the same as tanx/cosx and as x approaches 0 the function will equal 0/1 which is 0

If someone could point me in the right direction it would be much appreciated !
 
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The limit is correct, if the initial equation has P^2 instead of P in the denominator.

I would keep EI there, it gives the correct factors in the derivatives.
$$y=M \frac{\sec \left(\sqrt{\frac{P}{EI}}\frac{L}{2} \right)-1}{P^2}$$
 
So you want to replace \sqrt{P/EI}L/2 with x. Then P= (4x/L^2)(EI)= 4EIx/L^2. y= M/[(4EIx/L^2)(sec(x)- 1)= ML^2/(4EIx(sec(x)- 1)
 
I'm still a little bit confused have you said that the √x = √P/EI * L/2 ∴ x = P/EI * L^2/4 ∴ 4xEI/L^2 = P, then

I follow that you have placed P back into the equation I am confused why the sec x is on the bottom of the fraction now when it was on the top before.

Am i not right in saying that y=(ML^2*sec(x)-1)/4EIx
So d/dx(Ml^2*sec(x)-1) = ML^2secxtanx
d/dx 4EIx = 4Ei
Would i then subsitute x back into the equation ?

Also Mfb why have you replaced P with P^2 in the original equation
 
I think some brackets in HallsofIvy's post are wrong.
steve2510 said:
Also Mfb why have you replaced P with P^2 in the original equation
That is just a guess: with P^2, the given limit is correct, without, it is not.
 
So I've said that x = √(P/Ei) * L/2
∴P = (x2*EI *4)/L2

Subsituting back in i get

(ML2(sec(x)-1)/x24EI

D/dx ML2(sec(x)-1) = ML2sec(x)tan(x) Where x = √(P/Ei) * L/2 as P reaches 0 x = 0 aswell. Which means sec(x)tan(x) = zero which cancells the top complety so surely this isn't correct!?

The bottom would become 8EIx

Am i doing something wrong, this is really my first problem with limits and i don't even recall being taught L'Hôpital's rule. This is a MathCad problem, and i think it is meant to calculate mos this maths!

I'm just sturggling to arrange it in mathCAD without getting a catalogue of errors, so i thought it might be easier to generate a paper solution first !
 
I still don't see where the p^2 came from , if p is approaching 0 then what difference does it make if p is p^2, also if its p^2 on the bottom would you not have to times the top by another factor of p. whether p is p^2 or just p surely we have the correct intermediate form of 0/0 already as secx -1 = 0
 
steve2510 said:
I still don't see where the p^2 came from , if p is approaching 0 then what difference does it make if p is p^2, also if its p^2 on the bottom would you not have to times the top by another factor of p. whether p is p^2 or just p surely we have the correct intermediate form of 0/0 already as secx -1 = 0
That P^2 is not the result of any calculation.

The formula given in post 1 does not fit to the limit given there. Therefore, we have 3 options:
- the given limit is wrong. In that case, you already know the correct limit
- the given formula is wrong, but the limit is right. In that case, I found a new, different formula, with the correct limit
- both are wrong. We are lost.
 

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