Getting to this 4th order O.D.E.

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The discussion centers on confusion regarding the derivation of a fourth-order ordinary differential equation (O.D.E.) and the terms involved in the equations presented. Steve questions the presence of a t² term in the denominator of the k4 term after equation (3), suggesting it should only involve t¹. He notes discrepancies in the substitution used, particularly regarding the second moment of area and its impact on the final answer. Additionally, he seeks clarification on the final derivation of equation (4), expressing uncertainty about the expected form of the particular solution. The conversation highlights the complexities of deriving and understanding terms in higher-order O.D.E.s.
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Hi,

Try as I might, I cannot understand how equation (3) with the k4 term was derived. In equation (2), w is a function of t. But in the k4 term after equation (3), there is a t2 term in the denominator. Should it not be a t1 only? What am I missing? Please help.

Regards,

Steve
 

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The screenshot is not so clear. are these the equations?
##w = \sigma_x \frac{l}{R}=\frac{\sigma w t}{R} - \frac{Ety}{R^2} ##

and the substitution used is
##\frac{d^2y}{dx^2}=\frac{M}{D}##
##D=\frac{Et^2}{12(1-\mu^2)}##
##w=\frac{d^2M}{dx^2}=D\frac{d^4y}{dx^4}##

This leads indeed to a different answer with t instead of ##t^2##, but I do not know where the error is. It could be in the answer or in the substitution rule. You might need to check the second moment of area for your specific geometry to see if it is correct.
 
Hi,

I think the copy of the paper is unclear. I think there is a t3 term in D. Sorry for the confusion...

BUT, I could use your help with the final derivation of equation (4)-see attached. I know that the particular solution to the 4th order O.D.E should look something like this:
y(x) = k4(Rδ/E)x4/24
But, I don't see that in equation (4). What am I missing?

Regards,

Steve
 

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  • Deriving equation (4).PNG
    Deriving equation (4).PNG
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I have Mass A being pulled vertically. I have Mass B on an incline that is pulling Mass A. There is a 2:1 pulley between them. The math I'm using is: FA = MA / 2 = ? t-force MB * SIN(of the incline degree) = ? If MB is greater then FA, it pulls FA up as MB moves down the incline. BUT... If I reverse the 2:1 pulley. Then the math changes to... FA = MA * 2 = ? t-force MB * SIN(of the incline degree) = ? If FA is greater then MB, it pulls MB up the incline as FA moves down. It's confusing...

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