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Update with solution
Solution 1
Problem: Beam clamped at left side, free end on right side, point load pointing downwards. x is defined positive from the clamped end towards the free end.

Create FBD from a section x from the free end gives:
$M(x) = -P(L-x)$
$Q(x) = -P$

Differential equation for equilibrium is:
$-EI\frac{d\alpha}{dx} = M(x) = -P(L-x)$
Integrating this function results in:
$EI\alpha(x) = \frac{-Px(2L-x)}{2}+C_{1}$
Using the boundary condition that the beam is clamped at $x=0$:
$\alpha(0) = 0 \rightarrow 0 + C_{1} = 0 \rightarrow C_{1} = 0$
Now using the second differential equation for equilibrium:
$\frac{d}{dx}\left(\frac{dw}{dx}-\alpha(x)\right)+\frac{p(x)}{\kappa G A} = 0$
Integration of this equation leads to:
$\frac{dw}{dx} = \frac{Q(x)}{\kappa A G}+\alpha(x)$
Substituting in the previous found equation for $\alpha(x)$ gives:
$\frac{dw}{dx} = -\frac{P}{\kappa A G} - \frac{Px(2L-x)}{2EI}$
Integrating this once more to find the equation for $w(x)$ leads to:
$w(x) = -\frac{Px}{\kappa A G} - \frac{Px^2(3L-x)}{6EI} + C_{2}$
Using the boundary condition $w(0) = 0$ gives:
$w(0) = 0 = -0 - 0 + C_{2} \rightarrow C_{2} = 0$
Which gives:
$w(x) = -\frac{Px}{\kappa A G} - \frac{P}{EI}\left(-\frac{x^3}{6}+\frac{x^2L}{2}\right)$

Solution 2
Problem: Beam clamped at right side, free end on left side, point load pointing downwards.\

Create FBD from a section x from the free end gives:
$M(x) = -Px$
$Q(x) = -P$

Differential equation for equilibrium is:
$-EI\frac{d\alpha}{dx} = M(x) = -Px$
Integrating this function results in:
$EI\alpha(x) = \frac{Px^2}{2}+C_{1}$
Using the boundary condition that the beam is clamped at $x=L$:
$\alpha(L) = 0 \rightarrow \frac{PL^2}{2} + C_{1} = 0 \rightarrow C_{1} = -\frac{PL^2}{2}$
Which gives:
$\alpha(x) = \frac{P\left(x^2-L^2\right)}{2EI}$
Now using the second differential equation for equilibrium:
$\frac{d}{dx}\left(\frac{dw}{dx}-\alpha(x)\right)+\frac{p(x)}{\kappa G A} = 0$
Integration of this equation leads to:
$\frac{dw}{dx} = \frac{Q(x)}{\kappa A G}+\alpha(x)$
Substituting in the previous found equation for $\alpha(x)$ gives:
$\frac{dw}{dx} = -\frac{P}{\kappa A G} - \frac{P\left(x^2-L^2\right)}{2EI}$
Integrating this once more to find the equation for $w(x)$ leads to:
$w(x) = -\frac{Px}{\kappa A G} + \frac{P}{2EI}\left(\frac{x^3}{3}-L^2x\right) + C_{2}$
Using the boundary condition $w(L) = 0$ gives:
$w(L) = 0 = -\frac{PL}{\kappa A G} + \frac{P}{2EI}\left(\frac{L^3}{3}-L^3\right) + C_{2} \rightarrow C_{2} = \frac{PL}{\kappa A G} + \frac{PL^3}{3EI}$
Which gives:
$w(x) = \frac{P}{\kappa A G}\left(L-x\right) + \frac{P}{EI}\left(\frac{x^3}{6}-\frac{L^2x}{2}+\frac{L^3}{3}\right)$

Note that both solution now produce the same result, but in reverse.

Old message
I am trying to solve the simple problem of a left side supported beam, with a point load at the free end. Lets say the free end is located at L and the force is pointing downwards (negative z-direction). x is defined as positive towards the right (and thus positive towards the free end). I would arrive at equations:

$Q = -P = \kappa AG(-\phi + \frac{\partial w}{\partial x})$
$M = -P(L-x) = EI \frac{\partial \phi}{\partial x}$

After which I integrate the second equation, which results in:

$\frac{-P(L-x)}{EI} = \frac{\partial \phi}{\partial x} \rightarrow \phi(x) = \frac{Px(x-2L)}{2EI} + C_{1}$
Next I apply the boundary condition: $\phi = 0$ ax $x=0$, which gives:

$\phi(0) = 0 = \frac{P\cdot 0 (0-2L)}{2EI} + C_{1} \rightarrow C_{1} = 0$

Next I integrate the other equation and substitute the found solution and the other boundary condition into it which is $w(0) = 0$, which results in:

$-P = \kappa AG(-\phi + \frac{\partial w}{\partial x}) = \kappa AG(-\frac{Px(x-2L)}{2EI} + \frac{\partial w}{\partial x}) \rightarrow \frac{\partial w}{\partial x} = \frac{-P}{\kappa AG} + \frac{Px(x-2L)}{2EI}$
$\rightarrow w(x) = \frac{Px^{2}(x-3L)}{6EI} - \frac{Px}{\kappa AG} + C_{2}$
$w(0) = 0 = \frac{P \cdot 0^{2}(0-3L)}{6EI} - \frac{P \cdot 0}{\kappa AG} + C_{2} \rightarrow C_{2} = 0$

However when taking a look at the solution provide on Wikipedia: https://en.wikipedia.org/wiki/Timoshenko%E2%80%93Ehrenfest_beam_theory#Example:_Cantilever_beam, I would expect a solution that provides the same values, except in opposite direction, which is not the case. The difference between my solution and the one provided seems to be created in the first part, which finds the solution for $\phi$, but I am unsure what I did wrong. The solution at the extremities is the same, but not the solution throughout the beam itself.

Any help would be highly appreciated.

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  • $\begingroup$ Create FBD from a section x from the free end gives: M(x)=−P(L−x) ,---- How? I rather you deselect my answer, as I don't think we are on the same page. Now you have followed the wiki's pattern - mixing, jumping, and disorganized. Sorry, the change is worse than the original. $\endgroup$
    – r13
    Jan 2 at 15:08
  • $\begingroup$ I have deselected your answer as per your wish. I have not followed the wiki example but actually this paper: iieta.org/journals/ti-ijes/paper/10.18280/ti-ijes.630105, but their approach is indeed very similar. I arrived at the moment, because for that solution the problem is defined differently where x is defined from the clamped end towards the free end, instead of the other way around. Which I now realize I did not specify in my previous post $\endgroup$
    – Dion
    Jan 2 at 16:01
  • $\begingroup$ Note that steps 171 & 172 are in line with the Timoshenko solution shown in my answer. Case closed. $\endgroup$
    – r13
    Jan 2 at 17:38
  • $\begingroup$ Yes I never disagreed with the solution you provided. I just wanted to post the solution using a little more detail for other people that might stumble onto this thread and wanted to also show that you can indeed also arrive at the solution using x going from the fixed end towards the free end. Once again thank you for your help. $\endgroup$
    – Dion
    Jan 2 at 18:02
  • $\begingroup$ Note that we've never said you can't work out from the fixed end, but provided that there is a simple way to do it why bother to go around, then make unnecessary complications that potentially cause mistakes. Hope you will keep in mind in your future practices that, oftentimes, simple is the true beauty, and that's made Timoshenko one of the best. Wish you doing well. Good luck. $\endgroup$
    – r13
    Jan 2 at 18:30

2 Answers 2

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As addressed before, you should formulate the problem with the x-axis pointing from the free end to the fixed end, which is the conventional method, and was used by the wiki example as well.

On wiki example:

enter image description here

I've carefully reviewed the wiki example, the solution seems incorrect. However, I've failed to pinpoint the root problems, as the steps jump around quite a lot, with many steps containing problematic/questionable terms. Rather than make the line-by-line correction, which could lead to more confusion, the deflection, based on Timoshenko Beam Theory, of a cantilever beam with concentrate load at the free end is provided below for your information. (Per the textbook of Timoshenko & Gere)

enter image description here

Revised per updated info:

Total curvature of an elastic beam (per Timoshenko):

$\dfrac{d^2w}{dx^2} = \dfrac{d^2w_b}{dx^2} + \dfrac{d^2w_s}{dx^2}$ = $-\dfrac{M}{EI} - \dfrac{\kappa}{GA}\dfrac{dV}{dx}$

Where $M = Px$, $V = P$ , $\kappa =$ Shear (form) Factor ($\dfrac{_3}{_2}$ for rectangle shape; $\dfrac{_4}{_3}$ for circular shape)

Integrate the above equation twice to get the deflection due to $P$, thus

$w = \dfrac{PL^3}{3EI} + \dfrac{\kappa PL}{AG}$

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  • $\begingroup$ That would indeed solve the problem. I am however curious why you have to define x from the free end towards the fixed end and why I would not be able to do this the other way around? Similar to en.wikipedia.org/wiki/…, where x is defined from the fixed end towards the free end. $\endgroup$
    – Dion
    Dec 29, 2021 at 17:30
  • $\begingroup$ I have edited my original post, to now include the integration steps that I took as well as applying the boundary conditions which I believe should be $\phi(0) = 0$ and $w(0) = 0$, please correct me if I am wrong. And if you notice any mistake please let me know. Thank you for your help already $\endgroup$
    – Dion
    Dec 29, 2021 at 20:25
  • $\begingroup$ I do not agree with this simplyfication, neglecting the $q$ term, simplifies the equation to the euler-bernoulli equation. Furthermore as was shown in the equation on the wiki is that it should be possible to solve the beam equation for a point load. $\endgroup$
    – Dion
    Dec 30, 2021 at 11:05
  • $\begingroup$ This is not a simplification, it is responding to your own statement "...left side supported beam, with a point load at the free end. ". If you wish, you can integrate the shear deformation term ("q") too with q = 0, why don't you, but sit and complain? $\endgroup$
    – r13
    Dec 30, 2021 at 12:26
  • $\begingroup$ I am not trying to complain or to be rude. However the solution you posted above does not account shear deformation which is the added term from timoshenko compared to the euler-bernoulli beam equation. Also the wiki solution is clearly different from the one you provided above, which confuses me, as the problem is the same, with a different reference frame and a change of the frame of reference should not provided a different solution for a physical problem. Thank you for your help anyways. $\endgroup$
    – Dion
    Dec 30, 2021 at 12:37
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Your moment equation is missing the positive moment, $PL$ at the left support.

$$M= PL-P(L-x)$$

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