Solving Timoshenko beam equation for cantilever beam

Question

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.

Answer 1

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:

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)

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

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

Answer 2

Your moment equation is missing the positive moment, $PL$ at the left support.

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