Deformation of two perpendicular beams with parabolic properties

I want to calculate the deformation of the following structure:

I have numbered the vertical beam 1 and the horzontal beam 2. The hight of beam 2 is given by: $h(x)=h\sqrt{1-x/L_2}$

So I thought I could solve this with Castigliano's method. Where beam 1 undergoes an axial load $F$ and a constant bending moment of $Fa$. Beam 2 undergoes a bending moment of $0$ on the left and $Fa$ on the right. The energy is given by: $U=\int_0^L\frac{F(x)^2}{2AE}dx+\int_0^L\frac{M(x)^2}{2EI}dx$ So for beam 1 and 2 the energy would be:

$$U_1 = \int_0^{L_1}\frac{F^2}{2bhE}dx+\int_0^{L_1}\frac{F^2a^2}{2E\frac{1}{12}bh^3}dx = \frac{F^2L_1}{2bhE} + \frac{F^2a^2L_1}{2E\frac{1}{12}bh^3}$$

$$U_2 = \int_0^a\frac{F^2x^2}{2E\frac{1}{12}b(h\sqrt{1-x/L_2})^3}dx = \frac{4F^2L_2^2}{Ebh^3}\left(\frac{\sqrt{1-a/L_2}(a^2+4aL_2-8L_2^2)}{a-L_2}-8L_2\right)$$

Where $b$ is the thickness of the part

Then I applied Castigliano's method, which gives the deformation of the shape in the same direction as $F$:

$$\delta=\frac{\partial U}{\partial F}= \frac{F}{E}\left(\frac{L_1}{bh} + \frac{a^2L_1}{\frac{1}{12}bh^3} + \frac{8L_2^2}{bh^3}\left(\frac{\sqrt{1-a/L_2}(a^2+4aL_2-8L_2^2)}{a-L_2}-8L_2\right)\right)$$

But this equation gives me wrong results. For instance, with $F=1500, E=113.76*10^9, b=0.005, h=0.007, L_1=0.05, L_2=0.04$ and $a=0.032$ I get a deformation of 9.8 mm, where as a FEA simulation (Nastran In-Cad) gives a deformation of 6.2 mm (in the same direction as the force).

I tried this without the parabolic property and this does generate correct results. That equation is:

$$\frac{F}{E}\left(\frac{L_1}{A_1}+\frac{L_2^2L_1}{I_1}+\frac{L_2^3}{3I_3}\right)$$

So I think there something wrong with my parabolic equation, but I can't figure out what it is.

Edit: did some trial and error, and if I fill in $h(h/2)$ for $h$ in the last term I get pretty accurate results. To clarify, the height used (in this situation) in the last therm ($\frac{8L_2^2}{bh^3}$) is the hight of beam 2 where beam 1 ends, on the left. But I still have know idea why this is and if this works for every situation

• Check your coordinate system for the horizontal beam. Note that at $x = 0$, $h = h$ and at $x = L_2$, $h = 0$ which appears to be the opposite of what you have used for your moments (at $x = 0$, $M_H = Fx = 0$). Dec 20 '17 at 22:47

Thanks to Biswajit for his comment, which pointed out my mistake. My coordinates where wrong. The correct equation for $U_2$ is:
$$U_2 = \int_0^a\frac{F^2(x-a)^2}{2E\frac{1}{12}b(h\sqrt{1-x/L_2})^3} dx$$
So the correct equation for the deformation is: $$\delta= \frac{F}{E}\left(\frac{L_1}{bh} + \frac{a^2L_1}{\frac{1}{12}bh^3} + \frac{8L_2}{bh^3}\left(-3a^2+12aL_2+8L_2 \sqrt{1-\frac{a}{L_2}}(L_2-a)-8L_2^2\right)\right)$$