At every point along the beam, the curvature has to be such that the externally applied bending moment exactly counters the internal stress. This tells you that the curvature is not constant - it is a function of distance to the side (largest in the middle, zero at the wall). This means that your assumption of "circular section" is wrong.
See for example figure 3.16 in this link and associated derivations.
Simplifying the description found there:
From their equation 3.21, the curvature $\rho$ of a beam is related to the bending moment $M$ by
$$\rho = \frac{EI}{M}\tag1$$
Where $E$ is the Young's modulus and $I$ is the second moment of area. For a rectangular beam (not specified in your question, but that's what I am assuming) we can compute $I$ as
$$I = \frac{bd^3}{12}\tag2$$
(see for example this link)
Now we need an expression for the bending moment as a function of position. For points to the left of the center, bending moment is proportional to $Wx/2$ - half the weight (two supports) times the distance from the support.
Knowing that the radius of curvature is (for small deflections) inversely proportional to the second derivative of the shape, we can write
$$\frac{d^2 y}{dx^2} = -\frac{Wx}{2EI}\tag3$$
Integrating twice, we get
$$y = -\frac{Wx^3}{12EI} + Ax + B\tag4$$
If we set $y=0$ at $x=0$, we get $B=0$. Setting the slope of the curve =0 at $x = \frac{\ell}{2}$, we find
$$-\frac{W(\ell/2)^2}{4EI} + A = 0$$
$$A = \frac{W\ell^2}{16EI} \tag5$$
leading to an expression for the deflection
$$y = \frac{Wx}{12EI}\left(\frac{3\ell^2}{4}-x^2\right) \tag6$$
Substituting $x=\frac{\ell}{2}$ into (6), and using expression (2) for $I$, we obtain the deflection you were looking for.
This expression agrees with equation (7) at this reference.
There is no other force. The rod is being accelerated. The tension $T=F(x/L)$ in the rod will vary linearly from $F$ at the pulled end where $x=L$ to $0$ at the free end where $x=0$.
This situation is like a train of trucks being pulled by an engine. The forces on each truck are different, although if the mass of each is the same then the net force on each is also the same. However, the elastic energy stored in each truck (or in the coupling springs between trucks) depends on the balanced force at the two ends (ie the tension force in the spring) rather than the net force accelerating the truck/spring.
Best Answer
A thin slender beam (rod) with length $\ell$, 2nd area moment $I = \frac{1}{12} b d^3$ and Young's modulus $E$ obeys the following differential equation
$$ M(x) = E I \frac{\partial ^2 y(x)}{\partial x^2} $$ where $y(x)$ is the deflection shape and $M(x)$ the internal moment function.
Using statics split the beam into two sections and do a free body diagram on each one
For the first section from A to C the internal moment is $$M_1(x) = -x\,A_y = - x \,\frac{F}{2} $$
For the second section from C to B the internal moment is $$M_2(x) = -x \, A_y+ F (x-\frac{\ell}{2}) = - (\ell -x) \frac{F}{2}$$
Now you can integrate the moment to get slope and deflection and make sure you use the constants of integration for fitting into the boundary conditions (at $y_1(0)=0$ and $y_2(\ell)=0$) and continuity ($y_1(\frac{\ell}{2})=y_2(\frac{\ell}{2})$ and $y_1'(\frac{\ell}{2}) = y_2'(\frac{\ell}{2})$)
The deflection shape is $$ y_1(x) = \frac{1}{E I} \int \int M_1(x)\,{\rm d}x{\rm d}x + K_1 + K_2 x = -\frac{F x (4 x^2-3 \ell^2)}{48 E I} \\ y_2(x) = \frac{1}{E I} \int \int M_2(x)\,{\rm d}x{\rm d}x + K_3 + K_4 x = -\frac{F ( \ell^3 -9 \ell^2 x+12 \ell x^2 -4 x^3)}{4 8 EI }$$
In the middle use $x=\frac{\ell}{2}$ for $$ \boxed{\delta = -\frac{F \ell^3}{48 E I} } $$
Here the result is negative because the deflection is along the negative y direction
There is also a shorter method using what is called an energy method. After finding the internal moments, find total internal energy of the system as a function of the applied load
$$U(F) = \int \limits_{0}^{\ell/2} \frac{M_1^2(x)}{2 E I}\,{\rm d}x + \int \limits_{\ell/2}^{\ell} \frac{M_2^2(x)}{2 E I}\,{\rm d}x = \frac{F^2 \ell^3}{96 E I}$$
The deflection at the point $F$ is applied (the middle) use the following partial derivative
$$\boxed{ \delta = \frac{\partial U(F)}{\partial F} = \frac{F \ell^3}{48 E I} } $$
Here the result is positive because the deflection is along the direction of the force