\begin{aligned} &\Sigma M_{A}=0 \\ &-P \cdot 2+q \cdot 6 \cdot 3+M-V_{B} \cdot 9=0 \\ &V_{B}:=\frac{-P \cdot 2+q \cdot 6 \cdot 3+M}{9}=84.444 k N \\ &\Sigma Y=0 \\ &V_{A}+V_{B}+P-6 q=0 \\ &V_{A}:=-\left(V_{B}+P-6 q\right)=125.556 k N \\ &\Sigma_{B}:=\frac{-1}{3}\left(M-V_{B} \cdot 3\right)=51.111 k N \end{aligned} $$ \begin{aligned} &\Sigma X=0 \\ &H_{A}:=H_{B}=51.111 \quad k N \end{aligned} $$ Shape of a parabolic arc axis: $$ \begin{aligned} &y=\frac{4 f}{l^{2}} x \cdot(l-x) \\ &\frac{d y}{d x}=\tan (\varphi)=\frac{4 f}{l^{2}}(l-2 x) \end{aligned} $$ $$ \begin{aligned} &\sin (\varphi)=\frac{\tan (\varphi)}{\sqrt{1+\tan (\varphi)^{2}}} \\ &\cos (\varphi)=\frac{1}{\sqrt{1+\tan (\varphi)^{2}}} \end{aligned} $$ \begin{aligned} &f:=3 \quad l:=12 \\ &y^{\prime}=\tan \varphi=\frac{4 f}{l^{2}}(l-2 x)=1-\frac{x}{6} \\ &y=\frac{4 f}{l^{2}} x \cdot(l-x)=-\frac{x^{2}}{12}+x \\ &y(2)=-\frac{2^{2}}{12}+2=\frac{5}{3} \end{aligned} For a parabolic arc: $$ \begin{aligned} &N=-V_{A} \cdot \sin (\varphi)-H_{A} \cdot \cos (\varphi) \\ &T=V_{A} \cdot \cos (\varphi)-H_{A} \cdot \sin (\varphi) \end{aligned} $$

Section AC

$$ \begin{aligned} &x(0 ; 2) \\ &T(x)=125.556 \cdot \cos \varphi-51.111 \cdot \sin \varphi-40 x \cdot \cos (\varphi) \\ &N(x)=-125.556 \cdot \sin \varphi-51.111 \cdot \cos \varphi+40 x \cdot \sin (\varphi) \\ &M(x)=125.556 x-51.111 y-40 \cdot x \cdot \frac{x}{2} \\ &M(x)=125.556 x-51.111\left(x-\frac{x^{2}}{12}\right)-40 \cdot x \cdot \frac{x}{2} \end{aligned} $$