The load increases from $W(\frac{N}{m})$ at $x=0,$ to $kW(\frac{N}{m})$ at $x=L,$ where $k$ is a non$-$negative constant. Thus, the total beam loading is $\frac{WL}{2}(1+k).$

$[4$a$]$ Starting from first principles of engineering, show that the vertical reactions at A and B are:

$V_A=\frac{WL}{6}(2+k),$ and $,V_B=\frac{WL}{6}(1+2k).$

$[4$b$]$ Show that the bending moment at point $x$ is:

$M(x)=\frac{W{L}^2}{6}(\frac{x}{L})[(2+k)-3(\frac{x}{L})+(1-k)(\frac{x}{L}{)}^2].$

Notice that $M(0)=M(L)=0,$

$[4$c$]$ Hence, show that the location of maximum curvature $$ in the beam corresponds to the solution of the quadratic equation:

$3(1-k)x^2-6Lx+(2+k)L^2=0.$

$[4$d$]$ For the case where $k=1($i$.$e$.,$ a constant uniform loading$),$ use equations $4$ and $5$ to determine the position and value of the maximum bending moment.regardless of the value of $k.$