Please provide full workings to show how this problem is solved $($Answers are given$)$

The diagram below shows the lifting arrangement for a plate girder. The lifting cables $ab,ac$ and od have cross sectional areas of $169m{m}^2,120m{m}^2$ and $169m{m}^2$ respectively and they all have an $E$ value of $65k\frac{N}{m}{m}^2.$ Cable $ab$ is $8$ mm too long, i$.$e$.$ when cables $ac$ and adgo taught there is $8$ mm of slack in cable $ab.$ The girder weighs $100$ kN $.$

$($a$)$ Use the principle of minimum total potential energy to calculate the forces in the cables when the girder lifts clear of the ground. The girder can be assumed to be rigid.

$($b$)$ Demonstrate that the girder is in equilibrium under the action of self$-$weight and the forces calculated in part $($a$).$

$V=\frac{E{A}_{ab}}{2L_{ab}}(\frac{12}{13}{}_b-8{)}^2+\frac{E{A}_{ac}}{2L_{ac}}(\frac{A_a+A_d}{2}{)}^2+\frac{E{A}_{aa}}{2L_{ad}}(\frac{12}{13}{}_a{)}^2$

$-100(\frac{{}_0+{}_d}{2})$

$=0\mathrm{.}44125A_b^2+0\mathrm{.}44125A_z^2+0\mathrm{.}1625A_p{A}_d-56\mathrm{.}24A_p$

$-50{}_a+27\mathrm{.}04$

${}_b=55\mathrm{.}16mm$ and ${}_a=40\mathrm{.}52mm$

$F_{eb}=\frac{E{A}_{ab}}{L_{ab}}(\frac{12}{13}{}_b-8)=\frac{65169}{13000}(\frac{12}{13}55\mathrm{.}16-8)=36\mathrm{.}24kN$

$F_{ac}=\frac{E{A}_{es}}{L_{es}}(\frac{{}_p+A_d}{2})-\frac{65120}{12000}(\frac{55\mathrm{.}16+46\mathrm{.}52}{2})-33\mathrm{.}05kN$

$F_{ee}=\frac{E{A}_{a+1}}{L_{ed}}\frac{12}{13}{}_d=\frac{65169}{13000}\frac{12}{13}46\mathrm{.}52=36\mathrm{.}28kv$