$($a$)($I$)$ is case $2$ in App. F

$($b$)($II$)$ is not in App. F $($Note: From $9\mathrm{.}1($II$)$ is a complex beam, whie all cantilever beams in App. F are simple beams$)$

Slope and deflection for $(!)$

Using units of $(KN,m)$

$=>E=200{10}^6(\frac{kN}{m^2}),I=250{10}^{-6}(m^4),P=64(KN),L=9(m)=>EI=50000(kN*m^2),w=7\mathrm{.}111(\frac{kN}{m})$

For $($l$),$

$($a$)$ the deflection at $C$ is given as $y_{Cl}=-\frac{w{L}^4}{8EI}=-116\mathrm{.}64{10}^{-3}(m)=116\mathrm{.}64(mm)(darr)($using max. deflection from App. F$)$

$($b$)$ the slope $("$angle$)$ at Cis${}_{Cl}=-\frac{w{L}^3}{6EI}=-1\mathrm{.}73{10}^{-2}(rad)=0\mathrm{.}99($degree$)$ U $($using slope at end from App. F$)$

Superposition principle for $($II$)$

Dividing $($II$)$ into two separate beams as $($with continuity in moments and shear force$)$

$($a$)($Ila$)$ is case $1$ in App. F

$($b$)($Ilb$)$ is not in App. F $($Note: $($llb$)$ is in static equilibrium$)$

Slope and deflection for $($Ila$)$

For $($lla$),$

$($a$)$ the deflection at $B$ is given as $YBIl{a}_a=,(mm)(uarr)$

$($b$)$ the slope $({?}^{}$ angle$)$ at $B$ is ${}_{\text{B}\text{I}\text{l}\text{a}}=$

$]($degrees$)$ SOLVE FOR IIA AND IIB