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

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

Slope and deflection for $($IIa$)$

Using units of $(KN,m)$

$=>E=200{10}^6(\frac{kN}{m^2}),I=290{10}^{-6}(m^4),P=62(KN),L=5(m)=>EI=58000(kN*m^2),w=12\mathrm{.}4(\frac{kN}{(m)})$

For $($IIa$),$

$($a$)$ the deflection at B is given as $y_{\text{B}\text{I}\text{I}}=y_{B^{-}}=y_{B^{+}}=-\frac{(-P)(\frac{L}{2}{)}^3}{3EI}=5\mathrm{.}568{10}^{-3}(m)=5\mathrm{.}568(mm)($ uarr $)($using max. deflection from App.F$)$

$($b$)$ the slope $(($angle$)$ at B is ${}_{\text{B}\text{I}\text{a}}={}_{B^{-}}={}_{B^{+}}=-\frac{(-P)(\frac{L}{2}{)}^2}{2EI}=0\mathrm{.}334{10}^{-2}(rad)=0\mathrm{.}191($degrees$)$ v $($using slope at end from App. F$)$

Slope and deflection for $($IIb$)$

For $($IIb$)$

$($a$)$ the deflection at C is given as $yCllb=(mm)(uarr)$

$($b$)$ the slope at C is given as ${}_{Cllb}=$ degrees$){?}^{}$