From
Case No. 7, midspan deflection is $\delta = \dfrac{Pb}{48EI}(3L^2 - 4b^2) \text{ when } a \gt b$
From Case No. 8, midspan deflection is $\delta = \dfrac{5w_oL^4}{384EI}$
Midspan deflection of the given beam
EIδ = EIδ due to 2 kN concentrated load + EIδ due to 1 kN/m uniform loading
$EI \, \delta = \dfrac{Pb}{48}(3L^2 - 4b^2) + \dfrac{5w_oL^4}{384}$
$EI \, \delta = \dfrac{2(1)}{48}[ \, 3(4^2) - 4(1^2) \, ] + \dfrac{5(1)(4^4)}{384}$
$EI \, \delta = \frac{11}{6} + \frac{10}{3}$
$EI \, \delta = \frac{31}{6} \text{ kN}\cdot\text{m}^3$
$\delta = \dfrac{\frac{31}{6}}{EI}$
$\delta = \dfrac{\frac{31}{6}(1000^4)}{10000(20 \times 10^6}$
$\delta = 25.83 \text{ mm}$ answer
