From
Case No. 7, midspan deflection is
δ=Pb48EI(3L2−4b2) when a>b
From Case No. 8, midspan deflection is δ=5woL4384EI
Midspan deflection of the given beam
EIδ = EIδ due to 2 kN concentrated load + EIδ due to 1 kN/m uniform loading
EIδ=Pb48(3L2−4b2)+5woL4384
EIδ=2(1)48[3(42)−4(12)]+5(1)(44)384
EIδ=116+103
EIδ=316 kN⋅m3
δ=316EI
δ=316(10004)10000(20×106
δ=25.83 mm answer