ΣMto the left of B=0
3.5RA−2.5(3.5)(3.52)=0
RA=4.375 kN ← Answer for Part (1)
By symmetry:
RC=RA
RC=4.375 kN
ΣFV=0
RB+RA+RC=7(2.5)
RB=8.75 kN ← Answer for Part (2)
Δ=Δuniform load−Δreaction at B
Δ=5wL4384EI−PL348EI
Δ=5(2.5)(74)(10004)384(2.8×1011)−8.75(73)(10004)48(2.8×1011)
Δ=55.83 mm ← Answer for Part (3)