Let force shared by each aluminium pillar be Pa and that shared by steel pillar be Ps .
∴ The forces in vertical direction = 0 →
Pa + Ps + Pa = 250
2Pa + Ps = 250 ...(1)
From compatibility condition, we get
From eqns. (1) and (2), we get
Pa (2 + 1.111) = 250
∴ Pa = 80.36 kN
Hence from eqn. (1),
Ps = 250 – 2 × 80.36 = 89.28 kN
∴ Stresses developed are