Each of the segment of this composite bar is subjected to axial pull P =150 kN.

Axial Stress in the middle portion \(\sigma\)_{2} = Axial pull/Area = 150 × 10^{3} /[(\(\pi\)/4).(d_{2}^{2})]

\(\therefore\) d_{2} = 39.1 mm

Since stress is limited to 125 N/mm^{2} , in the middle portion

125 = 150 × 10^{3} /[(\(\pi\)/4).(d_{2}^{2} )]

Diameter of middle portion

d_{2} = 39.1mm

(ii) Stress in the end portions, \(\sigma\)_{1} = \(\sigma\)_{3}

= 150 × 10^{3} /[(\(\pi\)/4).(50^{2} )] = 76.43 N/m^{2}

Total change in length of the bar,

= change in length of end portions + change in length of mid portion

δL = δL_{1} + δL_{2} + δL_{3}

= \(\sigma\)_{1}L_{1}/E + \(\sigma\)_{2}L_{2}/E + \(\sigma\)_{3}L_{3}/E; Since E is same for all portions

= \(\sigma\)_{1}(L_{1} + L_{3})/E + \(\sigma\)_{2}L_{2}/E

L_{1} + L_{3} = 300 – L_{2}