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The bar shown in figure is subjected to an axial pull of 150 kN. Determine diameter of the middle portion if stress there is limited to 125N/mm2 . Proceed to determine the length of this middle portion if total extension of the bar is specified as 0.15 mm. Take modulus of elasticity of bar material E = 2 × 105N/mm2 .

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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 × 103 /[(\(\pi\)/4).(d22)] 

\(\therefore\) d2 = 39.1 mm

Since stress is limited to 125 N/mm2 , in the middle portion

125 = 150 × 103 /[(\(\pi\)/4).(d22 )]

Diameter of middle portion

d2 = 39.1mm

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

= 150 × 103 /[(\(\pi\)/4).(502 )] = 76.43 N/m2

Total change in length of the bar,

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

δL = δL1 + δL2 + δL3

= \(\sigma\)1L1/E + \(\sigma\)2L2/E + \(\sigma\)3L3/E; Since E is same for all portions 

= \(\sigma\)1(L1 + L3)/E + \(\sigma\)2L2/E 

L1 + L3 = 300 – L2

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