# The bar shown in figure is subjected to an axial pull of 150 kN.

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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 . +1 vote
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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