Question

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 .

Answer 1

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