Calculate the stress in the duralumin rod

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SUMMARY

The discussion focuses on calculating the stress in a duralumin rod and a mild steel tube under an axial load of 20 kN. The rod has a diameter of 50 mm, while the tube has an outer diameter of 60 mm, reduced to 55 mm over half its length. The Young's modulus for steel is 196 GN/m², and for duralumin, it is 126 GN/m². The initial calculation for stress in the duralumin rod is 101,859,163.6 N/m², but the participant expresses uncertainty about the approach and the application of the Young's modulus values.

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Homework Statement


a duralumin rod of 50mm diameter is a loose fit inside a mild steel tube of 60mm outside diameter. if the steel tube is turned down to 55mm diameter over one half its length, calculate the stress in the duralumin rod and the stress in each portin of the tube due to an akial load of 20KN. both rod and tube are the same length. E for steel=196GN/m^2 and E for duralumin = 126GN/m^2.

Homework Equations


force=stress in duralumin*area of duralumin+stress in steel*area of steel
strain in steel=strain in duralumin

The Attempt at a Solution


stress in duralumin=load/area of duralumin
=20000/0.001963495409
=101,859,163.6N/m^2
and I am stuck any help please
 
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Are you sure you reported the full question correctly? Or, more to the point, are you sure you reported all of it? I'm interested in why you need E for the two materials.

For example: It is not clear from what you stated if the force is applied to just the tube, just the rod, both equally, or just what.

Anyway, a naïve reading would suggest you simply have two different areas for the cross section of the tube. So the stress just has two values.
 
yes that's the whole question as it was given to our class.
 
i think my attempt above was wrong(wrong path)
 
i think my attempt above was wrong(wrong path)
 

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