Solve Axial Displacement: 80kN Load on Steel Rod Attached to Aluminum Tube

[mbaron]

Please give me some guidance on this problem. I can attach the figure if it will help.

The assembly shown in the figure consists of an aluminum tube AB having a cross sectional area of 400mm2. A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod. Take Est = 200 GPa, Eal = 70 GPa. [ΔL = 4.20 mm]


Here is what I have so far:

E_al = \frac{\sigma}{\epsilon}

70GPa = \frac{\frac{Force}{Area}}{\frac{\delta_length}{length_0}}

70GPa = \frac{\frac{80KN}{.0004m^2}}{\frac{\delta_length}{.4m}}

\delta_length = \left[ \frac{80KN}{.0004m^2} \right] .4m

\delta_length = 1.1428m


The displacement is already to large for the given answer. This does not yet include the displacement of the steel bar.

I attached the image that was provided with the homework problem.

 

[Pyrrhus]

Can you post the picture?

 

[Pyrrhus]

Hello again,

You have a superposition of displacements, which will give you the final the displacement of point C. First you have the displacement of the steel rod due to the tensile stress, and then the displacement of the aluminium tube due to compressive stress.

\delta_{aluminium} + \delta_{steel} = \delta_{C}

Use: (displacement of a prismatic bar based on Hooke's Law)

\delta = \frac{PL}{EA}

Becareful of your units.

 

[mbaron]

Thanks for the response. I worked it out.

-Morgan