Thermal + Compressive Stress in a Rod + Spring system

[maverick280857]

Hello everyone

First off, a very happy new year to all my PF friends. I need some help with a mathematical analysis of the following situation:

One end of a metallic rod of length L_{1} measured at temperature T_{1} is fixed to a rigid wall and the other end is connected to a spring with force constant k (the other end of the spring is anchored to a parallel rigid wall). The coefficient of linear expansion of the rod is \alpha. We have to find the compression of the spring when the temperature is raised to T_{2}.

I've worked out the details...

If we add the strains algebraically,

\epsilon = \epsilon_{thermal} + \epsilon_{spring}

with \epsilon_{thermal} = \alpha\Delta T, \epsilon_{spring} = -\frac{k\Delta L}{YA} and \epsilon = \frac{\Delta L}{L_{1}}, we do get an expression for \Delta L under the combined action of these forces. This does yield the correct answer but I want to be sure about it. Is it correct?

Thanks and cheers
Vivek

 

[Mindscrape]

I'm not sure I really understand the apparatus, but your work looks good.

 

[Astronuc]

I am not sure why one would add the strains.

The bar expands thermally against the spring, so some of the thermal expansion of the bar causes the spring to displace. Now I say some, because as the spring is displaced from its equilibrium position, the spring force increases - F = kx, where x is the displacement from no-load. The spring imposes a force in opposition to the bar which presumably responds with a mechanical strain according to Hooke's law (elastic region).

If the bar was unrestrained it would expand thermally in three dimensions with the displacement along the major axis being most significant.

If I read the problem correctly, the spring is just contacting the bar when it's temperature is T₁. So when the temperature is raised to T₂, the bar expands thermally. The spring force increases until a new equilibrium is achieved, and both the spring and bar are under compression (uniaxial compression for the bar).

 

[Pyrrhus]

Astronauc did a good complete physical analysis, so i am just going to add this:

Ok for linear elastic and homogenous materials the compatibility equation for longitudinal displacements:

\delta = \sum_{i=1}^{n} \delta_{i}

so if there aren't any loads in intermediate points and the bar has an uniform cross section.

\frac{\delta}{L} = \frac{1}{L} \sum_{i=1}^{n} \delta_{i}

\epsilon = \sum_{i=1}^{n} \epsilon_{i}

by the way, i agree with your work.

 

[maverick280857]

Thank you Cyclovenum and Astronuc!

Cheers
Vivek