# Axially loaded member with a spring / Mechanic of Materials

i've been struggling with this problem i made up. It is about a cylindrical copper member (diameter $$d=20$$ mm and length $$L=250$$ mm) with a spring (stiffness $$k=166,5$$ kN/mm) fixed on a wall, as the figure shows:

The copper member, initially at $$20 ºC$$, is submitted to a temperature difference of $$\Delta T=50ºC$$, and the elasticity modulus and the thermal expansion coefficient of copper are $$E=110$$ GPa and $$\alpha = 1,7\cdot 10^{-5} \frac{1}{ºC}$$, respectively.

The problem is about calculating the reactions at A and C.

Im not sure how to approach this problem with superposition or the 3 step method (equilibrium, geometric compatibility and force/displacement relation).

I know the temperature gradient of $$\Delta T=50ºC$$ will cause the member to expand $$\delta_T=0.2125$$ mm , exceeding the $$0,08$$ mm gap between the wall and the copper member. Leaving a "residual expansion" of $$\delta_{res}= \delta_T - gap = 0.2125 - 0,08 = 0,1325$$ mm, unable to expand beyond the wall, therefore, once the member touches the right wall, the spring starts compressing.

So the question is, how do i write down the correct geometrical compatibility of displacement?

Thank you for your patience (im not native speaker)

I'll try to be a bit more verbose than the other answers. Essentially, I believe we should arrive at the same results (although I haven't checked).

Like you have surmised, the temperature difference of $$\Delta T=50ºC$$ will

• initially cause the member to expand and bridge the gap -between the wall and the copper member- up to $$0,08$$ mm .
• Then $$\Delta T$$ will cause it try to continue to expand for the "residual expansion" of $$\delta_{res}= \delta_T - gap = 0.2125 - 0,08 = 0,1325$$ mm. However it is going to be resisted by the spring force.

So instead of increasing by $$0,1325 mm$$ it will expand by a smaller quantity (lets denote it $$\delta(<=0,1325 mm)$$.

Obviously this expansion will be equal to the compression of the spring. So the magnitude of the force from the spring will be $$F_s = k_s \delta$$

Now the tricky part is that this compression force will cause a reduction in length on the copper member. That compression is the unrealized expansion due to $$\Delta T$$. Namely that contraction of the copper due to the spring force will be: $$\delta_{contraction} = 0.1325 - \delta$$

That contraction will be the reason for a internal force ($$F_c$$) developing in the material which is equal to:

$$F_c = \frac{E A}{L}\delta_{contraction}= \frac{E A}{L}(0.1325 -\delta)$$

Those two forces (spring $$F_s$$ and internal force $$F_c$$) need to be equal, therefore:

$$F_s = F_c$$ $$k_s \delta= \frac{E A}{L}(0.1325 -\delta)$$ $$(k_s + \frac{E A}{L})\delta= \frac{E A}{L}(0.1325)$$ $$\delta= \frac{E A}{L\cdot (k_s + \frac{E A}{L})}(0.1325)$$ $$\delta= \frac{E A}{ (L\cdot k_s + E A)}(0.1325)$$

From there you can calculate the actual displacement of the copper bar (past the 0.08[mm]). You should find this to be $$\delta \approx 0.06mm$$.

If you substitute that into the equation for spring and copper bar you should get the same force (which is what you are after and its a nice round number).

Also the final expansion of the copper bar will be $$0.08[mm]+ \delta$$

• Yes, we are on the same page. This is a problem of two deformable bodies in series. I made mistake in my original response, which treated the copper rod as rigid body. – r13 Mar 11 at 12:50

I made mistake on my earlier response (deleted). Allow me to try it again. Please let me know if there is mistake.

We know that the spring and the copper bar reach an equilibrium at Fs =Fc.

$$Fc = KcXc$$ where:

• $$Kc =\frac{(\text{Young modulus of copper})A}{L}$$

$$Fs =KsXs$$

$$KcXc=KsXs$$

$$Xc/Xs= Ks/Kc$$

$$Xc-0.08-Xs= 1.325 -166.5*1.325/(Ks+Kc)$$