Axially loaded member with a spring / Mechanic of Materials

Question

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)

Answer 1

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 $0.08[mm]+ \delta$

Answer 2

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

Answer 3

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$

and we already know

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

I let you handle the rest.