Can I calculate tension in a preloaded string by superposition when the string is deformed?

I am trying to calculate the tension in a string with preloaded tension after a force is applied to the string as in the diagram below. The string would be under tension prior to the force $$F$$ being applied. The force should deform the string by an amount determined by Young's modulus and the other physical properties of the string. However, this does not account for all tension in the string. If we call the preloaded tension $$T$$, when the string is straight, can I simply add the tension caused by the deformation to the initial tension in the string?

My attempt to solve for the tension (call it $$T'$$) after the force $$F$$ is applied begins by calculating the change in length of the string. The length before bending is $$\sqrt{L^2+h^2}$$ and the length after bending is $$x+\sqrt{(L-x)^2+h^2}$$ Therefore, the change in length is calculated as $$x+\sqrt{(L-x)^2+h^2}-\sqrt{L^2+h^2}$$ and the strain can be calculated as $$\epsilon=\frac{x+\sqrt{(L-x)^2+h^2}-\sqrt{L^2+h^2}}{\sqrt{L^2+h^2}}=\frac{x+\sqrt{(L-x)^2+h^2}}{\sqrt{L^2+h^2}}-1$$ Calling the tension induced by deformation $$T_d$$ and the cross-sectional area of the string $$A$$, stress is calculated as $$\sigma=T_d/A$$ Given that Young's modulus is $$Y=\frac{\sigma}{\epsilon} \therefore \sigma = Y\cdot \epsilon$$ we can substitute and rearrange to get $$T_d=Y\cdot A\cdot \left(\frac{x+\sqrt{(L-x)^2+h^2}}{\sqrt{L^2+h^2}}-1\right)$$ Firstly, is this a even valid calculation given the angle $$\theta$$? I'm not very confident about it, but it's the best I can come up with. Secondly, given the increase in tension due to the change in overall length, is it valid to simply add this amount to the preloaded tension in the string. I'm mostly concerned with the tension in the section of the string labeled $$x$$, but would be interested in knowing the tension in both sections if possible. Having established that, calculating the tension on an ideal string needs us to find the strain on the spring. Strain is $$\epsilon = \frac{\Delta{L}}{L_0}$$ where $$L_0$$ is the length of the unstrained spring (i.e. without force).
This makes the calculations trivial: You first calculate $$L_0$$ from the first configuration with tension $$T$$, then use this length to find the final strain and tension $$T'$$.
• If I'm reading your answer correctly, my calculation for $T_d$ is valid. My question then is, is it also valid to say that $T'=T+T_d$? – FoitGuy May 12 at 15:44
• No no no, while calculating $T_d$ you used $\sqrt{L^2+h^2}$ as the initial length, yet it is not the unstrained (original) size. Therefore your $T_d$ is wrong. The two tensions $T$ and $T'$ both need to be calculated as if they are strained from free state There is no intrinsic $T_d$ in this system rather than the calculated value. – acarturk May 12 at 15:52
• I see. So I need to calculate the original length based on the preloaded tension. $$T=A\cdot Y\cdot \frac{\sqrt{L^2+h^2}-L_0}{L_0}$$ Solving for $L_0$ I get $$L_0=\frac{A\cdot Y\cdot \sqrt{L^2+h^2}}{T+A\cdot Y}$$ I would then use this in place of $\sqrt{L^2+h^2}$ in my equation for $T_d$ and that would be the overall tension. Is that correct? – FoitGuy May 12 at 16:00
• In other words $$T'=T_d=A\cdot Y\cdot \left(\frac{x+\sqrt{(L-x)^2+h^2}}{\left(\frac{A\cdot Y\cdot \sqrt{L^2+h^2}}{T+A\cdot Y}\right)}-1\right)$$ – FoitGuy May 12 at 16:15