The force in the rod when the temperature is 150 °F is 718.72 pounds-force.
How to determine the resulting the resulting force due to mechanical and thermal deformation
Let suppose that rod experiments a quasi-static deformation and that both springs have a linear behavior, that is, force ([tex]F[/tex]), in pounds-force, is directly proportional to deformation. Then, the elongation of the rod due to temperature increase creates a spring deformation additional to that associated with mechanical contact.
Given simmetry considerations, we derive an expression for the spring force ([tex]F[/tex]), in pounds-force, as a sum of mechanical and thermal effects by principle of superposition:
[tex]F = k\cdot (\Delta x + 0.5\cdot \Delta l)[/tex] (1)
Where:
- [tex]k[/tex] - Spring constant, in pounds-force per inch.
- [tex]\Delta x[/tex] - Spring deformation, in inches.
- [tex]\Delta l[/tex] - Rod elongation, in inches.
The rod elongation is described by the following thermal dilatation formula:
[tex]\Delta l = \alpha \cdot L_{o}\cdot (T_{f}-T_{o})[/tex] (2)
Where:
- [tex]\alpha[/tex] - Coefficient of linear expansion, in [tex]\frac{1}{^{\circ}F}[/tex].
- [tex]L_{o}[/tex] - Initial length of the rod, in inches.
- [tex]T_{o}[/tex] - Initial temperature, in degrees Fahrenheit.
- [tex]T_{f}[/tex] - Final temperature, in degrees Fahrenheit.
If we know [tex]k = 1000\,\frac{lb}{in}[/tex], [tex]\Delta x = 0.7\,in[/tex], [tex]\alpha = 6.5\times 10^{-6}\,\frac{1}{^{\circ}F}[/tex], [tex]L_{o} = 48\,in[/tex], [tex]T_{o} = 30\,^{\circ}F[/tex] and [tex]T_{f} = 150\,^{\circ}F[/tex], then the force in the rod at final temperature is:
[tex]F = \left(1000\,\frac{lb}{in} \right)\cdot \left[0.7\,in + 0.5\cdot\left(6.5\times 10^{-6}\,\frac{1}{^{\circ}F} \right)\cdot (48\,in)\cdot (150\,^{\circ}F-30\,^{\circ}F)\right][/tex]
[tex]F = 718.72\,lbf[/tex]
The force in the rod when the temperature is 150 °F is 718.72 pounds-force. [tex]\blacksquare[/tex]
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