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The force in the rod when the temperature is 150 °F is 718.72 pounds-force.
<h3>How to determine the resulting the resulting force due to mechanical and thermal deformation</h3>Let suppose that rod experiments a <em>quasi-static</em> deformation and that both springs have a <em>linear</em> behavior, that is, force (), in pounds-force, is directly proportional to deformation. Then, the elongation of the rod due to <em>temperature</em> increase creates a <em>spring</em> deformation additional to that associated with <em>mechanical</em> contact.
Given simmetry considerations, we derive an expression for the <em>spring</em> force (), in pounds-force, as a sum of mechanical and thermal effects by principle of superposition:
(1)
Where:
- Spring constant, in pounds-force per inch.
- Spring deformation, in inches.
- Rod elongation, in inches.
The <em>rod</em> elongation is described by the following <em>thermal</em> dilatation formula:
(2)
Where:
- Coefficient of linear expansion, in
.
- Initial length of the rod, in inches.
- Initial temperature, in degrees Fahrenheit.
- Final temperature, in degrees Fahrenheit.
If we know ,
,
,
,
and
, then the force in the rod at final temperature is:
The force in the rod when the temperature is 150 °F is 718.72 pounds-force.
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