Thermal Expansion Force Calculator

Thermal Expansion Force Calculator:

Enter the values of modulus of elasticity E_(PSI), coefficient of thermal expansion a_{(in/in * F)}, change in temperature T_(F), and cross-sectional area A_(in2)  to determine the value of Force, F_(lbf).

Enter Modulus of Elasticity :		PSI
Enter Coefficient of Thermal Expansion :		in/in*F
Enter Change in Temperature :		F
Enter Cross-Sectional Area :		in2
Result – Force :		N

Thermal Expansion Force Formula:

Thermal expansion force arises due to the expansion or contraction of materials as they are heated or cooled. The force generated by this change in dimension is significant in structural and mechanical engineering.

Force, F_(lbf) in pounds-force is calculated by multiplying the modulus of elasticity, E_(PSI) in pounds per square inch, by the coefficient of thermal expansion, a_{(in/in * F)}in inches per inch per degree Fahrenheit, by the change in temperature, T_(F) in degrees Fahrenheit, and by the cross-sectional area, A_(in2) in square inches.

Force, F_(lbf) = E_(PSI) * a_{(in/in * F)} * T_(F) * A_(in2)

F_(lbf) = force in pounds-force, lbf.

E_(PSI) = modulus of elasticity in pounds per square inch, PSI.

a_{(in/in * F)} = coefficient of thermal expansion in inches per inch per degree Fahrenheit,  in/in*F.

T_(F) =  change in temperature in degrees Fahrenheit, F.

A_(in2) = cross-sectional area in square inches, in².

Thermal Expansion Force Calculation:

1.Suppose a steel bar with a modulus of elasticity of 29,000,000 PSI and a coefficient of thermal expansion of 0.00000645 in/in*F undergoes a temperature increase of 100°F. If the cross-sectional area of the bar is 2 in^2, calculate the force produced.

Given: E_(PSI) = 29,000,000PSI, a_{(in/in * F)} =0.00000645in/in*F, T_(F) = 100degree,F,

A_(in2) = 2in².

Force, F_(lbf) = E_(PSI) * a_{(in/in * F)} * T_(F) * A_(in2)

F_(lbf) = 29,000,000 * 0.00000645 * 100 * 2

F_(lbf) = 37410lbf.

2.Consider an aluminum rod with a force of -6150lbf (The negative sign indicates contraction force due to cooling) and a coefficient of thermal expansion of 0.0000123 in/in*F. The rod experiences a temperature decrease of 50°F, and the cross-sectional area is 1 in^2.

Given: F_(lbf) = -6150lbf, a_{(in/in * F)} =0.0000123in/in*F, T_(F) = 50degree,F,

A_(in2) = 1in².

Force, F_(lbf) = E_(PSI) * a_{(in/in * F)} * T_(F) * A_(in2)

E_(PSI) = F_(lbf) / a_{(in/in * F)} * T_(F) * A_(in2)

E_(PSI) = -6150 / 0.0000123 * (-50) * 1

E_(PSI) = 10,000,000PSI.