Question

A bimetallic strip (brass/steel), which is straight at room temperature, will be immersed in boiling water and allowed to equilibrate. I will then remove the strip and quickly place it on a piece of graph paper to measure the deflection of the end, i.e. how far the end is away from the original line, measured perpendicular to the original line.
The bimetallic strip is 222 mm long. The thickness was measured to be 0.036”.
Predict how far the end will deflect, to the nearest half millimeter.

Answer 1

The thermal expansion of the materials allows to find the deflection of the bimetallist strip is Δy = 3.48 cm

given paramers

    * Bimetallic brass / steel tape

    * Initial temperature, room temperature T = 20ºC

    * Final temperature, boiling water  = 100ºC

    * initial length L₀ = 222mm (1cm / 10mm) = 22.2cm

    * thickness of bimetallic tape e = 0.036 inch (2.54 cm/1 inch) = 0.0914 cm

to find

    * perpendicular deviation or deflection (Δy)

Thermal expansion is the phenomenon of change in the length of a body due to the change in temperature, due to the increase in the length of the atomic and molecular bonds, macroscopically it is described by

        ΔL = α L₀ ΔT

ΔL and ΔT are the variation of the length and temperature respectively, L₀ is the initial length and α the coefficient of expansion ends.

In this case we have a strip formed by two materials with different coefficient of thermal expansion,

Brass       α_{brass}   = 19 10⁻⁶ ºC⁻¹

Steel       α_{steel}    = 11 10⁻⁶ ºC⁻¹

In the attached we can see a diagram of the process, as the temperature increases, the material with greater thermal expansion lengthens more, so the system must curve towards the center of the material with less

thermal expansion. Let's find the length of the strip for each material

brass          L_{f brass} - L₀ = α_{brass} L₀ ΔT

Steel           L_{f steel} - L₀ = \alpha_{steel} L₀ ΔT

Note that the initial length is the same for the two materials and that the strip is in thermal equilibrium at room temperature.

If we assume that we have an arc of circumference, we can write the length of the arc

        θ = L / r

where θ is the angle in radines, L the length of the arc and r the radius of curvature, let's write this equation for each material

brass     L_{f \ brass} =θ r₁

steel      L_{f \ steel} = θ r₂

we substitute in our equations

           θ r₁ - L₀ = α_{brass} L₀ ΔT

           θ r₂ - L₀ = α_{steel} L₀ ΔT

let's subtract the two equations

           θ (r₁- r₂) = L₀ ΔT (α_{brass} - α_{steel})

the thickness of the strip is

           e = r₁ -r₂

           θ = $Lo \ \Delta T \ \frac{\alpha_{brass} - \alpha_{steel}}{e}$

we calculate

           θ = $22.2 \ (100-20) \ \frac{(19-11) \ 10^{-6}}{0.0914}$

           θ = 0.155 rad

Let's use trigonometry to find the perpendicular deflection

          tan θ = Δy / L₀

          Δy = L₀ tan θ

          Δy = 22.2 tan 0.155

          Δy = 3.48 cm

Using the thematic expansion of the two materials we find the deflection of the bimetallist strip is 3.38 cm

Learn more about thermal expansion here: brainly.com/question/18717902