A Brass Rod Joined to a Steel Rod

solved ThermalExpansion

Q. A brass rod of length 50 cm $L_{1,b} = 50$ cm and diameter 3.0 mm is joined to $Δ l_{c} = Δ l_{b} + Δ l_{s}$ a steel rod of the same length $L_{1,s} = L_{1,b}$ and diameter. The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹ $α_{l,b} = 2.0 \times 10^{−5}$ K⁻¹, steel = 1.2 × 10⁻⁵ K⁻¹ $α_{l,s} = 1.2 \times 10^{−5}$ K⁻¹). What is the change in length of the combined rod at 250 °C $T_{2,b} = 250$ °C, if the original lengths are at $T_{1,s} = T_{1,b}$ , $T_{2,s} = T_{2,b}$ 40.0 °C $T_{1,b} = 40.0$ °C, in cm?

Inputs

length

L_{1,b}

50 cm

brass expansion

α_{l,b}

2e-05 K⁻¹

steel expansion

α_{l,s}

1.2e-05 K⁻¹

hot

T_{2,b}

250 °C

cool

T_{1,b}

40 °C

Outputs

change

Δ l_{c}

How it's solved

Solved with ThermalExpansion

Given

$L_{1,b} = 50$ cm
$α_{l,b} = 2.0 \times 10^{−5}$ K⁻¹
$α_{l,s} = 1.2 \times 10^{−5}$ K⁻¹
$T_{2,b} = 250$ °C
$T_{1,b} = 40.0$ °C
$L_{1,s} = L_{1,b}$ from “of the same length”
$T_{1,s} = T_{1,b}$ from “if the original lengths are at”
$T_{2,s} = T_{2,b}$ from “if the original lengths are at”
$Δ l_{c} = Δ l_{b} + Δ l_{s}$ from “joined to”

Find $Δ l_{c}$ .

Working

$L_{1,b} = 50$ cm $= 0.5$ m
$T_{2,b} = 250$ °C $= 523.15$ K
$T_{1,b} = 40.0$ °C $= 313.15$ K
linear expansion(brass) $L_{2,b} - L_{1,b} = α_{l,b} \cdot L_{1,b} \cdot (T_{2,b} - T_{1,b})$
$L_{2,b} - 0.5 = 2 \times 10^{−5} \cdot 0.5 \cdot (523.15 - 313.15)$
$L_{2,b} = 0.5021$ m
the change is the length after less the length at first(brass) $Δ l_{b} = L_{2,b} - L_{1,b}$
$Δ l_{b} = 0.5021 - 0.5$
$Δ l_{b} = 0.0021$ m
of the same length $L_{1,s} = L_{1,b}$
$L_{1,s} = 0.5$
$L_{1,s} = 0.5$ m
if the original lengths are at (temperature after of steel) $T_{2,s} = T_{2,b}$
$T_{2,s} = 523.15$
$T_{2,s} = 523.15$ K
if the original lengths are at (temperature at first of steel) $T_{1,s} = T_{1,b}$
$T_{1,s} = 313.15$
$T_{1,s} = 313.15$ K
linear expansion(steel) $L_{2,s} - L_{1,s} = α_{l,s} \cdot L_{1,s} \cdot (T_{2,s} - T_{1,s})$
$L_{2,s} - 0.5 = 1.2 \times 10^{−5} \cdot 0.5 \cdot (523.15 - 313.15)$
$L_{2,s} = 0.50126$ m
the change is the length after less the length at first(steel) $Δ l_{s} = L_{2,s} - L_{1,s}$
$Δ l_{s} = 0.50126 - 0.5$
$Δ l_{s} = 0.00126$ m
joined to $Δ l_{c} = Δ l_{b} + Δ l_{s}$
$Δ l_{c} = 0.0021 + 0.00126$
$Δ l_{c} = 0.336$ cm
$Δ l_{c} = 0.00336$ m $= 0.336$ cm
$Δ l_{c} \approx 0.34$ cmto 2 significant figures

Conclusion

With both ends free, each rod grows as it would alone, and the combined rod grows by the sum of the two.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_brass_rod_joined_to_a_steel_rod do4  kind :numerical5  describes "A brass rod of length %{length} and diameter 3.0 mm is joined to a steel rod of the same length and " \6            "diameter. The ends of the rod are free to expand (Co-efficient of linear expansion of brass = " \7            "%{brass_expansion}, steel = %{steel_expansion}). %{change} at %{hot}, if the original lengths are at " \8            "%{cool}, in cm?"9  given length: "50 cm"10  given brass_expansion: "2.0e-5 K-1"11  given steel_expansion: "1.2e-5 K-1"12  given hot: "250 °C"13  given cool: "40.0 °C"14  asks change: "What is the change in length of the combined rod"15end

solution.rb

1require_relative "../questions/a_brass_rod_joined_to_a_steel_rod"2require_relative "../science/thermal_properties_of_matter/thermal_expansion"34solve :a_brass_rod_joined_to_a_steel_rod, using: ThermalExpansion do5  objects :brass, :steel, :combined67  take length: brass.length_at_first8  take brass_expansion: brass.coefficient_of_linear_expansion9  take steel_expansion: steel.coefficient_of_linear_expansion10  take hot: brass.temperature_after11  take cool: brass.temperature_at_first1213  deduce_from "of the same length", steel.length_at_first => brass.length_at_first14  deduce_from "if the original lengths are at", steel.temperature_at_first => brass.temperature_at_first,15                                                steel.temperature_after => brass.temperature_after16  deduce_from "joined to", combined.change_in_length => brass.change_in_length + steel.change_in_length1718  change = combined.change_in_length19  answer change:, in: "cm", figures: 22021  conclusion "With both ends free, each rod grows as it would alone, and the combined rod grows by the " \22             "sum of the two."23end

playground.rb

1require_relative "../solutions/a_brass_rod_joined_to_a_steel_rod"23play_with :a_brass_rod_joined_to_a_steel_rod do4  title "A Brass Rod Joined to a Steel Rod"5end

The laws it drew on

quantity	written	units	within	from
length at first	$L_{1}$	m	—	ThermalExpansion
length after	$L_{2}$	m	—	ThermalExpansion
change in length	$Δ l$	m	—	ThermalExpansion
temperature at first	$T_{1}$	K	—	ThermalExpansion
temperature after	$T_{2}$	K	—	ThermalExpansion
coefficient of linear expansion	$α_{l}$	K⁻¹	—	ThermalExpansion
length after of ground	$L_{2,g}$	m	—
length at first of ground	$L_{1,g}$	m	—
coefficient of linear expansion of ground	$α_{l,g}$	K⁻¹	—
temperature after of ground	$T_{2,g}$	K	—
temperature at first of ground	$T_{1,g}$	K	—
length after of brass	$L_{2,b}$	m	—
length at first of brass	$L_{1,b}$	m	—
coefficient of linear expansion of brass	$α_{l,b}$	K⁻¹	—
temperature after of brass	$T_{2,b}$	K	—
temperature at first of brass	$T_{1,b}$	K	—
length after of steel	$L_{2,s}$	m	—
length at first of steel	$L_{1,s}$	m	—
coefficient of linear expansion of steel	$α_{l,s}$	K⁻¹	—
temperature after of steel	$T_{2,s}$	K	—
temperature at first of steel	$T_{1,s}$	K	—
length after of combined	$L_{2,c}$	m	—
length at first of combined	$L_{1,c}$	m	—
coefficient of linear expansion of combined	$α_{l,c}$	K⁻¹	—
temperature after of combined	$T_{2,c}$	K	—
temperature at first of combined	$T_{1,c}$	K	—
change in length of ground	$Δ l_{g}$	m	—
change in length of brass	$Δ l_{b}$	m	—
change in length of steel	$Δ l_{s}$	m	—
change in length of combined	$Δ l_{c}$	m	—

L_{2,b} - L_{1,b} = α_{l,b} \cdot L_{1,b} \cdot (T_{2,b} - T_{1,b})

linear expansion

Δ l_{b} = L_{2,b} - L_{1,b}

the change is the length after less the length at first

L_{2,g} - L_{1,g} = α_{l,g} \cdot L_{1,g} \cdot (T_{2,g} - T_{1,g})

linear expansion(ground)

L_{2,b} - L_{1,b} = α_{l,b} \cdot L_{1,b} \cdot (T_{2,b} - T_{1,b})

linear expansion(brass)

L_{2,s} - L_{1,s} = α_{l,s} \cdot L_{1,s} \cdot (T_{2,s} - T_{1,s})

linear expansion(steel)

L_{2,c} - L_{1,c} = α_{l,c} \cdot L_{1,c} \cdot (T_{2,c} - T_{1,c})

linear expansion(combined)

Δ l_{g} = L_{2,g} - L_{1,g}

the change is the length after less the length at first(ground)

Δ l_{b} = L_{2,b} - L_{1,b}

the change is the length after less the length at first(brass)

Δ l_{s} = L_{2,s} - L_{1,s}

the change is the length after less the length at first(steel)

Δ l_{c} = L_{2,c} - L_{1,c}

the change is the length after less the length at first(combined)

L_{1,s} = L_{1,b}

of the same length

T_{1,s} = T_{1,b}

if the original lengths are at (temperature at first of steel)

T_{2,s} = T_{2,b}

if the original lengths are at (temperature after of steel)

Δ l_{c} = Δ l_{b} + Δ l_{s}

joined to