A Brass Rod Joined to a Steel Rod

Q. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹, steel = 1.2 × 10⁻⁵ K⁻¹). What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C, in cm? expects 0.34 cm

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The web of concepts

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The equations it used

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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?

The playground

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}

The solution

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 Ruby

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

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

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

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