A Brass Wire Cooled Between Supports

solved ThermalExpansion · YoungsModulus

Q. A brass wire 1.8 m $L_{1} = 1.8$ m long at 27 °C $T_{1} = 27$ °C is held taut $L = L_{1}$ with little tension between two rigid supports $Δ L = - Δ l$ . Co-efficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹ $α_{l} = 2.0 \times 10^{−5}$ K⁻¹; Young's modulus of brass = 0.91 × 10¹¹ Pa $Y = 0.91 \times 10^{11}$ Pa. If the wire is cooled to a temperature of -39 °C $T_{2} = -39$ °C, what is the tension developed in the wire, if its diameter is 2.0 mm $r = 1.0$ mm, in newtons?

Inputs

length

L_{1}

1.80 m

warm

T_{1}

27 °C

expansion

α_{l}

2e-05 K⁻¹

modulus

Y

9.1e+10 Pa

cold

T_{2}

-39 °C

Outputs

tension

F

How it's solved

Solved with ThermalExpansion, YoungsModulus

Given

$L_{1} = 1.8$ m
$T_{1} = 27$ °C
$α_{l} = 2.0 \times 10^{−5}$ K⁻¹
$Y = 0.91 \times 10^{11}$ Pa
$T_{2} = -39$ °C
$r = 1.0$ mm from “its diameter is 2.0 mm”
$L = L_{1}$ from “held taut”
$Δ L = - Δ l$ from “between two rigid supports”

Find $F$ .

Working

$T_{1} = 27$ °C $= 300.15$ K
$T_{2} = -39$ °C $= 234.15$ K
$r = 1.0$ mm $= 0.001$ m
linear expansion(wire) $L_{2} - L_{1} = α_{l} \cdot L_{1} \cdot (T_{2} - T_{1})$
$L_{2} - 1.8 = 2 \times 10^{−5} \cdot 1.8 \cdot (234.15 - 300.15)$
$L_{2} = 1.797624$ m
the change is the length after less the length at first(wire) $Δ l = L_{2} - L_{1}$
$Δ l = 1.797624 - 1.8$
$Δ l = −0.002376$ m
between two rigid supports $Δ L = - Δ l$
$Δ L = - (−0.002376)$
$Δ L = 0.002376$ m
held taut $L = L_{1}$
$L = 1.8$
$L = 1.8$ m
strain is stretch over length(wire) $ε = \frac{Δ L}{L}$
$ε = \frac{0.002376}{1.8}$
$ε = 0.00132$
hookes law(wire) $Y = \frac{σ}{ε}$
$9.1 \times 10^{10} = \frac{σ}{0.00132}$
$σ = 1.2012 \times 10^{8}$ Pa
a round rod s cross section(wire) $A = π \cdot r^{2}$
$A = 3.14159265 \cdot {0.001}^{2}$
$A = 3.14159265 \times 10^{−6}$ m²
stress is load over area(wire) $σ = \frac{F}{A}$
$1.2012 \times 10^{8} = \frac{F}{3.14159265 \times 10^{−6}}$
$F = 377.4$ N
$F \approx 3.8 \times 10^{2}$ Nto 2 significant figures

Conclusion

Cooled, the wire would shrink, but the supports hold it at its old length; it is stretched by just what it would have shrunk, and that strain sets the tension.

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_wire_cooled_between_supports do4  kind :numerical5  describes "A brass wire %{length} long at %{warm} is held taut with little tension between two rigid supports. " \6            "Co-efficient of linear expansion of brass = %{expansion}; Young's modulus of brass = %{modulus}. If the " \7            "wire is cooled to a temperature of %{cold}, %{tension}, if its diameter is 2.0 mm, in newtons?"8  given length: "1.8 m"9  given warm: "27 °C"10  given expansion: "2.0e-5 K-1"11  given modulus: "0.91e11 Pa"12  given cold: "-39 °C"13  asks tension: "what is the tension developed in the wire"14end

solution.rb

1require_relative "../questions/a_brass_wire_cooled_between_supports"2require_relative "../science/thermal_properties_of_matter/thermal_expansion"3require_relative "../science/mechanical_properties_of_solids/youngs_modulus"45solve :a_brass_wire_cooled_between_supports, using: [ ThermalExpansion, YoungsModulus ] do6  objects :wire78  take length: wire.length_at_first9  take warm: wire.temperature_at_first10  take expansion: wire.coefficient_of_linear_expansion11  take modulus: wire.youngs_modulus12  take cold: wire.temperature_after1314  deduce_from "its diameter is 2.0 mm", wire.thickness => "1.0 mm"15  deduce_from "held taut", wire.original_length => wire.length_at_first16  deduce_from "between two rigid supports", wire.extension => -wire.change_in_length1718  tension = wire.load19  answer tension:, in: "N", figures: 22021  conclusion "Cooled, the wire would shrink, but the supports hold it at its old length; it is stretched " \22             "by just what it would have shrunk, and that strain sets the tension."23end

playground.rb

1require_relative "../solutions/a_brass_wire_cooled_between_supports"23play_with :a_brass_wire_cooled_between_supports do4  title "A Brass Wire Cooled Between Supports"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
pi	$π$	—	0 … ∞	YoungsModulus
mass	$m$	kg	0 … ∞	YoungsModulus
gravity	$g$	m/s²	—	YoungsModulus
load	$F$	N	—	YoungsModulus
thickness	$r$	m	0 … ∞	YoungsModulus
diameter	$D$	m	—	YoungsModulus
cross section	$A$	m²	—	YoungsModulus
original length	$L$	m	—	YoungsModulus
extension	$Δ L$	m	—	YoungsModulus
stress	$σ$	Pa	—	YoungsModulus
strain	$ε$	—	—	YoungsModulus
youngs modulus	$Y$	Pa	—	YoungsModulus
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 wire	$L_{2}$	m	—
length at first of wire	$L_{1}$	m	—
coefficient of linear expansion of wire	$α_{l}$	K⁻¹	—
temperature after of wire	$T_{2}$	K	—
temperature at first of wire	$T_{1}$	K	—
change in length of ground	$Δ l_{g}$	m	—
change in length of wire	$Δ l$	m	—
thickness of ground	$r_{g}$	m	0 … ∞
diameter of ground	$D_{g}$	m	—
thickness of wire	$r$	m	0 … ∞
diameter of wire	$D$	m	—
cross section of ground	$A_{g}$	m²	—
cross section of wire	$A$	m²	—
stress of ground	$σ_{g}$	Pa	—
load on ground	$F_{g}$	N	—
stress of wire	$σ$	Pa	—
load on wire	$F$	N	—
strain of ground	$ε_{g}$	—	—
extension of ground	$Δ L_{g}$	m	—
original length of ground	$L_{g}$	m	—
strain of wire	$ε$	—	—
extension of wire	$Δ L$	m	—
original length of wire	$L$	m	—
youngs modulus of ground	$Y_{g}$	Pa	—
youngs modulus of wire	$Y$	Pa	—
mass of wire	$m$	kg	0 … ∞

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

r_{r} = \frac{D_{r}}{2}

radius is half the diameter

A_{r} = π \cdot {r_{r}}^{2}

a round rod s cross section

σ_{r} = \frac{F_{r}}{A_{r}}

stress is load over area

ε_{r} = \frac{Δ L_{r}}{L_{r}}

strain is stretch over length

Y_{r} = \frac{σ_{r}}{ε_{r}}

hookes law

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

linear expansion(wire)

Δ l_{g} = L_{2,g} - L_{1,g}

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

Δ l = L_{2} - L_{1}

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

r_{g} = \frac{D_{g}}{2}

radius is half the diameter(ground)

r = \frac{D}{2}

radius is half the diameter(wire)

A_{g} = π \cdot {r_{g}}^{2}

a round rod s cross section(ground)

A = π \cdot r^{2}

a round rod s cross section(wire)

σ_{g} = \frac{F_{g}}{A_{g}}

stress is load over area(ground)

σ = \frac{F}{A}

stress is load over area(wire)

ε_{g} = \frac{Δ L_{g}}{L_{g}}

strain is stretch over length(ground)

ε = \frac{Δ L}{L}

strain is stretch over length(wire)

Y_{g} = \frac{σ_{g}}{ε_{g}}

hookes law(ground)

Y = \frac{σ}{ε}

hookes law(wire)

L = L_{1}

held taut

Δ L = - Δ l

between two rigid supports