A Steel Rod Stretched

solved YoungsModulus

Q. A structural steel rod has a radius of 10 mm $r = 10$ mm and a length of 1.0 m $L = 1.0$ m. A 100 × 10³ N $F = 100 \times 10^{3}$ N force stretches it along its length. Young's modulus of structural steel is 2.0 × 10¹¹ N/m² $Y = 2.0 \times 10^{11}$ N/m². What is the elongation of the rod, in mm?

Inputs

radius

r

10 mm

length

L

1.00 m

force

F

1e+05 N

modulus

Y

2e+11 N/m²

Outputs

elongation

Δ L

How it's solved

Solved with YoungsModulus

Given

$r = 10$ mm
$L = 1.0$ m
$F = 100 \times 10^{3}$ N
$Y = 2.0 \times 10^{11}$ N/m²

Find $Δ L$ .

Working

$r = 10$ mm $= 0.01$ m
$Y = 2.0 \times 10^{11}$ N/m² $= 2 \times 10^{11}$ Pa
a round rod s cross section(rod) $A = π \cdot r^{2}$
$A = 3.14159265 \cdot {0.01}^{2}$
$A = 3.14159265 \times 10^{−4}$ m²
stress is load over area(rod) $σ = \frac{F}{A}$
$σ = \frac{100000}{3.14159265 \times 10^{−4}}$
$σ = 3.18309886 \times 10^{8}$ Pa
hookes law(rod) $Y = \frac{σ}{ε}$
$2 \times 10^{11} = \frac{3.18309886 \times 10^{8}}{ε}$
$ε = 0.00159154943$
strain is stretch over length(rod) $ε = \frac{Δ L}{L}$
$0.00159154943 = \frac{Δ L}{1}$
$Δ L = 1.592$ mm
$Δ L = 0.00159154943$ m $= 1.59154943$ mm
$Δ L \approx 1.59$ mmto 3 significant figures

Conclusion

The force spread over the rod's cross-section is the stress; steel answers it with a strain of stress over Young's modulus, and the stretch is that strain of the length.

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_steel_rod_stretched do4  kind :numerical5  describes "A structural steel rod has a radius of %{radius} and a length of %{length}. A %{force} force stretches " \6            "it along its length. Young's modulus of structural steel is %{modulus}. %{elongation}, in mm?"7  given radius: "10 mm"8  given length: "1.0 m"9  given force: "100e3 N"10  given modulus: "2.0e11 N/m2"11  asks elongation: "What is the elongation of the rod"12end

solution.rb

1require_relative "../questions/a_steel_rod_stretched"2require_relative "../science/mechanical_properties_of_solids/youngs_modulus"34solve :a_steel_rod_stretched, using: YoungsModulus do5  objects :rod67  take radius: rod.thickness8  take length: rod.original_length9  take force: rod.load10  take modulus: rod.youngs_modulus1112  elongation = rod.extension13  answer elongation:, in: "mm", figures: 31415  conclusion "The force spread over the rod's cross-section is the stress; steel answers it with a " \16             "strain of stress over Young's modulus, and the stretch is that strain of the length."17end

playground.rb

1require_relative "../solutions/a_steel_rod_stretched"23play_with :a_steel_rod_stretched do4  title "A Steel Rod Stretched"5end

The laws it drew on

quantity	written	units	within	from
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
thickness of ground	$r_{g}$	m	0 … ∞
diameter of ground	$D_{g}$	m	—
thickness of rod	$r$	m	0 … ∞
diameter of rod	$D$	m	—
cross section of ground	$A_{g}$	m²	—
cross section of rod	$A$	m²	—
stress of ground	$σ_{g}$	Pa	—
load on ground	$F_{g}$	N	—
stress of rod	$σ$	Pa	—
load on rod	$F$	N	—
strain of ground	$ε_{g}$	—	—
extension of ground	$Δ L_{g}$	m	—
original length of ground	$L_{g}$	m	—
strain of rod	$ε$	—	—
extension of rod	$Δ L$	m	—
original length of rod	$L$	m	—
youngs modulus of ground	$Y_{g}$	Pa	—
youngs modulus of rod	$Y$	Pa	—
mass of rod	$m$	kg	0 … ∞

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

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

radius is half the diameter(ground)

r = \frac{D}{2}

radius is half the diameter(rod)

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(rod)

σ_{g} = \frac{F_{g}}{A_{g}}

stress is load over area(ground)

σ = \frac{F}{A}

stress is load over area(rod)

ε_{g} = \frac{Δ L_{g}}{L_{g}}

strain is stretch over length(ground)

ε = \frac{Δ L}{L}

strain is stretch over length(rod)

Y_{g} = \frac{σ_{g}}{ε_{g}}

hookes law(ground)

Y = \frac{σ}{ε}

hookes law(rod)