A Steel Rod Stretched

Q. A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 × 10³ N force stretches it along its length. Young's modulus of structural steel is 2.0 × 10¹¹ N/m². What is the elongation of the rod, in mm? expects 1.59 mm

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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?

The playground

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

The solution

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 Ruby

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

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

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

The laws

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)