An Aluminium Cube Sheared

solved ShearModulus

Q. The edge of an aluminium cube $w_{face,c} = l_{face,c}$ , $L_{c} = l_{face,c}$ is 10 cm $l_{face,c} = 10$ cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg $m_{h} = 100$ kg is then attached to the opposite face $F_{c} = m_{h} \cdot g$ of the cube. The shear modulus of aluminium is 25 GPa $G_{c} = 25$ GPa. Taking g as 9.8 m/s² $g = 9.8$ m/s², what is the vertical deflection of this face, in metres?

Inputs

edge

l_{face,c}

10 cm

mass

m_{h}

100 kg

modulus

G_{c}

25 GPa

gravity

g

9.8 m/s²

Outputs

deflection

Δ x_{c}

How it's solved

Solved with ShearModulus

Given

$l_{face,c} = 10$ cm
$m_{h} = 100$ kg
$G_{c} = 25$ GPa
$g = 9.8$ m/s²
$w_{face,c} = l_{face,c}$ from “aluminium cube”
$L_{c} = l_{face,c}$ from “aluminium cube”
$F_{c} = m_{h} \cdot g$ from “attached to the opposite face”

Find $Δ x_{c}$ .

Working

$l_{face,c} = 10$ cm $= 0.1$ m
$G_{c} = 25$ GPa $= 2.5 \times 10^{10}$ Pa
attached to the opposite face $F_{c} = m_{h} \cdot g$
$F_{c} = 100 \cdot 9.8$
$F_{c} = 980$ N
aluminium cube (face width of cube) $w_{face,c} = l_{face,c}$
$w_{face,c} = 0.1$
$w_{face,c} = 0.1$ m
a rectangular face(cube) $A_{c} = l_{face,c} \cdot w_{face,c}$
$A_{c} = 0.1 \cdot 0.1$
$A_{c} = 0.01$ m²
shearing stress over the face(cube) $σ_{s,c} = \frac{F_{c}}{A_{c}}$
$σ_{s,c} = \frac{980}{0.01}$
$σ_{s,c} = 98000$ Pa
modulus of rigidity(cube) $G_{c} = \frac{σ_{s,c}}{θ_{s,c}}$
$2.5 \times 10^{10} = \frac{98000}{θ_{s,c}}$
$θ_{s,c} = 3.92 \times 10^{−6}$
aluminium cube (height of cube) $L_{c} = l_{face,c}$
$L_{c} = 0.1$
$L_{c} = 0.1$ m
shearing strain is shift over height(cube) $θ_{s,c} = \frac{Δ x_{c}}{L_{c}}$
$3.92 \times 10^{−6} = \frac{Δ x_{c}}{0.1}$
$Δ x_{c} = 3.92 \times 10^{−7}$ m
$Δ x_{c} \approx 3.92 \times 10^{−7}$ mto 3 significant figures

Conclusion

The hanging weight drags the free face down along itself, a shear; aluminium is so rigid that the face drops by far less than a thousandth of a millimetre.

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 :an_aluminium_cube_sheared do4  kind :numerical5  describes "The edge of an aluminium cube is %{edge} long. One face of the cube is firmly fixed to a vertical wall. " \6            "A mass of %{mass} is then attached to the opposite face of the cube. The shear modulus of aluminium is " \7            "%{modulus}. Taking g as %{gravity}, %{deflection}, in metres?"8  given edge: "10 cm"9  given mass: "100 kg"10  given modulus: "25 GPa"11  given gravity: "9.8 m/s2"12  asks deflection: "what is the vertical deflection of this face"13end

solution.rb

1require_relative "../questions/an_aluminium_cube_sheared"2require_relative "../science/mechanical_properties_of_solids/shear_modulus"34solve :an_aluminium_cube_sheared, using: ShearModulus do5  objects :cube, :hanging_mass67  take edge: cube.face_length8  take mass: hanging_mass.mass9  take modulus: cube.shear_modulus10  take gravity: g1112  deduce_from "aluminium cube", cube.face_width => cube.face_length, cube.height => cube.face_length13  deduce_from "attached to the opposite face", cube.shearing_force => hanging_mass.mass * g1415  deflection = cube.shift16  answer deflection:, in: "m", figures: 31718  conclusion "The hanging weight drags the free face down along itself, a shear; aluminium is so rigid " \19             "that the face drops by far less than a thousandth of a millimetre."20end

playground.rb

1require_relative "../solutions/an_aluminium_cube_sheared"23play_with :an_aluminium_cube_sheared do4  title "An Aluminium Cube Sheared"5end

The laws it drew on

quantity	written	units	within	from
mass	$m$	kg	0 … ∞	ShearModulus
gravity	$g$	m/s²	—	ShearModulus
shearing force	$F$	N	—	ShearModulus
face	$A$	m²	—	ShearModulus
face length	$l_{face}$	m	—	ShearModulus
face width	$w_{face}$	m	—	ShearModulus
height	$L$	m	—	ShearModulus
shift	$Δ x$	m	—	ShearModulus
shearing stress	$σ_{s}$	Pa	—	ShearModulus
shearing strain	$θ_{s}$	—	—	ShearModulus
shear modulus	$G$	Pa	—	ShearModulus
face of ground	$A_{g}$	m²	—
face length of ground	$l_{face,g}$	m	—
face width of ground	$w_{face,g}$	m	—
face of cube	$A_{c}$	m²	—
face length of cube	$l_{face,c}$	m	—
face width of cube	$w_{face,c}$	m	—
face of hanging mass	$A_{h}$	m²	—
face length of hanging mass	$l_{face,h}$	m	—
face width of hanging mass	$w_{face,h}$	m	—
shearing stress of ground	$σ_{s,g}$	Pa	—
shearing force on ground	$F_{g}$	N	—
shearing stress of cube	$σ_{s,c}$	Pa	—
shearing force on cube	$F_{c}$	N	—
shearing stress of hanging mass	$σ_{s,h}$	Pa	—
shearing force on hanging mass	$F_{h}$	N	—
shearing strain of ground	$θ_{s,g}$	—	—
shift of ground	$Δ x_{g}$	m	—
height of ground	$L_{g}$	m	—
shearing strain of cube	$θ_{s,c}$	—	—
shift of cube	$Δ x_{c}$	m	—
height of cube	$L_{c}$	m	—
shearing strain of hanging mass	$θ_{s,h}$	—	—
shift of hanging mass	$Δ x_{h}$	m	—
height of hanging mass	$L_{h}$	m	—
shear modulus of ground	$G_{g}$	Pa	—
shear modulus of cube	$G_{c}$	Pa	—
shear modulus of hanging mass	$G_{h}$	Pa	—
mass of cube	$m_{c}$	kg	0 … ∞
mass of hanging mass	$m_{h}$	kg	0 … ∞

A_{s} = l_{face,s} \cdot w_{face,s}

a rectangular face

σ_{s,s} = \frac{F_{s}}{A_{s}}

shearing stress over the face

θ_{s,s} = \frac{Δ x_{s}}{L_{s}}

shearing strain is shift over height

G_{s} = \frac{σ_{s,s}}{θ_{s,s}}

modulus of rigidity

A_{g} = l_{face,g} \cdot w_{face,g}

a rectangular face(ground)

A_{c} = l_{face,c} \cdot w_{face,c}

a rectangular face(cube)

A_{h} = l_{face,h} \cdot w_{face,h}

a rectangular face(hanging mass)

σ_{s,g} = \frac{F_{g}}{A_{g}}

shearing stress over the face(ground)

σ_{s,c} = \frac{F_{c}}{A_{c}}

shearing stress over the face(cube)

σ_{s,h} = \frac{F_{h}}{A_{h}}

shearing stress over the face(hanging mass)

θ_{s,g} = \frac{Δ x_{g}}{L_{g}}

shearing strain is shift over height(ground)

θ_{s,c} = \frac{Δ x_{c}}{L_{c}}

shearing strain is shift over height(cube)

θ_{s,h} = \frac{Δ x_{h}}{L_{h}}

shearing strain is shift over height(hanging mass)

G_{g} = \frac{σ_{s,g}}{θ_{s,g}}

modulus of rigidity(ground)

G_{c} = \frac{σ_{s,c}}{θ_{s,c}}

modulus of rigidity(cube)

G_{h} = \frac{σ_{s,h}}{θ_{s,h}}

modulus of rigidity(hanging mass)

w_{face,c} = l_{face,c}

aluminium cube (face width of cube)

L_{c} = l_{face,c}

aluminium cube (height of cube)

F_{c} = m_{h} \cdot g

attached to the opposite face