An Aluminium Cube Sheared

Q. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. Taking g as 9.8 m/s², what is the vertical deflection of this face, in metres? expects 3.92 × 10⁻⁷ m

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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?

The playground

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}

The solution

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 Ruby

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

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

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

The laws

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