A Lead Slab Sheared

solved ShearModulus

Q. A square lead slab $L = l_{face}$ of side 50 cm $l_{face} = 50$ cm and thickness 10 cm $w_{face} = 10$ cm is subject to a shearing force of 9.0 × 10⁴ N $F = 9.0 \times 10^{4}$ N on its narrow face. The lower edge is riveted to the floor. Taking the shear modulus of lead as 5.6 × 10⁹ N/m² $G = 5.6 \times 10^{9}$ N/m², how much will the upper edge be displaced, in mm?

Inputs

side

l_{face}

50 cm

thickness

w_{face}

10 cm

force

F

9e+04 N

modulus

G

5.6e+09 N/m²

Outputs

shift

Δ x

How it's solved

Solved with ShearModulus

Given

$l_{face} = 50$ cm
$w_{face} = 10$ cm
$F = 9.0 \times 10^{4}$ N
$G = 5.6 \times 10^{9}$ N/m²
$L = l_{face}$ from “square lead slab”

Find $Δ x$ .

Working

$l_{face} = 50$ cm $= 0.5$ m
$w_{face} = 10$ cm $= 0.1$ m
$G = 5.6 \times 10^{9}$ N/m² $= 5.6 \times 10^{9}$ Pa
a rectangular face(slab) $A = l_{face} \cdot w_{face}$
$A = 0.5 \cdot 0.1$
$A = 0.05$ m²
shearing stress over the face(slab) $σ_{s} = \frac{F}{A}$
$σ_{s} = \frac{90000}{0.05}$
$σ_{s} = 1.8 \times 10^{6}$ Pa
modulus of rigidity(slab) $G = \frac{σ_{s}}{θ_{s}}$
$5.6 \times 10^{9} = \frac{1.8 \times 10^{6}}{θ_{s}}$
$θ_{s} = 3.21428571 \times 10^{−4}$
square lead slab $L = l_{face}$
$L = 0.5$
$L = 0.5$ m
shearing strain is shift over height(slab) $θ_{s} = \frac{Δ x}{L}$
$3.21428571 \times 10^{−4} = \frac{Δ x}{0.5}$
$Δ x = 0.1607$ mm
$Δ x = 1.60714286 \times 10^{−4}$ m $= 0.160714286$ mm
$Δ x \approx 0.16$ mmto 2 significant figures

Conclusion

The force spread over the narrow face is the shearing stress; lead answers it with a slant of stress over the shear modulus, and the top edge moves that slant of the slab's height.

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_lead_slab_sheared do4  kind :numerical5  describes "A square lead slab of side %{side} and thickness %{thickness} is subject to a shearing force of " \6            "%{force} on its narrow face. The lower edge is riveted to the floor. Taking the shear modulus of lead " \7            "as %{modulus}, %{shift}, in mm?"8  given side: "50 cm"9  given thickness: "10 cm"10  given force: "9.0e4 N"11  given modulus: "5.6e9 N/m2"12  asks shift: "how much will the upper edge be displaced"13end

solution.rb

1require_relative "../questions/a_lead_slab_sheared"2require_relative "../science/mechanical_properties_of_solids/shear_modulus"34solve :a_lead_slab_sheared, using: ShearModulus do5  objects :slab67  take side: slab.face_length8  take thickness: slab.face_width9  take force: slab.shearing_force10  take modulus: slab.shear_modulus1112  deduce_from "square lead slab", slab.height => slab.face_length1314  shift = slab.shift15  answer shift:, in: "mm", figures: 21617  conclusion "The force spread over the narrow face is the shearing stress; lead answers it with a slant " \18             "of stress over the shear modulus, and the top edge moves that slant of the slab's height."19end

playground.rb

1require_relative "../solutions/a_lead_slab_sheared"23play_with :a_lead_slab_sheared do4  title "A Lead Slab 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 slab	$A$	m²	—
face length of slab	$l_{face}$	m	—
face width of slab	$w_{face}$	m	—
shearing stress of ground	$σ_{s,g}$	Pa	—
shearing force on ground	$F_{g}$	N	—
shearing stress of slab	$σ_{s}$	Pa	—
shearing force on slab	$F$	N	—
shearing strain of ground	$θ_{s,g}$	—	—
shift of ground	$Δ x_{g}$	m	—
height of ground	$L_{g}$	m	—
shearing strain of slab	$θ_{s}$	—	—
shift of slab	$Δ x$	m	—
height of slab	$L$	m	—
shear modulus of ground	$G_{g}$	Pa	—
shear modulus of slab	$G$	Pa	—
mass of slab	$m$	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 = l_{face} \cdot w_{face}

a rectangular face(slab)

σ_{s,g} = \frac{F_{g}}{A_{g}}

shearing stress over the face(ground)

σ_{s} = \frac{F}{A}

shearing stress over the face(slab)

θ_{s,g} = \frac{Δ x_{g}}{L_{g}}

shearing strain is shift over height(ground)

θ_{s} = \frac{Δ x}{L}

shearing strain is shift over height(slab)

G_{g} = \frac{σ_{s,g}}{θ_{s,g}}

modulus of rigidity(ground)

G = \frac{σ_{s}}{θ_{s}}

modulus of rigidity(slab)

L = l_{face}

square lead slab