A Lead Slab Sheared

Q. A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force of 9.0 × 10⁴ 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, how much will the upper edge be displaced, in mm? expects 0.16 mm

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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?

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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

The solution

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 Ruby

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

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

1require_relative "../solutions/a_lead_slab_sheared"23play_with :a_lead_slab_sheared do4  title "A Lead Slab 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 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