Solveeit Logo
Login

Question

Question: The frame of brass plate of an outer door design has area \(1.60\,{m^2}\) and thickness \(1\,cm\) . ...

The frame of brass plate of an outer door design has area 1.60 m21.60\,{m^2} and thickness 1 cm1\,cm . The brass plate experiences shear force due to earthquake. How large a parallel force must be exerted on each of its edges, if the lateral displacement is 0.32 mm0.32\,mm?
(ηbrass=3.5×1010  Nm−2)\left( {{\eta _{brass}} = 3.5 \times {{10}^{10\,}}\,N{m^{ - 2}}} \right)

Explanation

Solution

Use the formula of the efficiency of the metal and rearrange it to find the shear force. Substitute the known values in the obtained rearranged shear force formula and simplification of it, provides the value of the shear force that affects the door due to earthquake.

Formula used:
The efficiency of the brass plate is given by
η=FtAx\eta = \dfrac{{Ft}}{{Ax}}
Where η\eta is the efficiency of the brass plate, FF is the shear force exerted, tt is the thickness of the plate, AA is the area of the brass plate and xx is the lateral displacement.

Complete step by step answer:
Given: Area of the brass plate, A=1.6 m2A = 1.6\,{m^2}
Thickness of the brass plate, t=1 cm=1×10−2  mt = 1\,cm = 1 \times {10^{ - 2\,}}\,m
The efficiency of the brass, η=3.5×1010  Nm−2\eta = 3.5 \times {10^{10\,}}\,N{m^{ - 2}}
The lateral displacement, x=0.32 mm=0.32×10−3 mx = 0.32\,mm = 0.32 \times {10^{ - 3}}\,m
Using the formula of the efficiency of the brass plate,
η=FtAx\eta = \dfrac{{Ft}}{{Ax}}
Rearranging the above formula to find the shear force.
F=ηAxtF = \dfrac{{\eta Ax}}{t}
Substitute the known values in the above formula,
F=3.5×1010×1.6×0.32×10−31×10−2F = \dfrac{{3.5 \times {{10}^{10}} \times 1.6 \times 0.32 \times {{10}^{ - 3}}}}{{1 \times {{10}^{ - 2}}}}
By simplifying the above step, we get
F=3.5×1010×1.6×0.32×10−1F = 3.5 \times {10^{10}} \times 1.6 \times 0.32 \times {10^{ - 1}}
By multiplying the values in the right side of the equation, we get
F=1.792×109  NF = 1.792 \times {10^{9\,}}\,N

Hence, the value of the shear force that affects the brass door due to the earthquake is F=1.792×109  NF = 1.792 \times {10^{9\,}}\,N.

Note: The brass is formed as the alloy of the copper and the zinc. It is also substituted in place of the use of the copper in the jewelry, of the costumes, fashion and other imitation. This is because it is named for its durability and hardness.