这是一份southampton南安普敦大学FEEG1040W1-01作业代写的成功案例

where $V$ is total volume of the element. Substituting for the stress and strain
$$
U_{e}=\frac{E}{2} \int_{V} y^{2}\left(\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}\right)^{2} \mathrm{~d} V
$$
which can be written as
$$
U_{e}=\frac{E}{2} \int_{0}^{L}\left(\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}\right)^{2}\left(\int_{A} y^{2} \mathrm{~d} A\right) \mathrm{d} x
$$
A gain recognizing the area integral as the moment of inertia $I_{z}$ about the centroidal axis perpendicular to the plane of bending, we have
$$
U_{e}=\frac{E I_{z}}{2} \int_{0}^{L}\left(\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}\right)^{2} \mathrm{~d} x
$$

FEEG1040W1-01 COURSE NOTES ：

Using the specified data, The cross-sectional area is
$$
A=1(1)=1 \mathrm{in}^{2}
$$
And the area moment of inertia about the $z$ axis is
$$
I_{z}=b h^{3} / 12=1 / 12=0.083 \text { in. }{ }^{4}
$$
The characteristic axial stiffness is
$$
A E / L=1\left(10 \times 10^{6}\right) / 20=\left(5 \times 10^{5}\right) \mathrm{lb} / \mathrm{in} .
$$
and the characteristic bending stiffness is
$$
E I_{z} / L^{3}=10 \times 10^{6}(0.083) / 20^{3}=104.2 \mathrm{lb} / \mathrm{in} .
$$