CBT by variational approach

Consider a straight beam under lateral loading q(x) and axial force P.

The classical beam theory (CBT) makes kinematic assumption as

\begin{align*}
u&=u_0(x)-z\frac{dw_0}{dx}\\
v&=0\\
w&=w_0(x)
\end{align*}

The strain therefore is

\begin{align*}
&\epsilon_x=\frac{du_0}{dx}-z\frac{d^2w_0}{dx^2},\epsilon_y=\epsilon_z=0\\
&\gamma_{xy}=\gamma_{xz}=\gamma_{yz}=0
\end{align*}

Let’s follow virtual work principle

\delta U=\delta W

where internal virtual work is

\delta U=\int_V\sigma_x\delta\epsilon_xdV

Let’s expand it

\begin{align*}
\delta U&=\int_0^L\int_0^b\int_{-h/2}^{h/2}E\left(\frac{du_0}{dx}-z\frac{d^2w_0}{dx^2}\right)\left(\frac{d\delta u_0}{dx}-z\frac{d^2\delta w_0}{dx^2}\right)dzdydx\\
&=Eb\int_0^L\int_{-h/2}^{h/2}\left[\left(\frac{d^2u_0}{dx^2}\right)\delta u_0+\left(z^2\frac{d^4w_0}{dx^4}\right)\delta w_0\right]dzdx\\
&=\int_0^L\left[EA\left(\frac{d^2u_0}{dx^2}\right)\delta u_0+EJ\left(\frac{d^4w_0}{dx^4}\right)\delta w_0\right]dx
\end{align*}

Here we use

\int_{-h/2}^{h/2}\{1,z,z^2\}dz=\left\{h,0,\frac{h^3}{12}\right\}

and

A=\int_AdA=bh,J=\int_0^b\int_{-h/2}^{h/2}z^2dzdy=\frac{bh^3}{12}

and external virtual work is

\delta W=\int_V q\delta wdV+\left[\int_AT_x\delta udA\right]_0^L

Or we can write it into (make uniform cross section assumption)

\delta W=A\int_0^Lq\delta wdx+P\delta u_0|_0^L

where P=\int_AT_xdA. Besides
Now we get governing equation for both deflection and axial displacement

\frac{d^2u_0}{dx^2}=0,\ EJ\frac{d^4w_0}{dx^4}=qA