User:Eas4200c.f08.blue.a/Lecture 40

This would seem to imply that:

However, this is not the case, this situation is not possible.

$$0=q^{(1)}+q^{(2)}+q^{(3)}+q^{(4)}$$

is actually correct, with:

$$q^{(e)}=n_{z}Q_{z}^{(e)}+n_{y}Q_{y}^{(e)}$$

Where the values of n are given by:

$$n_{z}=-(k_{y}V_{y}-k_{yz}V_{z})$$

and

$$n_{y}=-(k_{z}V_{z}-k_{yz}V_{y})$$



$$Q(\hat{z})=\int_{A(\hat{z)}}^{}{zdA}$$

and, by the Mean value theorem

$$Q_{y}=z_{c}\oint_{A}^{}{dA}=0$$

Read and Report - Bidirectional Bending
Under bidirectional bending, the displacements in the three orthogonal directions is given as

$$u=u_{0}(x)+z\psi _{y}(x)+y\psi _{z}(x)$$

$$v=v_{0}(x)$$

$$w=w_{0}(x)$$

Assume that $$\gamma _{xy}=\gamma _{xz}=0$$ and $$du_{0}/dx=0$$ and find

$$\psi _{z}=-\frac{dv_{0}}{dx}$$

$$\psi _{y}=-\frac{dw_{0}}{dx}$$

$$\varepsilon _{xx}=-y\frac{d^{2}v_{0}}{dx^{2}}-z\frac{d^{2}w_{0}}{dx^{2}}$$

The bending moments will be

$$M_{y}=-EI_{yz}\frac{d^{2}v_{0}}{dx^{2}}-EI_{y}\frac{d^{2}w_{0}}{dx^{2}}$$

Also

$$q=-\int_{A}^{}{\frac{d\sigma _{xx}}{dx}dA}$$

Where

$$\frac{d\sigma _{xx}}{dx}=\begin{bmatrix} x &y \end{bmatrix}I^{-1}\begin{Bmatrix} \frac{dM_{y}}{dx}\\\frac{dM_{z}}{dx}

\end{Bmatrix}$$

and

$$V_{y}=\frac{dM_{y}}{dx}$$

$$V_{z}=\frac{dM_{z}}{dx}$$

Also

$$Q_{y}=\int_{A}^{}{zdA}$$

$$Q_{z}=\int_{A}^{}{ydA}$$