User:Eas4200c.f08.blue.a/Lecture 26

Bidirectional Bending


$$M_{y}=\int_{aA}^{}{z\sigma _{xx}dA}$$

$$M_{z}=\int_{aA}^{}{y\sigma _{xx}dA}$$

Moment of Inertia Tensors:

$$I_{yy}=\int_{A}^{}{z^{2}dA}$$

$$I_{zz}=\int_{A}^{}{y^{2}dA}$$

$$I_{yz}=\int_{A}^{}{zydA}$$

Recall that, by Hooke's Law :

$$\sigma _{xx}=E\varepsilon _{xx}$$

With the above equations this becomes:

$$\sigma _{xx}=(\frac{I_{yy}M_{z}-I_{yz}M_{y}}{I_{yy}I_{zz}-(I_{yz})^{2}})y$$+$$(\frac{I_{zz}M_{y}-I_{yz}M_{z}}{I_{yy}I_{zz}-(I_{yz})^{2}})z$$

Where the denominator of both terms is the determinant of bottom right corner of the Moment of inertia matrix (shown below):

$$I=\begin{bmatrix} I_{11} &I_{22} &I_{13} \\ I_{21}&I_{22} &I_{23} \\ I_{31}& I_{32} & I_{33} \end{bmatrix}$$

In addition, the location of the neutral axis can be found by setting $$\sigma _{xx}=0$$.

$$\sigma _{xx}=(\frac{I_{yy}M_{z}-I_{yz}M_{y}}{I_{yy}I_{zz}-(I_{yz})^{2}})y$$+$$(\frac{I_{zz}M_{y}-I_{yz}M_{z}}{I_{yy}I_{zz}-(I_{yz})^{2}})z=M_{yy}+M_{zz}=0$$

Therefore, $$z=(-\frac{M_{y}}{M_{z}})y=\tan (\beta) y$$