Beam Elements

The beam element stiffness is

$$\mathbf{K} = E\begin{bmatrix} \frac{A}{l} & 0 & 0 & 0 & 0 & 0 & - \frac{A}{l} & 0 & 0 & 0 & 0 & 0 \\ & \frac{12I_{zz}}{l^{3}} & 0 & 0 & 0 & \frac{6I_{zz}}{l^{2}} & 0 & - \frac{12I_{zz}}{l^{3}} & 0 & 0 & 0 & \frac{6I_{zz}}{l^{2}} \\ & & \frac{12I_{yy}}{l^{3}} & 0 & - \frac{6I_{yy}}{l^{2}} & 0 & 0 & 0 & - \frac{12I_{yy}}{l^{3}} & 0 & - \frac{6I_{yy}}{l^{2}} & 0 \\ & & & \frac{GJ}{El} & 0 & 0 & 0 & 0 & 0 & - \frac{GJ}{El} & 0 & 0 \\ & & & & \frac{4I_{yy}}{l} & 0 & 0 & 0 & \frac{6I_{yy}}{l^{2}} & 0 & \frac{2I_{yy}}{l} & 0 \\ & & & & & \frac{4I_{zz}}{l} & 0 & - \frac{6I_{zz}}{l^{2}} & 0 & 0 & 0 & \frac{2_{zz}}{l} \\ & & & & & & \frac{A}{l} & 0 & 0 & 0 & 0 & 0 \\ & & & & & & & \frac{12I_{zz}}{l^{3}} & 0 & 0 & 0 & - \frac{6I_{zz}}{l^{2}} \\ & & & & & & & & \frac{12I_{yy}}{l^{3}} & 0 & \frac{6I_{yy}}{l^{2}} & 0 \\ & & & & & & & & & \frac{GJ}{El} & 0 & 0 \\ & & & & & & & & & & \frac{4I_{yy}}{l} & 0 \\ & & & & & & & & & & & \frac{4I_{zz}}{l} \\ \end{bmatrix}$$

These are modified for a shear beam as follows

$$\alpha = 12\frac{EI_{ii}}{l^{2}GA_{s}},\quad As = Ak_{jj}$$ $$\begin{aligned}\frac{2EI}{l} &\rightarrow \left( \frac{2 - \alpha}{1 + \alpha} \right)\frac{EI}{l}\\ \frac{6EI}{l^{2}} &\rightarrow \left( \frac{6}{1 + \alpha} \right)\frac{EI}{l^{2}}\\ \frac{4EI}{l} &\rightarrow \left( \frac{4 + \alpha}{1 + \alpha} \right)\frac{EI}{l}\\ \frac{12EI}{l^{3}} &\rightarrow \left( \frac{12}{1 + \alpha} \right)\frac{EI}{l^{3}}\end{aligned}$$

The consistent mass matrix is

$$\mathbf{M} = \frac{\rho Al}{420}\begin{bmatrix} 140 & 0 & 0 & 0 & 0 & 0 & 70 & 0 & 0 & 0 & 0 & 0 \\ & 156 & 0 & 0 & 0 & 22l & 0 & 54 & 0 & 0 & 0 & - 13l \\ & & 156 & 0 & - 22l & 0 & 0 & 0 & 54 & 0 & 13l & 0 \\ & & & 140\frac{J}{A} & 0 & 0 & 0 & 0 & 0 & 70\frac{J}{A} & 0 & 0 \\ & & & & 4l^{2} & 0 & 0 & 0 & - 13l & 0 & - 3l^{2} & 0 \\ & & & & & 4l^{2} & 0 & 13l & 0 & 0 & 0 & - 3l^{2} \\ & & & & & & 140 & 0 & 0 & 0 & 0 & 0 \\ & & & & & & & 156 & 0 & 0 & 0 & - 22l \\ & & & & & & & & 156 & 0 & 22l & 0 \\ & & & & & & & & & 140\frac{J}{A} & 0 & 0 \\ & & & & & & & & & & 4l^{2} & 0 \\ & & & & & & & & & & & 4l^{2} \\ \end{bmatrix}$$

The axial terms from the consistent mass matrix are

$$\mathbf{M} = \frac{\rho Al}{420}\begin{bmatrix} 140 & 70 \\ 70 & 140 \\ \end{bmatrix} = \frac{\rho Al}{6}\begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix}$$

The corresponding axial terms for a lumped mass matrix are

$$\mathbf{M} = \frac{\rho Al}{2}\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$$

These provide upper and lower bounds on frequencies, so a better approach is to average them, which GSA does.

$$\mathbf{M} = \frac{\rho Al}{12}\begin{bmatrix} 5 & 1 \\ 15 & 5 \\ \end{bmatrix} = \frac{\rho Al}{420}\begin{bmatrix} 175 & 35 \\ 35 & 175 \\ \end{bmatrix}$$

The geometric stiffness is

$$\mathbf{K_g} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ & \frac{6F_{x}}{5l} & 0 & \frac{M_{y1}}{l} & 0 & \frac{F_{x}}{10} & 0 & - \frac{6F_{x}}{5l} & 0 & \frac{M_{y2}}{l} & 0 & \frac{F_{x}}{10} \\ & & \frac{6F_{x}}{5l} & \frac{M_{z1}}{l} & - \frac{F_{x}}{10} & 0 & 0 & 0 & - \frac{6F_{x}}{5l} & \frac{M_{z2}}{l} & - \frac{F_{x}}{10} & 0 \\ & & & \frac{F_{x}I_{xx}}{Al} & - \frac{V_{y}l}{6} & - \frac{V_{z}l}{6} & 0 & - \frac{M_{y1}}{l} & - \frac{M_{z1}}{l} & - \frac{F_{x}I_{xx}}{Al} & \frac{V_{y}l}{6} & \frac{V_{z}l}{6} \\ & & & & \frac{2F_{x}l}{15} & 0 & 0 & 0 & \frac{F_{x}}{10} & \frac{V_{y}l}{6} & - \frac{F_{x}l}{30} & 0 \\ & & & & & \frac{2F_{x}l}{15} & 0 & - \frac{F_{x}}{10} & 0 & \frac{V_{z}l}{6} & 0 & - \frac{F_{x}l}{30} \\ & & & & & & 0 & 0 & 0 & 0 & 0 & 0 \\ & & & & & & & \frac{6F_{x}}{5l} & 0 & - \frac{M_{y2}}{l} & 0 & - \frac{F_{x}}{10} \\ & & & & & & & & \frac{6F_{x}}{5l} & -\frac{M_{z2}}{l} & -\frac{F_{x}}{10} & 0 \\ & & & & & & & & & & \frac{2F_x l}{15} & 0 \\ & & & & & & & & & & & \frac{2F_x l}{15} \\ \end{bmatrix}$$

where

$$I_{xx} = I_{zz} + I_{yy}$$

Non-symmetric Beam Sections

In a beam with a symmetric section the bending properties depend only on the $I_{zz}$ and $I_{yy}$ terms. If such a beam is loaded in the y or z axis directions the deflection is in the direction of the loading.

When the section is not symmetric and is loaded in the y or z direction there is a component of deflection orthogonal to the loading. This is because the bending properties depend on $I_{zz}$, $I_{yy}$ and $I_{yz}$.

By rotating the section to principal axes this cross term can be omitted and if the beam is loaded in the $u$ or $v$ (principal bending) axis the deflection is in the direction of the loading. In this case the stiffness matrix for the element is calculated using the principal second moments of area and is then rotated into the element local axis system.

For a beam with a non-symmetric section the user must consider if the beam is restrained (so that deflections are constrained to be in the direction of the loading) or if it will act in isolation (resulting in deflections orthogonal to the loading).

If the beam is to act as constrained the user should use the local option for the bending axes. In this case the $I_{zz}$ and $I_{yy}$ values are used and the $I_{yz}$ value is discarded.

If the beam is to act in isolation the user should use the principal option for the bending axes. In this case the stiffness matrix for the element is calculated using the principal second moments of area and is then rotated into the element local axis system.

The effect of shear is also a tensor quantity involving the inverse of the shear are factor $k$. So for symmetric section the $k_{yz}$ is infinite (no effect) and for non-symmetric sections there is a $k_{yz}$ term that should be considered. Note: the principal axes for shear are in general not aligned with the principal axes for bending. To simplify the calculation of the element stiffness the $k$ terms are rotated into the principal bending axes of the section and the effect of the $k_{yz}$ term is ignored.

Where the user has specified section modifiers these are specified in directions $1$ and $2$. If the bending axes are set to local then these correspond to $y$ and $z$ respectively. If the bending axes are set to principal, then $1$ and $2$ correspond to $u$ and $v$ respectively.

All catalogue and standard sections except angles are symmetric. Explicit sections are assumed to be defined such that the principal and local axes coincide so there is no $I_{yz}$. Geometric (perimeter and line segment) sections are assumed to be non-symmetric.