10.3.1 How to compute torque and bending moments on beams

Figure 10.3 of the textbook is reproduced below:

\includegraphics[]{./Images/Fig103.png}
Figure 10.2: Forces and moments on the (x,y,z) user c.s. Panels numbered 1 through 5. Nodes numbered 1 through 6. Panel c.s. denoted by (x,s,r) on each panel. Reproduced from the textbook. Click the Back button on your browser to return to your previous page.

Torque $M_ x$, and bending moments $M_ y, M_ z$, are positive following the conventional right-hand rule.

In CADEC, three moments $M_ x, M_ y, M_ z,$ can be input, which are free vectors (i.e., invariants w.r.t. translations). Also, a number of forces can be input. Each force is a vector described by its three components $F_ x, F_ y, F_ z,$ and the point of application x, y, z, of each force.

The horizontal (y) and vertical (z) components of the forces produce shear $V_ y, V_ z,$ bending w.r.t. the c.g., and torque w.r.t the shear center, calculated in Equation 10.1.

The axial component of the forces add up to a total axial force applied at the mechanical neutral axis of bending (c.g.),1 plus bending moments calculated in (Equation 10.2 10.4).

In summary,

  \begin{equation} \label{Eq:1} F_ j = \sum _{i=1}^ N{F_ j^ i}\qquad \text {with}\qquad j=x,y,z \end{equation}   (10.1)

and

  $\displaystyle  M_ x  $ $\displaystyle = $ $\displaystyle  M_ x - \sum _ i{F_ y^ i (z_ f^ i-z_ c^ i)} + \sum _ i{F_ z^ i (y_ f^ i-y_ c^ i)} \label{Eq:2:x} $   (10.2)
  $\displaystyle M_ y  $ $\displaystyle = $ $\displaystyle  M_ y + \sum _ i{F_ x^ i (z_ f^ i-z_ g^ i)} - \sum _ i{F_ z^ i x_ f^ i} \label{Eq:2:y} $   (10.3)
  $\displaystyle M_ z  $ $\displaystyle = $ $\displaystyle  M_ z - \sum _ i{F_ x^ i (y_ f^ i-y_ g^ i)} + \sum _ i{F_ y^ i x_ f^ i} \label{Eq:2:z}  $   (10.4)

Footnotes

  1. For inhomogeneous materials, the mechanical neutral axis of bending is not necessarily the center of gravity but it is abbreviated c.g. to follow tradition.