Shear and Moment Diagrams are used to help you understand what the shear and moment curve look like for different generic types of beams and loading conditions. Basically it's a shortcut!! No longer will you have to use statics to find a reaction then back up a shear and moment curve. Also with the law of superposition it is very useful to have these so they can be combined and find even more complex systems (like a distributed load with a concentrated load, e.g. a concrete floor with a car driving over it).

The Breakdown
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**Normally Shear and Moment Diagrams will provide the following:**

- Your Reactions (R
_{1}& R_{2}) - The Maximum Shear force and Shear along the Beam (V
_{max}& V_{x}) - The Maximum Moment and the Moment along the beam (M
_{max}& M_{x}) - The Maximum Deflection and the Deflection along the beam (Δ
_{max}and Δ_{x})

This is the most basic of Shear Moment Diagrams.

**Notice the following:**

- The beam is simply supported (pinned on one end and a roller on the other). There are numerous types of end conditions such as: fixed-pinned, fixed-fixed, fixed-free, and continuous beams (multiple supports).
- This Shear and Moment Diagram is differentiated by others by: it's supports and it's loading types. Therefore there will also be a simply supported beam with a concentrated (point) load acting on it.
- The first diagram in the image shows the load acting on the beam. The second image is the shear diagram of the beam. The third image is the moment diagram of the beam.

V = \int F
-M = \int V
Notice that if you integrate the shear diagram you will get your moment curve. In my examples you actually get the negative of the moment curve but you could just as easily get the positive of the moment curve depending on how you've set up your axis'. It's up to your own preference (I prefer for my shear curves to start in the negative). If you need to brush up on your math, here is more information on derivatives. |

For each moment diagram you will be provided with a grouping of equations. Here are the applicable equations for the simply supported distributed load diagram I've provided in Figure 1 above:

R = V | = \frac{wl}{2} |

V_x | = w\left(\frac{l}{2}-x\right) |

M_{max} | = \frac{wl^2}{8} |

M_x | = \frac{wx}{2}\left(l-x\right) |

{Δ_{max}} (@ center) | = {{5wl^4}\over{384EI}} |

{Δ_x} | = {{wx}\over{24EI}}{\left(l^3-2lx^2+x^3\right)} |

*where:*

I'm sure there is more, but that's enough for now. Now feel free to go back to The Breakdown at the top of the page and see what this shortcut has to offer.

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Page last modified on March 16, 2011