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Structural Analysis Fundamentals / Forces, moments and equilibrium

Supports and free-body diagrams

What each support allows, and how to draw the picture you actually calculate from.

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This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

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Why this matters

A complete free-body diagram helps prevent missing loads and incorrect support reactions. Draw it before writing the equilibrium equations.

By the end of this lesson you should be able to

  • Match each support type to the movements it prevents
  • Draw a complete free-body diagram
  • Replace distributed loads with resultants correctly

The rule for supports is simple, and it is worth saying out loud every time: if a support stops a movement, it must push back in that direction. If it allows the movement, there is no force.

A roller stops movement perpendicular to its surface only. It gives one reaction, and the beam is free to slide along and to rotate. A pin stops movement in both directions but allows rotation, so it gives a horizontal and a vertical reaction but no moment. A fixed support stops all three movements, so it gives two forces and a moment.

An eight metre simply supported beam carrying a uniform load of five kilonewtons per metre and a forty kilonewton point load three metres from the left, with reactions shown at both supports.5 kN/m40 kN45.00 kN35.00 kN3.0 m8.0 m
A free-body diagram: the beam on its own, with every applied load and both reactions drawn on.

A free-body diagram is the beam drawn on its own, cut away from everything it touches, with every force that acts on it put back as an arrow. Supports become reaction arrows. Nothing may be left out, and nothing invented.

Distributed loads need one extra step. A uniformly distributed load of w kN/m acting over a length L has a total of wL, and for the purposes of equilibrium it behaves as a single force of wL acting at the middle of that length. Use the resultant for finding reactions; go back to the distributed form when you work out internal forces.

Predict first

A UDL of 6 kN/m runs over the middle 4 m of a beam, from x = 2 m to x = 6 m. Where does its resultant act?

Practice

A uniformly distributed load of 8 kN/m acts over a 5.0 m length of beam. What is the total load it represents?

Practice

A partial UDL of 12 kN/m runs from x = 2.0 m to x = 6.0 m along a beam. Where does its resultant act, measured from the left-hand end of the beam?

Practice

A beam has a fixed support at one end and nothing else. How many reaction components must appear on its free-body diagram?

Summary

  • A support that stops a movement provides a reaction in that direction
  • Free-body diagram: the member alone, with every applied load and every reaction shown
  • A UDL becomes wL at the centre of the loaded length for equilibrium purposes