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Structural Analysis Fundamentals / Forces, moments and equilibrium

Finding the reactions

Using the three equations in the right order, and checking the answer.

Lesson review: expert verification remains open

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.

Checks needed for project use

  • An independent structural reviewer must check remaining derivations, diagrams, sign conventions and the interactive models across their full input ranges.
  • Model applicability, diagrams and implementation across all input cases remain subject to independent engineering review.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Why this matters

Reactions are the gateway to everything else. Get them wrong and every shear force, bending moment and deflection that follows is wrong too — usually without any obvious warning.

By the end of this lesson you should be able to

  • Choose a moment point that removes an unknown
  • Solve for both reactions on a simply supported beam
  • Check the result independently
  • Recognise and explain uplift

For these simply supported beam examples, draw the free body and replace distributed loads with resultants. Take moments about one support to find the other reaction, then use vertical equilibrium. Check the arithmetic by taking moments about the other support; this is a consistency check using the same equilibrium conditions.

Try it

Beam laboratory

Change the beam, then reveal one diagram at a time. Predict each shape before you reveal it.

Beam type

m

kN (0 to remove)

m from left

kN/m (0 to remove)

Reveal

Simply supported beam of span 6 metres20 kN6 m

Worked example

Reactions for a beam with a UDL and a point load

Reactions for a beam with a UDL and a point load5 kN/m40 kN3.0 m8.0 m

Given

  • Simply supported span of 8.0 m, pin at A (left), roller at B (right)
  • Uniformly distributed load of 5 kN/m over the whole span
  • Point load of 40 kN at 3.0 m from A

Find

The vertical reactions at A and B.

Assumptions

  • Self-weight is included in the UDL
  • The beam is straight and the supports are at its ends

    Practice

    A simply supported beam spans 6.0 m. A single point load of 24 kN acts 2.0 m from the left support. What is the reaction at the left support?

    Practice

    A simply supported beam spans 10.0 m. It carries 20 kN at 3.0 m and 30 kN at 7.0 m from the left support. What is the left-hand reaction?

    Practice

    A cantilever 3.0 m long carries a UDL of 10 kN/m over its whole length plus a 15 kN point load at the free end. What is the vertical reaction at the fixed support?

    Practice

    A beam 6.0 m long has a pin at x = 0 and a roller at x = 4.0 m, with a 20 kN load at the free end x = 6.0 m. What is the reaction at the pin? (A negative answer means uplift.)

    Summary

    • Moments about one support removes that reaction from the equation
    • Then vertical equilibrium gives the other
    • Always check with an equation you have not used
    • A negative reaction means uplift and needs a physical holding-down detail