What is not checking with a second moment equation?

Taking moments about the other support is a free check that catches arithmetic slips before they propagate.

QCAA QCAA 2022-style practice: A simply supported beam AB spans 6.0\ m, pinned at A and on a roller at B. A point load of 15\ kN acts 1.5\ m from A and a point load of 25\ kN acts 4.0\ m from A. Determine the magnitude of the reaction forces at A and B.

A 6 mark determine question rewards a clear moment equation, a force balance and a check. Take moments about A so R_A drops out (anticlockwise positive, R_B acts up at 6.0\ m): $sum M_A = R_B(6.0) - 15(1.5) - 25(4.0) = 0$ $6.0\,R_B = 22.5 + 100 = 122.5, R_B = 20.4\ kN.$ Vertical balance sum F_y = 0 gives R_A = 15 + 25 - 20.4 = 19.6\ kN. Markers reward taking moments about a support to remove an unknown, correct arithmetic, and a confirming moment check about B. The larger reaction at A is expected because the heavier load sits nearer mid-span but the lighter load adds to A's share.

§-Syllabus dot point

QLDEngineeringSyllabus dot point

How do engineers find the reaction forces at the supports of a loaded structure?

Calculate the moment of a force about a point and apply moment equilibrium together with force equilibrium to determine the support reactions of a simply supported structure

A QCE Engineering Unit 3 answer on moments and reactions. Covers the moment of a force, the principle of moments, pin and roller supports, and using moment equilibrium to solve for the reactions of a simply supported beam, with worked arithmetic.

Generated by Claude Opus 4.86 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

QCAA wants you to compute the turning effect of a force, the moment, and use it together with force balance to find the unknown reaction forces at the supports of a structure. Reactions are the forces the ground or supports push back with to hold the structure still. You cannot analyse a truss member or a beam section until you know them, so this is the gateway calculation of Unit 3.

The answer

The moment of a force

A force applied at a distance from a pivot tends to rotate the body. That turning effect is the moment:

M = F \times d

Here $F$ is the force and $d$ is the perpendicular distance from the pivot to the line of action of the force. The perpendicular distance matters: a force whose line passes through the pivot has zero moment because $d = 0$ . Moments are measured in newton metres (N m) and carry a sense, conventionally taken as positive anticlockwise or positive clockwise as long as you stay consistent.

The principle of moments

For a body in rotational equilibrium, the total clockwise moment about any point equals the total anticlockwise moment, which is the same as saying:

\sum M = 0

This is the third equilibrium condition, alongside $\sum F_x = 0$ and $\sum F_y = 0$ . The power of the moment equation is that you can take moments about any point you like. Choosing a point where an unknown force acts makes that force drop out (its $d$ is zero), leaving fewer unknowns.

Supports and the reactions they provide

Real structures rest on supports that supply reaction forces:

A roller support can only push perpendicular to its surface, so it gives one reaction (usually vertical).
A pin (hinge) support can push in any direction, so it gives two components, vertical and horizontal.
A fixed (built-in) support resists translation and rotation, giving two force components and a moment.

A simply supported beam has a pin at one end and a roller at the other. That is three reaction components for three equilibrium equations, so the structure is statically determinate and solvable by hand.

The strategy for finding reactions

Draw the free-body diagram with all loads and the unknown reactions.
Take moments about one support. Its own reaction has zero moment there, so the equation contains only the other reaction and the loads. Solve it.
Use $\sum F_y = 0$ to find the remaining vertical reaction.
Use $\sum F_x = 0$ for any horizontal reaction.

Reactions of a simply supported beam

Find the support reactions under two point loads

A beam $AB$ is $8.0\,\text{m}$ long, pinned at $A$ and on a roller at $B$ . A $12\,\text{kN}$ point load acts $2.0\,\text{m}$ from $A$ , and a $20\,\text{kN}$ point load acts $5.0\,\text{m}$ from $A$ . Both loads are vertical and downward. Find the reactions $R_A$ and $R_B$ .

Step 1. moments about A (take anticlockwise positive; $R_B$ acts up at $8.0\,\text{m}$ , loads act down):

\sum M_A = R_B(8.0) - 12(2.0) - 20(5.0) = 0

8.0\,R_B = 24 + 100 = 124

R_B = \frac{124}{8.0} = 15.5\,\text{kN}

Step 2. vertical force balance ( $\sum F_y = 0$ ):

R_A + R_B - 12 - 20 = 0

R_A = 32 - 15.5 = 16.5\,\text{kN}

Check: take moments about $B$ to confirm. $R_A(8.0) = 16.5 \times 8.0 = 132$ ; the loads give $12(6.0) + 20(3.0) = 72 + 60 = 132$ . They balance, so $R_A = 16.5\,\text{kN}$ and $R_B = 15.5\,\text{kN}$ are correct.

Why this matters for civil structures

Reactions are the starting point for everything that follows. The method of joints needs them before any truss member can be solved, and a shear-force and bending-moment diagram is built directly from the reactions and loads. A wrong reaction silently corrupts every later number, so the moment-about-a-support technique and its built-in check are core skills for the Unit 3 data test and external exam.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20226 marksA simply supported beam AB spans

6.0\ \text{m}

, pinned at A and on a roller at B. A point load of

15\ \text{kN}

acts

1.5\ \text{m}

from A and a point load of

25\ \text{kN}

acts

4.0\ \text{m}

from A. Determine the magnitude of the reaction forces at A and B.

Show worked answer →

A 6 mark determine question rewards a clear moment equation, a force balance and a check.

Take moments about A so $R_A$ drops out (anticlockwise positive, $R_B$ acts up at $6.0\ \text{m}$ ):

\sum M_A = R_B(6.0) - 15(1.5) - 25(4.0) = 0

6.0\,R_B = 22.5 + 100 = 122.5,\qquad R_B = 20.4\ \text{kN}.

Vertical balance $\sum F_y = 0$ gives $R_A = 15 + 25 - 20.4 = 19.6\ \text{kN}$ .

Markers reward taking moments about a support to remove an unknown, correct arithmetic, and a confirming moment check about B. The larger reaction at A is expected because the heavier load sits nearer mid-span but the lighter load adds to A's share.

QCAA 20233 marksAnalyse why an engineer takes moments about a support when solving for the reactions of a simply supported beam, rather than about an arbitrary interior point.

Show worked answer →

A 3 mark analyse question wants the reasoning, not just a rule.

Taking moments about a support sets the perpendicular distance $d$ of that support's reaction to zero, so its moment $M = F \times d = 0$ and it drops out of $\sum M = 0$ . The equation then contains a single unknown reaction, which is solved directly without simultaneous equations. An arbitrary interior point keeps both unknown reactions in the moment equation, forcing a second equation to be solved together with it.

Markers reward linking $M = F \times d$ to the zero moment arm and explaining the resulting reduction to one unknown.