SQA SQA AH style-style practice: A horizontal beam 6.0\ m long rests on supports at each end, A and B. A 1200\ N load hangs 2.0\ m from A. Calculate the reaction forces at A and B.

Take moments about A to find the reaction at B, then use vertical equilibrium for A. Moments about A (clockwise = anticlockwise): R_B × 6.0 = 1200 × 2.0. R_B = 24006.0 = 400\ N. Vertical equilibrium: R_A + R_B = 1200, so R_A = 1200 - 400 = 800\ N. Markers reward taking moments about a support to remove one unknown, R_B = 400\ N, and R_A = 800\ N from vertical equilibrium. The larger reaction is nearer the load, as expected.

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How do we find the support reactions of a loaded structure and the internal forces in its members?

Static equilibrium of structures with the conditions for equilibrium, finding support reactions by taking moments, and resolving the internal forces in pin-jointed frameworks as ties and struts.

An SQA Advanced Higher Engineering Science answer on structural equilibrium, covering the conditions for static equilibrium, finding support reactions by taking moments, and resolving the internal forces in pin-jointed frameworks into ties and struts.

Generated by Claude Opus 4.816 min answerUpdated 2026-06-16

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

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Quick answer

A structure is in static equilibrium when it is neither accelerating nor rotating, which requires two conditions: the resultant force is zero (so $\sum F_{up} = \sum F_{down}$ and $\sum F_{left} = \sum F_{right}$ ) and the resultant moment is zero (so the sum of clockwise moments equals the sum of anticlockwise moments about any point). A moment is $M = Fd$ , the force times the perpendicular distance. Taking moments about a support removes its unknown reaction, letting you solve for the other; vertical equilibrium then gives the first. In a pin-jointed framework every member carries an axial force only: a tie is in tension (it pulls its joints together) and a strut is in compression (it pushes them apart). Resolving forces into horizontal and vertical components at a joint finds these member forces.

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What this key area is asking
The conditions for equilibrium
Finding support reactions by taking moments
Internal forces: ties and struts
Examples in context
Try this

What this key area is asking

The SQA wants you to apply the conditions for static equilibrium to a loaded structure, to find support reactions by taking moments, and to resolve the internal forces in a pin-jointed framework into members that carry tension (ties) and members that carry compression (struts). This is how an engineer checks that a structure can carry its loads safely.

The conditions for equilibrium

These two conditions are all that structural statics needs. The first keeps the structure from translating (sliding or lifting), the second from rotating. Because the moment condition holds about any point, you choose the point cleverly, usually a support, so that an unknown reaction acts through it and drops out of the moment equation.

Finding support reactions by taking moments

To find the reactions of a simply supported beam, take moments about one support: this eliminates that support's reaction (its distance is zero) and leaves a single equation for the other reaction. Then apply vertical equilibrium (total up = total down) to find the first reaction. The reaction is always larger at the support nearer a load, which is a useful check.

Internal forces: ties and struts

To find the force in each member, isolate a joint where only two unknown members meet and apply equilibrium there: the horizontal components must balance and the vertical components must balance ( $\sum F_x = 0$ , $\sum F_y = 0$ ). Resolving the inclined member forces into components and solving gives each member force; the direction of the force then tells you whether the member is a tie (pulling) or a strut (pushing).

Reactions then a member force

A framework joint carries a $600\ \text{N}$ downward load, held by a member at $40^\circ$ to the horizontal and a horizontal member. Find the force in the inclined member and classify both members.

step 1 Resolve vertically at the joint

Only the inclined member has a vertical component, so it must support the full load: $F_{inc}\sin 40^\circ = 600$ .

step 2 Solve for the inclined member force

$F_{inc} = \dfrac{600}{\sin 40^\circ} = \dfrac{600}{0.643} = 933\ \text{N}$ .

step 3 Resolve horizontally and classify

The inclined member pulls up and outward on the joint, so it is in tension, a tie. Its horizontal component $F_{inc}\cos 40^\circ = 933 \times 0.766 = 715\ \text{N}$ must be balanced by the horizontal member pushing back, so that member is in compression, a strut.

Examples in context

A roof truss is a pin-jointed framework whose sloping rafters act as struts in compression and whose tie beam acts as a tie in tension, carrying the roof load to the walls. A bridge truss combines ties and struts so each member is sized for its axial force. A crane jib balances the load moment against a counterweight moment, a direct moments calculation. A shelf bracket has a diagonal strut in compression supporting a horizontal arm. Identifying reactions and which members pull or push is what lets each part be made strong enough.

Try this

Q1. State the two conditions for static equilibrium. [2 marks]

Cue. The resultant force is zero (forces balance) and the resultant moment about any point is zero (moments balance).

Q2. A force of $200\ \text{N}$ acts $0.30\ \text{m}$ from a pivot, perpendicular to the line to the pivot. Find its moment. [1 mark]

Cue. $M = Fd = 200 \times 0.30 = 60\ \text{N m}$ .

Q3. State whether a member in compression is a tie or a strut. [1 mark]

Cue. A strut (it is in compression, pushing its joints apart).

Exam-style practice questions

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

SQA AH style5 marksA horizontal beam

6.0\ \text{m}

long rests on supports at each end, A and B. A

1200\ \text{N}

load hangs

2.0\ \text{m}

from A. Calculate the reaction forces at A and B.

Show worked answer →

Take moments about A to find the reaction at B, then use vertical equilibrium for A.

Moments about A (clockwise = anticlockwise): $R_B \times 6.0 = 1200 \times 2.0$ .

$R_B = \dfrac{2400}{6.0} = 400\ \text{N}$ .

Vertical equilibrium: $R_A + R_B = 1200$ , so $R_A = 1200 - 400 = 800\ \text{N}$ .

Markers reward taking moments about a support to remove one unknown, $R_B = 400\ \text{N}$ , and $R_A = 800\ \text{N}$ from vertical equilibrium. The larger reaction is nearer the load, as expected.

SQA AH style5 marksAt a loaded joint of a pin-jointed framework, a

500\ \text{N}

vertical load is supported by one horizontal member and one member at

30^\circ

above the horizontal. Calculate the force in the inclined member and state whether it is a tie or a strut.

Show worked answer →

Resolve vertically at the joint: the vertical component of the inclined member must balance the load.

Relationship: $F \sin\theta =$ vertical load.

Substitution: $F \sin 30^\circ = 500$ , so $F = \dfrac{500}{0.5} = 1000\ \text{N}$ .

The inclined member must pull up on the joint to hold the load, so it is in tension: it is a tie.

Markers reward resolving vertically with $F\sin 30^\circ = 500$ , the force of $1000\ \text{N}$ , and identifying the member as a tie because it carries tension. (The horizontal member then balances the horizontal component and is a strut in compression.)

Related dot points

Sources & how we know this

Advanced Higher Engineering Science Course Specification (C823 77) — SQA (2019)
Advanced Higher Engineering Science Data Booklet — SQA (2019)