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How do the members of a structure carry load, and how do we find the support reactions of a beam?

Structures: tension and compression in members (ties and struts), the equilibrium of a beam, and using the principle of moments to find the support reactions.

An SQA National 5 Engineering Science answer on structures, covering tension and compression in members, the difference between a tie and a strut, the equilibrium of a loaded beam, and using the principle of moments and balanced forces to calculate the support reactions of a simply supported beam.

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  1. What this key area is asking
  2. Tension and compression
  3. Equilibrium of a structure
  4. Finding beam reactions
  5. Why structures matter
  6. Try this

What this key area is asking

The SQA wants you to know that structural members carry either tension or compression, identify a tie and a strut, and use the equilibrium conditions (balanced moments and balanced forces) to find the reaction forces at the supports of a loaded beam.

Tension and compression

A simple way to picture it: imagine cutting a member. If the structure would fly apart (the member was holding things together), it was in tension; if the structure would collapse inwards (the member was holding things apart), it was in compression.

Equilibrium of a structure

A structure that is not moving is in equilibrium. For a loaded beam this means two things at once:

  1. The moments balance. The total clockwise moment about any point equals the total anticlockwise moment (the principle of moments).
  2. The forces balance. The total upward force equals the total downward force, so there is no resultant force.

These two conditions are all you need to find the unknown reaction forces - the upward forces the supports provide to hold the beam up.

Finding beam reactions

The standard method for a simply supported beam (one support near each end) is:

  1. Take moments about one support. This removes that support's reaction from the equation (its distance is zero), leaving one unknown - the other reaction.
  2. Solve for that reaction.
  3. Balance the vertical forces (total up = total down) to find the remaining reaction.

Why structures matter

Structural analysis is how engineers make sure a bridge, crane or frame is strong enough and will not fail. Identifying ties and struts decides the shape and material of each member, and calculating reactions tells the engineer how heavily each support and foundation is loaded. This key area brings together moments and forces from earlier in the course.

Try this

Q1. What type of force does a strut carry? [1 mark]

  • Cue. Compression (it is being squashed).

Q2. State the two conditions for a loaded beam to be in equilibrium. [2 marks]

  • Cue. The moments balance (clockwise = anticlockwise) and the forces balance (total up = total down).

Q3. A 4.0 m4.0 \text{ m} beam on end supports carries a 400 N400 \text{ N} load at its centre. State the reaction at each support. [2 marks]

  • Cue. By symmetry each support carries half: 200 N200 \text{ N} each.

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 N5 style2 marksState the difference between a tie and a strut in a structure, and the type of force each carries.
Show worked answer →

Markers want each member matched to the force it carries.

A tie is a member in tension: it is being stretched (pulled) by forces acting to lengthen it.

A strut is a member in compression: it is being squashed (pushed) by forces acting to shorten it.

Markers reward tie with tension (stretching/pulling) and strut with compression (squashing/pushing). Swapping the two is a common error.

SQA N5 style5 marksA uniform beam of length 4.0 m rests on a support at each end (P on the left, Q on the right). A 200 N load hangs 1.0 m from the left support P. Calculate the reaction force at each support. Ignore the weight of the beam.
Show worked answer →

Use the two equilibrium conditions: balanced moments and balanced forces.

Take moments about P to find Q. The 200 N load is 1.0 m from P (anticlockwise); Q acts 4.0 m from P (clockwise).

Balancing moments about P: RQ×4.0=200×1.0R_Q \times 4.0 = 200 \times 1.0, so RQ=2004.0=50 NR_Q = \dfrac{200}{4.0} = 50 \text{ N}.

Now balance the vertical forces: total up equals total down, so RP+RQ=200R_P + R_Q = 200. Therefore RP=200−50=150 NR_P = 200 - 50 = 150 \text{ N}.

Markers reward taking moments about one support to find the other reaction, then using the balance of vertical forces to find the remaining reaction. The support nearer the load carries more of it.

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