National 5 · Mechanisms and Structures › Structures and Forces

🏗️ Structures and Forces — Course Notes

Forces, levers, moments, beams, reactions and vectors

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Section 1: Types of Force

Forces impact structures in a variety of different ways depending on how they are applied. Forces can move a structure slightly or cause damage by changing its shape. Sometimes the changes are almost impossible to see — for example, a bridge will sag slightly when a vehicle drives over it. Forces can also make an object move or change direction.

Forces are measured in Newtons (N).

There are five types of force you need to know:

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Compression
Forces acting to squash or shorten a structure — pushing inwards from both ends
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Tension
Forces acting to stretch or lengthen a structure — pulling outwards from both ends
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Bending
Forces acting across the length of a structure causing it to bend or deflect
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Shear
Forces acting in opposite directions across a section — like cutting or tearing
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Torsion
Forces that twist or turn a structure about its own axis

Compression

The supporting columns of the Queensferry Crossing experience a huge downward force from the cables. The ground exerts an upward reaction force on the bases of the columns. These two opposing forces try to squash or shorten the column — the column is said to be in compression. A weightlifter's body experiences compression in the same way: the weights push down, the ground pushes up.

Tension

Tension is the opposite of compression. A structure under tension is being pulled apart. In a tug of war, both teams pull the rope in opposite directions — the rope is in tension. Wire ropes, cables and chains are typically in tension.

Bending

Structures that carry loads across their length are subject to bending forces. A weightlifting bar bends due to the downward forces of the weights at each end. Beams in buildings and bridges experience bending.

Shear

Shear forces can be described as tearing or cutting forces. A pair of scissors cuts a ribbon using shear — the two blades push in opposite directions across the material. A lawn mower blade shears the grass.

Torsion

Torsion (also called torque) forces twist or turn a structure. A screwdriver being twisted to drive a screw applies torsion to the screw. A spanner turning a bolt applies torsion to the bolt.

✏️ Task 1 — Types of force
For each of the five types of force, write down two real-life examples where that force is present.
Compression ✓ Saved
Tension ✓ Saved
Bending ✓ Saved
Shear ✓ Saved
Torsion ✓ Saved

Section 2: Levers

A lever is a simple machine that can amplify force. By using a pivot point (called a fulcrum) and a rigid bar, a small effort force can move a large load. The further from the fulcrum the effort is applied, the easier it is to move the load.

A lever system changes an input force and input motion into an output force and output motion:

Effort = the input force applied to the lever
Load = the output force (what you are trying to move)
Fulcrum = the pivot point the lever rotates about

When a counterbalance weight is attached to one side of a lever to assist the user, it is known as a counterbalance. Cranes and diggers use counterbalances to prevent them toppling over.


Section 3: Moments

When a force acts at a distance from a pivot, it creates a turning effect called a moment. Moments are measured in Newton-metres (Nm).

Moment (Nm) = Force (N) × Distance (m)
Distance is measured at right angles from the line of action of the force to the pivot

Equilibrium and the Principle of Moments

When a system is in a steady, balanced state it is said to be in equilibrium. For a lever to be in equilibrium, the clockwise moments must equal the anticlockwise moments.

The Principle of Moments states:

∑CWM = ∑ACWM
Sum of Clockwise Moments = Sum of Anticlockwise Moments
F1 × d1 = F2 × d2
Worked Example 1 — Is the lever in equilibrium?

A lever has a load of 400 N at 1 m from the fulcrum on the right, and an effort of 200 N at 2 m from the fulcrum on the left. Is the lever in equilibrium?

1
Calculate the clockwise moment: Load × distance = 400 N × 1 m = 400 Nm
2
Calculate the anticlockwise moment: Effort × distance = 200 N × 2 m = 400 Nm
3
Compare: 400 Nm = 400 Nm  ⇒  ∑CWM = ∑ACWM  ⇒  The lever is in equilibrium.
Worked Example 2 — Car footbrake (Class 2 lever)

A car footbrake lever amplifies the driver's foot force. The driver applies 5000 N at 600 mm from the fulcrum. The braking system connects 100 mm from the fulcrum. Find the braking force.

1
Convert distances to metres: 600 mm = 0.6 m, 100 mm = 0.1 m
2
Apply the Principle of Moments: F1 × d1 = F2 × d2
3
5000 N × 0.6 m = braking force × 0.1 m
4
Braking force = (5000 × 0.6) ÷ 0.1 = 30 000 N = 30 kN
✏️ Task 2 — Moments calculations

Use the Principle of Moments to find the missing force or distance in each problem. Show all working.

1. A lever has an effort of 300 N applied 1.5 m from the fulcrum. A load is placed 0.5 m from the fulcrum on the other side. Calculate the load.
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2. A load of 600 N is placed 2 m from a fulcrum. Where must an effort of 400 N be applied to balance the lever?
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3. A lever has two loads: 200 N at 3 m and 400 N at 1.5 m on the right. An effort is applied 2 m to the left of the fulcrum. Calculate the effort needed for equilibrium.
Working ✓ Saved

Section 4: Free-Body Diagrams

A free-body diagram is a simplified diagram that shows all the forces acting on a structure or body. It isolates the structure from its surroundings and replaces real components with arrows representing the forces they exert.

To draw a free-body diagram:

  1. Draw a simple outline of the structure (or just a line or box)
  2. Replace all connections and supports with arrows showing the force they exert
  3. Label each force with a letter or value
  4. Show downward forces (loads, weight) acting through their point of application
  5. Show upward reaction forces at support points
Free-body diagrams make it easier to see and calculate all the forces acting on a structure. Every force must be accounted for — including the weight of the structure itself if stated.
✏️ Task 3 — Free-body diagrams
1. Draw a free-body diagram of a car resting on flat ground. Show the downward forces (weight acting through the centre of mass) and the reaction forces at each axle. Use suitable letter labels.
Sketch and labels ✓ Saved
2. A bridge is supported at two ends. A bus sits on the bridge. Draw and label the free-body diagram showing all forces including the bridge's own weight.
Sketch and labels ✓ Saved

Section 5: Beams and Reactions

A beam is a structural member that carries loads across its length. For a horizontal beam to be in equilibrium, two conditions must be satisfied simultaneously:

∑ upward forces = ∑ downward forces
∑ clockwise moments = ∑ anticlockwise moments

The upward forces at the support points are called reactions (R1 and R2). To find both reactions, we use the Principle of Moments and the force balance equation together.

Method for finding beam reactions

  1. Take moments about one reaction (R1) — this eliminates R1 from the equation, allowing you to solve for R2
  2. Use the force balance (∑ up = ∑ down) to find R1
  3. Check: R1 + R2 should equal the sum of all downward forces
Worked Example — Simply supported beam with three loads

A 5 m beam is supported at each end (R1 at left, R2 at right). Three downward loads are applied: 10 000 N at 2 m from R1, 500 N at 2.5 m, and 6 000 N at 4 m. Beam weight is ignored.

1
Take moments about R1 (so R1 drops out):
∑CWM = ∑ACWM
(10 000 × 2) + (500 × 2.5) + (6 000 × 4) = R2 × 5
2
20 000 + 1 250 + 24 000 = R2 × 5
45 250 = R2 × 5
R2 = 9 050 N
3
Use force balance:
R1 + R2 = 10 000 + 500 + 6 000 = 16 500 N
R1 = 16 500 − 9 050 = 7 450 N
4
Check: 7 450 + 9 050 = 16 500 N ✓
✏️ Task 4 — Beam reactions

Show all working for each problem. Remember: take moments about one reaction first, then use the force balance to find the other.

1. A 4 m beam is simply supported at each end. A single load of 8 000 N acts 1 m from the left support. Determine both reactions.
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2. A 6 m beam is supported at each end. Loads of 4 000 N at 1.5 m and 7 000 N at 4 m from the left support act downwards. Determine both reactions.
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3. A 7 m horizontal beam is part of a bridge structure, simply supported at A (left) and D (right). Loads of 6 000 N at 2 m, 3 000 N at 3.5 m, and 9 000 N at 5.5 m from A act downward. Determine the reactions at A and D.
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4. A car has a total weight of 12 000 N. The front axle (RF) is 0.8 m from the front and the rear axle (RR) is 2.7 m from the front. The centre of mass acts 1.2 m from the front. Calculate both axle reactions.
Working ✓ Saved
5. A contestant in a weight-lifting competition performs a barrel lift. The barrel weighs 800 N and is positioned 1.2 m from the pivot. The contestant's lifting force F acts 0.4 m from the pivot on the other side. Calculate the lifting force F.
Working ✓ Saved
📖 Once you can calculate the forces in structural members, the next step is to check whether the material chosen can withstand those forces. Stress and strain calculations are covered in the Materials course notes.

Vectors

Force is a vector quantity — it has both magnitude (size) and direction. Vectors can be represented as drawn lines where:

  • The length of the line (drawn to scale) represents the magnitude of the force
  • The direction and arrowhead show the direction of the force

When multiple forces act along the same line, they can simply be added or subtracted. For example: a cyclist pedalling with 800 N, assisted by a tailwind of 400 N, but with 200 N road friction, has a net force of 800 + 400 − 200 = 1 000 N.

Resultant of two angled forces

When two forces act at an angle to each other, their combined effect is a single force called the resultant. The resultant can be found graphically using the parallelogram of forces:

  1. Choose a suitable scale and draw both force vectors from a common point C
  2. From the tip of each vector, draw a line parallel to the other vector
  3. The point where these lines intersect (D) completes the parallelogram
  4. Draw the line from C to D — this is the resultant
  5. Measure the length of C to D and convert using your scale to find the magnitude
  6. Measure the angle to find the direction
The equilibrant is the single force that would exactly balance the resultant — it is equal in magnitude but opposite in direction to the resultant.

Conditions for Equilibrium

For a structure or body to be in equilibrium (a balanced, stable state), three conditions must all be satisfied:

ConditionMeaning
∑ upward forces = ∑ downward forcesVertical forces balance — no net vertical movement
∑ leftward forces = ∑ rightward forcesHorizontal forces balance — no net horizontal movement
∑ CWM = ∑ ACWMMoments balance — no net rotation

Graphical solution for forces in members

For a structure such as a crane or frame, the forces in individual members can be found graphically by drawing a triangle of forces:

  1. Identify the joint where the known load acts
  2. Draw the known force to scale
  3. From each end of this line, draw lines parallel to the unknown force members
  4. Where the lines intersect completes the triangle
  5. Measure each side to find the forces (following arrowheads round the triangle indicates compression or tension)
✏️ Task 5 — Vectors and equilibrium
1. State whether each of the following is TRUE or FALSE and explain your reasoning:
a) A body that is accelerating is in a state of equilibrium.
Answer ✓ Saved
b) For a body to be in equilibrium it is necessary only for the vector sum of forces to be zero.
Answer ✓ Saved
c) A resultant force is a single force that can replace two or more forces.
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d) If two or more forces are replaced by a resultant force, the effect on the body is changed.
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2. Describe the three conditions necessary for a structure or body to be in equilibrium.
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3. A crane fixed to a wall supports a load of 1 000 N. Using a graphical method (triangle of forces), find the forces in the compression and tension members. Use a scale of 10 mm = 200 N and sketch your triangle.
Sketch and answer ✓ Saved
4. A small crane on a fishing trawler lifts cases of fish. The lift weighs 1 200 N. Using a scale of 10 mm = 200 N, determine the size and direction of the forces in each crane member.
Working and answer ✓ Saved
5. A weight of 2 000 N hangs from a rope attached to a roof hook. A second rope pulls horizontally until the first rope makes 30° to the vertical. Determine the horizontal pull and the force on the hook.
Working and answer ✓ Saved
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