Structures & forces

Forces in structures

• Tension pulls a member (it would stretch); a member in tension is a tie.
• Compression pushes a member (it would squash); a member in compression is a strut.
• When showing the forces acting on a member: arrows pointing towards each other (pushing in) show compression; arrows pointing away from each other (pulling apart) show tension.

Equilibrium and free body diagrams

A structure is in equilibrium when it isn't moving: the forces balance. A free body diagram shows one object with all the forces acting on it. Three concurrent forces (acting through one point) in equilibrium form a closed triangle of forces when drawn to scale head to tail.

Triangle of forces by scale drawing

For three concurrent forces in equilibrium, a scale drawing finds two unknown forces when you know one force and the directions of the other two.

1. Choose a large scale, e.g. 10 mm = 200 N. A bigger scale is more accurate.
2. Draw the known force first, in its direction and to scale.
3. From one end draw a line parallel to the first unknown force; from the other end, a line parallel to the second.
4. Where the lines cross closes the triangle. Measure each side and convert using the scale.
5. The arrows follow each other round the triangle. Use them to decide whether each member is a strut or a tie.

To find the resultant of two forces, draw them to scale and complete the parallelogram: the diagonal is the resultant. The equilibrant is the same size as the resultant but acts in the opposite direction.

Frames are made of triangles because a triangle can't change shape under load (triangulation), so it stays rigid.

Moments

• Moment = force × perpendicular distance from the pivot (N m).
• Principle of moments: for equilibrium, the sum of clockwise moments = the sum of anticlockwise moments (ΣCWM = ΣACWM).
• Also, for equilibrium, upward forces = downward forces (ΣF_v = 0).

Simply supported beam: take moments about one support to find the other reaction, then use ΣF_v = 0 for the first. E.g. loads 650 N at 0.8 m and 890 N at 3 m balanced by 560 N at distance d: 520 + 2670 = 560 d, d = 5.7 m.

Balance beams and parallel forces

• Concurrent forces act through one point. Non-concurrent forces don't, so they can turn an object. Parallel forces, such as loads on a beam, are non-concurrent: use moments.
• For equilibrium: upward forces = downward forces, leftward = rightward, and clockwise moments = anticlockwise moments.
• Balance beam (a pivot in the middle, like a see-saw): a 400 N child sits 1.5 m left of the pivot. Where must a 600 N child sit? ACWM = 400 × 1.5 = 600 N m, so 600 × d = 600 and d = 1.0 m to the right. The pivot pushes up with 400 + 600 = 1000 N.

Interactive simulation: try this on the revision site, where you can change the values and see the result.

Weight and loads

Loads are often given as a mass. Convert to a force first: weight = m × g = m × 9.8. A 95 kg person: 95 × 9.8 = 931 N; a 500 kg container: 4900 N. Then draw the free body diagram with every force, reaction and distance before taking moments.

Design challenge: footbridge

The brief. A council in the Borders is building a 6 m footbridge over a burn. A 900 N person stands 2 m from support A, and a 400 N bike is 5 m from A.

1. Draw a free body diagram of the bridge.
2. Calculate the reaction at support B by taking moments about A.
3. Calculate the reaction at support A.
4. State whether each member of a supporting truss would be in tension or compression and how you would find out.

Resolving forces HIGHER

• Split an angled force into components: horizontal F cos θ, vertical F sin θ (θ measured from the horizontal).
• For equilibrium: ΣF_H = 0 and ΣF_V = 0. The resultant R = √(F_H² + F_V²) at angle tan⁻¹(F_V/F_H); the equilibrant is equal and opposite to the resultant.
• A uniformly distributed load (UDL) is replaced by a point load (load per metre × length) at the centre of the length it covers.
• Angled loads on beams: only the component at 90° to the beam creates a moment about the support (e.g. 3.8 sin 62).
• A hinge (pin) support can have horizontal and vertical reactions; a roller support has only a reaction at 90° to its surface.

Polygon of forces HIGHER

With more than three concurrent forces, draw them head to tail to scale: the line that closes the polygon is the equilibrant, and the resultant is equal and opposite. The quicker exam method is to resolve every force.

Example: 50 N at 0°, 80 N at 90° and 40 N at 210° (angles from the positive horizontal).

• ΣF_H = 50 + 0 + 40 cos 210° = 50 − 34.6 = 15.4 N
• ΣF_V = 0 + 80 + 40 sin 210° = 80 − 20 = 60.0 N
• R = √(15.4² + 60.0²) = 61.9 N at tan⁻¹(60.0 ÷ 15.4) = 75.6° above the horizontal.
• The equilibrant is 61.9 N acting in the opposite direction (255.6°).

Structures can be mass structures (dams, walls), frame structures (bridges, pylons) or shell structures (car bodies, domes).

Nodal analysis of frames HIGHER

Start at a node with no more than two unknown members. Resolve vertically and horizontally to find each member force, then move to the next node. State the size and the nature of each force: strut (compression) or tie (tension). E.g. at node A with a 47 kN reaction: AB = 47 ÷ sin 60 = 54 kN (strut); AE = 54 cos 60 = 27 kN (tie).

Key terms

Quick check