Frameworks

Introduction to Frameworks

Welcome to Frameworks (also known as pin-jointed trusses)! This topic forms an essential part of CCEA A2 2 Section A: Mechanics 1. Have you ever looked closely at a railway bridge, a construction crane, or the triangular timber roof of a house? These structures are made from straight beams joined together in triangles to carry heavy loads efficiently without bending. In this chapter, you will learn how to calculate the exact forces acting inside every single member of a framework and determine whether each rod is being stretched or squashed.

Don't worry if mechanics has felt intimidating before. Frameworks follow a reliable, step-by-step method based entirely on the equilibrium of forces and moments you have already met.

Key Takeaway: A framework is a network of rods pinned together. We analyse it by treating the whole structure as one rigid body in equilibrium, and then treating each joint as a particle in equilibrium.

Core Definitions & Modelling Assumptions

In A Level Further Mathematics, we simplify real-world structures into mathematical models using standard assumptions:

• Light Framework (Pin-jointed Truss): A rigid structure constructed from straight, light rods joined at their ends by smooth, frictionless pins or hinges.
• Light Rods: The mass and weight of the individual rods are negligible compared to the external loads. This means gravity acts only at joints where external masses or forces are explicitly applied.
• Smooth Pins: The joints cannot exert any rotational resistance (no bending moments). Rods can rotate freely around the pins.
• Two-Force Members: Because the rods are light and connected by smooth pins at their ends, forces can only be transmitted along the straight line of each rod. No bending or shear forces exist within the members.

Did you know? Triangles are the only geometric polygons that are naturally rigid. A rectangle pinned at four corners can easily wobble into a parallelogram, but a triangle pinned at three corners cannot change its shape without changing the length of its sides!

Internal Forces: Tension vs. Thrust

Every rod in a framework experiences one of two types of internal force:

1. Tension (A "Tie")

• What is happening? The rod is being pulled or stretched apart.
• Force on the rod: Forces at both ends pull outwards.
• Force on the joint: By Newton's Third Law, the rod pulls away from the joint.
• Arrow convention on diagrams: Draw the force arrow pointing away from the pin/joint along the rod.

2. Thrust or Compression (A "Strut")

• What is happening? The rod is being squashed or compressed.
• Force on the rod: Forces at both ends push inwards.
• Force on the joint: The rod pushes back, meaning it pushes towards the joint.
• Arrow convention on diagrams: Draw the force arrow pointing towards the pin/joint along the rod.

Memory Trick:
• Tie = Tension = Towards the middle of the rod (pulling away from the joint).
• Strut = Squash = Shoving towards the joint.

Key Takeaway: When drawing a Free-Body Diagram at a joint, Tension pulls away from the joint and Thrust pushes into the joint.

The Two-Stage Solving Strategy

To solve any framework problem, follow this structured two-stage method:

Stage 1: Overall (Global) Equilibrium

Before looking at the inside of the framework, treat the entire framework as a single rigid body to find the external reaction forces at the supports.

1. Identify Support Types:
• Smooth Pin / Hinge Support: Provides reaction forces in two perpendicular directions, horizontal \(R_x\) and vertical \(R_y\).
• Roller Support / Smooth Contact: Provides a reaction force purely perpendicular to the supporting surface (usually a vertical normal reaction \(N\) or \(R\)).

2. Apply the 3 Global Equilibrium Conditions:
• \(\sum F_x = 0\) (Sum of all horizontal forces is zero)
• \(\sum F_y = 0\) (Sum of all vertical forces is zero)
• \(\sum M_{\text{point}} = 0\) (Taking moments about a convenient support pin with multiple unknowns eliminates them).

Stage 2: The Method of Joints

Once external reactions are known, treat each joint (pin) as a particle in concurrent equilibrium:

1. Find a Starting Joint: Look for a joint that has at most two unknown member forces.
2. Resolve Forces: At the chosen joint, set up two equilibrium equations:
\(\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0\)
3. Assigning Directions: You can either assume every unknown member is in tension (pulling away from the joint) so that a negative calculated value indicates thrust, OR deduce the direction by inspection.
4. Progress Through the Framework: Substitute the newly found forces into neighbouring joints (remembering that a rod in tension pulls away from both its joints, and a rod in thrust pushes into both its joints) until every member is solved.
5. State the Nature: Always write down both the magnitude and the nature (Tension/Tie or Thrust/Strut) for every rod.

Key Takeaway: Global equilibrium first gives you support reactions. The method of joints then solves internal forces joint-by-joint using \(\sum F_x = 0\) and \(\sum F_y = 0\).

Fully Worked Example

Example: A light framework \(ABC\) consists of three light rods \(AB\), \(BC\), and \(AC\) smoothly hinged at their ends to form an equilateral triangle of side length \(2\text{ m}\). The framework stands in a vertical plane with \(A\) and \(B\) on a horizontal floor. Support \(A\) is a smooth hinge, and support \(B\) rests on a smooth roller. A vertical downward load of \(60\text{ N}\) is applied at the apex \(C\). Find the reactions at the supports and the magnitude and nature of the force in each of the three rods.

Step 1: Geometry and Angles

Since triangle \(ABC\) is equilateral, all interior angles are \(60^\circ\).
The horizontal distance between \(A\) and \(B\) is \(2\text{ m}\).
By symmetry, apex \(C\) lies horizontally halfway between \(A\) and \(B\) (distance \(1\text{ m}\) from \(A\)).

Step 2: Global Equilibrium (Finding Support Reactions)

Let the reaction at hinge \(A\) have horizontal component \(A_x\) (to the right) and vertical component \(A_y\) (upwards).
Let the reaction at roller \(B\) be purely vertical \(B_y\) (upwards).

• Horizontal equilibrium for the whole framework:
\(\sum F_x = 0 \implies A_x = 0\text{ N}\)

• Taking moments about point \(A\) (taking clockwise as positive):
\(\sum M_A = 0 \implies (60 \times 1) - (B_y \times 2) = 0\)
\(2B_y = 60 \implies B_y = 30\text{ N}\)

• Vertical equilibrium for the whole framework:
\(\sum F_y = 0 \implies A_y + B_y - 60 = 0\)
\(A_y + 30 - 60 = 0 \implies A_y = 30\text{ N}\)

Step 3: Joint Analysis at Joint A

Joint \(A\) is acted upon by:
• Upward support reaction \(A_y = 30\text{ N}\)
• Internal force \(F_{AC}\) along rod \(AC\) (inclined at \(60^\circ\) to the horizontal)
• Internal force \(F_{AB}\) along rod \(AB\) (horizontal)

Let us assume rod \(AC\) is in thrust (pushing into joint \(A\) at \(60^\circ\)) and rod \(AB\) is in tension (pulling away from joint \(A\) to the right).

• Resolving vertically at joint \(A\) (\(\sum F_y = 0\)):
\(30 - F_{AC} \sin 60^\circ = 0\)
\(F_{AC} \left(\frac{\sqrt{3}}{2}\right) = 30\)
\(F_{AC} = \frac{60}{\sqrt{3}} = 20\sqrt{3}\text{ N} \approx 34.6\text{ N}\)
Since the result is positive, our assumption is correct: \(F_{AC} = 20\sqrt{3}\text{ N}\) (Thrust).

• Resolving horizontally at joint \(A\) (\(\sum F_x = 0\)):
\(F_{AB} - F_{AC} \cos 60^\circ = 0\)
\(F_{AB} = (20\sqrt{3}) \times \frac{1}{2} = 10\sqrt{3}\text{ N} \approx 17.3\text{ N}\)
Since the result is positive, our assumption is correct: \(F_{AB} = 10\sqrt{3}\text{ N}\) (Tension).

Step 4: Joint Analysis at Joint B (or Joint C)

By symmetry across the vertical line through \(C\):
• Force in rod \(BC\) is equal in magnitude and nature to rod \(AC\):
\(F_{BC} = 20\sqrt{3}\text{ N}\) (Thrust).

We can verify this at Joint \(C\) vertically (\(\sum F_y = 0\)):
Both rods \(AC\) and \(BC\) are in thrust, pushing upwards into joint \(C\):
\(2 \times (20\sqrt{3} \sin 60^\circ) = 2 \times \left(20\sqrt{3} \times \frac{\sqrt{3}}{2}\right) = 2 \times 30 = 60\text{ N}\)
This perfectly balances the downward \(60\text{ N}\) load!

Final Summary of Member Forces:

• Rod \(AC\): \(20\sqrt{3}\text{ N}\) (Thrust / Strut)
• Rod \(BC\): \(20\sqrt{3}\text{ N}\) (Thrust / Strut)
• Rod \(AB\): \(10\sqrt{3}\text{ N}\) (Tension / Tie)

Common Pitfalls & Examiner Warnings

Examiner reports frequently highlight avoidable errors in this topic. Watch out for these:

• Forgetting the Nature of the Force: Giving an answer as just "\(34.6\text{ N}\)" without stating Tension or Thrust (or Tie / Strut) will lose final accuracy marks.
• Inconsistent Arrow Directions: Remember that if a rod is in tension, it pulls away from both joints at its ends. If it is in thrust, it pushes towards both joints at its ends.
• Skipping Support Reactions: Do not jump straight into resolving at a joint before finding the external reactions at supports \(A\) and \(B\); otherwise, you will have too many unknowns.
• Trigonometry Confusion: Always double check whether you need \(\cos\theta\) or \(\sin\theta\). If \(\theta\) is the angle with the horizontal, the horizontal component uses \(\cos\theta\) and the vertical component uses \(\sin\theta\).

Quick Revision Checklist

• Did you treat the whole framework as a rigid body first to find reactions using \(\sum F_x = 0\), \(\sum F_y = 0\), \(\sum M = 0\)?
• Did you choose a starting joint with at most 2 unknowns?
• Are tension forces drawn pointing away from the joint?
• Are thrust forces drawn pointing towards the joint?
• Did you clearly state both the magnitude and the nature (Tension/Thrust) for every member?