Trusses — Analysis by Method of Joints & Method of Sections

1. What is a Truss?

A truss is an assembly of straight structural members connected at their ends by frictionless pin joints (hinges), forming a rigid framework. Unlike a beam, which resists load through bending, a truss resists load through direct axial forces — each member either pulls apart (tension) or pushes together (compression) along its own length. Because there is no bending, material is used far more efficiently, making trusses ideal for long-span structures.

Trusses appear throughout civil and structural engineering. Roof trusses carry roofing loads to the walls of industrial buildings and warehouses. Bridge trusses (Pratt, Warren, Howe) span rivers and valleys, carrying traffic loads to the abutments. Tower cranes use truss booms to reach long horizontal distances. Electricity transmission towers are three-dimensional trusses. Space frames — the roofs of large stadiums, airports, and exhibition halls — are three-dimensional extensions of the same principle.

The key geometric property of a truss is its triangulation. A triangle is the only polygon that is inherently rigid — if you fix the lengths of three members and pin them at the corners, the shape cannot change without changing a member length. A square frame with four pinned corners, by contrast, can collapse into a parallelogram. Trusses achieve rigidity by subdividing the structure into triangles.

The most basic rigid truss has 3 members and 3 joints forming a single triangle. Each additional joint added to the truss requires exactly 2 new members to maintain rigidity. This gives the fundamental relationship: m = 2j − 3 for a simple (just-rigid) truss, where m = number of members and j = number of joints.

2. Assumptions in Ideal Truss Analysis

The classical truss analysis methods (Method of Joints and Method of Sections) are based on several simplifying assumptions. These assumptions are stated in every textbook but their practical implications are often not fully understood. Understanding what each assumption means — and when it is violated in real structures — is important for GATE conceptual questions.

3. Types of Trusses

Trusses are classified by the arrangement of their members. In a typical roof or bridge truss, the members along the top edge form the top chord, those along the bottom form the bottom chord, and the internal members connecting the two chords are verticals and diagonals. The arrangement of these internal members distinguishes the different truss types.

4. Determinacy of Trusses

Before analysing any truss, always check whether it is statically determinate, indeterminate, or a mechanism. An indeterminate truss cannot be solved by the methods described here — it requires compatibility equations involving member stiffness.

5. Zero-Force Members — Two Rules

A zero-force member is a member that carries no axial force under a particular loading condition. Zero-force members can be identified by inspection — without any calculation — using two simple rules. Identifying them before starting the analysis eliminates those members from consideration, simplifying the solution significantly. This is a favourite GATE CE topic and appears almost every year.

Important clarification: Zero-force members are zero-force under the given loading condition only. Under a different load arrangement, the same member may carry significant force. Zero-force members are also not useless — they provide lateral stability to compression members (preventing buckling) and become load-carrying members when the loading pattern changes (e.g., wind vs gravity loads).

6. Method of Joints — Step by Step

The Method of Joints analyses the truss joint by joint. At each joint, the pin is in equilibrium under the forces in the members meeting at that joint plus any external load at that joint. Since all forces at a joint are concurrent (they all pass through the joint), only two equilibrium equations are available: ΣFx = 0 and ΣFy = 0 (moment equilibrium is automatically satisfied for concurrent forces). This means you can solve for a maximum of two unknown member forces at any one joint.

Step-by-step procedure:

1. Step 1 — Check determinacy: Verify m + r = 2j before starting.
2. Step 2 — Find support reactions: Apply ΣFx = 0, ΣFy = 0, and ΣM = 0 to the whole truss to find all support reactions.
3. Step 3 — Identify zero-force members: Apply Rules 1 and 2 to eliminate zero-force members immediately.
4. Step 4 — Choose a starting joint: Select a joint where only two members meet with unknown forces (typically a support joint or a joint with one known external load). A support joint after finding reactions is ideal.
5. Step 5 — Assume all unknown members in tension: This is the standard convention. Draw each unknown member force pointing away from the joint (tension pulls the joint outward). If the result is positive, the member is in tension. If negative, it is in compression.
6. Step 6 — Apply ΣFx = 0 and ΣFy = 0: Write two equilibrium equations for the joint and solve for the two unknowns.
7. Step 7 — Move to the next joint: Use the member forces just found (now known) and move to an adjacent joint with no more than two unknowns. Repeat until all member forces are found.
8. Step 8 — Verify at the last joint: The last joint should satisfy equilibrium automatically with no new unknowns — this confirms the solution is correct.

7. Method of Sections — Step by Step

The Method of Sections finds the force in a specific member without solving the entire truss. It works by cutting the truss into two parts with an imaginary section that passes through the member of interest (and at most two other members). Each part is then treated as a free body in equilibrium under all external forces on that part plus the cut member forces. With three equilibrium equations (ΣFx = 0, ΣFy = 0, ΣM = 0) and at most three unknown cut member forces, the problem is always solvable.

Step-by-step procedure:

1. Step 1 — Find support reactions for the whole truss.
2. Step 2 — Draw the cut: Pass an imaginary section through the truss, cutting through exactly the member whose force you want AND at most two other members. The section should completely separate the truss into two independent parts. Choose the cut to pass through no more than three members total.
3. Step 3 — Choose one part: Analyse the simpler of the two parts — typically the one with fewer external loads.
4. Step 4 — Assume all cut members in tension: Draw the cut member forces pointing away from the free body (tension convention). Positive results confirm tension; negative results indicate compression.
5. Step 5 — Apply equilibrium strategically: The power of the Method of Sections lies in choosing the moment centre wisely. Take moments about a point where two of the three unknown cut member forces intersect — this eliminates those two unknowns and gives the third directly from a single equation. This is the method’s greatest advantage over the Method of Joints.
6. Step 6 — Solve for remaining unknowns using ΣFx = 0 and ΣFy = 0 as needed.

8. Worked Example 1 — Simple Truss by Method of Joints

Problem: A simply supported Pratt truss has joints A, B, C, D, E arranged as follows: A and E are the support joints (A = pin, E = roller). B and D are the top chord joints, C is the crown (top centre joint). The bottom chord runs A–F–E (F is the midpoint of the bottom chord). All panels are 2 m wide and the truss height is 2 m. A vertical load of 40 kN acts downward at F (midpoint of bottom chord). Find the forces in all members using the Method of Joints.

Member list: Bottom chord: AF, FE. Top chord: AB, BC (left half), CD, DE (right half). Verticals: BF. Diagonals: AF connection…

Let us use a cleaner, standard 5-joint truss for clarity:

Revised Problem: A simply supported truss ABCDE has: A (pin, bottom left), B (bottom right, roller), C (top left), D (top centre), E (top right). Geometry: AB = 4 m (bottom chord), CD = 2 m, DE = 2 m (top chord). Height = 2 m. Joints: A(0,0), B(4,0), C(0,2), D(2,2), E(4,2). Members: AC, CD, DE, EB (top and side), AB (bottom), AD, DB (diagonals). Load: 20 kN downward at D (top centre joint).

9. Worked Example 2 — Method of Sections (Finding Force in a Specific Member)

Problem: A Warren truss bridge has a span of 12 m with 4 equal panels of 3 m each. Top chord joints: P(0,3), Q(3,3), R(6,3), S(9,3), T(12,3). Bottom chord joints: A(0,0), B(3,0), C(6,0), D(9,0), E(12,0). A = pin support, E = roller support. A vertical load of 60 kN acts downward at joint C (midspan bottom chord). Using the Method of Sections, find the force in members QR (top chord), CD (bottom chord), and QC (diagonal).

10. Worked Example 3 — Zero-Force Member Identification

Problem: A truss has the following joints and members. Bottom chord: A–B–C–D–E (equally spaced). Top chord: F–G–H (F above B, G above C, H above D). Additional members: AF, BF, CF, CG, DG, EH, DH. A vertical load acts at joint C only. Identify all zero-force members.

11. Identifying Tension vs Compression

The sign of a member force reveals whether the member is in tension or compression. Understanding the physical meaning of these signs — and being able to identify the likely state of members by inspection — is a useful check on calculated results.

Quick memory rule for simply supported trusses under gravity loads: The top chord acts like a compressed arch — it is in compression. The bottom chord acts like a stretched string — it is in tension. This is directly analogous to the top fibre compression and bottom fibre tension in a sagging beam.

12. Common Mistakes Students Make

• Trying to solve a joint with more than two unknowns: The Method of Joints gives only two equations per joint (ΣFx = 0 and ΣFy = 0). If three or more unknown member forces meet at a joint, you cannot solve it directly — you must first solve adjacent joints to reduce the unknowns. Always start at a joint with exactly two unknowns. If no such joint exists at the start, find support reactions first (which converts support joints into known-force joints).
• Incorrect sign of reaction forces in the Method of Sections: When analysing the left portion of a cut truss, the support reaction at A acts upward on the left portion. Many students draw the reaction incorrectly or forget it entirely when writing equilibrium equations for the free body. Always draw the complete FBD of the chosen portion — showing all external loads, support reactions, and the three cut member forces — before writing any equations.
• Confusing the direction of cut member forces: Assuming all cut members in tension (forces pointing away from the free body, toward the cut) is the standard convention. If a result comes out negative, it simply means that member is in compression — do not redraw and re-solve. Changing the assumed direction mid-problem is a common source of sign errors.
• Applying zero-force rules incorrectly: Rule 1 (two members, no load → both zero) only works when there are exactly two members and no external load or reaction at that joint. If a reaction acts at the joint (as it does at support joints), the rules do not apply. Rule 2 requires exactly three members, two collinear, and no external load. Carefully count members and check for reactions before applying either rule.
• Not checking the final joint after Method of Joints: The last joint in a Method of Joints analysis should be in equilibrium using all previously found member forces — no new unknowns. If it is not, an error was made somewhere. This verification step takes thirty seconds and catches virtually every calculation mistake.

13. Frequently Asked Questions

When should you use Method of Joints vs Method of Sections?

Use the Method of Joints when you need forces in all members of the truss, or when the truss is small enough that solving joint by joint is practical. It is systematic and guaranteed to work for any determinate truss. Use the Method of Sections when you need the force in only one or a few specific members — particularly members in the interior of a large truss that would require solving through many joints to reach by the Method of Joints. In GATE CE, questions almost always ask for specific member forces — use the Method of Sections first and you will reach the answer in one or two equations.

Can a zero-force member become a load-carrying member under different loading?

Yes — and this is important for design. A member that is zero-force under one load case (e.g., a symmetric gravity load) may carry significant force under a different load case (e.g., an asymmetric live load, wind load, or load applied at a different joint). Real structures are designed for multiple load combinations, and a member that is zero-force in the governing gravity case must still be checked for all other load cases. Zero-force members also serve structural purposes even when carrying no force — they reduce the unsupported (buckling) length of compression members and maintain the geometric stability of the truss under all loading conditions.

What happens if a truss has m + r < 2j?

When m + r < 2j, the truss has fewer members and reactions than required for rigidity — it is a mechanism. This means the truss can deform (change shape) without any member changing length — it simply rotates or collapses as a kinematic chain. A mechanism is not a structure and cannot carry load in the intended way. In practice, this means the truss is under-designed and will collapse. This condition should be caught during the determinacy check before any analysis is attempted.

How do you handle a truss that is statically indeterminate?

A statically indeterminate truss (m + r > 2j) cannot be solved by equilibrium alone. The extra members are called redundant members — the truss would remain stable and determinate without them, but they provide additional paths for load transfer and increase structural redundancy (important for robustness and fatigue resistance). Solving an indeterminate truss requires the use of compatibility equations — each redundant member introduces one compatibility condition. The most common method is the force method (method of consistent deformations), where the redundant member forces are treated as unknowns and compatibility (the condition that member deformations are geometrically consistent) provides the additional equations. This requires knowledge of member axial stiffness (AE/L for each member).