Analysis of Trusses & Frames

Truss analysis is a dependable 1–2 marks in GATE ME and rewards a clean method choice. The whole topic reduces to two techniques — method of joints and method of sections — plus spotting zero-force members.

Assumptions & method

A truss has straight two-force members joined by frictionless pins, with loads applied only at joints. So every member carries pure axial force (tension = pulling away from joint, compression = pushing toward joint).

• Method of joints: apply ΣFx = 0, ΣFy = 0 at each pin (2 equations/joint). Start at a joint with ≤ 2 unknowns. Best when you need forces in most members.
• Method of sections: cut through ≤ 3 unknown members, treat one side as a body, apply ΣF and ΣM. Best when you need a few specific members quickly — take moments to isolate one.

Determinacy: m + r = 2j → statically determinate (m = members, r = reactions, j = joints). m + r > 2j → indeterminate.

Zero-force members: at an unloaded joint of two non-collinear members, both are zero; at an unloaded joint of three members where two are collinear, the third is zero.

Exam Tricks & Tips

• 🎯 Spot zero-force members first using the two joint rules — they simplify the count and are a common direct MCQ.
• 🎯 Need one specific member's force? Use method of sections and take moments about the intersection of the other two cut members.
• 🎯 Direction convention: assume tension (arrow away from joint); a negative result then automatically means compression.
• 🎯 Check determinacy with m + r = 2j before solving — an indeterminate truss cannot be done by statics alone.
• 🎯 A section cut can pass through non-parallel curved paths as long as it severs ≤ 3 unknown members.
• ❌ Common mistake: applying a member load mid-span — trusses take loads only at joints; a mid-span load turns the member into a beam (bending), invalidating the axial-only assumption.

Expected exam pattern

A 1-mark "identify zero-force members" or "tension/compression in member X," and a 2-mark method-of-sections force in a named diagonal or chord. Determinacy classification appears as a quick conceptual question.

Quick recap

Truss members are two-force (axial only). Joints: 2 eqns each, start with ≤2 unknowns. Sections: cut ≤3 members, use moments to isolate one. Determinate if m + r = 2j. Learn the two zero-force-member rules — they save time.

Cover the answer, recall, then check. 10 cards on truss analysis for GATE ME.

Q1. What are the standard idealised-truss assumptions?
A1. Straight members, frictionless pin joints, and loads applied only at joints — so every member is a two-force (axial) member.

Q2. State the determinacy condition for a plane truss.
A2. m + r = 2j → statically determinate; m + r > 2j → indeterminate; m + r < 2j → unstable (mechanism).

Q3. When is the method of joints preferred?
A3. When forces in most/all members are needed. Apply ΣFx = 0, ΣFy = 0 at each joint; start where ≤ 2 unknowns exist.

Q4. When is the method of sections preferred?
A4. When only a few specific member forces are needed. Cut through ≤ 3 unknown members and use ΣF and ΣM on one side.

Q5. State the two-member zero-force rule.
A5. At an unloaded joint with only two non-collinear members, both members carry zero force.

Q6. State the three-member zero-force rule.
A6. At an unloaded joint with three members where two are collinear, the third (non-collinear) member is a zero-force member.

Q7. How do you tell tension from compression by sign?
A7. Assume tension (force arrow away from the joint). A positive answer confirms tension; a negative answer means compression.

Q8. In method of sections, how do you isolate one member's force?
A8. Take moments about the point where the other two cut members intersect, so only the target member's force appears.

Q9. Why can a section not cut more than three unknown members (usually)?
A9. Only three equilibrium equations are available in 2D, so ≤ 3 unknowns keep the section solvable.

Q10. Why must truss loads act only at joints?
A10. A mid-span load bends the member, introducing shear and moment and violating the axial-only two-force assumption.

Trusses turn up in every structural GATE set because they are the cleanest test of equilibrium reasoning. The whole subject is two methods (joints and sections) plus the ability to spot zero-force members instantly.

Core concept: an ideal truss is a pin-jointed frame loaded only at joints, so every member is a two-force member carrying pure tension or compression — no bending.

Deep explanation

Beginner — assumptions and determinacy

Ideal-truss assumptions: members are straight, joints are frictionless pins, loads act only at joints, self-weight is neglected. A plane truss is statically determinate if
m + r = 2j
where m = members, r = support reactions, j = joints. If m + r > 2j it is indeterminate; if less, it is a mechanism (unstable).

Intermediate — the two methods

• Method of joints: isolate each joint (a concurrent force system): ΣFx = 0, ΣFy = 0 (2 equations/joint). Start at a joint with ≤ 2 unknown members. Best when you need all member forces.
• Method of sections: cut through the truss (≤ 3 unknown members) and apply ΣM, ΣFx, ΣFy to one portion. Best when you need one or two specific members deep inside — one equation can give the answer directly.

Advanced — zero-force members

Spot these before calculating; they simplify everything:

• At an unloaded two-member joint with non-collinear members, both are zero-force.
• At an unloaded three-member joint where two are collinear, the third (non-collinear) member is zero-force.
  Sign convention: assume tension (arrow pulling away from joint); a negative result means the member is in compression.

Worked example

Use method of sections to find the force in the top chord member of a simply-supported truss. Suppose a section exposes top-chord member U with the truss carrying support reaction R_A = 30 kN at the left, and U is 3 m above the bottom chord; take moments about the bottom-chord joint 4 m to the right of A where the other cut members pass through.
ΣM about that joint: 30(4) − F_U(3) = 0 → F_U = 120/3 = 40 kN. The sign (taking the assumed tension direction) tells tension vs compression; a top chord in a simply-supported truss is typically in compression, so F_U = 40 kN (C).

GATE relevance

Truss member-force questions appear regularly in Engineering Mechanics. Sections is the exam-efficient choice for "find force in member X"; zero-force recognition can turn a 3-minute problem into a 10-second one.

Exam tricks & shortcuts

• Scan for zero-force members first — they often include the exact member being asked about.
• Use sections for a specific interior member (moment about the right joint isolates it); use joints only if all forces are wanted.
• Mnemonic: "Positive pulls (tension), Negative nudges in (compression)."

Key formulas

• Perfect truss (2D): m = 2j − 3 (m = members, j = joints); deficient/redundant otherwise.
• Method of joints: ΣFₓ = 0, ΣF_y = 0 at each joint (≤ 2 unknowns).
• Method of sections: cut through ≤ 3 members, apply ΣF and ΣM.
• Zero-force members: identified by joint geometry.
• Members carry only axial force (tension +, compression −).

Use joints for full trusses and sections to find a few specific member forces quickly.

Worked Example

Problem: A simply supported truss has a pin support at A(0, 0), a roller at B(6, 0) m, and an apex joint C(3, 4) m. A 10 kN downward load acts at C. Using the method of joints, find the forces in members AB and AC (state tension or compression).

Solution:
Support reactions: the load is at the horizontal midpoint, so by symmetry each vertical reaction is
R_Ay = R_By = 10/2 = 5 kN (upward); R_Ax = 0.

Geometry of member AC: from A(0,0) to C(3,4), length = √(3² + 4²) = 5, so its direction cosines are (3/5, 4/5).

Analyse joint A (members AB horizontal and AC inclined). Take tension positive (member pulls the joint toward the far end):
Vertical: R_Ay + F_AC·(4/5) = 0
5 + 0.8·F_AC = 0 ⇒ F_AC = −6.25 kN.
The negative sign means AC is in compression, magnitude 6.25 kN.

Horizontal: F_AB + F_AC·(3/5) = 0
F_AB + (−6.25)(0.6) = 0 ⇒ F_AB = +3.75 kN (positive ⇒ tension).

By the symmetry of the truss, member BC carries the same as AC: 6.25 kN compression.

Answer: F_AB = 3.75 kN (tension); F_AC = 6.25 kN (compression) [and F_BC = 6.25 kN compression].