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Qwestrum Engineering360 · Mechanical Engineering · Engineering Mechanics

Free Body Diagram

A free body diagram isolates one body and shows every external force and reaction, after which equilibrium demands ΣF_x = 0, ΣF_y = 0, ΣM = 0. Correct support reactions and Newton's third-law pairs are the essence, as RK Bansal stresses.
Exam tip: draw a complete FBD first; resolve forces along chosen axes and check moment equilibrium about a convenient point.

Key formulas & points

Skim these first — then read the full notes below.

  • Isolate body; show all external forces and moments
  • Internal forces cancel in FBD of complete system
  • Statically determinate: equations = unknowns

Topic details

Introduction

The free body diagram is the single most important skill in statics; examiners in Indian universities routinely award separate marks for a correct FBD before any calculation. It converts a physical assembly into a solvable set of equilibrium equations.

Scope in B.Tech and GATE syllabus

RK Bansal insists on isolating the body, replacing each support by its reaction (roller → normal force, pin → two components, fixed → two forces and a moment), and including weight through the centroid. Internal forces between parts cancel when the whole system is taken as one body.

Why this topic matters in practice

A structure is statically determinate when the number of unknown reactions equals the available equilibrium equations (three in 2D). If unknowns exceed equations the problem is indeterminate and needs compatibility relations — recognising this distinction is a common conceptual question.

Key relations & formulas

ΣFx=0,ΣFy=0ΣF_{x} = 0, ΣF_{y} = 0
(2D equilibrium)
ΣMA=0ΣM_{A} = 0
(third equilibrium equation)
FAB=−FBAF_{AB} = -F_{BA}
(Newton III, action-reaction)
W=m⋅gW = m\cdot g
(weight force, g = 9.81 m/s²)

Notation and sign conventions

Relation 1 —
ΣFx=0,ΣFy=0ΣF_{x} = 0, ΣF_{y} = 0
ΣFx=0,ΣFy=0ΣF_{x} = 0, ΣF_{y} = 0
(2D equilibrium)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
ΣMA=0ΣM_{A} = 0
ΣMA=0ΣM_{A} = 0
(third equilibrium equation)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
FAB=−FBAF_{AB} = -F_{BA}
FAB=−FBAF_{AB} = -F_{BA}
(Newton III, action-reaction)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 4 —
W=m⋅gW = m\cdot g
W=m⋅gW = m\cdot g
(weight force, g = 9.81 m/s²)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

Equilibrium requires the net force and net moment on the isolated body to vanish: ΣF_x = 0, ΣF_y = 0, and ΣM_A = 0 about any convenient point A. Choosing A at an unknown-heavy joint eliminates those unknowns from the moment equation.

Governing relations in practice

Supports are modelled by the motion they prevent: a roller resists only the normal direction (one reaction), a hinge/pin resists translation in two directions (two reactions), and a fixed support resists translation and rotation (two forces and a moment).

Design and analysis considerations

Newton's third law governs interaction forces: the force of A on B equals and opposes the force of B on A. When two members are separated in an FBD, these action-reaction pairs must be drawn consistently.

Advanced theory and extensions

For a two-force member, equilibrium forces the internal force to lie along the line joining the two connection points — a shortcut that simplifies truss and frame analysis. Applying these rules yields exactly as many equations as unknowns for a determinate body.

Assumptions and validity limits

State assumptions explicitly before using any relation for free body diagram — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Engineering Mechanics viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Engineering Mechanics papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to free body diagram.
4. Use equation 1:
ΣFx=0,ΣFy=0ΣF_{x} = 0, ΣF_{y} = 0
.
5. Use equation 2:
ΣMA=0ΣM_{A} = 0
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Free Body Diagram appears in trusses, frames, and friction problems. In Indian mechanical curricula this topic is tested because it connects theory to force equilibrium and motion of rigid bodies.
GATE and semester exams often combine free body diagram with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use free body diagram?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Omitting a support reaction or drawing the wrong number of components for the support type
• Including internal forces when the whole assembly is taken as one free body
• Forgetting the weight acting at the centroid
• Attempting to solve a statically indeterminate body with only equilibrium equations

Quick revision checklist

Before attempting free body diagram problems, confirm you can:
1. Isolate body; show all external forces and moments
2. Internal forces cancel in FBD of complete system
3. Statically determinate: equations = unknowns
Revise the solved examples in Strength of Materials — RK Bansal and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Reactions of a simply supported beam

Problem

A 4 m simply supported beam carries a central point load W = 100 N. Find the two support reactions.

Solution

By symmetry (or ΣM = 0): R_A = R_B = W/2 = 50 N each; check ΣF_y = 50 + 50 − 100 = 0.

Conceptual check — Free Body Diagram

Problem

In a Engineering Mechanics semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of free body diagram." What should a complete answer include?

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

  1. 1
    What is Free Body Diagram, and why does it appear in B.Tech / GATE syllabi?

    Model answer

    A free body diagram isolates one body and shows every external force and reaction, after which equilibrium demands ΣF_x = 0, ΣF_y = 0, ΣM = 0. Correct support reactions and Newton's third-law pairs are the essence, as RK Bansal stresses.
  2. 2
    State the relation ΣF_x = 0, ΣF_y = 0 and name each symbol.

    Model answer

    The governing relation is ΣFx=0,ΣFy=0ΣF_{x} = 0, ΣF_{y} = 0. Write every symbol with SI units before substituting numbers.
  3. 3
    State the relation ΣM_A = 0 and name each symbol.

    Model answer

    The governing relation is ΣMA=0ΣM_{A} = 0. Write every symbol with SI units before substituting numbers.
  4. 4
    State the relation F_AB = −F_BA and name each symbol.

    Model answer

    The governing relation is FAB=−FBAF_{AB} = -F_{BA}. Write every symbol with SI units before substituting numbers.
  5. 5
    State the relation W = m·g and name each symbol.

    Model answer

    The governing relation is W=m⋅gW = m\cdot g. Write every symbol with SI units before substituting numbers.
  6. 6
    Explain: Isolate body; show all external forces and moments

    Model answer

    Isolate body; show all external forces and moments — state the assumption range and one exam trap linked to this point.
  7. 7
    Explain: Internal forces cancel in FBD of complete system

    Model answer

    Internal forces cancel in FBD of complete system — state the assumption range and one exam trap linked to this point.
  8. 8
    Explain: Statically determinate: equations = unknowns

    Model answer

    Statically determinate: equations = unknowns — state the assumption range and one exam trap linked to this point.
  9. 9
    How would you correct this error in a viva: Omitting a support reaction or drawing the wrong number of components for the support type?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  10. 10
    How would you correct this error in a viva: Including internal forces when the whole assembly is taken as one free body?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  11. 11
    How would you correct this error in a viva: Forgetting the weight acting at the centroid?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  12. 12
    How would you correct this error in a viva: Attempting to solve a statically indeterminate body with only equilibrium equations?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.

Exams & GATE

  • 1
    Draw FBD before writing equations — examiners award diagram marks.
  • 2
    Avoid: Omitting a support reaction or drawing the wrong number of components for the support type
  • 3
    Avoid: Including internal forces when the whole assembly is taken as one free body
  • 4
    Avoid: Forgetting the weight acting at the centroid

📖 Standard books (India)

  • Strength of Materials — RK Bansal

    Read: Syllabus unit

    SOM — beams, torsion, columns, and deflection

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