Free Body Diagrams

The Free Body Diagram (FBD)

A free body diagram isolates one body and shows every external force and moment acting on it. Draw the FBD before writing any equilibrium equation — if a force or reaction is missing, every answer that follows will be wrong.

A solid FBD workflow on the exam:

  1. Identify the body to isolate (whole structure, a single member, or a joint).
  2. Cut the body free and sketch its outline.
  3. Replace supports with their reactions (arrows on the body).
  4. Add all applied loads, weights, and known forces at correct locations.
  5. Label every unknown and choose a sign convention.

Equilibrium Equations

For a rigid body in 2-D static equilibrium:

\[ \sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_O = 0 \]

You have three independent equations. A statically determinate problem has at most three unknown support reactions solvable from them. Pick the moment center to eliminate unknowns when possible — often at a support where two reactions meet.

Support Reactions

Replace ideal supports with the reactions they can exert on the member:

Pin: two force components. Roller: one normal reaction. Fixed: two forces plus a reaction moment.

Exam habit

Draw reaction arrows on the body — the force the support exerts on the member. A roller on a horizontal surface carries vertical load only; a pin at a wall carries horizontal and vertical components.

Two-Force and Three-Force Members

A two-force member (straight cable, strut, or link) is loaded only at its two ends. The internal force acts along the line joining those points. Resolve \(T \sin\theta = W\) for a cable supporting a hanging weight.

The axial force \(T\) lies along the member; its vertical component balances \(W\).

A three-force member in equilibrium carries exactly three non-parallel forces — their lines of action must be concurrent. Use this to find an unknown force direction without full equilibrium algebra.

Common mistakes

Forgetting a reaction component at a pin. Drawing weight as a horizontal force. Taking moments about a point without counting every force that creates moment about that point. Mixing up the force the member exerts on a support vs. the reaction on the member.

Example: Simply Supported Beam (Symmetric Load)

Pin and roller at ends

A pin at \(A\) and a roller at \(B\) support a beam with a \(600\,\text{N}\) downward load at midspan (\(L = 4\,\text{m}\)). Find \(R_A\) and \(R_B\).

Pin at \(A\), roller at \(B\), point load \(P\) at midspan.

Solution. By vertical symmetry (load at center):

\[ R_A = R_B = \frac{600}{2} = 300\,\text{N} \]

Each support carries 300 N upward.

Example: Beam with Off-Center Load

Asymmetric loading

A \(6\,\text{m}\) beam has pin support at \(A\) and roller at \(B\). A \(12\,\text{kN}\) downward load acts \(2\,\text{m}\) from \(A\). Find \(R_B\) using moments about \(A\).

Take \(\sum M_A = 0\) to find \(R_B\) without knowing \(R_A\) first.

Solution. With \(P = 12\,\text{kN}\) at \(a = 2\,\text{m}\) and \(L = 6\,\text{m}\):

\[ \sum M_A = 0 \Rightarrow R_B \cdot L = P \cdot a \Rightarrow R_B = \frac{12 \times 2}{6} = 4\,\text{kN} \]

Then from \(\sum F_y = 0\): \(R_A = 12 - 4 = \mathbf{8\,\text{kN}}\) upward.

Example: Cantilever Reaction Moment

Fixed support at the wall

A cantilever beam (\(L = 3\,\text{m}\)) is built into a wall at \(A\). A downward load \(P = 800\,\text{N}\) acts \(1.5\,\text{m}\) from the wall. Find the reaction moment \(M_A\) at the fixed support.

The fixed support provides \(R_A\), horizontal restraint if needed, and \(M_A\).

Solution. Summing moments about \(A\) (CCW positive):

\[ M_A = P \cdot a = 800 \times 1.5 = 1200\,\text{N}\cdot\text{m} \]

The wall must provide a reaction moment of 1200 N·m (direction per your sign convention).

Work through six FBD problems — beam reactions, moment balances, cantilevers, and two-force members — with diagrams and full solutions.

Go to Practice →