Moments, Free-Body Diagrams and Equilibrium

Moments of Force & Varignon's Theorem

Definition of a Moment (Torque)

A moment of force measures the rotational tendency of a force about a specific pivot point or rotational axis. It is defined as the product of the magnitude of the force ($F$) and the perpendicular distance ($d_{\perp}$) from the pivot point to the line of action of the force:

$$\text{Moment } (M) = F \times d_{\perp}$$

• SI Unit: Newton-meter ($\text{N}\cdot\text{m}$). Note: Although dimensionally identical to the joule ($\text{J} = \text{N}\cdot\text{m}$), moments and torques are strictly stated in $\text{N}\cdot\text{m}$ to avoid confusion with work and energy.
• Sign Convention: Rotations in the counterclockwise (CCW) direction are conventionally assigned a positive (+) sign; clockwise (CW) rotations are assigned a negative (-) sign.

Varignon's Theorem (Principle of Moments)

Varignon's Theorem states that the moment of a force about any axis is equal to the algebraic sum of the moments of its components about the same axis. This principle allows engineers to break an angled force into horizontal and vertical components, compute their individual moments about a pivot, and sum them algebraically.

Conditions of Static Equilibrium

For a mechanical body (such as a pipe hanger, boiler support beam, or turbine skid) to remain at rest in static equilibrium, it must experience neither linear acceleration nor angular acceleration. According to Newton's laws, this requires satisfying three coplanar equations of equilibrium:

1. Horizontal Force Balance:

$$\sum F_x = 0$$

1. Vertical Force Balance:

$$\sum F_y = 0$$

1. Rotational Moment Balance:

$$\sum M_O = 0 \quad \text{(about any arbitrary point } O \text{)}$$

Center of Gravity & Equipment Stability

The center of gravity (CG) is the unique point through which the entire gravitational weight of a body acts. The position of the CG determines the stability of plant machinery during transport and lifting:

• Stable Equilibrium: A slight displacement causes the CG to rise, creating a restoring moment that returns the equipment to its resting base (e.g., broad-base pump skid).
• Unstable Equilibrium: A slight displacement causes the CG to drop, generating an overturning moment that topples the equipment (e.g., a tall vertical vessel hoisted without outriggers).
• Neutral Equilibrium: The CG remains at the same elevation when displaced (e.g., a cylindrical drum rolling along a flat surface).

Lifting geometry: The hook/load-centre alignment is one consideration in the approved lifting plan. Stability depends on the centre of gravity, attachment geometry and the approved lifting arrangement; a simple centre-of-gravity rule cannot establish stability.

Example 2: Simply Supported Beam Reactions

Problem: A horizontal structural steel beam $6.0\text{ m}$ long is simply supported at its left end ($A$) and right end ($B$). The beam supports an isolated valve weighing $18\text{ kN}$ located $1.5\text{ m}$ from support $A$, and a pipe strainer weighing $12\text{ kN}$ located $4.5\text{ m}$ from support $A$. Neglecting the weight of the beam, determine the vertical reaction forces at support $A$ ($R_A$) and support $B$ ($R_B$).

Solution:

1. Apply moment equilibrium about support $A$ ($\sum M_A = 0$): Clockwise moments created by loads = Counterclockwise moment created by $R_B$:

$$(18\text{ kN} \times 1.5\text{ m}) + (12\text{ kN} \times 4.5\text{ m}) - (R_B \times 6.0\text{ m}) = 0$$ $$27.0\text{ kN}\cdot\text{m} + 54.0\text{ kN}\cdot\text{m} = 6.0 R_B$$ $$81.0\text{ kN}\cdot\text{m} = 6.0 R_B \implies R_B = \frac{81.0}{6.0} = 13.5\text{ kN}$$

1. Apply vertical force equilibrium ($\sum F_y = 0$):

$$R_A + R_B - 18\text{ kN} - 12\text{ kN} = 0$$ $$R_A + 13.5\text{ kN} - 30.0\text{ kN} = 0 \implies R_A = 30.0 - 13.5 = 16.5\text{ kN}$$

1. Verification via moment equilibrium about support $B$ ($\sum M_B = 0$):

$$(R_A \times 6.0) - (18 \times 4.5) - (12 \times 1.5) = (16.5 \times 6.0) - 81.0 - 18.0 = 99.0 - 99.0 = 0$$

The calculated reactions of $R_A = 16.5\text{ kN}$ and $R_B = 13.5\text{ kN}$ satisfy all equilibrium criteria.

Build the free-body diagram before calculating

Isolate the beam, valve support or bracket from its surroundings. Show every external load and support reaction, including self-weight when supplied. Internal forces between parts of the isolated assembly cancel in the whole-assembly diagram; they reappear when a component is isolated separately. A pin support can supply two force components, while an ideal roller supplies a reaction perpendicular to its surface. Do not invent a fixing moment for a freely rotating pin.

Choose a pivot that eliminates an unknown reaction from the moment equation. Use the perpendicular distance from pivot to the force’s line of action, rather than the length of an angled handle. A 200 N force perpendicular to a 0.30 m wrench gives 60 N·m; the same force at 30 degrees to the wrench gives only 200 × 0.30 × sin 30 degrees = 30 N·m. Keep one sign convention throughout. The beam example above writes clockwise loads as positive; reversing all signs produces the same reactions.

After finding reactions, substitute them into both force balance and a moment balance about another point. Negative values indicate that an assumed reaction direction was reversed, not that a magnitude should be silently made positive. Equilibrium establishes the necessary load balance, but does not prove that a beam, anchor or foundation has adequate strength or stiffness.

Source notes

Official scope and technical references checked 2026-10-17: Reference 1; Reference 2. Archived DOE handbooks are used for physical principles, not current Canadian legal requirements.