Equilibrium of Rigid Bodies

Why it matters

Understanding the equilibrium of rigid bodies is crucial in designing structures and mechanical systems that remain stable under various loads. This knowledge ensures that bridges, buildings, and machinery operate safely and efficiently without collapsing or malfunctioning.

Key ideas

• Rigid Body: An idealization where deformation is negligible, and the distance between any two points remains constant.
• Equilibrium Conditions: For a rigid body to be in equilibrium, the sum of all forces and the sum of all moments acting on it must be zero.
  • Translational Equilibrium: ΣF = 0 (sum of all forces is zero).
  • Rotational Equilibrium: ΣM = 0 (sum of all moments about any point is zero).
• Free Body Diagram (FBD): A graphical representation used to visualize the forces and moments acting on a body.
• Support Reactions: Forces and moments developed at the supports or connections to maintain equilibrium.

Supports and scope

For planar static equilibrium there are three independent scalar equations, listed below. In three dimensions there are six: three force components and three moment components. Equilibrium is necessary but does not by itself establish structural stability or determine statically indeterminate reactions.

A smooth planar roller supplies one reaction normal to its supporting surface. A pin supplies two force components but no couple. A fixed planar support can supply two force components and a reaction couple. Replace each support with the actions it can actually exert.

A distributed load is replaced, for overall equilibrium, by its resultant acting at the centroid of the load diagram. This replacement does not reproduce internal shear and moment at every section.

Formulas

• ΣF_x = 0: Sum of horizontal forces.
• ΣF_y = 0: Sum of vertical forces.
• ΣM = 0: Sum of moments about a point.

Worked example

Problem: A simply supported beam of length 6 m carries a downward uniformly distributed load of 2 kN/m over its entire span. It has a pin at A and a roller on a horizontal surface at B; neglect self-weight beyond the stated load. Calculate the reactions at the supports.

Given:

• Length of beam, L = 6 m
• Uniform distributed load, w = 2 kN/m

1. Calculate total load:

   • Total load, W = w * L = 2 kN/m * 6 m = 12 kN

2. Assume reactions at supports A and B:

   • Let R_A and R_B be the reactions at supports A and B respectively.

3. Apply equilibrium equations:

   • ΣF_y = 0: R_A + R_B = 12 kN
   • ΣM_A = 0: R_B * 6 m - 12 kN * 3 m = 0
     • R_B = 6 kN
   • Substitute R_B in ΣF_y = 0: R_A + 6 kN = 12 kN
     • R_A = 6 kN

Answer: Reactions at supports are 6 kN upward at A and 6 kN upward at B. The horizontal reaction at A is zero.

Common mistakes

• Not drawing a complete Free Body Diagram (FBD) with all forces and moments.
• Forgetting to include support reactions in equilibrium equations.
• Incorrectly calculating the moment arm distances.

For GATE ME

Questions often involve calculating support reactions, analyzing trusses, or determining the stability of structures. Practice drawing FBDs and applying equilibrium equations to various configurations.

Quick check

1. What are the conditions for a rigid body to be in equilibrium?
2. How do you calculate the moment about a point?
3. What is the purpose of a Free Body Diagram?

Answers: 1. ΣF = 0 and ΣM = 0; 2. Moment = Force × Perpendicular distance; 3. To visualize forces and moments acting on a body.