Learning Outcomes

Upon completing this lesson, you will be able to:

• State the two conditions and three equations required for the equilibrium of a 2D rigid body.
• Identify common types of 2D supports and the reactions they provide.
• Differentiate between various types of loads, including distributed loads.
• Draw a complete and accurate Free-Body Diagram (FBD) for a rigid body.
• Identify two-force and three-force members to simplify analysis.

Topics Covered

• Supports and Reactions
• Loads
• Equations of Equilibrium
• Two- and Three-Force Members

I. Supports and Reactions

Supports are what connect a structure to the ground or to other objects.

They prevent motion by exerting reactions, which are the forces and moments that hold the structure in place.

To analyze a rigid body, we must replace its physical supports with the specific reactions they provide. In 2D, the three most common supports are:

1. Roller: A roller allows rotation and translation parallel to the surface it rests on. It only prevents translation perpendicular to the surface. It provides one reaction force that is perpendicular to the surface.

1. Pin (or Hinge): A pin prevents all translation but allows free rotation. It provides two reaction forces, typically represented by their horizontal $(A_x)$ and vertical $(A_y)$ components.

1. Fixed Support: A fixed support (like a beam embedded in a wall) prevents all motion. It prevents translation in both directions and prevents rotation. It provides three reactions: two force components $(A_x)$ and $(A_y)$ and a couple moment $(M_A)$.

II. Loads

Loads are the external forces and moments that are applied to a structure. These can be simple forces at a point or forces spread out over a length or area. Loads are the known forces and moments acting on a body.

$$ F_R=\int_Lw(x)\ dx =\int_AdA=A $$

Common types include:

• Concentrated Force: A force applied at a single point.
• Concentrated Moment (Couple): A pure turning effect applied to the body.
• Distributed Load: A force spread out over a length or area, measured in force per unit length (e.g., $N/m$). To analyze a distributed load, we must replace it with an equivalent single resultant force.
  • The magnitude of the resultant force is equal to the area under the distributed load curve.
  • The location of this resultant force is at the geometric center (centroid) of that area. For a rectangle, this is the midpoint; for a triangle, it is one-third of the way from the tall end.

Example 1

The granular material exerts the distributed loading on the beam as shown. Determine the magnitude and location of the equivalent resultant of this load.

The area of the loading diagram is a trapezoid, and therefore the solution can be obtained directly from the area and centroid formulas for a trapezoid listed on the inside back cover. Since these formulas are not easily remembered, instead we will solve this problem by using “composite” areas. Here we will divide the trapezoidal loading into a rectangular and triangular loading as shown. The magnitude of the force represented by each of these loadings is equal to its associated area,

$$ F_1=\frac{1}{2}(9\text{ ft})(50\text{ lb/ft})=225\text{ lb} \\ F_2=(9\text{ ft})(50\text{ lb/ft})=450\text{ lb} $$

The lines of action of these parallel forces act through the respective centroids of their associated areas and therefore intersect the beam at

$$ \bar{x}_1=\frac{1}{3}(9\text{ ft})=3\text{ ft} \\ \bar{x}_1=\frac{1}{2}(9\text{ ft})=4.5\text{ ft} $$

The two parallel forces $F_1$ and $F_2$ can be reduced to a single resultant $FR$. The magnitude of $F_R$ is

$$ F_R=225 + 450 = 675\text{ lb} $$

We can find the location of $F_R$ with reference to point $A$. We require

$$ \bar{x}(675)=3(225)+4.5(450) \\ \bar{x} =4 \text{ ft} $$

III. Equations of Equilibrium

For a rigid body to be in equilibrium (not moving or rotating), the sum of all forces in the x-direction must be zero, the sum of all forces in the y-direction must be zero, and the sum of all moments (turning effects) about any point must be zero.

The conditions for equilibrium of a rigid body provide us with three independent scalar equations for 2D problems. These are: