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4 Equilibrium of Bodies (2D)
Earlier chapters have introduced mechanical quantities and analytical tools for solving engineering problems: forces, moments, couples,
calculation of resultants.
Most engineering structures rest in state of equilibrium, i.e. resultants of
forces and moments are balanced - equal to zero.
Hence, very important equations in statics are Equilibrium Equations
(EE’s).
Resultant Force: ̅ = 𝟎
Resultant Moment: 𝑴 ̅ = 𝟎 – about any point
EE’s are necessary and sufficient conditions of equilibrium.
Despite their seeming simplicity, their application to a specific problem might be complex. By experience, it is one of the most difficult topics in statics.
There are two reasons for aforementioned complexity, apart from sheer number of bodies, forces and moments:
•
There are multiple ways a force can be applied to the body.
Examples: through frictionless contacts or contacts with friction
(rough surfaces), cables, roller supports, pin or built-in supports, etc. The type of support often determines the direction of support, hence reducing a number of unknowns.
•
The presence of action/reaction forces. From Newton’s third law one knows that forces always come in pairs. Hence, they should be carefully distinguished. The primary interest in statics – external
forces.

Types of Support and Reactions
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Additional Examples
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Basic Equilibrium Problems
In a typical problem, structural geometry and some external forces are given. One has to find unknown forces using Equilibrium Equations.
A technique to account for all forces is body isolation or construction of
Free Body Diagram:
•
Select the body or group of bodies
•
Isolate it (draw a complete external boundary)
•
Indicate all external forces
•
Choose coordinates
Construction of FBD is of an extreme importance for problem solving and should be mastered.
FBD, Example 1

Example 2
Example 3
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Example 4
FBD Exercises
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2D Equilibrium
In 2D two vector EE’s reduce to a system of three scalar EE’s for components:
∑ 𝑭 𝒙
= 𝟎, ∑ 𝑭 𝒚
= 𝟎, ∑ 𝑴
𝑶
= 𝟎
Note: 𝑶 is an arbitrarily selected point, on or even off the body.
Important: Instead of the set of equations above we can reformulate problem in one force and two moment equations, or three moment equations about different points.

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Example of equivalent systems of equations:
∑ 𝑭 𝒙
= 𝟎, ∑ 𝑭 𝒚
= 𝟎, ∑ 𝑴
𝑨
= 𝟎
∑ 𝑭 𝒙
= 𝟎, ∑ 𝑴
𝑨
= 𝟎, ∑ 𝑴
𝑩
= 𝟎
Point B cannot lie on a line parallel to axis y that passes through point A.
Nonetheless, the maximum number of independent
equations in 2D remains equal to three, i.e. suitable for finding no more than 3 unknowns. Problem with more unknowns is called statically indetermined.
Special Equilibrium Cases
1.
Collinear forces. Only one EE is needed: ∑ 𝑭 𝒙 satisfied by default.
= 𝟎 . Two others are
2.
Concurrent forces. Two EE’s: ∑ 𝑭 𝒙 balanced by default.
= 𝟎 , ∑ 𝑭 𝒚
= 𝟎 . Moments are
3.
Parallel forces. Two EE’s
𝑭 𝒙 default.
= 𝟎 , ∑ 𝑴
𝑶
= 𝟎 . ∑ 𝑭 𝒚
= 𝟎 by

Summary of Special Cases
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Techniques for Simplified Solution of Equilibrium Problems
Elimination of an unknown force or forces from EE’s can be done via two simple techniques:
1.
From force EE: analyze force equilibrium in a direction perpendicular to forces being eliminated.
2.
From moment EE: analyze moment equilibrium about a point on a line of action of the forces being eliminated.

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Often, application of these techniques results in breaking of a system of EE’s into independent/separate equations – significant help with solving for unknowns, particularly reducing probability of making computational errors.
Example: Figure to the right.
Equilibrium of forces along the X axis eliminates the responses at A and B from the equations. Equilibrium of moments around C eliminates reactions at A and D from the equations. Equilibrium of moments around D eliminates reactions at B and D from the equations.
The result: ∑ 𝑭 𝒙 single unknown.
= 𝟎, ∑ 𝑴
𝑪
= 𝟎, ∑ 𝑴
𝑫
= 𝟎 is a set of equations with a
2D Equilibrium, Example 1

Example 2
Example 3
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Example 4
Example 5
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Example 6
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Example 7
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Example 8

Example 9
Statically Indeterminate Reactions and Partial Constraints
When a body is given supports that result in a situation where under loading conditions the body cannot move, the body is considered
completely constrained.
When the number of unknown reactions correspond to the number of independent equilibrium equations and all the unknowns can be determined from these equations, the reactions are said to be statically
determined.
When the number of unknown reactions is greater than the number of independent equilibrium equations, the reactions that cannot be determined are said to be statically indeterminate.
In the figure to the right, the number of unknown reaction components (4) is greater than the number of independent equilibrium equations (3). While the
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vertical components of the reactions can be determined from moment equilibrium around A and B, only the sum of the horizontal components can be determined from the force equilibrium along the X axis.
In the figure to the right, the supports available cannot generally prevent movement along the X axis (except for cases where the forces P, Q and S are vertical, or their X components cancel each other out). This means that the equilibrium cannot be maintained under general loading conditions, and the body is said to be not completely constrained.
Note: In the above case, the number of unknowns (2) is smaller than the number of independent equations (3).
For the body to be completely constrained and for the reactions to be statically determined, the number of unknowns has to correspond to the number of independent equilibrium equations. However, while this condition is necessary, it is not sufficient.
In the example to the right, the number of unknown reactions (3) corresponds to the number of equations, but the truss is free to move horizontally, making it
improperly constrained. Since only two additional equations remain to determine three reactions, the reactions are also statically indeterminate.
An additional example of improperly constrained structure with statically indeterminate reactions can be seen to the right. Since there are four unknown reaction components and only three equations, the reactions will
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Conclusion: the body will be improperly constrained, even if the supports provide a sufficient number of reactions, if all the reactions are either concurrent or parallel.
For the body to be completely constrained and the reactions to be statically determined, the number of unknowns must correspond to the number of equations, and the reactions cannot be all either parallel or concurrent.
While improperly constrained or statically indeterminate designs should be used with care, such structures can sometimes maintain equilibrium under special conditions, and the reactions can be partially solved using statics.
Two-Force Bodies
Definition: Two-force body is a body with any number of forces applied at only two points in the absence of applied couples.
Consequences: Resultant forces at two points of forces applications are collinear, equal and opposite with line of their action passing through those two points.
Proof: Equilibrium of forces ̅ = 𝟎 leads to 𝑭
𝟏
= −𝑭
𝟐
.

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Equilibrium of moments ∑ 𝑴 = 𝒅𝑭 = 𝟎 leads to collinearity, otherwise forces form unbalanced moment
Three-Force Bodies
Definition: Three-force body is a body with any number of forces applied at only three points in the absence of applied couples.
Consequences: Resultant forces at two points of forces applications are concurrent – have lines of action intersection at one point.
Proof: From the opposite: let’s say lines of action do not intersect in one point. Balance of moments around A. ∑ 𝑴
𝑨
= 𝑭 means d has to be zero, i.e. they have to intersect.
𝟑 𝒅 = 𝟎 . That
Consequence: Knowing two lines of action leads to the third one;
Exception: Parallel forces (i.e. lines of action do not intersect at all)

Example 1
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Example 2
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