Applied Mechanics and Structural Equilibrium

Introduction

Structural analysis begins with a deceptively simple question: how do applied forces produce motion, deformation, or internal resistance in a structure? A beam, column, truss, slab, or frame remains safe only when its applied loads are transferred through a continuous load path to the supports and foundations. Applied mechanics provides the language for describing this process, while equilibrium provides the mathematical conditions required for a structure to remain at rest.

This chapter develops the essential tools used in structural analysis: vectors, force systems, free-body diagrams, equilibrium equations, support reactions, internal actions, compatibility, constitutive relationships, section properties, and sign conventions. The emphasis is on idealized two-dimensional and three-dimensional structural systems used in construction technology.

A structural model is always an approximation of a real object. The analyst must decide which loads are significant, how supports behave, whether members are rigid or deformable, and whether the response is linear. Good analysis therefore requires both mathematical competence and engineering judgment.

Learning Objectives

By the end of this chapter, a student should be able to:

1. Represent forces and moments as vectors in two and three dimensions.
2. Construct clear and complete free-body diagrams.
3. Calculate support reactions using equilibrium equations.
4. Distinguish equilibrium, compatibility, and constitutive relationships.
5. Determine centroids, areas, second moments of area, and section moduli for common sections.
6. Apply consistent sign conventions to shear force, bending moment, axial force, and torsion.
7. Recognize determinate, indeterminate, stable, and unstable structural systems.
8. Check structural calculations using units, alternative methods, and physical reasoning.
9. Relate analytical results to construction practice, safety, codes, and professional responsibility.

1. The Structural Idealization

A real structure contains materials, connections, foundations, imperfections, construction tolerances, and variable loads. Structural analysis simplifies this reality into an idealized model consisting of:

• Geometry: lengths, angles, member axes, and connection locations.
• Loads: concentrated forces, distributed loads, moments, temperature effects, and imposed displacements.
• Supports: idealized restraints that prevent selected translations or rotations.
• Members: beams, columns, truss bars, slabs, cables, or rigid bodies.
• Material behavior: relationships between stress, strain, force, and deformation.

The model must be sufficiently simple to solve but sufficiently accurate to represent the behavior relevant to the design decision. For example, a simply supported beam may be modeled with a pin at one end and a roller at the other. This model captures vertical reactions and bending but does not describe local connection deformation or foundation settlement.

External and Internal Actions

External actions include applied loads and support reactions. Internal actions develop within a member to maintain equilibrium. At a cut through a beam, the internal actions may include:

• axial force N,
• shear force V,
• bending moment M,
• torsional moment T.

The relationship between external loading and internal actions is central to structural analysis. Internal actions are not additional applied loads; they are the result of the material transmitting force across a section.

2. Scalars, Vectors, and Coordinate Systems

A scalar has magnitude only, such as mass, temperature, length, or area. A vector has both magnitude and direction. Force, displacement, velocity, acceleration, and moment are vector quantities.

In a Cartesian coordinate system, a force vector may be written as:

where Fₓ, Fᵧ, and F_z are the rectangular components of force and 𝐢, 𝐣, and 𝐤 are unit vectors in the x, y, and z directions. The magnitude is:

The formula applies to an orthogonal Cartesian coordinate system and is dimensionally consistent because all components have units of force.

For a two-dimensional force:

If a force F acts at angle θ measured counterclockwise from the positive x-axis, its components are:

The angle convention must be stated because measuring θ from the y-axis changes the component expressions.

Position Vectors

The position vector from point A to point B is:

Position vectors are used to calculate moments and define member geometry. Length units must be consistent with the units used for force and moment.

Unit Vectors Along a Line

If a force acts along the line from A to B, the unit vector is:

A force of magnitude F along that line is then:

This method is particularly useful for three-dimensional cables, braces, and struts.

3. Moments and Couples

The moment of a force about a point measures its tendency to cause rotation. In vector form:

where 𝐫 is the position vector from point O to any point on the force’s line of action, 𝐅 is the force vector, and × denotes the vector cross product. The units are force × length, such as N·m or kN·m.

The magnitude is:

where φ is the angle between 𝐫 and 𝐅. Equivalently:

where d is the perpendicular distance from O to the force’s line of action.

In two-dimensional analysis, a counterclockwise moment is commonly taken as positive and a clockwise moment as negative. The choice is conventional; consistency is essential.

A couple consists of two equal, opposite, and parallel forces separated by a distance. Its resultant force is zero, but its moment is not zero. The couple moment is:

A couple moment is a free vector: its effect is the same about any point, provided the forces form a true couple.

Varignon’s Theorem

The moment of a force about a point equals the sum of the moments of its components about that point:

This theorem permits a force to be resolved into components before calculating its moment.

4. Force Systems and Resultants

A system of forces can often be replaced by an equivalent resultant force and resultant moment. The resultant force is:

The resultant moment about point O is:

where 𝐫ᵢ is the position vector to the point of application of force 𝐅ᵢ and 𝐌ᵢ represents applied couples.

For a uniformly distributed load w over length L, the equivalent concentrated force is:

It acts through the centroid of the load distribution, which for a uniform rectangular distribution is at the midpoint of the loaded length. The units are consistent: force per length multiplied by length gives force.

For a triangular load varying from zero to maximum intensity w₀ over length L:

The resultant acts one-third of the base length from the larger-intensity end, or two-thirds from the zero-intensity end. This location is determined from the centroid of a triangle.

5. Free-Body Diagrams

A free-body diagram (FBD) isolates a body or structural component from its surroundings and replaces every external interaction with a force or moment. It is the most important communication and reasoning tool in elementary structural analysis.

A reliable procedure is:

1. Define the body being isolated.
2. Draw its geometry and dimensions.
3. Remove all supports and connections at the boundary.
4. Replace each removed interaction with the appropriate reaction.
5. Add applied loads, self-weight, and applied couples.
6. Establish coordinate axes and a positive moment direction.
7. Label unknowns clearly and include units in the calculation.

Common idealized supports in a planar system are:

A support reaction should never be omitted merely because its expected value is zero. It must first be included in the FBD and then determined from the equations.

Figure: Typical free-body diagram components for a simply supported beam. The arrows shown are assumed directions; a negative calculated value indicates the actual direction is opposite.

6. Equations of Equilibrium

A body is in static equilibrium when its translational and rotational accelerations are zero. For a coplanar rigid body, the independent equilibrium equations are:

These equations apply to a rigid body in a plane under static loading. They do not by themselves determine deformation or material stress distribution.

For a three-dimensional rigid body, six independent equations are available:

The equations may be applied to an entire structure, a member, or a portion exposed by an imaginary cut.

Determinacy and Stability

A planar rigid body with three independent reaction components can often be solved using the three equilibrium equations. Such a structure is statically determinate if all reactions and internal actions can be found using equilibrium alone.

A structure is unstable if its restraints do not prevent a possible rigid-body motion. A structure is statically indeterminate if it has more unknown reactions or internal actions than can be determined from equilibrium alone. Indeterminate structures require additional compatibility and constitutive relationships.

Counting unknowns alone is not sufficient. Geometry and support arrangement must also be checked. Three reactions may still be ineffective if their lines of action are parallel or concurrent in a way that permits rotation or translation.

7. Worked Example: Reactions of a Simply Supported Beam

Problem. A 6 m simply supported beam has a pin at A and a roller at B. A 12 kN downward point load acts 2 m from A, and a uniformly distributed load of 4 kN/m acts over the full span. Determine the support reactions.

Step 1: Replace the distributed load.

The equivalent load is:

It acts at the beam midpoint, 3 m from A.

Step 2: Identify unknown reactions.

At A: Aₓ and Aᵧ. At B: Bᵧ. Since there are no horizontal applied loads, Aₓ is expected to be zero, but it is retained initially.

Step 3: Take moments about A.

Step 4: Sum vertical forces.

Step 5: Sum horizontal forces.

Therefore, the reactions are Aₓ = 0, Aᵧ = 20 kN upward, and Bᵧ = 16 kN upward. A check confirms that upward reactions total 36 kN, equal to the total downward load of 36 kN.

8. Internal Forces and Sign Conventions

After support reactions are found, a member may be cut at a selected section. The exposed portion must be placed in equilibrium by introducing internal force and moment resultants.

A widely used beam convention is:

• positive axial force: tension,
• positive shear: upward on the left face of a cut section,
• positive bending moment: sagging, producing tension at the bottom fibers,
• positive torsion: defined by the right-hand rule about the member axis.

Conventions vary between textbooks and software. The important requirement is that the convention be stated and used consistently.

For a beam segment, the differential relationships are commonly written as:

Here w(x) is the distributed load intensity, V(x) is shear force, and M(x) is bending moment. The signs depend on the selected convention. Under the convention shown, a positive downward load is represented by positive w in the first equation.

These relationships imply that the slope of the shear-force diagram is related to load intensity, while the slope of the bending-moment diagram is related to shear. A concentrated force causes a jump in the shear diagram; a concentrated moment causes a jump in the bending-moment diagram.

9. Compatibility and Constitutive Relationships

Equilibrium describes force balance. It does not describe how much a structure deforms. Two additional ideas are required for deformable structures.

Compatibility

Compatibility requires that connected parts deform in a geometrically consistent manner. For example, a continuous beam has compatible rotations and deflections at a connection, while a truss joint must have one common displacement even though several members meet there.

Constitutive Relationships

A constitutive relationship connects an action to deformation through material or member properties. For a linearly elastic prismatic bar:

where δ is axial deformation, P is axial force, L is original length, A is cross-sectional area, and E is Young’s modulus. The assumptions are small deformation, uniform axial force, constant area, linear elasticity, and an axially aligned load. Units are consistent because N·m/(m²·N/m²) gives m.

For normal stress under concentric axial load:

This expression assumes the load is applied through the centroidal axis and that local effects, bending, and stress concentrations are neglected.

For a linearly elastic material:

where σ is normal stress and ε is normal strain. This law applies only within the elastic, approximately linear range of the material.

Structural analysis of indeterminate systems requires all three categories:

1. equilibrium equations,
2. compatibility equations,
3. constitutive relationships.

Ignoring compatibility can produce a force distribution that satisfies equilibrium but is physically impossible.

10. Section Properties

Section properties describe how area is distributed around an axis. They are essential for evaluating axial resistance, bending stiffness, deflection, and buckling behavior.

Centroid

For a composite area divided into parts:

where Aᵢ is the area of component i and xᵢ, yᵢ are the coordinates of its centroid. Areas removed from a section are assigned negative values.

Second Moment of Area

The second moment of area about the x-axis is:

and about the y-axis:

The units are length⁴, such as mm⁴ or m⁴. These are geometric properties, not material properties.

For a rectangle of width b and depth h about its centroidal horizontal axis:

The cubic dependence on depth explains why increasing beam depth is highly effective in resisting bending.

Parallel-Axis Theorem

If I_c is the second moment of area about a centroidal axis and d is the perpendicular distance to a parallel reference axis:

This theorem applies to plane areas and parallel axes. The term Ad² must have units of length⁴.

Bending Stress and Section Modulus

For elastic bending about a principal centroidal axis:

where M is bending moment, y is distance from the neutral axis, and I is the second moment of area. The maximum stress occurs at the greatest distance c from the neutral axis:

where the elastic section modulus is:

The formula assumes small deflection, linear elasticity, a prismatic member, and bending about a principal axis without significant axial force or shear interaction.

11. Three-Dimensional Equilibrium

Three-dimensional structural systems occur in space frames, towers, roof structures, crane systems, cable-supported structures, and foundations subject to eccentric loading. The same principles apply, but each force and moment has three components.

For a force applied at position 𝐫:

In determinant form:

The resulting components are:

The right-hand rule determines the positive direction of each moment component. A three-dimensional FBD should show a clear coordinate system, support reaction directions, member geometry, and all applied actions.

12. Applied Case: Preliminary Analysis of a Construction-Stage Beam

Case status: Illustrative educational case.

Background

A temporary steel beam supports formwork and construction materials during the casting of a floor slab. The beam spans 5 m between temporary supports. The estimated uniform construction load is 8 kN/m, including self-weight and formwork. A concentrated 10 kN material load acts 1.5 m from the left support.

Problem

Determine the support reactions and identify the maximum bending moment for preliminary design screening. This simplified calculation does not replace a code-based design review.

Analysis

The total distributed load is:

It acts at 2.5 m from the left support. Let the reactions be Aᵧ and Bᵧ.

Taking moments about A:

Vertical equilibrium gives:

For a uniformly loaded simply supported beam without the point load, the maximum moment would be wL²/8 = 25 kN·m. The point load modifies both the shear and moment diagrams. The maximum moment occurs where shear changes sign, so a complete analysis requires constructing the piecewise shear diagram.

For a section between A and the point load, the shear is:

At x = 1.5 m, V = 15 kN. Immediately to the right of the point load, the shear becomes 5 kN. For the region beyond the point load:

Setting V = 0 gives x = 2.125 m, which lies between 1.5 m and 5 m. The moment at that location is:

At x = 2.125 m:

Thus the estimated maximum sagging moment is approximately 34.6 kN·m.

Outcome and Professional Interpretation

The reaction calculation confirms vertical load balance, and the shear sign change identifies the location of maximum positive moment. Before the beam is used, an engineer must also check bending resistance, shear resistance, deflection, lateral stability, bearing at supports, connection capacity, construction-stage load combinations, and possible impact or accidental loads. Temporary works often present greater risk than permanent works because load paths and restraints change during construction.

Learning Points

1. Distributed loads must be replaced by forces acting at their load-distribution centroids.
2. Reactions are found from equilibrium, but member adequacy requires strength and serviceability checks.
3. The maximum moment is not always at midspan when concentrated loads are present.
4. Construction-stage conditions require separate structural models.

13. Common Errors and Best Practices

Common errors include:

• omitting self-weight or construction loads;
• placing a distributed-load resultant at the wrong location;
• using a force component with an incorrect sign;
• taking a moment using a non-perpendicular distance;
• including a support reaction that the support cannot provide;
• mixing N and kN, or mm and m, without conversion;
• confusing second moment of area with mass moment of inertia;
• changing sign convention between sections;
• treating a negative answer as an arithmetic failure rather than a reversed assumed direction;
• applying equilibrium equations to an incomplete or incorrect FBD.

Best practice is to draw the FBD before writing equations, keep units visible, use one consistent coordinate system, perform an independent equilibrium check, and assess whether the result is physically reasonable. Computer software should verify engineering reasoning, not replace it. A model can produce precise numbers while representing the wrong supports, loads, or geometry.

14. Safety, Codes, and Professional Responsibility

Structural calculations influence public safety. Engineers and construction professionals must work within applicable building codes, material standards, occupational safety regulations, and approved design procedures. Loads should be identified from the correct governing standard, including dead, imposed, wind, snow, seismic, thermal, construction, and accidental actions where relevant.

Ethical practice requires competent analysis, transparent assumptions, traceable calculations, independent checking where required, and prompt communication of unsafe conditions. A calculation should identify its scope and limitations. Preliminary equilibrium calculations must not be presented as a complete design if strength, stability, serviceability, durability, fire, connection, and foundation requirements have not been checked.

Digital structural-analysis and AI tools can assist with load organization, drawing interpretation, parametric studies, and error checking. However, they may misidentify supports, omit load cases, or generate plausible but unsupported results. Human review, code compliance, version control, and documented assumptions remain essential.

Chapter Summary

Applied mechanics provides the concepts and mathematical tools for representing structural actions. Forces are vectors with magnitude and direction; moments describe rotational effects and are calculated using the cross product or a perpendicular lever arm. Distributed loads are replaced by equivalent resultants acting through the centroids of their distributions.

Free-body diagrams isolate bodies and show all applied loads, support reactions, dimensions, and coordinate directions. For planar rigid bodies, equilibrium requires the sums of horizontal forces, vertical forces, and moments to be zero. Three-dimensional systems require three force and three moment equations.

Equilibrium alone determines statically determinate reactions, but indeterminate structures additionally require compatibility and constitutive relationships. Section properties such as centroid, second moment of area, and section modulus govern the distribution of stress and resistance to bending. Consistent sign conventions and unit checks are essential for reliable analysis.

In professional practice, structural analysis is inseparable from judgment. The analyst must select a suitable idealization, recognize construction-stage conditions, check stability and serviceability, follow applicable codes, and communicate assumptions clearly.

Key Terms

Vector: A quantity having both magnitude and direction.

Scalar: A quantity having magnitude but no direction.

Resultant: A single force or force–moment system equivalent to several actions.

Moment: The rotational effect of a force about a point or axis.

Couple: Two equal, opposite, parallel forces separated by a distance.

Free-body diagram: A diagram of an isolated body showing all external actions and reactions.

Equilibrium: A condition in which the resultant force and resultant moment are zero.

Reaction: An action exerted by a support or connection on a structure.

Support: An idealized restraint that limits one or more rigid-body motions.

Axial force: Internal force acting along a member’s longitudinal axis.

Shear force: Internal force acting transverse to a member’s longitudinal axis.

Bending moment: Internal moment associated with curvature of a member.

Torsion: Twisting action about a member’s longitudinal axis.

Compatibility: Geometric requirement that connected structural parts deform consistently.

Constitutive relationship: Relationship between force or stress and deformation or strain.

Centroid: The geometric center of an area or load distribution.

Second moment of area: A geometric measure of area distribution about an axis.

Section modulus: The ratio I/c used to relate bending moment to extreme-fiber stress.

Statically determinate: Solvable using equilibrium equations alone.

Statically indeterminate: Requiring equilibrium, compatibility, and constitutive relationships.

Review Questions

Short Answer Questions

1. What is the difference between a scalar and a vector?
2. What information must be included on a complete free-body diagram?
3. State the three independent equilibrium equations for a planar rigid body.
4. Where does the resultant of a uniformly distributed load act?
5. Distinguish equilibrium from compatibility.

Long Answer Questions

1. Explain how a force is resolved into Cartesian components and describe the importance of sign conventions.
2. Explain the procedure for determining support reactions in a simply supported beam.
3. Discuss the assumptions and limitations of the axial deformation relationship δ = PL/(AE).
4. Derive or explain the use of the parallel-axis theorem for a composite section.
5. Compare statically determinate and statically indeterminate structures, including the additional information required for indeterminate analysis.

Analytical / Application Questions

1. A 4 m beam carries a 6 kN point load at its midpoint and a 2 kN/m uniform load over the full span. Draw the FBD and determine the vertical reactions.
2. A rectangular section is 200 mm wide and 400 mm deep. Calculate its centroidal second moment of area about the horizontal axis and explain why depth strongly affects bending stiffness.
3. A three-dimensional brace runs between two known points and carries a tensile force. Describe how to determine its Cartesian force components and the moment it produces about a reference point.

Multiple Choice Questions with Answers

Question 1

Which set of equations describes static equilibrium of a planar rigid body?

A. ∑Fₓ = 0, ∑Fᵧ = 0, and ∑M = 0

B. ∑Fₓ = 1, ∑Fᵧ = 1, and ∑M = 1

C. Only ∑Fᵧ = 0

D. Only ∑M = 0

Correct Answer: A. ∑Fₓ = 0, ∑Fᵧ = 0, and ∑M = 0

Question 2

A uniform load of 5 kN/m acts over 4 m. What is its equivalent concentrated load?

A. 1.25 kN

B. 9 kN

C. 20 kN

D. 25 kN

Correct Answer: C. 20 kN

Question 3

Where does the resultant of a uniform load over a straight span act?

A. At the loaded span’s midpoint

B. At the support with the larger reaction

C. At the zero-load end

D. At one-third of the span from either end

Correct Answer: A. At the loaded span’s midpoint

Question 4

What does a roller support normally provide in a planar beam model?

A. A force reaction normal to its supporting surface

B. Two force reactions and a moment

C. A resisting couple only

D. No reaction

Correct Answer: A. A force reaction normal to its supporting surface

Question 5

The magnitude of the moment of a force about a point is calculated as:

A. Force divided by distance

B. Force multiplied by perpendicular distance

C. Distance divided by force

D. Force multiplied by area

Correct Answer: B. Force multiplied by perpendicular distance

Question 6

For a rectangular section of width b and depth h, the centroidal second moment of area about its horizontal axis is:

A. bh/12

B. b²h/12

C. bh³/12

D. bh²/12

Correct Answer: C. bh³/12

Question 7

Which additional concepts are required, besides equilibrium, for a statically indeterminate structure?

A. Compatibility and constitutive relationships

B. Only a larger load factor

C. Only the material density

D. Only the member length

Correct Answer: A. Compatibility and constitutive relationships

Question 8

If a calculated support reaction is negative, what does it usually indicate?

A. The equilibrium equations are automatically invalid

B. The assumed direction of the reaction is opposite to the actual direction

C. The load has no effect

D. The structure has zero weight

Correct Answer: B. The assumed direction of the reaction is opposite to the actual direction

Question 9

For an axially loaded elastic prismatic bar, which relationship gives axial deformation under the stated assumptions?

A. δ = AE/(PL)

B. δ = PL/(AE)

C. δ = PAE/L

D. δ = L/(PAE)

Correct Answer: B. δ = PL/(AE)

Question 10

In a beam, the location of maximum bending moment commonly occurs where:

A. The bending moment is always zero

B. The shear force changes sign or becomes zero

C. The distributed load is necessarily zero

D. The axial force is maximum

Correct Answer: B. The shear force changes sign or becomes zero