Complete Course Notes on Engineering Statics

Fundamentals and Governing Laws of Statics

• Scope and Definition of Statics:

  • Statics is the branch of mechanics that analyzes physical systems in static equilibrium.

  • All systems and components analyzed in statics are assumed to be in a state of rest (zero velocity and zero acceleration: $\mathbf{v} = \mathbf{0}$, $\mathbf{a} = \mathbf{0}$).

  • All components are idealized as rigid bodies, meaning deformations under applied loads are entirely neglected. The analysis of moving bodies is addressed in Dynamics, while the deformation of bodies is covered in Mechanics of Materials.

  • Primary objective: determine the external reaction forces and moments holding a system in equilibrium, identify load paths across interconnected members, and compute the internal loads experienced within structural elements.

• Newton's Laws of Motion (Principia Mathematica, 1687):

  • First Law (Law of Inertia): In an inertial frame of reference, an object remains at rest or continues to move at a constant velocity unless acted upon by a net external force.

  • Qualitative static implication: If an object is at rest, the resultant of all forces acting upon it must be exactly zero.

  • Second Law (Fundamental Law of Dynamics): In an inertial frame of reference, the resultant force on an object equals the product of its mass and its acceleration:     $\mathbf{F} = m\mathbf{a}$

  • Quantitative static implication: For an object at rest, its acceleration is identically zero ($\mathbf{a} = \mathbf{0}$), which requires the resultant force vector to vanish ($\mathbf{F} = \mathbf{0}$).

  • Third Law (Action and Reaction): When one body exerts a force on a second body, the second body simultaneously exerts a force on the first body that is equal in magnitude and opposite in direction:     $\mathbf{F}_{\text{action}} = -\mathbf{F}_{\text{reaction}}$

Equilibrium of a Particle

• Definition of a Particle:

  • A particle is an idealized object of zero physical dimensions, treated as existing at a single geometric point in space.

  • Because a particle has no geometric extent, forces acting on it are concurrent (their lines of action intersect at the particle's position), precluding any tendency to rotate.

• Conditions of Particle Equilibrium:

  • Two collinear opposing forces: A particle subjected to two forces $\mathbf{F}_1$ (directed left) and $\mathbf{F}_2$ (directed right) is in equilibrium if and only if their scalar magnitudes are equal:     $F_1 = F_2$

  • Orthogonal force pairs: For two horizontal forces ($F_1, F_2$) and two vertical forces ($F_3, F_4$), equilibrium requires simultaneous independent balance along each axis:     $F_1 = F_2 \quad \text{and} \quad F_3 = F_4$

  • General multi-force systems: Equilibrium under an arbitrary number of concurrent forces requires the vector sum of all applied forces to equal zero.

• Cartesian Coordinate System and Equations of Equilibrium:

  • A standard right-handed Cartesian coordinate system $Oxy$ is established where $O$ is the origin, the positive $x$-axis points horizontally to the right, and the positive $y$-axis points vertically upward.

  • The choice of coordinate frame is arbitrary; physical equilibrium is independent of the chosen reference axes. However, a consistent Cartesian convention ensures systematic formulation.

  • In two dimensions, Newton's first law yields two independent scalar equations of equilibrium (force balances):     $\sum F_x \stackrel{+}{\rightarrow} = 0$     $\sum F_y \uparrow+ = 0$

  • These conditions are both necessary and sufficient: if both equations are satisfied, the particle is in static equilibrium.

• Force Decomposition into Rectangular Components:

  • An arbitrary force vector $\mathbf{F}$ with magnitude $F$ oriented at an angle $\theta$ measured counterclockwise from the positive horizontal $x$-axis is decomposed into orthogonal components:     $F_x = F \cos(\theta)$     $F_y = F \sin(\theta)$

• Worked Example 2.1 — Particle Subjected to Three Forces:

  • System description: A particle is acted upon by an external force $F$ directed horizontally to the left, and is held in equilibrium by forces $F_1$ and $F_2$. Force $F_1$ acts in the first quadrant at an angle $\theta = 30^\circ$ relative to the vertical $y$-axis. Force $F_2$ acts in the fourth quadrant at an angle $\theta = 30^\circ$ below the horizontal $x$-axis.

  • Step 1: Formulate general equations of equilibrium:     $\sum F_x = -F + F_{1,x} + F_{2,x} = 0$     $\sum F_y = F_{1,y} + F_{2,y} = 0$

  • Step 2: Decompose forces into components:     $F_{1,x} = F_1 \sin(30^\circ) = \frac{1}{2}F_1$     $F_{1,y} = F_1 \cos(30^\circ) = \frac{1}{2}\sqrt{3}F_1$     $F_{2,x} = F_2 \cos(30^\circ) = \frac{1}{2}\sqrt{3}F_2$     $F_{2,y} = -F_2 \sin(30^\circ) = -\frac{1}{2}F_2$

  • Step 3: Substitute components into equilibrium equations:     $\sum F_x = -F + \frac{1}{2}F_1 + \frac{1}{2}\sqrt{3}F_2 = 0$     $\sum F_y = \frac{1}{2}\sqrt{3}F_1 - \frac{1}{2}F_2 = 0$

  • Step 4: Solve the system of two algebraic equations:

  • From $\sum F_y = 0$:       $F_2 = \sqrt{3}F_1$

  • Back-substitute into $\sum F_x = 0$:       $-F + \frac{1}{2}F_1 + \frac{1}{2}\sqrt{3}(\sqrt{3}F_1) = 0$       $-F + \frac{1}{2}F_1 + \frac{3}{2}F_1 = 0 \implies 2F_1 = F \implies F_1 = \frac{1}{2}F$

  • Solve for $F_2$:       $F_2 = \sqrt{3}\left(\frac{1}{2}F\right) = \frac{1}{2}\sqrt{3}F$

• Free Body Diagrams of a Particle:

  • A Free Body Diagram (FBD) is a schematic drawing that isolates the particle or object from its physical surroundings.

  • Imaginary cuts are made through all connecting elements (e.g., cables, rods), and the physical supports or constraints are replaced by their corresponding internal or reaction force vectors.

• Worked Example 2.2 — Lamp Suspended from Ceiling:

  • System description: A lamp of mass $m$ is suspended from a ceiling by two symmetric cables, each making an angle of $30^\circ$ with the horizontal dashed line. The acceleration due to gravity is $g$.

  • Step 1: Construct the FBD of the connection node where the two cables and lamp cable intersect. Replace the lamp with its gravitational weight vector $mg$ pointing downward, and replace the cables with tensile force vectors $F_1$ (acting up and to the left at $30^\circ$) and $F_2$ (acting up and to the right at $30^\circ$).

• Step 2: Formulate force equilibrium equations:   $\sum F_x = -F_1 \cos(30^\circ) + F_2 \cos(30^\circ) = -\frac{1}{2}\sqrt{3}F_1 + \frac{1}{2}\sqrt{3}F_2 = 0$   $\sum F_y = F_1 \sin(30^\circ) + F_2 \sin(30^\circ) - mg = \frac{1}{2}F_1 + \frac{1}{2}F_2 - mg = 0$

• Step 3: Solve the system:

  • From $\sum F_x = 0$:     $F_1 = F_2$

  • Substitute into $\sum F_y = 0$:     $\frac{1}{2}F_1 + \frac{1}{2}F_1 - mg = 0 \implies F_1 = mg$

  • Final cable tensions:     $F_1 = F_2 = mg$

Force Vectors and Trigonometric Vector Addition

• Vectors Versus Scalars:

  • Scalar: A physical quantity completely characterized by a single numerical value with appropriate units (e.g., mass, length, time, temperature).

  • Vector: A physical quantity possessing both a magnitude and a direction, obeying specific geometric laws of addition (e.g., force, displacement, velocity, acceleration).

  • Denoted graphically by an arrow whose length corresponds to the scalar magnitude ($F$) and whose orientation and arrowhead specify the line of action and sense ($\mathbf{F}$).

• Parallelogram Law of Vector Addition:

  • Two force vectors $\mathbf{F}_1$ and $\mathbf{F}_2$ acting on a point combine into a single resultant force vector $\mathbf{F}_R = \mathbf{F}_1 + \mathbf{F}_2$, representing the diagonal of a parallelogram formed with $\mathbf{F}_1$ and $\mathbf{F}_2$ as adjacent sides.

  • The magnitude of $\mathbf{F}_R$ is generally not the arithmetic sum of the individual magnitudes ($F_R \neq F_1 + F_2$), except in the degenerate case where both vectors are strictly parallel and unidirectional.

• Trigonometric Laws for General Triangles:

  • For a triangle with side lengths $A, B, C$ opposite internal angles $\alpha, \beta, \gamma$:

• Cosine Law:     $C = \sqrt{A^2 + B^2 - 2AB\cos(\gamma)}$

• Sine Law:     $\frac{A}{\sin(\alpha)} = \frac{B}{\sin(\beta)} = \frac{C}{\sin(\gamma)}$

  • Worked Example 2.3 — Resultant Force via Trigonometric Laws:

• System description: Two forces $\mathbf{F}_1$ and $\mathbf{F}_2$ act on a particle. Force $F_1 = 200\,\text{N}$ acts horizontally along the positive $x$-axis. Force $F_2 = 300\,\text{N}$ is directed in the first quadrant at an angle of $30^\circ$ relative to the vertical $y$-axis (hence at $60^\circ$ to the horizontal positive $x$-axis).

• Step 1: Construct the vector addition triangle using the parallelogram law. The angle between the tail of $\mathbf{F}_1$ and $\mathbf{F}_2$ is $60^\circ$. In the head-to-tail vector triangle, the internal angle $\gamma$ opposite the resultant $\mathbf{F}_R$ is supplementary:     $\gamma = 180^\circ - 60^\circ = 120^\circ$

• Step 2: Compute resultant magnitude using the cosine law:     $F_R = \sqrt{F_1^2 + F_2^2 - 2F_1F_2\cos(120^\circ)}$     $F_R = \sqrt{200^2 + 300^2 - 2(200)(300)(-0.5)} = \sqrt{40000 + 90000 + 60000} = \sqrt{190000} \approx 436\,\text{N}$

• Step 3: Compute the direction angle $\theta$ of $\mathbf{F}_R$ relative to the horizontal using the sine law:     $\frac{F_2}{\sin(\theta)} = \frac{F_R}{\sin(120^\circ)} \implies \frac{300}{\sin(\theta)} = \frac{436}{\sin(120^\circ)}$     $\sin(\theta) = \frac{300}{436}\sin(120^\circ) = \frac{300}{436}\left(\frac{1}{2}\sqrt{3}\right) \approx 0.60$     $\theta = \arcsin(0.60) \approx 36.9^\circ$

  • Worked Example 2.4 — Resultant Force via Rectangular Component Method:

• Re-solve the system of Example 2.3 using Cartesian decomposition:

• Step 1: Decompose each force along the Cartesian axes:     $F_{1,x} = 200\,\text{N}, \quad F_{1,y} = 0\,\text{N}$     $F_{2,x} = 300\sin(30^\circ) = 300(0.5) = 150\,\text{N}$     $F_{2,y} = 300\cos(30^\circ) = 300\left(\frac{1}{2}\sqrt{3}\right) \approx 260\,\text{N}$

• Step 2: Sum components to find resultant components $F_{R,x}$ and $F_{R,y}$:     $F_{R,x} = F_{1,x} + F_{2,x} = 200 + 150 = 350\,\text{N}$     $F_{R,y} = F_{1,y} + F_{2,y} = 0 + 260 = 260\,\text{N}$

• Step 3: Calculate resultant magnitude using the Pythagorean theorem:     $F_R = \sqrt{F_{R,x}^2 + F_{R,y}^2} = \sqrt{350^2 + 260^2} = \sqrt{122500 + 67600} = \sqrt{190100} \approx 436\,\text{N}$

• Step 4: Calculate orientation angle $\theta$ using the tangent function:     $\tan(\theta) = \frac{F_{R,y}}{F_{R,x}} = \frac{260}{350} \approx 0.74$     $\theta = \arctan(0.74) \approx 36.9^\circ$

• Comparison of methods: Both approaches yield identical results ($F_R = 436\,\text{N}$, $\theta = 36.9^\circ$). The Cartesian component method is algebraically superior when summing more than two forces.

Equilibrium of a Rigid Body

• Concept of a Rigid Body:

  • A body has finite spatial dimensions and geometry. A rigid body is an idealization where the distance between any two points remains invariant under applied forces (zero deformation).

  • When forces act on a body at points whose lines of action do not intersect at a single point, translational equilibrium alone does not guarantee overall equilibrium.

• The Concept and Mechanics of a Moment:

  • A force whose line of action does not pass through a reference point induces a tendency for the body to rotate about that point.

• The moment magnitude $M$ of a force $F$ about a reference point $A$ equals the magnitude of the force multiplied by the perpendicular distance (lever arm) from point $A$ to the line of action of the force:   $M = F L_\perp$

• Moment calculation via decomposition (Varignon's Theorem): If a force $\mathbf{F}$ applied at point $B$ (at horizontal distance $L$ from $A$) is inclined, it can be decomposed into $F_x$ and $F_y$. If the line of action of $F_x$ passes directly through $A$, its lever arm is zero, producing no moment. The moment is entirely produced by the perpendicular component $F_y$:     $M = F L' = F_y L$

  • Equations of Equilibrium for a Rigid Body in 2D:

• To ensure neither translation nor rotation occurs, three scalar equations of equilibrium must be satisfied simultaneously:     $\sum F_x \stackrel{+}{\rightarrow} = 0$     $\sum F_y \uparrow+ = 0$     $\sum M \stackrel{+}{\curvearrowleft} = 0$

• The reference point for the moment balance equation can be chosen arbitrarily. A strategic choice of reference point (e.g., an intersection point through which unknown reaction forces pass) simplifies algebra by eliminating those unknowns from the moment equation.

  • Worked Example 3.1 — Moment Equilibrium of a Rigid Bar:

• System description: A horizontal rigid bar of total length $L$ is subjected to three vertical forces perpendicular to the bar: upward force $F_1$ at left end $A$, downward force $F$ at distance $\frac{2}{3}L$ from $A$ (point $B$), and upward force $F_2$ at right end $C$ (distance $L$ from $A$).

• Step 1: Formulate equilibrium equations with reference point at $A$:     $\sum F_x = 0$     $\sum F_y = F_1 + F_2 - F = 0$     $\sum M_A = F_2 L - F\left(\frac{2}{3}L\right) = 0$

• Step 2: Solve the equations:

  • From $\sum M_A = 0$:       $F_2 L = \frac{2}{3}FL \implies F_2 = \frac{2}{3}F$

  • Back-substitute into $\sum F_y = 0$:       $F_1 + \frac{2}{3}F - F = 0 \implies F_1 = \frac{1}{3}F$

  • Worked Example 3.2 — Moment Equilibrium of an Equilateral Triangle:

• System description: An equilateral triangle with sides of length $L$ is subjected to horizontal force $F_1$ and vertical force $F_2$ at the lower-left vertex, vertical force $F_3$ at the lower-right vertex, and downward-left inclined load $F$ acting along the right slanted edge. The altitude of the triangle is $\frac{1}{2}\sqrt{3}L$.

• Step 1: Choose the lower-left vertex as the moment reference point to eliminate $F_1$ and $F_2$.

• Step 2: Formulate equilibrium equations:     $\sum F_x = F_1 + F = 0$     $\sum F_y = F_2 + F_3 = 0$     $\sum M = F_3 L - \left(\frac{1}{2}\sqrt{3}L\right)F = 0$

• Step 3: Solve unknowns directly:     $F_1 = -F$     $F_3 = \frac{1}{2}\sqrt{3}F$     $F_2 = -F_3 = -\frac{1}{2}\sqrt{3}F$

  • Couples and Pure Moments:

• A couple consists of two parallel forces of equal magnitude $F$ directed in opposite directions separated by a perpendicular distance $L$.

• The net force of a couple is identically zero: $\sum \mathbf{F} = \mathbf{0}$. It produces no linear translation.

• The net moment produced by a couple about any point in the plane is constant:     $M = 2\left(\frac{1}{2}FL\right) = FL$

• Limiting case: As the distance $L \to 0$ and magnitude $F \to \infty$ such that the product $FL = M$ remains finite, the couple becomes a concentrated pure moment (represented as a curved arrow).

• Location independence: Because a pure moment produces zero net force, it does not enter force balance equations and exerts the identical rotational effect regardless of its point of application on a rigid body.

  • Worked Example 3.3 — Rectangle Subjected to an External Moment:

• System description: A rectangular plate of width $a$ and height $b$ has vertices $A$ (bottom-left), $B$ (top-left), $C$ (top-right), and $D$ (bottom-right). An external clockwise moment $M$ acts on the plate. Forces are applied: $F_1$ horizontally to the right at $D$, $F_2$ vertically upward at $D$, and $F_3$ horizontally to the left at $C$.

• Step 1: Formulate equilibrium equations with moment taken about point $D$:     $\sum F_x = F_1 + F_3 = 0$     $\sum F_y = F_2 = 0$     $\sum M_D = F_3 a + M = 0$

• Step 2: Solve the equations:     $F_2 = 0$     $F_3 = -\frac{M}{a}$     $F_1 = -F_3 = \frac{M}{a}$

Support Types, Constraints, and Structural Determinacy

• Mechanical Constraints and Reaction Types:

  • Every physical support constrains motion in specific degrees of freedom (translation or rotation). For each restrained degree of freedom, the support develops a corresponding reaction force or moment.

  • Hinge / Pin Support:

  • Prevents translation in two orthogonal directions ($x$ and $y$); permits free rotation.

  • Unknowns introduced: Two reaction forces ($F_x$, $F_y$).

  • Roller Support / Smooth Sliding Surface:

  • Prevents translation perpendicular to the supporting surface; allows translation parallel to the surface and free rotation. Designed to accommodate thermal expansion without inducing thermal stresses.

  • Unknowns introduced: One reaction force ($F_n$) normal to the guide plane.

  • Clamped / Fixed Support (Cantilever):

  • Prevents translation in all directions and completely prevents rotation.

  • Unknowns introduced: Two reaction forces ($F_x$, $F_y$) and one reaction moment ($M_R$).

  • Sliding Clamp / Guided Roller:

  • Prevents rotation and prevents translation along one axis, while permitting translation along a guide track.

  • Unknowns introduced: One reaction force ($F_n$) and one reaction moment ($M_R$).

• Degrees of Freedom and Determinacy Classifications:

  • For a single rigid body in 2D, exactly $3$ independent equilibrium equations can be formulated ($\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$).

  • Statically Determinate: The number of independent reaction unknowns equals the number of independent equilibrium equations ($N_r = 3$), and constraints are properly configured.

  • Statically Indeterminate: The number of reaction unknowns exceeds the available equilibrium equations ($N_r > 3$). Reactions cannot be solved by statics alone; compatibility of deformations from Mechanics of Materials must be incorporated.

  • Unstable: The number of reaction unknowns is strictly less than 3 ($N_r < 3$), or the constraints are improperly arranged (e.g., three parallel rollers permit unconstrained translation; concurrent reaction lines permit rotation about the concurrency point).

• Worked Example 3.4 — Simply Supported Beam Reactions:

  • System description: Beam of length $L$ supported by a hinge at $A$ and a roller at $B$. Downward point loads of magnitude $F$ act at $x = \frac{1}{3}L$ and $x = \frac{2}{3}L$.

  • Step 1: Draw FBD with reaction unknowns $F_{A,x}$, $F_{A,y}$, and $F_{B,y}$.

  • Step 2: Formulate equilibrium equations:     $\sum F_x = F_{A,x} = 0$     $\sum F_y = F_{A,y} + F_{B,y} - 2F = 0$     $\sum M_A = F_{B,y}L - F\left(\frac{1}{3}L\right) - F\left(\frac{2}{3}L\right) = 0$

  • Step 3: Solve reactions:     $F_{A,x} = 0$     $F_{B,y}L = F L \implies F_{B,y} = F$     $F_{A,y} = 2F - F_{B,y} = F$

• Worked Example 3.5 — Cantilever Beam Reactions:

  • System description: Beam of length $L$ clamped at left end $A$ and subjected to a downward vertical point load $F$ at free end $B$.

  • Step 1: Draw FBD showing reactions $F_{A,x}$, $F_{A,y}$, and counter-clockwise moment $M_A$.

  • Step 2: Formulate equilibrium equations:     $\sum F_x = F_{A,x} = 0$     $\sum F_y = F_{A,y} - F = 0$     $\sum M_A = -M_A - FL = 0$

  • Step 3: Solve reactions:     $F_{A,x} = 0, \quad F_{A,y} = F, \quad M_A = -FL$

  • Negative sign indicates the physical reaction moment acts in the clockwise direction.

Multi-Body Systems and Inter-Component Coupling

• Principles of Multi-Body Static Analysis:

  • A complex mechanical system consists of multiple interconnected bodies. For the entire system to be in equilibrium, every individual subsystem and body must independently be in equilibrium.

  • When separating connected components, internal interaction forces appear at the interfaces as equal and opposite action-reaction pairs according to Newton's Third Law ($\mathbf{F}_{\text{action}} = -\mathbf{F}_{\text{reaction}}$).

  • Total equilibrium equations available for a system of $N_b$ distinct rigid bodies in 2D is $3N_b$.

• Worked Example 4.1 — Equilibrium of a Two-Member Frame:

  • System description: Two rigid members $AB$ and $BC$ are pinned to each other at $B$ and pinned to the foundation at $A$ and $C$. Each member carries a transverse point load $F$.

  • External reactions: Hinges at $A$ and $C$ introduce 4 unknown external reaction forces ($F_{A,x}, F_{A,y}, F_{C,x}, F_{C,y}$). The 3 equations of equilibrium for the entire structure cannot solve 4 unknowns.

  • Subsystem decomposition: Isolate member $AB$ and member $BC$. Disconnecting pin $B$ exposes internal coupling forces $F_{B,x}$ and $F_{B,y}$, acting in opposite directions on members $AB$ and $BC$.

  • Equation balance: Two bodies yield $2 \times 3 = 6$ equilibrium equations for 6 unknowns ($F_{A,x}, F_{A,y}, F_{B,x}, F_{B,y}, F_{C,x}, F_{C,y}$), rendering the system statically determinate.

• Internal Sectioning of a Continuous Member:

  • If a continuous structural member is sectioned by an imaginary cut, three internal forces must be exposed to maintain equilibrium of either half:

  • Normal force ($F_x$ or $N$): parallel to the member axis.

  • Shear force ($F_y$ or $S$): transverse to the member axis.

  • Bending moment ($M$): rotational couple resisting curvature.

  • Sectioning a single member adds three internal unknowns alongside three new equilibrium equations, leaving the net degree of determinacy unchanged.

Trusses

• Definition and Structural Idealization of Trusses:

  • A truss is a structural framework composed of slender members joined exclusively at their endpoints (joints/nodes), typically arranged in triangular units.

  • Governing modeling assumptions:

  • All joints are modeled as frictionless spherical or planar pins (no moments can be transferred at joints).

  • External loads and support reactions are applied strictly at the joints.

  • The self-weight of individual members is negligible or distributed equally to the adjacent joints.

  • All members act as two-force members, experiencing only axial tension or axial compression (bending moments and transverse shear forces are zero).

• Two-Force Members:

  • A member subjected to forces at only two points with no applied moments must have resultant force vectors at each joint that are equal in magnitude, opposite in direction, and collinear with the line passing through the two joints.

  • Sign convention: Tensile forces (pulling away from the joint) are defined as positive ($+$); compressive forces (pushing toward the joint) are defined as negative ($-$).

• Zero-Force Members:

  • Identification of unloaded members simplifies truss analysis prior to algebraic computation.

  • Case 1 (Two non-collinear members): If a joint connects only two members and has no external load or support reaction, both members are zero-force members.

  • Case 2 (Three members, two collinear): If a joint connects three members with no external load or support reaction, and two of the members are collinear, the third non-collinear member is a zero-force member.

• Static Determinacy of a Truss:

  • Let $N_j$ be the number of joints, $N_b$ the number of bars, and $N_r$ the number of external support reactions.

  • Because each joint provides two scalar equilibrium equations ($\sum F_x = 0, \sum F_y = 0$), the total number of available equations is $2N_j$.

  • Determinacy condition:     $N_b + N_r = 2N_j$

  • If $N_b + N_r > 2N_j$, the truss is statically indeterminate. If $N_b + N_r < 2N_j$, the truss is unstable.

• Method of Joints:

  • Step-by-step procedure:

  1. Verify static determinacy ($N_b + N_r = 2N_j$).

  2. Label all joints alphabetically ($A, B, C, \dots$).

  3. Draw FBD of the complete truss and solve external reaction forces using overall equilibrium.

  4. Construct FBDs of individual joints, assuming unknown bar forces to be tensile (directed outward from the joint).

  5. Solve equilibrium equations sequentially starting at a joint with at most two unknown forces.

  6. Verify calculations using unused equilibrium equations.

• Worked Example 5.4 & 5.5 — Triangular Truss via Method of Joints:

  • System description: Equilateral triangle truss ($N_b = 3, N_j = 3$) with bar length $L$. Pinned at $A$, roller at $C$. Horizontal force $F$ acts to the right at apex node $B$. Support reactions: $N_r = 3$. Check: $3 + 3 = 2(3) = 6$ (determinate).

  • Overall equilibrium:     $\sum F_x = F_{A,x} + F = 0 \implies F_{A,x} = -F$     $\sum M_A = F_{C,y}L - F\left(\frac{1}{2}\sqrt{3}L\right) = 0 \implies F_{C,y} = \frac{1}{2}\sqrt{3}F$     $\sum F_y = F_{A,y} + F_{C,y} = 0 \implies F_{A,y} = -\frac{1}{2}\sqrt{3}F$

  • Joint $B$ analysis:     $\sum F_x = -F_{AB}\cos(60^\circ) + F_{BC}\cos(60^\circ) + F = -\frac{1}{2}F_{AB} + \frac{1}{2}F_{BC} + F = 0$     $\sum F_y = -F_{AB}\sin(60^\circ) - F_{BC}\sin(60^\circ) = -\frac{1}{2}\sqrt{3}F_{AB} - \frac{1}{2}\sqrt{3}F_{BC} = 0$

  • From $\sum F_y = 0$: $F_{BC} = -F_{AB}$

  • Substitute into $\sum F_x = 0$: $-\frac{1}{2}F_{AB} - \frac{1}{2}F_{AB} + F = 0 \implies F_{AB} = F$

  • Result: $F_{AB} = F$ (Tension), $F_{BC} = -F$ (Compression)

  • Joint $C$ analysis:     $\sum F_x = -F_{AC} - F_{BC}\cos(60^\circ) = -F_{AC} - (-F)(0.5) = -F_{AC} + \frac{1}{2}F = 0 \implies F_{AC} = \frac{1}{2}F \quad (\text{Tension})$     $\sum F_y = F_{C,y} + F_{BC}\sin(60^\circ) = \frac{1}{2}\sqrt{3}F + (-F)\left(\frac{1}{2}\sqrt{3}\right) = 0 \quad (\text{Check satisfied})$

• Worked Example 5.6 — General Truss with Interior Node:

  • System description: Truss with $N_b = 5$ bars and $N_j = 4$ joints ($A, B, C, D$). Lengths $AB = BD = AD = L$; lengths $AC = BC = CD = \frac{1}{3}\sqrt{3}L$. Angles between bars are $30^\circ$. Pinned at $A$, roller at $D$. Downward vertical force $F$ applied at interior joint $C$.

  • Overall equilibrium:     $F_{A,x} = 0, \quad F_{A,y} = \frac{1}{2}F, \quad F_{D,y} = \frac{1}{2}F$

  • Joint $A$ equilibrium ($\sum F_x = 0, \sum F_y = 0$):     $\frac{1}{2}F_{AB} + \frac{1}{2}\sqrt{3}F_{AC} = 0$     $\frac{1}{2}\sqrt{3}F_{AB} + \frac{1}{2}F_{AC} + \frac{1}{2}F = 0$     $F_{AB} = -\sqrt{3}F_{AC} \implies \frac{1}{2}\sqrt{3}(-\sqrt{3}F_{AC}) + \frac{1}{2}F_{AC} + \frac{1}{2}F = 0$     $-F_{AC} + \frac{1}{2}F = 0 \implies F_{AC} = \frac{1}{2}F \quad (\text{Tension})$     $F_{AB} = -\frac{1}{2}\sqrt{3}F \quad (\text{Compression})$

  • Joint $D$ equilibrium (by symmetry with Joint $A$):     $F_{CD} = \frac{1}{2}F \quad (\text{Tension}), \quad F_{BD} = -\frac{1}{2}\sqrt{3}F \quad (\text{Compression})$

  • Joint $B$ vertical equilibrium:     $\sum F_y = -F_{AB}\sin(60^\circ) - F_{BD}\sin(60^\circ) - F_{BC} = 0$     $-\left(-\frac{1}{2}\sqrt{3}F\right)\left(\frac{1}{2}\sqrt{3}\right) - \left(-\frac{1}{2}\sqrt{3}F\right)\left(\frac{1}{2}\sqrt{3}\right) - F_{BC} = 0$     $\frac{3}{4}F + \frac{3}{4}F - F_{BC} = 0 \implies F_{BC} = \frac{3}{2}F \quad (\text{Tension})$

• Method of Sections:

  • Strategy: Determine internal forces in specific members directly without calculating all joint forces. An imaginary cut is passed through the truss dividing it into two completely separate parts. The cut must pass through no more than three members whose forces are unknown.

  • Equations: The isolated section is treated as a rigid body governed by three equilibrium equations ($\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$).

• Worked Example 5.7 — Overhanging Truss via Method of Sections:

  • System description: Truss of square panels with side length $L$ and diagonals. Cut through bars $CE, DE, DF$. Right-hand isolated section carries downward load $F$ at joint $F$.

  • Formulate equations of equilibrium on right section:     $\sum M_E = F_{DF} L - F L = 0 \implies F_{DF} = F \quad (\text{Tension})$     $\sum F_y = F_{DE}\sin(45^\circ) - F = \frac{1}{2}\sqrt{2}F_{DE} - F = 0 \implies F_{DE} = \sqrt{2}F \quad (\text{Tension})$     $\sum F_x = -F_{CE} - F_{DE}\cos(45^\circ) - F_{DF} = -F_{CE} - \frac{1}{2}\sqrt{2}(\sqrt{2}F) - F = 0$     $-F_{CE} - F - F = 0 \implies F_{CE} = -2F \quad (\text{Compression})$

• Worked Example 5.8 — Bridge Truss via Method of Sections:

  • System description: Truss of length $4L$ carrying downward vertical loads $F$ at lower joints $B, D, F$. Pinned at $A$, roller at $H$.

  • Global equilibrium: Total vertical load is $3F$. Symmetry yields reactions:     $F_{A,y} = \frac{3}{2}F, \quad F_{H,y} = \frac{3}{2}F, \quad F_{A,x} = 0$

  • Cut members $BD, CD, CE$. Isolate left section (joints $A, B, C$):     $\sum F_y = F_{A,y} - F + F_{CD}\sin(45^\circ) = 0$     $\frac{3}{2}F - F + \frac{1}{2}\sqrt{2}F_{CD} = 0 \implies \frac{1}{2}F + \frac{1}{2}\sqrt{2}F_{CD} = 0$     $F_{CD} = -\frac{1}{2}\sqrt{2}F \quad (\text{Compression})$

Pulleys and Cable Mechanics

• Ideal Cable and Pulley Assumptions:

  • Cables are perfectly flexible, inextensible, and frictionless; they support only tensile forces directed tangent to the cable path.

  • Friction between pulleys and cables is assumed negligible, and pulley mass/inertia is neglected relative to payload weight.

• Equilibrium of a Pulley:

  • Moment equilibrium about the central axis of an isolated frictionless pulley of radius $r$ with cable tensions $F_1$ and $F_2$:     $\sum M_O = r F_2 - r F_1 = 0 \implies F_1 = F_2$

  • The tension force in a continuous cable remains strictly constant along its entire length, even when wrapping around multiple frictionless pulleys.

• Worked Example 6.1 — Single Movable Pulley System:

  • System description: Mass $m$ suspended from a movable pulley of radius $r$, supported by two vertical cable branches.

  • Force balance:     $\sum F_y = F_1 + F_2 - mg = 2F_1 - mg = 0 \implies F_1 = F_2 = \frac{1}{2}mg$

  • Mechanical advantage: The force required to lift the mass is halved.

• Worked Example 6.2 — Multiple Pulley System:

  • System description: Three pulleys ($A, B, C$). Pulley $A$ is fixed to the ceiling by a hinge. Pulley $B$ is connected to $A$ via coupling cable $F_{AB}$. Pulley $C$ carries suspended mass $m$. A single continuous cable with tension $F$ winds through the system.

  • Equilibrium equations for pulleys:

  • Pulley $C$ (3 upward cable segments supporting mass $m$):       $\sum F_y = 3F - mg = 0 \implies F = \frac{1}{3}mg$

  • Pulley $B$ (supports coupling cable $F_{AB}$ upward and 2 downward branches of cable $F$):       $\sum F_y = F_{AB} - 2F = 0 \implies F_{AB} = 2F = \frac{2}{3}mg$

  • Pulley $A$ (fixed to ceiling, upward reaction $F_y$, downward forces from cable $F_{AB}$ and 2 segments of cable $F$):       $\sum F_y = F_y - F_{AB} - 2F = 0 \implies F_y = \frac{2}{3}mg + 2\left(\frac{1}{3}mg\right) = \frac{4}{3}mg$

Frames and Multi-Force Members

• Distinction Between Trusses and Frames:

  • While truss members are exclusively two-force members subjected only to axial loads at joint pins, frames contain at least one multi-force member.

  • Multi-force members are subjected to loads (forces or moments) applied at intermediate points along their span, or are connected to other members at three or more points.

  • Consequently, members in frames experience axial forces, transverse shear forces, and bending moments.

• Static Analysis Methodology for Frames:

  • Isolate each individual structural member by drawing its FBD.

  • For each member, formulate three independent equations of equilibrium.

  • Ensure interaction forces between touching bodies satisfy Newton's Third Law ($\mathbf{F}_{AB} = -\mathbf{F}_{BA}$).

  • A frame of $N$ members is statically determinate if the total number of unknown reaction and connection components equals $3N$.

• Worked Example 7.1 — Triangular Two-Member Frame:

  • System description: Members $AB$ and $BC$ of length $L$ are inclined at $60^\circ$ to the horizontal ground, pinned to the ground at $A$ and $C$, and pinned together at $B$. A vertical downward force $F$ acts on member $AB$ at mid-span; a horizontal force $F$ acts to the right on member $BC$ at mid-span.

  • Equations for Member $AB$:     $\sum F_x = F_{A,x} + F_{B,x} = 0$     $\sum F_y = F_{A,y} + F_{B,y} - F = 0$     $\sum M_A = -F_{B,x}L\sin(60^\circ) + F_{B,y}L\cos(60^\circ) - F\left(\frac{1}{2}L\cos(60^\circ)\right) = 0$     $-F_{B,x}\left(\frac{1}{2}\sqrt{3}L\right) + F_{B,y}\left(\frac{1}{2}L\right) - \frac{1}{4}FL = 0$

  • Equations for Member $BC$:     $\sum F_x = -F_{B,x} + F_{C,x} + F = 0$     $\sum F_y = -F_{B,y} + F_{C,y} = 0$     $\sum M_C = F_{B,x}\left(\frac{1}{2}\sqrt{3}L\right) + F_{B,y}\left(\frac{1}{2}L\right) - F\left(\frac{1}{2}L\sin(60^\circ)\right) = 0$     $F_{B,x}\left(\frac{1}{2}\sqrt{3}L\right) + F_{B,y}\left(\frac{1}{2}L\right) - \frac{1}{4}\sqrt{3}FL = 0$

  • Solving the 6 unknowns:

  • Add the two moment equations to eliminate $F_{B,x}$:       $F_{B,y}L - \frac{1}{4}FL - \frac{1}{4}\sqrt{3}FL = 0 \implies F_{B,y} = \left(\frac{1}{4} + \frac{1}{4}\sqrt{3}\right)F \approx 0.68F$

  • Subtract the moment equations to isolate $F_{B,x}$:       $\sqrt{3}F_{B,x}L - \frac{1}{4}\sqrt{3}FL + \frac{1}{4}FL = 0 \implies F_{B,x} = \left(\frac{1}{4} - \frac{1}{12}\sqrt{3}\right)F \approx 0.11F$

  • Compute remaining reaction forces via member force balances:       $F_{A,x} = -F_{B,x} = -0.11F$       $F_{A,y} = -F_{B,y} + F = -0.68F + F = 0.32F$       $F_{C,x} = F_{B,x} - F = 0.11F - F = -0.89F$       $F_{C,y} = F_{B,y} = 0.68F$

• Verification via Overall Frame Equilibrium (Example 7.2 & 7.3):

  • Equilibrium of entire system:     $\sum F_x = F_{A,x} + F_{C,x} + F = -0.11F - 0.89F + F = 0$     $\sum F_y = F_{A,y} + F_{C,y} - F = 0.32F + 0.68F - F = 0$     $\sum M_A = F_{C,y}L - \frac{1}{4}FL - \frac{1}{4}\sqrt{3}FL = 0.68FL - 0.25FL - 0.43FL = 0$

  • The global equilibrium equations are linearly dependent on the member equilibrium equations, serving as an effective consistency check.

• Worked Example 7.4 — Frame Supporting a Pulley:

  • System description: Member $AB$ (length $L$, inclined at $30^\circ$ to horizontal) pinned at $A$. Member $CD$ is pinned at $C$ (midpoint of $AB$) and pinned at $D$ to the foundation. Pulley at $B$ carries a cable supporting suspended mass $m$.

  • Member classification: Member $CD$ is pinned at only two points with no intermediate loads, functioning as a two-force member with collinear reaction $F_{CD}$.

  • Pulley equilibrium:

  • Cable tension: $F_m = mg$

  • Horizontal and vertical forces on pin $B$:       $F_{B,x} = F_m = mg, \quad F_{B,y} = F_m = mg$

  • Member $AB$ equilibrium:     $\sum M_A = \frac{1}{2}F_{B,x}L - \frac{1}{2}\sqrt{3}F_{B,y}L + \frac{1}{4}F_{CD}L = 0$     $\frac{1}{4}F_{CD} = \frac{1}{2}\sqrt{3}mg - \frac{1}{2}mg \approx 0.37mg \implies F_{CD} \approx 1.46mg$     $\sum F_x = F_{A,x} - F_{B,x} - F_{CD} = 0 \implies F_{A,x} = mg + 1.46mg = 2.46mg$     $\sum F_y = F_{A,y} - F_{B,y} = 0 \implies F_{A,y} = mg$

Machines and Mechanisms

• Static Equilibrium Analysis of Machines:

  • Machines are mechanical systems containing moving parts designed to modify, transmit, or exert forces.

  • Although configured for motion, a machine is analyzed in a specific instantaneous static configuration where an applied driving force or moment is balanced by a resisting load.

• Worked Example 8.1 — Sliding Mechanism:

  • System description: Member $AB$ of length $L$ is pinned to ground at $A$ and subjected to driving moment $M$. Member $CD$ of length $2L$ is pinned to $AB$ at midpoint $B$, supported by a horizontal roller at $C$, and carries vertical resisting force $F$ at end $D$. Angle $\angle ABC = 90^\circ$.

  • Member $CD$ equilibrium:     $\sum F_x = -F_{B,x} = 0 \implies F_{B,x} = 0$     $\sum M_B = F_{C,y}\left(\frac{1}{2}\sqrt{2}L\right) + F\left(\frac{1}{2}\sqrt{2}L\right) = 0 \implies F_{C,y} = -F$     $\sum F_y = -F_{B,y} + F_{C,y} - F = 0 \implies F_{B,y} = F_{C,y} - F = -2F$

  • Member $AB$ equilibrium:     $\sum F_x = F_{A,x} + F_{B,x} = 0 \implies F_{A,x} = 0$     $\sum M_A = M - F_{B,x}\left(\frac{1}{2}\sqrt{2}L\right) + F_{B,y}\left(\frac{1}{2}\sqrt{2}L\right) = 0$     $M + (-2F)\left(\frac{1}{2}\sqrt{2}L\right) = 0 \implies M - \sqrt{2}FL = 0$     $F = \frac{\sqrt{2}M}{2L}$

• Worked Example 8.2 — Water Pump Pliers:

  • System description: Pliers operated by closing grip force $F$ on handles. Distance from central hinge $A$ to grip force is $5L$; distance from hinge $A$ to bolt contact point $B$ is $L$. Clamping force on bolt acts at $45^\circ$.

  • Member equilibrium (moment balance about central pivot $A$):     $\sum M_A = 5FL - F_B L = 0 \implies F_B = 5F$

  • Reaction forces at hinge $A$:     $\sum F_x = -F_{A,x} + F_B\cos(45^\circ) = -F_{A,x} + \frac{1}{2}\sqrt{2}(5F) = 0 \implies F_{A,x} = \frac{5}{2}\sqrt{2}F$     $\sum F_y = F_{A,y} - F_B\sin(45^\circ) - F = F_{A,y} - \frac{5}{2}\sqrt{2}F - F = 0 \implies F_{A,y} = \left(\frac{5}{2}\sqrt{2} + 1\right)F$