Equilibrium of Particles and Rigid Bodies - Cheat Sheet
Your final-30-minutes reference for equilibrium analysis. Master FBDs, the three core equations, support reactions, and Lami's theorem. Every formula, every reaction type, every trap — all here.
Free-Body Diagram (FBD) Essentials
Important facts
- EVERY external force must be shown — missing a reaction invalidates the entire solution.
- Replace each support with the correct reaction type (see Support Reactions table).
- A correct FBD is 80% of the solution; most exam mistakes start with a wrong or incomplete FBD.
- Draw forces as arrows, label magnitude and direction (angle or components).
- Always include the weight (W = mg) if the body has mass.
Key definitions
Free-Body Diagram (FBD)
Example
A beam on two supports is cut free; you draw the load, the weight, and the two reactions at the supports.
Isolated sketch of a body showing all external forces (applied loads, reactions, weights) acting on it; no internal forces shown.
External Force
Example
Applied load, normal force, tension, weight.
Any force exerted on the body by the environment (loads, support reactions, gravity).
Diagrams to know
- Beam on two supports with point load and UDL — show reactions at each support.
- Cantilever beam with free end carrying load — show fixed-end reaction moment and force.
- Two-rope particle with weight — show tension vectors and weight vector.
- Pin-and-roller frame with inclined load — show two reactions at pin, one at roller.
Core Equilibrium Equations — 2D (Coplanar)
Formulas
ΣFₓ = 0
Meaning
Sum of all horizontal forces = 0 (horizontal equilibrium).
Watch out
Do NOT forget horizontal loads or inclined force components. If there are no horizontal forces, this equation is trivial (0 = 0), not a trick.
When to use
Always apply when solving for reactions or member forces; directly find horizontal unknowns.
ΣFᵧ = 0
Meaning
Sum of all vertical forces = 0 (vertical equilibrium).
Watch out
Forget the weight of the body itself and you fail. Every load component parallel to y must be included.
When to use
Always apply to find vertical reactions or forces; solved after moments equation.
ΣM = 0
Meaning
Sum of all moments about any point = 0 (rotational equilibrium).
Watch out
You can take moments about ANY point — choose wisely to eliminate unknowns. Sign convention: CCW positive is standard.
When to use
Choose a point that eliminates unknown forces (usually a support) to solve one reaction directly.
Important facts
- Three equations, three unknowns: a 2D rigid body has exactly 3 unknowns solvable from ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0.
- Choose the moment equation's pivot point to eliminate known unknowns — this is the fastest path to a solution.
- Distributed loads must be replaced by their resultant force (total load) acting at the centroid before applying ΣM = 0.
- Sign convention: take CCW as positive for moments; be consistent throughout.
- If ΣFₓ = 0 without any horizontal forces, that equation yields 0 = 0 and provides no new information.
Key definitions
Equilibrium
Example
A book resting on a table — no movement, no rotation.
State where net force and net moment are zero; body neither accelerates nor rotates.
Coplanar (2D) System
Example
Horizontal beam loaded vertically — all forces and reactions in the plane of the paper.
All forces and moments lie in the same plane; governed by three independent equations (ΣFₓ, ΣFᵧ, ΣM).
Moment (Torque)
Example
A 100 N force 2 m from a hinge creates a moment of 200 N·m.
M = F × d, rotational effect of a force about a point; SI unit = N·m (newton-meters).
Diagrams to know
- Force components in x–y plane — magnitude and angle.
- Moment vector and right-hand rule (CCW = positive, CW = negative).
- Resultant of distributed load — rectangular area = total; centroid = point of application.
Reactions or equations
ΣM_A = 0 ⟹ R_B = (ΣM_A + applied loads about A) / (distance to B)
Note
Taking moments about the left support A eliminates R_A, isolates R_B immediately.
Conditions
Applied to simply supported beam; A is left support, B is right support.
Particle Equilibrium (Concurrent Forces)
Formulas
ΣFₓ = 0 and ΣFᵧ = 0 (no moment equation — point body)
Meaning
Particle (forces meet at one point) requires only two force equilibrium equations; moment is automatically zero.
Watch out
A particle has no rotational equation. If forces don't meet at one point, it's NOT a particle — use rigid-body equations.
When to use
When all forces are concurrent (meet at a point, like ropes at a knot) — use this simplified approach.
F₁/sin(α₁) = F₂/sin(α₂) = F₃/sin(α₃) [LAMI'S THEOREM]
Meaning
For three concurrent forces in equilibrium: each force divided by the sine of the angle BETWEEN the other two forces are equal.
Watch out
CRITICAL: α is the angle BETWEEN the other two forces, NOT the angle the force makes with a horizontal axis. Most students get this wrong.
When to use
Exactly three concurrent forces (no more, no less) in equilibrium — direct one-step solution.
Important facts
- Particle equilibrium eliminates the moment equation — you have only TWO unknowns max (from ΣFₓ = 0, ΣFᵧ = 0).
- Lami's theorem is the FASTEST method for exactly three concurrent forces — one equation per force, no algebra.
- The angle in Lami's theorem is NOT the angle from horizontal; it is the angle BETWEEN the other two forces in the triplet.
- If you have four or more concurrent forces, use component method (ΣFₓ, ΣFᵧ); Lami does not apply.
- For a particle held in equilibrium by three forces, the three forces must close a triangle (graphically, head-to-tail).
Key definitions
Particle
Example
A knot holding three ropes; a point load junction in a pin-jointed truss.
A body where all forces act at a single point (concurrent); no moment equation needed; only ΣF = 0.
Concurrent Forces
Example
Three ropes tied to a ring — all forces pass through the ring's center.
Forces whose lines of action meet at a single point.
Lami's Theorem (3-Force Case)
Example
Weight 600 N held by two cords; find tensions using the angle between each pair.
For three concurrent forces in equilibrium, F₁/sin(α₁) = F₂/sin(α₂) = F₃/sin(α₃), where each α is the angle between the other two forces.
Diagrams to know
- Three concurrent forces forming a closed force triangle (head-to-tail diagram).
- Two-rope system with weight — angle geometry and force components.
- Lami's theorem diagram — label each force and the angle BETWEEN the other two.
Reactions or equations
ΣFₓ = 0: T₁cos(θ₁) + T₂cos(θ₂) + ... = 0
Note
Solve as a system with ΣFᵧ = 0. Lami is faster if exactly 3 forces.
Conditions
Horizontal equilibrium for concurrent forces; θᵢ measured from x-axis.
ΣFᵧ = 0: T₁sin(θ₁) + T₂sin(θ₂) - W = 0
Note
Combine with ΣFₓ = 0 to eliminate unknowns. Use Lami if 3 forces exactly.
Conditions
Vertical equilibrium; W is weight or applied downward load.
Support Reactions — Rigid Bodies
Important facts
- A roller support gives ONE vertical reaction perpendicular to the surface.
- A pin support gives TWO reactions (horizontal and vertical; no moment).
- A fixed support (built-in) gives THREE reactions (horizontal, vertical, moment).
- NSCP 2015 Section 201 (Design Loads) requires proper identification of support types for load path analysis.
- Always count unknowns vs. equations before starting: if > 3 equations needed, the structure is indeterminate (beyond this chapter's scope).
- The sum of all vertical reactions must equal the total downward load (check your math).
- The moment about any support from all other forces must equal the reaction moment at a fixed end.
Key definitions
Support Reaction
Example
A beam on two supports experiences normal forces at the supports (reactions) that balance the applied loads.
Force (and/or moment) exerted by a support on a structure in response to applied loads; equal and opposite to the action on the support.
Statically Determinate
Example
Simply supported beam (2 unknown vertical reactions + 1 moment equation) — 3 unknowns, 3 equations.
Structure where the number of unknowns equals the number of equilibrium equations (3 equations for 2D); reactions can be found from statics alone.
Statically Indeterminate
Example
Beam fixed at both ends (4 unknowns: 2 reactions + 2 moments) but only 3 equations — indeterminate.
Structure with more unknowns than equilibrium equations; requires additional equations (deflection, compatibility).
Diagrams to know
- Roller support symbol (cylinder on surface) — show single vertical reaction arrow.
- Pin support symbol (circle with pin) — show two reaction arrows (Rx horizontal, Ry vertical).
- Fixed support symbol (solid square at end) — show two force reactions and moment reaction arrow.
- Simply supported beam FBD — pin on left, roller on right.
- Cantilever beam FBD — fixed end on left, reactions at fixed end.
Reactions or equations
R_vertical (roller) = load perpendicular to support surface
Note
Roller is frictionless; cannot resist horizontal force or moment.
Conditions
Roller support; single contact point; no friction.
R_x (pin) and R_y (pin) = from ΣFₓ = 0 and ΣFᵧ = 0
Note
Pin cannot resist moment; it can resist force in any direction.
Conditions
Pin support; two unknowns; typically at one end of a beam.
R_x (fixed), R_y (fixed), M (fixed) = from ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0
Note
Fixed end resists all translations and rotation. M is reaction moment.
Conditions
Fixed (built-in) support; three unknowns; cantilever beams.
Two-Force and Three-Force Members
Important facts
- In a two-force member, the line of action of the internal force MUST pass through both load points; any deviation → bending moments exist.
- Truss members are always two-force members because they connect pin-to-pin with no intermediate loads.
- For a three-force member NOT in parallel, graphically the three forces must meet at a single point (concurrency rule) for equilibrium.
- If a three-force member has parallel forces, they must be in equilibrium (opposite sides, equal magnitude).
- Direction of force in a two-force member is determined by geometry alone — the line between the two joints.
Key definitions
Two-Force Member
Example
Truss member — loaded only at its two joints; carries pure axial force (tension or compression).
Member loaded (by external forces) only at two points and no other loads; the two forces must be equal, opposite, and collinear (along the line joining the two points).
Three-Force Member
Example
Bent bar loaded at three points — the line of action of each force must intersect at one point.
Member in equilibrium under three non-parallel forces; the three forces must be CONCURRENT (meet at a point) or all PARALLEL.
Diagrams to know
- Truss member (two-force) — two pins at ends, force along the member axis.
- Three-force member — three load points, forces meeting at one interior point.
- Force triangle (head-to-tail) for three concurrent forces — vectors close at the origin.
Reactions or equations
F₁ = F₂ (magnitude) and F₁ || F₂ (parallel) and F₁ ↑↓ F₂ (opposite direction)
Note
Force is along the line joining the two load points. This is why truss members are always straight lines in the geometry.
Conditions
Two-force member in equilibrium; no external moments, no distributed loads.
Three forces concurrent at point O OR all parallel
Note
Use this to find unknown directions and magnitudes. Graphically, the three force vectors form a closed triangle.
Conditions
Three-force member in equilibrium.
Distributed Loads and Resultants
Formulas
F_resultant = ∫ w(x) dx = area under load curve
Meaning
Total load (force) from a distributed load is the area under the load–distance diagram.
Watch out
The resultant force acts at the CENTROID of the load diagram, NOT at the geometric midpoint (unless uniform). For UDL, centroid is at mid-span.
When to use
Convert distributed load to a point load before solving reactions (ΣM = 0); area method is fastest.
x̄ = (∫ x · w(x) dx) / (∫ w(x) dx) = moment of area / total area
Meaning
Position of the resultant force (centroid of load diagram) measured from a reference point.
Watch out
If the load is not uniform, the centroid is NOT at the midpoint. For triangular load, centroid is at ⅓ or ⅔ depending on which end is higher.
When to use
Locate where to apply the resultant force when taking moments.
Important facts
- Always replace distributed loads with resultant forces BEFORE applying ΣM = 0.
- UDL: resultant = w × length; position = midpoint.
- Triangular load (zero at left, max at right): resultant = ½ × w_max × length; position = ⅔ from left.
- Triangular load (max at left, zero at right): resultant = ½ × w_max × length; position = ⅓ from left.
- The moment created by a distributed load = (resultant force) × (distance to centroid); do NOT integrate moment directly without first locating the resultant.
Key definitions
Uniformly Distributed Load (UDL)
Example
A floor slab carrying 10 kN/m over a 5 m beam — total load = 50 kN, acts at 2.5 m from either end.
Load of constant intensity w (N/m or kN/m) spread over a span; resultant = w × length, acts at the midpoint.
Linearly Varying Load (Triangular)
Example
Water pressure on a vertical dam increases from zero at top to maximum at bottom.
Load intensity varies linearly from zero to maximum; resultant = ½ × max intensity × length, acts at ⅓ or ⅔ of span.
Diagrams to know
- UDL diagram — rectangular shape, w (intensity) on y-axis, length L on x-axis.
- Triangular load diagram — zero at one end, peak at the other.
- Trapezoidal load — combination of uniform and triangular; split into components.
- Resultant force vector placed at the centroid.
Reactions or equations
F_UDL = w × L, acting at x = L/2
Note
For ΣM = 0, use this resultant and its centroid position.
Conditions
Uniformly distributed load of intensity w over length L.
Inclined Forces and Force Components
Formulas
Fₓ = F cos(θ) and Fᵧ = F sin(θ)
Meaning
Components of a force F at angle θ from the horizontal (x-axis).
Watch out
θ MUST be measured from the positive x-axis (horizontal right). If the force points up-left, use θ > 90°. If down-right, use θ < 0°. Consistent sign convention is critical.
When to use
Decompose inclined forces into x and y components before summing for ΣFₓ = 0 and ΣFᵧ = 0.
F = √(Fₓ² + Fᵧ²) and θ = arctan(Fᵧ / Fₓ)
Meaning
Magnitude and direction of a resultant force from its components.
Watch out
arctan gives angles in (−90°, +90°); use atan2(Fᵧ, Fₓ) to get the true quadrant (0° to 360°).
When to use
Recombine components to check your work or express the final reaction magnitude and angle.
Important facts
- Always decompose inclined loads into x and y components BEFORE applying equilibrium equations.
- Use cos(θ) for the component parallel to the angle's reference axis; sin(θ) for the perpendicular component.
- If a load is given as 'at 30° from the vertical,' convert to angle from horizontal: θ = 90° − 30° = 60°.
- Negative components indicate direction opposite the positive axis (left or down).
Key definitions
Inclined Force
Example
A rope at 30° above horizontal pulling a load with 100 N — has 86.6 N horizontal and 50 N vertical components.
Force not aligned with the x or y axis; must be decomposed into horizontal (x) and vertical (y) components.
Diagrams to know
- Force decomposition diagram — inclined force with x and y component vectors.
- Multiple inclined forces on a particle — all vectors from origin, components along axes.
Reactions or equations
ΣFₓ = 0: F₁cos(θ₁) + F₂cos(θ₂) + ... = 0
Note
Solve simultaneously with ΣFᵧ = 0.
Conditions
All horizontal components summed; each θ measured from positive x-axis.
Moment Calculation and Sign Convention
Formulas
M = F × d (scalar, 2D case)
Meaning
Moment (torque) of a force F about a point; d is the perpendicular distance from the point to the line of action of F.
Watch out
d MUST be the perpendicular distance. If the force line passes through the pivot, d = 0 and M = 0 (no moment).
When to use
Calculate the rotational effect of a force about a pivot point for ΣM = 0.
M = Fₓ × y − Fᵧ × x (moment components about origin O at (0,0))
Meaning
Moment of a force (Fₓ, Fᵧ) acting at point (x, y) relative to the origin.
Watch out
Sign depends on direction: CCW (counterclockwise) is positive by convention in most engineering texts. Be consistent.
When to use
Calculate moment about a reference point when force components and coordinates are known.
Important facts
- Moment is zero if the force line passes through the pivot (d = 0).
- Sign convention (standard): CCW moments are positive (+); CW moments are negative (−). Apply consistently throughout the problem.
- For a force at angle θ at point (x, y), the perpendicular distance d can be found by geometry or the component method (M = Fₓ·y − Fᵧ·x).
- Moment is a vector (3D) but in 2D analysis, moment is a scalar with sign indicating direction.
- Taking ΣM about a support often eliminates that support's reactions from the equation — choose the pivot wisely.
Key definitions
Moment (Torque)
Example
A 50 N force 2 m from a hinge creates a 100 N·m moment about the hinge.
Rotational effect of a force about a point; M = F × d, where d is perpendicular distance; SI unit = N·m.
Perpendicular Distance (Moment Arm)
Example
A force at an angle — drop a perpendicular from the pivot to the force line; measure that distance.
Shortest distance from the pivot point to the line of action of the force; always perpendicular to the force.
Diagrams to know
- Moment arm diagram — force vector with perpendicular distance d shown as a dashed line to the pivot.
- Sign convention illustration — CCW arrows labeled (+), CW arrows labeled (−).
Reactions or equations
ΣM_pivot = 0: Σ(F × d) = 0
Note
Choosing the pivot at a support point eliminates that support's reactions from this equation.
Conditions
All moments about a chosen pivot point; d is perpendicular distance for each force.
Solving Reaction Problems — Step-by-Step
Important facts
- STEP 1: Draw a clear, complete FBD. Include every external force, weight, and reaction. Neglect internal forces.
- STEP 2: Count unknowns vs. equations. For 2D: ≤3 unknowns use ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0. If > 3, structure is indeterminate.
- STEP 3: Replace distributed loads with resultants at their centroids.
- STEP 4: Choose a moment pivot (usually a support) that eliminates multiple unknowns. Apply ΣM = 0 to find one reaction.
- STEP 5: Apply ΣFᵧ = 0 to find the second vertical reaction.
- STEP 6: Apply ΣFₓ = 0 to find the horizontal reaction (if any).
- STEP 7: Verify by taking moments about a different point — all unknown reactions should satisfy the equilibrium equations.
- STEP 8: Check that ΣF_down = ΣF_up (sum of downward loads = sum of upward reactions).
Diagrams to know
- Complete FBD at each step — with loads, then with resultants, then with reactions labeled.
Statically Determinate vs. Indeterminate Structures
Important facts
- For 2D rigid bodies, max 3 independent unknowns solvable by statics: ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0.
- A simply supported beam has 3 unknowns if both supports are pins: one becomes indeterminate. If one is a roller (1 vertical) and one a pin (2 forces), that is 3 unknowns → determinate.
- A cantilever (fixed end) has 3 unknowns at that end: Rx, Ry, M → determinate.
- Two fixed ends = 6 unknowns (indeterminate by 3 degrees).
- This chapter covers DETERMINATE structures only. Indeterminate structures require additional methods beyond PRC Level 1 statics.
Key definitions
Statically Determinate
Example
Simply supported beam (2 vertical reactions) with no horizontal load — 2 unknowns solved by ΣM and ΣFᵧ.
Number of unknown reactions = number of independent equilibrium equations (typically 3 for 2D); reactions uniquely determined by statics alone.
Statically Indeterminate
Example
Beam fixed at both ends (4 unknowns: 2 forces + 2 moments) — 3 equations insufficient.
Number of unknown reactions > number of independent equilibrium equations; statics alone insufficient; requires deflection/compatibility equations.
Diagrams to know
- Simply supported beam (1 pin + 1 roller) — 3 unknowns (Rx at pin, Ry at pin, Ry at roller).
- Cantilever (1 fixed end) — 3 unknowns (Rx, Ry, M at fixed end).
- Beam with overhang and one pin, one roller — determine type (determinate or indeterminate) by counting unknowns.
Must remember
EVERY support must be drawn in the FBD with the correct number of reactions: roller = 1 (perpendicular), pin = 2 (Rx, Ry), fixed = 3 (Rx, Ry, M). A wrong reaction type invalidates the entire solution.
The three 2D equilibrium equations (ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0) give exactly 3 independent unknowns. If you have more unknowns, the structure is indeterminate — outside this chapter's scope.
For Lami's theorem (3 concurrent forces), the angle α in each ratio is the angle BETWEEN the other two forces, NOT the angle from horizontal. This is the #1 student mistake.
Replace distributed loads with their resultant force (area under the load curve) acting at the centroid BEFORE taking moments. Never integrate moment directly without locating the resultant first.
Choose your moment pivot wisely — take ΣM = 0 about a support to eliminate that support's reactions in one equation, isolating the reaction at the other support directly.
A two-force member (like a truss bar) carries force only along its axis (line joining the two joints). The direction is determined by geometry; no other forces act.
For a three-force member in equilibrium (non-parallel case), the three forces MUST meet at a single point. Use this concurrency rule to find unknown directions graphically.
Decompose all inclined forces into x and y components using cos(θ) and sin(θ), where θ is measured from the positive x-axis. Consistency in angle measurement is critical.
Sign convention for moments: CCW is positive (+), CW is negative (−) in standard engineering convention. Apply consistently or clearly state your alternative convention.
After solving reactions, VERIFY by checking that ΣF_down = ΣF_up and by taking moments about a point different from your pivot — all unknown reactions should satisfy equilibrium.
Last minute tips
If your reactions don't add up correctly (ΣF_down ≠ ΣF_up), stop and recheck your FBD. The most common error is a missing load or wrong reaction type at a support.
For Lami's theorem, draw the three forces emanating from a common point and label the angle BETWEEN each pair (not from horizontal). Use this sketch to avoid the angle trap.
When solving beam reactions, always take your first moment equation about a support to eliminate that support's reactions immediately. This reduces the system to 2 unknowns and 2 equations.
Distributed loads are your friend — converting them to resultants eliminates tedious integration. For UDL: resultant = w × L at the middle; for triangular load: resultant = ½ × w_max × L at ⅓ or ⅔ depending on orientation.
If you have a pin at one end and roller at the other (simply supported), you have 3 unknowns total (Rx and Ry at pin, Ry at roller) → exactly solvable. If both are pins or both are rollers, recount and check if the structure is determinate.
Comparison tables
Support Types and Their Reactions (2D)
| Support Type | Symbol | Reactions | Number of Unknowns | Example Use | |
|---|---|---|---|---|---|
| Roller | Frictionless wheel on surface | Circle on cylinder | 1 normal force (perpendicular to surface) | 1 | Right end of simply supported beam |
| Pin (Hinge) | Hinged joint; can rotate freely | Circle with pin through it | 2 forces: Rx (horizontal) + Ry (vertical) | 2 | Left end of simply supported beam; joints in frames |
| Fixed (Built-in) | Rigid connection; no rotation | Solid square at end | 3 reactions: Rx, Ry, M (moment) | 3 | Cantilever beam; fixed-end moment analysis |
| Smooth Surface | Frictionless contact | Arrow perpendicular to surface | 1 normal force perpendicular to surface | 1 | Block on an incline (no friction) |
2D Equilibrium Equations and When to Use Each
| Equation | What It Means | When to Apply | Typical Unknown | Order in Solution | |
|---|---|---|---|---|---|
| ΣFₓ = 0 | Sum of all horizontal forces = 0 | No net acceleration in x-direction | When there is a horizontal load or pin support with horizontal reaction | Horizontal reaction (Rx) or unknown horizontal force | Usually 3rd (after moments) |
| ΣFᵧ = 0 | Sum of all vertical forces = 0 | No net acceleration in y-direction | Always — every beam and frame has vertical loads | Vertical reaction (Ry) or unknown vertical force | Usually 2nd (after moment equation) |
| ΣM = 0 | Sum of all moments about any point = 0 | No net rotation | Most important for reactions — take about a support to eliminate its reactions | Reaction at other support or unknown force | Usually 1st (choose pivot wisely to isolate one unknown) |
Lami's Theorem vs. Component Method for 3 Concurrent Forces
| Aspect | Lami's Theorem | Component Method (ΣFₓ = 0, ΣFᵧ = 0) | When to Use |
|---|---|---|---|
| Condition | Exactly 3 concurrent forces in equilibrium | Any number of concurrent forces (≥2) | Lami if exactly 3; component if 4 or more |
| Formula | F₁/sin(α₁) = F₂/sin(α₂) = F₃/sin(α₃) | ΣFₓ = 0 and ΣFᵧ = 0 (two equations) | Lami is direct if 3 forces; component is always valid |
| Angle Definition | α₁, α₂, α₃ = angles BETWEEN the other two forces | θ₁, θ₂, ... = angles from a reference axis (e.g., horizontal) | Lami: visualize the force triangle; component: use x–y axes |
| Speed | Very fast — one ratio equation per force, often solvable in one step | Requires solving a 2×2 system; more steps but systematic | Lami saves time for 3 forces; component method is more general |
| Risk of Error | HIGH if you confuse the angle — students often use angle from horizontal instead of angle BETWEEN forces | LOW if you set up the axes clearly and are careful with signs | Lami: beware the angle trap; component: be careful with signs and angle measurement |
Distributed Load Types and Their Resultants
| Load Type | Load Diagram Shape | Resultant Force | Position of Resultant (from left end) | Example |
|---|---|---|---|---|
| Uniform (UDL) | Rectangle; constant intensity w | F = w × L | x̄ = L/2 (midpoint) | Floor slab dead load over a span |
| Triangular (zero left, max right) | Triangle rising to the right | F = ½ × w_max × L | x̄ = ⅔ × L | Hydrostatic pressure on dam (max at bottom) |
| Triangular (max left, zero right) | Triangle falling to the right | F = ½ × w_max × L | x̄ = ⅓ × L | Linearly decreasing load (e.g., wind profile) |
| Trapezoidal | Trapezoid (between two different heights) | F = area of trapezoid | x̄ = weighted average of component centroids | Load that is not zero at both ends but varying |
Diagrams
Equilibrium of Particles and Rigid Bodies — Concept Map
Comprehensive mind map showing how the main topics (FBD, particle equilibrium, rigid body equilibrium, supports, special members, distributed loads, and solution strategy) interconnect. Useful for seeing the big picture before diving into details.
Rigid Body Reaction Solution Flowchart
Step-by-step flowchart for solving rigid body reactions. Shows the critical decision point (determinate vs. indeterminate), the standard sequence of applying equilibrium equations, and the verification step. Emphasizes choosing the moment pivot wisely and checking the answer.
Support Type → Reaction State Diagram
State diagram showing the decision tree from identifying a support type to drawing the correct reactions. Clarifies that each support type (roller, pin, fixed) leads to a different number and direction of reactions, which then feed into the equilibrium equations.
Exam Error Distribution (Equilibrium Chapter)
Distribution of the most common errors students make in this chapter. The largest slice (35%) is FBD errors — emphasizing that a correct free-body diagram is 80% of the solution. Lami's angle mistake is the second largest, confirming it as a frequent trap.