First-year courses · Engineering

Engineering statics: forces, equilibrium and trusses

Engineering statics flashcards: force vectors, moments, couples, equilibrium, supports, trusses, friction and centroids, with worked numerical examples.

44 cards6 topicsAbout 20 minutes a roundTypical first-year engineering statics (rigid body mechanics)

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About this deck

This deck is for first-year engineering students taking statics, the course that teaches you to find the forces holding a structure still. It follows the usual order of the term: force vectors and components, moments and couples, equilibrium of particles and rigid bodies, support reactions, trusses analyzed by joints and sections, dry friction, and centroids with second moments of area.

You should already be comfortable with right-triangle trigonometry, solving two or three linear equations at once, and basic vector ideas such as adding components. A first physics course helps, but the language of statics (free-body diagram, two-force member, line of action) is introduced here.

Study with a pencil. Most cards carry a small worked example: cover it, set the problem up yourself with a sketch, then compare. Statics rewards a fixed routine: draw the free-body diagram, choose axes and a sign convention, write the equilibrium equations, and take moments about a point that removes as many unknowns as possible. Check units at every step and ask whether the direction of each answer makes physical sense.

Common mistakes: leaving a support reaction off the free-body diagram, or drawing a load that acts on a different body; mixing up sine and cosine when the angle is measured from the vertical instead of the horizontal; and forgetting that a negative truss force means you guessed the direction wrong (compression instead of tension), not that the arithmetic failed.

Before you start: Grade 12 physics and advanced functions (or Physics 30 and Math 30-1), including trigonometry and solving simultaneous linear equations.

All 44 cards

Cover the definitions with a sheet of paper and test yourself, or use the study modes above.

Forces and vectors

Resultant force: The single force whose effect equals the vector sum of all the forces acting on a body; add x, y and z components separately.Example: 30 N east plus 40 N north gives sqrt(30^2 + 40^2) = 50 N at atan(40/30) = 53.13 degrees north of east.
Resolving a force into components: Writing a force as perpendicular parts. With the angle theta measured from the +x axis, Fx = F cos(theta) and Fy = F sin(theta).Example: 200 N at 30 degrees above +x: Fx = 200 cos 30 = 173.2 N, Fy = 200 sin 30 = 100 N.
Unit vector: A vector of length 1 that shows direction only; divide a vector by its own magnitude to get it.Example: F = (3, 4, 12) kN has magnitude 13 kN, so its direction is (3/13, 4/13, 12/13) = (0.231, 0.308, 0.923).
Direction cosines: The cosines of the angles a force makes with the x, y and z axes: Fx/F, Fy/F and Fz/F. Their squares always add to 1.Example: For (3, 4, 12) kN: 0.231^2 + 0.308^2 + 0.923^2 = 1.00, and the angle with the x axis is acos(0.231) = 76.7 degrees.
Force directed along a line: Multiply the force magnitude by the unit vector pointing from the start point to the end point of the cable or member.Example: 260 N in a cable from A(0, 0, 0) to B(5, 0, 12) m: length 13 m, so F = 260(5/13, 0, 12/13) = (100, 0, 240) N.
Projection with the dot product: The part of a force acting along a direction u equals F dot u, where u is a unit vector in that direction.Example: F = (10, 20, 0) N and u = (0.6, 0.8, 0): F dot u = 6 + 16 = 22 N along u.
Principle of transmissibility: For a rigid body, a force may be slid anywhere along its line of action without changing the external reactions or the motion.Example: Pulling a cart with a rope or pushing it from behind along the same line gives the same support reactions.
Free-body diagram: A sketch of one body cut free from its surroundings, showing every external force and couple on it, including weights and support reactions.Example: For a ladder against a wall: weight at mid-length, normal and friction forces at the floor, normal force at the wall.

Moments and couples

Moment of a force about a point: The turning effect of a force: magnitude times the perpendicular distance from the point to the line of action, M = F d, in N·m.Example: 50 N acting 0.4 m (perpendicular) from a bolt gives 50 x 0.4 = 20 N·m.
Varignon's theorem: The moment of a force about a point equals the sum of the moments of its components about that same point.Example: Fx = 60 N and Fy = 80 N act at (3, 2) m: Mo = x Fy - y Fx = 3(80) - 2(60) = 120 N·m counterclockwise.
Moment as a cross product: In 3D, M = r x F, where r runs from the moment point to any point on the line of action of F.Example: r = (2, 1, 0) m, F = (0, 0, 10) N: M = (1(10) - 0, 0 - 2(10), 0) = (10, -20, 0) N·m.
Sign convention for 2D moments: Counterclockwise turning is usually taken as positive and clockwise as negative, which matches the right-hand rule with +z out of the page.Example: A 40 N·m clockwise moment is written -40 N·m in the sum of moments.
Couple: Two forces of equal size, opposite direction and parallel but separate lines of action. Its moment F d is the same about every point.Example: Two 40 N forces 0.25 m apart give 40 x 0.25 = 10 N·m, whatever point you take moments about.
Moment about an axis: The turning effect about a line: project r x F onto a unit vector u along the axis, Ma = u dot (r x F).Example: With r x F = (10, -20, 0) N·m, the part about the x axis is 10 N·m and the part about the z axis is 0.
Equivalent force-couple system: A force can be moved to a new point if a couple equal to its moment about that point is added, so the external effect is unchanged.Example: 500 N downward at the end of a 2 m bracket equals 500 N down at the wall plus a 1000 N·m couple.
Location of a resultant of parallel loads: Find the total load R, then place it at x = (sum of moments about the origin)/R so it produces the same moment.Example: 10 kN at x = 0 and 20 kN at x = 3 m: R = 30 kN at x = (20 x 3)/30 = 2 m.

Equilibrium and supports

Equilibrium of a particle: A point acted on by concurrent forces is at rest when the components sum to zero: sum Fx = 0, sum Fy = 0 (and sum Fz = 0 in 3D).Example: A 100 N weight hangs from two symmetric cables at 30 degrees above horizontal: 2T sin 30 = 100, so T = 100 N each.
Equilibrium of a rigid body in 2D: Three independent equations must hold: sum Fx = 0, sum Fy = 0 and sum M = 0 about any point, so at most three unknowns can be found.Example: Taking moments about a pin removes its two unknown reaction components from that equation.
Simply supported beam reactions: For a beam on a pin and a roller, take moments about one support to get the other reaction, then use vertical force balance.Example: 6 m beam, 12 kN at 2 m from A: RB = 12(2)/6 = 4 kN, RA = 12 - 4 = 8 kN.
Uniformly distributed load: A load spread evenly at w per unit length; replace it by a single force w L acting at the middle of the loaded length.Example: 5 kN/m over 4 m equals 20 kN acting 2 m from either end.
Triangular distributed load: A load that rises linearly from zero to a peak w; its resultant is w L/2, acting L/3 from the high end (2L/3 from the zero end).Example: Peak 6 kN/m over 3 m: 6(3)/2 = 9 kN acting 1 m from the high end.
Pin support: A connection that prevents translation in any direction but allows rotation, so it supplies two force components and no moment.Example: A pinned beam end gives unknowns Ax and Ay.
Roller support: A support that pushes only perpendicular to the surface it rests on, so it supplies a single force and allows sliding and rotation.Example: A roller on a horizontal floor gives one vertical reaction By.
Fixed support: A rigid connection that prevents translation and rotation, supplying two force components and a reaction moment in 2D.Example: A 3 m cantilever from a wall on its left with 4 kN down at the tip: the wall supplies 4 kN up and a 4 x 3 = 12 kN·m counterclockwise moment.
Two-force member: A weightless member loaded only at two points; the two forces must be equal, opposite and along the line joining those points.Example: Every member of an ideal pin-jointed truss carries force only along its own axis.
Three-force member: A body in equilibrium under exactly three forces; their lines of action must meet at one point or all be parallel.Example: A bar held by a pin and a cable with one load: the pin reaction passes through the point where the other two lines cross.
Statically indeterminate structure: A structure with more unknown support reactions than independent equilibrium equations, so deformation must be considered to solve it.Example: A beam pinned at both ends has 4 reaction unknowns but only 3 equations in 2D: one degree indeterminate.

Trusses and frames

Ideal truss assumptions: Members are straight, joined by frictionless pins, loaded only at the joints, and their own weight is ignored, so each one is a two-force member.Example: Gusset plates on a real bridge are treated as pins for a first analysis.
Method of joints: Draw a free-body diagram of each pin and apply sum Fx = 0 and sum Fy = 0, starting at a joint with no more than two unknown member forces.Example: Apex with 10 kN down and two members at 30 degrees to the horizontal: 2F sin 30 = 10, so each carries 10 kN in compression.
Method of sections: Cut the truss through no more than three members of unknown force, then use the three equilibrium equations on one piece, often taking moments where two cut members meet.Example: Taking moments about the joint where the top chord and diagonal meet gives the bottom chord force directly.
Zero-force member: A truss member carrying no load. Spot them at unloaded joints with two non-collinear members, or with three members where two are collinear.Example: An unloaded, unsupported corner joint joining one vertical and one horizontal member: both forces are 0.
Tension and compression sign convention: Assume every unknown member force pulls away from the joint (tension). A negative answer means the member pushes on the joint (compression).Example: Solving gives F = -8 kN, so the member carries 8 kN in compression.
Truss determinacy check: For a plane truss with m members, r reactions and j joints: m + r = 2j suggests determinate, more means indeterminate, fewer means unstable.Example: 7 joints, 11 members, 3 reactions: 11 + 3 = 14 = 2(7), so it can be statically determinate.
Frames and machines: Structures with at least one multi-force member; take them apart and draw each piece, using equal and opposite forces at shared pins.Example: Pliers: each handle carries the grip force, the hand force and the pin force.

Friction

Coefficient of static friction: mu_s sets the largest friction force before slipping: the friction force F satisfies F <= mu_s N, with equality only when slip is about to happen.Example: 50 kg crate, mu_s = 0.4: N = 50(9.81) = 490.5 N, so up to 0.4 x 490.5 = 196.2 N can be resisted.
Kinetic friction: Once surfaces slide, the resisting force is roughly constant at mu_k N, where mu_k is usually a bit smaller than the static coefficient.Example: Same crate with mu_k = 0.3: 0.3 x 490.5 = 147.2 N while it slides.
Angle of friction: The angle phi with tan(phi) = mu_s; a block on an incline is about to slip when the slope reaches this angle (angle of repose).Example: mu_s = 0.4 gives phi = atan(0.4) = 21.8 degrees.
Tipping versus sliding: A pushed block either slides when P reaches mu_s W or tips about its front edge when P h = W b/2. The smaller P governs.Example: W = 200 N, b = 0.5 m, P at h = 1.0 m, mu_s = 0.3: slides at 60 N, tips at 200(0.25)/1.0 = 50 N, so it tips first.
Flat belt or rope over a drum: For a belt about to slip around a fixed drum, T2 = T1 e^(mu beta), where beta is the wrap angle in radians and T2 is the larger tension.Example: mu = 0.3, half wrap beta = pi: T2/T1 = e^(0.942) = 2.57, so T1 = 100 N holds up to about 257 N.

Centroids and second moments

Centroid of a composite area: The geometric centre of a shape built from simple parts: x bar = sum(A x)/sum(A), with holes entered as negative areas.Example: A1 = 2000 mm^2 at x = 50 mm and A2 = 1200 mm^2 at x = 10 mm: x bar = (100000 + 12000)/3200 = 35 mm.
Centroid of a triangle: Located one third of the height above the base (two thirds from the vertex), at the intersection of the medians.Example: A triangle 90 mm tall has its centroid 30 mm above the base.
Centre of gravity: The point where the whole weight of a body can be taken to act; it matches the centroid when density and gravity are uniform.Example: A uniform 4 m beam has its weight acting at 2 m from either end.
Second moment of area of a rectangle: Resistance of a cross-section to bending about its centroidal axis parallel to the base: I = b h^3/12.Example: b = 100 mm, h = 200 mm: I = 100(200^3)/12 = 6.67 x 10^7 mm^4.
Parallel axis theorem: I about any axis equals I about the parallel centroidal axis plus A d^2, where d is the distance between the two axes.Example: Same rectangle about its base: 6.67 x 10^7 + 20000(100^2) = 2.67 x 10^8 mm^4, which matches b h^3/3.