Water Resources HubFE section 6 of 16 · free theory
Statics
Force resultants, equilibrium, trusses, centroids, moments of inertia, distributed loads, friction, and internal loadings — the statics core the FE Civil exam leans on.
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Force resultants and vector tools
Every statics question starts the same way: replace a system of forces with one equivalent force (the resultant) and one equivalent moment. Get fluent with components and you will rarely need to memorise a special case:
FR = ΣF Fx = F cos θ, Fy = F sin θ
| FR | resultant force — the single vector equivalent to the whole system |
| θ | angle of the force measured from the positive x-axis |
MO = r × F |M| = F d
| MO | moment about point O — the cross product of position vector and force |
| d | perpendicular distance from O to the force's line of action |
a · b = axbx + ayby + azbz = |a||b| cos θ
| a · b | dot product — use it for angles between vectors and for projecting one vector onto another |
F = Fxi + Fyj + Fzk cos²α + cos²β + cos²γ = 1
| cos α, cos β, cos γ | direction cosines — the components of the unit vector along F |
Take counterclockwise moments as positive and stick with it for the whole question. The cross product follows the right-hand rule: r × F points out of the page when the rotation from r to F is counterclockwise.
Equilibrium of particles and rigid bodies
A body in static equilibrium has no unbalanced force and no unbalanced moment. Draw the free-body diagram first — most statics errors are missing forces, not bad algebra:
ΣF = 0 ΣMO = 0
| ΣF = 0 | force equilibrium — 2 scalar equations in 2D (ΣFx, ΣFy), 3 in 3D |
| ΣMO = 0 | moment equilibrium — 1 scalar equation in 2D, 3 in 3D, about any point |
In 2D you get 3 equilibrium equations, so a body with more than 3 unknown reactions is statically indeterminate to the first degree — and the FE will not ask you to solve one by statics alone. Supports in 2D: roller = 1 reaction, pin = 2 reactions, fixed support = 3 (two forces plus a couple moment). A two-force member carries equal, opposite, collinear forces; a three-force member in equilibrium has its three forces concurrent at one point.
Trusses: joints, sections, and zero-force members
Truss members are two-force members, so each carries pure axial load — tension or compression. Loads act only at the joints. Two solution routes, and the exam rewards knowing which one is shorter:
Method of joints: ΣFx = 0, ΣFy = 0 at each joint
| Start | at a joint with at most two unknown member forces (usually a support joint after finding reactions) |
Method of sections: cut through ≤ 3 members, then ΣM = 0 about a joint
| Cut | through the members you want; taking moments about the joint where two cut members meet eliminates both, leaving the third |
Zero-force member rules
| Rule 1 | two non-collinear members meeting at an unloaded joint → both carry zero force |
| Rule 2 | three members at an unloaded joint, two of them collinear → the non-collinear member carries zero force |
Sign convention: assume every member in tension (pulling away from the joint). A negative answer means compression. Never skip the reactions — joints need them before you can start.
Frames and machines
Unlike trusses, frames and machines contain members that carry bending — multi-force members. The move is always the same: take the structure apart at the pins and draw a free-body diagram of each piece. Pin forces on mating members are equal and opposite (Newton's third law), which lets you carry unknowns from one diagram to the next. Machines differ from frames only in that they contain two-force members and are built to multiply a force — the mechanical advantage is output force over input force.
Worked example Simple truss by method of joints
Given:
- Triangular truss: pin support A at (0, 0), roller support B at (6, 0), apex joint C at (3, 4).
- A 12 kN load acts straight down at C.
- Find the force in the top chord AC and the bottom chord AB.
Solution:
- The truss and loading are symmetric, so the vertical reactions split the load evenly: Ay = By = 12/2 = 6.0 kN upward. No horizontal loads, so Ax = 0.
- Geometry of AC: run 3 m, rise 4 m, so sin θ = 4/5 = 0.80 and cos θ = 3/5 = 0.60.
- Joint A (assume members in tension, pulling away from the joint): ΣFy = 6.0 + FAC(0.80) = 0 → FAC = −7.5 kN. Negative means the assumption was wrong: AC is in compression, 7.5 kN.
- ΣFx = FAB + (−7.5)(0.60) = 0 → FAB = +4.5 kN, so AB is in tension, 4.5 kN.
Answer: FAC = 7.5 kN (C), FAB = 4.5 kN (T).
Centroids of composite bodies
The centroid is the geometric centre of a shape — the point where its area, length, or volume would balance. For anything the FE exam gives you, the shape is a composite of rectangles, triangles, and circles, so you never need to integrate:
x̄ = Σx̃iAi / ΣAi ŷ = ΣŷiAi / ΣAi
| x̃i, ŷi | centroid coordinates of piece i, measured from one fixed origin |
| Ai | area of piece i — enter holes as negative areas |
Lines: x̄ = Σx̃iLi / ΣLi Volumes: x̄ = Σx̃iVi / ΣVi
| Use | the same weighted-average form — only the measure changes (length, area, or volume) |
Centroids you should know cold: rectangle at (b/2, h/2); right triangle at (b/3, h/3) from the right-angle corner; semicircular area at 4r/(3π) above the diameter; quarter-circular area at 4r/(3π) from each straight edge.
Moments of inertia and the parallel-axis theorem
The second moment of area measures how far a shape's area sits from an axis — it controls bending stiffness later in Mechanics of Materials, but on the statics section you just need to compute it and move it between axes:
Ix = ∫y² dA
| Ix | second moment of area about the x-axis — the y-distance is measured perpendicular to the axis |
Rectangle: bh³/12 Triangle: bh³/36 Circle: πr⁴/4 = πd⁴/64
| Centroidal values | each formula is about the shape's own centroidal axis; h is the dimension perpendicular to that axis |
I = Ī + Ad² JO = Ix + Iy
| Ī + Ad² | parallel-axis theorem — shift from the centroidal axis to any parallel axis a distance d away |
| JO | polar moment of area about point O — the sum of the two perpendicular second moments |
The d in the parallel-axis theorem is the distance between the two parallel axes, not from the origin. A fast check for rectangles: the second moment about the base is bh³/3 — exactly four times the centroidal value.
Worked example Centroid and moment of inertia of a T-section
Given:
- T-section: bottom flange 200 mm × 40 mm; stem 160 mm tall × 40 mm wide, centred on the flange.
- Find the centroid height ŷ measured from the bottom, and Ixx about the horizontal centroidal axis.
- All dimensions in mm. (You can check this in the composite centroid & inertia calculator.)
Solution:
- Flange: A1 = 200 × 40 = 8,000 mm², centroid ŷ1 = 20 mm. Stem: A2 = 40 × 160 = 6,400 mm², centroid ŷ2 = 40 + 80 = 120 mm.
- ŷ = (8,000 × 20 + 6,400 × 120) / (8,000 + 6,400) = 928,000 / 14,400 = 64.4 mm from the bottom.
- Centroidal inertias: Ī1 = 200(40)³/12 = 1.067 × 10⁶ mm⁴; Ī2 = 40(160)³/12 = 13.653 × 10⁶ mm⁴.
- Shift each to the neutral axis: d1 = 64.4 − 20 = 44.4 mm → A1d1² = 15.802 × 10⁶; d2 = 120 − 64.4 = 55.6 mm → A2d2² = 19.753 × 10⁶.
- Ixx = 1.067 + 15.802 + 13.653 + 19.753 = 50.3 × 10⁶ mm⁴. The transfer terms dominate — a good reminder that where the area sits matters more than the local Ī.
Answer: ŷ = 64.4 mm from the bottom; Ixx ≈ 50.3 × 10⁶ mm⁴.
Distributed loads
A distributed load has an intensity w (force per unit length) that varies along the beam. Replace the whole distribution with its resultant before writing any equilibrium equation:
FR = ∫w dx x̄R = ∫x w dx / FR
| FR | magnitude of the resultant = area under the intensity diagram |
| x̄R | location of the resultant = centroid of the intensity diagram |
Uniform load: FR = wL at the midpoint. Triangular load: FR = wmaxL/2 acting one-third of the length from the maximum-intensity end. Draw the diagram and mark the centroid before touching the equilibrium equations.
Worked example Beam under a triangular distributed load
Given:
- Simply supported beam, span 6 m (pin at A, roller at B).
- Triangular distributed load: zero at A, maximum 12 kN/m at B.
- Find the support reactions.
Solution:
- Resultant: FR = ½(12)(6) = 36 kN. It acts at the centroid of the triangle — one-third of the span from the maximum end: 2 m left of B, i.e. 4 m right of A.
- Moment about A: ΣMA = 0 → By(6) − 36(4) = 0 → By = 24 kN upward.
- Vertical equilibrium: ΣFy = Ay + 24 − 36 = 0 → Ay = 12 kN upward. No horizontal loads, so Ax = 0.
Answer: Ay = 12 kN, By = 24 kN (both upward). The roller at the heavy end carries two-thirds of the load — worth remembering as a sanity check.
Dry friction
Dry (Coulomb) friction resists sliding between surfaces in contact. The exam tests two situations: impending motion (about to slip) and belt friction (a flexible belt over a rough drum):
F ≤ μsN impending: F = μsN slipping: Fk = μkN
| F | friction force — it only ever opposes the (impending) relative motion |
| N | normal force pressing the surfaces together |
Block on an incline impends when tan θ > μs
| Check | resolve weight into slope-parallel (W sin θ) and normal (W cos θ) components, then compare W sin θ with μsW cos θ |
T2 = T1eμβ
| T2 | tension on the tight side — the side that resists the impending motion |
| β | total angle of belt contact, in radians |
Static friction is an inequality until motion impends — do not set F = μN unless the question says slipping is impending (or occurring). μs is always ≥ μk for the same pair of surfaces.
Internal loadings at a section
Cut the member at the section of interest and keep one side. Equilibrium of that free body reveals three internal loadings at the cut: the normal force N (axial, positive in tension), the shear force V (transverse), and the bending moment M (positive when it causes sagging — compression in the top fibres). On this part of the exam the questions are conceptual: which internal loading is which, what sign convention is in play, and what equilibrium of the cut segment requires. The full shear-and-moment diagrams belong to Mechanics of Materials.