Water Resources HubFE section 10 of 16 · free theory
Structural Analysis
Determinacy and stability, beam and truss reactions, shear and moment diagrams, influence lines, approximate analysis of indeterminate frames, and beam deflections.
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Reactions from equilibrium
Every determinate structure starts the same way: draw the free-body diagram, then write equilibrium. For planar structures there are three equations, so a structure with more than three unknown reaction components (per free body) needs something extra — compatibility — and is indeterminate.
ΣFx = 0 ΣFy = 0 ΣM = 0
| Supports | roller: 1 reaction component (normal to surface); pin: 2 components; fixed: 3 components (2 forces + 1 moment) |
| Internal hinge | moment = 0 there — one extra equation per hinge, and the structure separates into determinate pieces |
Work the whole structure first for the reactions, then cut it into pieces (method of sections) or isolate joints (method of joints) for internal forces. Sign convention used on this page: shear is positive when it acts upward on the left face of a segment; bending moment is positive when it causes sagging (compression on the top fibres).
Determinacy and stability
Before solving anything, count. The count tells you whether equilibrium alone is enough — but a passing count does not guarantee stability. A truss that satisfies the equation can still be a mechanism if its members or reactions are arranged badly (all reactions parallel, or all concurrent at one point).
m + r = 2j (planar truss)
| m | number of members |
| r | reaction components |
| j | joints |
degree of indeterminacy = (m + r) − 2j (truss) / r − 3 per free body (beam), 3k + (r − 3) (rigid frame with k closed loops)
m + r < 2j means a mechanism is likely (unstable). m + r = 2j with stable geometry means determinate. m + r > 2j means indeterminate to that degree. A continuous beam over n supports is indeterminate to the (n − 2)th degree — each interior support adds one redundant.
Zero-force member rules (unloaded joints)
| Two members, not collinear | at a joint with no external load or support: both members are zero-force |
| Three members, two collinear | at a joint with no external load: the non-collinear member is zero-force |
Spot zero-force members before doing joint equilibrium — each one you find removes an unknown and often unlocks the next joint.
Shear and moment diagrams
The differential relationships are the whole game. If you can write them from memory and apply them, you can sketch any shear and moment diagram without cutting dozens of sections:
dV/dx = −w dM/dx = V
| w | distributed load intensity, positive downward |
| V | shear force |
| M | bending moment |
What the relationships mean in practice: a point load P makes the shear diagram jump by P; a point couple makes the moment diagram jump; the shear diagram’s slope at any point equals minus the load intensity there; the moment diagram’s slope equals the shear. Maximum (or minimum) moment occurs where the shear is zero — or under a point load if the shear jumps across zero there.
Simply supported, centre point load P: Mmax = PL/4
Simply supported, uniform load w: Mmax = wL²/8 (at midspan)
Cantilever, tip point load P: Mmax = PL (at the wall)
Cantilever, uniform load w: Mmax = wL²/2 (at the wall)
Memorise these four. A surprising number of exam questions are one of these with the numbers changed — and they are the first sanity check on any longer calculation.
Influence lines
An influence line shows how one quantity — a reaction, a shear, a moment — varies as a single unit load moves across the structure. The ordinate at a point is the value of that quantity when the unit load stands at that point.
Simply supported beam, span L, unit load at distance x from A:
ηA = (L − x)/L ηB = x/L
Moment at C (a from A, b from B): triangular influence line, peak η = ab/L at C
| η | influence ordinate — dimensionless for reactions and shear, units of length for moment |
To maximise an effect: place a concentrated load at the largest ordinate. To maximise with uniform live load: cover the regions where the influence line is positive (for maximum) or negative (for minimum). Shear at a point has a jump in its influence line at that point: just left of C the ordinate is −b/L, just right it is +a/L.
Approximate analysis of indeterminate frames
The FE exam will not ask you to run slope-deflection by hand on a big frame, but it does expect the two classical approximate methods for lateral load — they turn an indeterminate frame into a determinate one by assuming where the inflection points fall.
Portal method assumptions
| Inflection points | at mid-height of every column and mid-span of every beam |
| Column shears | each interior column carries twice the shear of an exterior column |
With the inflection points known, each column and beam segment becomes a free body with known moment arms — write equilibrium storey by storey, top down.
Cantilever method assumptions
| Inflection points | at the midpoint of every member |
| Column axial stress | varies linearly across the width of the frame, zero at the centroid of the column areas — the frame bends like a giant cantilever |
The cantilever method suits tall, slender frames where overturning dominates; the portal method suits low-rise frames where shear dominates. The exam usually tells you which to use — or the frame is one bay, where both are quick.
Beam deflections
Two routes to deflection on the FE exam: moment-area (geometric, fast for point loads) and virtual work (general, always works). Both need EI, and both reward a correct moment diagram — which is why this section follows the shear and moment section.
Moment-area, Theorem 1: θB − θA = area of M/EI diagram between A and B
Moment-area, Theorem 2: tB/A = moment of the M/EI area between A and B about B
Virtual work: 1 · Δ = ∫ (mv · M / EI) dx
| tB/A | tangential deviation of B from the tangent at A |
| mv | bending moment due to a dummy unit load applied at the point and in the direction of the desired deflection |
| M | real bending moment from the actual loads |
Cantilever, tip load P: Δ = PL³ / (3EI)
Simply supported, centre load P: Δ = PL³ / (48EI)
Simply supported, uniform w: Δ = 5wL⁴ / (384EI)
Deflection scales with the cube of the span for point loads and the fourth power for uniform load — doubling a span multiplies deflection by 8 to 16. That scaling alone answers several qualitative exam questions.
The Müller-Breslau principle
For sketching influence lines without computing a single ordinate: release the constraint corresponding to the function you want (remove the support for a reaction, cut the beam and insert a hinge for a moment), then impose a small unit displacement at the release. The resulting deflected shape — drawn to scale — is the influence line. It gives the shape immediately; the peak ordinate still comes from geometry (for example ab/L for moment).
Worked example Beam reactions, shear and moment
Given:
- Simply supported beam, span L = 8.0 m.
- Uniform load w = 6.0 kN/m over the full span, plus a point load P = 20 kN at a = 3.0 m from the left support.
Solution:
- Reactions: RA = wL/2 + P(L − a)/L = 6.0(8.0)/2 + 20(5.0)/8.0 = 24.0 + 12.5 = 36.5 kN. RB = wL/2 + Pa/L = 24.0 + 20(3.0)/8.0 = 24.0 + 7.5 = 31.5 kN. Check: 36.5 + 31.5 = 68.0 = 6.0(8.0) + 20 ✓.
- Shear: just right of A, V = 36.5 kN; just left of the point load, V = 36.5 − 6.0(3.0) = 18.5 kN; just right of it, V = 18.5 − 20 = −1.5 kN; just left of B, V = 36.5 − 6.0(8.0) − 20 = −31.5 kN = −RB ✓.
- Moment under the point load: M(3) = RA(3.0) − w(3.0)²/2 = 36.5(3.0) − 6.0(9.0)/2 = 109.5 − 27.0 = 82.5 kN·m.
- Where is the maximum? Shear is positive everywhere left of the load and negative everywhere right of it, and it never crosses zero inside a segment (zero-crossings would fall at x = 6.08 m and x = 2.75 m, both outside their segments), so the moment peaks under the point load. Check M at B: 36.5(8.0) − 6.0(64)/2 − 20(5.0) = 0 ✓.
Answer: RA = 36.5 kN, RB = 31.5 kN; Mmax = 82.5 kN·m at 3.0 m from A.
Worked example Truss determinacy and a joint
Given:
- Planar truss: pin support at A(0, 0), roller at B(12, 0); joints C(6, 4) and D(6, 0); members AD, DB, AC, CB, DC.
- A single 24 kN downward load at C.
Solution:
- Count: m = 5, r = 3, j = 4. m + r = 8 = 2j = 8 — internally and externally determinate (geometry is stable, so this is a real truss, not a mechanism).
- Reactions: symmetric, so RA = RB = 24/2 = 12 kN upward.
- Joint A: member AC rises at θ = atan(4/6). sin θ = 4/√52 = 0.5547, cos θ = 6/√52 = 0.8321. ΣFy = 0 gives 12 + FAC(0.5547) = 0, so FAC = −21.6 kN (compression). ΣFx = 0 gives FAD = −FAC(0.8321) = +18.0 kN (tension).
- Joint D: members AD and DB are collinear and horizontal, DC is vertical, no external load — by the zero-force rule, FDC = 0. Then ΣFx = 0 gives FDB = FAD = 18.0 kN (tension).
Answer: Determinate (m + r = 2j = 8). FAC = 21.6 kN compression, FAD = FDB = 18.0 kN tension, FDC = 0 (zero-force member).
Worked example Influence line for moment
Given:
- Simply supported beam, span L = 10 m. Point C is 4.0 m from A (6.0 m from B).
- A 15 kN load moves across the beam.
Solution:
- The influence line for MC is triangular with its peak at C: ηmax = ab/L = 4.0(6.0)/10 = 2.4 m. (Moment ordinates carry units of length.)
- Maximum moment at C occurs with the load placed exactly at C, where the ordinate is largest: MC,max = 15 kN × 2.4 m = 36 kN·m.
- Sanity check by statics with the load at C: RA = 15(6)/10 = 9.0 kN, MC = 9.0(4.0) = 36 kN·m ✓.
Answer: Peak influence ordinate 2.4 m at C; maximum MC = 36 kN·m with the load at C.