Study Companion: S14 — The Method of Sections

S13 solved a truss by treating every pin as a particle and walking from joint to joint. That finds every member force, but it is slow when only one member matters, and errors pile up along the way. A designer checking the most heavily loaded bar in the middle of a long bridge does not want to solve thirty joints to reach it. The method of sections cuts the truss in two with one imaginary line and treats the whole of one part as a single rigid body. A rigid body has three equilibrium equations, not two, and the moment equation is the useful one. With the moment centre chosen well, each unknown comes out of one equation of its own.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 4/4 ("Method of Sections"), with Sample Problems 4/3–4/4 and the Chapter Review (4/7). Harper, Solving Statics Problems in MATLAB, "Trusses. Method of Sections" (Problem 4/41). Meriam's Sample Problem 4/3 is in US units (tons, feet) and is converted to SI here: 1 US ton = 2000 lbf = 8.896\,\mathrm{kN} and 1 ft = 0.3048\,\mathrm m. Axes: x to the right, y up; moments counter-clockwise positive.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

Meriam's Sample Problem 4/3: a cantilever truss of six panels, each 3.658\,\mathrm m (12 ft) long, juts out from a wall where it is fixed at A and M. Its top chord is level at 4.877\,\mathrm m (16 ft) for three panels, then rises to 7.925\,\mathrm m (26 ft) at the wall. A load of 177.9\,\mathrm{kN} (20 US tons) hangs from the free end G.

The forces wanted are in three members near the wall: KL, CL and CB. By the method of joints that means eight joints solved in order, starting from G. Worse, the wall supports are both fixed, so the reactions at A and M cannot even be found: the truss as a whole is statically indeterminate. Can these three forces be found anyway, and each from a single equation? The puzzle is answered in Nugget 02 and revisited in the Synthesis.

Nugget 01: One Cut Turns Part of a Truss Into One Rigid Body

Claim. Pass an imaginary cut through a truss, across no more than three members whose forces are unknown. Either part is a rigid body in equilibrium under its loads, its reactions and the forces of the cut members. That body has three equations, so three cut members can be found.

The picture. A Pratt bridge truss, six panels of 4\,\mathrm m, 4\,\mathrm m deep, with 20\,\mathrm{kN} hanging at each of the five interior lower joints. The cut crosses the third panel.

On the right the left part is drawn alone. Every member inside it, and every force those members exchange, has disappeared: they are internal to the free body and cancel in pairs (S10). What remains is a nonconcurrent system: the 50\,\mathrm{kN} reaction, two 20\,\mathrm{kN} loads, and one force along each cut member. Each cut member is a two-force member, so its force acts along the member, as a tension (away from the cut part) or a compression (toward it).

Check with numbers. By symmetry each support carries half of 5 \times 20 = 100\,\mathrm{kN}: 50\,\mathrm{kN}. On the left part, with all three cut members taken in tension:

\Sigma M about L_3 (where the diagonal and the bottom chord meet): -50(12) + 20(8) + 20(4) - 4\,U_2U_3 = 0, so U_2U_3 = -90\,\mathrm{kN}: 90\,\mathrm{kN} C.
\Sigma M about U_2 (where the top chord and the diagonal meet): -50(8) + 20(4) + 4\,L_2L_3 = 0, so L_2L_3 = 80\,\mathrm{kN} T.
\Sigma F_y: 50 - 20 - 20 - U_2L_3\sin45^\circ = 0, so U_2L_3 = 10/0.7071 = 14.14\,\mathrm{kN} T.

Three equations, three answers, without touching a joint. A full solve of all twelve joints gives the same three values.

The idea. Meriam (4/4): the method of joints uses only two of the three equations, because the forces at a pin are concurrent. Taking a whole section as the free body puts the moment equation to work, and the force in almost any member can be found directly from a section that cuts it. Since a section gives three independent equations, in general no more than three members of unknown force should be cut. Meriam illustrates on the truss of its Fig. 4/12 (the truss of S13's Fig. 4/6): reactions first, from the whole truss; then a cut through BC, BE and EF. The left part is in equilibrium under L, R_1 and the three cut forces; the right part under R_2 and the same three forces reversed. Either part may be used; the one with fewer forces is usually simpler. The cut should pass through members, not through joints, so that it is clear which forces are external to the section.

The equation (physical law: equilibrium of a rigid body in a plane, S11, applied to a section):

\Sigma F_x = 0, \quad \Sigma F_y = 0, \quad \Sigma M_O = 0 \quad \text{for the section; at most three unknown cut members}

Assumptions: those of S13's ideal truss (two-force members, pinned joints, loads at joints).

Worked number. A second cut, through the second panel, finds the chords there: \Sigma M about L_2: U_1U_2 = -[50(8) - 20(4)]/4 = -80\,\mathrm{kN} (80 kN C).

Real example. Engineers inspecting an old truss bridge often need one member only, the one with a crack or corrosion. A section through it, and the known loads, give that member's force from three equations, wherever it sits.

Plausible wrong answer. "Members inside the section must also be in the equations." Their forces act in equal and opposite pairs within the free body and cancel; they never appear.

Check. Could a cut through four members of unknown force be solved from one section? (Not in general: three equations, four unknowns. Nugget 05.)

Narration spine. The loaded Pratt truss. A dashed blade slices the third panel; the right part slides away and the cut members on the left grow arrows along their lines. The inner members of the left part fade grey and one sentence appears: "inside forces cancel". Three equations light up, one per arrow.

Bridge. The order of the equations mattered: each moment centre removed two unknowns. Choosing those centres is the whole art.

Nugget 02: Pick the Moment Centre Where Two Unknowns Meet

Claim. Take moments about the point where the lines of two of the three unknown member forces cross. Both have zero moment there, so the equation holds the third unknown alone. The point need not lie on the section, or on the truss at all.

The picture. The puzzle's left part, after a cut through KL, CL and CB.

KL and CL both pass through L, so moments about L give CB. CL and CB meet at C, so moments about C give KL. KL and CB are not parallel: extend KL backwards and it meets the line of CB at a point P between G and C. Moments about P give CL.

Check with numbers (Meriam's solution, converted to SI).

Geometry. L is 21\ \text{ft} = 6.401\,\mathrm m above B (half way up the rise from 4.877 to 7.925\,\mathrm m). KL rises 5 in 12, so \cos\theta = 12/13. P lies where KL extended meets the bottom chord: PC/4.877 = 3.658/1.524, so PC = 38.4\ \text{ft} = 11.70\,\mathrm m and P is 2.926\,\mathrm m from G. CL makes \beta = \tan^{-1}(12/21) = 29.7^\circ with the vertical, \cos\beta = 0.868.
\Sigma M_L = 0: 177.9(18.29) - CB(6.401) = 0, so CB = 508.4\,\mathrm{kN} C.
\Sigma M_C = 0: 177.9(14.63) - \frac{12}{13}KL(4.877) = 0, so KL = 578.3\,\mathrm{kN} T.
\Sigma M_P = 0: 177.9(2.926) - CL(0.868)(11.70) = 0, so CL = 51.2\,\mathrm{kN} C.

In Meriam's units: 57.1, 65 and 5.76 tons. A force sum on the section checks all three at once: \Sigma F_x and \Sigma F_y come to zero.

The idea. Meriam (4/4): the moment equations are used to great advantage in the method of sections; choose a moment centre, on or off the section, through which as many unknown forces as possible pass. Before calculating, the senses can be read off the same moments. About L, the load's clockwise moment must be balanced by CB pushing toward C: compression. About C, KL must pull: tension. About P, only CL can balance the load's moment, and it must push: compression. In SP 4/3 Meriam notes that the supports' vertical reactions are statically indeterminate (both fixed), but every member except AM is determinate, because the section to the left of the cut never touches the supports. The hint adds that the method of joints would need eight joints to reach these three members.

The equation (mathematical identity: a moment sum about the intersection of two unknown lines contains only the third):

\Sigma M_{O_{12}} = 0 \;\Rightarrow\; F_3\,d_3 = \textstyle\sum(\text{known moments about } O_{12}), \qquad O_{12} = \text{line}_1 \cap \text{line}_2

Worked number. The answer to the puzzle: KL = 578.3\,\mathrm{kN} tension, CB = 508.4\,\mathrm{kN} compression, CL = 51.2\,\mathrm{kN} compression, each from one equation, without the reactions. The chords carry more than three times the load; the diagonal only 29\,\% of it.

Real example. A tower crane's jib or a balcony truss is checked this way at the section next to its support, where the moment of the load is greatest: one cut, two moment centres, and the chord forces are known.

Plausible wrong answer. "Moments must be taken about a point of the section." P lies off the cut members' part entirely, 2.9\,\mathrm m from the tip, and it is the only point that isolates CL by a moment.

Check. If KL and CB were parallel, where would P be, and how would CL be found? (At infinity; use the force sum perpendicular to the chords instead.)

Narration spine. The left part of the cantilever. A pivot marker drops on L: two arrows through it fade and CB glows with its value. The pivot jumps to C, then slides out along the extension of KL to P, where KL and CB both fade and CL alone remains.

Bridge. The two chord forces came from moments; the diagonal from a force sum or a far-off moment. That pattern is general, and it has a meaning.

Nugget 03: The Chords Carry the Bending, the Diagonals Carry the Shear

Claim. For a truss on two supports with vertical loads, a cut through a top chord, a diagonal and a bottom chord gives each chord force as the moment of all the forces on one side about the opposite joint, divided by the truss depth. The diagonal's vertical component equals the net vertical force on that side. So chord forces grow toward midspan, and diagonal forces grow toward the supports.

The picture. The Pratt truss of Nugget 01, solved in full.

Top chords are in compression and bottom chords in tension, the truss bending like a beam on two supports. The chord forces climb from 50 and 80\,\mathrm{kN} at the ends to 80 and 90\,\mathrm{kN} at midspan. The diagonals do the opposite: 70.7\,\mathrm{kN} in the end posts, 42.4\,\mathrm{kN} next, 14.1\,\mathrm{kN} at the centre.

Check with numbers. Panel by panel, the moment of the left-hand forces about the joint opposite each chord, and the net vertical force on the left part:

Cut through panel	Moment for top chord (kN·m)	Top chord (kN)	Net vertical force (kN)	Diagonal (kN)
1 (end post)	—	—	50	50\sqrt2 = 70.7 C
2	50(8) - 20(4) = 320	320/4 = 80 C	50 - 20 = 30	30\sqrt2 = 42.4 T
3	50(12) - 20(8) - 20(4) = 360	360/4 = 90 C	50 - 40 = 10	10\sqrt2 = 14.1 T

The idea. Meriam's illustration on Fig. 4/12 says it in words. Moments about B isolate the top chord EF; the load L exceeds the reaction R_1, so BE must supply the missing upward force and is in tension; and a glance at the truss shows the lower chord stretched "under the tension caused by bending". The moment about the opposite joint is what a beam would call the bending moment, and the net vertical force on one side is the shear force (both named and studied for beams in S21); the chords and the web share them out.

The equation (mathematical identity: Nugget 02's moment centres for a truss of depth h with parallel chords):

F_{\text{chord}} = \frac{M_{\text{opposite joint}}}{h}, \qquad F_{\text{diagonal}}\sin\phi = V_{\text{one side}}

with \phi the diagonal's angle to the chords.

Worked number. Put an extra 30\,\mathrm{kN} at L_2 (50\,\mathrm{kN} there). The left reaction becomes 70\,\mathrm{kN} (moments about L_6). In panel 3: M_{L_3} = 70(12) - 20(8) - 50(4) = 480\,\mathrm{kN\cdot m}, top chord 480/4 = 120\,\mathrm{kN} C; V = 70 - 20 - 50 = 0, so the diagonal U_2L_3 carries nothing.

Real example (one you can see). On many old truss bridges the end diagonals and end posts are visibly heavier than the diagonals near midspan, while the chords are heavier in the middle. The bridge is drawn in steel sizes exactly as the table above.

Plausible wrong answer. "Every member of a truss carries about the same force." The chord at midspan carries 90\,\mathrm{kN} and the centre diagonal 14\,\mathrm{kN}: over six times less.

Check. Which Pratt member carries nothing under these loads, and why? (U_3L_3: joint U_3 has two collinear chords and the vertical, no load: S13's special case.)

Narration spine. The Pratt truss lights up member by member in blue and orange, line widths proportional to force. A cut slides from the left support to midspan: one readout (moment over depth) rises, another (shear) falls, and the chord and diagonal at the blade swell and thin to match.

Bridge. So far every sense was known before calculating. Often it is not.

Nugget 04: Draw the True Sense If You Can, Otherwise Assume Tension

Claim. A cut member's force may be drawn in either sense. If the sense is evident, draw it so (Meriam's preference). If not, assume tension, pulling away from the section; a negative answer then means compression. The arithmetic is the same either way.

The picture. Meriam's Sample Problem 4/4, a Howe roof truss: six 4\,\mathrm m panels, 6\,\mathrm m high at the centre, with 10\,\mathrm{kN} at B, C and I. Section 1 cuts CD, CJ and JK.

The left part of section 1 holds A, B, C, L, K. The top chord ABCD is one straight line, and so is the bottom chord ALKJ: the lines of CD and JK both pass through A, so moments about A leave CJ alone. Moments about C eliminate CD and CJ; moments about J eliminate CJ and JK.

Check with numbers (Meriam's solution). Whole truss: \Sigma M_G = 0: 24A = 10(20) + 10(16) + 10(8), A = 18.33\,\mathrm{kN}.

\Sigma M_A = 0: 0.707\,CJ(12) - 10(4) - 10(8) = 0, CJ = 14.14\,\mathrm{kN} C (drawn pushing, as the moment demands). Its moment is taken from its components placed at J.
\Sigma M_J = 0 with CD assigned as tension although it is plainly compressed: 0.894\,CD(6) + 18.33(12) - 10(4) - 10(8) = 0, CD = -18.63\,\mathrm{kN}. The minus sign says the assumed sense was wrong: CD = 18.63\,\mathrm{kN} C. Its moment is taken from its components placed at D, 6\,\mathrm m above J.
\Sigma M_C = 0: 4\,JK - 18.33(8) + 10(4) = 0, JK = 26.67\,\mathrm{kN} T.

The idea. Meriam (4/4): it is not always possible to assign the right sense at first. Once a choice is made, a positive answer confirms it and a negative one reverses it. Some prefer to assign every unknown as tension (away from the section) and let the sign decide: plus is tension, minus compression. Meriam prefers drawing true senses wherever possible because that "emphasizes the physical action of the forces more directly". Its hint: the arrow may then be reversed on the diagram with the sign, or left as drawn with a note. Harper (P 4/41, Nugget 06) takes the other convention throughout: every member in tension.

The equation (definition: the sign convention):

F > 0 \text{ in the assumed sense}; \qquad \text{assumed tension: } F > 0 \Leftrightarrow \text{T},\ F < 0 \Leftrightarrow \text{C}

Worked number. The same truss by a cut through DE, DJ, CJ and JK with the left part (section 2 of Nugget 05) and moments about J: only DE survives, -6(0.894)DE - 18.33(12) + 10(8) + 10(4) = 0 with DE assumed tension, so DE = -18.63\,\mathrm{kN}: 18.63\,\mathrm{kN} C, the same as CD by the symmetry of joint D.

Real example. Truss design software uses the all-tension convention: it prints a signed number for every member, and a designer reads minus as "check buckling".

Plausible wrong answer. "A negative answer means the problem was set up wrong." It means only that the force acts opposite to the arrow drawn; the magnitude stands.

Check. A cut member assumed in compression gives +12\,\mathrm{kN}. Tension or compression? (Compression, 12\,\mathrm{kN}: the assumed sense was right.)

Narration spine. Section 1 of the Howe truss. The CD arrow is drawn pulling away; the equation runs and gives -18.63; the arrow flips round and turns orange, the readout becoming 18.63\,\mathrm{kN} C.

Bridge. The member Meriam actually wanted in this truss was DJ, and no single cut reaches it with only three unknowns.

Nugget 05: When Every Cut Meets Four Unknowns, Cut Twice

Claim. If every section through a member also cuts three other unknowns, first take a neighbouring section that finds one of them, then use the second section with that force known. A section of a joint (the method of joints) can serve as the second step too.

The picture. The Howe truss of Nugget 04 again: section 2 cuts DE, DJ, CJ and JK.

Check with numbers (Meriam's solution). With CJ = 14.14\,\mathrm{kN} C known from section 1, section 2 has three unknowns. The support G lies on the line of the top chord DEFG and of the bottom chord JIHG, so moments about G eliminate DE and JK:

12\,DJ + 10(16) + 10(20) - 18.33(24) - 14.14(0.707)(12) = 0 \;\Rightarrow\; DJ = 16.67\,\mathrm{kN}\ \mathrm T

DJ was assumed tension and came out positive. The moment of CJ was taken from its components at J.

The idea. Meriam (4/4): a cut through DJ always meets four unknowns. Three of those cut by section 2 are concurrent at J, so moments about J would give DE, but DJ cannot then come from the two force equations alone; section 1 is needed first. Its hint makes the converse point: a cut through CD, DJ and DE cuts only three members, but all three meet at D, so a moment about D says nothing and two force equations cannot find three unknowns. The count of three is necessary, not sufficient: the three lines must not meet in one point. An alternative route uses section 1 for CD and then the joint D for DJ: sections and joints combined. Meriam recommends such combinations for central members of large trusses.

The equation (mathematical identity: when a section works):

\text{solvable} \iff \text{at most three unknown cut members, not all concurrent and not all parallel}

Worked number. The joint route. Joint D has CD = 18.63\,\mathrm{kN} C (from section 1), DE, DJ, and no load. Horizontally, CD and DE have equal and opposite slopes, so DE = CD = 18.63\,\mathrm{kN} C. Vertically, both push D up with 18.63 \times 0.447 = 8.33\,\mathrm{kN} each: DJ = 2(8.33) = 16.67\,\mathrm{kN} T. Same answer.

Real example. Long-span roof trusses with crossing or K-shaped webs often have members no single section can isolate; their analysis is a planned chain of sections and joints.

Plausible wrong answer. "Three cut members, three equations: always solvable." Section CD–DJ–DE has three, all meeting at D, and it fails.

Check. Section 2 also gives DE in one equation. About which point? (J: DJ, CJ and JK all pass through it.)

Narration spine. A blade tries to reach DJ and every cut lights four red members. Section 1 slices first and CJ turns from red to a known value; the second blade now crosses only three unknowns, and moments about G isolate DJ.

Bridge. All of these examples had fixed geometry. Harper's problem asks what happens when the truss itself tilts.

Nugget 06: Keep the Geometry as a Parameter

Claim. Write the section equations with an angle or dimension left as a symbol, and the member forces become formulas. Dividing by the load makes them non-dimensional: forces per unit load, valid for any load.

The picture. Harper's version of Meriam's Problem 4/41: a four-panel truss of square d \times d panels carrying a ramp. It is tilted at an angle \theta (Meriam's 15^\circ), lower joints J, I, H, G, F and upper joints A–E, with the ramp's weight as vertical loads L at I, H, G and L/2 at the end joints J and F. J rests on a roller, F on a pin. A section cuts BC, BH and IH, leaving the left part J, A, B, I.

As the truss tilts, the forces change slowly: IH/L falls from 1.5 to 1.05, BH/L from 0.71 to 0.61, and the compression CD/L eases from 2 to 1.73.

Check with numbers. Whole truss, moments about F: the vertical loads have arms 4d\cos\theta, 3d\cos\theta, 2d\cos\theta, d\cos\theta, so J_y = 2L for every \theta (the roller carries 1.5L net beyond the L/2 applied at J). With all members in tension:

\Sigma M_B = 0: IH(d) - Ld\sin\theta - 1.5Ld\cos\theta + 1.5Ld\sin\theta = 0, so IH = 1.5L\cos\theta - 0.5L\sin\theta;
\Sigma M_H = 0: -BC(d) + Ld\cos\theta - 1.5L(2d\cos\theta) = 0, so BC = -2L\cos\theta;
\Sigma M_I = 0: -BH(d\sin45^\circ) - BC(d) - 1.5Ld\cos\theta = 0, so BH = \frac{\sqrt2}{2}L\cos\theta.

At \theta = 15^\circ: IH = 1.319L T, BC = CD = 1.932L C, BH = 0.683L T. A full solve of all ten joints gives the same formulas from 0 to 30^\circ.

The idea. Harper solves the three moment equations symbolically with MATLAB's solve (the unknowns renamed x, y, z, since solve wants short names) and plots IH/L, CD/L and BH/L over 0 \le \theta \le 30^\circ; setting L = 1 gives the same "force per unit load". It notes BC = CD, because BI is a zero-force member. That reason is a slip: BI carries 0.5L\cos\theta in compression. The idle member is CH: joint C has the collinear BC and CD and the unloaded CH across them (S13's special case), so CH = 0 and BC = CD.

The equation (mathematical identity: the section equations with \theta kept):

\frac{IH}{L} = 1.5\cos\theta - 0.5\sin\theta, \qquad \frac{CD}{L} = -2\cos\theta, \qquad \frac{BH}{L} = \frac{\sqrt2}{2}\cos\theta

Worked number. For a ramp weighing L = 10\,\mathrm{kN} per joint at Meriam's 15^\circ: IH = 13.19\,\mathrm{kN} T, CD = 19.32\,\mathrm{kN} C, BH = 6.83\,\mathrm{kN} T.

Real example. Conveyor galleries and pedestrian ramps between buildings are trusses built at a slope. One set of formulas in the slope angle serves every site, and shows at once that a steeper gallery relieves the chords a little.

Plausible wrong answer. "Tilting the truss tilts the loads with it, so the forces stay the same." The loads stay vertical while the members turn; every force changes with \cos\theta and \sin\theta.

Check. At what tilt would IH fall to zero? (\tan\theta = 3, \theta = 71.6^\circ; steeper than that the lower chord would be compressed there.)

Narration spine. The ramp truss rotates slowly from flat to 30^\circ about J. Three gauges on IH, BH and CD drift; beside them the three curves draw themselves, and a marker stops at 15^\circ with the book's values.

Bridge. Joints and sections together solve any statically determinate plane truss.

Synthesis

The puzzle's three forces came from three moment equations, one each, without the indeterminate reactions: about L for CB = 508.4\,\mathrm{kN} C, about C for KL = 578.3\,\mathrm{kN} T, and about the off-truss point P for CL = 51.2\,\mathrm{kN} C. The method of joints would have needed eight joints, and with the reactions unknown it could not have started at the wall at all.

The two methods are one idea applied to different free bodies. S13 isolated a pin: concurrent forces, two equations. A section isolates a whole piece of truss: three equations, and the moment equation, aimed at the crossing of two unknown lines, picks out one unknown at a time. The chord forces are the bending moment divided by the depth, and the diagonals carry the shear: a truss is a beam whose bending and shear are carried by separate bars. Cut where three unknown members cross the blade, take moments where two of them meet, and each force comes from one equation; when a cut meets four, cut twice or finish at a joint.

Central idea: a section of a truss is a rigid body with three equations; moment centres at the crossings of unknown lines isolate one force each.
Key relationship: \Sigma M_O = 0 about O = intersection of two cut members; F_{\text{chord}} = M/h, F_{\text{diag}}\sin\phi = V.
Most important assumption: at most three unknown cut members, not concurrent and not parallel; the ideal truss of S13.
Open question: a space truss has six equations per section but members pointing every way. Does the method of sections still pay? (S15.)

Computed views

The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Harper's programs; the Python is a translation, and both print the same numbers.

Python (numpy). S14_method_of_sections.py writes the free part beyond a cut as one rigid body: three unknown member forces and three equations, a 3\times3 system with no moment centres to choose. It solves the puzzle that way. It then solves Harper's ramp truss whole, all ten joints, with the shared truss.py, and checks the three section formulas from 0 to 30^\circ.

Part 1: Sample Problem 4/3, the cut through KL, CL and CB
  KL = 578.2 kN, CL = -51.2 kN, CB = -508.3 kN (tension +)
Part 2: Harper's ramp truss, forces per unit load L
  theta   IH/L   formula   CD/L   formula   BH/L   formula   BI/L
      0   1.500    1.500  -2.000   -2.000   0.707    0.707  -0.500
     15   1.319    1.319  -1.932   -1.932   0.683    0.683  -0.483
     30   1.049    1.049  -1.732   -1.732   0.612    0.612  -0.433
  L = 10 kN at 15 deg: IH 13.19, CD -19.32, BH 6.83 kN
  CH = 0.000: the zero-force member that makes BC = CD

The view uses Meriam's lengths in exact feet. They give KL = 578.2 and CB = 508.3\,\mathrm{kN} (65.0 and 57.14 tons). Nugget 02's 578.3 and 508.4 come from the SI lengths rounded to four figures. The BI/L column is Nugget 06's point about Harper's slip: BI carries 0.5L\cos\theta in compression, and CH is the idle member.

MATLAB (Octave-compatible). S14_method_of_sections.m does the same with the backslash operator and truss_solve.m.

KL = 578.2 kN, CL = -51.2 kN, CB = -508.3 kN (tension +)
 theta    IH/L    CD/L    BH/L
     0   1.500  -2.000   0.707
    15   1.319  -1.932   0.683
    30   1.049  -1.732   0.612

Traps

Cutting four unknown members and hoping. Three equations, three unknowns; find one first with another section or a joint. (From Nugget 05.)
Three unknowns that all meet at one point. A moment there gives nothing; the section fails. (From Nugget 05.)
Including internal members of the section. Their forces cancel in pairs. (From Nugget 01.)
Cutting through a joint. It becomes unclear which forces are external; cut through members. (From Nugget 01.)
Taking moments only about points of the truss. The best centre may lie off it (P in the puzzle). (From Nugget 02.)
Using the full force as the moment arm's partner. Resolve an inclined member force into components at a convenient point on its line, as Meriam does with CJ at J and CD at D. (From Nugget 04.)
Reading a negative answer as an error. It reverses the assumed sense. (From Nugget 04.)
Expecting equal forces along a truss. Chords peak at midspan, diagonals near the supports. (From Nugget 03.)
Turning the loads with a tilted truss. Gravity loads stay vertical. (From Nugget 06.)

Self-check

In the Pratt truss of Nugget 01 (six 4\,\mathrm m panels, 4\,\mathrm m deep, 20\,\mathrm{kN} at each interior lower joint), find U_1U_2, L_1L_2 and U_1L_2 with one section.
The same truss with 50\,\mathrm{kN} at L_2 instead of 20\,\mathrm{kN}: find the reactions and U_2U_3, L_2L_3 and U_2L_3.
Meriam's cantilever of the puzzle carries 100\,\mathrm{kN} at G. Find CB, KL and CL.
In the Howe truss (Nugget 04), find JK from section 1.
In the Howe truss, find DE from section 2 alone, and say why it needs no other result.
Why does a section through CD, DJ and DE of the Howe truss fail, although it cuts only three members?
S13's cantilever truss (equilateral panels of 5\,\mathrm m; 30\,\mathrm{kN} at the tip A, 20\,\mathrm{kN} at C; upper joints B, D, lower A, C, E): find CE, BD and CD with one section through those three members.
Harper's ramp truss with L = 10\,\mathrm{kN} at \theta = 15^\circ: find IH, CD and BH.
In Harper's truss, which member is the zero-force member that makes BC = CD? Show it.
At what tilt would IH in Harper's truss become zero? What would that mean?
In the Pratt truss, why is the force in the end post L_0U_1 larger than the force in any other diagonal?

Fully worked solutions

1. Cut panel 2 (through U_1U_2, U_1L_2, L_1L_2); left part, reaction 50\,\mathrm{kN} and 20\,\mathrm{kN} at L_1. \Sigma M_{L_2}: -50(8) + 20(4) - 4\,U_1U_2 = 0, U_1U_2 = -80\,\mathrm{kN} (80 kN C). \Sigma M_{U_1}: -50(4) + 4\,L_1L_2 = 0, L_1L_2 = 50\,\mathrm{kN} T. \Sigma F_y: 50 - 20 - U_1L_2\sin45^\circ = 0, U_1L_2 = 30/0.7071 = 42.43\,\mathrm{kN} T.

2. Loads 20, 50, 20, 20, 20\,\mathrm{kN} at x = 4, 8, 12, 16, 20\,\mathrm m. \Sigma M_{L_6}: 24A = 20(20) + 50(16) + 20(12) + 20(8) + 20(4) = 1680, A = 70\,\mathrm{kN}; B = 130 - 70 = 60\,\mathrm{kN}. Panel 3 cut: M_{L_3} = 70(12) - 20(8) - 50(4) = 480, so U_2U_3 = 480/4 = 120\,\mathrm{kN} C; M_{U_2} = 70(8) - 20(4) = 480, so L_2L_3 = 120\,\mathrm{kN} T; V = 70 - 20 - 50 = 0, so U_2L_3 = 0.

3. Every force scales with the load: multiply by 100/177.9. CB = 100(18.29)/6.401 = 285.7\,\mathrm{kN} C; KL = 100(14.63)/[(12/13)(4.877)] = 325.0\,\mathrm{kN} T; CL = 100(2.926)/[0.868(11.70)] = 28.8\,\mathrm{kN} C.

4. Section 1, left part, moments about C (CD and CJ pass through it): JK acts along the bottom chord, 4\,\mathrm m below C. 4\,JK - 18.33(8) + 10(4) = 0, JK = (146.7 - 40)/4 = 26.67\,\mathrm{kN} T.

5. Section 2, left part, moments about J, where DJ, CJ and JK all meet. DE assumed tension, its components placed at D (6\,\mathrm m above J; horizontal component 0.894DE toward E): -6(0.894)DE - 18.33(12) + 10(8) + 10(4) = 0, DE = -100/5.366 = -18.63\,\mathrm{kN}: 18.63\,\mathrm{kN} C. It needs nothing else because the other three cut members all pass through J.

6. All three meet at D. Moments about D give no equation in them, and two force equations cannot find three unknowns.

7. Height h = 2.5\sqrt3 = 4.330\,\mathrm m. The left part holds A, B, C. \Sigma M_D (BD and CD pass through D, at x = 7.5\,\mathrm m): 30(7.5) + 20(2.5) - 4.330\,CE = 0, CE = 275/4.330 = 63.5\,\mathrm{kN} C. \Sigma M_C (CD, CE pass through C): 30(5) - 4.330\,BD = 0, BD = 34.6\,\mathrm{kN} T. \Sigma F_y: only CD has a vertical component: CD\sin60^\circ = 30 + 20, CD = 57.7\,\mathrm{kN} T. All three agree with the joint-by-joint solution of S13.

8. \cos15^\circ = 0.9659, \sin15^\circ = 0.2588. IH = 10(1.5 \times 0.9659 - 0.5 \times 0.2588) = 13.19\,\mathrm{kN} T; CD = -20(0.9659) = -19.32\,\mathrm{kN} (19.32 kN C); BH = 7.071(0.9659) = 6.83\,\mathrm{kN} T.

9. CH. Joint C has BC and CD along the top chord, CH perpendicular to it, and no load. Summing forces perpendicular to the chord leaves only CH, so CH = 0, and then BC = CD along the chord. (BI is not idle: joint I carries the load L.)

10. 1.5\cos\theta = 0.5\sin\theta, \tan\theta = 3, \theta = 71.6^\circ. Beyond that the moment about B reverses and the lower chord at IH is compressed: a ramp that steep is far outside the formula's intended range, but the formula still says what would happen.

11. The end post is cut by the first section, where the net vertical force on the left part is the whole reaction, 50\,\mathrm{kN}: 50\sqrt2 = 70.7\,\mathrm{kN}. Every section further in has the loads already passed subtracted (30, then 10\,\mathrm{kN}), so its diagonal carries less.