Study Companion: S13 — Plane Trusses and the Method of Joints

Part 3 put one body at a time in equilibrium: isolate it, draw its free-body diagram, write its equations. A bridge, a roof or a crane jib is not one body. It is dozens of bars pinned together, and the question an engineer must answer is not "what holds the bridge up?" but "how hard is each bar pulled or pushed?", because that decides how thick each bar must be. Those are internal forces: they never appear on a free-body diagram of the whole structure. To see them, the structure has to be taken apart, and Newton's third law has to be applied at every connection. This lesson does that for the simplest structure, the plane truss. Every bar of a truss carries only a pull or a push along its own length. So each pin is a particle with two equilibrium equations, and the whole truss can be solved one joint at a time.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, articles 4/1 ("Introduction"), 4/2 ("Plane Trusses") and 4/3 ("Method of Joints"), with Sample Problems 4/1–4/2 and the Chapter Review (4/7, part a). Marghitu, Dupac and Madsen, Statics with MATLAB, section 4.5 ("Plane Trusses") and Example 4.6. Harper, Solving Statics Problems in MATLAB, "Trusses. Method of Joints" (Problem 4/40). 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 4/1: a cantilever truss sticks out from a wall. It is made of seven bars forming three equilateral triangles of side 5\,\mathrm m. A 30\,\mathrm{kN} load hangs from its tip A and 20\,\mathrm{kN} from the joint C. The lower end E is pinned to the wall; the upper joint D is held by a cable rising at 30^\circ to the wall.

Which bars are being stretched and which squeezed? Which bar works hardest, and with what force? A guess from experience with beams says "the top is in compression, the bottom in tension", and the heaviest load is the 30\,\mathrm{kN} at the tip, so perhaps no bar carries much more than 30\,\mathrm{kN}. Both guesses fail. The puzzle is answered in Nugget 03 and revisited in the Synthesis.

Nugget 01: A Truss Member Only Pulls or Pushes Along Its Own Line

Claim. A straight bar pinned at both ends, loaded only at its pins and light enough to ignore its weight, is a two-force member. The force it carries acts along the line joining its pins. It is either a tension (pulling the pins together) or a compression (pushing them apart). It has the same value at every cross-section.

The picture.

A two-force member in equilibrium has two forces acting on it, one at each pin. S11 showed that two forces balance only if they are equal, opposite and collinear, so both lie along the line through the pins. Cut the bar anywhere and look at one piece: it still has only two forces on it, the pin force and the force from the other piece at the cut. So the cut carries the same T (or C) as the pins. One number describes the whole member.

Check with numbers. A wall bracket: a horizontal tie AB, 2\,\mathrm m long, and a strut CB at 30^\circ to it, both pinned to the wall. A 400\,\mathrm{kg} load hangs from B: W = 400 \times 9.81 = 3924\,\mathrm N.

The pin B is a particle under three forces: the tie's force along BA, the strut's force along CB, and W. Vertically only the strut can help: C_{CB}\sin30^\circ = 3924, so C_{CB} = 7848\,\mathrm N = 7.85\,\mathrm{kN}, pushing up on the pin (compression). Horizontally the strut's push has a component 7848\cos30^\circ = 6797\,\mathrm N to the right, which the tie must cancel by pulling the pin toward the wall: T_{AB} = 6.80\,\mathrm{kN} (tension).

The idea. Meriam (4/1–4/2) defines a structure as any connected system of members built to carry or transfer loads, and studies three kinds in this chapter: trusses, frames and machines. The forces between members are internal to the structure, so the structure must be dismembered and each piece given its own free-body diagram, with the action on one piece equal and opposite to the reaction on the other. Only statically determinate structures are treated.

A truss is a framework of members joined at their ends to form a rigid structure; if the members lie in one plane it is a plane truss. Bridges use plane trusses in pairs, one on each side (Meriam's Fig. 4/1): the deck rests on longitudinal stringers, which rest on cross beams, which deliver the load to the joints of the two trusses. The analysis rests on four assumptions:

1. Two-force members. Each member is a straight link carrying tension or compression (Meriam's Fig. 4/4, Marghitu's Fig. 4.12).
2. Weight neglected, or split. If a member's weight W matters, replace it by W/2 at each end, applied to the pins as external loads. That gives the right average tension or compression but ignores the bending the weight causes.
3. Pinned joints. Welded or riveted joints act as pins if the members' centrelines meet at one point (Meriam's Fig. 4/5).
4. Loads at the joints. True for most trusses: a bridge deck is laid on cross beams that sit on the joints.

Large trusses sit on a roller, rocker or slip joint at one end, so that temperature changes and deflection do not build up forces; otherwise the truss is statically indeterminate (S11). Marghitu (4.5) states the same model, calling the members links and the joints nodes.

The equation (material model: the ideal truss member, from S11's two-force member):

Assumptions: rigid, straight, weightless members; frictionless pins; loads only at the pins.

Worked number. The tie of the bracket, T = 6.80\,\mathrm{kN}, is a steel bar with cross-section 100\,\mathrm{mm^2}: its stress is 6797/(100\times10^{-6}) = 68.0\,\mathrm{MPa}, the same at every section along it, which is why one number per member is enough to size it.

Real example. A bicycle wheel's spokes are thin wires pinned (hooked) at the hub and rim: each one can only pull along its own length. That is why spokes can be so thin, and why a spoke that is pushed simply buckles.

Plausible wrong answer. "The strut carries the weight, 3.92\,\mathrm{kN}." The strut carries 7.85\,\mathrm{kN}, twice the weight, because only its vertical component, half of it at 30^\circ, holds the load up. A shallow strut is pushed much harder than the load it supports.

Check. If the strut were at 45^\circ instead, what would it carry? (3924/\sin45^\circ = 5.55\,\mathrm{kN}.)

Narration spine. A bar between two pins: the pins pull and arrows point outward; a cut slides along the bar and the same T appears at every cut. The arrows flip inward for compression. Then the bracket: the load drops onto pin B, the pin is pulled out as a free particle, and the two member forces grow along their lines until the arrows close.

Bridge. A bracket of two bars is easy. Why are real trusses always built from triangles?

Nugget 02: Triangles Are Rigid, Four-Bar Frames Fold

Claim. Three bars pinned in a triangle cannot change shape, because three lengths fix a triangle. Four bars pinned in a square fold. A truss is rigid if it grows from a triangle by adding two bars and one new joint at a time; such a truss is a simple truss, and it has exactly m = 2j - 3 members.

The picture.

The pins let each bar turn freely, so a frame holds its shape only through the bars' lengths. Three lengths determine a triangle (side–side–side), so the triangle is rigid. Four lengths do not determine a quadrilateral: the square can lean into any parallelogram. One diagonal splits it into two triangles and locks it. Each new joint fixed by two bars to two existing joints is fixed in position (two lengths from two known points), so the structure stays rigid.

Check with numbers. Start with a triangle: j = 3 joints, m = 3 members, and 2j - 3 = 3. Each addition adds one joint and two members, so m rises by 2 while 2j - 3 rises by 2. Meriam's cantilever (the puzzle) has j = 5 and m = 7: 2(5) - 3 = 7. Harper's bridge truss (Nugget 07) has j = 10 and m = 17: 2(10) - 3 = 17.

The idea. Meriam (4/2, Fig. 4/3): the triangle is the basic element of a plane truss. Four or more bars pinned into a polygon form a nonrigid frame, made rigid by adding a diagonal. The structure is extended by adding units of two end-connected bars pinned to two fixed joints. Structures built this way are simple trusses. If more members are present than are needed to prevent collapse, the extras are redundant and the truss is statically indeterminate: equilibrium alone cannot find its forces. Meriam's Fig. 4/2 shows the common types, all made of triangles: Pratt, Howe, Warren, Fink and K roof trusses; Pratt, Howe, Warren, K and Baltimore bridge trusses. Here rigid means both "noncollapsible" and "deforming so little that the geometry does not change". Marghitu (4.5, Fig. 4.11) gives the same construction.

The equation (geometric relationship: counting a simple truss):

Worked number. A roof truss with j = 7 joints built as a simple truss needs m = 2(7) - 3 = 11 members. A twelfth member is redundant.

Real example (one you can see). Look at an electricity pylon, a tower crane's mast or a footbridge: every panel is a triangle. A flat-pack bookcase without its back panel wobbles sideways, a four-bar frame folding into a parallelogram; nailing on the thin back board adds the "diagonal" and it stops.

Plausible wrong answer. "Make the square frame stiffer with thicker bars." A square of pinned bars folds however thick they are; the shape, not the size, gives rigidity.

Check. A pinned frame has 6 joints. How many members does a simple truss on those joints have? (2(6) - 3 = 9.)

Narration spine. Three bars snap into a triangle and are shaken: nothing moves. Four bars form a square and are nudged: it sags into a parallelogram. A diagonal drops in and the square locks. Two new bars swing out from the triangle, meet at a new pin, and the counter beside it ticks j and m up together.

Bridge. A simple truss is rigid. Now the forces in its members.

Nugget 03: Every Joint Is a Particle in Equilibrium

Claim. Cut every member around a pin and the pin is a particle under concurrent forces: the member forces along the members and any load. Two equations, \Sigma F_x = 0 and \Sigma F_y = 0, hold at each pin. Start at a joint with at most two unknowns and walk through the truss joint by joint.

The picture. The puzzle truss solved, and the free-body diagram of its tip joint A.

On a joint's diagram each member force is drawn on the member's side of the pin. Then a tension, which pulls the pin toward the member's far end, is an arrow pointing away from the joint, and a compression is an arrow pointing toward it. Meriam stresses that "tension" and "compression" describe the member, not the joint.

Check with numbers (Meriam's solution; \sin60^\circ = 0.866, \cos60^\circ = 0.5).

• Whole truss first. Moments about E: the cable's moment arm about E is 5\,\mathrm m, so 5T - 20(5) - 30(10) = 0 and T = 80\,\mathrm{kN}. Then \Sigma F_x = 0: 80\cos30^\circ - E_x = 0, E_x = 69.3\,\mathrm{kN}; \Sigma F_y = 0: 80\sin30^\circ + E_y - 20 - 30 = 0, E_y = 10\,\mathrm{kN}.
• Joint A (two unknowns). \Sigma F_y = 0: 0.866AB - 30 = 0, AB = 34.6\,\mathrm{kN} T. \Sigma F_x = 0: AC - 0.5(34.6) = 0, AC = 17.32\,\mathrm{kN} C.
• Joint B (C still has three unknowns). BC must supply an upward component: 0.866BC - 0.866(34.6) = 0, BC = 34.6\,\mathrm{kN} C; BD - 2(0.5)(34.6) = 0, BD = 34.6\,\mathrm{kN} T.
• Joint C. 0.866CD - 0.866(34.6) - 20 = 0, CD = 57.7\,\mathrm{kN} T; CE - 17.32 - 0.5(34.6) - 0.5(57.7) = 0, CE = 63.5\,\mathrm{kN} C.
• Joint E. 0.866DE = 10, DE = 11.55\,\mathrm{kN} C.
• Joint D, the check. Every force on D is now known. Horizontally: 69.28 - 34.64 - 0.5(57.7) - 0.5(11.55) = 0.0; vertically: 40 - 0.866(57.7) + 0.866(11.55) = 0.0. Joint D balances, so the work is right.

The idea. Meriam (4/3) calls this the method of joints. The forces at a pin are concurrent, so only two independent equations are available there. Begin where a known load acts and no more than two forces are unknown; then move to the next such joint. Name each member force by its two joints (AB). Draw each force in its true sense where it is evident; where it is not, assume one, and a negative answer means the other sense. The reactions, found first from the whole truss, leave three equations spare, so the last joints are checks. Graphically, each joint's two equations are a closed force polygon (Meriam's Fig. 4/8).

Marghitu (4.5) writes the action and reaction explicitly: \mathbf F_{1A} is the force of member 1 on node A, and \mathbf F_{A1} = -\mathbf F_{1A} the force of node A on member 1. A tension \mathbf F_{1A} points away from A. Its Example 4.6 (a five-member truss with AB = CD = 1\,\mathrm m, 2F down at D and F to the left at C, F = 100\,\mathrm{daN} = 1000\,\mathrm N) works joint by joint in the same order, each pair of equations solved symbolically in MATLAB: AD = 2\sqrt2F = 2828\,\mathrm N C, DC = 2F = 2000\,\mathrm N C, BC = \sqrt2F = 1414\,\mathrm N C, AC = \sqrt5F = 2236\,\mathrm N T and AB = F = 1000\,\mathrm N. Marghitu labels AB "tension", but its own joint-B diagram has AB pushing on B: AB is in compression (see Self-check 4).

The equation (physical law: equilibrium of each pin, a particle):

Assumptions: those of Nugget 01; sign convention x right, y up.

Worked number. The answer to the puzzle. Tension: AB, BD (the top chord) and CD. Compression: AC, CE (the bottom chord), BC and DE. The hardest working bar is CE, at 63.5\,\mathrm{kN} in compression, more than twice the 30\,\mathrm{kN} tip load. A cantilever bends downward, so its top stretches and its bottom squeezes: the reverse of a beam resting on two supports.

Real example (one you can see). The Forth Bridge in Scotland is built from cantilever trusses reaching out from three towers. Their compression members, along the bottom, are huge steel tubes 3.7 m across; their tension members, along the top, are open lattice girders. The pattern is the puzzle's, made visible: compression needs bulk to resist buckling, tension does not.

Plausible wrong answer. "No bar carries more than the biggest load, 30\,\mathrm{kN}." Moments make the chords near the support carry the bending of the whole cantilever: CE = 63.5\,\mathrm{kN}, CD = 57.7\,\mathrm{kN}.

Check. At joint A, why did \Sigma F_y come first? (It contains only AB; AC is horizontal.)

Narration spine. The truss with its loads. Pin A is pulled out on its own: two arrows grow along its members until the polygon of three forces closes; the member AB lights blue (tension), AC orange (compression). The camera moves to B, C and E in turn, each lighting its members. At D every arrow is already known and the polygon closes by itself: the check.

Bridge. The joint order mattered: at joint C there were three unknowns at first. Is there always an order that works, and when is a truss solvable at all?

Nugget 04: Count Members and Joints Before You Solve

Claim. A plane truss with j joints gives 2j equations. Its unknowns are m member forces plus the support reactions, three for a determinate support. If m + 3 = 2j the count matches. If m + 3 > 2j some members are redundant; if m + 3 < 2j the truss collapses. A matching count is necessary but not sufficient.

The picture.

Each joint is a particle with two equations, so 2j equations in all, whatever order they are solved in. Each member adds one unknown. With a pin and a roller there are three reactions. The square on the left has too few members and folds. The middle one matches. The right one, with both diagonals, has one more member than equations: equilibrium cannot decide how the two diagonals share the load (that needs their stretch, a strength-of-materials question).

Check with numbers. Marghitu's Fig. 4.13 truss: five joints A, B, C on the bottom and E, D above, every member of length a except the diagonals AE and BD; a load F at E, a pin at A and a roller at C. Seven members: 7 + 3 = 10 = 2(5). Writing all ten joint equations as one linear system and solving them together gives the reactions F/2 at A and at C and

Marghitu finds the same values joint by joint (A, C, E, then B as a check).

The idea. Meriam (4/3, "Internal and External Redundancy"): a truss with more supports than needed for a stable position has external redundancy; one with more members than needed to prevent collapse once removed from its supports has internal redundancy. Either makes it statically indeterminate. For a truss that is determinate externally, 2j equations must match m + 3 unknowns. A simple truss satisfies m + 3 = 2j automatically (m = 2j - 3, Nugget 02). Some non-simple trusses, such as the K-truss, satisfy it too. But the count is only a necessary condition: members can be placed so that one panel has two diagonals while another has none, and the truss still folds. If m + 3 > 2j the truss is internally redundant; if m + 3 < 2j it is unstable and collapses under load. Marghitu writes the same condition as n + 3 = 2c (n links, c joints). Meriam (4/5) notes the alternative to walking joint by joint: write all the joint equations at once and solve them simultaneously by computer. For a determinate truss that system is square and has one solution.

The equation (mathematical identity: counting equations against unknowns):

with r = 3 reactions for a pin and a roller.

Worked number. Harper's bridge truss (Nugget 07): j = 10, m = 17, a rocker (one reaction) and a pin (two): 17 + 3 = 20 = 2(10). Add a second diagonal in one panel: 18 + 3 = 21 > 20, one redundant member. Replace the pin by a second rocker: 17 + 2 = 19 < 20, and nothing stops the truss sliding sideways.

Real example. Railway bridges are often built as two separate spans rather than one continuous truss over a middle pier. Each span on a pin and a roller is determinate, so settlement of the pier does not induce extra forces in it.

Plausible wrong answer. "Extra members always make a truss safer, and the equations still give their forces." They make it indeterminate: equilibrium no longer fixes the forces, which then depend on how each member stretches, including stretch from heat and fitting errors.

Check. A truss has 8 joints, 13 members, a pin and a roller. What can be said? (13 + 3 = 16 = 2(8): the count matches; it is determinate if no panel is left without bracing.)

Narration spine. A tally board: each pin lights two equation slots, each member and reaction one unknown slot. For the square, slots are left empty and the square folds. With one diagonal the slots match. With two diagonals one unknown has no slot, and the two diagonals flicker as their share of the load is left undecided.

Bridge. In Marghitu's truss one member, BC, carried nothing. Such members can be spotted before any arithmetic.

Nugget 05: Some Members Carry Nothing, and Cables Only Pull

Claim. At an unloaded joint, a force sum across a member that has no partner along its line shows that the member carries nothing. Two collinear members with one other member: the third is zero and the collinear pair equal. Two non-collinear members alone: both zero. And in a panel braced by two cables, only the cable that the load stretches works.

The picture. Meriam's special conditions (Fig. 4/9) and a cable-braced panel (Fig. 4/10).

(a) Take y perpendicular to the collinear pair: only F_3 has a y component, so F_3 = 0, and then F_1 = F_2 along x, whatever the angle \theta and whether the pair is in tension or compression. (b) Sum forces perpendicular to F_1: only F_2 appears, so F_2 = 0; likewise F_1 = 0. (c) Two crossing pairs: summing perpendicular to one pair gives the other pair equal, and vice versa. (d) Cables cannot push. The load P shears the panel, lengthening one diagonal and shortening the other; the shortened cable goes slack and is left out.

Check with numbers. In Meriam's Fig. 4/8 truss, joint C has a collinear bottom chord and one vertical member and no load, so CE = 0. In Marghitu's Fig. 4.13 truss, BC = 0 because joint C has only BC (horizontal), CD (vertical) and the vertical reaction. In Harper's bridge truss (Nugget 07):

• joint E has only DE and EF, at right angles, no load: DE = EF = 0 (case b);
• joint D has CD and DE collinear and DG across: DG = 0 and CD = DE = 0 (case a);
• joint B has AB and BC collinear and BI across: BI = 0, AB = BC (case a).

Five of seventeen members are idle under this loading, for every direction of the 6 kN load.

The idea. Meriam (4/3, "Special Conditions"): the third member in case (a) keeps the two compressed members aligned against buckling, so it is needed even though it carries nothing; it would carry force if a load with a component across the pair were applied at the joint. Cross-bracing (Fig. 4/10): with two diagonals able to carry tension or compression a panel is indeterminate; with cables, only the one in tension acts. The deflection usually shows which; if not, assume one. If its tension comes out negative, the other cable is the one working and the calculation is repeated with it.

The equation (mathematical identity: force sums at an unloaded joint):

Worked number. A panel 4\,\mathrm m wide and 3\,\mathrm m tall, pin A at the bottom left, roller C at the bottom right, braced by cables AD and CB; a 10\,\mathrm{kN} push to the right at the top-left joint B. Assume CB works: the joint equations give CB = -12.5\,\mathrm{kN}, a push, impossible for a cable. So AD works instead: AD = 12.5\,\mathrm{kN} T (its horizontal component 12.5 \times 4/5 = 10\,\mathrm{kN} carries the push), the top member BD = 10\,\mathrm{kN} C and the right post CD = 7.5\,\mathrm{kN} C.

Real example (one you can see). Look at the bracing on a steel-framed warehouse wall or a radio mast: crossed rods or cables in an X in each panel. Wind from either side is carried by whichever rod it stretches, so thin rods suffice.

Plausible wrong answer. "A zero-force member does nothing, so it can be removed." It holds the joint against buckling and carries load the moment the loading changes.

Check. Joint C in Meriam's Fig. 4/8 gets a 5\,\mathrm{kN} load straight down. Is CE still zero? (No: \Sigma F_y now gives CE = 5\,\mathrm{kN} in tension.)

Narration spine. A joint with three members: an axis swings to lie across the collinear pair, and the third member fades to grey. Two members at an angle, no load: both fade. Then the cross-braced panel is pushed sideways; one cable tightens and glows, the other wrinkles and goes slack.

Bridge. Zero-force members came from choosing the axis of the force sum. That choice helps at loaded joints too, and so does noticing that a member's shape is irrelevant.

Nugget 06: Pick Axes That Skip an Unknown, and Ignore the Shape of the Member

Claim. A force sum along an axis perpendicular to one unknown member force contains only the other unknown, so a joint can be solved one unknown at a time. And a curved member pinned at both ends is still a two-force member: its force acts along the straight line between its pins.

The picture. Meriam's Fig. 4/11, and a curved member from Sample Problem 4/2.

On the left, F_1 is perpendicular to the x-axis, so \Sigma F_x = 0 contains only F_2 and L; F_1 follows from a sum along x', perpendicular to F_2. On the right, the member DE is an arc. Only two forces act on it, at D and at E; they must be equal, opposite and collinear, so they lie along the chord DE. The arc's shape only decides how the member bends inside, not the force at its pins.

Check with numbers (Meriam's Sample Problem 4/2, a quarter-circle truss of curved members with loads L at F and E; outer radius 2R, inner radius R, joints every 22.5^\circ).

• Joint E. The chord DE makes \beta = 180^\circ - 11.25^\circ - 90^\circ = 78.75^\circ \approx 78.8^\circ with the horizontal. \Sigma F_y = 0: DE\sin78.8^\circ - L = 0, DE = 1.020L T; \Sigma F_x = 0: EF - DE\cos78.8^\circ = 0, EF = 0.1989L C.
• Joint F. The chord FD makes \gamma = \tan^{-1}[2R\sin22.5^\circ/(2R\cos22.5^\circ - R)] = 42.1^\circ. The equations -GF\cos67.5^\circ + DF\cos42.1^\circ - 0.1989L = 0 and GF\sin67.5^\circ + DF\sin42.1^\circ - L = 0 are solved together: GF = 0.646L T, DF = 0.601L T.
• Joint D. With \delta = 33.8^\circ and \epsilon = 2.92^\circ (the chords CD and DG), two more simultaneous equations give CD = 1.617L T and DG = -1.147L, that is 1.147L C.

The idea. Meriam notes first that every curved member of this truss is a two-force member, so each acts exactly like a straight one along its chord. Its hint adds that the 11.25^\circ angle could be used directly instead of \beta. For choosing axes (4/3, Fig. 4/11): a sum along x eliminates F_1, a sum along x' eliminates F_2; but when the angles are awkward to find, solving the two equations simultaneously in one set of axes may be easier, as at joints F and D here.

The equation (geometric relationship: projecting a joint's equilibrium on a chosen direction \mathbf n):

Worked number. A joint carries L = 10\,\mathrm{kN} along +x; member 1 leaves it at 30^\circ, member 2 at 135^\circ (both forces taken as tensions along those directions). Project on \mathbf n_1 = (-\sin30^\circ, \cos30^\circ), perpendicular to member 1: F_2\sin(135^\circ - 30^\circ) + 10(-\sin30^\circ) = 0, so F_2 = 5/\sin105^\circ = 5.18\,\mathrm{kN} T. Project on the normal to member 2: F_1 = -7.07/\sin105^\circ = -7.32\,\mathrm{kN}, that is 7.32\,\mathrm{kN} C. One equation, one unknown, each time.

Real example. Arched roof trusses and curved bridge arches made of pinned segments: a curved segment carries its force along its chord, so the designer can shape it for appearance or clearance and still compute it as a straight bar (then check its bending separately).

Plausible wrong answer. "A curved member pushes along its tangent at the pin." The force at a pin of a two-force member lies along the chord. At E the arc of Sample Problem 4/2 is vertical, but the force is at 78.8^\circ.

Check. Why could Meriam not avoid simultaneous equations at joint F by an axis choice? (It could, with an axis perpendicular to DF or GF, but finding those angles is as much work as solving the two equations.)

Narration spine. A joint with a load and two members. An axis rotates until it is perpendicular to member 1, whose arrow fades from the equation; member 2's arrow is all that is left. Then the curved member: its arc fades to a dashed chord, and the pin force slides onto the chord.

Bridge. Every load so far had a fixed direction. A designer must also ask how the member forces change when the load turns.

Nugget 07: Member Forces Follow the Direction of the Load

Claim. Keep the direction of a load as a parameter \theta and the method of joints gives each member force as a function of \theta. Some members change sign, some do not change at all, and the worst case is not always the load straight down. A one-sided support must also be checked over the whole range.

The picture. Harper's version of Meriam's Problem 4/40: a bridge truss 12\,\mathrm m long and 4\,\mathrm m deep in four 3\,\mathrm m panels, top joints A–E, bottom joints J, I, H, G, F. Loads 4, 10 and 8\,\mathrm{kN} hang at I, H, G; a 6\,\mathrm{kN} load at A points down to the right at an angle \theta below the horizontal (60^\circ in Meriam). J sits on a rocker, F on a pin. The diagonals JC and CF cross BI and DG without being connected to them.

Each curve is one member's force as the 6\,\mathrm{kN} load turns through a full circle. IC is flat: it carries only the 4\,\mathrm{kN} at I. BC changes sign: it is pushed by a load pointing right and pulled by one pointing left. JC is always in compression, worst when the load pushes horizontally to the left (\theta = 180^\circ).

Check with numbers. Whole truss, moments about F: 6\sin\theta(12) - 6\cos\theta(4) + 4(9) + 10(6) + 8(3) - J_y(12) = 0, so J_y = 10 + 6\sin\theta - 2\cos\theta\ \mathrm{kN}. Then, with every member assumed in tension (Harper's convention; a negative value is compression):

• joint A: AB + 6\cos\theta = 0 and -6\sin\theta - AJ = 0, so AB = -6\cos\theta, AJ = -6\sin\theta; BC = AB, since BI is a zero-force member;
• joint J (JC has slope 2/3, length factor \sqrt{13}): JC = -\frac{\sqrt{13}}{2}(AJ + J_y) = -\frac{\sqrt{13}}{2}(10 - 2\cos\theta), IJ = 15 - 3\cos\theta;
• joint I (IC is a 3-4-5 line): \frac45IC - 4 = 0, IC = 5\,\mathrm{kN}; IH = IJ - \frac35(5) = 12 - 3\cos\theta.

At Meriam's \theta = 60^\circ: J_y = 14.20\,\mathrm{kN}, BC = -3.00 (3.00 kN C), JC = -16.22 (16.22 kN C), IC = 5.00 T and IH = 10.50\,\mathrm{kN} T.

The idea. Harper first asks whether the truss can be in equilibrium for every \theta. The rocker at J can only push up, so J_y must stay positive. Its smallest value is 10 - \sqrt{6^2 + 2^2} = 10 - \sqrt{40} = 3.68\,\mathrm{kN} > 0, so it always can. Then it analyses joints A, J, I in that order, each with two unknowns, and lets MATLAB evaluate the expressions over 0 \le \theta \le 2\pi instead of substituting by hand. (Harper names the bottom chord member "IG"; it is IH, equal to HG because joint H carries only the vertical CH.)

The equation (mathematical identity: the joint equations with the load angle kept as a parameter):

Worked number. Load straight down, \theta = 90^\circ: J_y = 16\,\mathrm{kN}, BC = 0, JC = -5\sqrt{13} = -18.03\,\mathrm{kN}, IH = 12\,\mathrm{kN}. Load horizontal to the left, \theta = 180^\circ: JC = -6\sqrt{13} = -21.63\,\mathrm{kN}, larger. The diagonal JC must be sized for 21.6\,\mathrm{kN}, not the 18.0\,\mathrm{kN} of the "obvious" vertical load.

Real example. Wind on a sign gantry over a motorway can blow from any side. Each member of the gantry is sized for the worst direction for that member, found by exactly this kind of sweep over the wind angle.

Plausible wrong answer. "The worst case is the load pointing straight down." For JC the worst is a horizontal push (21.6 against 18.0\,\mathrm{kN}), and for BC a vertical load gives zero.

Check. For which \theta is BC in tension? (-6\cos\theta > 0: 90^\circ < \theta < 270^\circ, when the load points to the left.)

Narration spine. The bridge truss with an arrow at A turning like a clock hand. Beside it four curves draw themselves in step: IC flat, BC swinging through zero, JC dipping deepest when the arrow points left. The rocker's reaction readout never reaches zero.

Bridge. One idea has done all the work: every pin is a particle.

Synthesis

The puzzle's cantilever truss, joint by joint: the top members AB and BD and the diagonal CD are stretched; the bottom members AC and CE and the diagonals BC and DE are squeezed. The bar next to the wall at the bottom, CE, works hardest, at 63.5\,\mathrm{kN} in compression under a 30\,\mathrm{kN} tip load. Neither guess survived. The bending of the whole cantilever, stretching its top and squeezing its bottom, is what the chord forces carry, and moments grow toward the support.

What made it solvable is that every member of a truss is a two-force member. That turns a structure of many bodies into a set of particles, the pins, each with two equations, linked by one unknown per member. The truss is solvable by equilibrium when the count of equations matches the unknowns and the members are arranged in triangles. Marghitu's section 4.5 also cuts its truss into two pieces and takes moments about a joint (F_{D6} = F/2 in compression from a single equation): that is the method of sections, S14's subject. A truss is solved by treating every pin as a particle; the triangle makes it rigid, and the count m + 3 = 2j says when the pins' equations are just enough.

• Central idea: truss members are two-force members, so each pin is a particle with two equations, and joints are solved one after another.
• Key relationship: \Sigma F_x = \Sigma F_y = 0 at every joint; m + 3 = 2j for a determinate plane truss on three reactions.
• Most important assumption: pinned joints and loads only at joints; member weights split to the ends; bending ignored.
• Open question: to find the force in one member in the middle of a large truss, must one walk through every joint to reach it? (S14: the method of sections.)

Computed views

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

By hand, the method of joints goes one joint at a time, in an order that keeps two unknowns per joint. A computer writes all the joint equations at once, 2j of them in the m + r unknowns, and solves them in one step. That is Marghitu's "one linear system per truss". The shared solver truss.py (MATLAB: truss_solve.m) takes joint coordinates, member pairs, support directions and loads. It returns every member force, with tension positive, and the reactions. S14 and S15 use the same solver.

Python (numpy). S13_method_of_joints.py solves the puzzle as one 10\times10 system. It then sweeps the 6\,\mathrm{kN} load on Harper's bridge truss round a full circle and solves all twenty joint equations at each angle.

Part 1: Sample Problem 4/1, equilateral triangles of 5 m
  AB:   34.64 T
  AC:   17.32 C
  BC:   34.64 C
  BD:   34.64 T
  CD:   57.74 T
  CE:   63.51 C
  DE:   11.55 C
  cable T = 80.0 kN, E = (-69.3, 10.0) kN
Part 2: Harper's bridge truss, the 6 kN load at theta below horizontal
  theta    J_y      BC      JC      IH   (kN, tension +)
      0    8.00   -6.00  -14.42    9.00
     60   14.20   -3.00  -16.22   10.50
     90   16.00   -0.00  -18.03   12.00
    180   12.00    6.00  -21.63   15.00
    270    4.00    0.00  -18.03   12.00
  worst JC -21.63 kN at 180 deg; smallest J_y 3.68 kN (10 - sqrt 40 = 3.68)

The full solve agrees with Nugget 07's hand formulas at every angle, which also confirms BI is a zero-force member: nothing in the formulas needed it.

MATLAB (Octave-compatible). S13_method_of_joints.m calls truss_solve for both trusses.

AB    34.64 kN
AC   -17.32 kN
BC   -34.64 kN
BD    34.64 kN
CD    57.74 kN
CE   -63.51 kN
DE   -11.55 kN
cable T = 80.0 kN, E = (-69.3, 10.0) kN
 theta     J_y      BC      JC      IH
     0    8.00   -6.00  -14.42    9.00
    60   14.20   -3.00  -16.22   10.50
    90   16.00    0.00  -18.03   12.00
   180   12.00    6.00  -21.63   15.00
   270    4.00    0.00  -18.03   12.00

Traps

• Reading tension and compression from the joint instead of the member. An arrow away from the joint is tension in the member. (From Nugget 03.)
• Starting at a joint with three unknowns. Two equations cannot find them; find the reactions or start at another joint. (From Nugget 03.)
• Throwing away the last joints' equations. With the reactions known, the last three equations are checks; joint D caught nothing only because the work was right. (From Nugget 03.)
• Taking the member force equal to the load. The bracket strut carries 7.85\,\mathrm{kN} for a 3.92\,\mathrm{kN} load. (From Nugget 01.)
• Trusting the count alone. m + 3 = 2j is necessary, not sufficient; a panel without a diagonal folds. (From Nugget 04.)
• Removing zero-force members. They brace joints and carry load when it changes. (From Nugget 05.)
• Letting a cable push. A negative cable force means the other diagonal is the one working. (From Nugget 05.)
• Directing a curved member's force along its tangent. It acts along the chord between its pins. (From Nugget 06.)
• Checking only the vertical load. The worst direction differs member by member: JC peaks at \theta = 180^\circ. (From Nugget 07.)

Self-check

1. A roof truss has 7 joints and 11 members on a pin and a roller. Is the count right for a determinate truss? What if a twelfth member is added? Could the 11-member truss still collapse?
2. A three-member truss: pin A at (0, 0), roller B at (6, 0)\,\mathrm m, apex C at (2, 3)\,\mathrm m, and 12\,\mathrm{kN} hanging from C. Find the reactions and the three member forces.
3. Repeat the puzzle truss with 40\,\mathrm{kN} at A (still 20\,\mathrm{kN} at C). Find the cable tension and every member force. Which force did not change, and why?
4. Marghitu's Example 4.6 with F = 1000\,\mathrm N: A(0, 0) pin, B(1, 0) roller, D(1, 1), C(2, 1)\,\mathrm m; members AB, BC, DC, AD, AC; 2000\,\mathrm N down at D and 1000\,\mathrm N to the left at C. Find every member force, stating T or C.
5. In Harper's bridge truss, which members carry no force for any \theta? Justify each from a joint.
6. A truss member is a 6\,\mathrm m steel section of 30\,\mathrm{kg/m}. How is its weight included in a truss analysis, and what does that model miss?
7. The cable-braced panel of Nugget 05 is turned round: roller at A, pin at C, and a 6\,\mathrm{kN} push to the left at the top-right joint D. Which cable works? Find all member forces and reactions.
8. A joint carries 10\,\mathrm{kN} along +x; member 1 leaves it at 30^\circ and member 2 at 135^\circ. Find both member forces by projecting on axes perpendicular to each member.
9. In Harper's truss, find J_y, BC, JC and IH for \theta = 90^\circ and \theta = 180^\circ.
10. Harper concluded the rocker can hold the truss for every \theta. What is the largest load P at A (in place of 6\,\mathrm{kN}) for which that stays true? With P = 20\,\mathrm{kN}, over what range of \theta would the rocker have to pull?
11. A curved two-force member is pinned at A(0, 0) and B(3, 4)\,\mathrm m and carries 15\,\mathrm{kN} in compression. Give the force it exerts on pin B in components.
12. Marghitu's Fig. 4.13 truss with a = 2\,\mathrm m and F = 6\,\mathrm{kN} at E: find every member force, and check the count.

Fully worked solutions

1. m + 3 = 11 + 3 = 14 = 2(7): the count matches. With 12 members, 15 > 14: one redundant member, statically indeterminate. Yes, the 11-member truss could still collapse: the count is necessary, not sufficient. If one panel is a square with no diagonal while another has two, the count still reads 14 but the square panel folds.

2. Whole truss: \Sigma M_A = 0: 6B_y - 12(2) = 0, B_y = 4\,\mathrm{kN}; A_y = 12 - 4 = 8\,\mathrm{kN}; A_x = 0. Joint B, every member taken in tension: BC pulls B along (-4, 3)/5 and AB pulls it along (-1, 0). \Sigma F_y: 4 + \frac35BC = 0, BC = -6.667\,\mathrm{kN} (6.67 kN C). \Sigma F_x: -AB - \frac45BC = 0, so AB = -\frac45(-6.667) = 5.333\,\mathrm{kN} T. Joint A: AC along (2, 3)/\sqrt{13}; \Sigma F_y: 8 + \frac{3}{\sqrt{13}}AC = 0, AC = -8\sqrt{13}/3 = -9.615\,\mathrm{kN} (9.62 kN C). Check \Sigma F_x at A: 5.333 + \frac{2}{\sqrt{13}}(-9.615) = 5.333 - 5.333 = 0.

3. Moments about E: 5T - 20(5) - 40(10) = 0, T = 100\,\mathrm{kN}; E_x = 100\cos30^\circ = 86.60\,\mathrm{kN}, E_y = 20 + 40 - 100\sin30^\circ = 10\,\mathrm{kN}. Joint A: AB = 40/0.866 = 46.19\,\mathrm{kN} T, AC = 0.5(46.19) = 23.09\,\mathrm{kN} C. Joint B: BC = 46.19 C, BD = 46.19\,\mathrm{kN} T. Joint C: CD = (20 + 0.866 \times 46.19)/0.866 = 69.28\,\mathrm{kN} T; CE = 23.09 + 0.5(46.19) + 0.5(69.28) = 80.83\,\mathrm{kN} C. Joint E: 0.866DE = 10, DE = 11.55\,\mathrm{kN} C, unchanged, because E_y is unchanged: the extra 10\,\mathrm{kN} at A raises the cable's vertical pull by exactly 10\,\mathrm{kN}.

4. Joint D (members AD at 45^\circ and DC horizontal; 2000\,\mathrm N down). \Sigma F_y: the only vertical member force is AD's, which must push D up: AD\sin45^\circ = 2000, AD = 2828\,\mathrm N C. \Sigma F_x: AD's push on D has 2000\,\mathrm N toward +x, so DC pushes D back: DC = 2000\,\mathrm N C. Joint C: DC pushes C to the right with 2000; the load is 1000 to the left; take AC and BC as tensions, pulling C along (-2, -1)/\sqrt5 and (-1, -1)/\sqrt2. \Sigma F_y: -AC/\sqrt5 - BC/\sqrt2 = 0, so BC = -\sqrt2\,AC/\sqrt5. \Sigma F_x: 2000 - 1000 - 2AC/\sqrt5 - BC/\sqrt2 = 0, which becomes 1000 - AC/\sqrt5 = 0: AC = 1000\sqrt5 = 2236\,\mathrm N T and BC = -1000\sqrt2 = -1414\,\mathrm N (1414 N C). Joint B: BC pushes B along (-1, -1)/\sqrt2, that is 1000\,\mathrm N left and 1000\,\mathrm N down; the roller gives B_y = 1000\,\mathrm N up; so AB must push B 1000\,\mathrm N to the right: AB = 1000\,\mathrm N compression (Marghitu prints "tension"). Whole truss check: A_x = 1000\,\mathrm N, A_y = 1000\,\mathrm N, B_y = 1000\,\mathrm N, and at A: AD pushes (-2000, -2000), AC pulls (2000, 1000), AB pushes (-1000, 0), the reactions add (1000, 1000): sum zero.

5. DE and EF: joint E has only these two, at right angles, and no load (case b). DG: joint D has CD, DE collinear and DG across, no load (case a), so DG = 0 and CD = DE = 0. BI: joint B, same pattern, BI = 0. That is BI, CD, DE, DG and EF, for every \theta, since the 6\,\mathrm{kN} load acts at A, not at these joints.

6. W = 6 \times 30 \times 9.81 = 1766\,\mathrm N; apply W/2 = 882.9\,\mathrm N downward at each of its two joints, as external loads, and solve as usual. This gives the correct average tension or compression in the member but not the bending its own weight causes between the joints.

7. The push moves the top to the left: B moves away from C, so CB lengthens and works, while AD shortens and goes slack. Joint D: load 6 left; BD horizontal, CD vertical, and no cable now: \Sigma F_x: -6 - BD = 0 (a tension BD pulls D toward B, to the left), BD = -6\,\mathrm{kN} (6 kN C); \Sigma F_y: CD = 0. Joint B: BD pushes B left with 6; CB pulls B toward C along (4, -3)/5; AB vertical: \Sigma F_x: -6 + \frac45CB = 0, CB = 7.5\,\mathrm{kN} T; \Sigma F_y: -\frac35(7.5) - AB = 0, AB = -4.5\,\mathrm{kN} (4.5 kN C). Joint A: AB pushes A down with 4.5, so the roller gives A_y = 4.5\,\mathrm{kN} up, and AC = 0. Joint C: CB pulls C with (-6, 4.5), so the pin gives C_x = 6\,\mathrm{kN} to the right and C_y = 4.5\,\mathrm{kN} down: it holds that corner down. Assuming AD instead gives AD = -7.5\,\mathrm{kN}, a push, which confirms the choice. (Had the roller been at C, it would have had to pull down: a roller cannot, so the supports are part of the design.)

8. Tensions F_1 along \mathbf u_1 = (\cos30^\circ, \sin30^\circ) and F_2 along \mathbf u_2 = (\cos135^\circ, \sin135^\circ). Normal to member 1: \mathbf n_1 = (-\sin30^\circ, \cos30^\circ); \mathbf u_2\cdot\mathbf n_1 = \sin(135^\circ - 30^\circ) = 0.9659, \mathbf L\cdot\mathbf n_1 = -5; so F_2 = 5/0.9659 = 5.18\,\mathrm{kN} T. Normal to member 2: \mathbf n_2 = (-\sin135^\circ, \cos135^\circ); \mathbf u_1\cdot\mathbf n_2 = \sin(30^\circ - 135^\circ) = -0.9659, \mathbf L\cdot\mathbf n_2 = -7.071; so F_1 = -(-7.071)/(-0.9659) = -7.32\,\mathrm{kN}, that is 7.32\,\mathrm{kN} C.

9. \theta = 90^\circ: J_y = 10 + 6 - 0 = 16\,\mathrm{kN}; BC = 0; JC = -\frac{\sqrt{13}}{2}(10) = -18.03\,\mathrm{kN}; IH = 12\,\mathrm{kN} T. \theta = 180^\circ: J_y = 10 + 0 + 2 = 12\,\mathrm{kN}; BC = +6\,\mathrm{kN} (tension); JC = -\frac{\sqrt{13}}{2}(12) = -21.63\,\mathrm{kN}; IH = 15\,\mathrm{kN} T.

10. With load P: J_y = 10 + P\sin\theta - \frac{P}{3}\cos\theta, whose minimum is 10 - P\sqrt{1 + 1/9} = 10 - 1.054P. It stays positive for P < 10/1.054 = 9.49\,\mathrm{kN}. With P = 20\,\mathrm{kN} the minimum is 10 - 21.08 = -11.08\,\mathrm{kN}: the rocker would have to pull for 226.8^\circ < \theta < 350.1^\circ (the load pointing upward), and the truss would lift off at J.

11. The force acts along the chord AB, direction (3, 4)/5, at 53.13^\circ. In compression the member pushes pin B away from A: 15(0.6, 0.8) = (9, 12)\,\mathrm{kN}, whatever the member's curve.

12. The forces scale with F and do not depend on a: AB = F/2 = 3\,\mathrm{kN} T, AE = F\sqrt2/2 = 4.243\,\mathrm{kN} C, BE = 3\,\mathrm{kN} C, BC = 0, CD = 3\,\mathrm{kN} C, ED = 3\,\mathrm{kN} C, BD = 4.243\,\mathrm{kN} T; reactions 3\,\mathrm{kN} up at A and at C. Count: m + 3 = 7 + 3 = 10 = 2(5).