Study Companion: S15 — Space Trusses

S13 and S14 solved plane trusses: bars in one plane, two equations per joint, the triangle as the unit of rigidity. Many structures are not flat. A transmission tower, a crane's mast, the roof of an airport hall or the truss of a space station has bars pointing every way in space, and loads that do too. This lesson carries the plane-truss ideas into three dimensions. Every one of them survives with a change of number. The triangle becomes the tetrahedron, two equations per joint become three, m + 3 = 2j becomes m + 6 = 3j, and every member force has to be written as a vector. The method of sections, so powerful in a plane, turns out to be of little use in space.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 4/5 ("Space Trusses"), with Sample Problem 4/5 and the Chapter Review (4/7). Harper, Solving Statics Problems in MATLAB, "Space Trusses" (the same problem, numbered Sample Problem 4/4 in Harper's edition). Axes right-handed, z up; vectors in \mathbf i, \mathbf j, \mathbf k as in S07–S09 and S12.

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 4/5. A rigid tetrahedron ABCD stands on a ball-and-socket joint at A: B is 4\,\mathrm m above A, C is 3\,\mathrm m from A along y and D is 3\,\mathrm m along x. Three links, one at B and two at D, stop it turning about the three axes. A fifth joint E, at (3, 3, 4)\,\mathrm m, is fixed to the tetrahedron by three bars EB, EC and ED, and a horizontal load L pushes on E in the -x direction.

The load is horizontal and along x. Which of the three bars at E is pushed and which pulled, and by how much? A natural guess: EB is the only horizontal bar, so it carries the horizontal load, and the two sloping bars EC and ED, which only steady E vertically, carry little. The answer is in Nugget 03 and the Synthesis.

Nugget 01: In Space the Tetrahedron Is the Rigid Unit

Claim. A triangle of bars is rigid in its own plane but can swing out of it about any of its sides. Six bars on the edges of a tetrahedron cannot move at all. A space truss grows rigidly by adding three bars from three fixed joints to one new joint; such a truss is a simple space truss, with m = 3j - 6 members.

The picture. Meriam's Fig. 4/13 in three steps.

On the left, bars AD and BD pinned to the ground at A and B form a triangle that keeps its shape but swings about the line AB like a door. A third bar CD to another fixed point C stops that swing: three lengths from three fixed points fix a point in space. In the middle, three more bars AB, BC, CA replace the ground: six bars, four joints, a tetrahedron that is rigid on its own. On the right, three bars from three joints of a rigid structure meet at a new joint, which is then fixed in space too.

Check with numbers. A tetrahedron: j = 4, m = 6 = 3(4) - 6. Each new joint adds 1 to j and 3 to m, so 3j - 6 keeps pace. Meriam's Fig. 4/13c builds a simple space truss with joints A–H, j = 8: m = 3(8) - 6 = 18 bars. The puzzle's truss has j = 5 and m = 6 + 3 = 9 = 3(5) - 6.

The idea. Meriam (4/5): a space truss is the three-dimensional counterpart of the plane truss. Its idealised joints are ball-and-socket joints (S10, S12), which transmit force but no moment; in practice, joints whose members' centrelines meet at a point justify the same two-force model, as riveted and welded joints did in a plane. The basic noncollapsible unit is six bars on the edges of a tetrahedron. A new rigid unit is formed with three concurrent bars attached to three fixed joints of the existing structure; a structure built so is a simple space truss.

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

Worked number. A lattice tower built as a simple space truss with 12 joints needs 3(12) - 6 = 30 bars.

Real example (one you can see). Look up at the roof of an airport terminal, an exhibition hall or a petrol-station canopy: many are space frames, a layer of bars in triangles and pyramids joined at ball-shaped nodes, the tetrahedron repeated across the whole roof.

Plausible wrong answer. "Triangles make anything rigid, so a flat triangulated truss is rigid in space too." It is rigid in its plane only; out of plane it swings about its supports unless braced, which is why plane trusses are used in pairs joined by bracing.

Check. How many bars does a simple space truss with 6 joints have? (12.)

Narration spine. Two bars on the ground swing to and fro about the line between their feet; a third bar drops in and the swing stops. Three bars replace the ground and the tetrahedron lifts free, rigid. Three new bars reach out to a new node and lock.

Bridge. With three bars per joint, the count changes.

Nugget 02: Three Equations per Joint: m + 6 = 3j

Claim. Each joint of a space truss is a particle in space with three force equations: 3j in all. The unknowns are m member forces and, for a determinate support, six reactions. If m + 6 = 3j the count matches; it is necessary but not sufficient, as in a plane.

The picture. A tripod: apex P 3\,\mathrm m above the ground and three legs to feet on a circle of radius 2\,\mathrm m, 120^\circ apart.

Four joints give 3 \times 4 = 12 equations. The unknowns are the three leg forces and the support reactions. Here each foot sits on a ball-and-socket support, three components each, nine in all: 3 + 9 = 12. (The general rule takes six reactions; the tripod's supports supply nine, and its members three fewer than a tetrahedron's six, which balances: m + r = 3j.)

Check with numbers. A 12\,\mathrm{kN} load hangs from P. Each leg runs from P to a foot along a unit vector with vertical component -3/\sqrt{13} (leg length \sqrt{2^2 + 3^2} = \sqrt{13} = 3.606\,\mathrm m). By symmetry the three leg forces are equal. Vertically: 3F(3/\sqrt{13}) = 12, so F = 4\sqrt{13}/3 = 4.81\,\mathrm{kN}, compression in each leg. Horizontally the three leg forces cancel by symmetry.

The idea. Meriam (4/5): when a space truss is supported so that it is statically determinate as a whole, the equilibrium of each joint gives three scalar equations, 3j in all. The unknowns are the m member forces plus six support reactions in the general case. So m + 6 = 3j holds for an internally determinate space truss, and a simple space truss satisfies it automatically (starting from the tetrahedron, three members and one joint at a time). As in the plane, the count is necessary, not sufficient: members can be arranged so that some do not help stability. If m + 6 > 3j there are redundant members; if m + 6 < 3j the truss is unstable and collapses under load. Meriam stresses that the count is especially useful in preliminary design of space trusses, because the right arrangement is far less obvious by eye than in a plane. The six reactions must themselves be adequately placed (S12): in the puzzle, three at A and the links 1, 2, 3.

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

Worked number. The puzzle's truss: m = 9, j = 5: 9 + 6 = 15 = 3(5). Meriam checks this first, together with the six properly placed constraints.

Real example. A camera tripod with locked legs is a three-member space truss on three foot supports. Spreading the legs wider lowers the leg forces: at a foot radius of 2\,\mathrm m and a height of 3\,\mathrm m each leg carries 4.81\,\mathrm{kN} for 12\,\mathrm{kN}; set the legs nearly vertical and each carries just over a third of the load.

Plausible wrong answer. "Each leg carries a third of the load, 4\,\mathrm{kN}." The legs lean, so each must push harder than its share: 4.81\,\mathrm{kN}.

Check. A space truss has 10 joints, 24 members and six support reactions. What does the count say? (24 + 6 = 30 = 3(10): determinate, if well arranged.)

Narration spine. The tripod. Each joint lights three equation slots; legs and reactions fill unknown slots until the tally balances. The load drops on P and the three legs glow equally orange.

Bridge. Three equations per joint, in three unknown directions. Writing them needs vectors.

Nugget 03: Every Joint Is a Particle in Space

Claim. At a joint, write each member force as its magnitude times the unit vector from the joint toward the member's far end, so that a positive magnitude means tension. The vector equation \Sigma\mathbf F = \mathbf 0 then splits into three scalar equations. Start at a joint with a known force and at most three unknowns.

The picture. The free-body diagram of the puzzle's joint E.

All three member forces are drawn as tensions, pulling E toward B, C and D. Their unit vectors come from coordinates: E = (3, 3, 4), B = (0, 0, 4), C = (0, 3, 0), D = (3, 0, 0)\,\mathrm m, so \overrightarrow{EB} = (-3, -3, 0), \overrightarrow{EC} = (-3, 0, -4) and \overrightarrow{ED} = (0, -3, -4), lengths 3\sqrt2, 5 and 5\,\mathrm m.

Check with numbers (Meriam's solution).

With \mathbf L = -L\mathbf i, \Sigma\mathbf F = \mathbf 0 at E collects into

Each coefficient is zero. The \mathbf k equation gives F_{EC} = -F_{ED}. Adding the \mathbf i and \mathbf j equations then cancels the sloping bars: -L - 2F_{EB}/\sqrt2 = 0, so F_{EB} = -L/\sqrt2 = -0.707L. The \mathbf j equation gives F_{ED} = -\frac53F_{EB}/\sqrt2 = 5L/6, and F_{EC} = -5L/6.

The idea. Meriam (4/5): the method of joints extends directly to space by satisfying the complete vector equation \Sigma\mathbf F = \mathbf 0 (Eq. 4/1) at each joint. Begin at a joint where at least one known force acts and no more than three are unknown, then move to neighbours with no more than three unknowns. This keeps the simultaneous equations small. Meriam's hint 2: with every force assumed in tension, away from the joint, a negative value means compression. Vector notation is a considerable advantage in space. Harper (4.3) solves the same joint the same way, and notes that joint E alone answers the question: its three unknowns are exactly the forces wanted.

The equation (physical law: equilibrium of a particle, in vector form; Eq. 4/1):

with \mathbf P the applied load. Assumptions: two-force members, ball-and-socket joints, loads at joints.

Worked number. The answer to the puzzle: EB is pushed with 0.707L (compression), EC is pushed with 0.833L and ED is pulled with 0.833L. The two sloping bars carry more than the horizontal one. They form a pair: one pushes E along +x with \frac35(0.833L) = 0.5L, the other pulls it along -y with 0.5L, and their vertical components cancel.

Real example. Satellite antenna masts and the trusses of space stations are space trusses analysed joint by joint like this: every node a particle, every strut a vector.

Plausible wrong answer. "EB is the horizontal bar, so it carries the horizontal load." It carries 0.707L; the load is shared with EC and ED, which carry 0.833L each.

Check. Why could the \mathbf k equation be solved before the others? (Only EC and ED have z components; EB and L are horizontal.)

Narration spine. Joint E is lifted out of the truss. Three dashed lines run to B, C, D; three unit vectors grow along them, labelled with their components. The vector equation splits into three coloured rows; the \mathbf k row ties F_{EC} to F_{ED}, then the \mathbf i and \mathbf j rows give the rest, and two arrows flip to compression.

Bridge. From E, Meriam moves to joint C, then B, D, A. The supports appear at the last joints, and the whole truss can predict them.

Nugget 04: The Whole Truss Gives the Reactions and a Check

Claim. Treated as one rigid body, a space truss has six equilibrium equations, enough for six support reactions. Found first, they shorten the joint work; found last, from the final joints, they must agree with the whole-truss values, which checks everything.

The picture. The puzzle's truss as one free body.

The only external forces are the load at E and the six reactions: three at the ball joint A, one in link 1 at B (along y), and two in links 2 and 3 at D (along y and z).

Check with numbers. Moments about A, so the three reactions at A drop out:

• load: \mathbf r_E\times(-L\mathbf i) = (3, 3, 4)\times(-L, 0, 0) = (0,\ -4L,\ 3L);
• link 1 at B(0, 0, 4): (0, 0, 4)\times(0, B_y, 0) = (-4B_y, 0, 0);
• links 2, 3 at D(3, 0, 0): (3, 0, 0)\times(0, D_y, D_z) = (0,\ -3D_z,\ 3D_y).

\Sigma M_x: -4B_y = 0, B_y = 0. \Sigma M_y: -4L - 3D_z = 0, D_z = -4L/3. \Sigma M_z: 3L + 3D_y = 0, D_y = -L. Forces: A_x - L = 0, A_x = L; A_y + B_y + D_y = 0, A_y = L; A_z + D_z = 0, A_z = 4L/3. These are exactly Meriam's hint 1: \mathbf A_x = L\mathbf i, \mathbf A_y = L\mathbf j, \mathbf A_z = \frac43L\mathbf k, \mathbf B_y = \mathbf 0, \mathbf D_y = -L\mathbf j, \mathbf D_z = -\frac43L\mathbf k.

The idea. Meriam (4/5, SP 4/5): the external reactions can be calculated easily as a first step, though they also follow from solving the joints in succession. After joint E, the next joint is C (known F_{EC} and three unknowns F_{CB}, F_{CA}, F_{CD}), then B, D and A, each with no more than three unknowns. The reactions that come out of the last joints must agree with those from the truss as a whole. Carried through, all nine members are: AB = -\frac43L, AC = AD = -L, BC = BD = \frac56L, CD = \frac{\sqrt2}{2}L, with the three at E above.

The equation (physical law: S12's six equations for the truss as one body):

Worked number. With L = 12\,\mathrm{kN}: A_x = A_y = 12\,\mathrm{kN}, A_z = 16\,\mathrm{kN} up, D_y = 12\,\mathrm{kN} along -y, D_z = 16\,\mathrm{kN} down, B_y = 0. Link 3 holds D down: the load tends to tip the truss up at D.

Real example. Before a lattice tower is erected, its foundation bolts are sized from exactly this whole-structure calculation: the six reactions under wind, long before any member is checked.

Plausible wrong answer. "Link 1 must carry something, or it would not be there." Under this load B_y = 0; link 1 is needed for loads with a moment about x through A, and for stability.

Check. Why do the reactions at A not enter the moment equations about A? (They act at A: zero arm.)

Narration spine. The whole truss glows as one body. Six reaction arrows appear at A, B, D; the moment of the load about A appears as a vector and the reaction moments grow to cancel it, one component at a time. Then the joints are solved in order E, C, B, D, A, and the last reactions land on the same arrows.

Bridge. In a plane, the method of sections went straight to one member. In space, does it?

Nugget 05: Sections Rarely Pay in Space; Solve All the Joints at Once

Claim. A section of a space truss is a rigid body with six equations, so it can cut up to six unknown members. But a moment axis that meets or parallels all but one of six skew members is rarely available, so sections seldom isolate one force. The practical alternative is to write all 3j joint equations and solve them together by computer.

The picture. A plane slicing a space truss.

A moment about an axis eliminates a member only if the member's line meets the axis or is parallel to it (S08, S12). In a plane every line crosses every other, and two unknowns always meet at a point. In space the cut members are generally skew: no single axis meets five of them.

Check with numbers. The puzzle's truss has j = 5: 3j = 15 joint equations in m + 6 = 15 unknowns (nine member forces and six reactions). Solved as one 15 \times 15 linear system, they return every value of Nuggets 03 and 04 at once: F_{EB} = -0.707L, F_{EC} = -0.833L, F_{ED} = 0.833L, AB = -1.333L, AC = AD = -L, BC = BD = 0.833L, CD = 0.707L, A_x = A_y = L, A_z = 1.333L, B_y = 0, D_y = -L, D_z = -1.333L.

The idea. Meriam (4/5): the method of sections may be applied to space trusses, with the two vector equations \Sigma\mathbf F = \mathbf 0 and \Sigma\mathbf M = \mathbf 0, six scalar equations, so a section should not cut more than six members of unknown force. It is not widely used, because a moment axis can seldom be found that eliminates all but one unknown, as in plane trusses. The joint-by-joint technique keeps the simultaneous equations to three at a time, and Meriam recommends it, though it is not easily reduced to a routine. The alternative: write the 3j equations of all joints; if the truss is noncollapsible off its supports, has six support reactions and no redundant members, 3j = m + 6 and the whole system can be solved simultaneously, usually by computer. Then no starting joint needs to be chosen.

The equation (mathematical identity: all joints at once):

Each column of \mathbf A holds a member's unit vector at its two joints (with opposite signs) or a reaction's direction; \mathbf b holds the loads.

Worked number. A space roof with 200 joints has 600 joint equations. By hand, hopeless; as one linear system, a computer solves it in a fraction of a second. That is how every space frame is analysed in practice.

Real example. Structural analysis programs for stadium roofs and transmission towers assemble exactly this matrix, one column per member, and solve it in one step.

Plausible wrong answer. "Cut through three members and take moments, as in a plane." Three moment equations about point-pairs no longer isolate one force each; in space a section with three cut members can usually be solved only with all six equations together.

Check. A space section cuts seven members of unknown force. Can it be solved from that section alone? (No: six equations, seven unknowns.)

Narration spine. A plane slides through a space frame and five members glow where it cuts. An axis tries one orientation after another, never touching more than two glowing lines. The picture dissolves into a grid of numbers, the matrix, and every member colours in at once.

Bridge. Once forces are formulas, the geometry can be varied. Harper varies the height.

Nugget 06: Member Forces as Functions of the Height

Claim. Keep the truss height d as a symbol and the joint-E forces become formulas. The sloping bars' forces grow with height as their slope steepens; the horizontal bar's force does not change at all.

The picture. Harper's plot for L = 10\,\mathrm{kN} and 2 \le d \le 10\,\mathrm m.

F_{ED} (tension) and F_{EC} (compression) are mirror images and grow steadily. F_{EB} is a flat line at -7.07\,\mathrm{kN}.

Check with numbers. With height d, \overrightarrow{EC} = (-3, 0, -d) and \overrightarrow{ED} = (0, -3, -d), both of length \sqrt{9 + d^2}. The \mathbf k equation still gives F_{EC} = -F_{ED}, and adding the \mathbf i and \mathbf j equations still gives F_{EB} = -L/\sqrt2. The \mathbf j equation then gives

At d = 4\,\mathrm m: \frac{L}{6}(5) = 0.833L, Meriam's result. With L = 10\,\mathrm{kN}: F_{ED} = 6.01, 8.33, 11.18, 17.40\,\mathrm{kN} at d = 2, 4, 6, 10\,\mathrm m, and F_{EB} = -7.07\,\mathrm{kN} throughout.

The idea. Harper (4.3) needs no whole-truss diagram: joint E has the external force L and exactly the three unknowns of interest, so one free-body diagram answers the question. It writes the three forces as Cartesian vectors with d in them, solves the three scalar equations, and plots over 2 \le d \le 10\,\mathrm m. Its conclusion: ED is in tension, EC and EB in compression, for every height.

The equation (mathematical identity: joint E with the height kept as a parameter):

Worked number. Why does F_{ED} grow with height? The horizontal component of ED must stay 0.5L (it balances the y pull of EB), so F_{ED}\cdot 3/\sqrt{9 + d^2} = 0.5L. As d grows the bar points more steeply, its horizontal share 3/\sqrt{9 + d^2} falls, and the force must rise: 0.5L \times \sqrt{109}/3 = 1.74L at d = 10\,\mathrm m.

Real example. Tall, slender lattice towers carry horizontal wind loads with steep bracing that must be much stronger than the wind load itself, for exactly this reason; designers add intermediate bracing levels to keep the slopes shallow.

Plausible wrong answer. "A taller truss is stiffer, so its members carry less." The steep bars carry more; only EB, which is horizontal like the load, is unaffected.

Check. At what height is F_{ED} = L? (\sqrt{9 + d^2} = 6, d = \sqrt{27} = 5.20\,\mathrm m.)

Narration spine. The truss stretches upward as a slider moves d from 2 to 10\,\mathrm m. Bars EC and ED tilt toward vertical and their force gauges climb; EB's gauge does not move. The three curves trace themselves beside it.

Bridge. Joints in space are particles; a space truss is a set of them. The last lesson of the part turns to structures whose members are not two-force members.

Synthesis

The puzzle's three bars at E: EB, the horizontal one, is pushed with 0.707L; EC is pushed and ED pulled with 0.833L each. The guess that the horizontal bar takes the horizontal load failed. EB and EC each push E along x with 0.5L, sharing the load; EB's push also has a y component of 0.5L, which ED balances; and the vertical components of EC and ED cancel. Every bar is needed for the one load. The whole truss confirmed it: six reactions from six equations, then every member from joints in order, or all fifteen unknowns from one linear system.

Every plane-truss idea went into space with a change of number. The triangle became the tetrahedron; m = 2j - 3 became m = 3j - 6; two equations per joint became three; m + 3 = 2j became m + 6 = 3j. The one idea that did not travel well is the method of sections, because skew lines rarely cross. A space truss is a set of particles, one per joint, each with three equations; write every member force as a magnitude times a unit vector, start where three unknowns meet a known force, and let the count m + 6 = 3j say when the joints' equations are just enough.

• Central idea: each joint of a space truss is a particle with \Sigma\mathbf F = \mathbf 0 in three components; member forces are magnitudes times unit vectors.
• Key relationship: m + 6 = 3j; \mathbf F_k = F_k\mathbf n_k with F_k > 0 tension.
• Most important assumption: ball-and-socket joints with concurrent centrelines, loads at joints, six adequately placed support reactions.
• Open question: in a frame or machine, some members carry forces at three or more points, or couples. How are those structures taken apart? (S16.)

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). S15_space_trusses.py solves the puzzle whole: five joints give fifteen equations in nine bar forces and six link reactions, solved by the shared truss.py. It then runs Harper's height sweep with joint E alone, three equations in three bar forces.

Part 1: Sample Problem 4/5 with L = 1 (forces per unit load)
  EB:    0.71 C
  EC:    0.83 C
  ED:    0.83 T
  reactions per unit L: A = [1.    1.    1.333], B_y = 0.000, D_y = -1.000, D_z = -1.333
  with L = 12 kN: A = [12. 12. 16.], D = [-12. -16.] kN
Part 2: Harper's sweep, B and E at height d, L = 10 kN
  d (m)    EB      EC      ED   (kN, tension +)
    2.0   -7.07   -6.01    6.01
    4.0   -7.07   -8.33    8.33
    6.0   -7.07  -11.18   11.18
   10.0   -7.07  -17.40   17.40
  ED at d = 10 per unit L: 1.740 (0.5 sqrt(109)/3 = 1.740)

EB does not change with d: it stays horizontal, and its job (0.5L along y to balance ED) does not depend on the height.

MATLAB (Octave-compatible). S15_space_trusses.m calls truss_solve.m for the whole truss and the backslash operator for joint E.

EB -0.707, EC -0.833, ED 0.833 per unit L (tension +)
reactions per unit L: A = [1.000 1.000 1.333], B_y = 0.000, D_y = -1.000, D_z = -1.333
 d (m)      EB      EC      ED
   2.0   -7.07   -6.01    6.01
   4.0   -7.07   -8.33    8.33
   6.0   -7.07  -11.18   11.18
  10.0   -7.07  -17.40   17.40

Traps

• Treating a plane triangle as rigid in space. It swings about any side; the tetrahedron is the unit. (From Nugget 01.)
• Using m + 3 = 2j for a space truss. In space it is m + 6 = 3j. (From Nugget 02.)
• Trusting the count without checking the supports. Six reactions must also be adequately placed (S12). (From Nugget 02.)
• Writing a member's direction from the drawing instead of coordinates. Unit vectors come from coordinate differences divided by lengths. (From Nugget 03.)
• Mixing sign conventions. With every force away from the joint, negative means compression; drawing some toward the joint mixes the meanings. (From Nugget 03.)
• Starting at a joint with four unknowns. Three equations cannot find them. (From Nugget 03.)
• Expecting a space section to isolate one force. Skew members rarely meet one axis; use joints or the full linear system. (From Nugget 05.)
• Assuming taller means lighter loads. Steeper bars carry more for the same horizontal force. (From Nugget 06.)

Self-check

1. A simple space truss is built from a tetrahedron by adding joints one at a time until it has 8 joints. How many members has it? With six support reactions, is it determinate?
2. The tripod of Nugget 02 (apex 3\,\mathrm m high, feet on a 2\,\mathrm m circle at 120^\circ) carries 12\,\mathrm{kN} straight down. Find the leg forces.
3. The same tripod carries instead a horizontal 6\,\mathrm{kN} at the apex, pointing toward foot F. Find the leg forces. Then add the 12\,\mathrm{kN} vertical load as well.
4. Meriam's space truss (the puzzle) with L = 12\,\mathrm{kN}: find the forces in EB, EC, ED.
5. Verify Meriam's reactions for the puzzle's truss by moments about A and force sums.
6. After joint E, analyse joint C of the puzzle's truss: find F_{CB}, F_{CA} and F_{CD}.
7. Harper's truss with L = 10\,\mathrm{kN}: find the three forces at E for d = 2\,\mathrm m and d = 10\,\mathrm m.
8. Explain, from the joint equations, why F_{EB} does not depend on the height d.
9. A tetrahedron of six bars rests on the ground on its three base corners. What supports at the corners make it statically determinate? What if every corner has a ball-and-socket joint?
10. Why does Meriam say the method of sections is "not widely used" for space trusses? How many unknown members can one space section have at most?

Fully worked solutions

1. m = 3j - 6 = 3(8) - 6 = 18. With six reactions, m + 6 = 24 = 3j: the count matches, and a simple space truss is well arranged by construction, so it is determinate if the six reactions are adequately placed.

2. Leg length \sqrt{4 + 9} = \sqrt{13} = 3.606\,\mathrm m; each leg's unit vector from P has vertical component -3/\sqrt{13} and horizontal component 2/\sqrt{13} toward its foot. By symmetry the three forces are equal and their horizontal parts cancel. With tension positive, \Sigma F_z: 3F(-3/\sqrt{13}) - 12 = 0, F = -4\sqrt{13}/3 = -4.81\,\mathrm{kN}: each leg 4.81\,\mathrm{kN} C.

3. Feet F(2, 0, 0), G(-1, \sqrt3, 0), H(-1, -\sqrt3, 0); the push is 6\,\mathrm{kN} along +x. By symmetry about the x-z plane, F_{PG} = F_{PH} = T. Unit vectors from P: (2, 0, -3)/\sqrt{13} and (-1, \pm\sqrt3, -3)/\sqrt{13}. \Sigma F_z: -3(F_{PF} + 2T)/\sqrt{13} = 0, so F_{PF} = -2T. \Sigma F_x: (2F_{PF} - 2T)/\sqrt{13} + 6 = 0, so -6T/\sqrt{13} = -6, T = \sqrt{13} = 3.61\,\mathrm{kN} and F_{PF} = -7.21\,\mathrm{kN}. The leg toward which the apex is pushed is compressed with 7.21\,\mathrm{kN}; the other two are pulled with 3.61\,\mathrm{kN}. Both loads together (adding the results of question 2): F_{PF} = -7.21 - 4.81 = -12.02\,\mathrm{kN} (C); F_{PG} = F_{PH} = 3.61 - 4.81 = -1.20\,\mathrm{kN} (C).

4. F_{EB} = -12/\sqrt2 = -8.49\,\mathrm{kN} (C); F_{EC} = -\frac56(12) = -10\,\mathrm{kN} (C); F_{ED} = +10\,\mathrm{kN} (T).

5. Moments about A: the load gives (0, -4L, 3L); link 1 gives (-4B_y, 0, 0); links 2 and 3 give (0, -3D_z, 3D_y). So B_y = 0, D_z = -\frac43L, D_y = -L. Forces: A_x = L, A_y = -B_y - D_y = L, A_z = -D_z = \frac43L. These match Meriam's values.

6. At C(0, 3, 0): EC is compressed with \frac56L, so it pushes C away from E, along -(3, 0, 4)/5: (-0.5L, 0, -0.667L). Unit vectors from C: to B, (0, -3, 4)/5; to A, (0, -1, 0); to D, (3, -3, 0)/(3\sqrt2). \Sigma F_x: F_{CD}/\sqrt2 - 0.5L = 0, F_{CD} = 0.707L T. \Sigma F_z: 0.8F_{CB} - 0.667L = 0, F_{CB} = 0.833L T. \Sigma F_y: -0.6F_{CB} - F_{CA} - F_{CD}/\sqrt2 = 0: -0.5L - F_{CA} - 0.5L = 0, F_{CA} = -L (C).

7. d = 2\,\mathrm m: F_{ED} = \frac{10}{6}\sqrt{13} = 6.01\,\mathrm{kN} T, F_{EC} = 6.01\,\mathrm{kN} C, F_{EB} = 7.07\,\mathrm{kN} C. d = 10\,\mathrm m: F_{ED} = \frac{10}{6}\sqrt{109} = 17.40\,\mathrm{kN} T, F_{EC} = 17.40\,\mathrm{kN} C, F_{EB} = 7.07\,\mathrm{kN} C.

8. The \mathbf k equation, -d(F_{EC} + F_{ED})/\sqrt{9 + d^2} = 0, forces F_{EC} = -F_{ED} whatever d is. The \mathbf i equation contains -3F_{EC}/\sqrt{9 + d^2} and the \mathbf j equation -3F_{ED}/\sqrt{9 + d^2}; adding them, these cancel, leaving -L - 2F_{EB}/\sqrt2 = 0 with no d in it. Physically, EB and the load are horizontal, and the sloping pair's horizontal contributions to the sum of the x and y directions cancel.

9. j = 4, m = 6, so 3j = 12 needs 6 reactions. For example: a ball-and-socket joint at one corner (3), a support allowing sliding in one horizontal direction at the second (2), and a support allowing horizontal sliding at the third (1), placed so they stop all six motions. Ball joints at all three corners give 6 + 9 = 15 > 12: three redundant reactions, statically indeterminate (and the bars would fight any thermal expansion of the base).

10. A space section has six equations, so at most six unknown cut members. But a moment axis eliminates a member only if it meets its line or is parallel to it, and among skew members in space an axis meeting all but one is seldom available. So a section rarely gives one force per equation, and the joint method or the full linear system is used instead.