Structural Analysis Fundamentals / Pin-jointed trusses
Trusses as beams, tension coefficients and space frames
Reading a truss as a deep beam, a faster hand method, and stepping into three dimensions.
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Read the lesson register and outstanding review notesWhy this matters
Joint and section equilibrium solve stable, statically determinate trusses. A beam analogy provides useful force estimates, while tension coefficients express joint equilibrium through coordinates in two or three dimensions.
By the end of this lesson you should be able to
- Estimate chord and diagonal forces by treating the truss as a beam
- Use tension coefficients to solve a joint
- Recognise a compound truss and know why joints alone can stall on one
- Apply the determinacy count m + r = 3j to a pin-jointed space frame
What you should already know
- Method of joints and method of sections (this module)
- Determinacy of plane trusses (this module)
- Shear force and bending moment (Module 3)
A section through a parallel-chord truss must equilibrate the beam shear and moment from the same external loading. Horizontal chords contribute to the bending couple; inclined members contribute vertical shear.
The estimates F = M/d and F = V/sin θ need a suitable section and force geometry. For an exact chord force, take moments about the intersection of the other cut members. If several web members cross the section, their combined vertical components resist the shear.
What it calculates: chord and diagonal forces from the equivalent beam's M and V
- M
- bending moment of the equivalent beam at the section (kNm)
- V
- shear force of the equivalent beam at the section (kN)
- d
- depth between the chord centrelines (m)
- θ
- angle of the diagonal to the horizontal (degrees)
This assumes
- Parallel chords, so the chord forces form a couple of constant lever arm
- Pin joints and loads applied only at the joints
- One diagonal or vertical crossing the section
In plain terms: For the stated beam analogy, chord force is inversely proportional to depth at unchanged loading. Real member sizing also depends on buckling, connections and serviceability.
Predict first
A parallel-chord truss is redesigned twice as deep, with the same span and loading. What happens to the chord forces?
For a member from A to B carrying axial tension T over length L, define the tension coefficient t = T/L. Its components at A are t·Δx and t·Δy, using the coordinate differences from A to B.
Joint equilibrium becomes Σt·Δx + X = 0 and Σt·Δy + Y = 0. Recover the member force from T = tL. In three dimensions add Σt·Δz + Z = 0. This expresses the same joint equilibrium without explicitly calculating member angles.
Space frames are pin-jointed structures in three dimensions. Every joint now has three equations of equilibrium instead of two, so the determinacy condition becomes:
m + r = 3j
with m + r − 3j > 0 indicating indeterminacy and < 0 a mechanism — exactly the same logic as the plane case, and with the same caveat: the count is necessary but not sufficient. A badly arranged space frame can satisfy it and still be unstable.
The simplest useful case is a tripod: three legs from a common apex down to three anchors. It has m = 3, j = 4 and r = 9 (three pinned supports), so m + r − 3j = 3 + 9 − 12 = 0. Determinate, and if the arrangement is symmetric the three legs share the vertical load equally.
Worked example
A truss read as a beam, and a symmetric tripod
Given
- Parallel-chord truss: span 24.0 m, depth 3.00 m, simply supported
- Uniform load equivalent to 40.0 kN/m along the span
- Separately: a tripod with apex 3.00 m above three anchors on a circle of radius 2.00 m, carrying 60.0 kN vertically
Find
The bottom chord force at mid-span, and the force in each tripod leg.
Practice
A parallel-chord truss spans 24.0 m with a depth of 3.00 m and carries the equivalent of 40.0 kN/m. What is the force in the bottom chord at mid-span, in kN?
Practice
A member runs from (0, 0) to (4.00 m, 3.00 m) and carries a tensile force of 50.0 kN. What is its tension coefficient, in kN per metre?
Practice
A pin-jointed space frame has 12 members, 6 joints and 6 restraint components. What is m + r − 3j?
Worked example
A compound truss that stops the method of joints
Given
- Two stable triangulated sub-trusses, each with six joints and nine members, joined side by side
- They are connected by three members that are neither concurrent nor parallel
- Total: 21 members, 12 joints, 3 reaction components
Find
Whether the truss is determinate, why the method of joints stalls, and how to restart it.
Practice
A compound plane truss has 21 members, 12 joints and 3 reaction components. What is m + r − 2j?
Summary
- A parallel-chord truss is a beam: chords carry the moment as a couple, diagonals carry the shear
- Chord force = M/d, so doubling the depth halves the chord forces
- Diagonal force = V/sin θ
- Tension coefficient t = T/L turns joint equilibrium into Σt·Δx = 0 and Σt·Δy = 0, with no trigonometry
- The same method extends to three dimensions by adding Σt·Δz = 0
- Space frames: m + r = 3j for static determinacy
- A compound truss may need a section before the method of joints can start