7.1 Trusses, Braces, and Column Axial Force

Axial analysis determines whether a member pulls or pushes and by how much. It does not, by itself, determine whether the member is adequate. For a July 2026 PE Civil: Structural problem, use the April 2024 specification, the current PE Civil Reference Handbook, the AISC Steel Construction Manual 15th edition, and NDS 2018 with ASD only where applicable. Do not substitute the April 2027 reference set or PE Structural/SE exam content.

Model the System Before Solving

An ideal planar truss has straight members connected by frictionless pins, with external loads and reactions applied at joints. Each member is then a two-force member, so its end forces are equal, opposite, and collinear. A load between joints, a rigid connection, an eccentric gusset, or self-weight applied along a member can introduce bending; the pure-truss model is no longer exact unless the problem directs an idealization.

Start with a free-body diagram of the whole truss. Solve support reactions using ΣF_x = 0, ΣF_y = 0, and ΣM = 0. Then adopt one sign convention and keep it. A useful method assumes every unknown member pulls away from its joint. A positive answer confirms tension; a negative answer means the actual force is compression. Do not reverse arrows midway merely because compression seems likely.

Joints, Sections, and Zero-Force Members

The method of joints isolates one pin and uses the two planar force equations. Begin at a joint with no more than two unknown member forces. Resolve diagonals carefully: if angle θ is measured from horizontal, the components are F cos θ horizontal and F sin θ vertical. Carry a solved force to the adjacent joint with an equal-and-opposite arrow.

The method of sections cuts through selected members and applies the three planar equilibrium equations to either side. A useful cut usually crosses no more than three unknown forces. Taking moments about the intersection of two unknown force lines can eliminate both and solve the third directly. Sections are efficient when only a few interior forces are requested.

Two common zero-force observations accelerate either method:

• At an unloaded, unsupported joint with exactly two noncollinear members, both are zero-force members.
• At an unloaded, unsupported joint with three members, if two are collinear, the noncollinear member is zero.

These rules fail when an external load or support reaction acts at the joint, and a member that is zero for one load case may carry force in another. Zero force is an analysis result, not permission to delete a stability or construction member.

Worked Example: Force Sign and Capacity Route

A symmetric triangular truss has a 24 kip downward load at its apex. Two identical diagonal members meet there, each at 45 degrees to horizontal. A bottom chord ties the two supports. By symmetry, each vertical reaction is 12 kips. Assume both diagonal forces F are tensile, pointing away from the apex.

Vertical equilibrium at the apex gives:

ΣF_y = -24 - 2F sin 45° = 0

F = -24/(2 sin 45°) = -16.97 kips

The negative result means each diagonal carries 16.97 kips in compression. At either support joint, the diagonal's horizontal component is

16.97 cos 45° = 12.0 kips

so the bottom chord provides 12.0 kips of tension. The member-force check is complete because the signs and joint equilibrium agree.

Now suppose a problem gives an ASD allowable axial tension of 24 kips and allowable axial compression of 15 kips for the same diagonal configuration. The computed member is not checked against 24 kips merely because that number is larger. Compression controls, and 16.97/15 = 1.13, so the stated compression check fails. The allowable values are problem data here; an actual AISC or NDS calculation must apply the correct slenderness, stability, bracing, section, duration or adjustment factors, and connection provisions.

Braces and Columns

A single diagonal brace at angle θ from horizontal that alone resists story shear V has the ideal axial demand

P = V/cos θ

because its horizontal component must equal V. Its vertical component also enters the beam-column joints and can add or subtract column axial load. In an X-braced bay, a tension-only model may assign the active diagonal by load direction; if both braces resist compression and tension, stiffness and compatibility distribute force. Chevron braces can impose an unbalanced vertical force on the beam when brace strengths differ or one buckles. Do not assume every brace shares shear equally without a stiffness and compatibility basis.

Columns collect floor and roof reactions down the structure, but their axial demand can also include brace vertical components, frame overturning, transfer-member reactions, or restraint forces. At a splice, base plate, or footing, trace the axial force through the connection into the next element and ultimately the ground. For statically indeterminate trusses or multi-brace systems, equilibrium alone is insufficient; deformations and relative stiffness determine force sharing.

Analysis Result Versus Code Capacity

Tension members may be governed by yielding, rupture, net section, block shear, or connections. Compression members additionally face global, local, and flexural-torsional buckling, so unbraced length and end restraint matter. A force labeled C = 50 kips is only demand and does not prove a 50 kip member capacity. Pair LRFD demand with LRFD resistance or ASD demand with ASD resistance; for NCEES-listed wood design, remain on the NDS ASD route. Finally, check the gusset, fasteners, splice, and supporting member: an adequate axial member with an inadequate connection is not a complete load path.