Truss Tension and Compression Explained | Structural Engineering Basics

This section covers the core concepts distinguishing determinate from indeterminate structures. A statically determinate structure allows solving all unknown reaction and internal forces using only equilibrium equations. In 2D analysis, these include ∑Fx=0, ∑Fy=0, and ∑Mz=0. The degree of indeterminacy (d-o-i) equals unknowns minus usable equilibrium equations. When d-o-i=0, the system is determinate; when >0, it's indeterminate; when <0, the structure is unstable. For trusses specifically, the formula is Reactions + Elements - 2×Connections, since each connection provides two equilibrium equations (no moment equations at truss joints).

A statically determinate structure can have all unknown forces (reactions and member forces) determined using only equilibrium equations. When the number of unknowns equals the number of independent equilibrium equations, the structure is determinate. If there are more unknowns than independent equilibrium equations, the structure is statically indeterminate and requires additional methods beyond basic equilibrium to solve.

A structure is statically determinate if it can be analyzed using only the three equations of equilibrium (∑Fx=0, ∑Fy=0, and ∑M=0), meaning the number of unknown reactions equals the number of equilibrium equations; conversely, a structure is statically indeterminate when the number of unknown reactions exceeds the available equilibrium equations, requiring additional compatibility equations for analysis. For example, a simply supported beam with hinge and roller supports has three unknowns solvable by equilibrium, while a beam with two fixed supports has six unknowns that cannot be solved by equilibrium alone.

A structure is classified as statically determinate when the number of reactions equals the number of equilibrium equations (typically 3 for 2D structures), allowing solution through basic equilibrium analysis; if reactions exceed equations, the structure is statically indeterminate and requires additional compatibility equations; if reactions are fewer than equations, the structure is unstable and cannot maintain equilibrium under arbitrary loading.

The degree of static indeterminacy (i) is calculated using the formula i = m + r - 2j, where m is the number of members, r is the number of reaction forces, and j is the number of joints. This equation determines whether the truss is statically determinate (i = 0), statically indeterminate (i > 0), or unstable (i < 0). When i = 0, the truss has exactly the right number of members and reaction forces. When i > 0, the truss has redundant members or reaction forces that can be removed without affecting stability. When i < 0, the truss has deficient members or reaction forces that must be added to achieve stability.

Compression member design requires understanding three failure modes: flexural buckling (Euler buckling in elastic range), local buckling (thin elements flexing), and torsional-flexural buckling (cross-section rotation). Slenderness ratio (KL/r) controls stability, limited to 200 by NSR 10. Euler's critical load formula P_cr = π²EI/L² defines elastic buckling. Critical stress σ_cr = π²E/(KL/r)² depends on modulus, radius of gyration, and slenderness. Effective length factor K varies by support conditions (doubly fixed K=0.5, pinned-pinned K=1.0, fixed-free K=2.0). When applied stress is less than yield stress, Euler's formula applies for elastic design.

Compression members fail through flexural buckling, local buckling, or torsional-flexural buckling, with longer columns having greater buckling tendency. The design procedure uses Euler's critical stress formula: Fe = π²E / (KL/r)². When KL/r ≤ 4.71√(E/Fy), inelastic buckling applies: Fcr = 0.658^(Fy/Fe) × Fy. Ultimate loads are calculated using LRFD (φ = 0.9) and ASD (Ω = 1.67) factors. Multiple examples demonstrate this procedure for W8x24, W14x109, HCS 8×6×3/8, W1x152, and 10×10×3/8 tubes, with all lengths converted to inches for consistent calculations.

AISC Chapter E provides comprehensive requirements for compression member design. Nominal compressive strength Pn = fcr × Ag. LRFD uses φcPn with φc=0.9 resistance factor. ASD uses Pn/Ωc with Ωc=1.67 safety factor. The effective length factor (K) accounts for end conditions (K=1.0 for pinned-pinned). Three assumptions underlie Euler buckling: perfect straightness, axial loading, and pinned ends. These specifications ensure consistent design methodology across different structural applications. The design process involves calculating KL/r, determining whether elastic or inelastic buckling governs, computing fcr, and applying appropriate factors to obtain design strength.

The limit state method for compression member design is based on several key assumptions: (1) Plane sections normal to the axis remain plane after bending, (2) Maximum strain in concrete at the outermost compression fiber is 0.0035 for bending, (3) Tensile strength of concrete is ignored, (4) The stress block may be rectangular, trapezoidal, parabolic, or any shape that predicts strength in agreement with test results, (5) Stresses in reinforcement are derived from representative stress curves for the type of steel used, and (6) Maximum strain in steel depends on the steel grade.

Compression members are susceptible to geometric instability (buckling) even if strong in tension. Euler's buckling theory gives critical load Pcr = (π²EI)/(KL)², where E is modulus of elasticity, I is moment of inertia, K is effective length factor, and L is length. Diagonal members should be designed to be in tension when possible, as tension members are less susceptible to buckling than compression members.

Different types of trusses are designed for specific applications. For roof structures, common types include Pratt truss, Howe truss, and Fink truss. For bridge construction, Pratt truss, Howe truss, Warren truss, and K-truss are frequently used. The K-truss is named for its distinctive K-shaped configuration. Stadium and concert venue trusses are typically three-dimensional structures, often made of aluminum, that are designed to be disassembled. All these truss types share the common characteristic of having nodes where members connect and bars that support loads.

Three classic bridge truss designs emerged in the 1840s: (1) Howe truss—vertical members in tension, diagonal members in compression; less cost-effective because thick diagonal members are long; (2) Pratt truss—vertical members mostly in compression, inner diagonals in tension; more cost-effective since longer tension members can be thinner; (3) Warren truss—based on equilateral triangles, all members same length for construction efficiency, uses fewer members overall, but has long compression members requiring careful design.

Roof trusses are structural frameworks designed to support roof loads, with different types serving specific structural purposes based on span requirements, slope conditions, and load distribution needs. The Howe truss, patented in 1940, features diagonals running from top to bottom, with vertical members in tension and diagonal members in compression, making it suitable for bridges, roofs, and warehouses. The Pratt truss, similar to the Howe but with reversed diagonal arrangement, can span 18-30 meters with diagonals in tension and verticals in compression. The F truss is designed for spans exceeding 30 meters and slopes greater than 15 degrees, where most elements work in tension with few in compression, making it ideal for high-rainfall regions. The Warren truss, patented in 1848, has elements working in both tension and compression and is commonly used for bridges. The sawtooth truss provides natural illumination through its distinctive geometry. Each truss type offers specific advantages depending on the structural requirements and environmental conditions of the construction project.

The Howe truss is an engineering structure used to support weight in bridges, consisting of upper and lower parallel beams called chords. Connecting these chords are vertical posts and diagonally arranged braces and counter-braces, which rest on and are bolted to angled blocks. This design allowed panels to be completed offsite and transported or assembled with minimal tooling on location. According to the US Army Corps of Engineers, a Howe truss bridge should span no more than approximately 150 feet due to stress placed on the panels.