A truss is a structure made of straight members connected at joints, designed so that all members carry only axial force (pure tension or pure compression). The pin-jointed assumption means no member bends — its sole job is to push or pull along its length. This makes trusses extremely material-efficient: a Pratt truss spans 30 m with the same steel weight as a solid girder spanning 12 m.

What is the method of joints?

A simple equilibrium method that solves the axial force in every member of a statically-determinate truss. Procedure: (1) compute support reactions from global equilibrium; (2) pick a joint with at most two unknown member forces; (3) apply ΣF_x = 0 and ΣF_y = 0 to solve for those two; (4) move to the next joint that now has at most two unknowns; repeat until all member forces are found. The calculator on this page implements this iteratively.

What is a statically determinate truss?

A truss where the number of unknowns (member forces + reactions) equals the number of equilibrium equations. Algebraically: m + r = 2j, where m = members, r = reaction components, j = joints. A pin-roller-supported parallel-chord truss with the right member count is determinate. Add an extra brace and it becomes indeterminate — the method of joints can't solve it; you need stiffness or flexibility analysis.

When do I use Pratt vs Howe vs Warren?

Pratt — steel highway / railway bridges, industrial roof trusses. Diagonals in tension; vertical posts in compression. Standard since the 1850s. Howe — historical wooden bridges, modern timber roof trusses. Diagonals in compression (wood is strong here); verticals in tension (steel rods). Warren — pedestrian bridges, short to medium roof spans, transmission towers. Equilateral diagonals at ~60° give the fewest distinct member sizes. Warren + verticals — when intermediate top-chord support is needed for purlins or load points.

What does T (tension) and C (compression) mean for member sizing?

Tension members fail by yielding or fracture; size them by area: A = T / (φ × F_y) with F_y = yield stress, φ = 0.9 in AISC LRFD. Compression members fail by buckling, which depends on slenderness L/r where L is the unbraced length and r is the radius of gyration. Compression members usually need larger or differently-shaped sections (HSS or W) to keep slenderness low. The compression-to-tension capacity ratio of a typical W-shape drops from 1.0 (very short) to 0.3 (slender).

How does the load combine in a truss?

Convert distributed loads (kN/m on the roof or floor) to equivalent point loads at the joints by multiplying by bay spacing. The truss only carries loads applied at the joints — anything else (load on the middle of a chord segment) introduces bending in that segment, violating the pure-axial-force assumption. For real structures, secondary purlins or stringers transfer the distributed load to the truss joints.

What if my loads are unsymmetric?

The calculator accepts an equal load at every top joint (most common case). For unsymmetric or wind / snow drift loads, the support reactions and member forces will be asymmetric — Pratt diagonals can flip from tension to compression on the unloaded side. Real designs check multiple load cases (NEC, balanced + unbalanced snow, wind L→R + R→L, dead + live) and size every member for the worst case.

Are end posts diagonals or verticals?

In Pratt and Howe trusses, the end posts are inclined diagonals from the support up to the first interior top joint — they share the function of "diagonal at panel 0" and "vertical at the support." The calculator generates them automatically. Warren trusses (without verticals) have an inclined end post that is structurally identical to all the other diagonals.

Calculator · Structural · Method of joints · Beer & Johnston · AISC

Truss calculator (2D, method of joints)

Build a Pratt, Howe, Warren, or Warren-with-verticals parallel-chord truss with span, height, and bay count, apply equal vertical loads at the top chord, and get support reactions plus the axial force (tension / compression) in every member. With SVG truss diagram and PDF report. Reviewed by a licensed PE.

Use the calculator

Pick a topology, enter span, depth, and bay count, set the load applied at each top-chord joint (negative = downward gravity), and the calculator returns support reactions, every member force with tension / compression label, the maximum-stressed members, and an SVG diagram colour-coded by force magnitude.

CALC.017 Truss · Method of joints · Pratt / Howe / Warren · pin + roller

Topology[T]

Bays (panels)[n]

Span L[m]

Height h[m]

Panel load (per top joint)[kN]

Negative load = downward (gravity). Loads are applied at every TOP-chord joint. Supports: pin at left, roller at right (statically determinate). Geometry must satisfy m + r = 2j; the solver flags violations.

Support reactions

— kN

Set inputs to compute member forces.

FORMULA · ΣF_x = 0, ΣF_y = 0 (per joint) SOURCE · BEER & JOHNSTON · HIBBELER · AISC

Truss analysis — at a glance

What is the method of joints?: A joint-by-joint equilibrium walk: pick a joint with ≤ 2 unknowns, apply ΣF_x = 0 and ΣF_y = 0, solve, advance. Solves any statically determinate truss.
Determinacy condition: A truss is statically determinate when m + r = 2j (m = members, r = reactions, j = joints). m + r > 2j is indeterminate; m + r < 2j is a mechanism.
Pratt vs Howe vs Warren: Pratt: diagonals in tension (steel). Howe: diagonals in compression (timber). Warren: alternating T/C, fewest member types — pedestrian bridges and short-medium spans.
Pratt truss span range: Pratt parallel-chord trusses are practical from 30 to 150 ft (9–45 m). Beyond ~60 m, K-trusses or Howe with sub-bracing become more efficient.
Warren truss span range: Warren trusses (with or without verticals) work well from 30 to 100 ft (9–30 m); equilateral diagonals at ~60° give the minimum number of distinct member sizes.
Chord-couple shortcut: For parallel-chord trusses, max chord force ≈ F_chord = M(x) / h (M = bending moment, h = depth). Within ~1 % of method-of-joints exact value.
Wood vs steel trusses: Wood: pick Howe (compression diagonals); economical to ~12 m. Steel: pick Pratt (tension diagonals) for spans up to 45 m; HSS or W-shapes for compression members.
Depth-to-span ratio: Roof trusses use L/h = 10–15; floor trusses 15–20; pedestrian footbridges 8–10 (vibration governs); long-span industrial 15–25.

The four truss-analysis equations

Eq. 01 — Static determinacy condition — · Maxwell 1864 · Hibbeler

m + r = 2 \cdot j

m: number of members (each contributes one unknown axial force), —
r: number of reaction components (pin = 2, roller = 1), —
j: number of joints (each gives 2 equilibrium equations), —

If m + r = 2j the truss is determinate and the method of joints solves it. If m + r > 2j the truss is indeterminate — you need stiffness or flexibility methods (matrix analysis). If m + r < 2j it is a mechanism and unstable. Standard parallel-chord Pratt / Howe / Warren trusses with one pin and one roller satisfy this exactly.

Eq. 02 — Global equilibrium (support reactions) SI · Newton's third law

\sum F_{y} = 0, \qquad \sum M_{\text{left}} = 0 \implies R_{\text{right}} = \frac{\sum P_{i} \cdot x_{i}}{L}, \quad R_{\text{left}} = \sum P_{i} - R_{\text{right}}

P_i: load at joint i (positive up, negative down), kN
x_i: distance from left support to joint i, m
L: span between supports, m

Take moments about the left (pin) support — the pin\'s own moment is zero, so only loads and the right reaction contribute. For symmetric loading, R_left = R_right = total load ÷ 2 — a quick sanity check.

Eq. 03 — Joint equilibrium (method of joints) SI · Beer & Johnston, Statics § 6

\sum F_{x} = 0: \sum_{k} F_{k} \cos\theta_{k} + F_{x,\text{joint}} = 0 \\ \sum F_{y} = 0: \sum_{k} F_{k} \sin\theta_{k} + F_{y,\text{joint}} = 0

F_k: axial force in member k connected to the joint (T positive), kN
θ_k: angle of member k (away from joint), rad
F_x,joint: sum of horizontal applied loads + reaction at this joint, kN
F_y,joint: sum of vertical applied loads + reaction at this joint, kN

Two equations per joint, with at most 2 unknown member forces solvable per visit. Visit joints one at a time — start at a support (which has the fewest unknowns), solve, mark forces as known, move on. The calculator iterates this loop until every member is solved.

Eq. 04 — Sign convention and the chord-couple formula SI · Hibbeler · classical bridge engineering

F_{\text{chord}} \approx \frac{M(x)}{h}

F_chord: axial force in the top or bottom chord at section x, kN
M(x): bending moment at section x (treating truss as a beam), kN·m
h: truss depth (distance between top and bottom chord), m

A useful approximation for parallel-chord trusses: the truss carries bending moment as a couple between top (compression) and bottom (tension) chords, separated by depth h. So F_chord ≈ M / h. For a 12 m span with a 30 kN central load and 3 m depth, M_max ≈ 90 kN·m → F_chord ≈ 30 kN at midspan. The detailed method of joints gives the exact value plus all secondary forces.

How to use a truss, step by step

Pick the topology that matches your design. Pratt for steel — diagonals carry tension, the cheaper failure mode. Howe for timber and composite — diagonals carry compression, where wood is strong. Warren for short to medium spans — equilateral pattern, fewest member types, minimal verticals. Warren-with-verticals for longer spans needing intermediate panel points to support secondary loads.
Set span, height, and bay count. Span L is the distance between supports. Height h is the vertical depth of the truss. The depth-to-span ratio for parallel-chord trusses is typically L/10 for roof, L/15 for floor, L/8 for short wide-stance roof. Bays count: 4–8 panels is typical; more bays mean shorter members but more joints (more cost in connections).
Apply equal joint loads at the top chord. For a uniformly distributed dead-plus-live load w (kN/m), each top-joint load = w × bay-spacing. A 5 kN/m roof on a 6-bay 18 m span has bay spacing 3 m, so each top joint takes 15 kN downward. Enter that as −15 kN (negative = downward / gravity).
Read the support reactions. For symmetric loading, R_left = R_right = (total load) / 2. Pin support takes Fx + Fy; roller takes Fy only. The calculator returns reactions as the first output — quick check: their sum should equal the total applied load.
Identify the maximum-loaded member. Bottom chord at midspan carries maximum tension. Top chord at midspan carries maximum compression. The end-post diagonals also carry significant force. The calculator highlights both extremes — these are the members to size first when picking steel sections.
Convert axial forces to member sections. For tension: required area A = T_max × γ_M / f_y (AISC LRFD: A_required = T_u / (φ_t × F_y)). For compression: pick a section using the column buckling tables (AISC Table 4-22), accounting for the unsupported length L = bay × √2 typical for diagonals. Don't forget connection details — gusset plate sizes are often what controls cost.

Reference values

Topology comparison

The four common parallel-chord topologies and where they fit best.

Topology	Diagonals carry	Verticals carry	Best for
Pratt	Tension	Compression	Steel bridges and roof trusses; standard since 1850s
Howe	Compression	Tension (often steel rods)	Timber bridges and composite wood-steel trusses
Warren (no verticals)	Alternating T/C	—	Pedestrian bridges, short-to-medium spans, transmission towers
Warren + verticals	Alternating T/C	Carry purlin loads	Longer spans needing intermediate top-chord support

Typical depth-to-span ratios

The depth h of a parallel-chord truss is usually a fraction of the span L. Optimal values balance steel weight against connection cost.

Application	L/h ratio	Comment
Pedestrian footbridge	8–10	Stiffness governs (vibration), so use deeper
Roof truss (industrial / commercial)	10–15	Standard range; 12 typical
Floor truss (composite)	15–20	Deeper for vibration control under live load
Highway bridge truss	8–12	Low ratios for stiffness and impact loads
Stadium roof / long-span industrial	15–25	Optimised for material; complex connections

Worked example: 24 m roof truss, 6 bays

A 24 m clear-span industrial-roof Pratt truss, 4 m deep (L/h = 6), 6 bays of 4 m each, carries a uniform load of 5 kN/m (combined dead + live). Top-joint load = 5 × 4 = 20 kN per joint at T1, T2, T3, T4, T5 (five interior top joints). Total load = 100 kN downward.

Step	Calculation	Result
Total load	5 joints × 20 kN	100 kN ↓
Reactions (symmetric)	R_L = R_R = 100 / 2	50 kN each ↑
Bending moment at midspan (treating as beam)	50 × 12 − 20 × 4 − 20 × 8	M_max = 360 kN·m
Approx max chord force	F = M / h = 360 / 4	~90 kN
Bottom chord (tension) at midspan	method of joints solver	+90 kN T
Top chord (compression) at midspan	method of joints solver	−90 kN C
End-post diagonal (Pratt)	solver: ~71 kN compression at √2 angle	−71 kN C
Required steel A_T (Fy = 350 MPa, φ = 0.9)	A = 90 000 / (0.9 × 350) = 286 mm²	L 50×50×6 = 568 mm²
Required compression section (KL = 4 m diagonal)	HSS 89×89×4.8 from AISC Table 4-22	capacity 110 kN

The chord-couple shortcut F = M/h is within 1 % of the exact answer for parallel-chord trusses — a useful estimate. Diagonals carry roughly the local shear divided by sin θ, where θ is the diagonal angle (≈ 45° for L/h = 6 and bay = h). The compression check on diagonals is the binding constraint, so use HSS or W-sections with high radius of gyration.

Variants and special cases

Pitched (sloped-chord) trusses — Fink, scissors

Roof trusses for pitched roofs use sloped top chords. Fink truss — common residential roof truss with W-shaped web pattern; minimises member length for short spans. Scissors truss — sloped both chords, gives vaulted ceiling appearance. Both can be solved with the method of joints; the geometry generator just changes the joint coordinates.

Long-span K-trusses

For spans over 60 m, the diagonal members in a Pratt or Howe truss become too long for stable compression. A K-truss divides each panel\'s diagonal into two shorter pieces meeting at a vertical, halving the unsupported length and allowing thinner compression members. Used in long-span bridges and television transmission towers.

Indeterminate trusses

Adding even one extra brace to a determinate truss makes it indeterminate — the method of joints fails because every joint now has 3+ unknowns. Indeterminate trusses are stiffer and tougher (a single member failure isn\'t catastrophic) but require finite-element analysis (matrix stiffness method) to solve. Common in high-rise floor systems and stadium roofs.

Space trusses (3D)

Real-world structures often need 3D triangulation (geodesic domes, exhibition-hall roofs, industrial conveyor towers). The method of joints generalises directly: 3 equilibrium equations per joint instead of 2, m + r = 3j for determinacy. Solving by hand becomes impractical above ~10 joints — use Frame3DD, OpenSees, or commercial FEM software.

The original method of joints

The method of joints consists of considering the equilibrium of each joint of the truss in turn. Since each joint is acted on by a system of concurrent forces, two equilibrium equations may be written for each joint: ΣF_x = 0 and ΣF_y = 0. The joints are analysed in succession, starting with one at which only two members meet, where the two unknown forces in those members can be determined.

Beer, F.P. & Johnston, E.R. — Vector Mechanics for Engineers: Statics, 11th Edition → Chapter 6: Analysis of Structures

Related calculators and references

Frequently asked questions

What is a truss?: A truss is a structure made of straight members connected at joints, designed so that all members carry only axial force (pure tension or pure compression). The pin-jointed assumption means no member bends — its sole job is to push or pull along its length. This makes trusses extremely material-efficient: a Pratt truss spans 30 m with the same steel weight as a solid girder spanning 12 m.
What is the method of joints?: A simple equilibrium method that solves the axial force in every member of a statically-determinate truss. Procedure: (1) compute support reactions from global equilibrium; (2) pick a joint with at most two unknown member forces; (3) apply ΣF_x = 0 and ΣF_y = 0 to solve for those two; (4) move to the next joint that now has at most two unknowns; repeat until all member forces are found. The calculator on this page implements this iteratively.
What is a statically determinate truss?: A truss where the number of unknowns (member forces + reactions) equals the number of equilibrium equations. Algebraically: m + r = 2j, where m = members, r = reaction components, j = joints. A pin-roller-supported parallel-chord truss with the right member count is determinate. Add an extra brace and it becomes indeterminate — the method of joints can't solve it; you need stiffness or flexibility analysis.
When do I use Pratt vs Howe vs Warren?: Pratt — steel highway / railway bridges, industrial roof trusses. Diagonals in tension; vertical posts in compression. Standard since the 1850s. Howe — historical wooden bridges, modern timber roof trusses. Diagonals in compression (wood is strong here); verticals in tension (steel rods). Warren — pedestrian bridges, short to medium roof spans, transmission towers. Equilateral diagonals at ~60° give the fewest distinct member sizes. Warren + verticals — when intermediate top-chord support is needed for purlins or load points.
What does T (tension) and C (compression) mean for member sizing?: Tension members fail by yielding or fracture; size them by area: A = T / (φ × F_y) with F_y = yield stress, φ = 0.9 in AISC LRFD. Compression members fail by buckling, which depends on slenderness L/r where L is the unbraced length and r is the radius of gyration. Compression members usually need larger or differently-shaped sections (HSS or W) to keep slenderness low. The compression-to-tension capacity ratio of a typical W-shape drops from 1.0 (very short) to 0.3 (slender).
How does the load combine in a truss?: Convert distributed loads (kN/m on the roof or floor) to equivalent point loads at the joints by multiplying by bay spacing. The truss only carries loads applied at the joints — anything else (load on the middle of a chord segment) introduces bending in that segment, violating the pure-axial-force assumption. For real structures, secondary purlins or stringers transfer the distributed load to the truss joints.
What if my loads are unsymmetric?: The calculator accepts an equal load at every top joint (most common case). For unsymmetric or wind / snow drift loads, the support reactions and member forces will be asymmetric — Pratt diagonals can flip from tension to compression on the unloaded side. Real designs check multiple load cases (NEC, balanced + unbalanced snow, wind L→R + R→L, dead + live) and size every member for the worst case.
Are end posts diagonals or verticals?: In Pratt and Howe trusses, the end posts are inclined diagonals from the support up to the first interior top joint — they share the function of "diagonal at panel 0" and "vertical at the support." The calculator generates them automatically. Warren trusses (without verticals) have an inclined end post that is structurally identical to all the other diagonals.

Sources and methodology

Beer, F.P., Johnston, E.R., Mazurek, D.F., Cornwell, P.J. Vector Mechanics for Engineers: Statics, 11th Edition. McGraw-Hill, 2016. Chapter 6 — Analysis of Structures.
Hibbeler, R.C. Structural Analysis, 10th Edition. Pearson, 2017. Chapter 4 — Analysis of Statically Determinate Trusses.
AISC. Steel Construction Manual, 16th Edition, 2023. Chord and diagonal section design tables.
Maxwell, J.C. On the Calculation of the Equilibrium and Stiffness of Frames. Philosophical Magazine, 1864. Original formal derivation of static determinacy condition.
Pratt, T.W. & Pratt, C. U.S. Patent 3 523 — Pratt Truss Bridge Design, 1844.
Warren, J. British Patent 12 242 — Warren Girder, 1848.