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Truss Analysis Calculator

Geometry
Nodes
IDX (m)Y (m)SupportRemove
Members
IDStart nodeEnd nodeRemove
Member section & material (applied to all members)
Node loads (kN)
Node B0
Node B1
Node B2
Node B3
Node B4
Node T1
Node T2
Node T3

Truss Analysis

EN 1993-1-1 · HEA160 · S235
M1: 200.0TM2: 200.0TM3: 200.0TM4: 200.0TM5: 266.7CM6: 266.7CM7: 100.0TM8: 0.00M9: 100.0TM10: 250.0CM11: 250.0CM12: 83.3TM13: 83.3TB0B1B2B3B4T1T2T3
Governing checkMember M1173.3%ALL CHECKS PASS
Checks
CheckDemandCapacityUtil.Status
M1200.0 kN (tension)911.1 kN22.0%PASS
M2200.0 kN (tension)911.1 kN22.0%PASS
M3200.0 kN (tension)911.1 kN22.0%PASS
M4200.0 kN (tension)911.1 kN22.0%PASS
M5266.7 kN (compression)456.7 kN58.4%PASS
M6266.7 kN (compression)456.7 kN58.4%PASS
M7100.0 kN (tension)911.1 kN11.0%PASS
M80.0 kN (zero)911.1 kN0.0%PASS
M9100.0 kN (tension)911.1 kN11.0%PASS
M10250.0 kN (compression)340.8 kN73.3%PASS
M11GOVERNS250.0 kN (compression)340.8 kN73.3%PASS
M1283.3 kN (tension)911.1 kN9.1%PASS
M1383.3 kN (tension)911.1 kN9.1%PASS
  • Models an idealized pin-jointed truss — every member carries pure axial force only, and loads can only be applied at nodes, not along a member’s length.
  • Stiffness-method solve with 2-DOF/joint bar elements — mathematically equivalent to the method of joints for a statically determinate truss; a linear-elastic, small-deflection (first-order) analysis only.
  • Net-section rupture at connections is not checked — verify separately at any bolted/welded joint.
  • A distributed load must be approximated as statically-equivalent point loads at the nearest joints.
  • Geometry and results are in SI units (m, kN, kN·m) only.
  • If the structure actually has rigid, moment-resisting joints, use Frame Analysis instead.
Truss Analysis v1.0Validated against 16 benchmark cases →Learn: What does "statically determinate" actually mean? →Learn: Buckling: why slender steel members fail differently than stocky ones →Learn: Moment frames vs braced frames vs shear walls: how lateral systems actually differ →
Geometry & determinacy
Model summary
8 joints, 13 members, 3 reaction components
8 joints, 13 members, 3 reactions
StaticsDeterminacy count (necessary, not sufficient — the stiffness solve below is the rigorous check)
m + r = 2j ?
13 + 3 vs 2×8 = 16
m+r=16 → statically determinate
Stiffness-method solve
Bar-element assembly & solve
Kff · Uf = Pf (2 DOF/joint, axial-only elements)
16 total DOF, 13 free
Solved — joint displacements recovered
Global equilibrium check
ΣFx = ΣFy = 0 (applied loads + reactions)
ΣFx=-0.000 kN, ΣFy=0.000 kN
Satisfied
Member axial forces
Member M1 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = 200.0 kN (tension)
Member M2 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = 200.0 kN (tension)
Member M3 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = 200.0 kN (tension)
Member M4 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = 200.0 kN (tension)
Member M5 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = -266.7 kN (compression)
Member M6 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = -266.7 kN (compression)
Member M7 (HEA160)
N = (EA/L)·Δ, tension +ve
L=3.00 m
N = 100.0 kN (tension)
Member M8 (HEA160)
N = (EA/L)·Δ, tension +ve
L=3.00 m
N = 0.0 kN (zero)
Member M9 (HEA160)
N = (EA/L)·Δ, tension +ve
L=3.00 m
N = 100.0 kN (tension)
Member M10 (HEA160)
N = (EA/L)·Δ, tension +ve
L=5.00 m
N = -250.0 kN (compression)
Member M11 (HEA160)
N = (EA/L)·Δ, tension +ve
L=5.00 m
N = -250.0 kN (compression)
Member M12 (HEA160)
N = (EA/L)·Δ, tension +ve
L=5.00 m
N = 83.3 kN (tension)
Member M13 (HEA160)
N = (EA/L)·Δ, tension +ve
L=5.00 m
N = 83.3 kN (tension)
Support reactions
Reaction at B0
Rx = -0.0 kN, Ry = 150.0 kN
Reaction at B4
Rx = 0.0 kN, Ry = 150.0 kN
Governing member M11 — compression
Table 5.2Material coefficient
ε = √(235 / fy)
ε = √(235 / 235)
ε = 1.000
Table 5.2Flange (outstand in compression)
c = (b − tw − 2r)/2; limits 9ε | 10ε | 14ε
c/tf = 62.0/9 = 6.89 vs 9.00 | 10.00 | 14.00
Flange: Class 1
Table 5.2Web (internal, in bending)
c = h − 2tf − 2r; limits 72ε | 83ε | 124ε
c/tw = 104.0/6 = 17.33 vs 72.0 | 83.0 | 124.0
Web: Class 1
§5.5.2(6)Section class = least favourable element class
max(flange, web)
max(1, 1)
Section: Class 1
EN 1993-1-1 Table 6.2Buckling curve selection (h/b=0.95, tf=9.0 mm)
curve y-y = b, curve z-z = c
EN 1993-1-1 §6.3.1.3Elastic critical buckling force, both axes
Ncr = π²·E·I/Lcr²
Lcr,y=5000 mm, Lcr,z=5000 mm
Ncr,y=1387.0 kN, Ncr,z=510.7 kN
EN 1993-1-1 (6.50)Non-dimensional slenderness
λ̄ = √(Npl,Rd/Ncr)
Npl,Rd=911.1 kN
λ̄y=0.810, λ̄z=1.336
EN 1993-1-1 (6.49)Reduction factor χ (imperfection factor α from the curve)
χ = 1/(Φ+√(Φ²−λ̄²)) ≤ 1
αy=0.34, αz=0.49
χy=0.718, χz=0.374
EN 1993-1-1 (6.47)Buckling resistance, both axes — governing axis controls
Nb,Rd = χ·A·fy/γM1
γM1=1
Nb,Rd,y=654.1 kN, Nb,Rd,z=340.8 kN → governs z-z

Step values are shown in SI (m, kN, kN·m) regardless of the display-unit toggle.