EDUCATIONAL USE ONLY

1. Project information

Calculation report

Date31/10/2026

2. Overview of the structure

Structure

The structure with the numbers of the nodes (n), members (m) and supports (S), and the length of each member (mm).

Sections and materialsMembers
HEA180 · S2351

2 nodes, 1 members · consequence class CC1, new build · design code Eurocode · Z is the vertical axis · units: kN, kNm, mm.

3. Summary of the results

QuantityMax. valueCombinationWhere
Moment My (strong axis)31.8 kNmULS 6.10b (CC1 new)member b, x = 2075 mm
Normal force N (− compression)0.0 kNULS 6.10a (CC1 new)–
Shear force Vz (strong axis)30.7 kNULS 6.10b (CC1 new)member b, x = 0 mm
Vertical reaction30.7 kNULS 6.10b (CC1 new)node A
Displacement (max., incl. between nodes)8.38 mmSLS char.member b, x = 2075 mm
Deflection of a member (relative to chord)8.38 mm (L/495)SLS char.member b

Moments, forces and reactions: the governing ULS combination. Displacement: the governing SLS combination (the ULS combinations if there is no SLS combination).

Highest unity check: UC 0.50 in member b (HEA180): deflection (SLS).

4. Nodes, members and supports

Nodes

Nodex (mm)y (mm)z (mm)Support
A000S1
B415000S2

Members

MemberNodesL (mm)SectionMaterialRemarks
bA – B4150HEA180S235

Supports

SupportNodeRestraintSprings
S1Apinned (translations fixed)
S2Bpinned (translations fixed)

5. Loads

Loads: Permanent (G)

Loads: Permanent (G)
TypeOnValue
Line loadmember bX 0.00, Y 0.00, Z -2.08 kN/m

Loads: Variable (Q)

Loads: Variable (Q)
TypeOnValue
Line loadmember bX 0.00, Y 0.00, Z -9.00 kN/m

The self weight of members, walls and slabs is automatic (case SW) and is not drawn. Loads are shown as in the model; combination factors are in the table at the end.

6. Load combinations

Consequence class CC1, new build.

CombinationKindSWGQ
ULS 6.10a (CC1 new)ULS1.221.220.54
ULS 6.10b (CC1 new)ULS1.081.081.35
SLS char.SLS111
ULS 0.9·G favourable (CC1 new)ULS0.90.90
ULS 0.9·G + Variable (Q) (CC1 new)ULS0.90.91.35
ULS γG·G permanent only (CC1 new)ULS1.221.220
ULS 0.9·G + ψ0 variable (CC1 new)ULS0.90.90.54
SLS permanent (G)SLS110
SLS quasi-permanent (G + ψ2·Q)SLS110.3
SLS frequent (G + ψ1·Q)SLS110.5

SW = Self weight; G = Permanent (G); Q = Variable (Q). SW = self weight.

7. Moments, normal forces and shear forces

Bending moment

Maximum 31.8 kNm in member b, x = 2075 mm. (ULS 6.10b (CC1 new))

Bending moment

Normal force

Maximum 0.0 kN (− compression) in member , x = 0 mm. (ULS 6.10a (CC1 new))

Normal force

Shear force

Maximum 30.7 kN in member b, x = 0 mm. (ULS 6.10b (CC1 new))

Shear force

Largest section forces per member — ULS 6.10b (CC1 new)

MemberL (mm)|M| max (kNm)N at max |N| (kN)|V| max (kN)
b415031.80.030.7

The 12 members with the largest moment. All members are in the calculation; open a member for the full values.

Support reactions

Largest vertical reaction 30.7 kN at node A. Total vertical reaction 61.3 kN. (ULS 6.10b (CC1 new))

Support reactions

Support reactions (envelope of the ULS combinations)

SupportNodex, y, z (mm)FX (kN)FY (kN)FZ (kN)
S1A0, 0, 00.00.030.7
S2B4150, 0, 00.00.030.7

For every component the value of the largest magnitude over all ULS combinations (sign as in the global axes, Z up), so the horizontal reactions are included. The largest vertical reaction (total 61.3 kN) occurs in ULS 6.10b (CC1 new). Forces in kN; there are no horizontal reactions or support moments.

Vertical support reactions: envelope of the ULS combinations (min / max)

SupportNodeMin (kN)Max (kN)
S1A4.5530.67
S2B4.5530.67

Deformation

Maximum displacement 8.38 mm at node b. (SLS char.)

Deformation

Deflection and sway check (SLS)

MemberL (mm)δ (mm)LimitLimit (mm)δ / limit
b41508.38L/25016.600.50within the limit

δ is the largest deflection relative to the chord between the member ends (columns: the horizontal sway of the top relative to the bottom, h = height; the bending of the column itself is not part of this check) in SLS char.. Standard limit L/250; set a stricter limit (e.g. L/500 with masonry or a brittle finish) per member. The values are project requirements: check the National Annex.

Results per combination

CombinationM min (kNm)M max (kNm)V min (kN)V max (kN)N min (kN)N max (kN)Vertical reaction (kN): S1 / S2Max. displacement (mm)
ULS 6.10a (CC1 new)0.0016.86-16.2516.250.000.0016.25 / 16.255.74
ULS 6.10b (CC1 new)0.0031.82-30.6730.670.000.0030.67 / 30.6710.83
SLS char.0.0024.62-23.7323.730.000.0023.73 / 23.738.38
ULS 0.9·G favourable (CC1 new)0.004.72-4.554.550.000.004.55 / 4.551.61
ULS 0.9·G + Variable (Q) (CC1 new)0.0030.88-29.7629.760.000.0029.76 / 29.7610.50
ULS γG·G permanent only (CC1 new)0.006.40-6.176.170.000.006.17 / 6.172.18
ULS 0.9·G + ψ0 variable (CC1 new)0.0015.18-14.6314.630.000.0014.63 / 14.635.16
SLS permanent (G)0.005.24-5.055.050.000.005.05 / 5.051.78
SLS quasi-permanent (G + ψ2·Q)0.0011.06-10.6610.660.000.0010.66 / 10.663.76
SLS frequent (G + ψ1·Q)0.0014.93-14.3914.390.000.0014.39 / 14.395.08

Moment and shear force about the strong axis, section forces at 21 points per member; N − = compression. Displacement: the largest, including between the nodes.

8. Unity checks and calculations

Unity checks

Members coloured by their governing unity check: green up to UC 1.00, red above. The deflection check counts too.

MemberSectionUCGoverning check
bHEA1800.50deflection (SLS)

Member calculations

Detailed calculations of the members with a check (concrete EN 1992-1-1, steel EN 1993-1-1 / AISC 360).

Steel member b — HEA180, S235

EN 1993-1-1. Section properties are own calculations from nominal dimensions (see SOURCES.md).

1. Input

Sectionh × b × tw × tf (r)171.0 × 180.0 × 6.00 × 9.50 (15.0) mm
A, Iy, Iz45.27 cm², 2511.0 cm⁴, 924.6 cm⁴
Wel,y, Wpl,y293.7 cm³, 325.0 cm³
It, Iw15.10 cm⁴, 60290 cm⁶
Materialfy, E, G235 N/mm², 210000 N/mm², 81000 N/mm²
Length4150 mm
Lateral restraint pointsunrestrained lengthsnone (ends only): 4150 mm
Load position, k, kw, ky, kztop, 1, 1, 1, 1

2. Results per combination

ULS 6.10a (CC1 new)Lateral-torsional buckling (LTB), segment 1: x = 0–4150 mm, L = 4150 mm0.26
ULS 6.10b (CC1 new)Lateral-torsional buckling (LTB), segment 1: x = 0–4150 mm, L = 4150 mm0.49
ULS 0.9·G favourable (CC1 new)Lateral-torsional buckling (LTB), segment 1: x = 0–4150 mm, L = 4150 mm0.07
ULS 0.9·G + Variable (Q) (CC1 new)Lateral-torsional buckling (LTB), segment 1: x = 0–4150 mm, L = 4150 mm0.47
ULS γG·G permanent only (CC1 new)Lateral-torsional buckling (LTB), segment 1: x = 0–4150 mm, L = 4150 mm0.10
ULS 0.9·G + ψ0 variable (CC1 new)Lateral-torsional buckling (LTB), segment 1: x = 0–4150 mm, L = 4150 mm0.23

3. Calculation of every combination

ULS 6.10a (CC1 new) 0.26

0.22 Cross-section resistance x = 2075 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 0/196.545
0
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [16.859/76.365]^2 + [0/36.779]^1
0.048738
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 16.859/76.365 = 0.221; 0/36.779 = 0
0.22077
0.08 Cross-section resistance (maximum shear) x = 0 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 16.25/196.545
0.082676
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [2.709e-16/76.365]^2 + [0/36.779]^1
1.258e-35
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 2.709e-16/76.365 = 3.547e-18; 0/36.779 = 0
3.547e-18
0.26 Lateral-torsional buckling (LTB) segment 1: x = 0–4150 mm, L = 4150 mm

EN 1993-1-1 §6.3.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
W_y (class 1) [6.3.2.2]W_pl,y
= 324957.199 mm³
324960 mm³
π²·E·Iz/(k·L)² [NCCI SN003]π²·E·Iz/(k·L)²
= π²·210000·9.246e+6/(1·4150)²
1112.7 kN (N·mm/mm)
C1 [NCCI SN003]table value, uniform load (k = kw = 1,0)
= C1 = 1.127, C2 = 0.454
1.127
Load position z_g [NCCI SN003]z_g = z_a − z_s (positive: load above the shear centre)
= z_g = 85.5 mm
85.5 mm
Square-root term [NCCI SN003](k/kw)²·Iw/Iz + (kL)²·G·It/(π²·E·Iz) + (C2·zg)²
= (1/1)²·6.029e+10/9.246e+6 + (1·4150)²·81000·150965.805/(π²·210000·9.246e+6) + (0.454·85.5)²
19017
Elastic critical moment [NCCI SN003]Mcr = C1·π²EIz/(kL)²·{√[...] − C2·zg}
= 1.127·1112.719·(√19016.822 − 38.817)
124.26 kNm
Check: numerical eigenvalue analysisthin-walled beam FEM (Hermite elements)
= M_cr,num = 124.442 kNm; deviation from the formula = 0.15 %
information only
124.44 kNm
Relative slenderness λ̄_LT [6.3.2.2(1)]λ̄_LT = √(W_y·fy/Mcr)
= √(324957.199·235/1.243e+8)
0.78395
LTB curve [Table 6.5]Table 6.5: rolled I-section, h/b = 0.95 ≤ 2
= curve b, α_LT = 0.34
0.34
Φ_LT [6.3.2.3(1)]Φ_LT = 0,5[1 + α_LT(λ̄_LT − λ̄_LT,0) + β·λ̄_LT²], λ̄_LT,0 = 0,4, β = 0,75
= 0.5·[1 + 0.34·(0.784 − 0.4) + 0.75·0.784²]
0.79574
χ_LT [6.3.2.3(1)]χ_LT = 1/(Φ_LT + √(Φ_LT² − β·λ̄_LT²)) ≤ 1,0 and ≤ 1/λ̄_LT²
= 1/(0.796 + √(0.796² − 0.75·0.784²))
0.82591
Correction factor k_c [Table 6.6]Table 6.6
= k_c = 0.94
0.94
f [6.3.2.3(2)]f = 1 − 0,5(1 − k_c)[1 − 2(λ̄_LT − 0,8)²] ≤ 1,0
= 1 − 0.5·(1 − 0.94)·[1 − 2·(0.784 − 0.8)²]
0.97002
χ_LT,mod [6.3.2.3(2)]χ_LT,mod = χ_LT/f ≤ 1,0
= 0.826/0.97
0.85144
Buckling resistance moment [6.3.2.1(3)]M_b,Rd = χ_LT·W_y·fy/γM1
= 0.851·324957.199·235/1
65.02 kNm
LTB ratio [6.3.2.1(1)]M_Ed/M_b,Rd
= 16.859/65.02
0.25929

ULS 6.10b (CC1 new) 0.49

0.42 Cross-section resistance x = 2075 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 0/196.545
0
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [31.819/76.365]^2 + [0/36.779]^1
0.17361
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 31.819/76.365 = 0.417; 0/36.779 = 0
0.41667
0.16 Cross-section resistance (maximum shear) x = 0 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 30.669/196.545
0.15604
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [2.398e-16/76.365]^2 + [0/36.779]^1
9.861e-36
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 2.398e-16/76.365 = 3.140e-18; 0/36.779 = 0
3.140e-18
0.49 Lateral-torsional buckling (LTB) segment 1: x = 0–4150 mm, L = 4150 mm

EN 1993-1-1 §6.3.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
W_y (class 1) [6.3.2.2]W_pl,y
= 324957.199 mm³
324960 mm³
π²·E·Iz/(k·L)² [NCCI SN003]π²·E·Iz/(k·L)²
= π²·210000·9.246e+6/(1·4150)²
1112.7 kN (N·mm/mm)
C1 [NCCI SN003]table value, uniform load (k = kw = 1,0)
= C1 = 1.127, C2 = 0.454
1.127
Load position z_g [NCCI SN003]z_g = z_a − z_s (positive: load above the shear centre)
= z_g = 85.5 mm
85.5 mm
Square-root term [NCCI SN003](k/kw)²·Iw/Iz + (kL)²·G·It/(π²·E·Iz) + (C2·zg)²
= (1/1)²·6.029e+10/9.246e+6 + (1·4150)²·81000·150965.805/(π²·210000·9.246e+6) + (0.454·85.5)²
19017
Elastic critical moment [NCCI SN003]Mcr = C1·π²EIz/(kL)²·{√[...] − C2·zg}
= 1.127·1112.719·(√19016.822 − 38.817)
124.26 kNm
Check: numerical eigenvalue analysisthin-walled beam FEM (Hermite elements)
= M_cr,num = 124.442 kNm; deviation from the formula = 0.15 %
information only
124.44 kNm
Relative slenderness λ̄_LT [6.3.2.2(1)]λ̄_LT = √(W_y·fy/Mcr)
= √(324957.199·235/1.243e+8)
0.78395
LTB curve [Table 6.5]Table 6.5: rolled I-section, h/b = 0.95 ≤ 2
= curve b, α_LT = 0.34
0.34
Φ_LT [6.3.2.3(1)]Φ_LT = 0,5[1 + α_LT(λ̄_LT − λ̄_LT,0) + β·λ̄_LT²], λ̄_LT,0 = 0,4, β = 0,75
= 0.5·[1 + 0.34·(0.784 − 0.4) + 0.75·0.784²]
0.79574
χ_LT [6.3.2.3(1)]χ_LT = 1/(Φ_LT + √(Φ_LT² − β·λ̄_LT²)) ≤ 1,0 and ≤ 1/λ̄_LT²
= 1/(0.796 + √(0.796² − 0.75·0.784²))
0.82591
Correction factor k_c [Table 6.6]Table 6.6
= k_c = 0.94
0.94
f [6.3.2.3(2)]f = 1 − 0,5(1 − k_c)[1 − 2(λ̄_LT − 0,8)²] ≤ 1,0
= 1 − 0.5·(1 − 0.94)·[1 − 2·(0.784 − 0.8)²]
0.97002
χ_LT,mod [6.3.2.3(2)]χ_LT,mod = χ_LT/f ≤ 1,0
= 0.826/0.97
0.85144
Buckling resistance moment [6.3.2.1(3)]M_b,Rd = χ_LT·W_y·fy/γM1
= 0.851·324957.199·235/1
65.02 kNm
LTB ratio [6.3.2.1(1)]M_Ed/M_b,Rd
= 31.819/65.02
0.48937

ULS 0.9·G favourable (CC1 new) 0.07

0.06 Cross-section resistance x = 2075 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 0/196.545
0
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [4.719/76.365]^2 + [0/36.779]^1
0.0038179
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 4.719/76.365 = 0.062; 0/36.779 = 0
0.061789
0.02 Cross-section resistance (maximum shear) x = 0 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 4.548/196.545
0.02314
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [1.998e-16/76.365]^2 + [0/36.779]^1
6.848e-36
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 1.998e-16/76.365 = 2.617e-18; 0/36.779 = 0
2.617e-18
0.07 Lateral-torsional buckling (LTB) segment 1: x = 0–4150 mm, L = 4150 mm

EN 1993-1-1 §6.3.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
W_y (class 1) [6.3.2.2]W_pl,y
= 324957.199 mm³
324960 mm³
π²·E·Iz/(k·L)² [NCCI SN003]π²·E·Iz/(k·L)²
= π²·210000·9.246e+6/(1·4150)²
1112.7 kN (N·mm/mm)
C1 [NCCI SN003]table value, uniform load (k = kw = 1,0)
= C1 = 1.127, C2 = 0.454
1.127
Load position z_g [NCCI SN003]z_g = z_a − z_s (positive: load above the shear centre)
= z_g = 85.5 mm
85.5 mm
Square-root term [NCCI SN003](k/kw)²·Iw/Iz + (kL)²·G·It/(π²·E·Iz) + (C2·zg)²
= (1/1)²·6.029e+10/9.246e+6 + (1·4150)²·81000·150965.805/(π²·210000·9.246e+6) + (0.454·85.5)²
19017
Elastic critical moment [NCCI SN003]Mcr = C1·π²EIz/(kL)²·{√[...] − C2·zg}
= 1.127·1112.719·(√19016.822 − 38.817)
124.26 kNm
Check: numerical eigenvalue analysisthin-walled beam FEM (Hermite elements)
= M_cr,num = 124.442 kNm; deviation from the formula = 0.15 %
information only
124.44 kNm
Relative slenderness λ̄_LT [6.3.2.2(1)]λ̄_LT = √(W_y·fy/Mcr)
= √(324957.199·235/1.243e+8)
0.78395
LTB curve [Table 6.5]Table 6.5: rolled I-section, h/b = 0.95 ≤ 2
= curve b, α_LT = 0.34
0.34
Φ_LT [6.3.2.3(1)]Φ_LT = 0,5[1 + α_LT(λ̄_LT − λ̄_LT,0) + β·λ̄_LT²], λ̄_LT,0 = 0,4, β = 0,75
= 0.5·[1 + 0.34·(0.784 − 0.4) + 0.75·0.784²]
0.79574
χ_LT [6.3.2.3(1)]χ_LT = 1/(Φ_LT + √(Φ_LT² − β·λ̄_LT²)) ≤ 1,0 and ≤ 1/λ̄_LT²
= 1/(0.796 + √(0.796² − 0.75·0.784²))
0.82591
Correction factor k_c [Table 6.6]Table 6.6
= k_c = 0.94
0.94
f [6.3.2.3(2)]f = 1 − 0,5(1 − k_c)[1 − 2(λ̄_LT − 0,8)²] ≤ 1,0
= 1 − 0.5·(1 − 0.94)·[1 − 2·(0.784 − 0.8)²]
0.97002
χ_LT,mod [6.3.2.3(2)]χ_LT,mod = χ_LT/f ≤ 1,0
= 0.826/0.97
0.85144
Buckling resistance moment [6.3.2.1(3)]M_b,Rd = χ_LT·W_y·fy/γM1
= 0.851·324957.199·235/1
65.02 kNm
LTB ratio [6.3.2.1(1)]M_Ed/M_b,Rd
= 4.719/65.02
0.072571

ULS 0.9·G + Variable (Q) (CC1 new) 0.47

0.40 Cross-section resistance x = 2075 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 0/196.545
0
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [30.875/76.365]^2 + [0/36.779]^1
0.16347
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 30.875/76.365 = 0.404; 0/36.779 = 0
0.40431
0.15 Cross-section resistance (maximum shear) x = 0 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 29.759/196.545
0.15141
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [1.998e-16/76.365]^2 + [0/36.779]^1
6.848e-36
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 1.998e-16/76.365 = 2.617e-18; 0/36.779 = 0
2.617e-18
0.47 Lateral-torsional buckling (LTB) segment 1: x = 0–4150 mm, L = 4150 mm

EN 1993-1-1 §6.3.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
W_y (class 1) [6.3.2.2]W_pl,y
= 324957.199 mm³
324960 mm³
π²·E·Iz/(k·L)² [NCCI SN003]π²·E·Iz/(k·L)²
= π²·210000·9.246e+6/(1·4150)²
1112.7 kN (N·mm/mm)
C1 [NCCI SN003]table value, uniform load (k = kw = 1,0)
= C1 = 1.127, C2 = 0.454
1.127
Load position z_g [NCCI SN003]z_g = z_a − z_s (positive: load above the shear centre)
= z_g = 85.5 mm
85.5 mm
Square-root term [NCCI SN003](k/kw)²·Iw/Iz + (kL)²·G·It/(π²·E·Iz) + (C2·zg)²
= (1/1)²·6.029e+10/9.246e+6 + (1·4150)²·81000·150965.805/(π²·210000·9.246e+6) + (0.454·85.5)²
19017
Elastic critical moment [NCCI SN003]Mcr = C1·π²EIz/(kL)²·{√[...] − C2·zg}
= 1.127·1112.719·(√19016.822 − 38.817)
124.26 kNm
Check: numerical eigenvalue analysisthin-walled beam FEM (Hermite elements)
= M_cr,num = 124.442 kNm; deviation from the formula = 0.15 %
information only
124.44 kNm
Relative slenderness λ̄_LT [6.3.2.2(1)]λ̄_LT = √(W_y·fy/Mcr)
= √(324957.199·235/1.243e+8)
0.78395
LTB curve [Table 6.5]Table 6.5: rolled I-section, h/b = 0.95 ≤ 2
= curve b, α_LT = 0.34
0.34
Φ_LT [6.3.2.3(1)]Φ_LT = 0,5[1 + α_LT(λ̄_LT − λ̄_LT,0) + β·λ̄_LT²], λ̄_LT,0 = 0,4, β = 0,75
= 0.5·[1 + 0.34·(0.784 − 0.4) + 0.75·0.784²]
0.79574
χ_LT [6.3.2.3(1)]χ_LT = 1/(Φ_LT + √(Φ_LT² − β·λ̄_LT²)) ≤ 1,0 and ≤ 1/λ̄_LT²
= 1/(0.796 + √(0.796² − 0.75·0.784²))
0.82591
Correction factor k_c [Table 6.6]Table 6.6
= k_c = 0.94
0.94
f [6.3.2.3(2)]f = 1 − 0,5(1 − k_c)[1 − 2(λ̄_LT − 0,8)²] ≤ 1,0
= 1 − 0.5·(1 − 0.94)·[1 − 2·(0.784 − 0.8)²]
0.97002
χ_LT,mod [6.3.2.3(2)]χ_LT,mod = χ_LT/f ≤ 1,0
= 0.826/0.97
0.85144
Buckling resistance moment [6.3.2.1(3)]M_b,Rd = χ_LT·W_y·fy/γM1
= 0.851·324957.199·235/1
65.02 kNm
LTB ratio [6.3.2.1(1)]M_Ed/M_b,Rd
= 30.875/65.02
0.47486

ULS γG·G permanent only (CC1 new) 0.10

0.08 Cross-section resistance x = 2075 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 8.882e-16/196.545
4.519e-18
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [6.396/76.365]^2 + [0/36.779]^1
0.0070156
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 6.396/76.365 = 0.084; 0/36.779 = 0
0.083759
0.03 Cross-section resistance (maximum shear) x = 0 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 6.165/196.545
0.031367
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [2.709e-16/76.365]^2 + [0/36.779]^1
1.258e-35
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 2.709e-16/76.365 = 3.547e-18; 0/36.779 = 0
3.547e-18
0.10 Lateral-torsional buckling (LTB) segment 1: x = 0–4150 mm, L = 4150 mm

EN 1993-1-1 §6.3.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
W_y (class 1) [6.3.2.2]W_pl,y
= 324957.199 mm³
324960 mm³
π²·E·Iz/(k·L)² [NCCI SN003]π²·E·Iz/(k·L)²
= π²·210000·9.246e+6/(1·4150)²
1112.7 kN (N·mm/mm)
C1 [NCCI SN003]table value, uniform load (k = kw = 1,0)
= C1 = 1.127, C2 = 0.454
1.127
Load position z_g [NCCI SN003]z_g = z_a − z_s (positive: load above the shear centre)
= z_g = 85.5 mm
85.5 mm
Square-root term [NCCI SN003](k/kw)²·Iw/Iz + (kL)²·G·It/(π²·E·Iz) + (C2·zg)²
= (1/1)²·6.029e+10/9.246e+6 + (1·4150)²·81000·150965.805/(π²·210000·9.246e+6) + (0.454·85.5)²
19017
Elastic critical moment [NCCI SN003]Mcr = C1·π²EIz/(kL)²·{√[...] − C2·zg}
= 1.127·1112.719·(√19016.822 − 38.817)
124.26 kNm
Check: numerical eigenvalue analysisthin-walled beam FEM (Hermite elements)
= M_cr,num = 124.442 kNm; deviation from the formula = 0.15 %
information only
124.44 kNm
Relative slenderness λ̄_LT [6.3.2.2(1)]λ̄_LT = √(W_y·fy/Mcr)
= √(324957.199·235/1.243e+8)
0.78395
LTB curve [Table 6.5]Table 6.5: rolled I-section, h/b = 0.95 ≤ 2
= curve b, α_LT = 0.34
0.34
Φ_LT [6.3.2.3(1)]Φ_LT = 0,5[1 + α_LT(λ̄_LT − λ̄_LT,0) + β·λ̄_LT²], λ̄_LT,0 = 0,4, β = 0,75
= 0.5·[1 + 0.34·(0.784 − 0.4) + 0.75·0.784²]
0.79574
χ_LT [6.3.2.3(1)]χ_LT = 1/(Φ_LT + √(Φ_LT² − β·λ̄_LT²)) ≤ 1,0 and ≤ 1/λ̄_LT²
= 1/(0.796 + √(0.796² − 0.75·0.784²))
0.82591
Correction factor k_c [Table 6.6]Table 6.6
= k_c = 0.94
0.94
f [6.3.2.3(2)]f = 1 − 0,5(1 − k_c)[1 − 2(λ̄_LT − 0,8)²] ≤ 1,0
= 1 − 0.5·(1 − 0.94)·[1 − 2·(0.784 − 0.8)²]
0.97002
χ_LT,mod [6.3.2.3(2)]χ_LT,mod = χ_LT/f ≤ 1,0
= 0.826/0.97
0.85144
Buckling resistance moment [6.3.2.1(3)]M_b,Rd = χ_LT·W_y·fy/γM1
= 0.851·324957.199·235/1
65.02 kNm
LTB ratio [6.3.2.1(1)]M_Ed/M_b,Rd
= 6.396/65.02
0.098374

ULS 0.9·G + ψ0 variable (CC1 new) 0.23

0.20 Cross-section resistance x = 2075 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 1.776e-15/196.545
9.038e-18
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [15.181/76.365]^2 + [0/36.779]^1
0.039521
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 15.181/76.365 = 0.199; 0/36.779 = 0
0.1988
0.07 Cross-section resistance (maximum shear) x = 4150 mm

EN 1993-1-1 §5.5, 6.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (compression) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 33ε/38ε/42ε = 33/38/42 → class 1
20.333
Class (compression) [5.5]class = max(flange, web)
= max(1, 1)
1
Plastic axial resistance [6.2.4]N_pl,Rd = A·fy/γM0
= 4526.624·235/1
1063.8 kN
Bending resistance y-y (plastic) [6.2.5]M_c,y,Rd = Wpl,y·fy/γM0
= 324957.199·235/1
76.365 kNm
Bending resistance z-z (plastic) [6.2.5]M_c,z,Rd = Wpl,z·fy/γM0
= 156507.081·235/1
36.779 kNm
Plastic shear resistance (z) [6.2.6]V_pl,z,Rd = Avz·fy/(√3·γM0)
= 1448.624·235/(√3·1)
196.55 kN
Shear ratio [6.2.6]V_Ed/V_pl,z,Rd
= 14.632/196.545
0.074448
Axial ratio [6.2.4]|N_Ed|/N_pl,Rd
= 0/1063.757
0
n = N_Ed/N_pl,Rd [6.2.9.1]n
= 0/1063.757
0
a = (A − 2 b tf)/A ≤ 0,5 [6.2.9.1(5)]a
= (4526.624 − 2·180·9.5)/4526.624
0.24447
Reduced M_N,y,Rd [6.2.9.1(5)]M_pl,y,Rd(1−n)/(1−0,5a) ≤ M_pl,y,Rd
= 76.365·(1−0)/(1−0.5·0.244)
76.365 kNm
Reduced M_N,z,Rd [6.2.9.1(5)]n ≤ a: M_pl,z,Rd; else M_pl,z,Rd[1 − ((n−a)/(1−a))²]
= n = 0, a = 0.244
36.779 kNm
Interaction [6.2.9.1(6)][M_y,Ed/M_N,y,Rd]^α + [M_z,Ed/M_N,z,Rd]^β, α = 2, β = max(1; 5n)
= [7.105e-15/76.365]^2 + [0/36.779]^1
8.657e-33
Bending ratio per axis [6.2.9.1(2), 6.2.5]M_y,Ed/M_N,y,Rd and M_z,Ed/M_N,z,Rd (each ≤ 1,0)
= 7.105e-15/76.365 = 9.305e-17; 0/36.779 = 0
9.305e-17
0.23 Lateral-torsional buckling (LTB) segment 1: x = 0–4150 mm, L = 4150 mm

EN 1993-1-1 §6.3.2

ε [Table 5.2]ε = √(235/fy)
= √(235/235)
1
Flange outstand c/t [Table 5.2]c = (b − tw − 2r)/2
= (180 − 6 − 2·15)/2 = 72; c/tf = 7.579
limits 9ε/10ε/14ε = 9/10/14 → class 1
7.5789
Web c/t (bending) [Table 5.2]c = h − 2tf − 2r
= 171 − 2·9.5 − 2·15 = 122; c/tw = 20.333
limits 72ε/83ε/124ε = 72/83/124 → class 1
20.333
Class (bending) [5.5]class = max(flange, web)
= max(1, 1)
1
W_y (class 1) [6.3.2.2]W_pl,y
= 324957.199 mm³
324960 mm³
π²·E·Iz/(k·L)² [NCCI SN003]π²·E·Iz/(k·L)²
= π²·210000·9.246e+6/(1·4150)²
1112.7 kN (N·mm/mm)
C1 [NCCI SN003]table value, uniform load (k = kw = 1,0)
= C1 = 1.127, C2 = 0.454
1.127
Load position z_g [NCCI SN003]z_g = z_a − z_s (positive: load above the shear centre)
= z_g = 85.5 mm
85.5 mm
Square-root term [NCCI SN003](k/kw)²·Iw/Iz + (kL)²·G·It/(π²·E·Iz) + (C2·zg)²
= (1/1)²·6.029e+10/9.246e+6 + (1·4150)²·81000·150965.805/(π²·210000·9.246e+6) + (0.454·85.5)²
19017
Elastic critical moment [NCCI SN003]Mcr = C1·π²EIz/(kL)²·{√[...] − C2·zg}
= 1.127·1112.719·(√19016.822 − 38.817)
124.26 kNm
Check: numerical eigenvalue analysisthin-walled beam FEM (Hermite elements)
= M_cr,num = 124.442 kNm; deviation from the formula = 0.15 %
information only
124.44 kNm
Relative slenderness λ̄_LT [6.3.2.2(1)]λ̄_LT = √(W_y·fy/Mcr)
= √(324957.199·235/1.243e+8)
0.78395
LTB curve [Table 6.5]Table 6.5: rolled I-section, h/b = 0.95 ≤ 2
= curve b, α_LT = 0.34
0.34
Φ_LT [6.3.2.3(1)]Φ_LT = 0,5[1 + α_LT(λ̄_LT − λ̄_LT,0) + β·λ̄_LT²], λ̄_LT,0 = 0,4, β = 0,75
= 0.5·[1 + 0.34·(0.784 − 0.4) + 0.75·0.784²]
0.79574
χ_LT [6.3.2.3(1)]χ_LT = 1/(Φ_LT + √(Φ_LT² − β·λ̄_LT²)) ≤ 1,0 and ≤ 1/λ̄_LT²
= 1/(0.796 + √(0.796² − 0.75·0.784²))
0.82591
Correction factor k_c [Table 6.6]Table 6.6
= k_c = 0.94
0.94
f [6.3.2.3(2)]f = 1 − 0,5(1 − k_c)[1 − 2(λ̄_LT − 0,8)²] ≤ 1,0
= 1 − 0.5·(1 − 0.94)·[1 − 2·(0.784 − 0.8)²]
0.97002
χ_LT,mod [6.3.2.3(2)]χ_LT,mod = χ_LT/f ≤ 1,0
= 0.826/0.97
0.85144
Buckling resistance moment [6.3.2.1(3)]M_b,Rd = χ_LT·W_y·fy/γM1
= 0.851·324957.199·235/1
65.02 kNm
LTB ratio [6.3.2.1(1)]M_Ed/M_b,Rd
= 15.181/65.02
0.23349

Conclusion

Maximum unity check = 0.49 — ULS 6.10b (CC1 new): Lateral-torsional buckling (LTB), segment 1: x = 0–4150 mm, L = 4150 mm.

No lateral restraint points were given: LTB assumes restraint at the member ends only.

Calculation steps are in SI units (mm, N, kN, kNm, N/mm²).

For training, testing and educational use; not for the design of real structures.

For training, testing and educational use only. Not for design of real structures. Results are unverified. THE SOFTWARE IS PROVIDED 'AS IS', WITHOUT WARRANTY OF ANY KIND. Not a substitute for review by a licensed Professional Engineer (PE/SE) or equivalent.