Every tool, worked end to end

RA = RB = w·L / 2

RA = 10 × 6 / 2

RA = RB = 30 kN

By symmetry and vertical equilibrium (ΣV = 0) each support carries half of the total load w·L.

MEd = w·L² / 8 (midspan)

MEd = 10 × 6² / 8

MEd = 45 kNm

The moment diagram is parabolic with its peak at midspan — this is the value the cross-section design must resist.

VEd = w·L / 2 (supports)

VEd = 10 × 6 / 2

VEd = 30 kN

Shear is largest at the supports and zero at midspan — this value drives the shear design.

δ = 5·w·L⁴ / (384·E·I)

δ = 5 × 10 × 6000⁴ / (384 × 210000 × 83560000)

δ = 9.62 mm

Instantaneous midspan deflection from elastic beam theory; the relevant material Eurocode (§7.2) sets the limit it is checked against.

ε = √(235 / fy)

ε = √(235 / 275)

ε = 0.924

Scales every classification limit with yield strength: the higher the grade, the stricter the c/t limits for the same class.

c/t ≤ 72ε (Class 1) / 83ε (Class 2) / 124ε (Class 3)

c/t = hw/tw = 248.6/7.1 = 35.01 | 72ε = 66.6

Web is Class 1 (Plastic)

The web is an internal element in pure bending. Its slenderness decides whether it can sustain plastic rotation (Class 1), reach the plastic moment (Class 2), stay elastic (Class 3) or buckles locally (Class 4).

c/t ≤ 9ε / 10ε / 14ε, c = (b − 2r − tw)/2

c = (150 − 2×15 − 7.1)/2 = 56.5 mm | c/t = 5.28 | 9ε = 8.3

Flange is Class 1 (Plastic)

Each half-flange is an outstand element compressed by bending. The root radius r is excluded from the flat width c.

Overall class = the more critical (higher) of web and flange class

Section is Class 1 (Plastic)

The section's class is set by its most slender part, and it decides whether the plastic modulus Wpl or only the elastic modulus Wel may be used in the bending check.

Plastic section properties (Wpl) may be used.

Mc,Rd = Wpl·fy / γM0

Mc,Rd = 628,400 × 275 / 1.0

Mc,Rd = 172.81 kNm

Design bending resistance of the section. γM0 = 1.0 is the recommended partial factor for cross-section resistance — your National Annex may specify a different value.

MEd / Mc,Rd ≤ 1.0

45 / 172.81

Utilization = 0.26 ✓

The ULS verification for bending: the design moment must not exceed the design resistance. A utilization above 1.0 means the section is overloaded.

Av = A − 2b·tf + (tw + 2r)·tf (≥ η·hw·tw, η = 1.2)

Av = 5,381 − 2×150×10.7 + (7.1+2×15)×10.7

Av = 2568 mm²

Only the material near the web carries shear in an I-section. This rolled-section formula removes the flanges but adds back the material in the root radii; it may never be less than η·hw·tw.

Vpl,Rd = Av·(fy/√3) / γM0

Vpl,Rd = 2568 × (275/√3) / 1.0

Vpl,Rd = 407.72 kN

Plastic shear resistance: yielding in shear at fy/√3 over the shear area. For slender webs the reduced value Vbw,Rd from EN 1993-1-5 applies instead.

Not susceptible if hw/tw ≤ 72ε/η

hw/tw = 35.01 vs 72ε/η = 55.5

Web is NOT susceptible to shear buckling.

Stocky webs (hw/tw below the limit) yield before they buckle, so the plastic shear resistance can be used without checking EN 1993-1-5.

VEd / Vpl,Rd ≤ 1.0

30 / 407.72

Utilization = 0.074 ✓

The ULS verification for shear. Where bending and shear interact heavily (VEd > 0.5·Vpl,Rd at the same point), the moment resistance must also be reduced per §6.2.8.

δ ≤ L/250

δ = 9.62 mm vs limit = 24 mm

Utilization = 0.401 ✓

A serviceability (not safety) check: excessive deflection damages finishes and looks bad. The limit L/n is agreed for the project — EN 1993-1-1 §7.2 refers limits to the National Annex / client brief (L/250 is a common value for imposed load).

d = h − cover − φ/2

d = 500 − 35 − 25/2

d = 452.5 mm

Effective depth is measured to the centroid of the tension steel. The cover (nominal cover + allowance for deviation, §4.4.1) protects the bars against corrosion and fire.

fyd = fyk / 1.15

fyd = 500 / 1.15

fyd = 434.8 N/mm²

The characteristic yield strength fyk is divided by the partial factor γS = 1.15 (recommended value) to obtain the design strength used in all section checks.

MRdc = 0.167 · fck · b · d²

MRdc = 0.167 × 30 × 300 × 453²

MRdc = 307.75 kNm

With the neutral axis at its ductility limit x/d = 0.45 (from §5.5, no redistribution) and the rectangular stress block of §3.1.7 (αcc = 0.85, γC = 1.5), the compressed concrete can carry at most this moment without compression steel.

If MEd ≤ MRdc → singly reinforced. If MEd > MRdc → doubly reinforced.

MEd = 250 kNm vs MRdc = 307.75 kNm

Section is SINGLY reinforced — no compression steel needed.

Keeping x/d ≤ 0.45 guarantees the tension steel yields before the concrete crushes — a ductile failure with warning. If the moment exceeds MRdc, the excess is carried by compression steel instead of a deeper stress block.

z = (d/2) · [1 + √(1 − 3.53·MEd/(fck·b·d²))] (≤ 0.95d)

z = 389.6 mm

Distance between the tension steel and the centroid of the compression force. The closed form solves M = 0.8x·fcd·b·(d − 0.4x) of the rectangular stress block; the cap 0.95d is the code maximum lever arm.

As = MEd / (fyd·z)

As = (250×10⁶) / (434.8 × 389.6)

As = 1476 mm²

Moment equilibrium of the internal couple: the steel force As·fyd acting through the lever arm z must balance the applied moment.

Bars required = As / (area of one φ25 bar)

1476 / 491

Use 4 × φ25mm bars

Round up so the provided area is at least As, spread in at least two layers if needed. Detailing (spacing, anchorage, minimum steel per §9.2.1) must also be satisfied.

VRd,c = [0.12·k·(100ρ₁fck)^⅓]·bw·d (≥ vmin·bw·d)

k = 1.66, ρ₁ = 0.0109

VRd,c = 86.65 kN

Shear carried by the concrete alone: aggregate interlock and dowel action. k = 1 + √(200/d) ≤ 2 is a size effect, ρ1 (≤ 2%) is the tension steel ratio, and the floor value vmin = 0.035·k^1.5·√fck applies to lightly reinforced members.

Shear reinforcement is required if VEd > VRd,c

VEd = 150 kN vs VRd,c = 86.65 kN

YES — links must be designed.

If the applied shear exceeds the concrete's capacity, designed links are required; where links are used at all, the minimum link area of §9.2.2 applies.

vEd = VEd / (bw·z), z ≈ 0.9d

vEd = 1.23 N/mm²

Average shear stress over the effective web area bw·z — the working stress the truss model has to carry.

θ = 0.5·sin⁻¹[ vEd / (0.2·fck·(1 − fck/250)) ]

θ = 6.7° (cot θ = 2.5)

Strut inclination from the crushing limit of the concrete struts (VRd,max). Flatter struts (cot θ up to 2.5) need less steel but stress the struts more — the code limits cot θ to 1 ≤ cot θ ≤ 2.5.

Asw/s = vEd·bw / (fywd·cotθ)

Asw/s = 0.339 mm²/mm

From the variable-strut truss model: VRd,s = (Asw/s)·z·fywd·cotθ ≥ VEd. Choose a link diameter and spacing so the provided Asw/s per leg-set meets or exceeds this value.

Choose a link diameter and spacing so that Asw/s (for the chosen number of legs) meets or exceeds this value.

kmod = f(service class, load-duration class) [EC5 Table 3.1]

Service class 2, Medium-term load

kmod = 0.8

Timber strength depends on how long the load acts and how moist the wood is: short loads on dry timber can use more than the characteristic strength, while permanent loads in damp conditions must be reduced heavily.

γM = 1.30 (solid timber) / 1.25 (glulam)

γM = 1.3

Covers material variability and the effect of duration of load and moisture not already taken by kmod. These are the recommended values — several National Annexes (e.g. Denmark) specify different γM.

kdef = f(service class) [EC5 Table 3.2]

kdef = 0.8

Creep factor: timber keeps deforming under sustained load, and the damper the environment (higher service class), the more creep.

fm,d = kmod·kh·fm,k / γM

fm,d = 0.8 × 1 × 24 / 1.3

fm,d = 14.77 N/mm²

EC5's basic recipe for every design strength: characteristic strength adjusted for duration and moisture (kmod), size (kh) and material safety (γM).

W = b·h²/6, I = b·h³/12

b = — mm, h = — mm

W = 5333333 mm³, I = 1066666667 mm⁴

Elastic section modulus and second moment of area of the rectangular section — W converts moment into bending stress, I stiffens the beam against deflection.

RA = RB = w·L / 2

RA = 5 × 5 / 2

RA = RB = 12.5 kN

By symmetry and vertical equilibrium (ΣV = 0) each support carries half of the total load w·L.

MEd = w·L² / 8 (midspan)

MEd = 5 × 5² / 8

MEd = 15.63 kNm

The moment diagram is parabolic with its peak at midspan — this is the value the cross-section design must resist.

VEd = w·L / 2 (supports)

VEd = 5 × 5 / 2

VEd = 12.5 kN

Shear is largest at the supports and zero at midspan — this value drives the shear design.

δ = 5·w·L⁴ / (384·E·I)

δ = 5 × 5 × 5000⁴ / (384 × 11000 × 1066666667)

δ = 3.47 mm

Instantaneous midspan deflection from elastic beam theory; the relevant material Eurocode (§7.2) sets the limit it is checked against.

σm,d = MEd/W ≤ fm,d

σm,d = 15.63×10⁶ / 5333333 = 2.93 N/mm² vs fm,d = 14.77 N/mm²

Utilization = 0.198 ✓

The ULS bending verification: the design bending stress from the elastic section modulus must not exceed the design bending strength.

τd = 1.5·VEd/(b·h) ≤ fv,d, fv,d = kmod·fv,k/γM

τd = 1.5 × 12.5×10³ / (— × —) = 0.23 N/mm² vs fv,d = 2.46 N/mm²

Utilization = 0.095 ✓

The parabolic shear distribution is simplified to the average stress × 1.5. For members with cracks or notches at the support, §6.1.7 also requires an effective width kcr·b and load positioned within d from the support may be neglected.

winst ≤ L/300

winst = 3.47 mm vs limit = 16.7 mm

Utilization = 0.208 ✓

Immediate deflection under the rare (characteristic) load combination, computed with the mean modulus E0,mean. The limiting L/n value comes from the National Annex (L/300 is a common requirement for timber).

wfin = winst·(1 + kdef) ≤ L/150

wfin = 3.47 × (1 + 0.8) = 6.24 mm vs limit = 33.3 mm

Utilization = 0.187 ✓

Final deflection including creep: EC5 scales the instantaneous value by (1 + kdef), with the quasi-permanent load share contributing the creep part. The limit (commonly L/150) is set by the National Annex.

Σ gk = covering + joists + insulation + boarding + lining

Floor covering 0.08 + Joists 0.126 + Insulation 0.087 + Sec. spaced boarding 0.016 + Ceiling lining 0.117

Σ gk = 0.427 kN/m²

The joists enter as A·5 kN/m³ ÷ spacing; the insulation fills the joist depth at 0.5 kN/m³.

qk = imposed load + moveable partitions

category A1 (Rooms in residential buildings): qk = 1.5 · partitions: 0.5

Σ qk = 2 kN/m²

Ed = 1.0·gk + 1.5·qk → qjoist = Ed·spacing

1.0·0.427 + 1.5·2 = 3.43 kN/m² · × 0.5 m

Ed = 3.43 kN/m² → 1.71 kN/m on one joist

Vd = ½·q·L · Md = q·L²/8

Vd = ½·1.71·4.5 · Md = 1.71·4.5²/8

Vd = 3.86 kN · Md = 4.34 kNm

fmd, fvd from the quality and service class tables

C18 · service class 1 (Moisture ≤ 12 % — indoor.)

fmd = 10.7 N/mm² · fvd = 1.2 N/mm²

Wness = Md/fmd ≤ Wy

4.34·1000/10.7 = 405 ·×10³ mm³ vs 420 ·×10³ mm³

OK — 420 ≥ 405 ×10³ mm³

σvd = 1.5·Vd/A ≤ fvd

1.5·3.86·1000/12600 = 0.459 N/mm² vs fvd = 1.2

OK — 0.459 ≤ 1.2 N/mm²

uinst = 5·q·L⁴/(384·E·Iy) · ufin = uinst·(1 + ψ2·kdef)

frequent: 0.51 kN/m → 7.25 → 8.12 mm · quasi-perm.: 0.41 kN/m → 5.84 → 6.54 mm

governing u = 8.12 mm

E = 9000 N/mm², Iy = 42000·10³ mm⁴, kdef = 0.6 (class 1), ψ2 = 0.2.

limit = L/500 (rounded up)

4.5·1000/500

OK — 9.0 mm ≥ 8.12 mm

μ1 = 0.8 for α ≤ 30°; 0.8·(60 − α)/30 for 30° < α < 60°; 0 for α ≥ 60°

α = 30° → μ1 = 0.8

μ1 = 0.8

Snow slides less off shallow roofs: μ1 stays at 0.8 up to 30°, then falls linearly to zero at 60° — steeper roofs shed their snow.

Ce = 0.8 windswept · 1.0 normal · 1.2 sheltered

Sheltered → Ce = 1.2

Ce = 1.2

Windswept ground loses snow to the wind (load reduced); sheltered ground keeps it (load increased).

Ct = 1.0 (Danish national annex)

Ct = 1

Ct = 1

A warm roof melts its underside snow layer; the Danish annex fixes Ct at 1.0, so no reduction is taken here.

sk,0 from the national annex (terrain value)

sk,0 = 1 kN/m²

sk,0 = 1 kN/m²

The reference load on flat ground. The Danish annex takes sk,0 = 1.0 kN/m² for the whole country.

s = μi·Ce·Ct·sk,0

s = 0.8 × 1.2 × 1 × 1

s = 0.96 kN/m²

The uniformly distributed load to apply to the roof — multiply by the loaded width to get the line load w (kN/m) on each rafter or beam. For ULS design, combine it with the partial factor γQ from EN 1990.

z0, zmin from the terrain category; kt = 0.19·(z0/0.05)^0.07

z0 = 0.05 m · zmin = 2 m → kt = 0.19·(0.05/0.05)^0.07

kt = 0.19

II — Low vegetation, isolated obstacles. The annex defines five terrain categories with their own roughness lengths.

z = max(h, zmin)

h = 7 m · zmin = 2 m

z = 7 m

Below zmin the profile is capped at zmin — the wind speed never drops further.

qb = ½·ρ·vb²

vb = 24 m/s · Air density ρ = 1.25 kg/m³ (Danish NA). → 0.5·1.25·24²/1000

qb = 0.36 kN/m²

vb,0 = 24 m/s elsewhere in Denmark.

vm = kt·ln(z/z0)·vb

cr = 0.19·ln(7/0.05) = 0.939 → vm = 0.939·24

vm = 22.53 m/s

Iv = 1/ln(z/z0)

1/ln(7/0.05)

Iv = 0.202

qp = (1 + 7·Iv)·½·ρ·vm²

(1 + 7·0.202)·0.5·1.25·22.53²/1000

qp = 0.767 kN/m²

This is the pressure that carries all gusting — used directly for the roof zones.

Characteristic (50-year) wind — cprob = 1.0 in the Danish NA basis.

roof width = (b + 2f)/cos v · roof length = l + 2g

(10 + 2·0.5)/cos 30° · 20 + 2·0.5

12.7 m × 21 m

Measured on the roof plane — the overhangs belong to the loaded area.

e = min(b, 2h)

facade: min(20, 2·7) · gable: min(10, 2·7)

e = 14 m (facade) · 10 m (gable)

e sets the width of the edge zones F and G: e/4 strips at the corners, e/10 strips along the edges.

Ed = γQ·(cpe + cpi)·qp·A·cos v − γG·gk·A·cos v

γQ = 1.5 · cpi = +0.2 · Anec = Ed·1000/201 → ties = ⌈Anec/60 mm²⌉

cpi = +0.2 · γG = 0.9 · γQ = 1.5

Positive Ed is net uplift. Each zone is anchored with 40 mm × 2 mm steel ties in S235 (Anec = Ed/fyd).

z0, zmin from the terrain category; kt = 0.19·(z0/0.05)^0.07

z0 = 0.05 m · zmin = 2 m → kt = 0.19·(0.05/0.05)^0.07

kt = 0.19

II — Low vegetation, isolated obstacles. The annex defines five terrain categories with their own roughness lengths.

z = max(h, zmin)

h = 7 m · zmin = 2 m

z = 7 m

Below zmin the profile is capped at zmin — the wind speed never drops further.

qb = ½·ρ·vb²

vb = 24 m/s · Air density ρ = 1.25 kg/m³ (Danish NA). → 0.5·1.25·24²/1000

qb = 0.36 kN/m²

vb,0 = 24 m/s elsewhere in Denmark.

vm = kt·ln(z/z0)·vb

cr = 0.19·ln(7/0.05) = 0.939 → vm = 0.939·24

vm = 22.53 m/s

Iv = 1/ln(z/z0)

1/ln(7/0.05)

Iv = 0.202

qp = (1 + 7·Iv)·½·ρ·vm²

(1 + 7·0.202)·0.5·1.25·22.53²/1000

qp = 0.767 kN/m²

This is the pressure that carries all gusting — used directly for the roof zones.

Characteristic (50-year) wind — cprob = 1.0 in the Danish NA basis.

roof width = (b + 2f)/cos v · roof length = l + 2g

(10 + 2·0.5)/cos 30° · 20 + 2·0.5

12.7 m × 21 m

Measured on the roof plane — the overhangs belong to the loaded area.

e = min(b, 2h)

facade: min(20, 2·7) · gable: min(10, 2·7)

e = 14 m (facade) · 10 m (gable)

e sets the width of the edge zones F and G: e/4 strips at the corners, e/10 strips along the edges.

Ed = γQ·(cpe + cpi)·qp·A·cos v − γG·gk·A·cos v

γQ = 1.5 · cpi = +0.2 · Anec = Ed·1000/201 → ties = ⌈Anec/60 mm²⌉

cpi = +0.2 · γG = 0.9 · γQ = 1.5

Positive Ed is net uplift. Each zone is anchored with 40 mm × 2 mm steel ties in S235 (Anec = Ed/fyd).

Σ gk = covering + joists + insulation + boarding + lining

Floor covering 0.08 + Joists 0.059 + Insulation 0.118 + Sec. spaced boarding 0.016 + Ceiling lining 0.117

Σ gk = 0.39 kN/m²

The joists enter as A·5 kN/m³ ÷ spacing; the insulation fills the joist depth at 0.5 kN/m³.

qk = imposed load + moveable partitions

category A1 (Rooms in residential buildings): qk = 1.5 · partitions: 0.5

Σ qk = 2 kN/m²

Ed = 1.0·gk + 1.5·qk → qjoist = Ed·spacing

1.0·0.39 + 1.5·2 = 3.39 kN/m² · × 0.5 m

Ed = 3.39 kN/m² → 1.7 kN/m on one joist

Vd = ½·q·L · Md = q·L²/8

Vd = ½·1.7·4.5 · Md = 1.7·4.5²/8

Vd = 3.81 kN · Md = 4.29 kNm

fmd, fvd from the quality and service class tables

C24 · service class 1 (Moisture ≤ 12 % — indoor.)

fmd = 14.2 N/mm² · fvd = 1.5 N/mm²

Wness = Md/fmd ≤ Wy

4.29·1000/14.2 = 302 ·×10³ mm³ vs 399 ·×10³ mm³

OK — 399 ≥ 302 ×10³ mm³

σvd = 1.5·Vd/A ≤ fvd

1.5·3.81·1000/5852 = 0.978 N/mm² vs fvd = 1.5

OK — 0.978 ≤ 1.5 N/mm²

uinst = 5·q·L⁴/(384·E·Iy) · ufin = uinst·(1 + ψ2·kdef)

frequent: 0.5 kN/m → 4.64 → 5.19 mm · quasi-perm.: 0.4 kN/m → 3.7 → 4.15 mm

governing u = 5.19 mm

E = 11000 N/mm², Iy = 51826·10³ mm⁴, kdef = 0.6 (class 1), ψ2 = 0.2.

limit = L/500 (rounded up)

4.5·1000/500

OK — 9.0 mm ≥ 5.19 mm