The Problem

A 6 m simply supported internal beam in an office building, at 5 m centres, carries g_k = 35 kN/m (slab, finishes, self-weight) and q_k = 20 kN/m (office + movable partitions). Concrete C30/37, steel B500B. A compact bending pre-step gives 4H25 as the main steel. The task: design the vertical links, and walk through every EC2 shear check — V_Rd,c, V_Rd,max, required A_sw/s, minimum reinforcement and spacing limits — with each intermediate value shown.

The Relevant Clauses

  • §6.2.1(8): shear may be checked at distance d from the support face
  • §6.2.2(1), Eq (6.2a/b): V_Rd,c for members without shear reinforcement
  • §6.2.2(6), Eq (6.5)/(6.6N): cap on V_Ed, strength reduction factor ν
  • §6.2.3(2)/(3), Eq (6.7N)/(6.8)/(6.9): truss model — V_Rd,s, V_Rd,max, θ limits
  • §6.2.3(3), Eq (6.8): required link area for vertical links
  • §9.2.2(5)–(8), Eq (9.4), (9.5N), (9.6N), (9.8N): minimum links and spacing limits
  • Recommended values used throughout — check your National Annex

Input Data

Parameter Value Unit Notes
Span L 6000 mm Simply supported
Tributary width 5000 mm 5 m grid
b_w 300 mm Beam width
h 600 mm Overall depth
g_k 35 kN/m Permanent, incl. self-weight
q_k 20 kN/m Office + movable partitions
f_ck 30 MPa C30/37
f_yk 500 MPa B500B
c_nom 35 mm XC1, structural class S4
Main bars 4H25 — From bending pre-step (Step 3)
Links H8, two legs — To be sized

Step-by-Step Calculation

Step 1: Design Loads and Shear Force

Load combination per EN 1990 Eq (6.10), persistent design situation:

w_d = 1.35 × 35 + 1.5 × 20 = 77.25 kN/m

Shear at the support centreline: V = 77.25 × 6.0 / 2 = 231.8 kN.

Effective depth: d = 600 − 35 − 8 − 25/2 = 544.5 mm.

Because the beam carries a predominantly uniformly distributed load, §6.2.1(8) permits checking shear at distance d from the support face. With a 300 mm support width:

V_Ed = 231.8 − 77.25 × (0.150 + 0.5445) = 178.1 kN

If we skipped this rule and checked at the face (220 kN), we would be over-designing the links by about 25%. The reduction is a real clause, not a convenience.

Step 2: Material Properties

f_cd = 1.0 × 30 / 1.5 = 20.0 MPa (§3.1.6(1)P NOTE, α_cc = 1.0 — the recommended value and the French NA value; the UK NA adopts 0.85)

f_yd = f_ywd = 500 / 1.15 = 435 MPa (§3.2.7(2), §6.2.3(3))

f_ctm = 2.9 MPa (Table 3.1, C30/37)

Step 3: Effective Depth and Flexural Steel (Compact Pre-Step)

The V_Rd,c formula needs the longitudinal tension ratio ρ_l, so the bending design is a prerequisite. Here it is in compact form (it gets its own full post):

K = M_Ed / (b d² f_ck) = 347.6×10⁶ / (300 × 544.5² × 30) = 0.130 ≤ 0.167 — no compression steel needed

z = d × [0.5 + √(0.25 − K/1.134)] = 472 mm

A_s = M_Ed / (f_yd × z) = 347.6×10⁶ / (435 × 472) = 1692 mm² → provide 4H25 (1963 mm²)

Checks: A_s,min = 0.26 × 2.9 × 300 × 544.5 / 500 = 246 mm² ✓ (governs over the 0.0013·b·d = 212 mm² floor, §9.2.1.1(1)); A_s,max = 0.04 × 300 × 600 = 7200 mm² ✓ (§9.2.1.1(3))

ρ_l = 1963 / (300 × 544.5) = 1.20% ≤ 2% ✓

Step 4: Can the Concrete Work Alone? V_Rd,c

k = 1 + √(200/544.5) = 1.61 ≤ 2.0

V_Rd,c = [C_Rd,c × k × (100 × ρ_l × f_ck)^(1/3) + k₁ × σ_cp] × b_w × d (Eq 6.2a)

With C_Rd,c = 0.18/1.5 = 0.12, σ_cp = 0 (no axial force):

V_Rd,c = [0.12 × 1.61 × (100 × 0.0120 × 30)^(1/3)] × 300 × 544.5 = 104.0 kN

Minimum (Eq 6.2b, with v_min from Eq 6.3N): v_min = 0.035 × 1.61^1.5 × √30 = 0.390 MPa → 63.7 kN

V_Rd,c = 104.0 kN < V_Ed = 178.1 kN → shear links are required.

For completeness, the absolute cap on V_Ed from §6.2.2(6): 0.5 × b_w × d × ν × f_cd = 0.5 × 300 × 544.5 × 0.528 × 20.0 = 862 kN ✓

Step 5: Strut Check — V_Rd,max

EC2 truss model — concrete struts at θ = 21.8° fanning from the supports, vertical links, compression chord on top, tension chord at the bottom

The truss model we are working with. Note the struts: cot θ = 2.5 means each one runs 2.5 times further horizontally than the lever arm z — genuinely flat, which is why the links do most of the work.

Strength reduction factor (Eq 6.6N): ν = 0.6 × (1 − 30/250) = 0.528

Here is a trap worth knowing: ν₁ may be taken as 0.6 for f_ck ≤ 60 MPa only when the design stress of the links is below 0.8 f_yk (Note 2 to §6.2.3(3)). With f_ywd = 435 MPa that condition does not hold, so we use ν₁ = ν = 0.528. Taking 0.6 would overestimate V_Rd,max by 14%.

z = 0.9d = 490 mm (§6.2.3(1))

Calculate the strut angle from the demand by setting V_Ed equal to V_Rd,max (Eq 6.9) — the angle at which the concrete strut would just reach capacity:

sin 2θ = 2 × V_Ed / (α_cw × b_w × z × ν₁ × f_cd) = 2 × 178.1×10³ / (1.0 × 300 × 490 × 0.528 × 20.0) = 0.229 → θ = 6.6°

That is below the lower limit of Eq 6.7N (21.8°), so the code minimum governs: θ = 21.8°, cot θ = 2.5. The demand angle being far below the limit shows the section is over-strength in shear; the code minimum, not the demand, fixes the angle.

V_Rd,max = α_cw × b_w × z × ν₁ × f_cd / (cot θ + tan θ) (Eq 6.9)

= 1.0 × 300 × 490 × 0.528 × 20.0 / (2.5 + 0.4) = 535.3 kN > 178.1 kN ✓ (ratio 0.33)

Even at θ = 45° the ceiling is 776.2 kN — the section is comfortably sized for shear. §6.2.1(8) also requires the shear at the support to be checked against the strut capacity: V = 231.8 kN < 535.3 kN ✓.

A_sw/s = V_Ed / (z × f_ywd × cot θ) (Eq 6.8, rearranged)

= 178.1×10³ / (490 × 435 × 2.5) = 0.334 mm²/mm = 334 mm²/m

H8 two-leg links: A_sw = 2 × 50.3 = 100.5 mm² → s = 100.5 / 0.334 = 301 mm.

Try H8@300: V_Rd,s = 100.5/300 × 490 × 435 × 2.5 = 178.5 kN (Eq 6.8) — passes at a ratio of 0.998.

Technically that works. I would never detail it that tight. Rounding to H8@250 gives V_Rd,s = 214.2 kN (ratio 0.83) for about 0.5 kg of extra steel per metre of beam. When the ratio on a hand calc is 0.998, the field tolerance of your bar bending has already eaten your margin.

Minimum: ρ_w,min = 0.08 × √f_ck / f_yk = 0.000876 (Eq 9.5N) → A_sw,min/s = 0.000876 × 300 = 263 mm²/m < 402 mm²/m provided ✓

Longitudinal spacing: s_l,max = 0.75 × d = 408 mm > 250 mm ✓ (Eq 9.6N, α = 90°)

Transverse spacing: s_t,max = 0.75 × d = 408 mm ≤ 600 mm ✓ (Eq 9.8N)

Not covered here, but required in the full design: anchorage of the bottom bars at the end support (§9.2.1.4). Treat this post as the shear slice of a complete beam design, not the whole job.

Result Summary

Check Required Provided Ratio Pass?
V_Rd,s ≥ V_Ed (Eq 6.8) 178.1 kN 214.2 kN 0.83 ✅
V_Rd,max ≥ V_Ed (Eq 6.9) 178.1 kN 535.3 kN 0.33 ✅
V_support ≤ V_Rd,max (§6.2.1(8)) 231.8 kN 535.3 kN 0.43 ✅
A_sw/s (Eq 6.8) 334 mm²/m 402 mm²/m 0.83 ✅
A_sw,min/s (Eq 9.5N) 263 mm²/m 402 mm²/m 0.65 ✅
s ≤ s_l,max (Eq 9.6N) 408 mm 250 mm 0.61 ✅

Design: H8 links at 250 mm centres, two legs.

Download

📥 Download the full calculation spreadsheet

Four sheets: Inputs, Steps (every formula with its clause), Summary, and Notes. The file regenerates from a Python script, so the numbers in this post and the spreadsheet cannot drift apart. Fully free — no gate, no signup.

If you spot a mistake or want a variation of this example, write to me at hugo@civion.cloud — corrections to published calculations go straight to the top of my list.

What I’m Building

I’m building a tool that runs this exact sequence — section in, it checks V_Rd,c, V_Rd,max, sizes the links, and flags the traps before you hit them: the ν₁ shortcut, the NA differences, the 0.998 near-miss. Twenty minutes of arithmetic becomes twenty seconds. If that sounds useful, write to me at hugo@civion.cloud and I’ll tell you when it’s ready.

Verified against EN 1992-1-1 §6.2 by the author.