Learn · worked example

RC beam design to SANS 10100

A simply supported reinforced concrete beam from load to bars: bending, shear links, minimum steel, bar fit, span/depth, crack width and long-term deflection. The numbers come from the free STATIX RC beam calculator; the ultimate-limit-state steps were checked by hand.

1. The problem

Input	Value
Span	6.0 m, simply supported
Design load (ultimate, already factored)	w = 20 kN/m over the full span
Section	300 mm wide × 550 mm deep
Materials	f_cu = 30 MPa, f_y = 450 MPa
Cover to links	40 mm, links 10 mm

Beam loading, shear force, bending moment and deflected shape diagrams from the STATIX RC beam calculator — The free RC beam calculator with these inputs: loading, shear force (60 kN), bending moment (90 kNm) and deflected shape.

2. Design forces

M = wL²/8 = 20 × 6.0² / 8 = 90 kNm V = wL/2 = 20 × 6.0 / 2 = 60 kN

3. Bending (SANS 10100 cl. 4.3.3)

d = 550 − 40 − 10 − 20/2 = 490 mm (a 20 mm bar is assumed until the bars are chosen) K = M / (f_cu b d²) = 90×10⁶ / (30 × 300 × 490²) = 0.042 ≤ K′ = 0.156 → no compression steel z = d [0.5 + √(0.25 − K/0.9)] = 466 mm, limited to 0.95d = 465.5 mm A_s,req = M / (0.87 f_y z) = 90×10⁶ / (0.87 × 450 × 465.5) = 494 mm² Provide 5Y12 = 565 mm² OK

With 12 mm bars the real effective depth is 494 mm, a little more than the 490 mm assumed, so the check is on the safe side. The calculator's design result uses d = 494 mm: K = 0.041 and A_s,req = 490 mm². The 5Y12 then give a moment resistance of 103.3 kNm, so the beam works at 87% in bending (90 / 103.3).

4. Shear (SANS 10100 cl. 4.3.4)

v = V / (b d) = 60×10³ / (300 × 490) = 0.41 MPa 100A_s/(bd) = 100 × 565 / (300 × 490) = 0.38 v_c = (0.75/1.4) × (30/25)^1/3 × (0.38)^1/3 × (400/490)^1/4 = 0.39 MPa v ≤ v_c + 0.4 → nominal links are enough Nominal links: A_sv/s_v ≥ 0.4b / (0.87 f_yv); with 2 legs of Y10 (157 mm²): s_v ≤ 512 mm Maximum spacing 0.75d = 368 mm → provide Y10 @ 300 mm OK v_max = min(0.75√f_cu, 4.75) = 4.1 MPa ≥ 0.41 MPa OK

The calculator spaces the links a little wider. With d = 494 mm the limit is 0.75d = 370 mm, and it provides Y10 @ 350 mm, 18 links over the span. It also adds 2Y12 top bars as hangers, so each link has a bar to pass round in its top corners (cl. 4.11.6.4). Both spacings meet the nominal link rules; 300 mm is simply a rounder figure for site.

5. Detailing

A_s,min = 0.13% × b × h = 0.0013 × 300 × 550 = 214 mm² ≤ 565 mm² OK A_s,max = 4% × b × h = 6,600 mm² OK Five 12 mm bars need 260 mm of the 300 mm width OK

6. Serviceability

Span/depth (cl. 4.3.6), with d = 494 mm: Basic ratio 16 (Table 10, truly simply supported beam) Tension steel (Table 11): f_s = 0.87 f_y × (A_s,req/A_s,prov) / 1.2 = 391.5 × (490/565) / 1.2 = 283 MPa M/(bd²) = 90×10⁶ / (300 × 494²) = 1.23 factor = 0.55 + (477 − 283) / [120 × (0.9 + 1.23)] = 1.31 Compression steel (Table 12), the 2Y12 hangers: 100A_s′/(bd) = 100 × 226 / (300 × 494) = 0.15 → factor 1.05 L/d = 6,000 / 494 = 12.1 ≤ 16 × 1.31 × 1.05 = 22.0 OK

In the calculator this load is entered as a dead load G_k = 16.67 kN/m with no live load (1.2 × 16.67 = 20 kN/m), so for the service stress it takes the service load as G_k = 16.67 kN/m, the ultimate load divided by 1.2, which is on the safe side. If the ends of the beam are nominally restrained, Table 10 allows a basic ratio of 20 instead of 16 (the calculator has a setting for this). The beam passes either way.

The calculator also checks crack control by the bar spacing rules and works out the long-term deflection with the BS 8110-2 method, using the service load G_k = 16.67 kN/m (service moment 75.0 kNm), creep factor 2.0 and shrinkage strain 300 × 10⁻⁶:

Check	Result
Crack control: bar spacing	passes (86 % of the allowed spacing)
Long-term deflection	20.3 mm ≤ L/250 = 24 mm

Taking all of the load as service load is on the safe side. For a real design, work out the quasi-permanent load (permanent load plus ψ₂ times imposed load) and use that.

7. Result

Item	Result
Bottom steel	5Y12 (565 mm²), one layer; 87% in bending
Top steel	2Y12 hangers for the links
Links	Y10 @ 300 mm by hand, Y10 @ 350 mm in the calculator (both nominal, ≤ 0.75d)
Span/depth	12.1 ≤ 22.0
Reactions	60 kN at each support; total 120 kN = applied load

Run this example yourselfIn the STATIX RC beam calculator, enter the span, section and materials above, then dead load G_k = 16.67 kN/m and live load Q_k = 0: it gives M = 90 kNm, 5Y12 and Y10 links at 87 % in bending. It also does continuous beams, point loads, T-sections and torsion, with a calc sheet.
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A worked example for learning, not a design. Check every value against the current code and your own project, and have the design checked and signed by a registered professional engineer.

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