Concrete Bridge Design To Bs 5400 Pdf 🔥
BS 5400 provided a rigorous, practical framework for concrete bridge design that remains the reference standard for much of the UK’s bridge stock. Its limit state approach, distinctive load combinations, and conservative crack control rules are still taught and applied in bridge assessment today. For engineers revisiting a BS 5400 design, mastering Part 2 (loads) and Part 4 (concrete) is essential. However, for all new bridge designs, the Eurocodes (BS EN 1992-2) are mandatory in the UK and most of Europe.
Further Reading:
This article is intended as a technical summary. Always consult the original British Standard documents for legal and contractual purposes.
Q1: Can I use BS 5400 for a bridge outside the UK?
Yes, many countries (e.g., Nigeria, Sri Lanka, Hong Kong historically) adopted it. Check with local authorities if still permitted for new designs. concrete bridge design to bs 5400 pdf
Q2: Is there a PDF that includes all amendments?
The final consolidated version was BS 5400-4:1990 + Amendments 1, 2, and 3. The 2000 reprint (with amendments incorporated) is the definitive edition. Look for ISBN 0 580 27922 8.
Q3: Does BS 5400 cover integral bridges (no joints)?
Yes, Part 4 clause 5.9 gives advice on abutment flexibility and soil-structure interaction – well ahead of its time.
Q4: Where can I get a free PDF without violating copyright?
Try the Standards for Highways website (UK) for DMRB BA 44/96, or consult your institutional library. Avoid pirated scanned copies from unknown websites – they often contain missing tables or pages 28–31 (the critical shear table). BS 5400 provided a rigorous, practical framework for
Given:
– Span = 10 m simply supported
– Slab thickness = 500 mm, cover = 50 mm, d = 440 mm
– f_cu = 40 N/mm², f_y = 460 N/mm²
– ULS moment (Combination 1) = 320 kNm/m
Step 1 – Flexural reinforcement
[
K = M / (b d^2 f_cu) = 320 \times 10^6 / (1000 \times 440^2 \times 40) = 0.041
]
Since K < 0.156 → singly reinforced.
[
z = d [0.5 + \sqrt0.25 – K/0.9] = 440 [0.5 + \sqrt0.25 – 0.041/0.9] = 418 \text mm
]
[
A_s = M / (0.87 f_y z) = 320 \times 10^6 / (0.87 \times 460 \times 418) = 1916 \text mm^2/m
]
Provide T20 @ 150 mm c/c (2094 mm²/m).
Step 2 – Shear check
Maximum shear at support (V) = 180 kN/m.
[
V_c = 0.27/1.25 \times (100 \times 2094 / (1000 \times 440))^1/3 \times 40^1/3 \times 1000 \times 440 / 1000
]
≈ 112 kN < 180 kN → Provide links: A_sv/s_v = (180 – 112) / (0.87 × 460 × 0.440) = 0.42 mm²/mm → Use T10 @ 150 mm (0.52 mm²/mm). This article is intended as a technical summary
If you cannot obtain the official standard, these books summarize BS 5400 concrete design:
Flexure: ( K = 2100\times10^6 / (1000\times600^2\times40) = 0.146 ) ( z = 600[0.5 + \sqrt0.25 - 0.146/0.9] = 456 , mm ) (≤ 0.95d = 570 mm) ( A_s = 2100\times10^6 / (0.87\times500\times456) = 10,580 , mm²/m ) Use 32 mm dia @ 75 mm c/c (As = 10,720 mm²/m)
Shear check: Shear at support = 800 kN/m → v = 800×10³/(1000×600) = 1.33 N/mm² v_c ≈ 0.72 N/mm² → provide links: A_sv/s_v = 1000(1.33-0.72)/(0.87×500) = 1.40 mm²/mm Use 12 mm links @ 150 mm c/c (A_sv/s_v = 1.50)
Deflection: Service moment = 1200 kNm/m → short-term modulus E_c = 30 kN/mm² → deflection ≈ span/800 < span/250 → OK.
