BS5400:Pt2
LOADING
Click on the Clause No. for the commentary.
INDEX
CLAUSE No
SUBJECT
Type HA loading (Example)
Problem:
How do you work out the HA loading and bending moment for a bridge deck ?
Example:
| Carriageway = 7.3m wide Deck span = 34m (centre to centre of bearings for a simply supported single span) Design for a metre width of deck : |
|
Clause 3.2.9.3.1. |
Number of notional lanes = 2 Notional lane width = 7.3/2 = 3.65m |
| Clause 6.2.1. | Loaded length = 34m W = 336(1/L)0.67 kN/m (per notional lane) W = 31.6 kN/m (per notional lane) |
| Clause 6.2.2. | Knife Edge Load = 120 kN (per notional lane) |
| Clause 6.4.1.1. | a2 = 0.0137[bL(40-L)+3.65(L-20)] a2 = 0.0137[3.65(40-34.0)+3.65(34.0-20)] = 1.0 Note: For loaded lengths less than 20m the load is proportioned to a standard lane width of 3.65m, i.e. 0.274bL = bL/3.65. For a metre width of deck : W = (31.6 x 1.0)/3.65 = 8.66 kN/m KEL = (120 x 1.0)/3.65 = 32.88 kN |
| Clause 6.2.7. | gfL = 1.50 (Ultimate limit state - combination 1) Design HA loading for a metre width of deck : W = 1.5 x 8.66 = 12.99 kN/m KEL = 1.5 x 32.88 = 49.32 kN Maximum mid span Bending Moment with KEL at mid span = Mult Mult = (12.99 x 342)/8 + (49.32 x 34)/4 Mult = 1877 + 419 = 2300 kNm |
| Note: Use of gf3 BS 5400 Pt.3 & Pt.5 - gf3 is used with the design strength so Mult = 2300 kNm. BS 5400 Pt.4 - gf3 is used with the load effect so Mult = 1.1 x 2300 = 2530 kNm. |
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Last Updated : 18/04/2008
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