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BD 31/01

DESIGN OF BURIED CONCRETE BOX STRUCTURES

Click on the Clause No. for the commentary.
 

INDEX
 

CLAUSE No

SUBJECT

Loads due to temperature.

Reinforcement Details at Corners.

 

Problem:

What are the maximum and minimum effective temperatures for cover depths between 0.2m and 0.6m as they are not included in BD 37?

Solution:

BD 31 makes no distinction for temperatures on structures with cover depths between 0.2m and 0.6m. BD 91 clauses 3.3.4 to 3.3.6 make allowances for different fill depths over the structure which resolves the anomoly in BD 31.
Background information to the development of the National Annex to BS EN 1991-1-5: 2003 says that there is very limited research on the temperature in buried structures, and no evidence to justify the difference in approach to the treatment of thermal actions in BD31/01 and BD91/04.
Both Standards were reviewed in drafting the NA and the approach given in BD91/04 was adopted as there was little physical justification for the way BD31/01 treated temperature differences.
It would therefore be reasonable to adopt the BD 91 solution if the values obtained from BD 31 were too onerous.

 

Problem:

How to determine the ultimate moment of resistance of the corner reinforcement for opening moments.

Solution:

A method for the design is given in Appendix D.
Example:

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Type 2 corner detail shown. Refer to Figure D/2 of BD 31/01 for Notations.
Assume:
fcu = 40 N/mm2
Main bar reinforcement = B16@125 c/c (As = 1608mm2/m)

BD 31/01 cl. D.3

Fillet bar reinforcement = B12@125 c/c (As = 905mm2/m > 1608/2 = 804mm2/m)
Cover to reinforcement = 60mm throughout.
then:

BS 5400 Pt 4 cl. 5.8.6.9

Maximum allowable bearing stress = 1.5fcu/(1+2Φ/ab) = 1.5×40/(1+2×16/125) = 47.8 N/mm2
Allowable tensile force due to ultimate loads in a bend in a bar = Fbt = rΦ×Bearing stress
r = 300/2-(60+16) = 74
Fbt = 74×16×47.8×10-3 = 56.6kN

Table 14

For type 2 deformed bars in tension allowable bond stress fb = 4.0 N/mm2

BD 31/01 cl. D.4

Consider maximum allowable tension in main bars at C
Lst = 300/2 - 60 - 16/2 = 82mm
Maximum allowable tension = fbπΦLst + Fbt = 4×π×16×82×10-3+56.6 = 73.1 kN
Stress due to this force = 73.1×103/82π = 363.6N/mm2

BS 5400 Pt 4 cl. 4.3.3.3

γms = 1.15
fyms = 500/1.15 = 434.8N/mm2 > 363.6 hence use 363.6N/mm2
Hence maximum allowable tensile forces Fsh and Fsv = 363.6×82×π×10-3 = 73.1 kN

BD 31/01 Fig D/2

df = 300/√2 + 200/√2 - 60 - 6 = 288mm

BS 5400 Pt 4 cl. 5.3.2.3

Laf = [1-(1.1fyAs)/(fcubd)]d = [1-(1.1×500×905)/(40×1000×288)]d = 0.957d > 0.95d
Hence Laf = 0.95d = 0.95×288 = 274mm
La = [300 -2(60+8)]/√2 - (288-274) = 116 - 14 = 102mm

BD 31/01 Fig D/2

Design strength of fillet bars = Fsf = fyAsm = 500×905×10-3/1.15 = 393.5kN
Mr = FsfLaf + (Fsh+Fsv)La/√2 = 393.5×274×10-3 + 2×73.1×102×10-3/√2 = 118kNm

 

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Last Updated : 14/07/2017
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