Click on the Clause No. for the commentary.

*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.

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*Problem:*

How to determine the torsional moment from the earth pressues due to skew.

*Solution:*

This is not a definitive solution, but has been proposed to promote discussion.

Example:

2.5 × 2.0m box with 0.3m thick concrete walls

Joints at 10m centres with 45° skew

0.2m road construction (γ = 24kN/m^{3})

0.8m fill (γ = 20kN/m^{3})

2.0m backfill (γ = 19kN/m^{3})

Design angle of base friction = 20°

Use BD 31/01 Diagram A/6a loading combination.

cl.4.1.2(d)(ii)

(Excluding live load surcharge)

Minimum superimposed dead load on roof:

Surfacing = 0.2 × 24 = 4.8kN/m^{2}

Fill = 0.8 × 20 = 16kN/m^{2}

Total = 20.8kN/m^{2}

(γ_{fL} = γ_{f3} = 1.0)

Self weight of box = [(2.5 × 2.0) - (1.9 × 1.4)] × 25 = 58.5kN/m

Active earth pressures on box (γ_{fL} = 1.5 γ_{f3} = 1.1):

Pressure at base due to backfill = 1.1 × 1.5 × K × 19 × 2.0 = 62.7K kN/m^{2}

Active Force F1 = 62.7K × 2.0 / 2 = 62.7K kN/m

Pressure at base due to surcharge = 1.1 × 1.5 × K × 20.8 = 34.3K kN/m^{2}

Active Force F2 = 34.3K × 2.0 = 68.6K kN/m

cl.3.2.1(a)(ii)

Fill depth > 0.6m hence use 30 units of HB

Wheel load = 30 × 10 / 4 = 75kN

BD37/01 cl.6.10.2

Traction Force = 25% × 30 × 10 × 4 = 300kN

BD31/01 cl.3.2.7(e)

K_{t} = (L_{L} - H)/(L_{L} - 0.6) = (2.5 - 1.0)/(2.5 - 0.6) = 0.789

(γ_{fL} = 1.1 γ_{f3} = 1.1)
Traction Force applied to the roof = 1.1 × 1.1 × 0.789 × 300 = 286.4kN

BD31/01 cl.3.2.7(g)

For plan rotation effects distribute the traction force along L_{j}

Assume a worst case that the full traction force is applied to the roof but only the wheels over the structure are contributing to the frictional resistance.

From Diagram A/6a for active pressures K = 0.33

Total active force P_{active} = 286.4 + F1 + F2 = 286.4 + (62.7 + 68.6) × 0.33 × 10.0 = 719.7kN

Offset between P_{active} and P_{passive} = 2.5m hence applied torque = 719.7 × 2.5 = 1799kNm

Friction on the base of the structure will resist the torque created by the offset between the Passive and Active forces represented in the sketch by P_{passive} and P_{active}

Total vertical force on base (γ_{fL} = γ_{f3} = 1.0) = (20.8 × 2.5 + 58.5) × 10 + 6 × 75 = 1555kN

Assume rotation at mid length of structure then frictional torque resistance is provided by two halves of structure at 5.0 / √2m

Hence frictional torque = 0.5 × 1555 × tan20° × 3.535 = 1000kNm

Net torque to be resisted by passive pressure = 1799 - 1000 = 799kNm

cl.4.1.2(d)(ii)

L_{L}tanθ = 2.5 × tan45° = 2.5m ≥ L_{j}/6 = 10/6 = 1.67 hence increase horizontal earth pressure to resist unbalanced torque.

Passive earth pressures on box (γ_{fL} = γ_{f3} = 1.0):

Pressure at base due to backfill = K × 19 × 2.0 = 38K kN/m^{2}

Passive Force F3 = 38K × 2.0 / 2 = 38K kN/m

Pressure at base due to surcharge = K × 20.8 = 20.8K kN/m^{2}

Passive Force F4 = 20.8K × 2.0 = 41.6K kN/m

F3 + F4 = 79.6K kN/m

Using the pressure diagram for an eccentrically loaded slab then the maximum pressure = P/A + M/Z

P/A = 719.7 / 2.0 / 10.0 = 36kN/m^{2}

M/Z = 799 / 2.0 / 10.0^{2} = 4kN/m^{2}

Maximum Pressure = 36 + 4 = 40kN/m^{2}

Maximum Force = 40 × 2.0 = 80kN/m

Hence maximum K required = 80 / 79.6 = 1.01 < 0.5K_{p} = 1.5 ∴ OK

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*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:

Type 2 corner detail shown. Refer to Figure D/2 of BD 31/01 for Notations.

Assume:

f_{cu} = 40 N/mm^{2}

Main bar reinforcement = B16@125 c/c (A_{s} = 1608mm^{2}/m)

BD 31/01 cl. D.3

Fillet bar reinforcement = B12@125 c/c (A_{s} = 905mm^{2}/m > 1608/2 = 804mm^{2}/m)

Cover to reinforcement = 60mm throughout.

then:

BS 5400 Pt 4 cl. 5.8.6.9

Maximum allowable bearing stress = 1.5f_{cu}/(1+2Φ/a_{b}) = 1.5×40/(1+2×16/125) = 47.8 N/mm^{2}

Allowable tensile force due to ultimate loads in a bend in a bar = F_{bt} = rΦ×Bearing stress

r = 300/2-(60+16) = 74

F_{bt} = 74×16×47.8×10^{-3} = 56.6kN

Table 14

For type 2 deformed bars in tension allowable bond stress f_{b} = 4.0 N/mm^{2}

BD 31/01 cl. D.4

Consider maximum allowable tension in main bars at C

L_{st} = 300/2 - 60 - 16/2 = 82mm

Maximum allowable tension = f_{b}πΦL_{st} + F_{bt} = 4×π×16×82×10^{-3}+56.6 = 73.1 kN

Stress due to this force = 73.1×10^{3}/8^{2}π = 363.6N/mm^{2}

BS 5400 Pt 4 cl. 4.3.3.3

γ_{ms} = 1.15

f_{y}/γ_{ms} = 500/1.15 = 434.8N/mm^{2} > 363.6 hence use 363.6N/mm^{2}

Hence maximum allowable tensile forces F_{sh} and F_{sv} = 363.6×8^{2}×π×10^{-3} = 73.1 kN

BD 31/01 Fig D/2

d_{f} = 300/√2 + 200/√2 - 60 - 6 = 288mm

BS 5400 Pt 4 cl. 5.3.2.3

L_{af} = [1-(1.1f_{y}A_{s})/(f_{cu}bd)]d = [1-(1.1×500×905)/(40×1000×288)]d = 0.957d > 0.95d

Hence L_{af} = 0.95d = 0.95×288 = 274mm

L_{a} = [300 -2(60+8)]/√2 - (288-274) = 116 - 14 = 102mm

BD 31/01 Fig D/2

Design strength of fillet bars = F_{sf} = f_{y}A_{s}/γ_{m} = 500×905×10^{-3}/1.15 = 393.5kN

M_{r} = F_{sf}L_{af} + (F_{sh}+F_{sv})L_{a}/√2 = 393.5×274×10^{-3} + 2×73.1×102×10^{-3}/√2 = **118kNm**