Click on the Clause No. for the commentary.

CLAUSE No. | SUBJECT |
---|---|

9.6.4.1.3 | Live Load rating of a plate girder with U-frame restraint. |

*Problem:*

Based on the moment capacity, what is the live load rating of a riveted plate girder with U-frame restraint. ?

*Example:*

Carriageway = 6m wide

Verges = 1m and 0.64m wide

Construction Depth at Verges = 0.9m

Construction Depth at Carriageway = 0.8m

Main Girders - Mid Span Section:

Top and Bottom Flanges: 508 × 38mm plate

Web: 2286 × 9 mm plate

Main Girders - End Section:

Top and Bottom Flanges: 508 × 25mm plate

Web: 2286 × 11 mm plate

Flange Angles: 102 × 102 × 14mm

Cross Girders :

Top Flange: 254 × 19mm plate

Flange: 356 × 12 mm plate

Web: 508 × 9mm plate

Flange Angles: 89 × 89 × 12mm

Web stiffeners are positioned at each cross girder and consist of 12mm gusset plates with 89 × 89 × 12mm angles alternating with 127 × 76 × 11mm T's.

The main girders are bedded on 508 × 1118 × 12mm thick plates on large masonry pad stones.

Records show that the bridge was built in 1925 so, from BD 21/01 Clause 4.3, the matal shall be assumed to be of steel with a characteristic yield strength of 230 N/mm^{2}.

**Dead Load :**

Cl. 4.1. Table 4.1.

Main Girder self weight :

Cross sectional area of plates and angles (from spreadsheet ″Sectionprop5610.xls″)
= 69822mm^{2}/m

Allowing 10% for web stiffeners and gusset plates then:

Self weight = 69822 × 1.1 × 7850 × 9.81 × 10^{-9} = 5.91kN/m

Cross Girder self weight :

Cross sectional area of plates and angles (from spreadsheet ″Sectionprop5610.xls″)
= 21638mm^{2}/m

Allowing 10% for fittings then:

Self weight = 21638 × 1.1 × 7850 × 9.81 × 10^{-9} = 1.83kN/m

Infill concrete :

Radius of jack arch = (635^{2} + 405^{2})/(2 × 405) = 700mm

ψ = 2Sin^{-1}(635/700) = 130°

Area of arch = 0.5 × r^{2}( ψ° / 180° - Sinψ°) = 0.368m^{2}

Area of infill concrete = (1.626 - 0.009)(0.508 + 0.019) - 0.368 = 0.484m^{2}

Weight of infill concrete = 0.484 × 2300 × 9.81 × 10^{-3} = 10.92kN/m

Depth of Road construction above jack arch = (800 - 12 - 508 - 19) = say 260mm

From Note to Table 3.1: surfacing depth = 100mm

Therefore fill depth = 260 - 100 = 160mm

Weight of fill = 0.16 × 2200(miscellaneous fill) × 9.81 × 10^{-3} = 3.45kN/m^{2}

Weight of surfacing = 0.1 × 2300(Asphalt) × 9.81 × 10^{-3} = 2.26 kN/m^{2}

Weight of verges (above fill) = 0.2 × 2300(Asphalt) × 9.81 × 10^{-3} = 4.51 kN/m^{2}

Cl. 3.7. Table 3.1

Applying Partial Load Factors γ_{fL} we get:

Main girder self weight = 1.05 × 5.91 = 6.21 kN/m

Cross girder self weight = 1.05 × 1.83 = 1.92 kN/m

Concrete = 1.15 × 10.92 = 12.56 kN/m

Fill = 1.2 × 3.45 = 4.14 kN/m^{2}

Surfacing = 1.75 × 2.26 = 3.96 kN/m^{2}

Verges = 1.2 × 4.51 = 5.41 kN/m^{2}

The load path for distributed loads on the deck to the main girders is via the cross girders. These loads could be represented more accurately by point load reactions at the end of the cross girders, however it will be sufficiently accurate to represent them as uniformly distributed loads to simplify the calculation.

Reactions from the loads applied as a unifomly distributed load to each main girder:

Self Weight = 6.21 kN/m

Cross Girders = 1.92 × 8.148 / (2 × 1.626) = 4.81 kN/m

Concrete = 12.56 × 8.148 / (2 × 1.626) = 31.47 kN/m

Fill = 4.14 × 8.148 / 2 = 16.85 kN/m

**Sub-Total = 59.34 kN/m**

Uniformly distributed reactions from the surfacing and verge loads to each main girder:

North Main Girder:

Surfacing = 3.96 × 6.0 × 3.894 / 8.148 = 11.36 kN/m

North Verge = 5.41 × 1.0 × 7.394 / 8.148 = 4.91 kN/m

South Verge = 5.41 × 0.64 × 0.574 / 8.148 = 0.24 kN/m

**Sub-Total = 16.51 kN/m**

South Main Girder:

Surfacing = 3.96 × 6.0 × 4.254 / 8.148 = 12.40 kN/m

North Verge = 5.41 × 1.0 × 0.754 / 8.148 = 0.50 kN/m

South Verge = 5.41 × 0.64 × 7.574 / 8.148 = 3.22 kN/m

**Sub-Total = 16.12 kN/m**

Total Assessment Dead Load on each girder:

North Main Girder = 59.34 + 16.51 = **75.85 kN/m**

South Main Girder = 59.34 + 16.12 = **75.46 kN/m**

Cl. 16A Fig. 16.1A

The Effective Span (not to be confused with the effective length
_{e}) of the main girders, for calculating the load effects ** only** , may be determined by calculating the position of the 'Centroid of Pressure' in accordance with the pressure diagram shown in Fig. 16.1A of BD 56/10.

Effective Span (from spreadsheet ″EffectiveSpan.xls″) = 17.725m

Maximum Mid span Dead Load Moment = 75.85 × 17.725

Restraint to the compression flange is only provided by U-frames.

Assume only web stiffeners with gusset plates, at 3.251m centres, are effective in restraining the compression flange.

Note: The 'T' stiffeners will act as web stiffeners but are unlikely to be strong enough to resist U-frame forces F_{R} and F_{c} (calculated below).

Effective width of web plate in stiffener = 16t = 16 × 9 = 144mm

I_{1} = I_{XX}:

Gusset = 12 × 508^{3} / 12 = 131.097×10^{6}mm^{4}

Angles = 2(12 × 187^{3} + 77 × 33^{3}) / 12 = 13.540×10^{6}mm^{4}

Remaining web = 2 × 49 × 9^{3} / 12 = 0.006×10^{6}mm^{4}

I_{1} = (131.097 + 13.540 + 0.006)×10^{6} = 144.643×10^{6}mm^{4}

I_{2} = second moment of area of cross girder (from spreadsheet ″EffectiveSpan.xls″)
= Ixx_{net} = 1.056×10^{9}mm^{4}

d_{1} = 38/2 + 2286 - (12 + 508 + 19) = 1766mm

d_{2} = 1766mm + depth to centroid of cross girder (yc_{net})(from spreadsheet ″EffectiveSpan.xls″)
= 1766 + 251 = 2017mm

u = 0.5 for an outer girder

B = 8148mm

f = 0.5×10^{-10} for riveted cleat joint

E = 205000 N/mm^{2} (BD 21/01 Table 4.2)

δ_{R} = (d_{1}^{3}/3EI_{1}) + (uBd_{2}^{2}/EI_{2}) + fd_{2}^{2}

δ_{R} = [1766^{3} / (3 × 205000 × 144.643 × 10^{6})] + [0.5 × 8148 × 2017^{2} / (205000 × 1.056 × 10^{9})] + 0.5 × 10^{-10} × 2017^{2}

δ_{R} = 6.192×10^{-5} + 7.656×10^{-5} + 20.341×10^{-5} = **34.189×10 ^{-5}**

Cl. 9.6.4.1.1.2

From Clause 9.6.4.1.3:

_{R} = 3251mm

δ_{e} = δ_{e, max} = δ_{R} = 34.189×10^{-5} (all U-frame stiffeners are the same)

a)

_{e} = k_{2}k_{3}k_{5}_{1} but not less than k_{3}_{R} nor greater than L

L = Distance between end U-frames = 19304 - (2 × 1118) = 17068mm

k_{2} = 1.0 (load applied via cross girder to bottom flange)

k_{3} = 1.0 (initial conservative assumption)

I_{c} = 38 × 508^{3} / 12 = 415 × 10^{6}mm^{4}

_{1} = (EI_{c}_{R}δ_{R})^{0.25}

_{1} = (205000 × 415 × 10^{6} × 3251 × 34.189×10^{-5})^{0.25} = 3118mm

*Χ* = _{1}^{3} / (√2EI_{c}δ_{e, max})

*Χ* = 3118^{3} / (√2 × 205000 × 415 × 10^{6} × 34.189×10^{-5}) = 0.737

k_{5} = 2.22 + 0.69 / (*Χ* + 0.5) = 2.22 + 0.69 / (0.737 + 0.5) = 2.78

_{e} = k_{2}k_{3}k_{5}_{1}

_{e} = 1.0 × 1.0 × 2.78 × 3118 = 8668mm

Minimum _{e} = k_{3}_{R} = 1.0 × 3251mm < 8668mm ∴ _{e} = 8668mm (<17068)

b)

_{e} = k_{2}(EI_{c} (δ_{e1} + δ_{e2}) / L)^{1/2}

_{e} = × 1.0 × (205000 × 415 × 10^{6} × 2 × 34.189×10^{-5} / 17725)^{1/2} = 5691mm

_{e} = The greater of (a) and (b) = **8668mm**

Cl. 9.7.1c)

Where intermediate restraints are not fully effective then, in accordance with Cl. 9.6.4.1.1.2 the definition for _{w} (the half-wavelength of buckling) is:

L / _{w} = next integer below L / _{e} but not less than 1.

L = Distance between end U-frames = 19304 - (2 × 1118) = 17068mm

L /_{e} = 17068 / 8668 = 1.97

Hence L / _{w} = Integer(1.97) = 1

_{w} = L = **17068mm**

Cl. 9.7.2

λ_{LT} = _{e} k_{4}η / r_{y}

Section properties (from spreadsheet ″EffectiveSpan.xls″)

Z_{pe} (net) = 58.815×10^{6} mm^{3}

I_{x} (gross) = 74.270×10^{9} mm^{4}

I_{y} (gross) = 0.854×10^{9} mm^{4}

A (gross) = 69822 mm^{2}

r_{y} (gross) = 111 mm

h = 2286 + 38 = 2324 mm

k_{4} = [{4Z^{2}_{pe}(1-I_{y}/I_{x})}/(A^{2}h^{2})]^{1/4}

k_{4} = [{4 × (58.815×10^{6})^{2} × (1-0.854/74.270)} / (69822^{2} × 2324^{2})]^{1/4} = 0.85

Figure 10

A value of η < 1.0 can be obtained from Figure 10b) using Example 4 then:

M_{B} / M_{A} = 0

M_{A} / M_{M} = 0

From Figure 10b) η = 0.94

I_{c} = I_{t} ∴ i = I_{c} / (I_{c} + I_{t}) = 0.5 & ψ_{i} = 0.8(2i-1) = 0

λ_{F} = (_{w} / r_{y})( t_{f} / D) = (17068 / 111)(38 / {2286 + 2×38}) = 2.474

Table 9 Note 3

= [{4i(1-i)+0.05λ_{F}^{2}+ ψ_{i}^{2}}^{0.5}+ψ_{i}]^{-0.5}

= [{4 × 0.5 (1 - 0.5) + 0.05 × 2.474^{2}}^{0.5}]^{-0.5} = 0.935

λ_{LT} = 8668 × 0.85 × 0.94 × 0.935 / 111 = 58.34

Cl. 9.8

The section is Non-Compact (from spreadsheet ″EffectiveSpan.xls″)

Hence M_{ult} = the least of Z_{xc} σ_{yc} or Z_{xt} σ_{yt} (Web and flanges assumed to be of same steel grade ∴ Z_{xw} σ_{yw} will not be critical)

Z_{xc (net)} σ_{yc} = 59.922×10^{6} × 230 × 10^{-6} = 13782 kNm

Z_{xt (net)} σ_{yt} = 50.671×10^{6} × 230 × 10^{-6} = 11654 kNm

∴ M_{ult} = 11654 kNm

M_{pe} = Z_{pe (net)} σ_{yc} = 58.815×10^{6} × 230 × 10^{-6} = 13527 kNm

λ_{LT}[(σ_{yc}/355)(M_{ult}/ M_{pe})]^{0.5}= 58.34[(230/355)(11654/13527)]^{0.5} = 43.59

Annex G G.8

β = λ_{LT}[(σ_{yc}/355)(M_{ult}/ M_{pe})]^{0.5} = 43.59

For Figure 11b): η = 0.0035(β - 30) = 0.0035(43.59 - 30) = 0.048

M_{R}/M_{ult} = 0.5[{1+(1+η)5700/β^{2}}- ({1+(1+η)5700/β^{2}}^{2}-22800/β^{2})^{0.5}]

M_{R}/M_{ult} = 0.5[{1+(1+0.048)(5700/43.59^{2})}-({1+(1+0.048)(5700/43.59^{2})}^{2} - 22800/43.59^{2})^{0.5}] = 0.935

M_{R} = 0.933M_{ult} = 0.935 × 11654 = 10896 kNm

9.9.1.2

M_{D} = M_{R} / γ_{m}γ_{f3}

4.3.3

γ_{m} = 1.05

γ_{f3} = 1.1

M_{D} = 10896 / (1.05 × 1.1) = **9434 kNm**

Cl. 9.6.4.1.3

*NOTE:* The effective length _{e} has been calculated assuming that the U-frames comply with Clause 9.12.3, this will need to be confirmed later.

**BD 21/01**

The live load capacity of the main girder for bending at mid span = M_{D} - Dead Load Moment

Live Load Capacity = 9434 - 2980 = **6454 kNm**

Cl. 5.6

Notional Lane Width b_{L} = 6.0 / 2 = 3.0 m (2 Notional Lanes)

Cl. 5.18

HA UDL and KEL:

W = 336(1/L)^{0.67} = 336(1/17.725)^{0.67} = 49 kN/m (of 3.65m lane width)

KEL = 120 kN

Cl. 5.22

Assume worst conditions for reduction factor:

High Traffic Flows and Poor Road Surfacing (Hp Traffic Condition)

From Figure 5.2 K Factor = 0.91 for 40 tonne level.

Cl. 5.23

Adjustment Factor (AF) = a_{L} / 2.5 = 3.65 / 2.5 = 1.46

Cl. 5.25

Adjusted HA UDL and KEL for a 2.5 metre width of lane:

UDL W = 0.91 × 49 / 1.46 = 30.54 kN/m

KEL = 0.91 × 120 / 1.46 = 74.8 kN

Cl. 5.36(a)

Footway Live Load = 5.0 kN/m^{2}

Cl. 5.37

Footway Live Load for elements supporting footways and carriageway = 0.8 × 5 = 4.0 kN/m^{2}

Cl. 5.26

Adjusted HA UDL and KEL shall be applied to a 2.5m width within the lane for the worst effect:

Proportion the live load UDL's to each main girder as uniformly distributed reactions.

W1 = 4.0 × 1.0 = 4 kN/m

W2 = 30.54 kN/m

W3 = 4.0 × 0.64 = 2.56 kN/m

Cl. 3.7. Table 3.1

Applying Partial Load Factor γ_{fL} we get:

W1 = 1.5 × 4.0 = 6.0 kN/m

W2 = 1.5 × 30.54 = 45.81 kN/m

W3 = 1.5 × 2.56 = 3.84 kN/m

KEL = 1.5 × 74.8 = 112.2 kN

HA UDL and KEL positioned in the notional lane as shown above will produce the worst load effects in the South Girder.

Hence total live load UDL on South Girder:

W1 = 6.0 × 0.754 / 8.148 = 0.56 kN/m

North Lane W2 = 45.81 × 3.004 / 8.148 = 16.89 kN/m

South Lane W2 = 45.81 × 6.004 / 8.148 = 33.76 kN/m

W3 = 3.84 × 7.574 / 8.148 = 3.57 kN/m

Total Live Load UDL = 54.78 kN/m

Similarly total point load from KEL's on South Girder:

KEL = 112.2 × (3.004 + 6.004) / 8.148 = 124.04 kN

Maximum Mid span Live Load Moment:

M = 54.78 × 17.725^{2} / 8 + 124.04 × 17.725 / 4 = **2701 kNm**

Live Load Capacity of Main Girder = 6454 kNm > 2701 kNm Hence mid span section OK for bending effects providing that the U-frames comply with Clause 9.12.3.

Note: If the moment capacity of the girder is insufficient for the dead + live load moment then a suitable K factor (Clause 5.25) will need to be applied to the live live moment. The live load moment of 2701 kNm already includes a K factor of 0.91 so the revised K factor needs to be applied to the full moment of 2701 / 0.91 = 2968 kNm.

Cl. 9.12.3.1

The cross members of each U frame may be assumed to be restrained by the concrete infill against movement in a direction normal to their compression flange.

Cl. 9.12.3.2

*Strength*: U-frames need to be checked for force F_{R} (plus wind load if critical) calculated from Clause 9.12.2. Also a force F_{c}, from Clause 9.12.3.3, resulting from loads on the cross member, needs to be added to F_{R}.

Cl. 9.12.2

F_{R} = [ σ_{fc} / ( σ_{ci} - σ_{fc} )] × _{w} / (667δ_{R}) but not greater than

[ σ_{fc} / ( σ_{ci} - σ_{fc} )] × (n + 1)EI_{c} / (16.7n _{R}^{2})

_{w} = 17068 mm

δ_{R} = 34.189 × 10^{-5}

I_{c} = 415 × 10^{6} mm^{4}

_{R} = 3251 mm

Z_{pe net} = 58.815 ×10^{6} mm^{3}

Z_{xc net} = 59.922 ×10^{6} mm^{3}

λ_{LT} = 58.34

n = 5 (number of restraints in the half wavelength of buckling)

M = M_{Dead} + M_{Live} = 2980 + 2701 = 5681 kNm

S = Z_{pe} / Z_{xc} = 58.815 / 59.922 = 0.982

σ_{ci} = ^{2}ES / λ_{LT}^{2} = ^{2} × 205000 × 0.982 / 58.34^{2} = 584 N/mm^{2}

σ_{fc} = M / Z_{xc} = 5681 / 59.922 = 94.8 N/mm^{2}

F_{R} = [ σ_{fc} / ( σ_{ci} - σ_{fc} )] × _{w} / (667δ_{R}) = [ 94.8 / (584 - 94.8)] × 17068 × 10^{-3} / (667 × 34.189 × 10^{-5}) = 14.5 kN

[ σ_{fc} / ( σ_{ci} - σ_{fc} )] × (n + 1)EI_{c} / (16.7n _{R}^{2}) = [ 94.8 / (584 - 94.8)] × (5 + 1) × 205000 × 415 ×10^{6} × 10^{-3} / (16.7 × 5 × 3251^{2}) = 112 kN

∴ F_{R} = 14.5 kN (< 112 kN)

Cl. 9.12.3.3

F_{c} = θd_{2} / [ 1.5δ_{R} + {_{R}^{3} / (12EI_{c})} ]

θ is the difference in rotation between a cross member and the mean rotations of the cross members on either side of it.

As the dead load and HA UDL will have a similar effect on the rotaion for all cross members then θ for these loads = 0.

The KEL however will be applied to only one cross member so we need to determine θ for the KEL only (average θ of adjacent members = 0).

To calculate θ we can use the Moment Area theorem which states: The change in slope of the deflected shape of a beam between points A and B (A at end and B at mid-span) is equal to the area under the M/EI diagram between those two points.

The bending moment M at the centre of the cross girder with the two KEL in the lanes at W2 shown above:

Reaction_{north} = 112.2(5.144 + 2.144) / 8.148 = 100.4 kN

Mid-span moment M = [100.4 × 4.074] - [112.2 × {1.25 + (4.074 - 3.004)}^{2} / (2.5 × 2)] = 288 kNm

It will be sufficiently accurate to assume a parabolic moment diagram as produced by a UDL.

θ = Area of M/EI diagram between mid-span and end = 2 × 4074 × 288 × ^{6} / (3 × 205000 × 1.056 × 10^{9}) = 3.613 × 10^{-3}

d_{2} = 2017 mm

δ_{R} = 34.189 × 10^{-5}

I_{c} = 415 × 10^{6} mm^{4}

_{R} = 3251 mm

F_{c} = 3.613 × 10^{-3} × 2017 × 10^{-3} / [ 1.5 × 34.189 × 10^{-5} + { 3251^{3} / (12 × 205000 × 415 × 10^{6})} ] = 13.3 kN

Cl. 9.12.3.2

F_{R} + F_{c} = 14.5 + 13.3 = 27.8 kN

Wind loading should be calculated in accordance with Clause 5.3 of BD 37/01 however a value of

6 kN/m^{2} may be used for general cases (see Clause 5.3.1 of BD 37/01).

Wind load on face of girder (per U-frame) = 6 × 2.362 × 3.251 = 46.1 kN

Note: Wind loading is Combination 2 loading which has a γ_{fL} = 1.25 for HA loading instead of 1.5 as used above, so F_{R} and F_{c} may be reduced when combined with wind load. However continue with the calculated values for the example.

Moment in stiffener at top of cross girder = (27.8 × 1.766) + (46.1 × 0.604) = **77 kNm**

Check bending capacity of stiffener:

I_{XX} = 144.643×10^{6}mm^{4}

I_{YY} = 9 × 288^{3}/12 + 24 × 190^{3}/12 + 154 × 36^{3}/12 + 321 × 12^{3}/12 = 32.279 × 10^{6} mm^{4}

Area = 288 × 9 + 499 × 12 + 4(89 + 77) × 12 = 16548 mm^{2}

Cl. 9.7.3.2

34t_{w} /m × √(355/σ_{yw}) = [34 × 12 / 0.5] × [√(355/230)] = 1014 > 508 hence section is compact

Cl. 9.6.4 Table 8

_{e} = 1.4L = 1.4 × 1766 = 2472mm

Cl. 9.7.2

λ_{LT} = _{e} k_{4}η / r_{y}

t_{ f} = 0 ∴ λ_{F} = 0 and = 1.0

k_{4} = 1.0

Figure 10b)

M_{A} = -M and M_{M} = M_{B} = 0:

M_{A} / M_{M} = -∞ and M_{B} / M_{A} = 0 then η = 0.76

r_{y} = √(I_{y}/A) = √ ( 32.279 × 10^{6} / 16548) = 44 mm

λ_{LT} = 2472 × 0.76 / 44 = 43

Cl. 9.8

_{e} / _{w} = 1

M_{ult} = M_{pe} (section compact)

Z_{pe} = 2 × [ (12 × 254^{2} / 2) + (24 × 93.5^{2} / 2) + (154 × 16.5^{2} / 2) + (98 × 16.5^{2} / 2) = 1.053 × 10^{6} mm^{3}

M_{pe} = Z_{pe}σ_{yc} = 1.053 × 10^{6} × 230 × 10^{-6} = 242 kNm

λ_{LT} = 2472 × 0.76 / 44 = 43

λ_{LT}√[(σ_{yc}/355)(M_{ult}/M_{pe})] = 43 × √(230 / 355) = 35

Figure 11

M_{R} / M_{ult} = 0.94

Cl. 9.9.1.2

M_{D} = M_{R} / (γ_{m}γ_{f3} = 0.94 × 242 / (1.05 × 1.1) = **197 kNm**

197 > 77 ∴ the intermediate stiffeners are strong enough to restrain the main girder compression flange by U-frame action.

Cl. 9.12.5.1

The main girders also need to be restrained at their supports.

Note: Where the restraint provided is less than required to resist force F_{S} then the slenderness parameter λ_{LT} must be modified as described in the addition to Clause 6.1 in BD 56/10.

Cl. 9.12.5.2.1

Restraining force F_{S} = F_{S1} + F_{S2} + F_{S3} + F_{S4}

Cl. 9.12.5.2.2

F_{S1} = 0.005 M / [d_{f} {1 - ( σ_{fc} / σ_{ci} )^{2} } ]

M = 5681 kNm

d_{f} = 2286 + 25 = 2311 mm (25mm thick flanges at support)

σ_{fc} = 94.8 N/mm^{2}

σ_{ci} = 584 N/mm^{2}

F_{S1} = 0.005 × 5681 × 10^{3} / [2311 × {1 - ( 94.8 / 584 )^{2} }] = 12.7 kN

Cl. 9.12.5.2.3

F_{S2} = β( Δ_{e1} + Δ_{e2} ) σ_{fc} /({σ_{ci} - σ_{fc}} Σδ )

β = 1.0

Δ_{e1} = Δ_{e2} = D / 200 = (2286 + 50) / 200 = 11.7 mm

Σδ = δ_{t1} + δ_{t2}

δ_{t} = d_{1}^{3} / 3EI_{1} = 1766^{3} / (3 × 205000 × 144.643×10^{6}) = 6.2 × 10^{-5}

Σδ = 2 × 6.2 × 10^{-5} = 12.4 × 10^{-5}

F_{S2} = 1.0 × 2 × 11.7 × 94.8 × 10^{-3}/ ({584 - 94.8} × 12.4 × 10^{-5}) = 36.6 kN

Cl. 9.12.5.2.4

F_{S3} = 0 as the load is applied directly through the bottom flange and bearing plate

Cl. 9.12.5.2.5

F_{S4} = 0 as there is no skew.

Cl. 9.12.5.2.6

F_{L} = d_{2}θ / [2.5δ_{R} + (δ_{e} / 2) + { _{R}^{3} / (3EI_{c})}]

d_{2} = 2017 mm

θ = 3.613 × 10^{-3} radians

δ_{R} = δ_{e} = 34.189 × 10^{-5} mm

_{R} = 3251mm

I_{c} = 415 × 10^{6} mm^{4}

F_{L} = 2017 × 3.613 × 10^{-3} × 10^{-3} / [(2.5 × 34.189 × 10^{-5}) + (34.189 × 10^{-5} / 2) + {3251^{3} / (3 × 205000 × 415 × 10^{6})}] = 6.3 kN

F_{S} + F_{L} = 12.7 + 36.6 + 0 + 0 + 6.3 = 55.6 kN

Add in wind effects as for the internal U-frame, this is over-estimated as the end U-frames are closer together.

Moment in stiffener at top of cross girder = (55.6 × 1.766) + (46.1 × 0.604) = **126 kNm**

Capacity of end stiffener is the same as the internal stiffener = 197 kNm > 126 ∴ end U-frame is adequate.

Having established that the U-frames are adequate then the main girder mid span section is able carry the full range of vehicles up to 40/44 tonnes gross weight as described in BD 21/01.

Note: All sections of the girder where plate sizes change (flange curtailment for example) or where corrosion has reduced the plate thickness, need to be checked for adequacy. Web plates also need to checked for shear effects; joints need to considered to ensure they can transfer loads to the relevant elements.

The live load capacity is determined by the weakest element in the bridge and a comprehensive check of every element needs to be caried out.