bs 5400 part 2 : 2006 clause 6.2 type ha loading problem ... · pdf fileclause 6.2 type ha...

BS 5400 Part 2 : 2006

Clause 6.2 Type HA Loading Problem: How do you work out the HA loading and bending moment for a bridge deck ?

Example: Carriageway = 6m wide

Deck span = 34m (centre to centre of bearings for a simply supported single span)

Design for a metre width of deck : Cl 3.2.9.3.1.

Number of notional lanes = 2

Notional lane width = 6.0/2 = 3.0m Cl 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) Cl 6.2.2.

Knife Edge Load = 120 kN (per notional lane) Cl 6.4.1.1.

α2 = 0.0137[bL(40-L)+3.65(L-20)]

α2 = 0.0137[3.0(40-34.0)+3.65(34.0-20)] = 0.947

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 0.947)/3.0 = 10.0 kN/m

KEL = (120 x 0.947)/3.0 = 37.88 kN

Maximum mid span Bending Moment with KEL at mid span:

M = (10.0 x 342)/8 + (37.88 x 34)/4 = 1767 kNm Cl 6.2.7.

γfL = 1.20 (Serviceability limit state - combination 1)

γfL = 1.50 (Ultimate limit state - combination 1)

Design HA moment for a metre width of deck :

Msls = 1767 x 1.2 = 2120 kNm

Mult = 1767 x 1.5 = 2650 kNm

Note: Use of γf3

BS 5400 Pt.3 & Pt.5 - γf3 is used with the design strength so Mult = 2650 kNm.

BS 5400 Pt.4 - γf3 is used with the load effect so Mult = 1.1 x 2650 = 2915 kNm.

Clause 6.3 Type HB Loading

Assume the road over the bridge is not a Principal Road then we need to check for 30 units

type HB loading (see BD 37/01 Chapter 4).

Cl 6.3.1

Nominal load per axle = 30units x 10kN = 300kN

The maximum bending moment will be achieved by using the shortest HB vehicle i.e. with

6m spacing (see BS 5400-2:2006 Fig 12).

The maximum moment for a simply supported span occurs under the inner axle when the

vehicle is positioned such that the mid span bisects the distance between the centroid of

the load and the nearest axle.

With a 34m span and the 6m HB vehicle with equal axle loads, the inner axle is placed at

1.5m from the mid span.

RL = 300(10.7+12.5+18.5+20.3)/34 = 547 kN

RR = 4x300-547 = 653kN

Moment at X = 547x15.5 - 300x1.8 = 7939kNm Cl 6.4.2

The HB vehicle occupies one lane with HA load in the adjacent lane. Assume for the

example that the HB load is carried by a standard lane width of 3.65m.

Hence the moment per metre width of deck = 7939/3.65 = 2175kNm

Cl 6.3.4.



Design HB moment for a metre width of deck :

Msls = 1.1 x 2175 = 2393 kN/m (compared to 2120 for HA load)

Mult = 1.3 x 2175 = 2828 kN/m (compared to 2650 for HA load)

Hence in this case HB load effects would govern although a grillage or finite element type

distribution would reduce the HB moment considerably.

3/7/2016

1

3/7/2016

2

3/7/2016

3

3/7/2016

4

3/7/2016

5

3/7/2016

6

3/7/2016

7

3/7/2016

8

3/7/2016

9

-

3/7/2016

10

3/7/2016

11

3/7/2016

12

3/7/2016

13

3/7/2016

14

3/7/2016

15

3/7/2016

16

3/7/2016

17

3/7/2016

18

3/7/2016

19

3/7/2016

20

3/7/2016

21

3/7/2016

22

3/7/2016

23

3/7/2016

24

3/7/2016

25

BS 5400 Part 2 : 2006

Clause 6.2 Type HA Loading Problem: How do you work out the HA loading and bending moment for a bridge deck ?

Example: Carriageway = 6m wide


Design for a metre width of deck : Cl 3.2.9.3.1.


Notional lane width = 6.0/2 = 3.0m Cl 6.2.1.

Loaded length = 34m


W = 31.6 kN/m (per notional lane) Cl 6.2.2.

Knife Edge Load = 120 kN (per notional lane) Cl 6.4.1.1.

α2 = 0.0137[bL(40-L)+3.65(L-20)]

α2 = 0.0137[3.0(40-34.0)+3.65(34.0-20)] = 0.947

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.


W = (31.6 x 0.947)/3.0 = 10.0 kN/m

KEL = (120 x 0.947)/3.0 = 37.88 kN

Maximum mid span Bending Moment with KEL at mid span:

M = (10.0 x 342)/8 + (37.88 x 34)/4 = 1767 kNm Cl 6.2.7.



Design HA moment for a metre width of deck :

Msls = 1767 x 1.2 = 2120 kNm

Mult = 1767 x 1.5 = 2650 kNm

Note: Use of γf3



Clause 6.3 Type HB Loading

Assume the road over the bridge is not a Principal Road then we need to check for 30 units

type HB loading (see BD 37/01 Chapter 4).

Cl 6.3.1

Nominal load per axle = 30units x 10kN = 300kN

The maximum bending moment will be achieved by using the shortest HB vehicle i.e. with

6m spacing (see BS 5400-2:2006 Fig 12).

The maximum moment for a simply supported span occurs under the inner axle when the

vehicle is positioned such that the mid span bisects the distance between the centroid of

the load and the nearest axle.

With a 34m span and the 6m HB vehicle with equal axle loads, the inner axle is placed at

1.5m from the mid span.

RL = 300(10.7+12.5+18.5+20.3)/34 = 547 kN

RR = 4x300-547 = 653kN

Moment at X = 547x15.5 - 300x1.8 = 7939kNm Cl 6.4.2

The HB vehicle occupies one lane with HA load in the adjacent lane. Assume for the

example that the HB load is carried by a standard lane width of 3.65m.

Hence the moment per metre width of deck = 7939/3.65 = 2175kNm

Cl 6.3.4.



Design HB moment for a metre width of deck :

Msls = 1.1 x 2175 = 2393 kN/m (compared to 2120 for HA load)

Mult = 1.3 x 2175 = 2828 kN/m (compared to 2650 for HA load)

Hence in this case HB load effects would govern although a grillage or finite element type

distribution would reduce the HB moment considerably.

Clause 6.2

Problem:

How do you work out the HA loading and bending moment for a bridge deck ?

Example:

Carriageway = 7.3m wide


Design for a metre width of deck : Cl.3.2.9.3.1


Notional lane width = 7.3/2 = 3.65m Cl. 6.2.1.

Loaded length = 34m


W = 31.6 kN/m (per notional lane) Cl. 6.2.2.

Knife Edge Load = 120 kN (per notional lane) Cl. 6.4.1.1.

α2 = 0.0137[bL(40-L)+3.65(L-20)]

α2 = 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.


W = (31.6 x 1.0)/3.65 = 8.66 kN/m

KEL = (120 x 1.0)/3.65 = 32.88 kN Cl. 6.2.7.


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 γf3



Influence Line Tutorial for HA UDL and HB Loading

Bending Effects

Example: Three span deck with continuity over pier supports.

Step1: Determine the position of the point of maximum bending moment in each element for a single

point load.

Point A − maximum sagging moment in span 1

Point B − maximum hogging moment over pier 1

Point C − maximum sagging moment in span 2

Note: as end spans are equal then critical points over pier 2 and in span 3 can be obtained from point

A and B by symmetry.

Step 2: Determine influence line diagram for point A:

Step 3: Determine influence line diagram for point B:

Step 4: Determine influence line diagram for point C:

Step 5: Determine loading for critical cases:

Point A

The maximum sagging moment is achieved by loading spans 1 and 3, however we also need to check

HA UDL for loading in span 1 only.

HA Span 1 only: loaded length = 10m hence udl = 71.8 kN/m (BD37-table 13)

HA Span 1 and 3: loaded length = 20m hence udl = 45.1 kN/m (BD37-table 13)

KEL: = 120 kN (BD37- Clause 6.2.2)

HB loading will produce worst sagging moment with an axle at the maximum ordinate (2.15). Any one

of the 4 axles can be located at this position; the vehicle is however positioned with the other 3 axles

to achieve the maximum total ordinates:

Note: The HB vehicle has a range of spacings between the centre axles, in this case the 26m spacing

gives the worst effect.

Point B

The maximum hogging moment is achieved by loading spans 1 and 2, however we also need to check

HA UDL for loading in span 2 only.


HA Span 1 and 2: loaded length = 30m hence udl = 34.4 kN/m (BD37-table 13)

KEL: = 120 kN (BD37- Clause 6.2.2)

Usually HB loading will produce the worst hogging moment with an axle at the maximum ordinate

(2.051). Any one of the 4 axles can be located at this position; the vehicle is however positioned with

the other 3 axles to achieve the maximum total ordinates.

In the case below the sum of the ordinates is 0.677 + 0.71 + 2.051 + 1.966 = 5.404

Other cofigurations of HB loading need be checked, and in this case the 6m vehicle will produce a

greater value with the vehicle in the position shown below. The sum of the ordinates for this

configuration = 1.604 + 1.943 + 1.651 + 1.305 = 6.503

Point C

The maximum sagging moment is achieved by loading span 2 only.


KEL: = 120 kN (BD37- Clause 6.2.2)

HB loading will produce worst sagging moment with an axle at the maximum ordinate (3.125). Any

one of the 4 axles can be located at this position; the vehicle is however positioned with the other 3

axles to achieve the maximum total ordinates:

Note: The HB vehicle has a range of spacings between the centre axles, in this case the 6m spacing

gives the worst effect.

Step 6: Determine load effects on deck.

The following assumptions will be made to demonstrate principles of influence lines:

• Assume loads applied to 1 notional lane width of deck (3.65m wide).

• Assume ultimate limit state hence use load factor γfL of 1.5 for HA loading and 1.3 for HB loading.

• Assume 30 units of HB.

Span 1

Maximum sagging moment due to HB loading:

M = 1.3 x 30 x 10 x (2.172 + 1.345 + 0.088 + 0.108) = 1448 kNm

Maximum sagging moment due to HA loading at point A:

Case 1 − Span1 loaded

M = 1.5 x (71.8 x 10.285 + 120 x 2.172) = 1499 kNm (HA critical)

Case 2 − Span1 and 3 loaded

M = 1.5 x [45.1 x (10.285 + 0.695) + 120 x 2.172] = 1134 kNm

Pier 1

Maximum hogging moment due to HA loading at point B:

Case 1 − Span 2 loaded

M = 1.5 x (45.1 x 24.776 +120 x 2.051) = 2045 kNm

Case 2 − Span1 and 2 loaded

M = 1.5 x [34.4 x (24.776 + 4.641) + 120 x 2.051] = 1887 kNm

Maximum hogging moment due to HB loading:

M = 1.3 x 30 x 10 x (1.604 + 1.943 + 1.651 + 1.305) = 2536 kNm (HB critical)

Span 2

Maximum sagging moment due to HA loading at point C:

Case 1 − Span 2 loaded

M = 1.5 x (45.1 x 25.25 + 120 x 3.125) = 2271 kNm

Maximum sagging moment due to HB loading:

M = 1.3 x 30 x 10 x (3.125 + 2.286 + 0.8 + 0.366) = 2565 kNm (HB critical)

Note: HB loading is shown to be critical for two of the cases, however if the loads are distributed using

a computer analysis, such as a grillage analysis, then the HB moments will be reduced considerably

bs 5400 part 2 : 2006 clause 6.2 type ha loading problem ... · pdf fileclause 6.2 type ha...

Documents