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MOMENTS GENERATED BY HEAVY TRUCK

SCENARIO 4

Horse 0 Trailer

Cargo= 100t

Cargo+Trailer = 130t

Truck = 16t Trailer = 30t

3t 6t 5t 3t 13t 13t 13t 13t 13t 13t

2.40M 1.40M 1.40M 1.60M 1.60M 1.60M 1.60M 1.60M 1.60

22.30M

2.30

-0.50

0.90

2.30M

5.00M

6.60M

8.20M

9.80M

11.40M

13.00M

14.60M

16.20M

13.4

Load (t) a (m) b (m) RA (t) RB (t) Mc (t.m)

6t -.50M 13.90M 5.7052239 -0.205224 -1.375

5t .90M 12.50M 4.5708955 0.329104 2.205

3t 2.30M 11.10M 2.1537313 0.446269 2.99

13t 5.00M 8.40M 8.1492537 4.850746 32.513t 6.60M 6.80M 6.5970149 6.402985 44.2

13t 8.20M 5.20M 5.0447761 7.955224 33.8

13t 9.80M 3.60M 3.4925373 9.507463 23.4

13t 11.40M 2.00M 1.9402985 11.0597 13

13t 13.00M .40M 0.3880597 12.61194 2.6

13t 14.60M -1.20M -1.164179 14.16418 -7.8

145.52 (t.m)

Moments 1427.551 knm

3219.581

55.66034

Moments per meter is obtained by 407.8718

Hb moments obtained 407.8718 KNm

2.70M

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KIPRANYA BEAM 6,7,8,9 HB LOADING

The carriage way width is 7.0m therefore the notional lanes are two.

HB Loading is taken to occupy any transverse position on the carriageway,lying either wholly

within one notional lane or straddling two notional lanes

No other primary live loading should be considered 25m in front to 25m behind the HB Vehicle

DIMENSIONS OF A HB VEHICLE :

Axle Axle Axle Axle

BS 5400 3.5m

Pt II 6.4.2

1.8 6,11,16,21 or 26 1.8

Considering 40 units of HB Vehicle :

1 unit = 10 Kn per axle

40 units = 400 Kn per axle

Divide by No of wheels (4) = 100 Kn per wheel

Only one HB Vehicle is considered per bridge

HB Vehicle is considered as occupying one notional lane or straddling two notional lanes,

whichever produces the most adverse effect.

CASE 1 : HB Vehicle in one notional lane

Loaded length for intensity of HA UDL

Overall Vehicle length for

axle spacing having most

25m severe effect 25m

1 1 1HA UDL&KEL

1m

1m

1m

BS 5400 Pt II

:Fig 12-Type HA

and HB HighwayLoading in

combination

NO LOADING NO LOADINGNotional Lane (3.5m)

Notional Lane (3.5m)

HB VEHICLE

FULL HA

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Loaded length for intensity of HA UDL

Overall Vehicle length for

axle spacing having most

25m severe effect 25m

1

1 1

b2 b3

By observation,case 1 gives the most adverse effect

Configuration of HB Vehicle for maximum reaction :

1.8`m 6m 1.8m

17

1 2

Substituting,P= 100

M1=1.8*P+7.8*P+9.6*P-15V2

V2= 451.764706 451.764706

V1= 400*4-V2 1148.23529 KN

Configuration of HB Vehicle for maximum Moments :

For a simple span HB Vehicle with 6m axle spacing is considered for maximum moment with

the second axle 1.5m from centre of span thus :

Calculation of HB moments =

P P P

1.5 4.5

3.4

5.2

11.2

13 P=300

Moment at the Center of beam :

M1=3.4*P+5.2*P+11.2*P+13*P-15V2 =0

V2= 771.7647059

V1= 828.2352941

Notional Lane

NO LOADINGNotional Lane

NO LOADINGHB VEHICLE

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Moment Max at inner axle from centre line therefore =708*5.2-300*1.8

3766.82353

1076.235294 KNM 1076.23529 per metre

BMD Diagram for the effect of HB Vehicle

1076.23529

Factored =1.3 1399.10588

Beam Spacing= 2.4 m

Total HB Moment on beam 1= 3357.854118 KNm

Total Self weight moment of beam 1 = 971.04 KNm

Total Self weight moment for 350mm Deck= 1019.592 KNm

Total moments acting on Beam= 5348.486118 KNm

Effective depth, d = 1305 mm

k = M / (bd2fcu)

= (6650x106 /(1000x13002x30)

= 0.105 < 0.156Hence no compression r/f is required

z = d (0.5+(0.25-k/0.9)1/2

z = d [0.5+(0.25-0.044/0.9)1/2

= 1129.64 < 0.95d

Take Z as 0.95d

Z = 0.95d

= 1239.75 mm

Design of main reinforcement

As = M / 0.95fyz

= 5350 x106

/ 0.95x410x1239

= 11510 mm /m As req

= 12868 mm2/m As pro (16Y32)

Minimum area of main reinforcement for beams

100As / bad= 100x13510/(500x1305) 1.76 0.13

Hence o.k

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Design for Shear Reinforcement

Check shear in U.L.S.

Take Load case 02

Shear across support

Design shear force, V design = kN/m

Effective depth, d = mm

Tension steel across shear plane = Y12 -250 c/c

100 As/bd =

=

Effective depth = mm

vc = 0.79x{(100As/bd)1/3.

(400/d)1/4

/1.25

= 0.60376

Design shear stress v = V/bd= (1350x10

3)/(500x1305)

= N/mm2

v > vc Hence shear reinforcement required

Asv =bv (v-vc)/(0/95fyv =500(1.76-0.603)/0.95/410

Sv

=1.48

hence provide Y12@275c/c ( 4 legs)

6.2

1240.00

100 x 11510

500x1305

1.76

1305

0.568191044

1.90

1305

mailto:Y12@200c/c%20(%204%20legs)mailto:Y12@200c/c%20(%204%20legs)

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Sub-loads

L80 80130

1 L-20 5000 350 430-L 300

2 L-40

P= 27.5

w= 1.925

L= 20

Mc =

1.925 t/m

27.5 t

Mc = 233.75 tm

Mc = 2293.088 Knm

Class of

Bridge Loading

Line load

P (kg/m)

Uniform load, P (kg/m2)

P = 5t/m x 5.5m =

50% of

main

loads

70% Of those of 1st class

WL2/8 +PL/4

W= 0.35t/mx 5.5m =

Main loads (upto 5.5m width)

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STEEL MEMBER DESIGN

Column1 Column2 Column3 Column4

Summary Page

Remarks

Scenario 1 407.871771 KNm Most critical

Scenario 2 699.142857 KNm ignored

Scenario 3 553.428571 KNm ignoredScenario 4 407.714286 KNm ignored

STRUT/COLUMN DESIGNMc=25% of 772

Mc=193KNm

Mc=PyZ

Py=275N/mm2

Z=193x103/275

Z=701cm3

Hence adopt 254x254x19.2 UC

LE=1.0L

COMPRESSION CHECK LE=3.5

Py=250N/mm2

ry=6.79cm

Ag=212cm2

Design capacity load= 250x212x103

5300KN

Design capacity=5300KN

Section OK

Actual load is 65KN

6M 24 BOLTS PER CONNECTION

CONNECTION DESIGNPy=375N/mm2

ry=6.79cm

Ag=353cm2

DESIGNAPPLIED LOAD 65 KN

No. of bolts=6

DESIGN LOAD PER BO 16.25KN

M24 AREA 353mm2LESS 10% ps=375N/mm

2

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=318mm2

Bolt strength=353x375x0.6x6 =476kN

Axial load on member is 325KN

Hence 6M24 bolts are sufficient

400x400x25mm CLEAT PLATE TO BE PROVIDED

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Design of main reinforcement

As = M / 0.87fyz

= 24.15 x106

/ 0.87x460x1

=

Use T @

Minimum area of main reinforcement for slabs

100As / bad = 100x452/(1000x149)

#REF!

6.1.1

#REF!

#REF!

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2 As req

mm2/m

( As = mm2/m As pro

= 0.13 Main r/f

T

#REF!

160 #REF!

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=

mm2/m

=

mm2/m

@

#REF!

#REF! 160

#REF!

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z = d (0.5+(0.25-k/0.9)1/2

z = d [0.5+(0.25-0.044/0.9)1/2

= #REF! d#REF! 0.950

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