Hb Loading Kipranya 2
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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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