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7/28/2019 calculation of live load reaction
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TYPICAL CALCULATION OF LIVE LOAD REACTION FOR PIER SUBSTRUCTURE
FOR SIMPLY SUPPORTED SPANS OF A THREE LANE BRIDGE STRUCTURE
Centre line of pier w.r.t. the bearings :-
Rb = 0.3 m
Rc = 0.3 m
Reaction has been calculated for the following cases
1. One lane of class 70-R(W)
2. One lane of class - A
3. Two lane of class - A
4. Three lane of class - A
5. One lane of class 70-R(W) + One lane of class - A
Condition A: MAXIMUM LONGITUDINAL MOMENT CASE
Case 1: One lane of class 70-R(W)Cg of 100 t
5.12
0.3 m 18.80 m Rb Rc 18.80 m 0.30 m
Ra 0.30 m 0.30 m Rd
Rb = 100*(18.8-5.12+0.3)/18.8 = 81.3 t
Rc = = 0.0 tRa= = 18.7 t
Vert.Reaction= 81.3 + 0 = 81.3 t
Braking Force, B = 0.2*100 = 20.0 t
Dead load reaction on the pier , Rg = 410.0 t
Value of " m " = = 0.00
Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 t
Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 20.0 t
( neglecting shear rating of elastomeric bearing in the adjacent span, which is on the conservative side )
CL of 70-R CL of c/w
2.595 2.905
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B) One lane of class-A
Cg of 55.4t
9.7 0.0
0.3 m 18.80 m Rb Rc 18.80 m 0.3 mRa 0.30 m 0.30 m Rd
Rc = 0*(18.8-0.3)/18.8 = 0.0 t
Rb = 55.4*(18.8-9.7+0.3/2)/18.8 = 27.7 t
Ra= = 27.7 t
Vert.Reaction = 0 + 27.7 = 27.7 t
Braking Force, B = 0.2*(0+55.4) = 11.1 t
Dead load reaction on the pier , Rg = 410.0 t
Value of " m " = = 0.00
Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 t
Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 11.1 t( neglecting shear rating of elastomeric bearing in the adjacent span, which is on the conservative side )
CL class A(1L) CL of c/w
1.30
11 m
Transverse eccentricity = 4.20 m
Transverse moment = 4.2*27.7 = 116.3 t.m
Long. moment = 27.7*0.3-0*0.3 = 8.3 t.m
Long. Eccentricity ( for input) = 0.300 m
Case 3 : Two lane of class-A
Rc = 2*0 = 0.0 t
Rb = 2*27.7 = 55.4 t
Ra= = 55.4 t
Vert.Reaction = 0 + 55.4 = 55.4 t
Braking Force(For single lane only) = 11.1 t
Dead load reaction on the pier , Rg = 410.0 tValue of " m " = = 0.00
Horizontal force due to temperature T = m*(Rg+Ra) = 0 0 t
4.20
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Case 4 : Three lane of class-A
Rc = 90% of 3*0 = 0.0 t
Rb = 90% of 3*27.7 = 74.8 t
Ra= = 1.3 t
Vert.Reaction = 0 + 74.8 74.8
Braking Force, B = (0.2)*55.4+0.05*55.4 = 13.9 t
(5% extra taken for third lane)Dead load reaction on the pier , Rg = 410.0 t
Value of " m " = = 0.00
Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 t
Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 13.9 t
( neglecting shear rating of elastomeric bearing in the adjacent span, which is on the conservative side )
CL class A(3L) CL of c/w
0.7
11 mTransverse eccentricity = 0.70 m
Transverse moment = 0.7*74.8 = 52.4 t.m
Long. moment = 74.8*0.3-0*0.3 = 22.4 t.m
Long. Eccentricity ( for input) = 0.300 m
Case 5 : One lane of class-70R(W)+One lane of class-A
Rc = 90% of(0+0) = 0.0 t
Rb = 90% of(27.7+81.28) = 98.1 t
Ra= = 41.8 t
Braking Force = 20 + 5% of 55.4 = 22.8 t
(5% extra taken for class A)
Dead load reaction on the pier , Rg = 410.0 t
Value of " m " = = 0.00
Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 t
Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 22.8 t
( neglecting shear rating of elastomeric bearing in the adjacent span, which is on the conservative side )
CL class 70-R CL of c/w CL class A(1L)
2.595 0.84
4.80
2.905
11 0 m
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Condition B: MAXIMUM TRANSVERSE MOMENT / REACTION CASE
CASE 1: ONE LANE OF CLASS 70-R(W)
cg 100.0 t Cg of 51.0
Cg of 49.0 t
0.3 m 18.80 m Rb Rc 18.80 m
Ra 0.30 m 1.60 m Rd
Rb = 49*(18.8 - 3.33 + 0.3)/18.8 = 41.10 t
Rc = 51*(18.8-3.19+1.6)/18.8 = 38.01 t
Ra= = 11.0 t
Vert. Reaction = 41.1 + 38 = 79.0 t
Braking Force, B = 0.2*100 = 20.0 t
Dead load reaction on the pier , Rg = 410.0 t
Value of " m " = = 0.00
Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 t
Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 20.0 t
( neglecting shear rating of elastomeric bearing in the adjacent span, which is on the conservative side )
CL of 70-R CL of c/w
2.595
11 m
Transverse eccentricity = 2.905 m
Transverse moment = 2.905*(41.1 + 38) = 229.5 t.m
Long. moment = 41.1*0.3-38.01*0.3 = 0.9 t.m
Long. Eccentricity ( for input) = 0.012 m
Case 2: One lane of class-A
Cg of 28.2 55.4 t Cg of 27.20 t
9.09 9.71 m
4.07 5.02 5.21 4.5 m
5.12
3.19
3.33
1.60m
2.905
Cg of
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Case 3 : Two lane of class-A
Rc = 2*20.1 = 40.2 t
Rb = 2*21.1 = 42.2 t
Ra= = 14.2 t
Vert.Reaction = 40.2 + 42.2 = 82.4 tBraking Force(For single lane only) = 11.1 t
Dead load reaction on the pier , Rg = 410.0 t
Value of " m " = = 0.00
Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 t
Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 11.1 t
( neglecting shear rating of elastomeric bearing in the adjacent span, which is on the conservative side )
CL class A(2L) CL of c/w
3.05
11 m
Transverse eccentricity = 2.45 m
Transverse moment = 2.45*82.4 = 202.0 t.m
Long. moment = 42.2*0.3-40.2*0.3 = 0.6 t.m
Long. Eccentricity ( for input) = 0.007 m
Case 4 : Three lane of class-A
Rc = 90% of 3*20.1 = 54.3 t
Rb = 90% of 3*21.1 = 57.0 t
Ra= = 19.1 t
Vert.Reaction = 54.3 + 57 111.3
Braking Force, B = (0.2)*55.4+0.05*55.4 = 13.9 t
(5% extra taken for third lane)
Dead load reaction on the pier end , Rg = 410.0 t
Value of " m " = = 0.00
Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 t
Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 13.9 t
( neglecting shear rating of elastomeric bearing in the adjacent span, which is on the conservative side )
CL class A(3L) CL of c/w
0 7
2.45
4 80
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Case 5 : One lane of class-70R(W)+One lane of class-A
Rc = 90% of(20.1+38.01) = 52.3 t
Rb = 90% of(21.12+41.1) = 56.0 t
Ra= = 20.1 t
Braking Force = 20 + 5% of 55.4 = 22.8 t
(5% extra taken for class A)
Dead load reaction on the pier , Rg = 410.0 tValue of " m " = = 0.00
Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 t
Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 22.8 t
( neglecting shear rating of elastomeric bearing in the adjacent span, which is on the conservative side )
CL class 70-R CL of c/w CL class A(1L)
2.595 0.84
Transverse ecc.(class 70 R) = 2.905 mTransverse ecc.(class A) = -0.84 m
Trans. moment = 0.9*(81.3*2.9-0*-0.8) = 175.4
Net transverse ecc. (for input) = 1.620 t.m
Long. moment = 56*0.3-52.3*0.3 = 1.1 t.m
Long. Eccentricity ( for input) = 0.010 m
2.905
11.0 m
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first span
SPAN LOAD CG
8.28 49 3.33
5.04 58 2.18
19.40
second span
4.4 34 3.715
5.12 51 3.19
22.00
two span length load cg6.8 end cg2.7 end9 27.2 4.5 4.5
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28.2 4.07 18.2 1.81
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span load cg
4.42 51 1.93
5.79 68 2.8957.92 80 3.65
9.44 92 4.4
13.4 100 5.12
19.23
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SPAN LOAD CG
5.5 29.6 1.73
8.5 36.4 2.99
11.5 43.2 4.33
14.5 50 5.71
24 50 5.71
19.23
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second span
SPAN LOAD CG
3 80 3.65
4.52 92 4.4
8.48 100 5.1224 100 5.12
19.40
first span
3 17 0.87
4.52 29 1.75
8.48 41 2.56
24 49 3.53
19.40
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Summary of Loads
Max.
vertical
reaction (t)
81.3 2.905 0.300
27.7 4.200 0.300
55.4 0.700 0.30074.8 0.700 0.300
98.1 1.953 0.300
Max.
vertical
reaction (t)
Transverse
moment
(t.m)
Longitudinal
moment (t.m)
1L class 70 - R 79.0 229.5 0.9 20.0 2.905 0.012
1L class - A 41.2 173.1 0.3 11.1 4.200 0.0072L class - A 82.4 202.0 0.6 11.1 0.614 0.007
3L class - A 111.3 77.9 0.8 13.9 0.700 0.007
108.3 175.4 1.1 22.8 1.620 0.010
Transverse
ecc. (m)
Longitudinal
ecc. (m)
Transverse
ecc. (m)
Longitudinal
ecc. (m)
20.0
22.8
11.1
11.113.9
135.7
116.3
1L class 70 - R +
1L class - A
191.6
Load case
29.4
Max.Transverse Moment
52.4 22.4
8.3
Vertical reaction d ue to braking has been neglected.
Transverse moment
(t.m)
16.6
24.4
Design horizontal force
(t)
Max. Longitudinal Moment
Design
horizontal
force (t)
Longitudinal moment (t.m)
236.1