calculation of live load reaction for pier substructure
DESCRIPTION
CALCULATION OF LIVE LOAD REACTION FOR PIER SUBSTRUCTURETRANSCRIPT
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 cases1. One lane of class 70-R(W)2. One lane of class - A3. Two lane of class - A4. Three lane of class - A5. 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 t5.12
0.3 m 18.80 m Rb Rc 18.80 m 0.30 mRa 0.30 m 0.30 m Rd
Rb = 100*(18.8-5.12+0.3)/18.8 = 81.3 tRc = = 0.0 tRa= = 18.7 tVert.Reaction= 81.3 + 0 = 81.3 tBraking Force, B = 0.2*100 = 20.0 tDead load reaction on the pier , Rg = 410.0 t
= 0.00= 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
11 mTransverse eccentricity = 2.905 mTransverse moment = 2.905*(81.3 + 0) = 236.1 t.mLong. moment = 81.3*0.3-0*0.3 = 24.4 t.mLong. Eccentricity ( for input) = 0.300 m
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
B) One lane of class-ACg 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 tRb = 55.4*(18.8-9.7+0.3/2)/18.8 = 27.7 tRa= = 27.7 tVert.Reaction = 0 + 27.7 = 27.7 tBraking Force, B = 0.2*(0+55.4) = 11.1 tDead load reaction on the pier , Rg = 410.0 t
= 0.00= 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 4.20
11 mTransverse eccentricity = 4.20 mTransverse moment = 4.2*27.7 = 116.3 t.mLong. moment = 27.7*0.3-0*0.3 = 8.3 t.mLong. Eccentricity ( for input) = 0.300 m
Case 3 : Two lane of class-A
Rc = 2*0 = 0.0 tRb = 2*27.7 = 55.4 tRa= = 55.4 tVert.Reaction = 0 + 55.4 = 55.4 tBraking Force(For single lane only) = 11.1 tDead load reaction on the pier , Rg = 410.0 t
= 0.00= 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 2.45
11 mTransverse eccentricity = 2.45 mTransverse moment = 2.45*55.4 = 135.7 t.mLong. moment = 55.4*0.3-0*0.3 = 16.6 t.mLong. Eccentricity ( for input) = 0.300 m
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
Case 4 : Three lane of class-A
Rc = 90% of 3*0 = 0.0 tRb = 90% of 3*27.7 = 74.8 tRa= = 1.3 tVert.Reaction = 0 + 74.8 74.8Braking 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
= 0.00= 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
4.80 0.7
11 mTransverse eccentricity = 0.70 mTransverse moment = 0.7*74.8 = 52.4 t.mLong. moment = 74.8*0.3-0*0.3 = 22.4 t.mLong. 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 tRb = 90% of(27.7+81.28) = 98.1 tRa= = 41.8 tBraking 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
= 0.00= 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 2.905 0.84
11.0 mTransverse ecc.(class 70 R) = 2.905 mTransverse ecc.(class A) = -0.84 mTrans. moment = 0.9*(81.3*2.905-27.7*-0.84) = 191.6Net transverse ecc. (for input) = 1.953 t.mLong. moment = 98.1*0.3-0*0.3 = 29.4 t.mLong. Eccentricity ( for input) = 0.300 m
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
Condition B: MAXIMUM TRANSVERSE MOMENT / REACTION CASE
CASE 1: ONE LANE OF CLASS 70-R(W)
cg 100.0 t Cg of 51.0Cg of 49.0 t 3.33 5.12
3.19
0.3 m 18.80 m Rb Rc 18.80 m 1.60mRa 0.30 m 1.60 m Rd
Rb = 49*(18.8 - 3.33 + 0.3)/18.8 = 41.10 tRc = 51*(18.8-3.19+1.6)/18.8 = 38.01 tRa= = 11.0 tVert. Reaction = 41.1 + 38 = 79.0 tBraking Force, B = 0.2*100 = 20.0 tDead load reaction on the pier , Rg = 410.0 t
= 0.00= 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
11 mTransverse eccentricity = 2.905 mTransverse moment = 2.905*(41.1 + 38) = 229.5 t.mLong. moment = 41.1*0.3-38.01*0.3 = 0.9 t.mLong. Eccentricity ( for input) = 0.012 m
Case 2: One lane of class-A Cg of 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
0.3 m 18.80 m Rb Rc 18.80 m 0.3Ra 0.30 m 0.30 m m Rd
Rc = 27.2*(18.8-5.21+0.3)/18.8 = 20.10 tRb = 55.4*(18.8-5.02+0.3)/18.8 = 21.12 tRa= = 7.1 tVert.Reaction = 20.1 + 21.12 = 41.22 tBraking Force, B = 0.2*(55.4) = 11.1 tDead load reaction on the pier , Rg = 410.0 t
= 0.00= 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 4.2
11.0 mTransverse eccentricity = 4.20 mTransverse moment = 4.2*41.2 = 173.1 t.mLong. moment = 21.1*0.3-20.1*0.3 = 0.3 t.mLong. Eccentricity ( for input) = 0.007 m
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
Case 3 : Two lane of class-A
Rc = 2*20.1 = 40.2 tRb = 2*21.1 = 42.2 tRa= = 14.2 tVert.Reaction = 40.2 + 42.2 = 82.4 tBraking Force(For single lane only) = 11.1 tDead load reaction on the pier , Rg = 410.0 t
= 0.00= 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 2.45
11 mTransverse eccentricity = 2.45 mTransverse moment = 2.45*82.4 = 202.0 t.mLong. moment = 42.2*0.3-40.2*0.3 = 0.6 t.mLong. Eccentricity ( for input) = 0.007 m
Case 4 : Three lane of class-A
Rc = 90% of 3*20.1 = 54.3 tRb = 90% of 3*21.1 = 57.0 tRa= = 19.1 tVert.Reaction = 54.3 + 57 111.3Braking 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
= 0.00= 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
4.80 0.7
11 mTransverse eccentricity = 0.70 mTransverse moment = 0.7*111.3 = 77.9 t.mLong. moment = 57*0.3-54.3*0.3 = 0.8 t.mLong. Eccentricity ( for input) = 0.007 m
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
Case 5 : One lane of class-70R(W)+One lane of class-A
Rc = 90% of(20.1+38.01) = 52.3 tRb = 90% of(21.12+41.1) = 56.0 tRa= = 20.1 tBraking 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
= 0.00= 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 2.905 0.84
11.0 mTransverse ecc.(class 70 R) = 2.905 mTransverse ecc.(class A) = -0.84 mTrans. moment = 0.9*(81.3*2.9-0*-0.8) = 175.4Net transverse ecc. (for input) = 1.620 t.mLong. moment = 56*0.3-52.3*0.3 = 1.1 t.mLong. Eccentricity ( for input) = 0.010 m
Value of " m " =Horizontal force due to temperature, T = m*(Rg+Ra)
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
13.3 38.6 7.1 6.214.5 50 8.79 5.7118.7 52.7 9.24 9.4618.8 55.4 9.71 9.0938.5 55.4 9.71 9.09
load Span2load cg 6.8 load Span2 load cg 6.827.2 13.6 1.5 55.4 27.2 4.5
38.6 20.4 4.14 52.7 27.2 4.5
50 20.4 4.14 50 20.4 4.1452.7 27.2 4.5 38.6 20.4 4.1455.4 27.2 4.5 27.2 13.6 1.5
span2 19.23
load 1 Cg 2.7 end load 1 Cg 2.7 end13.6 1.5 28.2 4.0718.2 1.81 25.5 3.425.5 3.4 29.6 1.73
28.2 4.07 18.2 1.81
span load cg4.42 51 1.935.79 68 2.8957.92 80 3.65
9.44 92 4.413.4 100 5.12
19.23
SPAN LOAD CG5.5 29.6 1.738.5 36.4 2.99
11.5 43.2 4.3314.5 50 5.71
24 50 5.7119.23
second spanSPAN LOAD CG
3 80 3.654.52 92 4.48.48 100 5.12
24 100 5.1219.40
first span3 17 0.87
4.52 29 1.758.48 41 2.56
24 49 3.53
19.40
Summary of Loads
Max. Longitudinal Moment
Longitudinal moment (t.m)
81.3 236.1 24.4 20.0 2.905 0.30027.7 116.3 8.3 11.1 4.200 0.30055.4 135.7 16.6 11.1 0.700 0.30074.8 52.4 22.4 13.9 0.700 0.30098.1 191.6 29.4 22.8 1.953 0.300
Max.Transverse Moment
Load case
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 0.0073L 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
Vertical reaction due to braking has been neglected.
Design horizontal force (t)
Transverse ecc. (m)
Longitudinal ecc. (m)
Max. vertical reaction
(t)
Transverse moment (t.m)
Design horizontal force (t)
Transverse ecc. (m)
Longitudinal ecc. (m)
Max. vertical reaction
(t)
Transverse moment
(t.m)
Longitudinal moment (t.m)
1L class 70 - R + 1L class - A