calculation of live load reaction

TYPICAL CALCULATION OF LIVE LOAD REACTION FOR PIER SUBSTRUCTURE SUBSTRUCT URE FOR SIMPLY SUPPORTED SPANS OF A THREE LANE BRIDGE STRUCTURE Centre line of pier w.r.t. the bearings :Rb Rc

=

0.3 m

=

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 Cond Condit itio ion n A: Case Case 1:

MAX MAXIMUM IMUM LONG ONGITUD ITUDIN INA AL MOM MOMENT ENT CASE CASE One One lan lane e of of cla class ss 70-R 70-R(W (W)) Cg of  100 t 5.12

0.3 m Ra

18.80 m

Rb 0.30 m

Rc 0.30 m

18.80 m

0.30 m Rd

Rb = 100*(18.8-5.12+0.3)/18.8 = 81.3 Rc = = 0.0 Ra= = 18.7 Vert.Reaction= 81.3 + 0 = 81.3 Braking Force, B = 0.2*100 = 20.0 Dead load reaction on the pier , Rg = 410.0 Value of " m " = = 0.00 Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 20.0 ( 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

t t t t t t t t

B)

One lane of class-A Cg of

0.3 m Ra

18.80 m

55.4t 9.7

0.0

Rb

Rc 0.30 m

18.80 m

0.30 m

0.3 m Rd

Rc = 0*(18.8-0.3)/18.8 = 0.0 Rb = 55.4*(18.8-9.7+0.3/2)/18.8 = 27.7 Ra= = 27.7 Vert.Reaction = 0 + 27.7 = 27.7 Braking Force, B = 0.2*(0+55.4) = 11.1 Dead load reaction on the pier , Rg = 410.0 Value of " m " = = 0.00 Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 11.1 ( neglecting shear rating of elastomeric bearing i n the adjacent span, which is on the conservative side ) CL class A(1L) 1.30

t t t t t t t t

CL of c/w 4.20

11 m Transverse eccentricity Transverse moment = 4.2*27.7 Long. moment = 27.7*0.3-0*0.3 Long. Eccentricity ( for input)

Case 3 :

= = = =

4.20 116.3 8.3 0.300

m t.m t.m m

= = = = = = = = =

0.0 55.4 55.4 55.4 11.1 410.0 0.00 0.0 11 1

t t t t t t

Two lane of class-A Rc = 2*0 Rb = 2*27.7 Ra= Vert.Reaction = 0 + 55.4 Braking Force(For single lane only) Dead load reaction on the pier , Rg Value of " m " = Horizontal force due to temperature, T = m*(Rg+Ra) Design horizontal force is higher of either ( B/2+T ) or ( B T )

t t

Case 4 :

Three lane of class-A Rc = 90% of 3*0 = 0.0 Rb = 90% of 3*27.7 = 74.8 Ra= = 1.3 Vert.Reaction = 0 + 74.8 74.8 Braking Force, B = (0.2)*55.4+0.05*55.4 = 13.9 (5% extra taken for third lane) Dead load reaction on the pier , Rg = 410.0 Value of " m " = = 0.00 Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 13.9 ( neglecting shear rating of elastomeric bearing i n the adjacent span, which is on the conservative side ) CL class A(3L)

t t t t t t t

CL of c/w 4.80

0.7

11 m Transverse eccentricity Transverse moment = 0.7*74.8 Long. moment = 74.8*0.3-0*0.3 Long. Eccentricity ( for input) Case 5 :

= = = =

0.70 52.4 22.4 0.300

m t.m t.m m

One lane of class-70R(W)+One lane of class-A Rc = 90% of(0+0) = 0.0 Rb = 90% of(27.7+81.28) = 98.1 Ra= = 41.8 Braking Force = 20 + 5% of 55.4 = 22.8 (5% extra taken for class A) Dead load reaction on the pier , Rg = 410.0 Value of " m " = = 0.00 Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 22.8 ( neglecting shear rating of elastomeric bearing i n the adjacent span, which is on the conservative side ) CL class 70-R 2.595

CL of c/w 0.84

2.905

t t t t t t t

CL class A(1L)

11.0 m Transverse ecc (class 70 R)

=

2 905

m

Condition B:

MAXIMUM TRANSVERSE MOMENT / REACTION CASE

CASE 1:

ONE LANE OF CLASS 70-R(W) Cg of 

cg 100.0 t

Cg of  49.0 t

51.0

5.12

3.33

3.19

0.3 m Ra

18.80 m

Rb

Rc 0.30 m

18.80 m

1.60 m

1.60m Rd

Rb = 49*(18.8 - 3.33 + 0.3)/18.8 = 41.10 Rc = 51*(18.8-3.19+1.6)/18.8 = 38.01 Ra= = 11.0 Vert. Reaction = 41.1 + 38 = 79.0 Braking Force, B = 0.2*100 = 20.0 Dead load reaction on the pier , Rg = 410.0 Value of " m " = = 0.00 Horizontal force due to temperature, T = m*(Rg+Ra) = 0.0 Design horizontal force is higher of either ( B/2+T ) or ( B-T ) = 20.0 ( neglecting shear rating of elastomeric bearing in the adjacent span, which is on the conservative side ) CL of 70-R

t t

CL of c/w 2.595

2.905

11 m Transverse eccentricity Transverse moment = 2.905*(41.1 + 38) Long. moment = 41.1*0.3-38.01*0.3 Long. Eccentricity ( for input) Case 2:

t t t t t t

One lane of class-A Cg of

= = = =

Cg of  55.4 t

28.2

Cg of 27.20 9.71 m

9.09 4.07

5.02

5.21

4.5 m

2.905 229.5 0.9 0.012

t

m t.m t.m m

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 t Braking 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 i n the adjacent span, which is on the conservative side ) CL class A(2L) 3.05

CL of c/w 2.45

11 m Transverse eccentricity Transverse moment = 2.45*82.4 Long. moment = 42.2*0.3-40.2*0.3 Long. Eccentricity ( for input)

Case 4 :

= = = =

2.45 202.0 0.6 0.007

m t.m t.m m

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 i n the adjacent span, which is on the conservative side ) CL class A(3L)

CL of c/w 4.80

0.7

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 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 i n the adjacent span, which is on the conservative side ) CL class 70-R 2.595

CL of c/w 0.84

2.905

CL class A(1L)

11.0 m Transverse ecc.(class 70 R) Transverse ecc.(class A) Trans. moment = 0.9*(81.3*2.9-0*-0.8) Net transverse ecc. (for input) Long. moment = 56*0.3-52.3*0.3 Long. Eccentricity ( for input)

= = = = = =

2.905 -0.84 175.4 1.620 1.1 0.010

m m t.m t.m m

first span SPAN

LOAD 8.28 5.04

CG 49 58

3.33 2.18

34 51

3.715 3.19

19.40 second span 4.4 5.12

22.00

two span length 9 13.3 14.5

load 27.2 38.6 50

cg6.8 end 4.5 7.1 8.79

cg2.7 end 4.5 6.2 5.71

28.2

4.07

18.2

1.81

span

load

cg

4.42 5.79 7.92

51 68 80

1.93 2.895 3.65

9.44 13.4

92 100

4.4 5.12

19.23

SPAN 5.5 8.5 11.5 14.5 24 19.23

LOAD CG 29.6 36.4 43.2 50 50

1.73 2.99 4.33 5.71 5.71

second span SPAN LOAD CG 3 80 4.52 92 8.48 100 24 100 19.40 first span 3 4.52 8.48 24 19.40

17 29 41 49

3.65 4.4 5.12 5.12

0.87 1.75 2.56 3.53

Summary of Loads Max. Longitudinal Moment Max. vertical reaction (t)

Transverse moment (t.m)

Longitudinal moment (t.m)

Design horizontal force (t)

Transverse ecc. (m)

Longitudinal ecc. (m)

81.3 27.7 55.4 74.8 98.1

236.1 116.3 135.7 52.4 191.6

24.4 8.3 16.6 22.4 29.4

20.0 11.1 11.1 13.9 22.8

2.905 4.200 0.700 0.700 1.953

0.300 0.300 0.300 0.300 0.300

Max.Transverse Moment Load case 1L class 70 - R 1L class - A 2L class - A 3L class - A 1L class 70 - R + 1L class - A

Design

Max. Transverse Longitudinal horizontal vertical moment moment (t.m) force (t) reaction (t) (t.m)

Transverse ecc. (m)

Longitudinal ecc. (m)

79.0 41.2 82.4 111.3

229.5 173.1 202.0 77.9

0.9 0.3 0.6 0.8

20.0 11.1 11.1 13.9

2.905 4.200 0.614 0.700

0.012 0.007 0.007 0.007

108.3

175.4

1.1

22.8

1.620

0.010

Vertical reaction d ue to braking has been neglected.