7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

1/21

DESIGN OF ELASTOMERIC BEARINGS

TYPE OF SUPERSTRUCTURE : 19.00 m GIRDER

max min

Dead Load = 46.12 t 45.919SIDL = 5.18 t 5.285

FF LL = 0.48 0.473

Maximum Live Load = 50.155 t

Minimum Live Load = 37 t 4.73

Impact Factor = 1.18 1.18

SUMMARY OF VERTICAL REACTION :

say 112 t

say 50 t

110.96Total Max. (Normal Case)

Total Min. (Normal Case) 56.79

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

2/21

.

FOR MAXIMUM LOAD (NORMAL CASE)

Nmax Design vertical load on bearing 112.000 t 1098.8 KN

Fh External horizontal force 10.00 t

Hbd Horizontal force along b (long. Direction) 3.33 t 32.7 KN

Hld Horizontal force along l (trans. Direction) 0.00 t 0.0 KN

Dbd Translation along b (long. Direction) 5.00 mmDld Translation along l (trans. Direction) 0.0 mmabd Rotation along b (long. Direction) 0.000544 radiansald Rotation along l (trans. Direction) 0.00002 radiansDimension details of bearings

0 vera rea ong. rec on sa s y ng ser es mm

l0 Overall length (trans. Direction) of IS:1076 360 mm

i Thickness of each layer of elastomer 10 mm

n Number of internal elastomer layer 5 Nos.

e c ness o ou er e as omer ayer = i

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

3/21

Check for plan dimensions (Clause 916.3.3 of IRC : 83 part II )

a) l0/b0 = 360/360 = 1.000 6

= 8.70 < 12

Thus OK

Check for elastomer stress IRC : 83 part II

smax = 1098.8*1000/121104 = 9.08 < 10MPaThus OK

Check for translation (Clause 916.3.4 of IRC : 83 part II )

gdb = Hb/A+Dbd/h =32.7*1000/121104+5/60 = 0.354gdl = Hl/A+Dld/h =0*1000/121104+0/60 = 0.000gd = sqrt(gdb2+gdl2) =sqrt(0.354*0.354+0*0) 0.354

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

4/21

Check for friction (Clause 916.3.6 of IRC : 83 part II )

gd < 0.2+0.1*sm0.2+0.1*sm = 0.2+0.1*9.08

= 1.11 > gd calculated above = 0.354

Thus OK

Check for total shear stress (Clause 916.3.7 of IRC : 83 part II )

tc+tg+ta < 5 Mpatc = 1.5*sm/S = 1.5*9.08/8.7= 1.566 Mpatg = gd = 0.354 Mpata = 0.5*(b2*abi+l2*ali)/hi2

= 0.5*(348*348*0.0005+348*348*0.00002)/10/10

0.342 Mpa

tc+tg+ta = 1.566+0.354+0.342 2.262 Mpa

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

5/21

FOR MINIMUM LOAD (NORMAL CASE)

Nmax Design vertical load on bearing 50.000 t 490.5 KN

Fh External horizontal force 10.00 t

Hbd Horizontal force along b (long. Direction) 4.44 t 43.6 KN

Hld Horizontal force along l (trans. Direction) 0.00 t 0.0 KN

Dbd Translation along b (long. Direction) 5.00 mmDld Translation along l (trans. Direction) 0.0 mmabd Rotation along b (long. Direction) 0.000544 radiansald Rotation along l (trans. Direction) 0.000020 radiansDimension details of bearings

b0 Overall breadth (long. Direction) satisfying R20 series 360 mm

l0 Overall length (trans. Direction) of IS:1076 360 mm

hi Thickness of each layer of elastomer 10 mm

n Number of internal elastomer layer 5 Nos.

he Thickness of outer elastomer layer (=hi/2

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

6/21

Check for plan dimensions (Clause 916.3.3 of IRC : 83 part II )

a) l0/b0 = 360/360 = 1.000 6

= 8.70 < 12

Thus OK

Check for elastomer stress IRC : 83 part II

smax = 490.5*1000/121104 = 4.06 < 10MPaThus OK

Check for translation (Clause 916.3.4 of IRC : 83 part II )

gdb = Hb/A+Dbd/h =43.6*1000/121104+5/60 = 0.444gdl = Hl/A+Dld/h =0*1000/121104+0/60 = 0.000gd = sqrt(gdb2+gdl2) =sqrt(0.444*0.444+0*0) 0.444

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

7/21

Check for friction (Clause 916.3.6 of IRC : 83 part II )

gd < 0.2+0.1*sm0.2+0.1*sm = 0.2+0.1*4.06

= 0.61 > gd calculated above = 0.444

Thus OK

Check for total shear stress (Clause 916.3.7 of IRC : 83 part II )

tc+tg+ta < 5 Mpatc = 1.5*sm/S = 1.5*4.06/8.7= 0.700 Mpatg = gd = 0.444 Mpata = 0.5*(b2*abi+l2*ali)/hi2

= 0.5*(348*348*0.0005+348*348*0.00002)/10/10

0.342 Mpa

tc+tg+ta = 0.7+0.444+0.342 1.486 Mpa

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

8/21

DESIGN OF ELASTOMERIC BEARINGS

TYPE OF SUPERSTRUCTURE : 19.0 m pretensioned T girder

max min

Dead Load = 46.12 t 45.919SIDL = 5.18 t 5.285

Maximum Live Load = 25.0775 t

Minimum Live Load = 18.5 t 2.365

Impact Factor = 1.18 1.18

Horizontal Seismic Coefficient = 0.08 0.08

Vertical Seismic Coefficient = 0.04 0.04

SUMMARY OF VERTICAL REACTION :

say 85 t

say 47.0 t53.99

80.89Total Max. (Normal Case)

Total Min. (Normal Case)

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

9/21

.

FOR MAXIMUM LOAD (LONG SEISMIC CASE)

Nmax Design vertical load on bearing 87.052 t 854.0 KN

Fh External horizontal force 9.10 t

Hbd Horizontal force along b (long. Direction) 4.14 t 40.6 KN

Hld Horizontal force along l (trans. Direction) 0.00 t 0.0 KN

Dbd Translation along b (long. Direction) 5.00 mm

Dld Translation along l (trans. Direction) 0.00 mm

abd Rotation along b (long. Direction) -0.000056 radians

ald Rotation along l (trans. Direction) 0.00002 radians

Dimension details of bearings

b0 Overall breadth (long. Direction) satisfying R20 series 360 mm

l0 Overall length (trans. Direction) of IS:1076 360 mm

hi Thickness of each layer of elastomer 10 mm

n Number of internal elastomer layer 5 Nos.

e Thickness of outer elastomer layer = i

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

10/21

Check for plan dimensions (Clause 916.3.3 of IRC : 83 part II )

a) l0/b0 = 360/360 = 1.000 6

= 8.70 < 12

Thus OK

Check for elastomer stress IRC : 83 part II

smax = 854*1000/121104 = 7.06 < 10MPaThus OK

Check for translation (Clause 916.3.4 of IRC : 83 part II )

gdb = Hb/A+Dbd/h =40.6*1000/121104+5/60 = 0.419

gdl = Hl/A+Dld/h =0*1000/121104+0/60 = 0.000

gd = sqrt(gdb2+gdl

2) =sqrt(0.419*0.419+0*0) 0.419

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

11/21

Check for friction (Clause 916.3.6 of IRC : 83 part II )

gd < 0.2+0.1*sm

0.2+0.1*sm = 0.2+0.1*7.06

= 0.91 > gd calculated above = 0.419

Thus OK

Check for total shear stress (Clause 916.3.7 of IRC : 83 part II )

tc+tg+ta < 5 Mpa

tc = 1.5*sm/S = 1.5*7.06/8.7= 1.218 Mpa

tg = gd = 0.419 Mpa

ta = 0.5*(b2*abi+l

2*ali)/hi

2

= 0.5*(348*348*-0.0001+348*348*0.00002)/10/10

-0.022 Mpa

tc+tg+ta = 1.218+0.419+-0.022 1.615 Mpa

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

12/21

FOR MINIMUM LOAD (LONG SEISMIC CASE)

Nmax Design vertical load on bearing 44.952 t 441.0 KN

Fh External horizontal force 9.10 t

Hbd Horizontal force along b (long. Direction) 4.13 t 40.6 KN

Hld Horizontal force along l (trans. Direction) 0.00 t 0.0 KN

Dbd Translation along b (long. Direction) 5.00 mm

Dld Translation along l (trans. Direction) 0.00 mm

abd Rotation along b (long. Direction) -0.000056 radians

ald Rotation along l (trans. Direction) 0.000020 radians

Dimension details of bearings

b0 Overall breadth (long. Direction) satisfying R20 series 360 mm

l0 Overall length (trans. Direction) of IS:1076 360 mm

hi Thickness of each layer of elastomer 10 mm

n Number of internal elastomer layer 5 Nos.

he Thickness of outer elastomer layer (=hi/2

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

13/21

Check for plan dimensions (Clause 916.3.3 of IRC : 83 part II )

a) l0/b0 = 360/360 = 1.000 6

= 8.70 < 12

Thus OK

Check for elastomer stress IRC : 83 part II

smax = 441*1000/121104 = 3.65 < 10MPa

Thus OK

Check for translation (Clause 916.3.4 of IRC : 83 part II )

gdb = Hb/A+Dbd/h =40.6*1000/121104+5/60 = 0.419

gdl = Hl/A+Dld/h =0*1000/121104+0/60 = 0.000

gd = sqrt(gdb2+gdl

2) =sqrt(0.419*0.419+0*0) 0.419

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

14/21

Check for friction (Clause 916.3.6 of IRC : 83 part II )

gd < 0.2+0.1*sm

0.2+0.1*sm =

= 0.57 > gd calculated above = 0.42

Thus OK

Check for total shear stress (Clause 916.3.7 of IRC : 83 part II )

tc+tg+ta < 5 Mpa

tc = 1.5*sm/S = 1.5*3.65/8.7= 0.630 Mpa

tg = gd = 0.419 Mpa

ta = 0.5*(b2*abi+l

2*ali)/hi

2

= 0.5*(348*348*-0.0001+348*348*0.00002)/10/10

-0.022 Mpa

tc+tg+ta = 0.63+0.419+-0.022 1.027 Mpa

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

15/21

DESIGN OF ELASTOMERIC BEARINGS

TYPE OF SUPERSTRUCTURE : 18.0 m pretensioned T girder

Dead Load in one girder = 46.12 t 45.919 tSIDL in one girder = 5.18 t 5.285 t

Maximum Live Load in one girder = 25.0775 t t

Minimum Live Load in one girder = 18.5 t 2.365 t

Impact Factor = 1.18 1.18

Horizontal Seismic Coefficient = 0.08 0.08

Vertical Seismic Coefficient = 0.04 0.04

SUMMARY OF VERTICAL REACTION :

say 85 t

say 47.0 t53.99

80.89Total Max. (Normal Case)

Total Min. (Normal Case)

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

16/21

.

FOR MAXIMUM LOAD (TRANS SEISMIC CASE)

Nmax Design vertical load on bearing 88.400 t 867.3 KN

Fh External horizontal force 5.00 t

Hbd Horizontal force along b (long. Direction) 2.77 t 27.2 KN

Hld Horizontal force along l (trans. Direction) 6.11 t 60.0 KN

Dbd Translation along b (long. Direction) 5.00 mm

Dld Translation along l (trans. Direction) 0.00 mm

abd Rotation along b (long. Direction) -0.000056 radians

ald Rotation along l (trans. Direction) 0.000020 radians

Dimension details of bearings

b0 Overall breadth (long. Direction) satisfying R20 series 360 mm

l0 Overall length (trans. Direction) of IS:1076 360 mm

hi Thickness of each layer of elastomer 10 mm

n Number of internal elastomer layer 5 Nos.

e Thickness of outer elastomer layer = i

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

17/21

Check for plan dimensions (Clause 916.3.3 of IRC : 83 part II )

a) l0/b0 = 360/360 = 1.000 6

= 8.70 < 12

Thus OK

Check for elastomer stress IRC : 83 part II

smax = 867.3*1000/121104 = 7.17 < 10MPa

Thus OK

Check for translation (Clause 916.3.4 of IRC : 83 part II )

gdb = Hb/A+Dbd/h =27.2*1000/121104+5/60 = 0.308

gdl = Hl/A+Dld/h =60*1000/121104+0/60 = 0.496

gd = sqrt(gdb2+gdl

2) =sqrt(0.308*0.308+0.496*0.496) 0.5838493

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

18/21

Check for friction (Clause 916.3.6 of IRC : 83 part II )

gd < 0.2+0.1*sm

0.2+0.1*sm = 0.2+0.1*7.17

= 0.92 > gd calculated above = 0.5838493

Thus OK

Check for total shear stress (Clause 916.3.7 of IRC : 83 part II )

tc+tg+ta < 5 Mpa

tc = 1.5*sm/S = 1.5*7.17/8.7= 1.237 Mpa

tg = gd = 0.584 Mpa

ta = 0.5*(b2*abi+l

2*ali)/hi

2

= 0.5*(348*348*-0.0001+348*348*0.00002)/10/10

-0.022 Mpa

tc+tg+ta = 1.237+0.584+-0.022 1.799 Mpa

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

19/21

DESIGN OF ELASTOMERIC BEARINGS

TYPE OF SUPERSTRUCTURE : 19.0 m pretensioned T girder

FOR MINIMUM LOAD (TRANS SEISMIC CASE)

Nmax Design vertical load on bearing 45.120 t 442.7 KN

Fh External horizontal force 5.00 t

Hbd Horizontal force along b (long. Direction) 2.77 t 27.2 KN

Hld Horizontal force along l (trans. Direction) 4.29 t 42.1 KN

Dbd Translation along b (long. Direction) 5.00 mm

Dld Translation along l (trans. Direction) 0.00 mm

abd Rotation along b (long. Direction) -0.000056 radians

ald Rotation along l (trans. Direction) 0.00002 radians

Dimension details of bearings

b0 Overall breadth ( long. Direction) satisfying R20 series 360 mm

l0 Overall length (trans. Direction) of IS:1076 360 mm

hi Thickness of each layer of elastomer 10 mm

n Number of internal elastomer layer 5 Nos.

he Thickness of outer elastomer layer (=hi/2

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

20/21

Check for plan dimensions (Clause 916.3.3 of IRC : 83 part II )

a) l0/b0 = 360/360 = 1.000 6

= 8.70 < 12

Thus OK

Check for elastomer stress IRC : 83 part II

smax = 442.7*1000/121104 = 3.66 < 10MPa

Thus OK

Check for translation (Clause 916.3.4 of IRC : 83 part II )

gdb = Hb/A+Dbd/h =27.2*1000/121104+5/60 = 0.308

gdl = Hl/A+Dld/h =42.1*1000/121104+0/60 = 0.348

gd = sqrt(gdb2+gdl

2) =sqrt(0.308*0.308+0.348*0.348) 0.465

7/29/2019 Brg Jammu 1 Span Curved (at the END 19m)

21/21

Check for friction (Clause 916.3.6 of IRC : 83 part II )

gd < 0.2+0.1*sm

0.2+0.1*sm = 0.2+0.1*3.66

= 0.57 > gd calculated above = 0.465

Thus OK

Check for total shear stress (Clause 916.3.7 of IRC : 83 part II )

tc+tg+ta < 5 Mpa

tc = 1.5*sm/S = 1.5*3.66/8.7= 0.632 Mpa

tg = gd = 0.465 Mpa

ta = 0.5*(b2*abi+l

2*ali)/hi

2

= 0.5*(348*348*-0.0001+348*348*0.00002)/10/10

-0.022 Mpa

tc+tg+ta = 0.632+0.465+-0.022 1.075 Mpa