# elastomeric bearing - 15m span

of 34 /34
NOTE NO. DN-18 SHEET DESIGN OF ELASTOMERIC BEARING CALCULATION OF BEARING LOADS Normal Case Max. Reaction on a bearing (Refer STAAD output) On outer Bearing On inner Bea Due to DL of RCC girder and slab (from design calc.) = 25.00 t 25.00 Due to Diaphragm = 0.92 t 3.06 25.92 t 28.06 Due to SIDL = 9.67 t 0.54 Total DL + SIDL = 35.59 t 28.60 Due to Live load (impact -1.1 Max. 37.99 t 52.89 Min. -1.20 t -0.35 Total Maximum load = Max. 73.58 t 81.49 Total Minimum load = Min. 34.39 t 28.25 Calculation of horizontal forces :-- Horizontal force from superstructure 3 Nos. elastomeric bearings are proposed on each support. Braking force = (100 x 0.2 ) = 20.00 t Total Horizontal force transferred from superstructure = 20.00 / 2 = 10.00 t This force is resisted by three bearing provided at each end Thus longitudinal force on each bearin 10.00 / 3 = 3.33 t Seismic Transverse Case = 0.075 ( Hor. seismic coeff.) Seismic force on Dead load = 5.99 t, say 5.99 t Its lever arm above bearing l 1.01 m (approx.) Thus moment at bearing level 5.99 1.01 = 6.02 Tm Seismic force on SIDL = 19.88 0.075 = 1.49 t The bridge is having simply support span. each support has three no. of elastomeric a h x x

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Design of Elastomeric Bearing For 15m Span bridge

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NOTE NO. DN-18 SHEET NO.

DESIGN OF ELASTOMERIC BEARING

The bridge is having simply support span. each support has three no. of elastomeric bearings.

Normal CaseMax. Reaction on a bearing(Refer STAAD output) On outer Bearing On inner Bearing

Due to DL of RCC girder and slab (from design calc.) = 25.00 t 25.00 tDue to Diaphragm = 0.92 t 3.06 t

25.92 t 28.06 tDue to SIDL = 9.67 t 0.54 tTotal DL + SIDL = 35.59 t 28.60 t

Due to Live load (impact -1.1) = Max. 37.99 t 52.89 tMin. -1.20 t -0.35 t

Total Maximum load = Max. 73.58 t 81.49 tTotal Minimum load = Min. 34.39 t 28.25 t

Calculation of horizontal forces :--

Horizontal force from superstructure

3 Nos. elastomeric bearings are proposed on each support.

Braking force = (100 x 0.2 ) = 20.00 t

Total Horizontal force transferred from superstructure =

20.00 / 2 = 10.00 t

This force is resisted by three bearing provided at each end

Thus longitudinal force on each bearing = 10.00 / 3 = 3.33 t

Seismic Transverse Case

= 0.075 ( Hor. seismic coeff.)

Its lever arm above bearing level = 1.01 m (approx.)

Thus moment at bearing level = 5.99 1.01 = 6.02 Tm

Seismic force on SIDL = 19.88 0.075 = 1.49 t

a h

x

x

NOTE NO. DN-18 SHEET NO.

Its lever arm above bearing level = 1.80 m

Thus moment at bearing level = 1.49 1.80 = 2.68 Tm

Total live load reaction at support= 67.0 t

Seismic force on LL (50%) = 33.5 0.075 = 2.51 t

Its lever arm above bearing level = 2.765 m

Thus moment at bearing level = 2.51 2.76 = 6.95 Tm

Total transverse force = 10.00 t

This force is resisted by three bearing provided at each end

Thus transverse force on each bearing = 10.00 / 3 = 3.33 t

Total transverse moment = 15.65 Tm

x ^2 = 2 x ( 2.90^2 + 0^2 ) = 16.82 m^2

Hence, Vertical load on inner bearing = 15.65 x 0.000 = (+/-) 0.00 t

Hence, Vertical load on outer bearing = 15.65 x 0.172 = (+/-) 2.70 t

Thus,Maximum load on outer bearing (50% LL) =

25.92 9.67 19.00 2.70 = 57.29 t

Minimum load on outer bearing =25.92 9.67 -0.60 -2.70 = 32.29 t

Maximum load on inner bearing =28.06 0.54 26.44 0.00 = 55.04 t

Minimum load on inner bearing =28.06 0.54 -0.18 0.00 = 28.42 t

The design vertical load (minimum) = 28.42 t (Governing)

Horizontal force from superstructure

3 Nos. elastomeric bearings are proposed on each support.

Braking force (50%) =(100 x 0.2) x 0.5 = 10.00 t

Total Horizontal force transferred from superstructure =

x

x

x

+ + +

+

+ + +

+

+

+

+

+

x

NOTE NO. DN-18 SHEET NO.

10.00 / 2 = 5.00 t

This force is resisted by three bearing provided at each end

Thus longitudinal force on each bearing = 5.00 / 3 = 1.67 t

Seismic Longitudinal Case

Maximum load on outer bearing =25.9 9.67 19.00 = 54.59 t

Minimum load on outer bearing =25.92 9.67 -0.60 = 34.99 t

Maximum load on inner bearing =28.06 0.54 26.44 = 55.04 t

Minimum load on inner bearing =28.06 0.54 -0.18 = 28.42 t

The design vertical load (minimum) = 28.42 t (Governing)

Horizontal force from superstructure (seismic)

5.99 1.49 = 7.48 t

Braking force (50%) =(100 x 0.2)x0.5 = 10.00 t

Total Horizontal force transferred from superstructure =

7.48 10.0 / 2 = 12.48 t

This force is resisted by three bearing provided at each end

Thus longitudinal force on each bearing = 12.48 / 3 = 4.16 t

Movement at bearing (@ 0.5 x 10^-3 )

= 0.5 x 10^-3 x 14.0 x 10^3 x0.5 = 3.50 mm

Rotation as per clause :916.3.5 IRC:83(part-II)

= 83.78 t-m (refer design calculations of superstructure)

Grade of concrete = M 30

ad 400 Mmax L /(Ec I )

Mmax

+ +

+

+ +

+

+

+

+

NOTE NO. DN-18 SHEET NO.

E 5000x = 2791654

I = Moment of inertia = 0.2882 (refer design of superstructure)

= 400 x 83.78 x 14.000 x 0.0010.5 x 2791654 x 0.2882

= 149.41 t-m (refer design calculations of superstructure)

= 0.00104

Total rotation = 0.0011663 + ### = 0.0022 rad

fck t/m2

Mmax

NOTE NO. DN-18 SHEET NO.

1. STAAD SPACE ANALYSIS OF SUPERSTRUCTURE FOR SIDL 2. INPUT WIDTH 79 3. UNIT MTON METRE 4. * 5. JOINT COORDINATE 6. * 7. 101 0.0 0.0 0.000 109 14.0 0.0 0.000 8. 201 0.0 0.0 1.450 209 14.0 0.0 1.450 9. 301 0.0 0.0 4.350 309 14.0 0.0 4.350 10. 401 0.0 0.0 7.250 409 14.0 0.0 7.250 11. 501 0.0 0.0 8.700 509 14.0 0.0 8.700 12. * 13. 701 0.0 0.0 0.725 709 14.0 0.0 0.725 14. 801 0.0 0.0 2.900 809 14.0 0.0 2.900 15. 901 0.0 0.0 5.800 909 14.0 0.0 5.800 16. 1001 0.0 0.0 7.975 1009 14.0 0.0 7.975 17. * 18. 110 -0.50 0.0 0.00; 111 14.50 0.0 0.0 19. 210 -0.50 0.0 1.450; 211 14.50 0.0 1.450 20. 310 -0.50 0.0 4.350; 311 14.50 0.0 4.350 21. 410 -0.50 0.0 7.250; 411 14.50 0.0 7.250 22. 510 -0.50 0.0 8.700; 511 14.50 0.0 8.700 23. * 24. 710 -0.50 0.0 0.725; 711 14.50 0.0 0.725 25. 810 -0.50 0.0 2.900; 811 14.50 0.0 2.900 26. 910 -0.50 0.0 5.800; 911 14.50 0.0 5.800 27. 1010 -0.50 0.0 7.975; 1011 14.50 0.0 7.975 28. * 29. MEM INCIDENCE 30. 101 101 102 108 31. 201 201 202 208 32. 301 301 302 308 33. 401 401 402 408 34. 501 501 502 508 35. * 36. 701 701 702 708 37. 801 801 802 808 38. 901 901 902 908 39. 1001 1001 1002 1008 40. * 41. 109 110 101; 110 109 111 42. 209 210 201; 210 209 211 43. 309 310 301; 310 309 311 44. 409 410 401; 410 409 411 45. 509 510 501; 510 509 511 46. * 47. 709 710 701; 710 709 711 48. 809 810 801; 810 809 811

NOTE NO. DN-18 SHEET NO.

49. 909 910 901; 910 909 911 50. 1009 1010 1001; 1010 1009 1011 51. * 52. 2101 101 701 2109 53. 2201 201 801 2209 54. 2301 301 901 2309 55. 2401 401 1001 2409 56. * 57. 3101 701 201 3109 58. 3201 801 301 3209 59. 3301 901 401 3309 60. 3401 1001 501 3409 61. * 62. MEMBER PROPERTIES 63. *DUMMY MEMBER 64. 101 TO 110 501 TO 510 PRIS YD 0.05 ZD 0.05 65. 701 TO 710 801 TO 810 901 TO 910 PRIS YD 0.05 ZD 0.05 66. 1001 TO 1010 PRIS YD 0.05 ZD 0.05 67. 209 210 309 310 409 410 PRI YD 0.05 ZD 0.05 68. * 69. 301 TO 308 401 TO 408 201 TO 208 - 70. PRIS AX 1.3645 IX 1E-10 IY .4673 IZ .2882 71. * DIAPHRAGM 72. 2201 TO 2301 BY 100 2209 TO 2309 BY 100 - 73. 3201 TO 3301 BY 100 3209 TO 3309 BY 100 - 74. PRIS AX 0.52932 IX 1E-10 IY 0.01429 IZ 0.05910 75. 2101 2401 2109 2409 - 76. 3101 3401 3109 3409 PRIS AX 0.17732 IX 1E-10 IY 0.009599 IZ 0.000715 77. * SLAB 78. 2102 TO 2108 2202 TO 2208 2302 TO 2308 2402 TO 2408 - 79. 3102 TO 3108 3202 TO 3208 3302 TO 3308 3402 TO 3408 - 80. PRIS AX 0.3850 IX 1E-10 IY 0.09826 IZ 0.001553 81. * 82. SUPPORTS 83. 201 301 401 PINNED 84. 209 309 409 FIXED BUT FX FZ MX MY MZ 85. * 86. CONSTANTS 87. E 3.0E6 88. DEN 2.4 89. * 90. LOAD 1 SIDL 91. MEMBER LOAD 92. ***WEARING COAT 93. * 0.2*(2.9/2+1.225) = 0.535 T/M 94. * 0.2*2.9 = 0.58 T/M 95. 201 TO 210 401 TO 410 UNI GY -0.535 96. 301 TO 310 UNI GY -0.58

NOTE NO. DN-18 SHEET NO.

97. **CRASH BARRIER 98. 101 TO 110 501 TO 510 UNI GY -0.50 99. * 100. LOAD 2 SELF WT OF DIAPHRAGM (WT. 0.88X0.40X2.4=0.845 T/M) 101. MEMBER LOAD 102. 2201 TO 2301 BY 100 3201 TO 3301 BY 100 2209 TO 2309 BY 100 - 103. 3209 TO 3309 BY 100 UNI GY -0.845 104. * 105. PERFORM ANALYSIS 119. LOAD LIST 1 120. PRINT SUPPORT REACTION JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z

201 1 .00 9.67 .00 .00 .00 .00 301 1 .00 .54 .00 .00 .00 .00 401 1 .00 9.67 .00 .00 .00 .00 209 1 .00 9.67 .00 .00 .00 .00 309 1 .00 .54 .00 .00 .00 .00 409 1 .00 9.67 .00 .00 .00 .00

121. LOAD LIST 2 122. PRINT SUPPORT REACTION JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z

201 2 .00 .92 .00 .00 .00 .00 301 2 .00 3.06 .00 .00 .00 .00 401 2 .00 .92 .00 .00 .00 .00 209 2 .00 .92 .00 .00 .00 .00 309 2 .00 3.06 .00 .00 .00 .00 409 2 .00 .92 .00 .00 .00 .00

123. FINISH

NOTE NO. DN-18 SHEET NO.

1. STAAD SPACE ANALYSIS OF SUPERSTRUCTURE FOR LIVE LOAD 70R WHEELED 2. INPUT WIDTH 79 3. UNIT MTON MET 4. * 5. JOINT COORDINATE 6. * 7. 101 0.0 0.0 0.000 109 14.0 0.0 0.000 8. 201 0.0 0.0 1.450 209 14.0 0.0 1.450 9. 301 0.0 0.0 4.350 309 14.0 0.0 4.350 10. 401 0.0 0.0 7.250 409 14.0 0.0 7.250 11. 501 0.0 0.0 8.700 509 14.0 0.0 8.700 12. * 13. 701 0.0 0.0 0.725 709 14.0 0.0 0.725 14. 801 0.0 0.0 2.900 809 14.0 0.0 2.900 15. 901 0.0 0.0 5.800 909 14.0 0.0 5.800 16. 1001 0.0 0.0 7.975 1009 14.0 0.0 7.975 17. * 18. 110 -0.50 0.0 0.00; 111 14.50 0.0 0.0 19. 210 -0.50 0.0 1.450; 211 14.50 0.0 1.450 20. 310 -0.50 0.0 4.350; 311 14.50 0.0 4.350 21. 410 -0.50 0.0 7.250; 411 14.50 0.0 7.250 22. 510 -0.50 0.0 8.700; 511 14.50 0.0 8.700 23. * 24. 710 -0.50 0.0 0.725; 711 14.50 0.0 0.725 25. 810 -0.50 0.0 2.900; 811 14.50 0.0 2.900 26. 910 -0.50 0.0 5.800; 911 14.50 0.0 5.800 27. 1010 -0.50 0.0 7.975; 1011 14.50 0.0 7.975 28. * 29. MEM INCIDENCE 30. 101 101 102 108 31. 201 201 202 208 32. 301 301 302 308 33. 401 401 402 408 34. 501 501 502 508 35. * 36. 701 701 702 708 37. 801 801 802 808 38. 901 901 902 908 39. 1001 1001 1002 1008 40. * 41. 109 110 101; 110 109 111 42. 209 210 201; 210 209 211 43. 309 310 301; 310 309 311 44. 409 410 401; 410 409 411 45. 509 510 501; 510 509 511 46. * 47. 709 710 701; 710 709 711 48. 809 810 801; 810 809 811

NOTE NO. DN-18 SHEET NO.

49. 909 910 901; 910 909 911 50. 1009 1010 1001; 1010 1009 1011 51. * 52. 2101 101 701 2109 53. 2201 201 801 2209 54. 2301 301 901 2309 55. 2401 401 1001 2409 56. * 57. 3101 701 201 3109 58. 3201 801 301 3209 59. 3301 901 401 3309 60. 3401 1001 501 3409 61. * 62. MEMBER PROPERTIES 63. *DUMMY MEMBER 64. 101 TO 110 501 TO 510 PRIS YD 0.05 ZD 0.05 65. 701 TO 710 801 TO 810 901 TO 910 PRIS YD 0.05 ZD 0.05 66. 1001 TO 1010 PRIS YD 0.05 ZD 0.05 67. 209 210 309 310 409 410 PRI YD 0.05 ZD 0.05 68. * 69. 301 TO 308 401 TO 408 201 TO 208 - 70. PRIS AX 1.3645 IX 1E-10 IY .4673 IZ .2882 71. * DIAPHRAGM 72. 2201 TO 2301 BY 100 2209 TO 2309 BY 100 - 73. 3201 TO 3301 BY 100 3209 TO 3309 BY 100 - 74. PRIS AX 0.52932 IX 1E-10 IY 0.01429 IZ 0.05910 75. 2101 2401 2109 2409 - 76. 3101 3401 3109 3409 PRIS AX 0.17732 IX 1E-10 IY 0.009599 IZ 0.000715 77. * SLAB 78. 2102 TO 2108 2202 TO 2208 2302 TO 2308 2402 TO 2408 - 79. 3102 TO 3108 3202 TO 3208 3302 TO 3308 3402 TO 3408 - 80. PRIS AX 0.3850 IX 1E-10 IY 0.09826 IZ 0.001553 81. * 82. SUPPORTS 83. 201 301 401 PINNED 84. 209 309 409 FIXED BUT FX FZ MX MY MZ 85. * 86. CONSTANTS 87. E 3.0E6 ALL 88. * 89. DEFINE MOVING LOAD FILE DML.TXT 90. ** 91. TYP 1 CLA 1.0 92. TYP 2 CL70R 1.0 93. * 94. **** CASE 1 : CLASS 70R MOST ECCENTRIC 95. LOAD GENERATION 100 96. TYPE 2 -13.9 0.0 6.855 XINC 0.30

NOTE NO. DN-18 SHEET NO.

97. * 98. **** CASE 2 : CLASS 70R ONE WHEEL OVER G2 99. LOAD GENERATION 100 100. TYPE 2 -13.9 0.0 6.280 XINC 0.30 102. **** CASE 3 : CLASS 70R TRAIN SYMMETRIC TO G2 103. LOAD GENERATION 100 104. TYPE 2 -13.9 0.0 5.315 XINC 0.30 105. * 106. *** CASE 4 : CLASS A MOST ECCENTRIC 107. LOAD GENERATION 100 108. TYPE 1 -19.30 0.0 8.075 XINC 0.35 109. * 110. PERFORM ANALYSIS 111. LOAD LIST 50 51 52 112. PRINT SUPPORT REACTION LIST 409 JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z

409 50 .00 33.51 .00 .00 .00 .00 51 .00 34.54 .00 .00 .00 .00 52 .00 26.64 .00 .00 .00 .00

113. LOAD LIST 64 65 66 114. PRINT SUPPORT REACTION LIST 209 JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z

209 64 .00 -.99 .00 .00 .00 .00 65 .00 -1.09 .00 .00 .00 .00 66 .00 -.60 .00 .00 .00 .00

115. LOAD LIST 250 251 252 116. PRINT SUPPORT REACTION LIST 309 JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z

309 250 .00 46.17 .00 .00 .00 .00 251 .00 48.08 .00 .00 .00 .00 252 .00 35.48 .00 .00 .00 .00

117. LOAD LIST 200 201 202 118. PRINT SUPPORT REACTION LIST 309 JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z

309 200 .00 .00 .00 .00 .00 .00 201 .00 -.32 .00 .00 .00 .00 202 .00 -.13 .00 .00 .00 .00

NOTE NO. DN-18 SHEET NO.

119. FINISH

NOTE NO. DN-18 SHEET NO.

A. Design Data - Normal case

814880 N (refer calculation of bearing loads)

282480 N (refer calculation of bearing loads)

33333 N (refer calculation of bearing loads)

0 N

0.0022 refer (refer calculation of bearing loads)

3.50 mm

0 mmConcrete grade of pedestal M 30 Mpa

7.5 Mpa

10.00 MpaB. Bearing Data

380 mm

380 mmSide cover c 6 mm

10 mm

5 mm

Total no. of internal layers n 6

3 mm

368 mm

368 mm

Effective plan area A = l * b 135424Modulus of rigidity of ealstomer G 1.0 Mpa

3. Design of Bearing1. Check for base Pressure

6.02 Mpa < 10.00 Mpa O.K

2.09 Mpa > 2.00 Mpa O.K 2. Checks to be made if standard size is not used as per cl.916.3.3

1.000 < 2

70 mm 76

Horz. force in Long. dir. hl from supstr. =

Horz. force inTrans. dir. ht from supstr. =

Rotation in Long. dir.abd =

Rotation in Trans. dir.ald=

Translation in long. dir. Dbd =

Translation in Trans. dir. Dld =

Permissble stress in bearing so =

Increased permissible stress as per cl. 307.1 of IRC:21 subject to a maximum value f 10 Mpa as per cl. 916.3.5

Assuming that condition of required area will be satisfied

Overall length of bearing in trans. dir.lo

Overall width of bearing in long. dir. bo

Thickness of individual layer of elastomer hi

Thickness of top/bottom layer of elastomer he

should be hi/2 subject to max of 6mm

Thickness of steel laminate hs

Effective width b = bo -2c

Effective length l = lo -2c

mm2

Maximum Base pressure on pedestal sm(max)= Nmax/A

Minimum Base pressure on pedestal sm(min)=

Nmin/A

1. Ratio of length to width lo/bo

2. Height of elastomer h = n*hi+2*he £ bo/5 =

NOTE NO. DN-18 SHEET NO.

38

91 mm

9.20 should be >6 and <12 O.K

3. Check for translation as per cl. 916.3.4

1488.17582 N/mm

38542 N

0 N

0.33

0.00

0.33

O.K4. Check for Rotation as per cl. 916.3.5

0.0022

0.002

0.602

0.002206 0.005796 O.K

5. Check for Friction as per cl. 916.3.6

2.09 Mpa

0.335 0.409 O.K(calculated above)

6. Check for Total Shear stress as per cl. 916.3.7

Here,

0.98 Mpa

0.33 Mpa

³ bo/10 =

Also total ht. Of bearing ho = n*hi+2*he+ (n+1)*hs

3. Shape factor S = A/(l+b)*2*hi

Shear strain of bearing gd= gbd= Dbd/h+tmd or gd=(gbd2+gld

2)1/2 if gld is co-exixting

Total horizontal force on bearing as per clause 214.5.4 of IRC:6 = hl+ Vr * lic

where Vr is shear rating of elastomer bearing = G * A/ho

& lic is movement of bearing equal to Dbd or Dld for either direction

Total horz. Force in long. direction Hl = hl + Vr * Dbd

Total horz. Force in trans. direction Ht = ht + Vr * Dld

Shear strain in long. dir. gbd= Dbd/h+Hl/A

Shear strain in trans. dir. gld= Dld/h+Ht/A

Shear strain of bearing gd= gbd or (gbd2+gld

2)1/2 if gld is co-exixting = £ 0.7

As per provisions of this clause total angle of rotation ad £ b*abi,max*n

where ad = abd or incase of ald co-existing ad = (abd *b+ald *l)/b

abi,max = 0.5*sm*hi/b*S2 =

for which sm = 10 Mpa as per codal provision

b =sm (max)/10

Total angle of rotation ad = £ b*abi,max*n =

As per provisions of this clause total shear strain gd< 0.2 + 0.1 sm

where sm is minimum bearing presssure sm(min)= Nmin/A ³ 2 Mpa

Now shear strain gd= £ 0.2+0.1 sm=

As per provision of his clause tc + tg + ta £ 5 Mpa

tc is shear stress due to axial compression = 1.5 * sm(max) /S

tg is shear stress due to horizontal defromation = gd

ta is shear stress due to rotation=0.5*(b/hi)2*abi or ta = 0.5*(b2*abi+l2*ali)/hi2 where ali co-exists

NOTE NO. DN-18 SHEET NO.

0.002

0.000

1.09 Mpa

2.40 5.0 O.K

Check for pedestal sizeSide of pedestal in long. dir. 680Side of pedestal required in trans. dir. 680

114

now abi is as calculated earlier 0.5 * sm * hi / (b*S2)=

also ali is as given by 0.5 * sm * hi / (l*S2)=

Therefore ta = 0.5*(b2*abi+l2*ali)/hi2 =

Now total shear stress tc + tg + ta = £

307.1 of IRC:21

NOTE NO. DN-18 SHEET NO.

A. Design Data - Seismic trans. Case

550440 N (refer calculation of bearing loads)

284240 N (refer calculation of bearing loads)

16667 N (refer calculation of bearing loads)

33320 N

0.0017 refer (refer calculation of bearing loads)

3.50 mm

0 mmConcrete grade of pedestal M 30 Mpa

7.5 Mpa

10.00 MpaB. Bearing Data

380 mm

380 mmSide cover c 6 mm

10 mm

5 mm

Total no. of internal layers n 6

3 mm

368 mm

368 mm

Effective plan area A = l * b 135424Modulus of rigidity of ealstomer G 1.0 Mpa

3. Design of Bearing1. Check for base Pressure

4.06 Mpa < 10.00 Mpa O.K

2.10 Mpa > 2.00 Mpa O.K 2. Checks to be made if standard size is not used as per cl.916.3.3

1.000 < 2

70 mm 76

Horz. force in Long. dir. hl from supstr. =

Horz. force inTrans. dir. ht from supstr. =

Rotation in Long. dir.abd =

Rotation in Trans. dir.ald=

Translation in long. dir. Dbd =

Translation in Trans. dir. Dld =

Permissble stress in bearing so =

Increased permissible stress as per cl. 307.1 of IRC:21 subject to a maximum value f 10 Mpa as per cl. 916.3.5

Assuming that condition of required area will be satisfied

Overall length of bearing in trans. dir.lo

Overall width of bearing in long. dir. bo

Thickness of individual layer of elastomer hi

Thickness of top/bottom layer of elastomer he

should be hi/2 subject to max of 6mm

Thickness of steel laminate hs

Effective width b = bo -2c

Effective length l = lo -2c

mm2

Maximum Base pressure on pedestal sm(max)= Nmax/A

Minimum Base pressure on pedestal sm(min)=

Nmin/A

1. Ratio of length to width lo/bo

2. Height of elastomer h = n*hi+2*he £ bo/5 =

NOTE NO. DN-18 SHEET NO.

38³ bo/10 =

NOTE NO. DN-18 SHEET NO.

91 mm

9.20 should be >6 and <12 O.K

3. Check for translation as per cl. 916.3.4

1488.17582 N/mm

21875 N

33320 N

0.21

0.25

0.32

O.K4. Check for Rotation as per cl. 916.3.5

0.0017

0.002

0.406

0.001686 0.003915 O.K

5. Check for Friction as per cl. 916.3.6

2.10 Mpa

0.32 0.41 O.K(calculated above)

6. Check for Total Shear stress as per cl. 916.3.7

Here,

0.66 Mpa

0.32 Mpa

0.002

Also total ht. Of bearing ho = n*hi+2*he+ (n+1)*hs

3. Shape factor S = A/(l+b)*2*hi

Shear strain of bearing gd= gbd= Dbd/h+tmd or gd=(gbd2+gld

2)1/2 if gld is co-exixting

Total horizontal force on bearing as per clause 214.5.4 of IRC:6 = hl+ Vr * lic

where Vr is shear rating of elastomer bearing = G * A/ho

& lic is movement of bearing equal to Dbd or Dld for either direction

Total horz. Force in long. direction Hl = hl + Vr * Dbd

Total horz. Force in trans. direction Ht = ht + Vr * Dld

Shear strain in long. dir. gbd= Dbd/h+Hl/A

Shear strain in trans. dir. gld= Dld/h+Ht/A

Shear strain of bearing gd= gbd or (gbd2+gld

2)1/2 if gld is co-exixting = £ 0.7

As per provisions of this clause total angle of rotation ad £ b*abi,max*n

where ad = abd or incase of ald co-existing ad = (abd *b+ald *l)/b

abi,max = 0.5*sm*hi/b*S2 =

for which sm = 10 Mpa as per codal provision

b =sm (max)/10

Total angle of rotation ad = £ b*abi,max*n =

As per provisions of this clause total shear strain gd< 0.2 + 0.1 sm

where sm is minimum bearing presssure sm(min)= Nmin/A ³ 2 Mpa

Now shear strain gd= £ 0.2+0.1 sm=

As per provision of his clause tc + tg + ta £ 5 Mpa

tc is shear stress due to axial compression = 1.5 * sm(max) /S

tg is shear stress due to horizontal defromation = gd

ta is shear stress due to rotation=0.5*(b/hi)2*abi or ta = 0.5*(b2*abi+l2*ali)/hi2 where ali co-exists

now abi is as calculated earlier 0.5 * sm * hi / (b*S2)=

NOTE NO. DN-18 SHEET NO.

0.000

1.09 Mpa

2.07 5.0 O.K

Check for pedestal sizeSide of pedestal in long. dir. 680Side of pedestal required in trans. dir. 680

114

also ali is as given by 0.5 * sm * hi / (l*S2)=

Therefore ta = 0.5*(b2*abi+l2*ali)/hi2 =

Now total shear stress tc + tg + ta = £

307.1 of IRC:21

NOTE NO. DN-18 SHEET NO.

A. Design Data - Seismic long. Case

550440 N (refer calculation of bearing loads)

284240 N (refer calculation of bearing loads)

41612 N (refer calculation of bearing loads)

0 N

0.0017 refer (refer calculation of bearing loads)

3.500 mm

0 mmConcrete grade of pedestal M 30 Mpa

7.5 Mpa

10.00 MpaB. Bearing Data

380 mm

380 mmSide cover c 6 mm

10 mm

5 mm

Total no. of internal layers n 6

3 mm

368 mm

368 mm

Effective plan area A = l * b 135424Modulus of rigidity of elastomer G 1.0 Mpa

3. Design of Bearing1. Check for base Pressure

4.06 Mpa < 10.00 Mpa O.K

2.10 Mpa > 2.00 Mpa O.K 2. Checks to be made if standard size is not used as per cl.916.3.3

1.000 < 2

70 mm 76

Horz. force in Long. dir. hl from supstr. =

Horz. force inTrans. dir. ht from supstr. =

Rotation in Long. dir.abd =

Rotation in Trans. dir.ald=

Translation in long. dir. Dbd =

Translation in Trans. dir. Dld =

Permissble stress in bearing so =

Increased permissible stress as per cl. 307.1 of IRC:21 subject to a maximum value f 10 Mpa as per cl. 916.3.5

Assuming that condition of required area will be satisfied

Overall length of bearing in trans. dir.lo

Overall width of bearing in long. dir. bo

Thickness of individual layer of elastomer hi

Thickness of top/bottom layer of elastomer he

should be hi/2 subject to max of 6mm

Thickness of steel laminate hs

Effective width b = bo -2c

Effective length l = lo -2c

mm2

Maximum Base pressure on pedestal sm(max)= Nmax/A

Minimum Base pressure on pedestal sm(min)=

Nmin/A

1. Ratio of length to width lo/bo

2. Height of elastomer h = n*hi+2*he £ bo/5 =

NOTE NO. DN-18 SHEET NO.

38³ bo/10 =

NOTE NO. DN-18 SHEET NO.

91 mm

9.20 should be >6 and <12 O.K

3. Check for translation as per cl. 916.3.4

1488.17582 N/mm

41612 N

0 N

0.36

0.00

0.36

O.K

4. Check for Rotation as per cl. 916.3.5

0.001686

0.002

0.406

0.0017 0.003915 O.K

5. Check for Friction as per cl. 916.3.6

2.10 Mpa

0.357 0.410 O.K(calculated above)

6. Check for Total Shear stress as per cl. 916.3.7

Here,

0.66 Mpa

0.36 Mpa

Also total ht. of bearing ho = n*hi+2*he+ (n+1)*hs

3. Shape factor S = A/(l+b)*2*hi

Shear strain of bearing gd= gbd= Dbd/h+tmd or gd=(gbd2+gld

2)1/2 if gld is co-exixting

Total horizontal force on bearing as per clause 214.5.4 of IRC:6 = hl+ Vr * lic

where Vr is shear rating of elastomer bearing = G * A/ho

& lic is movement of bearing equal to Dbd or Dld for either direction

Total horz. Force in long. direction Hl = hl + Vr * Dbd

Total horz. Force in trans. direction Ht = ht + Vr * Dld

Shear strain in long. dir. gbd= Dbd/h+Hl/A

Shear strain in trans. dir. gld= Dld/h+Ht/A

Shear strain of bearing gd= gbd or (gbd2+gld

2)1/2 if gld is co-exixting = £ 0.7

As per provisions of this clause total angle of rotation ad £ b*abi,max*n

where ad = abd or incase of ald co-existing ad = (abd *b+ald *l)/b =

abi,max = 0.5*sm*hi/b*S2 =

for which sm = 10 Mpa as per codal provision

b =sm (max)/10

Total angle of rotation ad = £ b*abi,max*n =

As per provisions of this clause total shear strain gd< 0.2 + 0.1 sm

where sm is minimum bearing presssure sm(min)= Nmin/A ³ 2 Mpa

Now shear strain gd= £ 0.2+0.1 sm=

As per provision of his clause tc + tg + ta £ 5 Mpa

tc is shear stress due to axial compression = 1.5 * sm(max) /S

tg is shear stress due to horizontal defromation = gd

ta is shear stress due to rotation=0.5*(b/hi)2*abi or ta = 0.5*(b2*abi+l2*ali)/hi2 where ali co-exists

NOTE NO. DN-18 SHEET NO.

0.002

0.000

1.09 Mpa

2.11 5.0 O.K

Check for pedestal sizeSide of pedestal in long. dir. 680Side of pedestal required in trans. dir. 680

114

now abi is as calculated earlier 0.5 * sm * hi / (b*S2)=

also ali is as given by 0.5 * sm * hi / (l*S2)=

Therefore ta = 0.5*(b2*abi+l2*ali)/hi2 =

Now total shear stress tc + tg + ta = £

307.1 of IRC:21

hi hs8 3 Print out not to be taken

10 312 416 6

grade stress20 525 6.230 7.535 8.540 8.5