LRB = Lead Rubber Bearing

1. LRB (Lead Rubber Bearing)
"by : Tony Hartono Bagio, PhD"
Building Details
Lokasi :
Plan
x = 8.5 m
y = 13.5 m
z = 37.2 m
na = 11
nb = 1
za = 3.15 m
zb = 2.55 m
z total = (11*3.15 + 1*2.55)
z total = 37.2 m

2. Material Properties
fc' = 25 Mpa
fy = 500 Mpa
Beam
b = 230 mm
h = 530 mm
Slab
t = 150 mm
Gravity Load
γConcrete = 25 kN/m³
Slab = 0.15*8.5*13.5*25
Slab = 430.3125 kN
DL Slab = 3.75 kN/m²
Design parameter
Design LRB
SM1 = 0.36
SD1 = 0.54
PuColMax = W = 1825.5 kN

3. Column Size
b = 300 mm
h = 900 mm
g = 9.81 m/sec²
β = 5 % Effective Dumping
G = 0.7 Mpa Shear Modulus
K = 2000 MPa Bulk modulus
n = 0.1 Post Yield Stiffness ratio, n = Kd/Ku

4. 1. Calculate design displacement , Dd
Assume Design Time Period
Td =Tmax = 2.5 sec
Bd = -0.0000006*β^4 + 0.0000678*β^ 3 -0.0029788*β^ 2 + 0.0788416*β + 0.6590773
note : β in % (not decimal); β = 5
Bd = 1
Dd = g/(4 p)² * SD1 * Td/Bd
Dd = 0.335461774 m
2. Effective stiffness , Keff
Keff = W/g * (2p/Td)²
Keff =1825.5/9.81 * (2*3.14/2.5)²
Keff = 1175.418574 kN/m
3. Energy dissipated per energy cycle, Wd
Wd = 2 π * Keff * Dd² * β
Wd = 2*3.14* 1175.42 * 0.335² * 0.05
Wd = 41.55549886 kNm

5. 4 Force at design displacement, Qd
Qd = Wd/(4*Dd)
Qd = 41.555/(4*0.335)
Qd = 30.9688779 kN
5 Stiffness in rubber
Kd = Keff - (Qd/Dd)
Kd = 1175.42 - (30.97/0.335)
Kd = 1083.1 kN/m
6 Yield displacement (Dγ)
Dγ = Qd/(Ku - Kd)
n = Kd/Ku
Ku = Kd/n
Ku = 1083.1/0.1 = 10831 kN/m
Dγ = 30.97/(9*1083.1)
Dγ = 0.003176979 m

6. 7 Yield Strength, Fy
Fy = Qd + Kd*Dy
Fy = 30.969 + 1083.1*0.0032 = 34.41 kN
8 Recalculation of force Q to Qr
Qr = Wd/[4(Dd-Dγ)]
Qr = 41.555/[4*(0.335 - 0.003)] = 31.26497176 kN
9 Calculation of area and diameter of lead plug
Yield strength of lead, FyL
FyL = 10 MPa = 10000 kPa
The area of lead plug needed
AL = Qr/Fs = 31.265 / 10000
AL = 0.003126497 m²
dL = √(4/ p * AL)
dL = √(4/ p * 0.003126497) = 0.063093422 m
dL = 65 mm (diameter Lead core)

7. 10 Revising rubber stiffness Keff to Krub
Krub = Keff - Qr/Dd
Krub = 1175.42 - 31.26/0.335 = 1082.21877 kN/m
Total thickness of rubber layer tγ = Dd/γ
Where γ = 100.00% (maximum shear strain of rubber)
tγ = Dd/γ = 0.335/1 , tγ = 0.335461774 m
11 BEARING
Area of bearing
ALRB = Krub * tγ / G
ALRB = 1082.219 * 0.335 / 700 =0.518632897 m²
Diameter of bearing
d = √(4/p * ALRB)
d = √(4/3.14 * 0.519) = 0.813 = 815 mm

8. 12 Shape Factor
S = (1/2.4) * (fV/ fH)
fH = 1/Td = 1/2.5 = 0.4 Hz horizontal frequency
fV ≈ 20 * fH = 8 Hz vertical frequency
S = (1/2.4) * (fV/fH) = (1/2.4) * (8/0.4)
S = 8.333333333
Rubber Layers
tr = d/4 S = 815/(4*8.3333)
tr = 24.45 ≈ 25 mm /layer
Number of rubber layers , Nn
Nn = tγ/t = 0.335/0.025
Nn = 13.41847096
Nn = 14 layers

9. 13 Dimensions of lead rubber bearing
Thickness of shim plates , tsh
tsh = 2.8 mm
Number of shim plates , nsh
nsh = Nn – 1 = 14 – 1 = 13
Thickness of Sealing plate, tplate
tplate = 25 mm
Total height of LRB, hLRB
hLRB = Nn*tr + nsh*tsh + 2*tplate
hLRB = 14*25 + 13*2.8 + 2*25 = 436.4 mm = 0.436 m
Cover Rubber, cR
cR = 25 mm
Bond Diameter, dB
dB = d - 2*cR = 815 - 2*25 = 765 mm

10. 14 End Plate (upper & bottom plate)
Thickness of end plate, tep ≥ tplate
tep = 25 mm
Lep ≥ d + 2*100 = 815 + 2*100 ≈ 1100 mm
Size of End plate = 1100 x 1100 x 25
15 Compression Modulus Ec
Ec = 6*G*S²*[1 - (6*G*S²/K)]
Ec = 6*0.7*8.333²*[1 - ( 6*0.7*8.333²/2000)] = 249.132 Mpa
16 Horizontal stiffness
KH = G * ALRB/tγ
KH = 700* 0.519/0.335 = 1082.21877 kN/m
17 Vertical stiffness
Kv = Ec * ALRB/tγ
Kv = 249132 * 0.519/0.335 = 385164.7522 kN/m