Challenge 6
Challenge-6
Ans of Question-1
Requirement of dimensioning of elastomeric bearings
|
Overall width of bearing longitudinal direction |
a |
= |
420 |
Mm |
|
Overall width of bearing transverse direction |
b |
= |
420 |
Mm |
|
Thickness of individual layers of elastomer |
ti |
= |
12 |
Mm |
|
Thickness of top/bottom layers of elastomer |
te |
= |
6 |
Mm |
|
Total no of internal layers |
n |
= |
5 |
|
|
Thickness of steel laminate |
ts |
= |
4 |
Mm |
|
Total plan area, A=a*b |
A |
= |
176400 |
Mm2 |
|
Effective plan area, A’=(a-u)*b |
A’ |
= |
164496.35 |
Mm2 |
Ans of Question-2
Design of elastomeric bearings
- Mean pressure - Serviceability state (cl 4.3.3.1):
Sigma m =P/A’<=11 MPA
So we know, P= Pc + Ps = 585544 N & A’=169320.25 mm2
Sigma m =585544/169320.25=3.45 MPA, which is less than 11 MPA.
(Condition verified)
- No slip serviceability state (cl 4.3.3.2):
B1
Sigma m min=Pm min/A >=2MPA
Sigma m min= 954324/176400=5.41 MPA, which is larger than 2 MPA.
(Condition verified)
B2
H < f*P
Where, f= friction, H= horizontal forces, p= vertical forces,
F = 0.12 + 0.2/ sigma m
For slow loads-
T > G*Uc * A / (0.12 Pc + 0.2 A)
Where, T= total translation
T= 0.9*16.86*176400 / (0.12*585544+0.2*176400)
=25.36, which is lesser than T
For rapid loads-
T> G*A(Uc+2Us) / (0.12P+ 0.2A)
T=0.9*176400(16.86+2*0) / (0.12*585544 + 0.2*176400)
=25.36, which is lesser than T.
- Limitation of distortion- serviceability state (clause 4.3.3.3):
T> (Uc+ Us)/ 0.7
=16.86+0/0.7
=24.08, which is lesser than T
- No uplift at the edge with the least load- fracture state (clause 4.3.3.4):
D1
Tanαc<6 summation of eic/a
For internal layer
Si= ab/2ti(a+b)
=420*420 / 2*12(420+420)
=8.75
Sigma nc=Pc/A’
=4.66 MPA
Eic1=ti sigma mc / (4GSi^2 + 3 sigma mc)
=0.97 mm
For top & bottom layer,
Si= a*b / 2te (a + b)
=420*420/ 2*6 (420 + 420)
=17.5
Eic2=te sigma mc / (4 GSi^2+3 sigma mc)
=0.05
Summation of Eic= Eic1+Eic2=0.97+0.05+1.02
6*summation of Eic/a=6*1.02/420=0.01457
Tanαc= (αc+αo) = (0.01+0) =0.01 I,e <0.01457
(Condition verified)
D2
Tanαc+1.5tanαs<6summation ei/a
Sigma m=(P0+Pa)/A’
=4.66MPA
For internal layers, ei1=ti*sigma m/ (4GSi^2+3 sigma m)
=0.97
For top and bottom layer, ei2=te*sigma m/ (4GSi^2+3 sigma m)
=0.05
6*summation of Eic/a=6*1.02/420=0.01457
Tanαc=(αc+α0) =0.01
Tanαs=(αs+α0) =0
Tanαc+1.5tanαs=0.01, I,e <0.01457
(Condition verified)
shear and strength of plates
- No buckling (clause 4.3.3.5):
a/10<=T<=a/5
a/10=420/10=42 mm
T=72 mm
a/5=420/5=84 mm
(Condition verified)
- shear serviceability state (clause 4.3.3.6):
tP+ th+ ts < 5G
tp=1.5(Pc+1.5Ps)/(Si*a*b)
=1.5(585544+1.5*0) / (8.75*420*420)
=0.60 MPA
th = (G*Uc + Gs*us)/T
= (0.90*16.86 + 1.8*0)/72
= 0.21 MPA
ts= G*a^2(tanαc+1.5tanαs/2Tti
= 0.9*420^2(0.01+0)/2*72*12
= 0.92
tp+th+ts= (0.60+0.21+0.92) =1.73 MPA
5G=5*0.9=4.5 MPA, which is gratter than 1.73 MPA.
(Condition verified)
- strength of plates fracture state (clause 4.3.3.7):
Ts>(a/si) * (sigma m/ sigma s) * gs
a/si=420/8.75=48 mm
sigma m/ sigma s=4.66/240=0.0194
gs= 1.20
(a/si) * (sigma m/ sigma s) * gs
=48*0.0194*1.20
=1.11744 mm
ts= 4 mm, which is gratter than 1.11744
(Condition verified)