Challenge

Challenge 6

Aim:-

1) List the requirements for dimensioning of Elastomeric Bearing. Also, mention the notations: “a”, “b”, “ti”, “te”, “n” and “ts”

2) Design an elastomeric bearing considering the following design parameters and design data:

Design Parameter
Span Length	L	=	27	m
Bearing to Bearing Length	Lo	=	25	m
Young's modulus of concrete for M60	E	=	36000	Mpa
Static Shear modulas of elastomer	G	=	0.9	Mpa
Dynamic Shear modulus of elastomer	Gs		1.8	Mpa
Inertia os section ( I Beam only ) - At mid span	I	=	0.27835	m4
Inertia os section ( I Beam+Slab ) - At mid span	I	=	0.62468	m4
Max reaction due to total dead load ( From Grillage Model)	PDL	=	51	T
Max reaction due SIDL ( From Grillage Model)	PSIDL	=	39	T
Max reaction due Live Load + Impact ( From Grillage Model)	PLL	=	33	T
Maximum rotation due to live load ( From Grillage Model)	αLL	=	0.002	rad
Rotation due to ( DL + Slab + X-girder) ( From Grillage Model ).	αDL	=	0.004	rad
Rotation due to ( SIDL ) ( From Grillage Model ).	αSIDL	=	0.003	rad
Creep Coefficient	ecr	=	4.30E-05	/Mpa
Translation due to Creep	ucr	=	7.28	mm
Shrinkage coefficient	eSH	=	3.00E-04
Translation due to shrinkage	USH	=	4.05	mm
Coefficient of thermal expansion	eT	=	1.17E-05
Translation due to thermal expansion	UT	=	5.53	mm
Translation due to BR/TR= F/K ( K = Stiffness of bearing = G*A/T)	UBR/TR	=	11.33	mm
Max translation (+) = UT + UBR/TR	Umax (+)	=	16.86	mm
Max translation (+) = Ucr + USH + UT + UBR/TR	Umax (-)	=	28.19	mm

Design Data
Dead Load	PDL	=	440782	N
SIDL	PSIDL	=	144762	N
Total Slow Vertical Load	PC	=	585544	N
Rapid Max Vertical Load	Psmax	=	0.00	N
Rotation in Longitudinal direction. due to slow loads	αc	=	0.01	rad
Rotation in Longitudinal direction. due to rapid loads	αs	=	0.00	rad
Rotational due to position tolerance	αo	=	0.00	rad
Translation in Longitudinal direction. due to slow loads	UC	=	16.86	mm
Translation in the Longitudinal direction. due to rapid loads	US	=	0.00	mm
Total Translation in Longitudinal dir.	U	=	16.86	mm

Dimensioning of Bearing
Overall width of bearing in longitudinal direction	a	=	420	mm
Overall width of bearing in Translation direction	b	=	420	mm
Thickness of Individual layer of elastomer	ti	=	12	mm
Thickness of top/bottom layer of elastomer	te	=	6	mm
Total number of internal layer	n	=	5
Thickness of steel laminate	ts	=	4	mm
Total plan area, A=a*b	A	=	176400	mm2
Effective plan area, A' = (a-U) x b	A'	=	169318.8	mm2
Total Thickness of Elastomer Bearing	T	=	72	The factor

Factor of sefserviceability
Factor for servicibility state (SLS)
Dead Load		=	1
SIDL		=	1.2
Live Load		=	1
Factors for fracture state (ULS)
Dead Load		=	1.2
SIDL		=	1.8
Live Load		=	1.3
Partial safety factor of steel at Limite State	gs	=	Yield
Yeild Strength of Steel	σa	=	240	Mpa
Max. Permissible compressive stress in bearing	σo max	=	11	Mpa
Min. Permissible compressive stress in bearing	σo min	=	2	Mpa

DESIGN OF BEARING

A. Mean PressurServiceabilitylity State (Cl.4.3.31)
σm = P/A' ≤ 11 Mpa	=	3.458233817	Mpa
Where
P = PC + PS

CONDITION VERIFIED

B. No SliServiceabilitylity State ( Cl.4.3.3.2)
B1 σm mini = Pmini / A ≥ 2.0 Mpa		=	3.319410431	Mpa



B.2 H<f*p
Enter the type of contact of bearing face with the structure
f=0.12+0.2/Sigma m
For Slow loads:
T>(G AUc) / (0.12 Pc + 0.2A)	T	>	25.36061868	mm
For Rapid Loads:
T>GA(Uc + 2Us) / (0.12 Pc + 0.2A)	T	>	25.36061868	mm

CONDITION VERIFIED

C. Limitation of Distortion - Serviceability StatClauseasue 4.3.3.3)
tH / G ≤ 0.7
or T > ( Uc +Us) / 0.7	T	>	24.08571429	mm

CONDITION VERIFIED

Fracture State

D. No uplift at the edge with the least load - Fracture load ( Clause 4.3.3.4)
D.1 tanα < 6∑ eic / a
*For Internal layer*
Si = ab / 2ti (a+b)	Si	=	8.75
σmc = Pc / A'	σmc	=	3.458233817	Mpa
eic1 = tiσmc/(4GSi2 + 3σmc)	∑eic1	=	0.145100871	mm
*For top & bottom*
Si = ab / 2te (a+b)	Si	=	17.5
eic2 = tiσmc/(4GSi2 + 3σmc)	∑eic2	=	0.037289738
Total eic ( ∑eic1 + ∑eic2)	∑eic	=	0.182390609

Therefore	6∑eic/a	=	0.00260558
Tanαc = (αc + αo)	tanαc	=	0.01

CONDITION VERIFIED

D.2 tan αc + 1.5 tan αs < 6∑e1 / a
αm = ( Pc + Ps ) / A'	σm	=	3.458233817	Mpa
For internal layer ei1 = tiσmc/(4GSi2 + 3σm)	∑ei1	=	0.145100871	mm
For top & bottom layer ei2 = tiσmc/(4GSi2 + 3σm)	∑ei2	=	0.037289738
	6∑ei/a	=	0.013027901
tanαc + 1.5 tanαs		=	0.01

CONDITION VERIFIED

F. Shear - Servicebility State ( Clause 4.3.3.6)
tp + th + ta < 5G
tp = 1.5 ( PC + 1.5 Ps ) /(Si a b)	tp	=	0.569041788	Mpa
tH = ( GUC +GsUs) / T	tH	=	0.21075	Mpa
ta = G a2 ( tanαc +1.5tanαs) / (2 T ti )	ta	=	0.91875	Mpa
tp+tH+ta		=	1.698541788	Mpa
	5G	=	4.5	Mpa

CONDITION VERIFIED

G. Strenght of Plate - Fracture State ( Clause 4.3.3.7)
ts > a/Si . Sm/Sa. Gs
	a/Si	=	48	mm
	σm / σs	=	0.014409308
	gs	=	1.2
a/Si.Sm/Sa.gs		=	0.829976116	mm
	ts	=	4	mm
CONDITION VERIFIED

Ship work like this
in your own course.