ELASTOMERIC BEARING DESIGN Spreadsheet_Theengineeringcommunity.xlsx

	A	B	C	D	E	F	G	H	I
1		ELASTOMERIC BEARING DESIGN
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4									AASHHTO LRFD 2007
5									METHOD B
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19	System and material input data :
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21	Expandable span length						Ls =		mm
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23	Constant amplitude fatigue threshold for Category A						DFt =	165	Mpa
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25	Elastomer hardness:						Hshore =	50
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27	Shear modulus of elastomer { (0,68 - 0,93) SELECT }						G =	1	Mpa
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29	Steel reinforcement yield strength:						fy =	240	Mpa
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31	Pad length (bridge longitudinal direction):						Lpad =	250	mm
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33	Pad width (bridge transverse direction):						Wpad =	250	mm
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35	Elastomer cover thickness:						hc =	2,5	mm
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37	Elastomer internal layer thickness:						hri =	8	mm
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39	Number of steel reinforcement layers:						Nst =	4
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41	Steel reinforcement thickness:						hs =	3	mm
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43	System and material output data :
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45	Elastomer creep deflection at 25 years divided by the instantaneous deflection:						Cd =	0,25
46	Number of elastomer internal layers						Nel =	3
47	total elastomer thickness						hrt =	29	mm
48	Total steel plate heigth						hst =	12	mm
49	Total bearing heigth						ht =	41	mm
50	Bearing surface area						Area =	62500	mm2
51	Check Nst (14.7.6.1) :
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53	Nst =	4				Nst > 2 ise;
54	hc =	2,5
55	0.70 hri =	5,6				hc ≤ 0.70 hri		OK.
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57	Compute Shape Factor (14.7.5.1-1) :
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59	Sint =	7,8125				Si = L.W / (2.hri.(L+W))
60	Scov =	25				Si = L.W / (2.hc.(L+W))
61	S =	7,8125				S=min(Sint,Scov)
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63	Check Compressive Stress (14.7.5.3.2) :
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65	DLs =	106	kN			DL reaction/girder
66	LLs =	111	kN			LL reaction /girder
67	σs =	3,472	MPa			σs = (DLs+LLs) /Area
68	σL =	1,776	MPa			σL = LLs / Area
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70	Shear deformation? -YES- (14.7.5.3.2-2) :
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72	1.66 G.S =	12,96875	Mpa			σs ≤ 1,66 G.S		OK.
73	0.66 G.S =	5,15625	Mpa			σs ≤ 11		NO
74						σL ≤ 0,66 G.S		OK.
75	Shear deformation? -NO- (14.7.5.3.2-4) :
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77	G.S =	7,8125	Mpa			σs ≤ 2 G.S		OK.
78	2G.S =	15,625	Mpa			σs ≤ 12		OK.
79						σL ≤ G.S		OK.
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81	Check Compressive Deflection (14.7.5.3.3) :
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83	εi =	0,0309284016		durometer	50	60	70
84	δLi =	0,2474272128	mm	C	0,01675950175	0,01815612689	0,01396625146
85	δLt =	0,8969236465	mm	x	0,2980503409	0,2725697913	0,3111086313
86	δcr =	0,007732100401	mm	εi = Cs^x	0,0309284016	0,01696858222	0,01292851005
87	Σδ =	0,9046557469	mm
88	0.07 hri =	0,56	mm			δLi ≤ 0.07 hri		OK.
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90	Check Shear Deformation (14.7.5.3.4) :
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92	α =	0,0000117	/°C
93	tset =	20	°C
94	γTU =	1,2
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96	Δco =	0	mm			Δco = α . tset . Ls
97	Δs =	0	mm			Δs = Δco . γTU
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99	2.Δs =	0	mm			2. Δco ≤ hrt		OK.
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