Seismic Design of Geosynthetic-Reinforced Soil Bridge Abutments with Modular Block Facing (2012)

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CHAPTER 4 BEARING PAD DESIGN INTRODUCTION The design of the bearing pads for the shake table experiment was based on Method B from AASHTO LRFD Bridge Design Specifications (2007). The bearing pads chosen to support and transfer vertical and horizontal loads from the bridge superstructure to the substructure are 305 mm x 457 mm x 52 mm steel reinforced elastomeric pads. The steel reinforced elastomeric bearing pad type was chosen based on its ability to be extremely forgiving of loads and translations exceeding those considered in design as well as being the preferred bearing type by numerous departments of transportation in seismic areas. The elastomeric bearing pads are vulcanized to top and bottom steel plates. The bottom steel plate or sole plate is mechanically connected to the bridge sill using two 25.4 mm dia. anchor bolts cast in the sill. The MC10x28.5 bridge girders are bolted directly to the steel top plate using (6) – 19.05 mm dia. threaded studs. BEARING PAD DESIGN Applied Forces Total bridge weight: DL + LL = 445 + 0 = 445 kN (100,000 lb) Dead load reaction: RDL = 445/4 = 111.25 kN Horizontal force per girder: FS = (445/2) x 0.2 = 44.5 kN Where 0.2 is the horizontal ground acceleration It should be noted that each of the two bridge girders are supported by an elastomeric bearing pad at one end and a roller (slide bearing) at the other. The rollers do not substantially resist horizontal forces; therefore, the bearings are designed to withstand the full bridge horizontal inertial force.

89 Design Calculations Trial Pad Initially, the dimensions of the steel reinforced elastomeric pad are assumed to be 305 mm x 457 mm x 52 mm (See Figures 4.1 and 4.2) The shape factor of a layer of an elastomeric bearing, Si, shall be taken as the plan area of the layer divided by the area of perimeter free to bulge. ( )WLh LW S ri i + = 2 (Eqn.14.7.5.1-1) Where: L = Length of rectangular elastomeric bearing (parallel to longitudinal bridge axis) W = Width of the bearing in the transverse direction hri = Thickness of ith elastomeric in layer in elastomeric bearing ( ) 4.645730529.142 457305 = +× × =iS Shear Modulus, G = 689 kPa for 50 hardness durometer 552 kPa < G = 689 kPa < 1207 kPa Okay (Article 14.7.5.2) The shear modulus for this bearing pad was given by the manufacturer, Tobi Engineering, Inc. Compressive Stress For bearings subject to shear deformation, the average compressive stress at the service limit shall satisfy Eqn. 14.7.5.3.2-1 in any elastomeric layer. 031,1166.1 ≤≤ GSsσ kPa (1.6 ksi) (Eqn. 14.7.5.3.2-1) Where: sσ = Service average compressive stress due to the total load G = Shear modulus of elastomer S = Shape factor of the thickest layer in the bearing 798 457.0305.0 025.111 = × + = + = A LLDL sσ kPa 73204.668966.166.1 =××=iGS kPa 11031kPa732066.1kPa798 <=<= GSsσ kPa Okay

90 Initial Dead Load Compressive Deflection ∑= ridid hεδ (Eqn.14.7.5.3.3-2) ( ) 20.029.14 4.66896 798 3 6 3 22 =      ×× =     = rid hGS σδ mm Where: diε = Initial dead load compressive strain in the ith elastomer layer of a laminated bearing 26GSdi σε = (Eqn. C14.7.5.3.3-1) σ = Instantaneous dead load compressive stress in an individual layer of a laminated bearing rih = thickness of ith elastomeric layer in a laminated bearing Maximum Shear Force at Slippage Using a shear modulus, G, given by the manufacturer of 689 kPa (100 psi) for 50 hardness durometer, the design shear force can be calculated as follows: rt s Sdesign h AG F Δ×× = Where: A = Plan area Δs = Maximum total shear deformation of the elastomer at the service limit state hrt = Total elastomer thickness Without knowing actual deflections, the design shear force, FSdesign, is calculated using the maximum allowable deflection of the pad which is given as half the thickness of the pad or hrt/2 given by Eqn. 14.7.5.3.4-1. ( ) 0.48 2 457.0305.06892/ = ×× = ×× = rt rt Sdesign h hAG F kN Applied shear force, FS = 44.5 kN < FSdesign = 48.0 kN Okay

91 Although the maximum allowable deflection of half the pad thickness was used in calculating the design shear force, laboratory tests reviewed show negligible damage to elastomeric bearings translated 100 percent of their design thickness (100 percent shear strain). Combined Compression and Rotation                    −< 2 200.01875.1 ri s s h B n GS θ σ (Eqn. 14.7.5.3.5-2) Where: n = Number of interior layers of elastomer θs = Maximum service rotation due to the total service load (rad.) hri = thickness of ith elastomer layer σs = Stress in elastomer B = Width of pad 661,6 29.14 305 3 0064.0 200.014.6689875.1kPa798 2 =                    −××< kPa Okay Stability of Elastomeric Bearings Bearings satisfying Eqn. 14.7.5.3.6-1 shall be considered stable and no further investigation is required. BA2 ≤ (Eqn. 14.7.5.3.6-1) in which: W L L hrt 0.2 1 92.1 A + = (Eqn. 14.7.5.3.6-2) ( )       ++ = W L S 0.4 10.2 67.2 B (Eqn. 14.7.5.3.6-3) Where: G = Shear modulus of elastomer L = Length of a rectangular elastomeric bearing (parallel to longitudinal bridge axis)

92 W = Width of the bearing in the transverse direction 067.0 457 3050.2 1 305 29.14 92.1 A = × + = ( ) 272.0 4570.4 305 10.24.6 67.2 B =       × ++ = 272.0B134.0A2 =≤= Okay, bearings are considered stable; no further investigation of stability is required. Reinforcement At the service limit state: y s s F h h σmax3≥ (Eqn. 14.7.5.3.7-1) Where: hmax = Thickness of thickest elastomer layer 14.0 248211 79829.143 29.14 = ×× ≥=sh mm Okay Use Illinois DOT Type 1, 12-a. Specifications for bearing pad given in Figure 3.7.4-21 from Page 3-273 Illinois DOT Bridge Manual (See Appendix B) Anchor Bolt Design Given: Factored shear force per girder, FSU ( ) AWF EQSU ××= 2/γ ( ) 5.4420.02/4450.1 =××=SUF kN Allowable shear force per bolt, F ubFAF 48.0φ= Where:

93 =φ 0.75 to nominally account for tension ksi)(60kPa685,413=uF for F1554 Gr.36 anchor bolt 46.75685,413 4 0254.0 48.075.0 2 =×      × ××= π F kN →=<= kN46.75kN5.44 FFSU Use 2 anchor bolts per bearing pad Top Bearing Plate Design Reference Figure 3.7.4-19 on Page 3-271 Illinois DOT Bridge Manual (See Appendix B) Given: Fy = 248,211 kPa (36 ksi)→ C = 0.183 Top plate reaction, R 25.1114/445 ==R kN (24.25 kips) Elastomeric pad length, Le 457=eL mm Top bearing plate width, Wt 356=tW mm (14 in) Top bearing plate thickness, Tt in1.02 14 1824.25 0.183 = × = × = t e t W LR CT = 25.9 mm Minimum Tt = 38.1 mm (1 ½ in) Use minimum, Tt = 38.1 mm Bottom Bearing Plate Design Reference Figure 3.7.4-19 on Page 3-271 Illinois DOT Bridge Manual (See Appendix B) Given: Fy = 248,211 kPa (36 ksi)→ C = 0.183

94 Bottom plate reaction, R 25.1114/445 ==R kN (24.25 kips) Elastomeric pad length, Le 457=eL mm Bottom bearing pad length, Lb Lb = 660 mm Bottom bearing plate width, Wb 356=bW mm Bottom bearing plate thickness, Tb ( ) ( ) 380 1426 2524 18261830 . . . WL R LLCT bb ebb =× −= × −= in = 9.65 mm Minimum Tb = 25.4 mm (1 in) Use minimum, Tb = 25.4 mm

95 Figure 4.1: Elevation View of Bearing Pad Details

96 Figure 4.2: Plan View of Bearing Pad Details BEARING PAD NATURAL FREQUENCY The natural frequency of the bearing pads significantly effect the performance of a GRS bridge abutment. The design procedure of the bearing pads presented in Section 4.2, adopted from Method B in AASHTO LRFD Bridge Design Specifications (2007), excludes the natural frequency as a design aspect. However, if the bearing pads are properly designed such that the natural frequency of the bridge-pad system is below the dominant frequency of the ground motion, the superstructure inertia force can be isolated from the bridge abutment. Isolating this motion greatly reduces the potential for the sill to slide, overturn or have bearing capacity failure. The following calculations are based on the bridge loads and bearings used in the shake table test.

97 Calculations In reference to Figure 4.3, the shear force, F, can be calculated as: T GA F Δ = Where: G = Shear modulus of elastomer A = Plan area of elastomer T = Total thickness of elastomer Δ = Shear deformation The elastomeric pad used in the shake table test has the following characteristics: G = 689 kPa, A = 0.14 m2, and T = 0.043 m Figure 4.3: Shear Deformation of Elastomeric Pad The shear stiffness, K, of the elastomeric pad can be determined as K = F/Δ, where Δ = 1 unit of displacement. 243,2 043.0 14.0689 = × == ∆ = T GAF K kN/m Therefore, the horizontal natural frequency, f, of the bearing pad-bridge system can be determined using: M K f π2 1 = Where: M = Mass supported by the bearing ( ) ( ) 340,11m/s81.9 N250,111 2 ==M kg

98 ( ) ( ) 24.2/340,11 /000,243,2 2 1 2 = ⋅ = msN mN f π Hz