Bachman and Bachman, Robert and Dowty, Susan, “Nonstructural Component or Dowty Nonbuilding Structure?”, Building Safety Journal, International Code Council, 2008. Jacobsen Jacobsen, L.S., “Impulsive Hydrodynamics of Fluid Inside a Cylindrical Tank and of Fluid Surrounding a Cylindrical Pier,” Bulletin of the Seismological Society of America, 39(3), 189-204, 1949. Morison Morison, J.R., O’Brien, J.W. and Sohaaf, S.A., “The Forces Exerted by Surface Waves on Piles,” Petroleum Transactions, AIME, Vol. 189; 1950. RMI Rack Manufacturers Institute, Specification for the Design, Testing and Utilization of Industrial Steel Storage Racks, MH16.1, 2012 Soules Soules, J. G., “The Seismic Provisions of the 2006 IBC – Nonbuilding Structure Criteria,” Proceedings of 8th National Conference on Earthquake Engineering, San Francisco, CA, April 18, 2006. 17.1 NONBUILDING STRUCTURES VERSUS NONSTRUCTURAL COMPONENTS Many industrial structures are classified as either nonbuilding structures or nonstructural components. This distinction is necessary to determine how the practicing engineer designs the structure. The intent of the Standard is to provide a clear and consistent design methodology for engineers to follow regardless of whether the structure is a nonbuilding structure or a nonstructural component. Central to the methodology is how to determine which classification is appropriate. Table 17-1 provides a simple method to determine the appropriate classification. Additional discussion on this topic can be found in Bachman and Dowty (2008). The design methodology contained in Chapter 13 of the Standard focuses on nonstructural component design. As such, the amplification by the supporting structure of the earthquake-induced accelerations is critical to the design of the component and its supports and attachments. The design methodology contained in Chapter 15 of the Standard focuses on the direct effects of earthquake ground motion on the nonbuilding structure. Table 17-1 Applicability of the Chapters of the Standard Supporting Structure Supported Item Nonstructural Component Nonbuilding Structure Building Chapter 12 for supporting structure; Chapter 13 for supported item Chapter 12 for supporting structure; Chapter 15 for supported item Nonbuilding Chapter 15 for supporting structure; Chapter 13 for supported item Chapter 15 for both supporting structure and supported item
Description A cruise ship company is developing a pier in Long Beach, California, to service ocean liners. The pier will be accessible by the general public. The pier contains a large warehouse owned by the cruise ship company. In the north-south direction, the pier is tied directly to an abutment structure supported on grade. In the east-west direction, the pier resists seismic forces using moment frames. Calculations for the abutment are not included in this example, but it is assumed to be much stiffer than the moment frames. The design live load for warehouse storage is 1,000 psf. Figure 17-5 Pier plan and elevation (1.0 ft = 0.3048 m) Provisions Parameters Risk Category (Standard Sec. 1.5.1) = II (The pier serves cruise ships that carry no hazardous materials.) Importance Factor, Ie (Standard Table 1.5-2) = 1.0 Site Class (Standard Chapter 20) = D (dense sand) Design Spectral Acceleration Response Parameters SDS = 1.167 SD1 = 0.60 Seismic Design Category (Standard Sec. 11.6) = D Seismic Force-resisting System (Standard Table 15.4-1) = Intermediate concrete moment frame with permitted height increase Response Modification Coefficient, R = 3 System Overstrength Factor, Ω0 = 2 Deflection Amplification Factor, Cd = 2.5 Height Limit (Standard Table 15.4-1) = 50 ft If the structure was classified as a building, an intermediate reinforced concrete moment frame would not be permitted in Seismic Design Category D. Design of the System 17.5.3.1 Seismic response coefficient. Using Standard Equation 12.8-2:
(Standard Eq. 13.3-1) Fpmax = 1.6(1.0)(1.5)(706 lb) = 1,694 lb (Standard Eq. 13.3-2) Fpmin = 0.3(1.0)(1.5)(706 lb) = 318 lb (Standard Eq. 13.3-3) Eq. 13.3-3 governs, and Fp = 318 lb Eh = ρQE (Standard Eq. 12.4-3) Ev = 0.2SDSD (Standard Eq. 12.4-4a) where: QE (due to horizontal application of Fp) = 318 lb (Standard Sec. 12.4.2.1) ρ = 1.0 (because egress stairs are nonstructural components) (Standard Sec 13.3.1) D = dead load effect (due to vertical load application) Substituting, the following is obtained: Eh = ρQE = (1.0)( 318 lb) = 318 lb (horizontal earthquake effect) Ev = 0.2SDSD = 0.2SDSD = (0.2)(1.0)(318 lb) = 64 lb (vertical earthquake effect) The above terms are then substituted into the following Basic Load Combinations for Strength Design from Standard Section 12.4.2.3 to determine the design member forces to be used in conjunction with seismic and gravity loads. 1.2D + Ev + Eh + L + 0.2S (Standard Basic Load Combination 6) 0.9D - Ev + Eh (Standard Basic Load Combination 7) 18.3.3.2 Landing. D = Wp = 513 lb (vertical gravity effect)
0.9D - Ev + Eh (Standard Basic Load Combination 7) These seismic load effects can be used to determine bolt shear, Vu and tension, Tu (where a negative value indicates tension). In the calculations below, the signs of SDS and Fp have been selected to result in the largest value of Tu. Similar calculations would be performed to determine the maximum compressive force Cu. U = 1.2D + 0.2SDSD + ρQE Tu = = -2,405 lb Cu = = 4,204 lb Vu = = 3,346 lb U = 0.9D - 0.2SDSD + ρQE Tu = = 2,630 lb Cu = = -3,980 lb Vu = = 3,346 lb The vibration isolator would be designed to resist these forces. 18.4.4.2 Proportioning and Details. In this example, there is no component or a support in the load path leading to the structure that undergoes ductile yielding at a load level less than the design strength of the corresponding anchor, so the anchors are designed to resist the load combinations considering Ω0 in accordance with Standard Section 2.3.6. In the calculations below, the signs of SDS and Fp have been selected to result in the largest value of Tu. The geometry of the vibration isolators are shown in Figure 18.4-4. By inspection, the load combination that results in net tension on the anchors governs. Applying the load combinations of Standard Section 2.3.6, Load Combination 7, the vertical design tension force is: U = 0.9D – Ev + Emh (Standard Basic Load Combination 7) Emh = Ω0QE (Standard Eq. 12.4-7)
(at roof) Ip = 1.5 (Standard Sec. 13.1.3) Gravity (non-seismic) supports provided every 10'-0" (given) System working pressure (P) = 200 psi (given) ASTM A53 Pipe Fy = 35 ksi, threaded connections (given) D = Dead Load = Wp = 16.4 = 31.7 Longitudinal brace spacing = 80 feet (given) According to Standard Section 13.3.1 (and repeated in Sec. 12.3.4.1), the redundancy factor does not apply to the design of nonstructural components. 18.5.2.2 Seismic Design Forces. (Standard Eq. 13.3-1) (Standard Eq. 13.3-2) (Standard Eq. 13.3-3) Standard Eq. 13.3-1 governs the horizontal seismic force on the piping system. 18.5.2.3 Performance Criteria. System failure must not cause failure of an essential architectural, mechanical, or electrical component (Standard Sec. 13.2.3). Component seismic attachments must be bolted, welded, or otherwise positively fastened without considering the frictional resistance produced by the effects of gravity (Standard Sec. 13.4). The effects of seismic relative displacements must be considered in combination with displacements caused by other loads as appropriate (Standard Sec. 13.3.2).
ϕrn = 10,400 lb (tension) Therefore, the bolts are adequate. The connection plates are 1/4 inch thick and 3 inches wide. = 0.0469 in3 The maximum moment in the plate is: MU = 534 lb (1.5 in.) = 801 in-lb The bending stress is: fb = MU /Z = 801/0.0469 = 17,088 psi OK Prying action can have the effect of increasing the tensile forces in the bolts. AISC 360 permits prying action to be neglected if the plate meets minimum thickness requirements, given by: where p = 3 inches is the tributary length per pair of bolts b' = (b-db/2) = (1.5-0.625/2) = 1.1875 in. = 0.13 in. This is less than the 0.25inch thickness provided, so prying need not be considered further. The welds of the vessel leg to the vessel body and of the leg to the upper connection plate are proportioned in a similar manner. The calculation can be simplified by assuming the weld is of unit thickness. This yields a demand per inch of weld and an appropriate weld thickness can then be selected. The vessel leg has an outer diameter, d, of 2.38 inches. The weld properties for a weld of unit thickness are: = 2.25 in3 A = πd =3.14(2.38) = 7.45 in. The shear in the weld is: v = 532 lb/7.45 in. = 714 lb/in.