Tools › Civil and Geotechnical › Design Loads › Seismic Equipment Anchorage › Method

Seismic Equipment Anchorage: Method

Open the Seismic Equipment Anchorage tool

Force Equation, Anchor-Group Statics, Direction Scan and Sources

What the tool does

Mechanical and electrical equipment, tanks, panels and suspended services are "nonstructural components": the building code does not ask them to resist the earthquake for the building, but it does ask that they stay where they are. The designer's job is to find the horizontal force the code assigns to the component, combine it with the weight and the vertical earthquake effect, and size the anchors, isolators, braces and hanger rods for the result. This tool does that arithmetic for five common mountings and prints the demand on each anchor. It does not design the anchor in concrete; the Anchor and Base Plate Design tool does.

The design force

The horizontal force at the centre of gravity of the component is (ASCE 7-16 Section 13.3.1, Equations 13.3-1 to 13.3-3; printed in FEMA P-751 page 14-4 and, as Equations 6.2-1, 6.2-3 and 6.2-4, in FEMA 450)

Fp = 0.4 ap SDS Wp (1 + 2 z/h) / (Rp / Ip),   0.3 SDS Ip Wp ≤ Fp ≤ 1.6 SDS Ip Wp.

0.4 SDS is the peak ground acceleration implied by the design spectrum. The factor 1 + 2 z/h raises it linearly from the base of the structure to three times as much at the roof. ap is the amplification of the component itself: 1.0 if it is rigid and rigidly attached, 2.5 if it is flexible or flexibly mounted. Rp allows for the ductility and overstrength of the component and its supports, and Ip is 1.5 for components that must work after the earthquake or whose failure is a life-safety hazard. SDS is entered, or computed as (2/3) Fa SS. A vertical force of ± 0.2 SDS Wp acts with the horizontal one. The redundancy factor is 1.0.

The tool applies the equation of ASCE 7-16 and earlier editions. ASCE 7-22 replaced it with an equation in which the height factor depends on the period of the building and the component factors are different; that equation is not implemented.

Where ap, Rp and Ω0 come from

They are tabulated in ASCE 7 Tables 13.5-1 (architectural) and 13.6-1 (mechanical and electrical), which are copyrighted and not reproduced here; the three factors are entered. The component list on the form fills ap and Rp with pairs that are printed in public documents: the values FEMA P-751 quotes from ASCE 7-05 for its worked examples (HVAC fan unit 2.5 and 6.0; spring-isolated equipment 2.5 and 2.0; threaded piping not built to ASME B31 2.5 and 4.5; a vessel not on skirts 1.0 and 2.5) and rows of Table 6.4-1 of the 2003 NEHRP Provisions (FEMA 450). Check them against the table of the code edition in force before relying on them.

The overstrength factor Ω0 exists for one purpose: anchors in concrete or masonry whose strength is not governed by a ductile steel element are designed for the horizontal earthquake force multiplied by Ω0, so that the brittle concrete failure does not come first. NIST GCR 18-917-43 (page 4-29) records that the values assigned were 2.5 and 2.0 in ASCE 7-10 Supplement 1 and 2.0 and 1.5 in ASCE 7-16. The default is 2.0. Every table gives the anchor forces with and without it. For the supports and attachments Rp may not be taken above 6 (ASCE 7-16 Section 13.4.1; NIST GCR 17-917-44 page 3-33); the tool applies that limit to the anchor, isolator, brace and rod forces and shows the component force beside it.

Load combinations

With the vertical effect substituted (FEMA P-751 pages 14-6 and 14-34):

Strength design: (1.2 + 0.2 SDS) D + Eh and (0.9 − 0.2 SDS) D + Eh.
Allowable stress design: (1.0 + 0.14 SDS) D + 0.7 Eh and (0.6 − 0.14 SDS) D + 0.7 Eh.

Eh is the effect of Fp (Ω0 Fp for the non-ductile anchor check). The second combination of each pair leaves the least weight to hold the component down and governs uplift; the first governs bearing and, on a wall, the anchor shear.

Anchor forces of a floor-mounted or wall-mounted component

The component is taken as a rigid body on N anchors of equal stiffness lying in one plane, with coordinates x and y in the plane measured from the centroid of the pattern and n normal to it. A force (Fx, Fy, Fn) acting at the centre of gravity, at (ex, ey) in plan and a distance d from the plane, gives

ti = Fn / N + b xi + c yi,   with Σ t x = Fn ex − Fx d and Σ t y = Fn ey − Fy d,

which for a pattern symmetric about both axes is b = (Fn ex − Fx d) / Σx² and c = (Fn ey − Fy d) / Σy². A positive t is tension; a negative one is bearing on the support. The shear of anchor i is (Fx, Fy) / N plus the twisting moment ex Fy − ey Fx shared in proportion to the distance from the centroid, (−yi, xi) / Σ(x² + y²). On a floor the weight is Fn = −P and the earthquake force lies in the plane; on a wall the weight is Fy = −P and the earthquake force lies in the plane of x and n.

For four corner anchors and a force along x this is the familiar T = Fp h / (2 Lx) − P / 4. The pattern on the form is a rectangle with nx anchors on each side parallel to x and ny on each side parallel to y, corners counted on both.

The model is the linear one used for bolt groups. It ignores the shift of the compression resultant toward the edge of the base and any prying, and it lets anchors "carry" compression that in reality goes through bearing. For a compact base on a stiff support the difference is small; for a flexible base frame the anchors near the load path take more.

Why the direction is scanned

The code requires the force in whichever horizontal direction is worst. Checking the two principal axes is not enough for a corner anchor: with the force at an angle θ its tension is P-independent part plus Fp h (cos θ / (2 Lx) + sin θ / (2 Ly)), which peaks at tan θ = Lx / Ly with the value Fp h √(1/Lx² + 1/Ly²) / 2. This is the "diagonal" analysis that FEMA P-751 takes from the ASHRAE Handbook for its isolator example. In the default case (the P-751 fan unit, directly attached) a force along either axis leaves every anchor in bearing, as the book finds, but the diagonal force puts 64 lb of tension on one corner.

The tension of every anchor is a sinusoid in θ, so its maximum is found in closed form. The shear magnitude and the combined tension-shear ratio are not sinusoids when the centre of gravity is off centre; they are scanned every 0.25° and refined. Figure 2 of the results plots all three against θ.

Vibration isolators

Equipment on springs has Rp = 2.0 or 2.5 and ap = 2.5, and needs a snubber in each horizontal direction. Where the nominal air gap at the snubber exceeds 0.25 in the equipment gathers speed before it hits the stop, and the design force is doubled (ASCE 7 Table 13.6-1 footnote, quoted in FEMA P-751 page 14-23). The tool applies the factor 2 or 1 by the gap select and reports the tension, bearing and shear per isolator, which the housing anchors then share.

Suspended trapeze

A brace at an angle α from the horizontal resists a horizontal force H with an axial force H / cos α. Its vertical component H tan α has to come out of the hanger rod next to it. A rigid strut works both ways: in compression it adds H tan α to the rod tension, in tension it lifts the trapeze and puts the rod in compression of H tan α less the least factored dead load. A cable works only in tension, so it produces the compression case only. The rod compression is compared with the flexural buckling strength of a solid round (AISC 360 Section E3, used as mathematics): Fe = π² E / (K L / r)², Fcr = 0.658Fy/Fe Fy up to K L / r = 4.71 √(E / Fy) and 0.877 Fe beyond, with 0.9 Pn or Pn / 1.67. The longest unbraced length that still carries the compression is the spacing a rod stiffener has to achieve.

Unanchored items

For an item standing free the tool takes the least-weight combination and reports the factor of safety against sliding, μ P / H, and against overturning about each edge of the base, P × (distance from the centre of gravity to the edge) / (H h), together with the position of the resultant: beyond one sixth of the base width from the centre the far edge lifts. The item slides before it rocks when μ is below the ratio of that distance to h. ASCE 7 does not allow friction to be counted for a component that requires seismic design, so this mode is a screening of exempt or movable items, not a compliance check.

Relative displacement

A component attached at two levels must also follow the building: Dp = δxA − δyA, not more than (hx − hy) ΔaA / hsx (FEMA P-751 page 14-8). The tool evaluates the upper limit when a level difference is entered.

What was checked

The force equation with its bounds and the overstrength factor reproduces the 90 cells of Tables 4-8 to 4-10 of NIST GCR 18-917-43 and the ASCE 7-10 and 7-16 rows of Table 3-3 of NIST GCR 17-917-44. The anchor statics reproduce FEMA P-751 Section 14.3: Fp = 2,231 lb, 558 lb of shear per bolt and 46 lb of bearing for the directly attached unit; Fp = 6,692 lb, the 51.8° direction and 3,346 lb of shear for the isolated unit. The isolator tension and compression the book prints (2,630 and 4,204 lb) are 13 lb below what its own expression gives (2,643 and 4,218 lb, which the tool prints), and the 299 lb it prints for the first combination of the direct attachment is not the value of the expression beside it (271 lb). The trapeze reproduces Section 14.5.4: brace 4,110 lb, hanger +3,359 and −2,681 lb. Twenty randomly generated units were solved a second time with numpy by six equilibrium equations and a 0.01° scan.