Base Isolation and Energy Dissipation Questions and Answers

Answer: A

Correct answer: 2.24 m

Rearranging \( T = 2\pi\sqrt{R/g} \) gives \[ R = \frac{gT^2}{4\pi^2} = \frac{9.81 \times 9}{39.48} = 2.24\ \mathrm{m}. \] The radius grows with the square of the period, so pushing from 2 s to 3 s requires more than double the radius - and the plan size of the bearing grows with it, which is what limits how long a period this device can practically deliver.

Answer: B

Damping reduces every spectral ordinate, and at long periods the ordinate is essentially a displacement, so the design value falls. Codes supply correction factors for damping values other than five per cent, and the reduction here would be of the order of a third.

Answer: C

Multiplying gives \( 10{,}000 \times 0.3 = 3{,}000 \) kN, which is the base shear the structure above must carry. Consistency of units between the stiffness and the displacement is the whole difficulty in this calculation.

Answer: D

The first two are the whole point of the technique. The third is the opposite of the truth: displacement across the isolation plane is large and is the quantity the design has to accommodate with a moat and flexible services.

Answer: A

The classification in the first two statements is correct. The third is wrong for viscous devices, whose force peaks as the structure passes through its neutral position, out of phase with the elastic force in the members.

Answer: B

Correct answer: 1-R, 2-P, 3-Q

Three different mechanisms. The lead rubber bearing dissipates by yielding its lead core in shear, cycle after cycle, the lead recrystallising as it goes. The viscous damper forces fluid through an orifice. And the friction damper dissipates through slip between clamped surfaces. Only the second is velocity-dependent, which is the point of the previous item.

Answer: C

The shims stop the rubber bulging and so make the bearing stiff vertically, the rubber shears freely in the horizontal plane, and the lead core yields to absorb energy. One component doing three jobs is what makes the bearing so compact.

Answer: D

Correct answer: 1, 2, 3, 4

The procedure is iterative because the properties depend on the answer. A displacement is assumed; the effective stiffness and damping at that displacement follow; the spectrum is read at the resulting period to give a new displacement; and the loop is repeated until the assumed and computed values agree. Nothing later can be evaluated before the step above it, which fixes the order.

Answer: A

Uplift at the edge of the isolation plane is the governing difficulty for a slender building, and laminated rubber bearings delaminate rather than stretch. A tall building is also flexible already, so there is less period shift to be gained in the first place.

Answer: B

Every gravity load path runs through the isolation plane, and vertical softness would produce rocking and settlement. The consequence is that vertical ground motion is not isolated, which has to be dealt with separately.