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EXAMPLE 2: Determine damping ratio, damping coefficient, and displacement at 100 cycles

__Situation:__

A 16 ft x 20 ft x 6" concrete canopy is supported by four columns at each corner

Column size: 16 in x16 in

Column height: 10 ft

Thickness of roof slab: 6"

Compression strength of concrete: 4000 psi, normal weight

__Assumption: __

1. columns are fixed at top and bottom

2. Neglect weight of columns

__Requirement__: Determine natural frequency, natural period of
structure and critical damping coefficient.

__Solution:__

Weight of roof: W = (150 lb/ft^{3})(16 ft)(20 ft)(0.5 ft) =
24000 lb = 24 kip

Mass of roof: m = W/g = W/ (32.2 ft/sec^{2}) = 746 lb-sec^{2}/ft
= 0.746 kip-sec^{2}/ft

Elastic modulus of concrete: E = 57*(4000)^{1/2} ksi
= 3605 ksi

Moment of inertia of column: I = (16/12 ft)^{4}/12 = 0.263
ft^{4}.

Stiffness of 4 columns: k = 4 [12EI/(10 ft)^{3} ]
= 6563 kip/ft

Natural frequency, f = (k/m)^{1/2} /2p =
14.9/sec

Natural period, T = 1/f = 0.067 sec.

Critical damping coefficient:

c_{c} = 2 m w_{ } =
2 m (k/m)^{1/2} =
140 kip-sec/ft

__Situation__:

For the structure in example 1, it was measured that lateral displacement reduces from 1" to 0.5" in 10 cycles.

__Requirement__: Determine damping ratio, damping coefficient,
and displacement at 100 cycles

Logarithmic decrement: d = ln (0.3"/1" )/10 = 0.12

Damping ratio: z = d/2p = 0.02

Damping coefficient: c = z c_{c} =
2.68 kip-sec/ft

Displacement at 100 cycle: X_{100} =
(0.3"/1")^{100/10} (1")
= 6 x 10^{-6 }in

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