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Structural dynamics

EXAMPLE 1: Determine natural frequency, natural period of structure and critical damping coefficient.

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

EXAMPLE 1

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/ft3)(16 ft)(20 ft)(0.5 ft) = 24000 lb = 24 kip

Mass of roof: m = W/g = W/ (32.2 ft/sec2) = 746 lb-sec2/ft = 0.746 kip-sec2/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 ft4.

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:

cc = 2 m w  = 2 m (k/m)1/2 = 140 kip-sec/ft

EXAMPLE 2:

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 cc = 2.68 kip-sec/ft

Displacement at 100 cycle: X100 = (0.3"/1")100/10 (1") = 6 x 10-6 in

 

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