Prob. 1 SDOF Structure subjected to Ground Shaking

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Aileen Hancock
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1 Prob. 1 SDOF Structure subjected to Ground Shaking What is the maximum relative displacement and the amplitude of the total displacement of a SDOF structure subjected to ground shaking? magnitude of ground acceleration v'' g0 0.5 g period of ground motion T g 0.3 sec circular frequency of ground motion ω g 20.9 rad/sec = 2 * PI() / T g v T v rel v weight of structure Wt 55 k mass of structure m k-s 2 /in = Wt / 386 stiffness of structure k 75 k/in natural frequency of structure ω 22.9 rad/sec = SQRT( k / m ) damping ratio of structure ξ 0.10 frequency ratio β = ω g / ω v g k/2 m c k/2 ground surface dynamic amplification factor D 4.05 = 1 / SQRT( ( 1 - β^2 )^2 + ( 2 * ξ * β )^2 ) amplitude of relative displacement v in = ( m * v'' g0 *386 ) / k * D amplitude of total displacement v T g = v'' g0 * SQRT( 1 + ( 2 * ξ * β )^2 ) * D earth core

2 Prob. 2 SDOF Structure subjected to Ground Shaking What is the maximum force transmitted by a vibrating piece of machinery to the floor (ground)? magnitude of sinusoidal force p lb p o sinωt frequency of sinusoidal force f machine 5 Hz ω machine 31.4 rad/sec m +v weight of machine Wt 3000 lb mass of machine m 7.8 lb-s 2 /in = Wt / 386 k/2 c k/2 stiffness of machine k 1000 lb/in natural frequency of machine on spring supports ω 11.3 rad/sec = SQRT( k / m ) frequency ratio β 2.77 = ω machine / ω damping ratio ξ 0.02 dynamic amplification ratio D = 1 / SQRT( ( 1 - β^2)^2 + ( 2 * ξ * β )^2 ) magnitude of force from machine to foundations f max 90 lb = p 0 * D * SQRT( 1 + (2 * ξ * β)^2 ) magnitude of machine displacement v in = p 0 / k * D

3 Prob. 3 Vibrating Machinery Mounted to RC Foundation Is the Vibration of a Piece of Machinery Mounted to a Reinf. Conc. Foundation Acceptable? rotating machinery magnitude of sinusoidal force p lb frequency of sinusoidal force f machine 5 Hz f machine 300 rpm = f machine * 60 ω machine 31.4 rad/sec foundation weight of machine W machine 3000 lb soil width of reinf. conc. foundation B 3 ft length of reinf. conc. foundation L 7 ft depth of reinf. conc. foundation D 2 ft weight of reinf. conc. foundation W fnd 6300 lb p o sinωt v( t) = ρ sin( ωt θ ) mass of machine and foundation m 24.1 lb-s 2 /in = ( W machine + W fnd ) / 386 soil foundation k/2 k/2 c assumed shear modulus of stiff clay G 2950 psi = ( ) / 2 assumed poisson's ratio µ 0.45 assumed unit weight of soil UW 120 pcf UW pci = UW / 1728 mass density of soil ρ lb-s 2 /in 4 = UW / 386 effective radius of foundation footprint r o 2.59 ft = SQRT( B * L / PI() ) r o 31.0 in = r o * 12 elastic stiffness of soil K 665,634 lb/in = 4 * G * r o / (1 - µ) mass ratio B z = (1 - µ) * m / ( 4 * ρ * r o^3 )

4 Problem 3 -cont'd Prob. 3 Vibrating Machinery Mounted to RC Foundation damping ratio of soil ξ = * SQRT( B z ) natural frequency of machine + foundation on soil ω 166 rad/sec = SQRT( K / m ) frequency ratio β = ω machine / ω dynamic amplification factor D 1.01 = 1 / SQRT( (1 - β^2)^2 + (2 * ξ * β)^2 ) magnitude of displacement of machine + foundation v in = p 0 / K * D magnitude of velocity of machine + foundation v' in/sec = v 0 * ω machine allowable v' in/sec Therefore vibrations are not likely to damage the machine itself For frequency = 300 rpm and v0 = , Figure 5 shows the vibration to be barely noticable to persons

Prob. 4 Sensitive Equipment Isolated from Floor magnitude of floor displacement v g0 0.

5 Prob. 4 Sensitive Equipment Isolated from Floor magnitude of floor displacement v g in frequency of floor motion f g 15 Hz ω g 94 rad/sec = f g * 2 * PI() weight of equipment mounted on springs W 1000 lb mass of equipment m 2.59 lb-s 2 /in = W / 386 stiffness of springs supporting equipment k 5000 lb/in natural frequency of equipment on springs ω 43.9 rad/sec = SQRT( k / m ) frequency ratio β 2.15 = ω g / ω damping ratio ξ 0.01 dynamic amplification factor D = 1 / SQRT( (1-β^2)^2 + (2 * ξ * β)^2 ) amplitude of total displacement v T in = v g0 * SQRT( 1 + (2 * ξ * β)^2 ) * D

6 Prob. 5 Car on Undulating Road Surface What is the speed of the car traveling on an undulating road surface to maxize the total displacement of the car? What is the total displacement? v g0 L magnitude of ground displacement v g0 2 in distance between peaks of undulating road surface L 60 ft weight of car W 3000 lb stiffness of car suspension k 900 lb/in = 100 / 0.08 damping ratio of car suspension ξ 0.50 Max. response occurs when the driving frequency (ω g ) equals the natural frequency (ω) of the car/suspension system (resonance) Therefore, set ω g = ω mass of car m 7.77 lb-s 2 /in = W / 386 natural frequency of car on suspension ω 10.8 rad/sec = SQRT( k / m ) frequency of ground motion ω g 10.8 rad/sec = ω f g 1.71 cy/sec = ω g / ( 2 * PI() ) speed of car speed ft/sec = f g * L speed 70.1 mph frequency ratio β (resonance) dynamic amplification factor D 1.00 = 1 / SQRT( (1 - β^2)^2 + (2 * ξ * β)^2 ) magnitude of total displacement v T in = v g0 * SQRT( 1 + (2 * ξ * β)^2 ) * D

Lecture 9: Harmonic Loads (Con t)

Lecture 9: Harmonic Loads (Con t) Reading materials: Sections 3.4, 3.5, 3.6 and 3.7 1. Resonance The dynamic load magnification factor (DLF) The peak dynamic magnification occurs near r=1 for small damping