Structural Dynamics 3DOF

STRUCTURAL DYNAMICS Problem Set No. M.S. Civil Engineering Forced Motion of Shear Building 1. Determine the response as a function of time for the two-story shear building shown when a constant force of 5,000 lb is suddenly applied at the level of the second floor as shown in the figure below. Bays are 15 ft. apart. Use E = 30 x 106 psi, IW10X21 = 106.30 in4, and IW10X45 = 248.60 in4 Figure:

W2 = 26,500 lb 20 psf W 10 X 21

5000 lb 10'

W1 = 52,500 lb 20 psf W 10 X 45

15'

30'

Solution: a. Natural Frequencies and Modal Shapes Computing for the weights and masses of each member: W1= 52,500 lbs + 20 psf x 12.5 ft. x 15 ft. x 2 = 60,000 lbs. W2= 26,500 lbs + 20 psf x 5 ft. x 15 ft. x 2 = 29,500 lbs. m1= 60,000 lbs. / (32.2 ft/sec2 x 12 in/ft.) = 155.280 lb-sec2/in m2= 29,500 lbs. / (32.2 ft/sec2 x 12 in/ft.) = 76.346 lb-sec2/in Computing for the weights stiffness of each story (assume rigid and fixed at two ends): k = 12E(2I) / L3 k1= 12 x 30 x 106 x 2 x 248.60 / (15 x 12)3 = 30,691.358 lb/in k2= 12 x 30 x 106 x 2 x 106.30 / (10 x 12)3 = 44,291.667 lb/in Computing for the natural frequencies of the structure (using matrix analysis):

(155.280 x 76.346) ω4 – [(30,691.358 + 44,291.667) x (76.346) + 155.28 x 44,291.667] ω2 + 30,691.358 x 44,291.667 = 0 11,855.01 ω4 – 12,602,264.08 ω2 + 1,359,371,408 = 0 ω12 = 941.203; ω1 = 30.679 rad/sec ω22 = 121.830; ω2 = 11.038 rad/sec Computing for corresponding frequencies and natural periods:

= 4.8827 cycles per sec. = 1.7568 cycles per sec.

= 0.2048 sec. = 0.5692 sec. Computing for the modal frequencies: (k1 + k2 – ω2 m1) a11 – k2a21 = 0 -k2a11 + (k2 - ω2 m2) a21 = 0 ---------------------------------------------– 71,166.977 a11 – 44,291.667 a21 = 0 if a11 = 1.0, a21 = – 1.607

– 1.607 56,065.263 a11 – 44,291.667 a21 = 0

if a11 = 1.0, a21 = 1.266

1.266

-

a 21 = 1.607

a 21 = 1.266

a 11= 1.000

a 11= 1.000

1

= 30.679 rad/sec

first mode

= 11.038 rad/sec

2

second mode NORMAL MODES

b. Orthogonality Properties of Normal Modes and Modal Forces Computing for the Orthogonality Property of the Normal Modes:

orthogonality relationship between modal shapes of a two degree-of-freedom system:

for diagonal mass matrix:

√ (155.28)(1.00)2 + (76.346)(1.266)2 = √ 277.644 √ (155.28)(1.00)2 + (76.346)(- 1.607)2 = √ 352.440 √ (155.28)(1.00)2 + (76.346)(1.266)2 = √ 277.644 √ (155.28)(1.00)2 + (76.346)(- 1.607)2 = √ 352.440 1.00 Ø11 = ---------------- = 0.06001 ; √ 277.644

1.00 Ø12 = --------------- = 0.05327 √ 352.440

1.266 Ø21 = ---------------- = 0.07598 ; √ 277.644

Ø22 =

Computing for modal force P(t) using the normal modes:

-1.607 --------------= - 0.0856 √ 352.440

P1 (t) = (1.00 x 0) + (1.266 x 5000) P1 (t) = 6,330 lbs. P2 (t) = (1.00 x 0) + (-1.607 x 5000) P2 (t) = 8,035 lbs. Computing for modal force using normalized modes, P: P1 (t) = (0.06001 x 0) + (0.0798 x 5000) P1 (t) = 379.90 lbs. P2 (t) = (0.05327 x 0) + (-0.0856 x 5000) P2 (t) = 428 lbs. c. Participation Factor and Response Equation of Motion Computing for the participation factors:

Γ1 = - (155.28 x 0.06001 + 76.346 x 0.07598) = - 15.11912 Γ2 = - (155.28 x 0.05327 + 76.346 x -0.0856) = - 1.73655 Computing for the response equation of motion, u(t):

u1(t) = a11

[ P (t) / ωω xt)(1 – cos] + a [ P (t) / ωω xt)(1 – cos ] 1

2 1

12

2

2 2

2

1

= 1.000 [379.90 / 11.0382 x (1 – cos 11.038t)] + 1.00 [-428 / 30.6792 x (1 – cos 30.679t)] u1(t) = 3.1181 – 3.1181cos11.038t – 0.4547 + 0.4547cos30.679t u1(t) = 2.6634 – 3.1181cos11.038t + 0.4547cos30.679t

u2(t) = a21

[ P (t) / ωω xt)(1 – cos] + a [ P (t) / ωω xt)(1 – cos ] 1

2 1

22

2

2 2

2

1

2

= 1.266 [379.90 / 11.038 x (1 – cos 11.038t)] – 1.607 [-428 / 30.6792 x (1 – cos 30.679t)]

u1(t) = 3.9475 – 3.9475cos11.038t + 0.7308 – 0.7308cos30.679t u1(t) = 4.6783 – 3.9475cos11.038t + 0.7308cos30.679t Computing for the maximum values of response, setting the maximum value of cosine as (-1): u1(t) = 1.000 [379.90 / 11.0382 x (1 – cos π)] + 1.00 [-428 / 30.6792 x (1 – cos π)] u1(t)max = 5.3267 in. u2(t) = 1.266 [379.90 / 11.0382 x (1 – cos π)] – 1.607 [-428 / 30.6792 x (1 – cos π)] u2(t) max = 9.3565 in.

2. Repeat the problem above if the excitation is applied to the base of the structure in the form of a suddenly applied acceleration of magnitude 0.5 g. Solution: ..

us = 0.5 x 32.2 ft / sec2 x 12 in / sec2 .. us = 193.20 in / sec.2

Computing for the modal equations:

Computing for the modal response:

Computing for the maximum modal response, setting the cosine value of maximum (-1):

Computing for the response equation, u(t):

Computing for the maximum response, setting the cosine value of maximum (-1):

Computing for the maximum response: