Problems on Earthquake Engineering

Problems on Earthquake Engineering

1. (i) Use the standard surface-wave formula to calculate the magnitude of an earthquake if it originates at a focal depth of 500 km, the maximum amplitude of ground vibration recorded at an epicentral distance of 5 km is 10 cm and the frequency of surface-wave is 0.05 Hz. (ii) For this earthquake, calculate the ground vibration amplitude at an epicentral distance of 50 km.

Solution (i) Using the standard surface-wave formula, with

A = 10 cm = 10 5 m, T = 1/f = 1/0.05 = 20 sec, D = d/h = 5/500 = 0.01 rad = 0.573 M 5

S = log 10 (A/T) + 1.66 log 10 (D) + 3.30 = log 10 (10 /20) + 1.66 log 10 (0.573) + 3.30 = 6.60

(ii) Using M S = log 10 (A/T) + 1.66 log 10 (D) + 3.30

6.60 = log 10 (A/20) + 1.66 log 10 (50/500) + 3.30

A = 0.219 cm

2. Use the BNBC response spectrum for Dhaka to calculate the elastic peak deformation and base shear

for the (20 20 ) floor system described in # 2 of Problem set 2, for k f equal to

6 4 (i) 2 10 lb/in, and (ii) 2 10 lb/in [assume = 5%]. Solution

4 For this floor system, k 2

1 = 4.02 10 lb/in, m 1 = 207.25 lb-sec /in

6 (i) As calculated earlier, k 4

k eff =k 1 k f /(k 1 +k f ) = 3.94 10 lb/in n = 13.79 rad/sec, T n =2/ n = 0.456 sec

f = 2 10 lb/in

2/3 Using BNBC response spectrum, C = 1.25/T = 1.25/(0.456) = 2.110 Elastic base shear V b(e) = ZCW = 0.15 2.110 (20 20) 200/1000 = 25.32 kips

4 Elastic maximum deformation u

0 =V b /k eff = 25.32 1000/(3.94 10 ) = 0.643

4 (ii) k 4

f = 2 10 lb/in

k eff =k 1 k f /(k 1 +k f ) = 1.34 10 lb/in

n = 8.03 rad/sec, T n =2/ n = 0.783 sec

2/3 Using BNBC response spectrum, C = 1.25/T = 1.25/(0.783) = 1.471 Elastic base shear V b(e) = ZCW = 0.15 1.471 (20 20) 200/1000 = 17.65 kips

4 Elastic maximum deformation u

0 =V b /k eff = 17.65 1000/(1.34 10 ) = 1.317

3. Answer Question 2 using the response spectrum for El Centro earthquake. Solution

4 For this floor system, k 2

1 = 4.02 10 lb/in, m 1 = 207.25 lb-sec /in

(i) For T n = 0.456 sec, C = 2.695 Elastic base shear V b(e) = ZCW = 0.313 2.695 (20 20) 200/1000 = 67.48 kips

0 =V b /k eff = 67.48 1000/(3.94 10 ) = 1.713 (ii) For T n = 0.783 sec, C = 1.563

4 Elastic maximum deformation u

Elastic base shear V b(e) = ZCW = 0.313 1.563 (20 20) 200/1000 = 39.14 kips

4 Elastic maximum deformation u

0 =V b /k eff = 39.14 1000/(1.34 10 ) = 2.921

4. A 12 long vertical cantilever pipe (made of steel, with E = 29 10 6 psi) supports a 5200 lb weight attached at the tip. Determine the peak deformation and bending stress in the cantilever due to the El

Centro data, assuming = 2%, with the properties of the pipe being

(i) d 0 = 4.5 , d i = 4.026 , t = 0.237 , (ii) d 0 = 6.75 , d i = 6.039 , t = 0.356 .

Solution

4 4 (i) For this system, I = [(4.5) 4 (4.026) ]/64 = 7.23 in

3 6 Lateral stiffness, k = 3EI/L 3 = 3 (29 10 ) 7.23/(12 12) = 211 lb/in Mass, m = W/g = 5200/386 = 13.47 lb-sec 2 /in

n = (211/13.47) = 3.958 rad/sec, T n =2/ n = 1.59 sec From El Centro response spectrum with = 2%, C = 0.655 Elastic base shear V b(e) = ZCW = 0.313 0.655 5200/1000 = 1.066 kips Maximum bending moment M = 1.066 (12 12) = 153.52 k-in

Maximum bending stress max = Mc/I = 153.52 (4.5/2)/7.23 = 47.80 ksi

4 4 (ii) For the new system, I = [(6.75) 4 (6.039) ]/64 = 36.62 in

3 6 Lateral stiffness, k = 3EI/L 3 = 3 (29 10 ) 36.62/(12 12) = 1066.82 lb/in

Mass, m = W/g = 5200/386 = 13.47 lb-sec 2 /in n = (1066.82/13.47) = 8.900 rad/sec, T n =2/ n = 0.706 sec

From El Centro response spectrum with = 2%, C = 2.538

Elastic base shear V b(e) = ZCW = 0.313 2.538 5200/1000 = 4.131 kips Maximum bending moment M = 4.131 (12 12) = 594.84 k-in

Maximum bending stress max = Mc/I = 594.84 (6.75/2)/36.62 = 54.83 ksi

5. Answer Question 3 assuming yield deformation u y = 0.50 . Solution

(i) Using u 0 = 1.713 and u y = 0.50 , R y = 1.713/0.50 = 3.426 Inelastic base shear V b =V b(e) /R y = 67.48/3.426 = 19.70 kips (also = k eff u y ) For T n = 0.456 sec, Ductility ratio = 4.13, Maximum deformation u m = u y = 2.065 (ii) Using u 0 = 2.921 and u y = 0.50 , R y = 2.921/0.50 = 5.842 Inelastic base shear V b =V b(e) /R y = 39.14/5.842 = 6.70 kips (also = k eff u y ) For T n = 0.783 sec, Ductility ratio = 5.842, Maximum deformation u m = u y = 2.921

6. Answer Question 4 assuming yield strength f y = 36 ksi. Solution

(i) Using f 0 = 47.80 ksi and f y = 36 ksi, R y = 47.80/36 = 1.33

Inelastic base shear V b =V b(e) /R y = 1.066/1.33 = 0.803 kips

Maximum bending moment M = 0.803 (12 12) = 115.61 k-in Maximum bending stress max = Mc/I = 115.61 (4.5/2)/7.23 = 36 ksi For T n = 1.59 sec, Ductility ratio = R y = 1.33 Also, u y =V b /k = 803/211 = 3.805 , Maximum deformation u m = u y = 5.053 (ii) Using f 0 = 54.83 ksi and f y = 36 ksi, R y = 54.83/36 = 1.52

Inelastic base shear V b =V b(e) /R y = 4.131/1.52 = 2.712 kips

Maximum bending moment M = 2.712 (12 12) = 390.57 k-in Maximum bending stress max = Mc/I = 390.57 (6.75/2)/36.62 = 36 ksi For T n = 0.706 sec, Ductility ratio = R y = 1.52 Also, u y =V b /k = 2712/1066.82 = 2.542 , Maximum deformation u m = u y = 3.872