POJAM STABILNOSTI KONSTRUKCIJA · 2020. 11. 22. · Rigle tavanica su fleksiono beskonačno krute. Podaci: 𝑀. 1= 50 kNs2 m 𝑀2= 30 kNs2 m 𝑀3= 20 kNs2 m 𝐸𝐼= 20000 𝑘𝑁𝑚2

Zbirka zadataka - Dinamika konstrukcija i zemljotresno inženjerstvo 95

Zadatak 6

Za nosač na slici odrediti efektivne tonske mase.

Rigle tavanica su fleksiono beskonačno krute.

Podaci:

𝑀1 = 50kNs2

m 𝑀2 = 30

kNs2

m 𝑀3 = 20

kNs2

m

𝐸𝐼 = 20000 𝑘𝑁𝑚2

Rešenje

Matrica krutosti i matrica masa

𝐊 = �3.56 −1. 3̇ 0−1. 3̇ 2. 23̇ −0. 8̇

0 −0. 8̇ 0. 8̇� 𝐸𝐼 𝐌 = �

50 0 00 30 00 0 20

� kNs2

m

Oblici oscilovanja i modalna matrica

det (K − ω2M) = 0

𝐀 = [𝐀1 𝐀2 𝐀3] = �1.0 1.0 1.0

2.1937 0.4361 −1.74813.0714 −1.2811 1.0589

�

Generalisane mase 𝑀𝑖 = 𝐴𝑖𝑇𝑀 𝐴𝑖

𝑀1 = {1.0 2.1937 3.0714} �50

3020� �

1.02.19373.0714

� = 383.0485

𝑀2 = {1.0 0.4361 −1.2811} �50

3020� �

1.00.4361

−1.2811� = 88.5297

𝑀3 = {1.0 −1.7481 1.0589} �50

3020� �

1.0−1.7481

1.0589� = 164.1034

Skalarne veličine 𝐿𝑖 = 𝐴𝑖𝑇𝑀𝐵

𝐿1 = {1.0 2.1937 3.0714} �50

3020� �

1.01.01.0

� = 177.2408

𝐿2 = {1.0 0.4361 −1.2811} �50

3020� �

1.01.01.0

� = 37.4592

𝐿3 = {1.0 −1.7481 1.0589} �50

3020� �

1.01.01.0

� = 18.7346

Faktori participacije 𝛤𝑖 = 𝐿𝑖𝑀𝑖

𝛤1 =177.2408383.0485

= 0.4627 𝛤2 =37.459288.5297

= 0.4231 𝛤3 =18.7346

164.1034= 0.1142

Efektivne tonske mase 𝑀𝑖,𝑒𝑓 = 𝛤𝑖 𝐿𝑖

𝑀1,𝑒𝑓 = 0.4627 ∙ 177.2408 = 82.0093 𝑀2,𝑒𝑓 = 0.4231 ∙ 37.4592 = 15.8489 𝑀3,𝑒𝑓 = 0.1142 ∙ 18.7346 = 2.1395

96 Ratko Salatić, Marko Marinković

Koeficijenti 𝛼𝑖 =𝑀𝑖,𝑒𝑓∑𝑀𝑖

�𝑀𝑖 = 100

𝛼1 =82.0093

100= 0.8201 𝛼2 =

15.8489100

= 0.1585 𝛼3 =2.1395

100= 0.0214

Zadatak 7 Za četvorospratni ram, zadate krutosti i raporeda masa, primenom spektralne analize za dati spektar odgovora odrediti maksimalna pomeranja i maksimalne seizmičke horizontalne sile.

𝐌 = 1500 �

12

23

� [kg] 𝐊 = � 800−800

−800 2400−1600

−1600 4000−2400

−2400 5600

� [kN/m]

Rešenje Kružne frekvencije, periodi oscilovanja i glavni oblici oscilovanja sistema

𝝎 = �

𝜔1𝜔2𝜔3𝜔4� = �

10.854124.217133.540645.6274

� 𝑟𝑎𝑑𝑠

𝐓 = �

𝑇1𝑇2𝑇3𝑇4

� = �0.57890.25950.18730.1377

� 𝑠

𝐀1 = �4.25423.31452.1124

1.0� 𝐀2 = �

−2.2851

0.2277 1.2337

1.0� 𝐀3 = �

1.2733− 1.4125 0.2240

1.0� 𝐀4 = �

−0.2424 0.7037 −1.5702

1.0�

𝐀 = [𝐀1 𝐀2 𝐀3 𝐀4] = �4.25423.31452.1124

1.0

−2.2851

0.2277 1.2337

1.0

1.2733 −0.2424−1.4125 0.2240

1.0

0.7037 −1.5702

1.0

�

i T [s] Sd [cm] 1 0.0 1.5 2 0.1 1.8 3 0.2 2.5 4 0.3 3.8 5 0.4 5.0 6 0.5 5.0 7 0.6 4.0 8 0.8 3.2 9 1.0 2.5 10 1.2 1.9 11 1.4 1.4 12 1.6 1.0 13 1.8 0.7


Generalisane mase 𝑀𝑖 = 𝐀𝑖𝑇𝐌 𝐀𝑖

𝑀1 = {4.2542 3.3145 2.1124 1.0} �1500

30003000

4500

� �4.25423.31452.1124

1.0� = 78139.1852 kg

𝑀2 = {−2.2851 0.2277 1.2337 1.0} �1500

30003000

4500

� � −2.2851

0.2277 1.2337

1.0� = 17054.112 kg

𝑀3 = {1.2733 −1.4125 0.2240 1.0} �1500

30003000

4500

� � 1.2733− 1.4125 0.2240

1.0� = 13067.9360 kg

𝑀4 = {−0.2424 0.7037 −1.5702 1.0} �1500

30003000

4500

� � −0.2424 0.7037 −1.5702

1.0� = 13470.3018 kg

Koeficijenti participacije 𝛤𝑖 = 𝐀𝑖𝑇𝐌𝐁𝑀𝑖

𝛤1 =1

78139.1852{4.2542 3.3145 2.1124 1.0} �

15003000

30004500

� � 1.01.01.01.0

� = 0.3481

𝛤2 =1

17054.112{−2.2851 0.2277 1.2337 1.0} �

15003000

30004500

� � 1.01.01.01.0

� = 0.3200

𝛤3 =1

13067.9360 {1.2733 −1.4125 0.2240 1.0} �

15003000

30004500

� � 1.01.01.01.0

� = 0.2177

𝛤4 =1

13470.3018 {−0.2424 0.7037 −1.5702 1.0} �

15003000

30004500

� � 1.01.01.01.0

� = 0.1141

Sa dijagrama spektra određuje se za određene periode ocilovanja:

𝑆𝑑,1 = 4.211 cm 𝑆𝑑,2 = 3.2735 cm 𝑆𝑑,3 = 2.4111 cm 𝑆𝑑,4 = 2.0639 cm 𝑢𝑖,𝑚𝑎𝑥 = 𝐀𝑖𝛤𝑖𝑆𝑑,𝑖


𝑢1,𝑚𝑎𝑥 = �

𝑢1,𝑚𝑎𝑥 (4)𝑢1,𝑚𝑎𝑥 (3)𝑢1,𝑚𝑎𝑥 (2)𝑢1,𝑚𝑎𝑥 (1)

� = {0.3481}{4.2111} �4.25423.31452.11241.0000

� = �6.11784.76653.05451.4381

� cm


𝑢2,𝑚𝑎𝑥 (4)𝑢2,𝑚𝑎𝑥 (3)𝑢2,𝑚𝑎𝑥 (2)𝑢2,𝑚𝑎𝑥 (1)

� = {0.3200}{3.2735} � −2.2851

0.2277 1.2337 1.0000

� = �−2.3478 0.2339 1.2675 1.0274

� cm


𝑢3,𝑚𝑎𝑥(4)𝑢3,𝑚𝑎𝑥(3)𝑢3,𝑚𝑎𝑥(2)𝑢3,𝑚𝑎𝑥(1)

� = {0.2177}{2.4111} � 1.2733− 1.4125 0.2240 1.0000

� = � 0.6536−0.7250 0.1150 0.5133

� cm


𝑢4,𝑚𝑎𝑥(4)𝑢4,𝑚𝑎𝑥(3)𝑢4,𝑚𝑎𝑥(2)𝑢4,𝑚𝑎𝑥(1)

� = {0.1141}{2.0639} � −0.2424 0.7037 −1.5702 1.0000

� = � − 0.0575 0.1670 − 0.3727 0.2373

� cm

Maksimalna rezultujuća pomeranja po spratovima

𝑢𝑘,𝑚𝑎𝑥 ≅ ��𝑢𝑖,𝑚𝑎𝑥2𝑚

𝑖=1

𝑢4,𝑚𝑎𝑥 = �𝑢1,𝑚𝑎𝑥(4)2 + 𝑢2,𝑚𝑎𝑥(4)

2 + 𝑢3,𝑚𝑎𝑥(4)2 + 𝑢4,𝑚𝑎𝑥(4)

2 = √6.11782 + 2.34782 + 0.65362 + 0.05752 = 6.5856 cm

𝑢3,𝑚𝑎𝑥 = �𝑢1,𝑚𝑎𝑥(3)2 + 𝑢2,𝑚𝑎𝑥(3)

2 + 𝑢3,𝑚𝑎𝑥(3)2 + 𝑢4,𝑚𝑎𝑥(3)

2 = √4.76652 + 0.23392 + 0.72502 + 0.16702 = 4.8289 cm

𝑢2,𝑚𝑎𝑥 = �𝑢1,𝑚𝑎𝑥(2)2 + 𝑢2𝑚𝑎𝑥(2)

2 + 𝑢3,𝑚𝑎𝑥(2)2 + 𝑢4,𝑚𝑎𝑥(2)

2 = √3.05452 + 1.26752 + 0.11502 + 0.37272 = 3.3299 cm

𝑢1,𝑚𝑎𝑥 = �𝑢1,𝑚𝑎𝑥(1)2 + 𝑢2,𝑚𝑎𝑥(1)

2 + 𝑢3,𝑚𝑎𝑥(1)2 + 𝑢4,𝑚𝑎𝑥(1)

2 = √1.43812 + 1.02742 + 0.51332 + 0.23732 = 1.8557 cm

Rezultujuća razmicanja spratova (relativna sptatna pomeranja)

𝑟3−4=�(𝑢1,max(4) − 𝑢1,max(3))2 + (𝑢2,max(4) − 𝑢2,max(3))2 + (𝑢3,max(4) − 𝑢3,max(3))2 + (𝑢4,max(4) − 𝑢4,max(3))2 = 3.2314 cm

𝑟2−3 = �(𝑢1,max(3) − 𝑢1,max(2))2 + (𝑢2,max(3) − 𝑢2,max(2))2 + (𝑢3,max(3) − 𝑢3,max(2))2 + (𝑢4,max(3) − 𝑢4,max(2))2

= 2.2352 cm 𝑟1−2=�(𝑢1,max(2) − 𝑢1,max(1))2 + (𝑢2,max(2) − 𝑢2,max(1))2 + (𝑢3,max(2) − 𝑢3,max(1))2 + (𝑢4,max(2) − 𝑢4,max(1))2

= 1.7892 cm

𝑟1−0 = �(𝑢1,𝑚𝑎𝑥 (1))2 + (𝑢2,𝑚𝑎𝑥 (1))2 + (𝑢3,𝑚𝑎𝑥 (1))2 + (𝑢4,𝑚𝑎𝑥 (1))2 = 1.8557 cm


Maksimalne sile po tonovima

𝐹𝑖,𝑚𝑎𝑥 = 𝐌𝐀𝑖𝛤𝑖𝜔𝑖2𝑆𝑑,𝑖 = 𝐌𝐀𝑖𝛤𝑖𝑆𝑝𝑎,𝑖

𝐅1,𝑚𝑎𝑥 =

⎣⎢⎢⎢⎡𝐹1,𝑚𝑎𝑥 (4)𝐹1,𝑚𝑎𝑥 (3)𝐹1,𝑚𝑎𝑥 (2)𝐹1,𝑚𝑎𝑥 (1)⎦

⎥⎥⎥⎤

= {0.3481}{10.8541}2{4.2111} �1500

30003000

4500

� �4.25423.31452.1124

1.0� = �

10.811316.846410.7366

7.6240

� kN



⎥⎥⎥⎤

= {0.3200}{24.2171}2{3.2735} �1500

30003000

4500

� � −2.2851

0.2277 1.2337

1.0� = �

−20.65334.1160

22.301027.1148

� kN



⎥⎥⎥⎤

= {0.2177}{33.5406}2{2.4111} �1500

30003000

4500

� � 1.2733− 1.4125 0.2240

1.0� = �

11.0648−24.5489

3.893126.0697

� kN


⎣⎢⎢⎡𝐹4,𝑚𝑎𝑥(4)𝐹4,𝑚𝑎𝑥(3)𝐹4,𝑚𝑎𝑥(2)𝐹4,𝑚𝑎𝑥(1)⎦

⎥⎥⎤

= {0.1141}{45.6274}2{2.0639} �1500

30003000

4500

� � −0.2424 0.7037 −1.5702

1.0� = �

−1.7965 10.4306

−23.2743 22.2338

� kN

Rezultujuće spratne sile 𝐹𝑘,𝑚𝑎𝑥 ≅ �∑ 𝐹𝑖,𝑚𝑎𝑥2𝑚𝑖=1

𝐹4,𝑚𝑎𝑥 = �𝐹1,𝑚𝑎𝑥(4)2 + 𝐹2,𝑚𝑎𝑥(4)

2 + 𝐹3,𝑚𝑎𝑥(4)2 + 𝐹4,𝑚𝑎𝑥(4)

2 = �10.81132 + 20.65332 + 11.06482 + 1.79652 = 25.8670 kN

𝐹3,𝑚𝑎𝑥 = �𝐹1,𝑚𝑎𝑥(3)2 + 𝐹2,𝑚𝑎𝑥(3)

2 + 𝐹3,𝑚𝑎𝑥(3)2 + 𝐹4,𝑚𝑎𝑥(3)

2 = �16.84642 + 4.11602 + 24.54892 + 10.43062 = 31.8149 kN

𝐹2,𝑚𝑎𝑥 = �𝐹1,𝑚𝑎𝑥(2)2 + 𝐹2,𝑚𝑎𝑥(2)

2 + 𝐹3,𝑚𝑎𝑥(2)2 + 𝐹4,𝑚𝑎𝑥(2)

2 = �10.73662 + 22.30102 + 3.89312 + 23.27432 = 34.1973 kN

𝐹1,𝑚𝑎𝑥 = �𝐹1,𝑚𝑎𝑥(1)2 + 𝐹2,𝑚𝑎𝑥(1)

2 + 𝐹3,𝑚𝑎𝑥(1)2 + 𝐹4,𝑚𝑎𝑥(1)

2 = �7.62402 + 27.11482 + 26.06972 + 22.23382 = 44.3544 kN


Podaci: 𝐸𝐼𝑠 = 64000 kNm2 𝑀1 = 120.0 kNs2/m 𝑀2 = 60.0 kNs2/m Koeficijenti: 𝑘𝑜 = 1.0 𝑘𝑠 = 0.1 𝑘𝑝 = 1.0

Napomena: Rigle tavanica smatrati da su beskonačno krute.

Rešenje

Tipovi 2D ramova konstrukcije


Određivanje krutosti rama u lokalnom sistemu

𝐙1 = 𝐙3 = 𝐙𝐼 = � 0.9375 −0.375−0.375 0.375� 𝐸𝐼


𝐙2 = 𝐙𝐼𝐼 = � 1.125 −0.375−0.375 0.375� 𝐸𝐼

𝐙4 = 𝐙7 = 𝐙𝐼𝐼𝐼 = [0.375]𝐸𝐼

𝐙5 = 𝐙6 = 𝐙𝐼𝑉 = � 1.125 −0.5625−0.5625 0.5625� 𝐸𝐼

Obeležavanje ramova

Matrice transformacije ramova

𝐚 = �cos𝜑1 sin𝜑1 𝑟1cos𝜑2 sin𝜑2 𝑟2

�

𝑅1 ∶ 𝜑 = 0° 𝑟1 = 3.0 𝐚1 = �1.0 0.0 3.01.0 0.0 3.0

�

𝑅2 ∶ 𝜑 = 0° 𝑟1 = 0 𝐚2 = �1.0 0.0 0.01.0 0.0 0.0

�

𝑅3 ∶ 𝜑 = 0° 𝑟1 = −3.0 𝐚3 = �1.0 0.0 −3.0 1.0 0.0 −3.0

�

𝑅4 ∶ 𝜑 = 90° 𝑟1 = −12.0 𝐚4 = �0.0 1.0 −12.0 0.0 0.0 0.0

�

𝑅5 ∶ 𝜑 = 90° 𝑟1 = −4. 0 𝐚5 = �0.0 1.0 −4.0 0.0 1.0 −4.0

�

𝑅6 ∶ 𝜑 = 90° 𝑟1 = 4. 0 𝐚6 = �0.0 1.0 4.0 0.0 1.0 4.0

�

𝑅7 ∶ 𝜑 = 90° 𝑟1 = 12. 0 𝐚7 = �0.0 1.0 12.0 0.0 0.0 0.0

�

Matrica krutosti konstrukcije

𝐊𝑖 = 𝐚𝑖𝑇 ∙ 𝐙𝑖 ∙ 𝐚𝑖


▫ Matrica krutosti rama 1

𝐊1 = 𝐚1𝑇 ∙ 𝐙1 ∙ 𝐚1 =

⎣⎢⎢⎢⎢⎡0.9375 0.0 2.8125 −0.375 0.0 −1.125

0.0 0.0 0.0 0.0 0.08.4375 −1.125 0.0 −3.375

0.375 0.0 1.1250.0 0.0

3.375⎦⎥⎥⎥⎥⎤

𝐸𝐼


𝐊2 = 𝐚2𝑇 ∙ 𝐙2 ∙ 𝐚2 =

⎣⎢⎢⎢⎢⎡1.125 0.0 0.0 −0.375 0.0 0.0

0.0 0.0 0.0 0.0 0.00.0 0.0 0.0 0.0

0.375 0.0 0.00.0 0.0

0.0⎦⎥⎥⎥⎥⎤

𝐸𝐼


𝐊3 = 𝐚3𝑇 ∙ 𝐙3 ∙ 𝐚3 =

⎣⎢⎢⎢⎢⎡0.9375 0.0 −2.8125 −0.375 0.0 1.125

0.0 0.0 0.0 0.0 0.08.4375 1.125 0.0 −3.375

0.375 0.0 −1.1250.0 0.0

3.375⎦⎥⎥⎥⎥⎤

𝐸𝐼


𝐊4 = 𝐚4𝑇 ∙ 𝐙4 ∙ 𝐚4 =

⎣⎢⎢⎢⎢⎡0.0 0.0 0.0 0.0 0.0 0.0

0.375 −4.5 0.0 0.0 054.0 0.0 0.0 0.0

0.0 0.0 0.00.0 0.0

0.0⎦⎥⎥⎥⎥⎤

𝐸𝐼


𝐊5 = 𝐚5𝑇 ∙ 𝐙5 ∙ 𝐚5 =

⎣⎢⎢⎢⎢⎡0.0 0.0 0.0 0.0 0.0 0.0

1.125 −4.5 0.0 −0.5625 2.2518.0 0.0 2.25 −9.0

0.0 0.0 0.00.5625 −2.25

9.0⎦⎥⎥⎥⎥⎤

𝐸𝐼


𝐊6 = 𝐚6𝑇 ∙ 𝐙6 ∙ 𝐚6 =

⎣⎢⎢⎢⎢⎡0.0 0.0 0.0 0.0 0.0 0.0

1.125 4.5 0.0 −0.5625 −2.2518.0 0.0 −2.25 −9.0

0.0 0.0 0.00.5625 2.25

9.0⎦⎥⎥⎥⎥⎤

𝐸𝐼


𝐊7 = 𝐚7𝑇 ∙ 𝐙7 ∙ 𝐚7 =

⎣⎢⎢⎢⎢⎡0.0 0.0 0.0 0.0 0.0 0.0

0.375 4.5 0.0 0.0 054.0 0.0 0.0 0.0

0.0 0.0 0.00.0 0.0

0.0⎦⎥⎥⎥⎥⎤

𝐸𝐼


▫ Matrica krutosti konstrukcije

𝐊 =

⎣⎢⎢⎢⎢⎡3.0 0.0 0.0 −1.125 0.0 0.0

3.0 0.0 0.0 −1.125 0160.875 0.0 0.0 −24.75

1.125 0.0 0.01.125 0.0

24.75⎦⎥⎥⎥⎥⎤

𝐸𝐼

Matrica masa

𝑀1 = 120.0kNs2

m 𝑀2 = 60.0kNs2/m

𝐼𝑜 = 𝑀𝐴�𝐼𝑥 + 𝐼𝑦�

𝐼𝑜1 = 12096

(288 + 3584) = 4840

𝐼𝑜2 =𝑀𝐴�𝐼𝑦 + 𝐼𝑥� =

𝑀𝑎𝑏

�𝑎𝑏3

12+𝑏𝑎3

12� =

𝑀12

(𝑎2 + 𝑏2) =6012

(82 + 62) = 500

𝐌 =

⎣⎢⎢⎢⎢⎡120.0

120.04840.0

60.060.0

500.0⎦⎥⎥⎥⎥⎤

Kružne frekvencije i periodi oscilovanja

(𝐊 − 𝜔2 ∙ 𝐌) ∙ 𝐀 = 0 𝑇 = 2𝜋𝜔

𝜔1 = 𝜔2 = 22.9830 rad/ s 𝑇1 = 𝑇2 = 0.27338 s 𝜔3 = 38.7833 rad/sec 𝑇3 = 0.1620 s 𝜔4 = 𝜔5 = 47.6632 rad/ s 𝑇4 = 𝑇5 = 0.1318 s 𝜔6 = 61.5722 rad/s 𝑇6 = 0.1020 s

Oblici oscilovanja

𝐀1 =

⎣⎢⎢⎢⎢⎡0.4885

0.00.0

0.87260.00.0⎦

⎥⎥⎥⎥⎤

𝐀2 =

⎣⎢⎢⎢⎢⎡

0.00.4885

0.00.0

0.87260.0⎦

⎥⎥⎥⎥⎤

𝐀3 =

⎣⎢⎢⎢⎢⎡

0.00.0

0.46500.00.0

0.8853⎦⎥⎥⎥⎥⎤

𝐀4 =

⎣⎢⎢⎢⎢⎡

0.00.6610

0.00.0

−0.74580.0⎦

⎥⎥⎥⎥⎤

𝐀5 =

⎣⎢⎢⎢⎢⎡−0.6661

0.00.0

0.74580.00.0⎦

⎥⎥⎥⎥⎤

𝐀6 =

⎣⎢⎢⎢⎢⎡

0.00.0

0.19300.00.0

−0.9812⎦⎥⎥⎥⎥⎤

Određivanje koeficijenta dinamičnosti

▫ Oscilacije u podužnom pravcu prema oblicima oscilovanja to su oblici 𝐴1,𝐴5

𝐾𝑑 = 𝐾 𝑇⁄ → 𝐾𝑑 je veće kada je 𝑇 manje, odnosno 𝜔 veće

𝐾𝑑 = 0.5𝑇5

= 3.7936 > 1.0 → 𝐾𝑑 = 1.0

▫ Oscilacije u poprečnom pravcu oblici oscilovanja gde su dominantna pomeranja u Y pravcu 𝐴2,𝐴4

𝐾𝑑 = 0.5𝑇4

= 3.7936 > 1.0 → 𝐾𝑑 = 1.0


Određivanje seizmičkih sila ukupna težina objekta

𝐺1 = 𝑔 ∙ 𝑀1 = 9.81 ∙ 120 = 1177.2 kN 𝐺2 = 𝑔 ∙ 𝑀2 = 9.81 ∙ 60 = 588.6 kN 𝐺 = ∑ 𝐺𝑖 = 1765.82

𝑖=1 kN

ukupni seizmički koeficijent i ukupna seizmićka sila

𝐾 = 𝑘𝑜 ∙ 𝑘𝑠 ∙ 𝑘𝑑 ∙ 𝑘𝑝 = 1.0 ∙ 0.1 ∙ 1.0 ∙ 1.0 = 0.1 𝑆 = 𝐾 ∙ 𝐺 = 0.1 ∙ 1765.8 = 176.58 kN

seizmičke sile po tavanicama

𝑆𝑖 = 𝑆𝐺𝑖 ∙ 𝐻𝑖𝛴𝐺𝑖 ∙ 𝐻𝑖

𝑆1 = 176.581177.2 ∙ 4

9417.6= 88.29 kN

𝛴𝐺𝑖 ∙ 𝐻𝑖 = 1177.2 ∙ 4 + 588.6 ∙ 8 = 9417.6 𝑆2 = 176.58588.6 ∙ 89417.6

= 88.29 kN

Pomeranja tavanica podužni pravac poprečni pravac

𝐊 ∙ 𝐮 =

⎣⎢⎢⎢⎢⎡88.29

0.00.0

88.290.00.0⎦

⎥⎥⎥⎥⎤

→ 𝐮 =

⎣⎢⎢⎢⎢⎡

94.1760.00.0

172.6560.00.0⎦

⎥⎥⎥⎥⎤

1𝐸𝐼

𝐊 ∙ 𝐮 =

⎣⎢⎢⎢⎢⎡

0.088.29

0.00.0

88.290.0⎦

⎥⎥⎥⎥⎤

→ 𝐮 =

⎣⎢⎢⎢⎢⎡

0.094.176

0.00.0

172.6560.0⎦

⎥⎥⎥⎥⎤

1𝐸𝐼

Seizmičke sile po ramovima

𝚫𝑚 = 𝐚𝑚 ∙ 𝐮

𝐑𝑚 = 𝐙𝑚 ∙ 𝚫𝑚

Vektori pomeranja i sila rama 1

podužni pravac poprečni pravac

𝚫1 = � 94.176172.656�

1𝐸𝐼

𝐑1 = �23.544029.4300� 𝚫1 = �0.0

0.0� 𝐑1 = �0.00.0�



𝚫2 = � 94.176172.656�

1𝐸𝐼

𝐑2 = �41.20229.430� 𝚫2 = �0.0

0.0� 𝐑2 = �0.00.0�



𝚫3 = � 94.176172.656�

1𝐸𝐼

𝐑3 = �23.544029.4300� 𝚫3 = �0.0

0.0� 𝐑3 = �0.00.0�



𝚫4 = �0.00.0� 𝐑4 = �0.0

0.0� 𝚫4 = �94.1760.0�

1𝐸𝐼

𝐑4 = �35.31570.0�



𝚫5 = �0.00.0� 𝐑5 = �0.0

0.0� 𝚫5 = � 94.176172.656�

1𝐸𝐼

𝐑5 = � 8.829344.1450�


Sile po ramovima i zidovima pri delovanju seizmičke sile u poprečnom pravcu

▫ Pomeranja tavanica

𝐊

⎣⎢⎢⎢⎢⎡𝑢1𝑣1𝜃1𝑢2𝑣2𝜃2⎦⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡

0𝑆100𝑆20 ⎦⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡

0.088.29

0.00.0

88.290.0⎦

⎥⎥⎥⎥⎤

→

⎣⎢⎢⎢⎢⎡𝑢1𝑣1𝜃1𝑢2𝑣2𝜃2⎦⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡

0.019.6837

0.00.0

39.93310.0⎦

⎥⎥⎥⎥⎤

1𝐸𝐼

▫ Raspodela sila po ramovima i zidovima

𝚫1 = 𝚫2 = 𝚫3 = �0.00.0�

𝐑1 = 𝐑2 = 𝐑3 = �0.00.0�

𝚫4 = �19.68370.0�

1𝐸𝐼

𝚫5 = �19.68370.0�

1𝐸𝐼

𝚫6 = �19.683739.9331�

1𝐸𝐼

𝚫7 = �19.683739.9331�

1𝐸𝐼

𝐑4 = �7.38140.0� 𝐑5 = �7.3814

0.0�

𝐑6 = �36.764244.1447� 𝐑7 = �36.7642

44.1447�

Zadatak 11 Odrediti najpovoljniji i najnepovoljniji slučaj rasporeda montažnih ramova, prema kriterijumu momenta koji se javlja u ramovima. Proveru uraditi za 𝑥 i 𝑦 pravac dejstva zemljotresa. Masu ramova zanemariti.

𝑑𝑝 = 30 cm 𝑝 = 5 kNm2 ∆𝑔 = 10

kNm2 𝐾 = 0.1

𝑏𝑑

=5050

cm 𝛾𝑏 = 25 kNm3


Rešenje

Koncentrisanje mase konstrukcije u nivou tavanice

težina ploče M’

𝐺′ = 8.5 ∙ 8.5 ∙ 0.3 ∙ 25 + 8.5 ∙ 8.5 ∙52

= 722.5 kN

težina ploče M’’

𝐺′′ = 722.5 + 8.5 ∙ 8.5 ∙ 10 = 1445 kN ukupna težina ploče

𝐺 = 2 ∙ 𝐺′ + 𝐺′′ = 2890 kN

masa tavanice M’ : 𝐺′

9.81= 73.6493 kNs

2

m

masa tavanice M” : 𝐺"

9.81= 147.2987 kNs

2

m

ukupna masa M : 𝐺

9.81= 294.5973 kNs

2

m

Određivanje položaja centra mase

𝑥𝑀 =4.25 ∙ 73.6493 + 4.25 ∙ 147.2987 + 12.75 ∙ 73.6493

294.5973= 6.375 m

𝑦𝑀 =4.25 ∙ 147.2987 + 12.75 ∙ 2 ∙ 73.6493

294.5973= 8.5 m

Matrice krutosti ramova

▫ Ortogonalni ram

𝐸𝐼𝛿11 = 10. 6̇

Matrica fleksibilnosti za ceo ram

𝐷 = 2 ∙ 10. 6̇ ∙1𝐸𝐼

= 21. 3̇1𝐸𝐼


Matrica krutosti za ceo ram

𝑍 =1𝐷

= 4.6875 ∙ 10−2𝐸𝐼

▫ Kosi ram

𝐸𝐼𝛿11 = 12.8759

Matrica fleksibilnosti za ceo ram

𝐷 = 2 ∙ 12.87591𝐸𝐼

= 25.75181𝐸𝐼

Matrica krutosti za ceo ram

𝑍 =1𝐷

= 3.8832 ∙ 10−2𝐸𝐼

1. slučaj rasporeda ramova

matrice transformacija pojedinih ramova

𝐚𝑚 = [cos𝜑𝑚 sin𝜑𝑚 𝑟𝑚] 𝐚1 = [0 1 −6.375] 𝐚2 = [0 1 10.625] 𝐚3 = [1 0 −8.5] 𝐚4 = [1 0 8.5]

matrice krutosti ramova u globalnom koordinatnom sistemu

𝐤𝑚 = 𝐚𝑚T 𝐙𝑚𝐚𝑚

𝐤1 = 𝐚1T𝐙1𝐚1 = 𝐸𝐼 �01

−6.375� ∙ [4.6875 ∙ 10−2] ∙ [0 1 −6.375 ] = 4.6875 ∙ 10−2𝐸𝐼 �

0 0 00 1 −6.3750 −6.375 40.6406

�

𝐤2 = 4.6875 ∙ 10−2𝐸𝐼 �0 0 00 1 10.6250 10.625 112.8906

� 𝐤3 = 4.6875 ∙ 10−2𝐸𝐼 �1 0 −8.50 0 0

−8.5 0 72.25�

𝐤4 = 4.6875 ∙ 10−2𝐸𝐼 �1 0 8.50 0 0

8.5 0 72.25�

𝐊 = �𝐤i

4

i=1

= 4.6875 ∙ 10−2𝐸𝐼 �2 0 00 2 4.250 4.25 298.0312

�

Proračun seizmičkih uticaja

𝑆𝑥 = 𝑆𝑦 = 𝐾 ∙ 𝐺 = 0.1 ∙ 2890 = 289 kN

raspodela seizmičkih sila na pojedine ramove

▫ 𝑋 pravac

𝐒 = 𝐊 ∙ 𝐮


�289

00� = 𝐸𝐼 �

0.09375 0 00 0.09375 0.199220 0.19922 13.97021

� ∙ �𝑢𝑣𝜃� 𝐮 =

1𝐸𝐼�3082. 6̇

00

�

∆i= 𝐚𝑖 ∙ 𝐮

∆3= 𝐸𝐼[1 0 −8.5] ∙ �3082. 6̇

00

� =1𝐸𝐼

∙ 3082. 6̇ = ∆4

𝐑i = 𝐙i ∙ ∆𝑖

𝐑3 = 𝐑4 = 𝐙3 ∙ ∆3= 4.6875 ∙ 10−2 ∙ 3082. 6̇ = 144.5 kN

▫ Y pravac

�0

2890� = 𝐸𝐼 �

0.09375 0 00 0.09375 0.199220 0.19922 13.97021

� ∙ �𝑢𝑣𝜃� 𝐮 =

1𝐸𝐼�

03179.0013−45.3337

�

∆1= 𝐸𝐼[0 1 −6.375] ∙ �0

3179.0013−45.3337

� =1𝐸𝐼

∙ 3468.0033 ∆2=1𝐸𝐼

∙ 2697.3312

𝐑1 = 𝐙1 ∙ ∆1= 4.6875 ∙ 10−2 ∙ 3468.0033 = 162.5625 kN

𝐑2 = 𝐙2 ∙ ∆2= 4.6875 ∙ 10−2 ∙ 2697.3312 = 126.4375 kN

2. slučaj rasporeda ramova 𝐙1 = 𝐙2 = 4.6875 ∙ 10−2 𝐙3 = 3.8832 ∙ 10−2


𝐚1 = [0 1 10.625] 𝐚2 = [1 0 −8.5] 𝐚3 = [−0.7071 0.7071 1.5026]


𝐤1 = 4.6875 ∙ 10−2𝐸𝐼 �0 0 00 1 10.6250 10.625 112.8906

�

𝐤2 = 4.6875 ∙ 10−2𝐸𝐼 �1 0 −8.50 0 0

−8.5 0 72.25�

𝐤3 = 3.8832 ∙ 10−2𝐸𝐼 �0.5 −0.5 −1.0625−0.5 0.5 1.0625

−1.0625 1.0625 2.2578�

𝐊 = 𝐸𝐼 �0.06629 −0.01942 0.4397−0.01942 0.06629 0.4568

0.4397 0.4568 8.76614�


𝑆𝑥 = 𝑆𝑦 = 289 kN raspodela seizmičkih sila na pojedine ramove ramove

▫ 𝑋 pravac

�289

00� = 𝐸𝐼 �

0.06629 −0.01942 0.4397−0.01942 0.06629 0.4568

0.4397 0.4568 8.76614� ∙ �

𝑢𝑣𝜃� 𝐮 =

1𝐸𝐼�

6.8039−1.56890.4378

� ∙ 103

∆2=1𝐸𝐼

∙ 3082.6635 ∆3= −1𝐸𝐼

∙ 5262.5678

𝐑2 = 𝐙2 ∙ ∆2= 144.5 kN 𝐑3 = 𝐙3 ∙ ∆3= −204.356 kN

𝐑3x = 0.7071 ∙ (−204.356) = −144.5 kN

𝐑2 + 𝐑3x = 289 kN


▫ Y pravac

�0

2890� = 𝐸𝐼 �

0.06629 −0.01942 0.4397−0.01942 0.06629 0.4568

0.4397 0.4568 8.76614� ∙ �


1𝐸𝐼�−1.56899.0898−0.6379

� ∙ 103

∆1= −1𝐸𝐼

∙ 2311.996 ∆3= −1𝐸𝐼

∙ 6578.2097

𝐑1 = 𝐙1 ∙ ∆1= 108.3748kN 𝐑3 = 𝐙3 ∙ ∆3= 255.445 kN

𝐑3y = 0.7071 ∙ 255.445 = 180.6252kN

𝐑1 + 𝐑3y = 289 kN


𝐙1 = 𝐙2 = 4.6875 ∙ 10−2 𝐙3 = 3.8832 ∙ 10−2


𝐚1 = [0 1 10.625] 𝐚2 = [1 0 8.5] 𝐚3 = [0.7071 0.7071 −4.5078]


𝐤1 = 4.6875 ∙ 10−2𝐸𝐼 �0 0 00 1 10.6250 10.625 112.8906

�

𝐤2 = 4.6875 ∙ 10−2𝐸𝐼 �1 0 8.50 0 0

8.5 0 72.25�

𝐤3 = 3.8832 ∙ 10−2𝐸𝐼 �0.5 0.5 −3.18750.5 0.5 −3.1875

−3.1875 −3.1875 20.3203�

𝐊 = 𝐸𝐼 �0.06629 0.01942 0.274660.01942 0.06629 0.374270.27466 0.37427 9.46754

�


𝑆𝑥 = 𝑆𝑦 = 289 kN

raspodela seizmičkih sila na pojedine ramove ramove

▫ 𝑋 pravac

�289

00� = 𝐸𝐼 �

0.06629 0.01942 0.274660.01942 0.06629 0.374270.27466 0.37427 9.46754

� ∙ �𝑢𝑣𝜃� 𝐮 =

1𝐸𝐼�

5079.2417−844.5444−113.9659

�

∆2=1𝐸𝐼

∙ 4110.5314 ∆3=1𝐸𝐼

∙ 3508.09

𝐑2 = 𝐙2 ∙ ∆2= 192.6812 𝑘𝑁 𝐑3 = 𝐙3 ∙ ∆3= 136.2262 kN

𝐑3x = 0.7071 ∙ 136.2262 = 96.3146 𝑘𝑁

𝐑2 + 𝐑3x = 289 kN

▫ Y pravac

�0

2890� = 𝐸𝐼 �

0.06629 0.01942 0.274660.01942 0.06629 0.374270.27466 0.37427 9.46754

� ∙ �𝑢𝑣𝜃� 𝐮 =

1𝐸𝐼�−844.54445752.6874−202.9139

�

∆1=1𝐸𝐼

∙ 3596.7269 ∆3=1𝐸𝐼

∙ 4385.2433

𝐑1 = 𝐙1 ∙ ∆1= 168.5966 kN 𝐑3 = 𝐙3 ∙ ∆3= 170.2878 kN


𝐑3y = 0.7071 ∙ 170.2878 = 120.4116 kN

𝐑1 + 𝐑3y = 289 kN



𝐚1 = [0 1 −6.375] 𝐚2 = [0 1 2.125] 𝐚3 = [1 0 −8.5] 𝐚4 = [1 0 0]


𝐤1 = 4.6875 ∙ 10−2𝐸𝐼 �0 0 00 1 −6.3750 −6.375 40.6406

� 𝐤2 = 4.6875 ∙ 10−2𝐸𝐼 �0 0 00 1 2.1250 2.125 4.5156

�

𝐤3 = 4.6875 ∙ 10−2𝐸𝐼 �1 0 −8.50 0 0

−8.5 0 72.25� 𝐤4 = 4.6875 ∙ 10−2𝐸𝐼 �

1 0 00 0 00 0 0

�

𝐊 = 4.6875 ∙ 10−2𝐸𝐼 �2 0 −8.50 2 −4.25

−8.5 −4.25 117.4062�


𝑆𝑥 = 𝑆𝑦 = 289 kN

raspodela seizmičkih sila na ramove

𝐗 pravac

�289

00� = 𝐸𝐼 �

0.09375 0 −0.398440 0.09375 −0.19922

−0.39844 −0.19922 5.5034� ∙ �


1𝐸𝐼�4624.0011770.6672362.6669

�

∆3=1𝐸𝐼

∙ 1541.3323 ∆4=1𝐸𝐼

∙ 4624.0011

𝐑3 = 𝐙3 ∙ ∆3= 72.25 kN 𝐑4 = 𝐙4 ∙ ∆4= 216.75 kN

Y pravac

�0

2890� = 𝐸𝐼 �

0.09375 0 −0.398440 0.09375 −0.19922

−0.39844 −0.19922 5.5034� ∙ �


1𝐸𝐼�

770.66723468.0003181.3335

�

∆1=1𝐸𝐼

∙ 2311.9995 ∆2=1𝐸𝐼

∙ 3853.3339

𝐑1 = 𝐙1 ∙ ∆1 = 108.3745 kN 𝐑2 = 𝐙2 ∙ ∆2 = 180.625 kN




𝐚1 = 𝐚2 = [0 1 2.125] 𝐚3 = 𝐚4 = [1 0 0]


𝐤1 = 𝐤2 = 4.6875 ∙ 10−2𝐸𝐼 �0 0 00 1 2.1250 2.125 4.5156

�

𝐤3 = 𝐤4 = 4.6875 ∙ 10−2𝐸𝐼 �1 0 00 0 00 0 0

�

𝐊 = 4.6875 ∙ 10−2𝐸𝐼 �2 0 00 2 4.250 4.25 9.0312

�


𝑆𝑥 = 𝑆𝑦 = 289 kN

raspodela seizmičkih sila na ramove ▫ 𝑋 pravac

�289

00� = 𝐸𝐼 �

0.09375 0 00 0.09375 0.199220 0.19922 0.42334

� ∙ �𝑢𝑣𝜃� 𝐮 =

1𝐸𝐼�3082. 6̇

00

�

∆3=1𝐸𝐼

∙ 3082. 6̇ = ∆4

𝐑3 = 𝐙3 ∙ ∆3= 144.5 kN = 𝐑4

▫ Y pravac

�0

2890� = 𝐸𝐼 �

0.09375 0 00 0.09375 0.199220 0.19922 0.42334

� ∙ �𝑢𝑣𝜃� 𝐮 =

1𝐸𝐼�

03234.2761−71.3452

�

∆1= ∆2=1𝐸𝐼

∙ 3082. 6̇

𝐑1 = 𝐙1 ∙ ∆1= 144.5 kN = 𝐑2 Drugi način proračuna raspodele seizmičke sile 1. slučaj rasporeda ramova

određivanje centra krutosti

𝐙1 = 𝐙2 = 𝐙3 = 𝐙4 = 4.6875 ∙ 10−2𝐸𝐼

�𝐙x = 9.375 ∙ 10−2𝐸𝐼 �𝐙y = 9.375 ∙ 10−2𝐸𝐼

𝑥𝑐 =∑ 𝐙𝑦𝑖 ∙ 𝑥𝑖4i=1

∑ 𝐙𝑦𝑖li=1

=4.6875 ∙ 10−2 ∙ 17

9.375 ∙ 10−2= 8.5 m

𝑦𝑐 =∑ 𝐙𝑥𝑖 ∙ 𝑦𝑖4i=1

∑ 𝐙𝑥𝑖4i=1

=4.6875 ∙ 10−2 ∙ 17

9.375 ∙ 10−2= 8.5 m

𝑒𝑥 = 2.125 m 𝑒𝑦 = 0.0 m

𝑆𝑦 = 𝑆𝑥 = 289 kN


raspodela seizmičkih sila

𝐑𝑦𝑖𝑇 =

𝑆𝑦∑ 𝐙𝑦𝑖4i=1

𝐙𝑦𝑖 = 𝐑𝑥𝑖𝑇 =

𝑆𝑥∑ 𝐙𝑥𝑖4i=1

𝐙𝑥𝑖 =289 ∙ 4.6875 ∙ 10−2

9.375 ∙ 10−2= 144.5 kN

𝐑𝑦1𝑇 = 𝐑𝑦2

𝑇 = 𝐑𝑥3𝑇 = 𝐑𝑥4

𝑇 = 144.5 kN

𝑀𝑐𝑖 = 𝑆𝑖 ∙ 𝑒𝑖

𝐑𝑥𝑖𝑅 = 𝐙𝑥𝑖 ∙ 𝜑 ∙ 𝑦𝑖𝑐 𝐑𝑦𝑖

𝑅 = 𝐙𝑦𝑖 ∙ 𝜑 ∙ 𝑥𝑖𝑐 𝐑𝑖 = 𝐑𝑖𝑅 + 𝐑𝑖

𝑇

�𝐙𝑦𝑖 ∙ 𝜑 ∙ (𝑥𝑖𝑐)2l

i=1

+ �𝐙𝑥𝑖 ∙ 𝜑 ∙ (𝑦𝑖𝑐)2k

i=1

= 𝑀𝑐

X pravac

𝑀𝑐𝑦 = 0.0 kNm 𝜑 = 0.0

𝑅𝑦1𝑅 = 𝑅𝑦2𝑅 = 𝑅𝑥3𝑅 = 𝑅𝑥4𝑅 = 0.0 kN

Ukupne horizontalne sile koje deluju na vertikalne elemente jednake su silama pri translaciji i pri rotaciji tavanice:

𝑅𝑦1 = 0.0 kN 𝑅𝑦2 = 0.0 kN 𝑅𝑥3 = 144.5 kN 𝑅𝑥4 = 144.5 kN

Y pravac

𝑀𝑐𝑥 = 289 ∙ 2.125 = 614.125 kNm

𝜑 ∙ 4.6875 ∙ 10−2𝐸𝐼[8.52 ∙ 4] = 614.125 𝜑 =614.12513.5469

=1𝐸𝐼

45. 3̇

horizontalne sile koje se prenose na vertikalne elemente

𝐑𝑦1𝑅 = 4.6875 ∙ 10−2𝐸𝐼 ∙

1𝐸𝐼

∙ 45. 3̇ ∙ 8.5 = 18.0625 kN

𝐑𝑦2𝑅 = −18.0625 kN 𝐑𝑥3

𝑅 = 18.0625 kN 𝐑𝑥4𝑅 = −18.0625 kN

Ukupne horizontalne sile koje deluju na vertikalne elemente jednake su silama pri translaciji i pri rotaciji tavanice:

𝐑𝑦1 = 144.5 + 18.0625 = 162.5625 kN 𝐑𝑦2 = 144.5 − 18.0625 = 126.4375 kN

𝐑𝑥3 = 18.0625 kN 𝐑𝑥4 = 18.0625 kN

merodavne sile u ramovima 𝐑𝟏 = 162.5625 kN 𝐑𝟐 = 126.4375 kN 𝐑𝟑 = 144.5 kN 𝐑𝟒 = 144.5 kN

Uticaji u ramovima

Dijagram momenata usled jediničnog opterećenja ortogonalnog rama (levo) i kosog rama (desno)

S obzirom da su maksimalni momenti pri dejstvu jedinične sile isti za oba rama, mogu se porediti merodavne sile u ramovima, kako bi se došlo do zaključka koji je raspored ramova najpovoljniji odnosno najnepovoljniji.

Iz priložene tabele vidi se da je najpovoljniji raspored ramova u 5. slučaju, a najnepovoljniji raspored ramova je u 2. slučaju.

ram 1. slučaj 2. slučaj 3. slučaj 4. slučaj 5. slučaj R [kN] R [kN] R [kN] R [kN] R [kN] 𝑅1 162.5625 108.3748 168.5966 108.375 144.5 𝑅2 126.4375 144.5 192.6812 180.625 144.5 𝑅3 144.5 255.445 170.2878 72.25 144.5 𝑅4 144.5 - - 216.75 144.5


Zadatak 12 Primenom modalne analize odrediti odgovor neprigušenog sistema. Sistem je bio u mirovanju pre početka dejstva dinamičkog opterećenja. Zanemariti uticaj aksijalne deformacije. Rigle rama su beskonačno fleksiono krute.

𝐸𝐼 = 330000 kNm2 𝐿 = 5 m 𝐻 = 3 m 𝑀 = 20 kNs2/m 𝑝 = 0.7𝜔1

Rešenje

Matrica krutosti sistema

Dati ram je “smičući ram”, što značajno pojednostavljuje određivanje matrice krutosti. S obzirom da su rigle rama beskonačno fleksiono krute, svaki stub rama je obostrano uklješten na svojim krajevima. Elementi matrice krutosti se mogu odrediti zadavanjem jediničnog pomeranja u nivou rigli i određivanjem reakcija oslonaca na svakom spratu. Te relacije predstavljaju članove matrice krutosti.


𝐊 =

⎣⎢⎢⎢⎢⎢⎢⎢⎡4

12𝐸𝐼(1.1𝐻)3

+ 312𝐸𝐼𝐻3 −3

12𝐸𝐼𝐻3 0.0 0.0

−312𝐸𝐼𝐻3 3

12𝐸𝐼𝐻3 + 2

12𝐸𝐼(2𝐻)3

+12𝐸𝐼(3𝐻)3

−212𝐸𝐼(2𝐻)3

−12𝐸𝐼(3𝐻)3

0.0 −212𝐸𝐼(2𝐻)3

212𝐸𝐼(2𝐻)3

+ 212𝐸𝐼𝐻3 −2

12𝐸𝐼𝐻3

0.0 −12𝐸𝐼(3𝐻)3

−212𝐸𝐼𝐻3 2

12𝐸𝐼𝐻3 +

12𝐸𝐼(3𝐻)3⎦

⎥⎥⎥⎥⎥⎥⎥⎤

𝐊 = �

8.8007 −4.4 0.0 0.0−4.4 4.821 −0.3667 −0.05430.0 −0.3667 3.3 −2.93330.0 −0.0543 −2.9333 2.9877

� 105

Matrica masa sistema

𝐌 = �80

6040

60

�

Iz jednačine (𝐊 − 𝜔2𝐌)𝐀 = 𝟎 potrebno je prvo odrediti svojstvene vektore odnosno modalnu matricu:

𝐀 = �

1.0 1.0 1.0 1.01.9389 1.4470 −0.3280 −0.9237

10.7430 −0.1370 −10.4015 0.172011.3503 −0.2782 6.5019 −0.0682

�

Kružne frekvencije sistema su:

𝜔1 = 18.3513 rad

s 𝜔2 = 55.1574

rads

𝜔3 = 113.1594 rad

s 𝜔4 = 126.8118

rads

Dijagonalna matrica 𝐌∗ se dobija množenjem:

𝐌∗ = 𝐀𝑇𝐌𝐀 = �

1.2652 0 0 00 0.0211 0 00 0 0.6951 00 0 0 0.0133

� ∙ 104

Kad postoji prinudna sila, diferencijalne jednačine postaju nehomogene, a slobodni član se određuje preko izraza:

𝐏∗ = 𝐀𝑇P𝑜 = �14.28922.16887.17380.0081

�P𝑜

Diferencijalne jednačine oblika: �̈�𝑖 + 𝜔𝑖2𝑌𝑖 = 𝑃𝑜𝑖

𝑀𝑖 𝑖 = 1,2,3,4 imaju rešenje:

𝑌𝑖 =1

𝜔𝑖2 − 𝑝2

𝑃𝑜𝑖𝑀𝑖

sin𝑝𝑡 𝑝 = 0.7𝜔1 = 12.8459 rad

s

𝑌1 =1

18.35132 − 12.8459214.2892𝑃𝑜

1.2652 ∙ 104sin𝑝𝑡 = 0.6576 ∙ 10−5𝑃𝑜 sin𝑝𝑡

𝑌2 =1

55.15742 − 12.845922.1688𝑃𝑜

0.0211 ∙ 104sin𝑝𝑡 = 0.3572 ∙ 10−5𝑃𝑜 sin𝑝𝑡

𝑌3 =1

126.81182 − 12.845927.1738𝑃𝑜

0.6951 ∙ 104sin𝑝𝑡 = 0.0065 ∙ 10−5𝑃𝑜 sin𝑝𝑡

𝑌4 =1

113.15942 − 12.845920.0081𝑃𝑜

0.0133 ∙ 104sin𝑝𝑡 = 0.0005 ∙ 10−5𝑃𝑜 sin𝑝𝑡

𝐘 = 10−5 �0.65760.35720.00650.0005

� 𝑃𝑜 sin𝑝𝑡

Odgovor sistema u osnovnim koordinatama je: 𝐲 = 𝐀 𝐘 = 10−5 �1.02181.78936.94817.4068

� 𝑃𝑜 sin(12.8459 𝑡)

Documents

POJAM STABILNOSTI KONSTRUKCIJA · 2020. 11. 22. · Rigle tavanica su fleksiono beskonačno krute. Podaci: 𝑀. 1= 50 kNs2 m 𝑀2= 30 kNs2 m 𝑀3= 20 kNs2 m 𝐸𝐼= 20000 𝑘𝑁𝑚2