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Structural Stability

SOLUTION Problem 1.1a Find the critical load Pcr of the bar – spring systems, using the bifurcation approach. Assume that all the bars are rigid L

P

k s1

ks2

Solution  Assume that the rotation  at node A is small L



k s1

A

B

P

k s2 L L sin   L L cos  L



Using Bifurcation approach Equilibrium condition M A  0  PL sin   k s1  k s 2 ( L sin  )( L cos )  0 With small deflection assumption sin    , cos  1

 PL  k s1  k s 2 L2  0





  PL  k s1  k s 2 L2  0 (1)   0 P  0 ~  (trivial solution) (2) PL  k s1  k s 2 L2 =0 Hence, Pcr 

k s1  ks2 L L

   ~ 

1

Structural Stability

Problem 1.1c Find the critical load Pcr of the bar – spring systems, using the bifurcation approach. Assume that all the bars are rigid 0.5L

0.5L

L

0.5L

A B

E

P

E

P

D

C

k s1

ks2

Solution  Assume that the rotation  at node A is small P

A

RA



B

B

C

C  

D

D

E

k s 2 D

k s1 B

RE



Using Bifurcation approach We have

   C  L sin   L   E 

C



L 2   1.5 L 3

1.5 L 1 1 1 1   B   C  L ;  D   C  L 2 2 3 3 Equilibrium condition M A  0

 k s1 B (0.5L)  k s 2 D (2 L)  RE (2.5L)  0 1 1  k s1 L (0.5L)  k s 2 L (2 L)  RE (2.5L)  0 2 3 4 1   R E   k s1  k s 2  L 15  10 

(1)

2

Structural Stability

Consider member CE

D

C

P

P

 ks 2 D

C’

RC

L

RE

0.5L

Equilibrium condition  M C'  0  k s 2 D ( L)  RE (1.5 L)  P  0

1  k s 2 L ( L)  RE (1.5L)  PL  0 3

(2)

From (1) and (2) we have 1 4 1  k s 2 L2   k s1  k s 2  L (1.5 L)  PL  0 3 15  10  3 1     P  k s1 L  k s 2 L   0 20 15   (a)   0 P  0 ~  (trivial solution) 3 1 (b) P  k s1 L  k s 2 L =0 20 15 Hence, P

3 1 k s1 L  k s 2 L 20 15

   ~ 

3

Structural Stability

Problem 1.4 Investigate the stability behavior of the snap through spring – bar model shown P

L

ks



Solution P

  ks  A

A’

B’



L sin 



L sin(   )

L

B

RB

B

A

 A RA

L cos  L cos(   )



Using energy approach Assume that the bar rotates angle of   A  L cos(   )  L cos   B  L sin   L sin(   )  The strain energy 1 1 2 U  k s 2A  k s L cos(   )  L cos  2 2 1 U  k s L2 cos2 (   )  cos2   2 cos(   ) cos 2  The potential energy V   P B   PLsin   sin(    ) The total potential energy





4

Structural Stability





1   U  V  k s L2 cos2 (   )  cos2   2 cos(   ) cos  PLsin   sin(   ) 2 

For equilibrium d 0 d 1  k s L2 2 cos(   ) sin(   ) 1  2 cos  sin(   ) 1  PL cos(   ) 1  0 2  k s L2 cos(   )  cos sin(   )  PL cos(   )  0

Hence, P  k s Lcos(   )  cos  tan(   ) 

Stability condition d 2  2 1  k L cos 2(   ) 2  cos cos(   ) 1  PL sin(   ) 1 s 2  d 2 

d 2  k s L2  cos 2(   )  cos cos(   )  PL sin(   ) 2 d When

d 2 0 d 2  k s L2  cos 2(   )  cos cos(   )  k s Lcos(   )  cos  tan(   ) L sin(   )  0  sin 2 (   )  cos2 (   )  cos cos(   )  sin 2 (   )  cos tan(   ) sin(   )  0  cos2 (   )  cos cos(   )  cos tan(   ) sin(   )  0

If cos(   ) #0, we have d 2  0   cos3 (   )  cos cos2 (   )  cos sin 2 (   )  0 2 d  cos  cos3 (   )  0



  cos1 cos 

1/ 3

          cos cos   1

d 2  If  0  cos  cos3 (   )  0 2 d 1/ 3 1/ 3    cos1 cos       cos1 cos  The system is unstable







1/ 3



5

Structural Stability

d 2  If  0  cos  cos3 (   )  0 2 d 1/ 3 1/ 3      cos1 cos  or     cos1 cos  The system is stable









6

Structural Stability

DISCUSSION A. Snap through We will investigate stability behavior with the given angle  of 300 P  k s Lcos(   )  cos  tan(   )



  ks  A



A’

B’

L sin 



L sin(   )

L



B

d 2  L2 k s  cos2 (   )  cos cos(   )  cos tan(   ) sin(   ) 2 d P RB B

A

 A RA

L cos  L cos(   ) d2/d2





P





(ksL)

0

45

0.00000

0.50000

Stable

5

40

0.04945

0.33624

Stable

10

35

0.07845

0.19221

Stable

15

30

0.09175

0.06650

Stable

17.986

27.014

0.09370

0.00000

Critical

20

25

0.09289

-0.04119

Unstable

25

20

0.08465

-0.13053

Unstable

30

15

0.06935

-0.20096

Unstable

35

10

0.04897

-0.25183

Unstable

40

5

0.02529

-0.28260

Unstable

45

0

0.00000

-0.29289

Unstable

50

-5

-0.02529

-0.28260

Unstable

55

-10

-0.04897

-0.25183

Unstable

60

-15

-0.06935

-0.20096

Unstable

65

-20

-0.08465

-0.13053

Unstable

70

-25

-0.09289

-0.04119

Unstable

72.014

-27.014

-0.09370

0.00000

Critical

75

-30

-0.09175

0.06650

Stable

80

-35

-0.07845

0.19221

Stable

85

-40

-0.04945

0.33624

Stable

90

-45

0.00000

0.50000

Stable

State of the structure

7

Structural Stability

Applied Load vs Rotation 0.15

Load P (k sL)

0.10 0.05 0.00 -0.05

0

10

20

30

40

50

60

70

80

90

100

70

80

90

100

-0.10 -0.15 Rotaion  (degree)

d2 /d 2 vs Rotation 0.60 0.50 0.40

d2 /d 2

0.30 0.20 0.10 0.00 -0.10 0

10

20

30

40

50

60

-0.20 -0.30 -0.40 Rotation (degree)

From the above graph, we can see that: For given angle  of 450, firstly, the system is in stable state when the load P less than (0.0937k sL). After P reaches that critical value, the system is changed into unstable (and the rotation is of 17.9860). In this stage, the rotation of bar  increases very quickly so that the system comes to stable state. The rotation  rises continuously until it reaches the value of 72.0140. At that time, the system begins being stable again. In conclusion, the system is stable for 17.9860 or 72.0140, but is unstable for 17.9860 < 72.0140

8

Structural Stability

B. Snap through We will investigate how do applied load P and rotation  change when snap through occurs P  k s Lcos(   )  cos  tan(   )



    cos1 cos 1/ 3



P

  ks  A

A’



B’

L sin 



L sin(   )

L

B

RB

B

A

 A RA

L cos  L cos(   )



cos

 

 

P (ksL)

0

1.000

0.000

0.000

0.00000

5

0.996

2.112

2.888

0.00013

10

0.985

4.217

5.783

0.00102

15

0.966

6.306

8.694

0.00345

20

0.940

8.373

11.627

0.00819

25

0.906

10.407

14.593

0.01599

30

0.866

12.399

17.601

0.02765

35

0.819

14.336

20.664

0.04394

40

0.766

16.205

23.795

0.06568

45

0.707

17.986

27.014

0.09370

50

0.643

19.658

30.342

0.12891

55

0.574

21.188

33.812

0.17232

60

0.500

22.533

37.467

0.22510

65

0.423

23.629

41.371

0.28872

70

0.342

24.373

45.627

0.36521

75

0.259

24.589

50.411

0.45765

80

0.174

23.911

56.089

0.57160

85

0.087

21.319

63.681

0.72014

90

0.000

0.000

90.000

1.00000

9

Structural Stability

Angle  vs Rotation  When Snap Through Occurs

Rotation  (degree)

30 25 20 15 10 5 0 0

10

20

30

40

50

60

70

80

90

100

Angle  (degree)

Angle  vs Load P When Snap Through Occurs 1.2

Load P (k sL)

1.0 0.8 0.6 0.4 0.2 0.0 0

10

20

30

40

50

60

70

80

90

100

Angle  (degree)

From the above graph, we can see that when snap through occurs, the load only increases lightly with small angle  (0~400), but very dramatically with large angle  (>400). On the contrary, the rotation goes up so quickly until it reaches the maximum value of 24.5890 (at that time, the angle of bar And when the angle of bar , the higher the angle of bar  is, the smaller the rotation  is.

10