Advanced Steel Structures
SOLUTION
Problem 1.3
Investigate the stability behavior of this asymmetric spring – bar model
L
P
A
B
ks
450
C
Solution

B’
P

L sin 
A
B
L(1  cos )
os
Lc
0
45
C’
450
C
L

Determine Δ
  L sin  cos 45o  L(1  cos ) cos 45o 

2
sin   cos  1 L
2
The strain energy
2

1
1  2
U  k s 2  k s 
Lsin   cos  1
2
2  2

1
2
U  k s L2 sin   cos   2sin   cos   2
4
1
1
U  k s L2 1  sin 2  2sin   cos   1  k s L2 sin 2  2sin   cos   2
4
4

The potential energy
V   P B   PL1  cos  
The total potential energy


1
Advanced Steel Structures
1
  U  V  k s L2 sin 2  2sin   cos   2  PL1  cos 
4

For equilibrium
d
1
 0  k s L2 2 cos2  2cos  sin    PL sin   0
d
4
 Fundamental path θ =0
The (1) satisfies itself P  0 ~ 
 Postbuckling equilibrium path
1
P
k s Lcos2  cos  sin  
2 sin 
For small deformation
sin   
 0
cos  1
Thus,
1
1
P  Pcr 
k s L1  1     k s L
2
2

(1)
For stability
d 2 1
 k s L2  2 sin 2  sin   cos   PL cos
2
2
d
d 2 1
1
2



k
L

2
sin
2


sin


cos


k s Lcos 2  cos  sin  L cos
s
2 sin 
d 2 2
d 2 1
1

cos2  cos  sin  cos 
 k s L2  2 sin 2  sin   cos 
2
2
2 sin 
d


Establish the table to investigate the stability of the system
2
Advanced Steel Structures

P
d2/d2
State of

(ksL)
-90
1.000
-0.500
Unstable
-80
1.065
-0.249
Unstable
-70
1.090
-0.029
Unstable
-68.529
1.090
0.000
???
-60
1.077
0.144
Stable
-50
1.033
0.259
Stable
-40
0.961
0.310
Stable
-30
0.866
0.299
Stable
-20
0.754
0.233
Stable
-10
0.630
0.127
Stable
0
0.500
0.000
Neutral
10
0.370
-0.127
Unstable
20
0.246
-0.233
Unstable
30
0.134
-0.299
Unstable
40
0.039
-0.310
Unstable
50
-0.033
-0.259
Unstable
60
-0.077
-0.144
Unstable
68.529
-0.090
0.000
???
70
-0.090
0.029
Stable
80
-0.065
0.249
Stable
90
0.000
0.500
Stable
the structure
From the table we can conclude that
0
0
 -90 ≤ θ < -68.5 : the structure is unstable
0
 -68.5 < θ < 0 : the structure is stable
0
 0 < θ < 68.5 : the structure is unstable
0
0
 -68.5 < θ ≤ 90 : the structure is stable
3
Advanced Steel Structures
Load P vs Rotation 
1.2
1.0
0.8
Load P (k sL)
0.6
0.4
0.2
0.0
-90 -80 -70 -60 -50 -40 -30 -20 -10
-0.2 0
10
20
30
40
50
60
70
80
90
30
40
50
60
70
80
90
-0.4
-0.6
-0.8
Rotation  (degree)
d  /d vs Rotation 
2
2
0.6
0.4
d2 /d 2
0.2
0.0
-90 -80 -70 -60 -50 -40 -30 -20 -10 0
10
20
-0.2
-0.4
-0.6
Rotation  (degree)
4
Advanced Steel Structures
Problem 2
The system shown below is composed of 2 rigid bars and 2 rotational springs with the
same spring constant ks
(1) Determine the buckling models and buckling shapes of the system
(2) If you can add one extra spring with the same spring constant to the system, where
will you install the spring to maximize the critical load? Explain your decision.
L
P
L
B
A
Solution
L
P
2
1
L
B
A
(1)

The strain energy
1
1
2
U  k s12  k s  2  1 
2
2

The potential energy
V   PL  L cos  2  L  L cos 1    PL 2  cos 1  cos  2 
The total potential energy
1
1
2
  U  V  k s12  k s  2  1   PL2  cos1  cos 2 
2
2
5
Advanced Steel Structures

For equilibrium
 d
 d  0
 1

 d  0
 d 2
k s1  k s  2  1   PLsin 1   0

k s  2  1   PLsin  2   0
For small deformation
sin   
 0
cos  1
Thus,
k s1  k s  2  1   PL1   0

k s  2  1   PL 2   0
2k s  PL1  k s 2  0

 k s1  k s  PL 2  0
The matrix form of the above two equations
 k s  1  0
2k s  PL
 k
    0
k

PL
s
s

 2   
(1)
Fundamental path θ1 = θ2 =0; P  0 ~ 
 Postbuckling equilibrium path
For non-trivial solution, we need
2k s  PL
 ks
0
 ks
k s  PL

 2k s  PLk s  PL  k s2  0
 2k s2  3PLks  P 2 L2  k s2  0
 P 2 L2  3PLks  k s2  0
3 5
ks
2L
The smallest value of P will be the critical load
3  5  ks 
k 
Pcr 
   0.382 s 
2 L
L
P
6
Advanced Steel Structures
Determine the buckling modes

The first mode
3  5  ks 
P1 
 
2 L
For θ1 = 1, the from (1) we have

3  5  ks  
 2k s 
  L 1  k s 2  0
2
L 

1 5
 1.618(rad )
2

The second mode
3  5  ks 
P1 
 
2 L
For θ1 = 1, the from (1) we have

3  5  ks  
 2k s 
  L 1  k s 2  0
2
L 

2 
1 5
 0.618(rad )
2
The two buckling modes are as following
2 
P
L
L
P
A
The first mode
1  1
B
L
1  1
 2  0.618
B
L
 2  1.618
A
The second mode
(2)
(a) Case 1: add one spring at point A
7
Advanced Steel Structures
P
L
L
P
2
B
ks
ks
2k s
L
L
1
A
2k s

The strain energy
1
1
2
U  2k s12  k s  2  1 
2
2

The potential energy
V   PL  L cos  2  L  L cos 1    PL 2  cos 1  cos  2 
The total potential energy
1
1
2
  U  V  2k s12  k s  2  1   PL2  cos1  cos 2 
2
2

For equilibrium
 d
 d  0
 1

 d  0
 d 2
2k s1  k s  2  1   PLsin 1   0

k s  2  1   PLsin  2   0
For small deformation
sin   
 0
cos  1
Thus,
2k s1  k s  2  1   PL1   0

k s  2  1   PL 2   0
3k s  PL1  k s 2  0

 k s1  k s  PL 2  0
8
Advanced Steel Structures
The matrix form of the above two equations
 k s  1  0
3k s  PL
 k
    0
k

PL
s
s

 2   
For non-trivial solution, we need
3k s  PL
 ks
0
 ks
k s  PL
 3k s  PLk s  PL  k s2  0
 3k s2  4 PLks  P 2 L2  k s2  0
 P 2 L2  4 PLks  2k s2  0
k
P 2 2 s
L
The smallest value of P will be the critical load
k
k
Pcr  2  2 s  0.58 s
L
L
(b) Case 2: add one spring at point B




P
L
L
P
(2)
2
B
2k s
2k s
A
ks
L
L
1
ks

The strain energy
1
1
2
U  k s12  2k s  2  1 
2
2

The potential energy
V   PL  L cos  2  L  L cos 1    PL 2  cos 1  cos  2 
The total potential energy
1
1
2
  U  V  k s12  2k s  2  1   PL2  cos1  cos 2 
2
2
9
Advanced Steel Structures

For equilibrium
 d
 d  0
 1

 d  0
 d 2
k s1  2k s  2  1   PLsin 1   0

2k s  2  1   PLsin  2   0
For small deformation
sin   
 0
cos  1
Thus,
k s1  2k s  2  1   PL1   0

2k s  2  1   PL 2   0
3k s  PL1  2k s 2  0

 2k s1  2k s  PL 2  0
The matrix form of the above two equations
 2k s  1  0
3k s  PL
  2k
    0
2
k

PL
s
s

 2   
For non-trivial solution, we need
3k s  PL
 2k s
0
 2k s
2k s  PL
 3k s  PL2k s  PL  4k s2  0
 6k s2  5 PLks  P 2 L2  4k s2  0
 P 2 L2  5 PLks  2k s2  0
5  17 k s
2
L
The smallest value of P will be the critical load
k
5  17 k s
Pcr 
 0.438 s
(3)
2
L
L
From (2) and (3) we can see that adding one extra spring with the same spring constant
at the hinge A will maximize the critical load
k
Pcr  0.438 s
L
P
10
Advanced Steel Structures
DISCUSSION
Problem 2
We will investigate two cases as following
a) Case 1: k s1  nk s 2 (n≥1)
P
L
L
P
ks2  ks
2
L
1
L
B
A
k s1  nk s

The strain energy
1
1
2
U  nks12  k s  2  1 
2
2

The potential energy
V   PL  L cos  2  L  L cos 1    PL 2  cos 1  cos  2 
The total potential energy
1
1
2
  U  V  nks12  k s  2  1   PL2  cos1  cos 2 
2
2

For equilibrium
 d
 d  0
 1

 d  0
 d 2
nks1  k s  2  1   PLsin 1   0

k s  2  1   PLsin  2   0
For small deformation
sin   
 0
cos  1
Thus,
11
Advanced Steel Structures
nks1  k s  2  1   PL1   0

k s  2  1   PL 2   0
(n  1)k s  PL1  k s 2  0

 k s1  k s  PL 2  0
The matrix form of the above two equations
 k s  1  0
(n  1)k s  PL

    0

k
k

PL
s
s

 2   
For non-trivial solution, we need
(n  1)k s  PL
 ks
0
 ks
k s  PL
 (n  1)k s  PLk s  PL  k s2  0
 (n  1)k s2  (n  2) PLks  P 2 L2  k s2  0
 P 2 L2  (n  2) PLks  nks2  0
(n  2)  n 2  4 k s
2
L
The smallest value of P will be the critical load
P
 Pcr 
(n  2)  n 2  4 k s
2
L
b) Case 2: k s 2  nk s1 (n≥1)
P
L
L
P
k s 2  nk s
2
L
1
L
B
A
k s1  k s

The strain energy
1
1
2
U  k s12  nks  2  1 
2
2
12
Advanced Steel Structures

The potential energy
V   PL  L cos  2  L  L cos 1    PL 2  cos 1  cos  2 
The total potential energy
1
1
2
  U  V  k s12  nks  2  1   PL2  cos1  cos 2 
2
2

For equilibrium
 d
 d  0
 1

 d  0
 d 2
k s1  nks  2  1   PLsin 1   0

nks  2  1   PLsin  2   0
For small deformation
sin   
 0
cos  1
Thus,
k s1  nks  2  1   PL1   0

nks  2  1   PL 2   0
(n  1)k s  PL1  nks 2  0

 nks1  nks  PL 2  0
The matrix form of the above two equations
 nks  1  0
(n  1)k s  PL
  

 nks
nks  PL  2  0

For non-trivial solution, we need
(n  1)k s  PL
 nks
0
 nks
nks  PL
 (n  1)k s  PLnks  PL  n 2 k s2  0
 (n  1)nks2  (2n  1) PLks  P 2 L2  n 2 k s2  0
 P 2 L2  (2n  1) PLks  nks2  0
(2n  1)  4n 2  1 k s
P
2
L
The smallest value of P will be the critical load
13
Advanced Steel Structures
 Pcr 
(2n  1)  4n 2  1 k s
2
L
The dependence of Pcr on n for each case is expressed as follows
n
ks1 = nks2
ks2 = nks1
Pcr (ks/L)
Pcr (ks/L)
1
0.382
0.382
1.5
0.500
0.419
2
0.586
0.438
2.5
0.649
0.450
3
0.697
0.459
3.5
0.734
0.464
4
0.764
0.469
4.5
0.788
0.472
5
0.807
0.475
5.5
0.824
0.477
6
0.838
0.479
6.5
0.850
0.481
7
0.860
0.482
7.5
0.869
0.483
8
0.877
0.484
8.5
0.884
0.485
9
0.890
0.486
9.5
0.896
0.487
10
0.901
0.488
14
Advanced Steel Structures
The Critical Load Pcr vs n for Case 1
Critical Load P cr (ks/L)
1.0
0.8
0.6
0.4
0.2
0.0
0
1
2
3
4
5
6
7
8
9
10
11
n
The Critical Load Pcr vs n for Case 2
Critical Load P cr (ks/L)
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
5
6
7
8
9
10
11
n
From the above graphs, we can see that: when we increase n, the value of the critical
load Pcr for case 1 goes up more quickly than for case 2. Therefore, to obtain the higher
critical load Pcr, increase of spring constant at node A (at hinge) is more effective than
at node B. In practices, if we need to have higher critical load Pcr, we should raise the
spring constant at hinge
15