# HW Problems by sanmelody

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```									HW#9 Problems 6.5 and 6.6, Chajes

Using the energy method, determine the critical load for the one-degree-of- freedom model of a

flat plate shown in Fig. 6-26. The model consists of four rigid bars pin connected to each other

and to the supports. At the center of the model two linear rotational springs of stiffness

C  M /  connect opposite bars to each other. Also, each of the two transverse bars contains a

linear extensional springs of stiffness K. For small lateral deflections the energy in the

extensional springs can be neglected. ( Pcr  4C / a )

Fig. 6-26

Using the same model, obtain and plot relationships for the load P versus the lateral deflection d

when

(a) the lateral deflection is large,

(b) the lateral deflection is large and the loads are applied eccentrically to the plane of the

undeformed model.

Which fundamental buckling characteristics of an actual plate are demonstrated by these models?

(Note: for large deflections the energy in the extensional springs must be considered.)

1
Model Analysis I

Using the energy method, determine the critical load for the one-degree-of- freedom model of a

flat plate shown in Fig. 6-26. The model consists of four rigid bars pin connected to each other

and to the supports. At the center of the model two linear rotational springs of stiffness

C  M /  connect opposite bars to each other (  is the total rotation of the rotational spring to

cause the moment M.). Also, each of the two transverse bars contains a linear extensional spring

of stiffness K. For small lateral deflections the energy in the extensional springs can be

neglected.

Solution


2
                      P
     d
a
2a
                                     a

d
a                                       
Fig. 6-27

Strain energy of the deformed body is that associated with the rotationa l spring only as the strain

energy stored in the extensional spring is neglected.

1
U  C  2   2  4C 2
2

2

where  is the deformation angle of the rigid bar at the support.

The loss of potential energy of the applied load P is

V   P

2
where  is the horizontal displacement of the load point given by

  2a  2a cos   2a 1  cos   V  2 Pa 1  cos  

  U  V  4C 2  2 Pa 1  cos  


 8C  2 Pa sin       8C  2Pa  0 for a small 


4C
Pcr 
a

Model Analysis II

Using the model, obtain and plot relationships for the lateral load P versus the lateral deflection d

when

(a) the lateral deflection is large,

(b) the lateral deflection is large and the loads are applied eccentrically to the plane of the

undeformed model.          Which fundamental buckling characteristics of an actual plate are

demonstrated by these models? (Note: For large deflections the energy in the extensional springs

must be considered.)

Solution (a)

As the lateral deflection becomes large, the strain energy stored in the extensional spring must be

considered as well as that of the rotational spring. It is noted that the horizontal projection of the

rigid rods perpendicular to the direction of the applied load remains unchanged due to the

boundary condition given in the model (pin-pin). Therefore, the angle of rotation of these bars

with respect to the undeformed horizontal plane is given by the relationship   tan  d / a  .
1

The total strain energy stored in the deformed body is

U  Ur  Ue

3
The strain energy due to rotational springs is

U r  C  2   C  2   2C 2  2C 2  2C  2   2 
1        2 1        2
(a)
2          2

where

  sin 1  d / a                                                                                (b)

From the Pythagoras theorem applied to the deformed model, the length of the extensional bar is

given by

                 d
2                   2    2

 a     d 2  a 1     
2
(c)
2               2a   a 

where  is given by

    a2  d 2                  d 
2

  2a 1  cos    2a  2 a  d  2a 1 
2   2
  2a 1  1    
                   (d)
      a                      a 
                                   

2
        d   d 
2       2

 a 1  1  1                                                                    (e)
        a  a
              

Hence, the extension of each spring is

       2             2
2    
    d          d        
   a  a  1+    1  1          1                                                 (f)
    a         a       
                              

 
2                                           2
d
If   is neglected since it is small compared to   , then
 2a                                          a

         2  
 1     1
d
     a                                                                                    (g)
     a    
            

1     
Ue   K 2   2                                                                            (h)
2     

4
Using equation (f) for  ,

       2               2
2
2             2
2
 d               d           d            d     
U e  Ka 2 2     1  1      2 1     1  1                                 (i)
 a             a         a          a  
 


Using equation (g) for  ,

  d 2          d 
2

U e  Ka 2     2 1    
2
                                                                   (j)
 a            a 
                      

The total strain energy is then

U  Ur  Ue

U =2C 2  2C 2  2 Ka 2
                       2                           2
 d          d          d           d   
2             2             2              2
(k)
 Ka 2    1  1      2 1     1  1     
 a         a         a          a  
 


Or approximate total strain energy is then

 d  2     d 
2

U  2C 2  2C 2  2 Ka 2  Ka 2    2 1                                               (l)
 a        a 
                 

Substituting the expressions for the deformed angle into equation (l), yields

 1  d  
2
 1  d  
2              d  2     d  
2

U  2C  tan     2C sin     2 Ka 2  Ka 2    2 1                             (m)
      a           a                  a        a 
                  

        d  
2

The loss of potential energy of the applied load is given by (since   2a 1  1     )
        a 
              

        d  
2

V   P  2 Pa 1  1                                                                   (n)
        a 
              

The total potential energy is the sum of equation (m) and equation. (n).

5
 1  d  
2
 1  d  
2            2
d 
2
2 d 
  2C  tan     2C sin     2 Ka  Ka    2 1    
2

       a             a           a        a 
                 
(o)
        d  
2

 2 Pa 1  1    
        a 
               

For the principle of minimum potential energy, the derivative of the total potential energy with

respect to the displacement amplitude must be equal to zero. Hence,

                                                      
                                                      
                d      1          1              1  d        1     1 
 2C  2sin 1                     a   2C  2 tan  a              2    
d                a      d 
2
                              d   a 
1                                        1    

    a       
                               a 
                         d 1                                  
             2  2     
                  2   d   1  
  
  d  1              a  a    2 Pa          a  a   0
 Ka 2  2                                                   2 
   
a a                      2
d  
2 1                       2 1  d  
                                                         
                        a              
           a 

Solving the above equation for P, gives

                                                          
2
  sin 1  d  tan 1  d  
                                              
 2C       d            a          a    Kda 1     1               
P     1                                2                                               (p)
 d        a          d 
2
d          2C          d 
2

1                
 1   a          a             
1  

                                        a             

It should be noted here that the approximate expression for the extension of the spring results in

a slightly smaller value of extension. However, neglecting the quantity  / 2 of the spring

extension results in a slightly greater rotation of the extension bars, thereby alleviating the effect

of the approximation. All subsequent derivation will be based on this approximate expression.

The error involves would be less than 0.5% for a value of d up to 50% of the plate dimension.

6
It should be of interest to note that the bifurcation buckling load, Pcr  4C / a , can be obtained by

taking the limiting value of equation (p) (applying the L’Hpital rule) when d approaches to

zero.

                                                                                         
                                                                                         
                                                                  1                      
                                                 Kda 1                                    
2 
                                                                   d                   

1 d
tan 1
d                     1                         
d          sin
                                         
           a  
  
               
a               a 
              
 2                                                                                           
2
d            d  1  d                           2C              
2              2
              
 a 1    1                                                                          
       a           a              a                                                   

                                                                                           
                                                                                           
               
                                                              
              
Pcr  2C lim                                                                                               
d 0
                                              1 d                                        
                                       d sin                                              
1                           a                 1
                                                                                        
 a 1   d  
3/2                    2
  d 2                 d 2 
2
                                                                                           
   a   a 1   a                        a 1    
2
                                                                                       
          2
                          
                       a 
                          
d                                                                                 
  1                                                                                 
      a                                                                                 
2d tan 1
d                                 
                                                                                          
a  Ka 1                         
2
1                    Kd
                                                                                    3/2 

              2   d 2 
2
2C                    2 
d                d 2   
      1  
(q)
              a 1                       
2aC 1      

             
       
      a                           a               a   
         

 1 1  4C
Pcr  2C         as expected
a a a

Solution (b)


2       1

P
e



7
Fig. 6-28

From Fig. 6-28, it becomes evident that

         d 
                 
2

1  2  a  a cos    2 a  a  d
2    2
 2a 1  1    
 
        a 
              

2ed
 2  2e sin           2e (This is justified for a small e, i.e., if e  0.2a , error < 0.7%. It is
a

assumed here that the eccentricity occurs at both ends.)

In the case when e is introduced to account for geometric imperfection of the structure, the

eccentricity remains a very small quantity.

  1   2

The loss of potential energy of applied load is

        d  
2                              d  
2
d
V   P  2 Pa 1  1      2 Pe sin   2 Pa 1  1      2 Pe                               (r)
        a                               a         a
                                                

Equation (r) can be derived alternately as

ed
M end  Pe and the loss of potential energy is VM end  2M end  2 Pe  2 P
a

The strain energy stored in the deformed structure is the same as in equation (m). Hence, the

total potential energy  is

 1  d  
2
 1  d  
2                2
d 
2
2 d 
  U  V  2C  tan     2C sin     2 Ka  Ka
2
   2 1    
 
      a           a               a        a 
                  
(s)
        d  
2
ed
 2 Pa 1  1      2 P
        a         a
              

8
In order to minimize the total potential energy  , set the derivative of  with respect to the

arbitrary displacement d, equal to zero and solve the resulting equation for P.

                                                                   
                                                d                
2 
     1      4C         1  d    4C         1  d     a  Ka 2  2 Kad 
0                sin                tan                           
d     a             2
a      d 
2
a           2
 1  d                1                     d 
1                 
                         a                                     
     a                                         a               
d 
2 
    a  P  2P e
d 
2         a
1  
a

                                                 
 sin 1  d  tan 1  d 
             adK                     
        a         a                1       
                        2         1             
 1  d 
2
d         2C
       d
2

      1                     1  
 4C  
      a         a             
       a     

Pcr                                                                                          (t)
 a                         2d                           
                       a      
2e                 
                            2   a                  
                       d                         
1  
                       a                         

                                                   


Equations (p) and (t) are plotted comparatively in Fig. 6-29a It can be shown by applying the

L’Hpital rule that the limit value of equation (t) yields 4C / a , the bifurcation buckling load,

when d approaches to zero.

Discussion

Examination of the elastic responses of this model clearly illustrates the different load versus

deflection characteristics between columns and plates. The elastic response of this model is

analogous to that of a column when the spring constant K is small, and to that of a plate when the

spring constant K is stiff enough. When K is small, the force developed in the extensional spring

is not strong enough to restore the deformed model to the stable state (original state). Instability

9
of the model is further accelerated due to the finite lateral displacement of the model when K is

small (illustrated in Fig. 6-30a for small K). As a result, the column buckles. The load carrying

capacity of the column is considered exhausted.

that are known as postbuckling strengths of plates if the boundary conditions of such structures

are conducive to develop a considerable degree of tensile stresses within the structures. Consider

the model with a large spring constant K. As the lateral displacement increases, a considerable

amount of restoring force develops in the extensional spring. This large restoring force in the

spring enables the model to resist the load beyond the elastic buckling load, P=4C/a. Note the

boundary condition of the two rigid rods with extensional springs. They are supported by pins at

their ends enabling the springs to stretch. If there is a roller, the spring will not be stretched and

no postbuckling strength is realized. Also, note that the model is still within the elastic range,

and no material nonlinearity is involved. Hence, this postbuckling strength is governed by

geometric nonlinearity.      In a real structure, however, the displacement simply cannot be

infinitely large before encountering material nonlinearity. Hence, postbuckling behavior in a real

structure is governed by both geometric and material nonlinearity, thereby rendering a

formidable computational challenge. An incremental nonlinear analysis algorithm implemented

in modern finite element codes, such as NASTRAN, ABAQUS, ADINA, etc., makes such an

analysis possible.

When the load is applied with an eccentricity “e” (frequently introduced to account for initial

imperfection of structural elements), the model exhibits deflection amplification type behavior

with no bifurcation type buckling behavior. For a very small value of initial imperfection, the

ultimate load of the model asymptotically approaches to the elastic buckling load if very large

10
deflections are permitted. This illustrates the fact that the elastic bifurcation buckling load

indeed establishes an upper-bound of the ultimate load of a real structure made of components

with code permitted geometric imperfections.

C    PLATE MODEL ANALYSIS-CHAJES BOOK PROBLEM 6.5 AND 6.6 (A) AND (B)
C    SOLUTION BASED ON LARGE DISPLACEMENT THEORY
IMPLICIT REAL*8 (A-H,O-Z)
DIMENSION D(100),P(100)
WRITE(6,20)A,C,EK,E
20 FORMAT(/' ','P-D RELATIONSHIP(',' A =',F5.2,' C =',F5.2,' EK =',
1 F5.2,' ECC =',F5.2,')')
WRITE(6,21)
21 FORMAT(/' ','     DEFLECTION','  APPLIED LOAD')
10 FORMAT(3X,D15.5,3X,D15.5)
D(1)=A/20000.0
DO 100 I=1,50
IF (I.EQ.1) GO TO 12
D(I)=A/100.0*(I-1)
12 DA=D(I)/A
DA2=(DA)**2
OMDA2=1.-DA2
OPDA2=1.+DA2
DUM1=DASIN(DA)/DSQRT(OMDA2)
DUM2=DATAN(DA)/OPDA2
DUM3=EK*D(I)*A/(2.*C)*(1.-1./DSQRT(OPDA2))
P(I)=4.*C/A*(DUM1+DUM2+DUM3)/(2.*D(I)/A/DSQRT(OMDA2)+2.*E/A)
WRITE(6,10) D(I),P(I)
100 CONTINUE
GO TO 99
98 STOP
END

The same problem may be solved based on the small displacement theory so that simplified

relationships permitted in the microgeometry can be utilized.

Model Analysis II

Using the model, obtain and plot relationships for the lateral load P versus the lateral deflection d

when

11
(a) the lateral deflection is large,

(b) the lateral deflection is large and the loads are applied eccentrically to the plane of the

undeformed model.            Which fundamental buckling characteristics of an actual plate are

demonstrated by these models? (Note: For large deflections the energy in the extensional springs

must be considered.)

Solution (a)

As the lateral deflection becomes large, the strain energy stored in the extensional spring must be

considered as well as that of the rotational spring. It is noted that the horizontal projection of the

rigid rods perpendicular to the direction of the applied load remains unchanged due to the

boundary condition given in the model (pin-pin). Therefore, the angle of rotation of these bars

with    respect       to   the   undeformed    horizontal   plane   is   given   by   the   relationship

  tan 1 d / a d / a (assuming small displacement theory). The total strain energy stored in the

deformed body is

U  Ur  Ue

The strain energy due to rotational springs is

U r  C  2   C  2   2C 2  2C 2  2C  2   2 
1        2 1        2
(a)
2          2

where

  sin 1 d / a d / a                                                                                     (b)

From the Pythagoras theorem applied to the deformed model, the length of the extensional bar is

given by

                 d
2                    2       2

 a     d 2  a 1     
2
(c)
2               2a   a 

where  is given by

12
      2             d2
  2a 1  cos   2a 1  1             a 2                                             (d)
       2             a

Since  is so small, it is neglected in equation (c).

Hence,

d 
2
 1  d 2 
 a 1                a 1                                                            (e)
a                 2 a  
          

Hence, the extension of each spring is

2
ad 
  a                                                                                     (f)
2 a 

1     
Ue   K 2   2                                                                             (g)
2     

Using equation (g) for  ,

4
a2  d  K d4
Ue  K   K   
2
(h)
4 a    4 a2

The total strain energy is then

U  Ur  Ue

K d4         K d4
U =2C  2C 
2
 4C 
2  2
(i)
4 a2         4 a2

Substituting the expressions for the deformed angle into equation (i), yields

K d 4 4Cd 2 K d 4
U  4C 2             2                                                                    (j)
4 a2   a    4 a2

        d  
2

The loss of potential energy of the applied load is given by (since   2a 1  1     )
        a 
              

13
       1d  
2
d2
V   P  2 Pa 1  1       P                                                              (k)

       2 a      a

The total potential energy is the sum of equation (j) and equation (k).

4Cd 2 K d 4    d2
              P                                                                                (l)
a2    4 a2    a

For the principle of minimum potential energy, the derivative of the total potential energy with

respect to the displacement amplitude must be equal to zero.

Hence,

 8Cd 4 Kd 3 2 Pd
 2           0
d   a   4a 2   a

Solving the above equation for P, gives

4C Kd 2
P                                                                                               (m)
a   2a

It should be noted here that the approximate expression for the extension of the spring results in

a slightly smaller value of extension. However, neglecting the quantity  / 2 of the spring

extension results in a slightly greater rotation of the extension bars, thereby alleviating the effect

of the approximation. All subsequent derivation will be based on this approximate expression.

The error involves would be less than 0.5% for a value of d up to 50% of the plate dimension.

It should be of interest to note that the bifurcation buckling load, Pcr  4C / a , can be obtained by

taking the limiting value of equation (m) when d approaches to zero.

Solution (b)

From the sketch, it becomes evident that

          2  d2
1  2  a  a cos   2a 1  1      
          2 a

14
2ed
 2  2e sin           2e   2e (This is justified for a small e, i.e., if e  0.2a , error < 0.7%. It
a

is assumed here that the eccentricity occurs at both ends.)

In the case when e is introduced to account for geometric imperfection of the structure, the

eccentricity remains a very small quantity.

  1   2

The loss of potential energy of applied load is

d2     d
V   P        P  2Pe                                                                               (n)
a      a

Equation (n) can be derived alternately as

ed
M end  Pe and the loss of potential energy is VM end  2M end  2 Pe  2 P
a

The strain energy stored in the deformed structure is the same as in equation (j). Hence, the total

potential energy  is

4Cd 2 K d 4     d2     d
  U V          2
     2
 P  2Pe                                                                 (o)
a     4 a      a      a

In order to minimize the total potential energy  , set the derivative of  with respect to the

arbitrary displacement d, equal to zero and solve the resulting equation for P.

 8Cd Kd 3 2 Pd 2 Pe
 2  2          0
d   a   a    a    a

 e  4C Kd
2
 4C Kd 2   1 
P 1         P                                                                              (p)
 d a     2a      a   2a   1  e / d 

Equations (m) and (p) are plotted comparatively in Figs. 6-29b and Fig. 6-30b. It can be shown

that the limit value of equation (t) yields 4C / a , the bifurcation buckling load, when d

approaches to zero.

15
P-D RELATIONSHIP (LARGE DISPL)
WHEN K IS LARGE
6

5

FORCE (*C/a)   4                                      a=1.0 in.
K=30.0 kip/in.
3                                      C=1.0 k-in./in.

2                                      e=0.00*a, prob 6.6a
e=0.01*a, prob 6.6b
1                                      e=0.02*a, prob 6.6b

0
0.0   0.1         0.2      0.3        0.4           0.5
DEFLECTION (*a)

Fig. 6-29a Postbuckling of plate

P-D RELATIONSHIP (SMALL DISPL)
WHEN K IS LARGE
6

5
FORCE (*C/a)

4                                      a=1.0 in.
K=30.0 kip/in.
3                                      C=1.0 k-in./in.
e=0.00*a, prob 6.6a
2
e=0.01*a, prob 6.6b
1                                      e=0.02*a, prob 6.6b

0
0.0    0.1        0.2       0.3        0.4           0.5
DEFLECTION (*a)

Fig. 6-29b Postbuckling of plate

16
P-D RELATIONSHIP (LARGE DISPL)

WHEN K IS SMALL
6
a=1.0 in.
5                                    K=1.0 kip/in.

FORCE (*C/a)
4                                    C=1.0 k-in./in.

3
e=0.00*a, prob 6.6a
2
e=0.01*a, prob 6.6b
1                                 e=0.02*a, prob 6.6b
0
0.00   0.10     0.20     0.30       0.40        0.50
DEFLECTION (*a)

Fig. 6-30a Postbuckling of plate

P-D RELATIONSHIP (SMALL DISPL)
WHEN K IS SMALL
6
a=1.0 in.
5                                               K=1.0 kip/in.
C=1.0 k-in./in.
FORCE (*C/a)

4

3

2                                           e=0.00*a, prob 6.6a
e=0.01*a, prob 6.6b
1                                           e=0.02*a, prob 6.6b

0
0.00             0.10    0.20      0.30        0.40         0.50
DEFLECTION (*a)

Fig. 6-30b Postbuckling of plate

17

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