Solution Manual For Introduction to Aeroelasticity, 1E James DeLaurier

Solution Manual For
Introduction to Aeroelasticity, 1E James DeLaurier
Chapters 1-3

Chapter 1

Problem 1:

Determine the elastic centre and the equivalent spring constants, K z and K for the
rigid rod supported by four springs of arbitrary constants and spacing:




Recall that the elastic centre is defined to be that point where an applied force causes
no rotation. Upon naming the applied force Q z , the situation is illustrated below:




All displacements are equal, by definition, so one has that

Q z  K1 z  K 2 z  K 3 z  K 4 z  K z z
Therefore,
4
K z   Ki
1
Also, one may find the location of the elastic centre by noting that

Q z d  ( K z z ) d  ( K1 z ) d1  ( K 2 z ) d 2  ( K 3 z ) d 3
Therefore

, 3
Kzd   K jd j
1
One may then calculate that

3


1 3 K j dj
d K j dj  1
4

K
Kz 1
i
1
Now, apply a moment Q to find the equivalent torsion spring constant K :




 
Q  K1 (d1  d ) 2  K 2 (d 2  d ) 2  K 3 (d 3  d ) 2  K 4 (d 4  d ) 2 z  K 

where sin    and cos  1. Therefore, one has that

4
K    K i (d i  d ) 2
1


Problem 2:

Three thin “disks” are connected by torsion springs of constants k 1 , k 2 , and k 3 . With
the displacement variables being the rotations of the disks:  1 ,  2 , and  3 , find the
stiffness and flexibility matrices of this system. Note that the disks are seen
edgewise:

,(a) In order to find the flexibility influence-coefficient matrix, one sequentially
applies a unit torque at each station:




Step 1: A unit torque applied at station 1 gives the following deformations:

1 1 1 1 1 1
C3,1   3  , C 2,1   2   , C1,1  1   
k3 k 2 k3 k1 k 2 k 3

Step 2: A unit torque applied at station 2 gives the following deformations:

1 1 1 1 1
C 3, 2   3  , C 2, 2   2   , C1, 2  1  
k3 k 2 k3 k 2 k3

Step 3: A unit torque applied at station 3 gives the following deformations:

1 1 1
C 3, 3   3  , C 2,3   2  , C1,3  1 
k3 k3 k3

Therefore, the flexibility influence-coefficient matrix is:

1 1 1 1 1 1
    
 k1 k 2 k 3 k 2 k3 k3 
C    1  1 1

1 1
k 2 k3 k 2 k3 k3 
 1 1 1 

 k3 k3 k 3 

Next, the stiffness influence-coefficient matrix is found by determining the torques
required to give unit displacements at each station:

, Step 1: The torques necessary to make  1  1 and  2   3  0 are:

k1,1  T1  k1 , k 2,1  T2  k1 , k3,1  T3  0

Step 2: The torques necessary to make  1  0 ,  2  1 , and  3  0 are:

k1, 2  T1  k1 , k 2, 2  T2  k1  k 2 , k3, 2  T3  k 2

Step 3: The torques necessary to make  1   2  0 and  3  1 are:

k1,3  T1  0 , k 2,3  T2  k 2 , k3,3  T3  k 2  k3

Therefore, the stiffness influence-coefficient matrix is:

 k1  k1 0 
K    k1 k1  k 2  k 2 
 0  k2 k 2  k 3 

One may check on the correctness of the solutions by

C K   I  ?
1 1 1 1 1 1
    
 k1 k 2 k3 k 2 k3 k3   k1  k1 0  1 0 0
C K    1  1 1 1 1  k k  k  k 2  = 0 1 0 , O.K.
  1 1 2
k k k 2 k3 k3 
 21 3  0  k2 k 2  k3  0 0 1
 1 1 
 k3 k3 k3 


(b) Because C K   I  , then one has that det C  det K   det I   1 . Thus:
.