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

d

1 3  K jd j  Kz 1

K d j

j

1 4

K

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: C3,1   3 

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

Step 2: A unit torque applied at station 2 gives the following deformations: C 3, 2   3 

1 1 1 1 1 , 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: C 3, 3   3 

1 1 , C 2,3   2  , k3 k3

C1,3  1 

1 k3

Therefore, the flexibility influence-coefficient matrix is: 1 1 1     k1 k 2 k 3 C    1  1 k 2 k3  1  k3 

1 1  k 2 k3 1 1  k 2 k3 1 k3

1  k3  1 k3  1  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 K    k1  0

 k1

  k 2  k 2  k 3  0

k1  k 2  k2

One may check on the correctness of the solutions by

C K   I  ? 1 1 1     k1 k 2 k3 C K    1  1 k k  21 3  k3 

1 1  k 2 k3 1 1  k 2 k3 1 k3

1  k3   k 1 1   k1 k3    1   0 k3 

 k1 k1  k 2  k2

 1 0 0  k 2  = 0 1 0 , O.K. k 2  k3  0 0 1 0

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

det C   0 and det K   0

which means that C  and K  are not singular. In fact, detC  

1 , detK   k1k 2 k 3 k1 k 2 k 3

(c) If k 3  0 , then det K   0 . Therefore, K  is singular. Note that k 3  0 gives T3  0 , no matter what  3 is. Therefore,  cannot be determined uniquely by T1 , T2 , T3 . Also, observe that K  is likewise singular if k1  0 and k 2  0 .

Problem 3:

For the elastically-supported airfoil section, a stiffness matrix K  is defined at the chosen reference point for freedoms in vertical translation and pitch:

K    zz k

kz

k z  k 

The following may be expressed in terms of the elements of the K  matrix: (a) The location of the elastic centre relative to the reference point may be found as follows: Define the force, moment, and coordinates at the elastic centre as Fe , M e , z e ,  e , such that Fe  k z z e and M e  k  e . Likewise, the force, moment, and coordinates at the reference point are Fr , M r , z r ,  r and, by definition, these are related by:

 Fr  k zz   M r  kz

k z   z r    k   r 

(1)

First of all, consider the situation where z e  1 and  e  0 . To achieve this, Fe  k z , and M e  0 . Likewise, at the reference point, z r  1 and  r  0 .

From Equation (1), one has that Fr  k zz and M r  kz .

Now, from the equilibrium equations, one obtains that Fe  Fr  k z  k zz and M e  M r  e Fr  0  kz  e k zz

Therefore, one has that k e   z k zz

Next, consider the situation where z e  0 and  e  1 . To achieve this, Fe  0 and M e  1. Likewise, at the reference point, z r  e   e  e and  r  1 . Therefore,

Equation (1) gives that Fr  e k zz  k z and M r  e kz  k .

Again, one has the equilibrium relationships:

Fe  Fr  0  e k zz  k z and M e  M r  e Fr  k  e kz  k This gives: k e   z , which is the same as before k zz

(b) Also, these equations give expressions for k z and k as functions of terms in the K  matrix: k  k 

( k z ) 2 k zz

And, from before, k z  k zz

Next, e , k z , and k will be expressed in terms of the flexibility influencecoefficient matrix: C C C    zz z  Cz C  The simplest way is to calculate C   K 1 . However, the longer way is as follows:

Apply Fe  1 and M e  0 at the elastic centre. This gives the displacements z e  1 / k e and  e  0 . The equivalent loads at the reference point are Fr  1 and M r  e. The relationship between displacement and load is:

 z r  C zz    r  Cz

C z   Fr    C  M r 

(2)

So, one has for this case that z r  C zz  e C z and  r  Cz  e C . One may thus obtain that 1 1  e   r  0  Cz  e C  0 and z e  z r   C zz  e C z  kz kz This gives that C e  z C C zz 

C z 2 C



1 kz

Now, apply Fe  0 and M e  1 at the elastic centre. The resulting displacements are  e  1 / k e and z e  0 . The equivalent loads at the reference point are Fr  0 and M r  1 . Therefore, from Equation (2), one obtains that z r  C z  1  C z and  r  C  1  C

These relationships give that

r  e 

1 1 1  C   k  k k C

z r  e  e  C z  e

C 1  e C  e  k C z  z k C

Thus, as before, one has that e

C z C

Also, the spring constants, k z and k as functions of the terms in the C  matrix are

kz 

1  C z 2  C   zz  C  

, k 

1 C

Problem 4: For a uniform cantilever beam of length L and rigidity EI is subjected by a force F and moment M as shown:

Given that the stiffness influence coefficients for the right-hand side are:  12 EI  6 EI  2 k k  3  K     L6EI 4LEI    zz z   2  kz k  L   L Extend the interpretation of the stiffness coefficient matrix to a 4 by 4 matrix representing an unsupported (free-free) beam element:

The column i of K  are the forces and moments required to give a unit displacement at station i , with no displacements or rotations elsewhere, as shown below:

From the given 2 by 2 K  matrix, the last two elements of column 3 are immediately found as F2z2  k 3,3  12 EI / L3 , M 2z2  k 4,3  6 EI / L2 . The other two elements of the column are found from the equilibrium equations:

F1z2   F2z2  12 EI / L3  k1,3 M 1z2   M 2z2  L F2z2  6 EI / L2  12 EI / L2  6 EI / L2  k 2,3 A similar procedure may be followed for station 4, where the last two elements of column 4 are F22  k 3, 4  6 EI / L2 , M 22  k 4, 4  4 EI / L . From the equilibrium equations, one obtains:

F1z2   F2z2  6 EI / L2  k1, 4 M 1z2   M 2z2  L F2z2  4 EI / L  6 EI / L  2 EI / L  k 2, 4 The first two columns of the matrix are determined by the loadings below:

By inspection, one sees that the first two elements of the first column are

F1z1  k1,1  F2z2  12 EI / L3 and M 1z1  k 2,1  M 2z2  6 EI / L2 And the equilibrium equations, as above, give that

F2z1   F1z1  12 EI / L3  k 3,1

M 2z1   M 1z1  L F1z1  6 EI / L2  12 EI / L2  6 EI / L2  k 4,1 Finally, for column 2, inspection of the loading for column 4 allows one to readily see that

F11  k1, 2   F22  6 EI / L2 and M 11  k 2, 2  M 22  4 EI / L The last two terms are found from the equilibrium equations:

F21   F11  6 EI / L2  k 3, 2 M 21   M 11  LF11  4 EI / L  6 EI / L  2 EI / L  k 4, 2 The complete matrix is  12 EI  L3  6 EI  L2 K    12  EI  L3  6 EI   L2

6 EI L2 4 EI L 6 EI  2 L 2 EI L

12 EI L3 6 EI  2 L 12 EI L3 6 EI  2 L



6 EI  L2  2 EI   L  6 EI  2  L 4 EI   L 

Problem 5: A wing is mounted on the end of a system consisting of two rigid beams connected together with a torsion spring of stiffness k 1 . Also, the whole assembly is connected to a wall with a torsion spring of stiffness k 2 :

(a) These two configurations will be important for determining the stiffness influence coefficients:

(b) Now, the flexibility influence coefficients may be found as functions of k 1 , k 2 , L1 , and L2 :

First, apply L  1 and M  0 . This gives the following displacements:

2 

1 ( L1  L2 ) , k2

1   2   1 L1 k1

Therefore, C h   1   2 

Also,

L 1 1  L1  L2   1 k1 k 2 k1

h  L1 1  L2  2 (Note the small-angle approximation)

So that 2

L L L C hh  h  1 L1  L2   1  2 ( L1  L2 ) k2 k1 k 2

Next, apply L  0 and M  1 . This gives the following displacements:

1 

1 , k2

1   2   1

k1

Therefore, C  1 

1 1 1  2   k1 k1 k 2

Also, as before,

h  L1 1  L2  (Small-angle approximation)

So that

1 1 1 1 1 C h   h  L1     L2  L1  L1  L2  k2 k1 k 2  k1 k 2  Thus, the complete flexibility influence-coefficient matrix is:

C C    hh C h

1 2 1 2 L  L1  L2  C h    k1 1 k2   1 1 C   L1  L1  L2   k1 k 2

L1

1 1   L1  L2  k1 k 2  1 1    k1 k 2

(c) Next, the deformations shown in (a) guide the derivation of the stiffness influence-coefficient matrix.

First, set h  1 and  1  0 . For small values of  , one has that

 2  sin  2 

h 1  L2 L2

Also, note that L L1  L2   M  k 2  2 

k2 L2

(1)

Further, one has that

1   2    2  1

L2

So that L  L1  M  k1 (1   2 )  

k1 L2

Now, Equation (1) minus Equation (2) gives: L  L2 

1 k1  k 2   K hh  L  12 k1  k 2  L2 L2

Also, Equation (2) gives that M 

k1 L L  L  L1  k1 22  k1  k 2  12  L2 L2 L2

(2)

K h  M  

L1  L  1     k 2 1  k1 1  L2   L2  L2 

The next step is to set h  0 and  1  1 . Again, the  values are assumed to be “small”, such that sin 1  1 and sin  2   2 . Therefore, one has that

L1 1  L2   2    2  

L1 L 1   1 L2 L2

The moment balance about k 2 is L L1  L2   M  k 2 2   k 2

L1 L2

(3)

Also, the moment balance about k 1 gives that

 L  L  L1  M  k1 1   2   k1 1  1   L2  Now, Equation (3) minus Equation (4) gives:   L  L  L  L2   k1 1  1   k 2 1   L2  L2    K h  L  

L1  L  1     k 2 1  k1 1  L2   L2  L2 

Finally, one obtains from Equation (4) that   L  L   L  L  L  M   L  L1  k1 1  1   1 k1 1  1   k 2 1   k1 1  1   L2  L 2   L2  L2  L2   

2

 L  L  K   M  k1 1  1   k 2  1  L2    L2 

2

Therefore, in summary, the stiffness influence-coefficient matrix is

(4)

K  K h

K    hh

 L1  L1   1 1       k  k  k 1   k   1  1 2 2 2 K h   L2   L2  L2   L2  2 2  K    1         L L L L 1 1 1 1  k1 1    k 2     k1  1    k 2   L2   L2  L2   L2   L2  

Solutions for “Exercises Pertaining to Divergence and Control Reversal” Problem 1: An un-swept rectangular wing of semi-span L (from root to tip) is fixed at the root (cantilevered). The torsion stiffness GJ , lift-curve slope a 0 , and distance between the aerodynamic centre and elastic axis e are constant along the span. Also, the moment coefficient about the aerodynamic centre C mac is zero and the un-deformed angle of attack

 is zero. Further, the flow speed is low subsonic.

(a) The divergence speed U div will be found for two cases of assumed elastic modes: (i)  ( y)  y / L (ii) 1 ( y )   y / L   2   y / L 

 2 ( y)   y / L2  3  2 y / L

The solution uses the “Assumed Mode” or “Rayleigh-Ritz” method:

A  qBqi   0 Where q is the divergence dynamic pressure and q i are the generalized coordinates.

Case (i):

1 ( y )  y / L , 1' ( y)  1 / L 2

2

L  d  GJ GJ 1 A1,1   GJ  1  dy   GJ   dy  2 l  L L L  dy  0 0 L

2

a ec L a ecL  y 2 B1,1   a0 ec1  dy   a0 ec  dy  0 2  0 3 L 3  L 0 0 L

L

3

Now, one has that

A  qBqi   A1,1  qB1,1 q1  0  A1,1  qB1,1  0  GJ  q a0 ecL  0 L

3

B1,1

So, the divergence dynamic pressure is given by q div 

3GJ a0 ecL2

If on defines the area of the semi-span as S '  Lc , then the solution becomes  2q  1 GJ q div   Vdiv   div  3 S ' ea 0   

1/ 2

 6 GJ      L S ' ea 0  

1/ 2

 GJ    2.45  LS ' ea 0  

1/ 2

Case (ii):

1 ( y ) 

y y 1 y y 2 y '  2   ,  1 ( y )   2    2  1   L L L L L L L 2

2y  2y  2y2 6y  y 2y   y  '  2 ( y )     3   ,  2 ( y )  2  3    3  2 1   L  L L  L  L L   L 

This gives that 2

2

2

L L  d1  2  y  4GJ  y  y  dy   GJ  1   dy  A1,1   GJ  1   d     L 0 L L  L  L   dy  0 0 L

L

4GJ 1  x 2 dx  4GJ  1 1  x 3   4GJ  L 0 L  3 3L  0 1

L  d1  d 2  2  y   6 y  y    dy   GJ  1    2 1   dy  A1.2  A2,1   GJ   L  L   L  L   dy  dy  0 0 L

2





12 GJ y  12 GJ y   y  12 GJ x (1  x) 2 dx  (1  x) 2  (1  x) 3 dx  1   d       L 0 L L   L  L 0 L 0 L

1

1

1

12 GJ  1 1 GJ   (1  x) 3  (1  x) 4    L  3 4 L  0 2

2

2

L L  d 2  6y  36 GJ  y   y  y  y  dy   GJ  2 1   dy  A2, 2   GJ    1   d     L 0L  L L  L  L   dy  0 0 L

1

1

2





36 GJ 2 36 GJ x (1  x) 2 dx  (1  x) 2  2(1  x)  1 (1  x) 2 dx    L 0 L 0 1





36 GJ (1  x) 4  2 (1  x) 3  (1  x) 2 dx   L 0 1

36 GJ  1 1 1 6 GJ   (1  x) 5  (1  x) 4  (1  x) 3    L  5 2 3 5L 0

Also, one obtains: 2

y y  2 B1,1   a 0 e c1  dy   a 0 e c   2   dy   a 0 e cx 2 2  x  dx  L  L  0 0 0 L

L

1

2



1



1  8 4 a 0 e cL  4 x  4 x  x dx  a 0 e cL  x 3  x 4  x 5   a 0 e cL 5  0 15 3 0 L

2

3

4

2 L L y y   y   y  B1, 2  B2,1   a 0 e c1  2  dy   a 0 e c   2      3  2  dy  L   L   L  L  0 0

1

1

a0 e cL  x (2  x)(3  2 x)dx  a0 e cL  (6 x 3  7 x 4  2 x 5 )dx  3

0

0

1

7 1  3  45  42  10  13 a 0 e cL  x 4  x 5  x 6   a 0 e cL  a 0 e cL  5 3  0 30 2   30 2

1  y  2  y  2 B2, 2   a0 e c 2  dy   a0 e c    3  2  dy  a0 e cL  x 4 3  2 x  dx  L   L   0 0 0 L

L

2



1



4  13 9 a 0 e cL  9 x 4  12 x 5  4 x 6 dx  a 0 e cL  x 5  2 x 6  x 7   a 0 e cL 5 7 35   0 0 L

Now, upon applying

A  qBqi   0 one obtains:  4  GJ  3    L 1  

 8 1   qa0 e cL  15  6 13   5  30

13   30    q1   0  13     q 2  35  

Therefore,

4 8 13  s 1 s 2 ' 3 15 30  0 , s  a 0 e cL q  a 0 eL S q 13 6 13 GJ GJ 1 s  s 30 5 35 Upon expanding this out, one has that 0.0103175 s 2  0.2685714 s  0.6  0

which gives:

 GJ  GJ s min  2.4680  q div  2.4680  Vdiv  2.2217  '  ' L S ea 0  L S ea 0  

1/ 2

(b) In order for these assumed modes to be a suitable choice, these must satisfy the end conditions for a cantilevered beam.

Case (i):

1 ( y )  y / L  1 (0)  0 Okay

1' ( y)  1 / L  1' ( L)  1 / L Not Okay The assumed mode shape does not satisfy the zero-slope condition at the wing tip.

Case (ii):

1 ( y ) 

y y  2    1 (0)  0 , Okay L L

1' ( y ) 

2 y ' 1    1 ( L)  0 , Okay L L 2

y  y   2 ( y)     3  2    2 (0)  0 , Okay L  L 

 2' 

6y  y 1     2' ( L)  0 , Okay 2  L  L

The assumed modes are suitable. Also, the solution for q div is very close to that from the exact differential-equation solution: q div  2.4674

GJ L S ' ea 0

This clearly shows the robustness of the assumed-mode method. For that matter, q div for Case (i) differs only by 21.6%. (c) Now, the torsional deformation shape (for Case (ii)) may be found: At V  Vdiv , q  q div and thus A  qBqi   0  (1 / GJ )A  qBqi   0   4  3   1  

 8 1   2.4680  15  6 13   5  30

13   30    q1   0   0.01707   0.06947 13      q 2  35  

 0.06947   q1   0 0.2833  q 2 

0.01707 q1  0.06947 q 2  q1  4.0697 q 2  4.07 q 2

Since q1 is 4.07 larger than q 2 , one may arbitrarily chose q1  1 , so that q2  1 / 4.07  0.2457 . Therefore, the generalized-coordinate vector is

 q1   1     q 2  0.2457  The complete mode shape is thus:

2

y

1

 L 

 ( y )    i qi  Const   2 

2 y y   y   0 .

2457     3  2   L L  L 

2 3 y  y  y  2  Const   0.13145    0.2457    L  L    L

Note that Const is an arbitrary value.

Problem 2: A wing is mounted with elastic restraints in heaving (vertical displacement) and pitching, as shown:

Also, the aerodynamic lift and moment about the wing’s mass centre are given by: L  q S C L  , M  L d  q d S C L 

One may find the divergence velocity, U div by in two ways: Method 1: The equilibrium equation at the torsion spring is M  l  L  k 2

Also, M and L are given by

L  qSCL  and M  qSdCL 

So that





(1)

qSCL   k1  k1l  (qSCL  k1l )  k1 z  0

(2)

qSdCL   qSCL   k 2  qSCL (d  l )  k 2  0

Further, the vertical-equilibrium equation is L  k1 ( z  l ) , where   sin 

Therefore, one has that

Equations (1) and (2) combine to give:

qSCL (d  l )  k 2   (qSC  k l ) L 1 

0     0 k1   z 

The solution for q div is given by:

qSCL (d  l )  k 2

0

 (qSCL  k1l )

k1





 0  k1 qSCL (d  l )  k 2  0

Since k1  0 , qSCL (d  l )  k 2  0  q div 

k2 SCL (d  l )

Method 2: Again, start with the equilibrium equations: L  k1 ( z  l ) M  L  l  k 2

(3) (4)

Equation (3) into (4) gives that M  k 2  L  l  k 2  k1 ( z  l )l  (k 2  k1l 2 )  k1l z

Upon combining Equations (3) and (5), one has:

(5)

 k1 l   z   L   k1   2   M   k1l k 2  k1l   

(6)

Note that

 L  0 qSCL   z      M  0 qSdCL   

(7)

Next, Equations (6) and (7) combine to give

 (k1l  qSCL )   z   k1  k l (k  k l 2  qSdC     0 2 1 L     1 The solution for q div comes from the non-trivial solution of:

k1

 (k1l  qSCL )

 k1l

(k 2  k1l 2  qSdCL q div 

 0  k 2  qSCL (d  l )  0 

k2 SCL (d  l )

This is the same solution as before.

Problem 3: The system shown consists of two rigid bars upon which a pivoting wing is mounted at its aerodynamic centre:

Note that the bars are connected with torsion springs k 1 and k 2 . Also, the wing’s aerodynamic lift is given by L  qSCL 

(a) The control effectiveness, L / Lr , may be found in terms of k 1 , k 2 , l1 , l 2 , C L , S, and the dynamic pressure, q . The first step is to note that

    1   2      1   2

(1)

Now, the equilibrium equation at spring k 1 is: L (l1  l 2 )  T1  k11  1 

L (l1  l 2 ) (l1  l 2 )  qSCL  k1 k1

(2)

And the equilibrium equation at spring 2 is: L l 2  T2  k 2 2   2 

L l2 l2  qSCL  k2 k2

(3)

When Equations (2) and (3) are substituted into (1), one obtains:



  l  l2 l2    qSCL      k2   k1 

  1   1 

 l l l  1   1 2  2  qSCL k2   k1

Therefore, the wing’s lift as a function of deflection angle  is:

L  qSCL  

qSCL  l l l  1   1 2  2  qSCL k2   k1

Now, the rigid case is when k 1 , k 2   , so that  1   2  0     :

Lr  qSCL  Therefore, the control effectiveness is given by

L r

L



1 l l l  1   1 2  2  qSCL k2   k1

(b) For control reversal, the definition is that L Lr

0

However, to achieve this, q  q rev   . Therefore, there is no reversal. (c) Next, the divergence dynamic pressure is obtained from the previously-derived expression:



 l l l  1   1 2  2  qSCL k2   k1

With the denominator set equal to zero, one has the equation with which to find q div :

l l l  1 1   1 2  2  qSCL  0  q div   0 k2   l1  l 2 l 2   k1    qSCL k k2   1 Therefore, divergence is impossible.

Problem 4: Consider the wing mounted to a uniform cantilever beam as shown:

The coordinates are z and  , and the constant elastic properties along the beam of length l are EI and GJ . Also, the wing produces a lift L and pitching moment M about the midchord reference point. (a) The matrix of flexibility influence coefficients may be found as functions of EI , GJ , d , and l as follows: z

Ll3 Ll3 ld  d   (L d  M ) 3EI 3EI GJ

  (L d  M )

l GJ

(1)

(2)

This gives:

 l 3 d 2l  l d      z   3EI GJ  GJ   L         ld l  M      GJ GJ 

(3)

So that one has the flexibility influence-coefficient matrix:

C    zz C

Cz

 l 3 d 2l  l d    C z   3EI GJ  GJ      C   ld l   GJ GJ 

(4)

(b) The matrix of stiffness influence coefficients is now found: From Equations (1) and (2), one has that L

Ld  M 

3EI d 3EI z 3  3 l l

3EI d 3EI d 2 GJ GJ  M    3 z  l l l l3

This gives:

 3EI  L   l3     3EI d M    l3

3EI d  3  z  l 2    GJ 3EI d        3 l l   

So, the stiffness influence-coefficient matrix is

(5)

 3EI K z   l 3  K    3EI d  l3

K    zz K

 K z

3EI d  3  l   GJ 3EI d 2     l 3   l 

(6)

(c) The divergence dynamic pressure may now be found: First, note that L and M may be written as

C L qS   z   L  0     M  0 C L qS c / 4   From Equations (3) and (4), this gives that

 z  C zz     Cz

1 0  z  0 0 1    0    

C L qS   z  C z  0     C zz  0 C L qS c / 4  

C zz  C z c / 4CL qS   z   Cz  C c / 4CL qS    



1  C zz  C z c / 4C L qS   z  0 1  C  C c / 4C qS     0 z  L     For divergence the determinant should equal zero, which gives that 1  (Cz  C c / 4) C L qS  0  q 

q div 

1 l C L S  c d   GJ  4



1  (Cz  C c / 4) C L S

GJ l C L S (d  c / 4)

Alternatively, from Equations (5) and (6), one has that

 K zz K  z

C L qS   z  K z   z   L  0        K     M  0 C L qS c / 4  

 K zz K  z

( K z  C L qS )   z   0 ( K   C L qS c / 4)  

For divergence, the determinant should equal zero:

K zz K  K z  C L qS ( Kz  K zz c / 4)  0  2

K z  K zz K  C L S ( Kz  K zz c / 4) 2

qdiv  2

2

GJ  3EI   EI  2  EI  9d 2  3    3   9d  3  GJ l  l  l  l   q div   3EI c  l C L S (d  c / 4)   3EI C L S  3 d   3 4 l  l

This is the same as before.

Solutions for “Exercises Pertaining to Modes of Vibration” Problem 1: A cantilever beam has constant values of EI and m (mass per unit length) along its length, l . With the differential-equation solution technique described in the text, one may determine the following: (a) The first two frequencies: The fundamental beam dynamic differential equation is 4w 2w   m y 4 t 2 Since EI and m are constant, this may be solved by separation of variables; EI

m 2 d 4W 4 4 w( y, t )  W ( y ) sin t  ( y )  k W ( y ) , where k  EI dy 4 The general solution is W ( y)  A sin( ky )  B cos(ky )  C sinh( ky )  D cosh(ky )

And the boundary conditions at y  0 are W (0)  0  B  D  0  D   B , and W ' (0)  0  kA  kC  0  C   A

This gives: W ( y )  Asin( ky )  sinh( ky )  Bcos( ky )  cosh( ky )

(1)

Further, the boundary conditions at y  l are that W ' ' (l )  0 (moment free) and W ' ' ' (l )  0 (shear free). Therefore, Equation (1) becomes: 0  W ' ' (l )  Ak 2  sin( kl )  sinh( kl )  Bk 2  cos(kl )  cosh(kl ) 

Asin( kl )  sinh( kl )  Bcos( kl )  cosh( kl )  0

(2)

And 0  W ' ' '{l}  Ak 3  cos(kl )  cosh(kl )  Bk 3 sin( kl )  sinh( kl )  Acos( kl )  cosh( kl )  B sin( kl )  sinh( kl )  0

(3)

For a non-trivial solution of W ( y ) , the determinant formed from Equations (3) and (4) must be equal to zero: sin( kl )  sinh( kl )

cos(kl )  cosh(kl )

cos(kl )  cosh(kl )  sin( kl )  sinh( kl )

0

 sin 2 (kl )  sinh 2 (kl )  cos2 (kl )  2 cos(kl ) cosh(kl )  cosh2 (kl )  0

From the trigonometric identities:  sin 2 (kl )  cos2 (kl )  1, sinh 2 (kl )  cosh2 (kl )  1

One has that  2  2 cos(kl ) cosh(kl )  0  cos(kl ) cosh(kl )  1  0

The solution for this gives two positive roots: 1 1 (kl )1  1.8751  k1  1.8751 , (kl ) 2  4.6941  k 2  4.6941 l l

Further, from the definition of k , one has that m 2  EI  k     k2  EI m

1/ 2

4

Therefore, the first and second frequencies are  EI  1  k   m 2 1

1/ 2

3.516  EI   2   l m

(b) The first two modes: From Equation (2), one has that

1/ 2

 EI  , 2  k   m 2 2

1/ 2

22.03  EI   2   l m

1/ 2

A  B

cos(kl )  cosh(kl ) sin( kl )  sinh( kl )

(4)

Aside: One may also obtain from Equation (3) that sin( kl )  sinh( kl ) cos(kl )  cosh(kl )

AB

End Aside.

When Equation (4) is substituted into Equation (1), one obtains:

  cos(kl )  cosh(kl )   sin( ky )  sinh( ky )  cos(ky )  cosh(ky ) W ( y )  B      sin( kl )  sinh( kl )   Upon substituting the values for k1l and k 2 l , one has that the first two modes are

   y y    y y     W1 ( y )  B  0.7341sin 1.8751   sinh 1.8751   cos1.8751   cosh1.8751   l l    l l       

   y y    y y     W2 ( y )  B  1.0185 sin  4.6941   sinh  4.6941   cos 4.6941   cosh 4.6941   l l    l l        Problem 2: The Rayleigh-Ritz (Assumed-Mode) method may be applied to the cantilever beam of Problem 1 to obtain an approximation for the first two uncoupled bending modes, with the assumed vibrational shape described by the function: W ( y)   2 q1 (t )   3 q2 (t ) ,

  y/l

(a) Upon following the terminology and methodology in the text, one has that Z1 

y2 , l2

Z2 

y3 l3

So that l

M zi , j   mZ i Z j dy  0

l

M 1,1   m 0

l

y2 y2 m 4 m l5 1 dy  y dy   ml l2 l2 l 4 0 l4 5 5 l

l

y2 y3 m 5 m l6 1 M 1, 2  M 2,1   m 2 3 dy  5  y dy  5  ml l l l 0 l 6 6 0 l

M 2, 2   m 0

l

y3 y3 m 6 m l7 1 dy  y dy   ml l3 l3 l 6 0 l6 7 7

Next, one has that l

K zi , j   EI Z i'' Z 'j' dy  0

l

l

2 2 4 EI 4 EI 4 EI K1,1   EI 2 2 dy  4  dy  4 l  3 l l l 0 l l 0 l

l

2 6y 12 EI 12 EI l 2 6 EI K1, 2  K 2,1   EI 2 3 dy  5  y dy 5  3 2 l l l 0 l l 0 l

K 2, 2   EI 0

l

6y 6y 36 EI 36 EI l 3 12 EI 2 dy  y dy   3 l3 l3 l 6 0 l6 3 l

Therefore,

M z   ml 

1 / 5 1 / 6  , 1 / 6 1 / 7

K z   EI3 

6  l 6 12 4

Now, assume that q1 (t )  q1 sin t ,

q 2 (t )  q 2 sin t

So, the dynamic equations given by

K   M q   0  M  1 K q   0 2

z

z

z



z

2

z



Upon defining:



1 EI  2 ml 4

one obtains that

 1 / 5 1 / 6 4 6    q1      6 12   q   0   2   1 / 6 1 / 7

For a non-trivial solution, one has that the determinant must equal zero: 1 / 5  4

1 / 6  6

1 / 6  6 1 / 7  12

 0  (1 / 5  4 )(1 / 7  12 )  (1 / 6  6 ) 2  0

This may be rewritten as (4  1 / 5)(12  1 / 7)  (6  1 / 6) 2  0 which expands to give: (4  12  36)2  (12 / 5  4 / 7  2)  (1 / 35  1 / 36) 

122  34 / 35  1 /(36  35)  0  122  0.971429  0.00079365  0 The roots are then found from



0.971429  0.951619  1  0.0008254 ,  2  0.080127 24

From the definition for  ,



1 1 2 l

EI m

one thus obtains that

1 

3.533 l2

EI 34.81 EI , 2  2 m m l

Upon comparison with the exact solution from Problem 1, it is seen that the approximate first-mode frequency is fairly close, within 5%. However, the approximate second-mode frequency is far off, by 58%. (b) The shapes of the approximate modes may be found as follows: For the first mode, substitute  1 above into 1    M z   2 K z q   0  

which will give the q  vector. One may also, equivalently, work with  and the previously-obtained equation:

 1 / 5 1 / 6 4 6    q1       q   0 1 / 6 1 / 7 6 12    2   In this case 1  0.080127 and this becomes

 q1   1  0.1205  0.3141   q1  0.3836 1   0   0         0.3141  0.8187  q 0.3837 1   2    q 2   0.3837  From the original equation of the assumed vibration shape, one obtains the first mode: y2 y3 y2 y3 W1 ( y )  q1 2  q 2 3  2  0.3837 3 l l l l This may be normalized to give W1 (l )  1 , in which case the mode equation becomes W1 ( )   2 (1.623  0.623 ) ,   y / l This plots out to give:

Note that this mode possesses the expected characteristics of: (1) W1' (0)  0 , and W1 (0)  0 (2) There is a single vibration node (at the origin).

For the second mode shape, when

2 

34.81 EI m l2

This is equivalent to   0.0008254 , so that one obtains

 q1   .8220  0.1967 0.1617   q1  1 0.8221 0.1617 0.1329  q   0  1 0.8219   0  q    1    2      2  The second mode equation is thus W2 ( y )  q1

y2 y3 y3 y2  q   0 .

8220 2 3 l2 l l3 l2

This may be likewise normalized to give W2 (1)  1 , so that the mode equation becomes W2 ( )   2 (5.618  4.618) ,   y / l which plots out to give

This possesses the expected characteristics of: (1) W2' (0)  0 , W2 (0)  0 (2) This mode has two vibration nodes. (c) In general, improved estimates for the frequencies and modes may be obtained by using additional terms in the analyses: N

w( y, t )   Z i ( y )qi (t ) , where N  2 i 1

However, if only two vibration shapes are assumed, then these must be selected such that they satisfy all boundary conditions. For example, observe that the previous two assumed shapes: Z1 ( y )  y 2 / l 2   2 , Z 2 ( y )  y 3 / l 3   3

do not satisfy the free-end boundary conditions at y  l (which is W (l )  0) .

Improved functions would be 4

y  y Z1 ( y)  1  4  1    1  4  (1   ) 4 l  l 5

y  y Z 2 ( y)  1  5  1    1  5  (1   ) 5 l  l

which satisfy all boundary conditions. Z 2 will give a more accurate result for the frequency and shape of the second mode.

Problem 3: Consider the free-free beam shown below, with a concentrated mass, 2M , at the centre:

One may use the fundamental mode obtained in the previous problem as an assumed dynamic mode. Further, for mathematical convenience, only half of the beam and mass need be considered in the calculation. Therefore, the vertical motion may be expressed as: w( y, t )   1 ( y )q1 (t )   2 ( y )q 2 (t ) where  1 ( y )  1 , rigid-body mode  2 ( y)   2  0.3837 3 ,   y / l Therefore, from:

L

M i , j   m( y )  i  j dy 0

one has that L

L

0

0

M 1,1   m( y )  1  1 dy  M   M  12 dy  M  mL If M  2mL , then M 1,1  3mL

The additional integrals are: L

L

0

0

M 2,1  M 1, 2   m( y )  1  2 dy   m  1  ( 2  0.3837 3 )dy 

 1 0.3837  mL ( 2  0.3837 3 )dy  mL    0.2374 mL 4  3 0 L

L

L

0

0

M 2, 2   m( y )  2  2 dy   m ( 2  0.3837 3 ) 2 dy 

 1 2  0.3837 0.3837 2    0.0931 mL mL ( 4  2  0.3837 5  0.3837 2  6 )d  mL   6 7 5  0 L

Further, from: L

K i , j   EI ( y )  i'' 'j' dy 0

and

 1'' ( y)  0 ;  2'' ( y ) 

1 (2  2.302 ) L2

one has that L

K 1,1   EI  1'' 1'' dy  0 0 L

K 1, 2  K 2,1   EI  1'' 2'' dy  0 0 2

EI 1  K 2, 2   EI   dy   EI  2 (2  2.302 ) dy  3  (4  4  2.302  2.302 2 2 )d  L 0 L  0 0 L

L

'' 2

1

'' 2

EI  4  2.302 2.302 2  EI    1.162 3 4   3  2 3  L  L Therefore, the matrices are

M   mL

0.2374  , 0.2374 0.0931 

Now, assume that

3.0

K   EI3 

0   L 0 1.162 

q1 (t )  q1 sin(  t ) ;

q 2 (t )  q 2 sin(  t )

0

The dynamic equations are given by 1    M   2 K q   0  

which become

  3.0 0.2374  1 EI 0 0   q1   2 3  mL       0  0.2374 0.0931  L 0 1.162  q 2  Define:



1 1 EI  2 L4 m

so that the equations become

  3.0 0.2374  0   q1  0  mL       0 0 1.162  q 2   0.2374 0.0931  Setting the determinant equal to zero gives the characteristic equation: