2.314/1.56/2.084/13.14 Fall 2006
Problem Set IX Solution
Solution:
The pipe can be modeled as in Figure 1.
R
Lumped mass
mB PB
B
mA PA
A
Pipe Beam with the same support
Geometry and properties:
L=3m, R=0.105m and t=0.007m
E=200GPa, ρ=8500kg/m3 and ρwater=750kg/m3
m A = mB =
π (R 2 − Ri 2 )Lρ + (πRi 2 + 1.1πR 2 )Lρ water
2
= 133.724Kg
The governing motion eqution for dynamic response of this pipe can be expressed as:
[M ]{u&&} + [D]{u&} + [K ]{u} = {F}
where:
0
m
[M]=mass matrix, M =  A

 0 m B 
[D]=damping matrix [K]=stiffness matrix
[F]=vector of loads, earthquake load in our case, {F } = [ M ] ⋅{S a }
{u}=vector of nodal displacements First of all, we should calculate the stiffnes matrix of the pipe using beam theory:
Suppose the displacements of point A (L/3) and B (2L/3) are uA and uB, respectively.
Assuming that there is only a force PA acting on point A (at L/3), we can calculate the
corresponding displacements of points A and B, satisfying the boundary conditions:
1
u(0) = 0
u(L) = u(3) = 0
du(z)
=0
dz z=0
For L>z>L/3:
dM y
M y = Vx (3 − z) = −
(3 − z)
dz
Solving this equation with the boundary condition My(L)=0, we get
M y = C1 (3 − z)
For z<L/3:
dM y
(3 − z)
M y = 2PA + V x (3 − z) = 2PA −
dz
Solving this equation:
M y = C2 (3 − z) + 2PA
θ y (0) =
Because of continuity at z=L/3, we have
C1 = C2 + PA
Meanwhile,
M y = − ∫ σ z xdA = − ∫ Eε z xdA = EK y ∫ x 2 dA = EK y I , where K y =
A
A
A
π
0.105
0
0.098
where I = ∫ x 2 dA = 2 ∫ cos 2 θ dθ ∫
A
r 3dr = 2.3023 ×10 −5
So that
C1

(3 − z),1 < z < 3

EI
Ky =
=
EI  C 2 (3 − z) + 2PA ,0 < z < 1
 EI
EI
 C1 (3 − z) 2 9C2 4PA
+
+
 − EI
EI at z=0, θy=0
2
2EI
θ y = ∫ K y dz = 
2
C
2P
9C
(3
−
z)
− 2
+ A z + 2
 EI
2
EI
2EI
3
 C1 (3 − z)
4P  13PA 9C 2
 9C
−
+  2 + A z −
 EI
6
2
EI
EI
3
EI
2EI


u(z) = ∫ θ y dz = 
3
C
P
9
C
9C
(3
z)
−
2

+ A z2 + 2 z − 2

EI
6
EI
2EI
2EI
At last, we have another boundary condtion, u(L)=0
So that
My
2
dθ d 2 u
=
dz dz 2
4P  13PA 9C2
 9C
3 2 + A  −
−
=0
EI  3EI 2EI
 2EI
23
C2 = −
PA
27
4
C1 = C 2 + PA =
PA
27
Then, we obtain
C 2 8 PA 9C 2 9C 2
P 
 

+
+
−
− 0.1358 A 


u A  
EI 6 EI 2EI 2EI
EI
=

 u  =  C 1 9C
8P
13P
9C
P
A
 B  1 + 2 + A −
− 2   − 0.142 A 
3EI 2EI  
EI
EI 
 EI 6 EI
Similarly, assuming that there is only a force PB acting on point B (at 2L/3), we can
calculate the corresponding displacements of points A and B, satisfying the boundary
conditions:
For L>z>2L/3:
dM y
(3 − z)
M y = Vx (3 − z) = −
dz
Solving this equation with the boundary condition My(L)=0, we get
M y = C1 (3 − z)
For z<2L/3:
M y = PB + V x (3 − z) = PB −
dM y
dz
(3 − z)
Solving this equation:
M y = C2 (3 − z) + PB
Because of continuity at z=2L/3, we have
C1 = C2 + PB
Meanwhile,
M y = − ∫ σ z xdA = − ∫ Eε z xdA = EK y ∫ x 2 dA = EK y I , where K y =
A
A
A
π
0.105
0
0.098
where I = ∫ x 2 dA = 2 ∫ cos 2 θ dθ ∫
A
r 3dr = 2.3023 ×10 −5
So that
 C1
 EI (3 − z),2 < z < 3
=
Ky =
EI  C2 (3 − z) + PB , z < 2
 EI
EI
 C1 (3 − z) 2 9C 2 5PB
−
+
+

2
2 EI 2 EI at z=0, θy=0
θ y = ∫ K y dz =  EI
2
− C 2 (3 − z) + PB z + 9C 2
 EI
2
EI
2EI
My
3
dθ d 2 u
=
dz dz 2
 C1 (3 − z) 3  9C 2 5PB 
9C 19PB
z− 2 −
+
+
 EI

6
2 EI 6EI
 2 EI 2 EI 
u(z) = ∫ θ y dz = 
3
 C 2 (3 − z ) + PB z 2 + 9C 2 z − 9C 2

EI
6
2EI
2EI
2EI
At last, we have another boundary condtion, u(L)=0
So that
5P  9C
19PB
 9C
3 2 + B  − 2 −
=0
 2EI 2EI  2EI 6EI
13
C2 = −
PB
27
14
C1 = C2 + PB =
PB
27
Then, we obtain
9C
9C
C2 8 PB
P 
 

+
+ 2 − 2
− 0.142 B 


u A  
EI 6 2EI 2EI 2EI
EI
=
 u  =  C 1 9C
5P
9C
19P
P 
B
 B  1 + 2 + B − 2 −
  − 0.247 B 
EI 2EI 6EI  
EI 
 EI 6 EI
So that, according to superposition when PA and PB act on the system simultaneously,
u A  1  − 0.1358 − 0.142  PA 
u  = EI  − 0.142 − 0.247  P 

 B
 B
In the meantime,
u A 
 PA 
 P  = K u 
 B
 B
We can get
−1
0.1358 0.142
18.46232 − 10.614 
18.46232 − 10.614 
K = EI 
= EI 
= 4.6046 ×10 6 



 0.142 0.247
 − 10.614 10.15054
 − 10.614 10.15054
Secondly, we should compute the damping matrix: The damping matrix has only two non-zero elements, located on the diagonal. These elements are equal and give two percent critical damping for vibration at the system
fundmental frequency. Thus, we should calculate the undamped fundmental frequency first. [M ]{u&&} + [K ]{u} = 0
0  u&&A 
18.46232 − 10.614  u A 
133.724
+ EI 


 =0
 0

133.724 u&&B 
 − 10.614 10.15054 u B 

Solving this equation by assuming u A = Aeω t and u B = Beω t , we get
4
133.724ω 2 + 18.46232EI
  A
− 10.614EI

  = 0
2
− 10.614EI
133.724ω + 10.15054EI   B 

(133.724ω 2 + 18.46232EI )(133.724ω 2 + 10.15054EI ) − 112.657(EI ) 2 = 0
17882.1ω 4 + 3826.226EIω 2 + 74.7455(EI ) 2 = 0
ω 2 = −1.00126 ×105 or − 8.851×105
Then, the fundamental undamped frequencys are
ω1 = 940.8, ω 2 = 316.43 in radius/s; note that ω = 2πf
and corresponding eigenvectors are
 0.826 
0.56364
u1 = 
, u2 = 


 − 0.56364
 0.826 
Then, we should also calculate the critical damping matrix,
Suppose that the critical damping of lumped mass is considered seperately:
0
C
Dcr =  1

 0 C2 
[M ]{u&&} + [D]{u&} + [K ]{u} = 0
N
and the solution is u(t) = ∑ An cos ω n tun
n =1
Multiply the above equation with ANT leftly, and also replace {u} with AN {u}, where
0.56364
 0.826
AN = [u1 u2 ] = 

 − 0.56364 0.826 
we get
133.724ω 2 + C1′ω + 25.7 EI
  A1 
0

  = 0
2
′
0
133.724
2.9
C
EI
ω
ω
+
+
2

  A2 
If there are nontrivial solutions for A1 and A2, we obtain
133.724ω 2 + C1′ω + 25.7 EI = 0
133.724ω 2 + C2′ω + 2.9 EI = 0
So that critical damping matrix:
0 
C ′ 0  251592.2
=
Dcr′ =  1

0
84514.2
 0 C2′  
 198513 − 77788 
T
Dcr = AN Dcr′ AN = 

 − 77788 137593.5
Damping matrix in this case
0
5031.844

D ′ = 0.02Dcr′ = 
0
1690.284

5
At last, the force due to earthquake
{F} = −[M ] ⋅{u&&g }
Natural frequency
f1 =
ω1
ω
= 50.36 and f 2 = 2 = 149.7
2π
2π
According to the response spectrum of Fig 8 in note M-32, extrapolating to the calculated
frequencies, we obtain the maximum displacements corresponding to Sa=0.33g are
S
0.33 ⋅ 9.8
S d 1 ≈ a2 =
= 3.6538 ×10 −6 m
2
940.8
ω1
Sd 2 ≈
Sa
ω2
2
=
0.33 ⋅ 9.8
= 3.23 ×10 −5 m
2
316.43
So
0 
133.724
{F} = − 
u&&g
133.724
 0
Finally, we obtain
0  u&&A  5031.844
0
0  u A 
 u& A  25.7EI
133.724
T  − 133.724 
A
=
+
+
N
 − 133.724u&&g
 u 
 u&   0
 0
 u&&  
133.724
0
1690.284
2.9EI


 B
 B 

 B 
0  u& A 
0  u A 
u&&A  37.63
0.2624
25.7EI
1
= −
+




u&&  +  0
u&&g



12.64 u& B  133.724  0
2.9EI  u B 
 1.39 
 B 
Then, maximum displacement
 7.92 × 10 −7 
u1,max = 0.2624 ⋅ S d 1 ⋅ u1 = 
−7 
 − 5.4 × 10 
2.53 × 10 −5 
u2,max = 1.39 ⋅ S d 2 ⋅ u2 = 
−5 
3.71 × 10 
[
P1,max = (Cωu) 2 + (Ku) 2
]
0.5
2
2

198513 − 77788   7.92 ×10 −7    18.46232 − 10.614   7.92 ×10 −7   

=  0.02 ⋅ 940.8
 − 5.4 ×10 −7   +  EI  − 10.614 10.15054  − 5.4 ×10 −7   
−
77788
137593.5




  
  

  3.75  2  93.72  2 
= 
 
 +
  − 2.56  − 63.95 
0.5
93.795
=

 64 
6
0.5
[
P2,max = (Cωu) 2 + (Ku) 2
]
0.5
2
2

198513 − 77788  2.53 ×10 −5    18.46232 − 10.614  2.53 ×10 −5   

=  0.02 ⋅ 316.43
 

  +  EI

− 77788 137593.5 3.71×10 −5     − 10.614 10.15054 3.71×10 −5   




 13.52 2 337.6 2 
= 
+

19.85 497.4 


0.5
337.87
=

 497.8 
Now, given the forces of PA and PB, we calculate the forces acted on supports:
FL
PB
B
PA
A
F0
Assuming the forces that act on the supports are F0 and FL, respectively, as in the above
figure.
dM y
dK y
4
14
FL =
= EI
=
PA +
PB
dz z=L
27
dz z=L 27
F0 = PA + PB − FL =
23
13
PA +
PB
27
27
For P1,max:
 F  1 23 13
110.71
F1 =  0  =
P1,max = 



 47.08 
 FL  27  4 14
Likewise, for P2,max:
 F  1 23 13
527.5
F2 =  0  =
P2,max = 



308.2
 FL  27  4 14
[
]
 539 
=
 Newton
311.8
So, the peak forces that act on the support z=0 and z=L are 539 and 311.8 Newton,
respectively.
2
Again, F = F1 + F2
2 0.5
7
0.5