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Rocket Engine Ignition Structural Shock
21 January 2014
by C. P. Hoult
Introduction
An elementary calculation provides the steady state axial load along the rocket
body during flight. But, transient loading at ignition is often much larger. Following the
sudden increase in thrust at ignition, an elastic compression wave is sent up the structural
“stack”. This wave produces significant shock loading, especially on fragile payload
elements. General experience with large rockets (satellite launch vehicles) shows that
this is the primary source of payload (satellite) structural loading.
Also, if the thrust vector is misaligned with the forebody center of mass, large
bending torques are introduced. Even with small sounding rockets (e.g., NIRO) it is a
major loading consideration. More complex models with variable properties along the
rocket body, or with solid propellant motors can be readily modeled using finite element
techniques.
Nomenclature
_____Mnemonic_____________Definition_____________________________________
A ......................Cross section area of the load-bearing part of the column
Acc ...................Steady state axial acceleration
a ( x, t ) ...............Axial acceleration of a column element
E ......................Young’s modulus of the load bearing material
 ( x, t ) ..............Axial stress, positive in compression
 ( x, t ) ...............Axial strain, positive in tension
L ......................Column length (or length of a column element)
 ......................Column mass per unit length
c .. ....................Wave speed of axial elastic displacements
x ... ...................Nominal distance forward of the aft end of the rocket
 ( x, t ) ...............Elastic displacement, positive in the + x direction
 SS (x) ................Steady State elastic displacement
TR ( x, t ) .............Transient vibrational elastic displacement
P ..................... Axial force, positive in tension
s.......................Laplace transform variable
T ......................Thrust force
t .......................Time after rocket ignition
b ......................Laplace transform of b(t)
1
Two Mass Model of Ignition Shock
This memo documents a highly simplified analysis of a rocket structure with a
thrust chamber at the extreme aft end. We start by considering a model having two
discrete masses connected by a spring:
x1
m1
k
m2
x2
T
The two coupled equations of motion are
m2 x2  T  k ( x1  x2 )  m2 g
m1x1  k ( x1  x2 )  m1 g
Here g = acceleration due to gravity.
The initial conditions are
x1 (0)  0, x2 (0)  0, x2 (0)  0, x1 (0)  
m1 g
k
Next, take the Laplace transform (s = the transform variable)
T
mg
 k ( x1  x2 )  2
s
s
, or
m1 g
m1 g
2
m1 ( s x1  s
)  k ( x1  x2 ) 
k
s
m2 s 2 x2 
(T  m2 g )
s
2
sm g m g
mg
(m1s 2  k ) x1  kx2   1  1   1 (m1s 2  k )
k
s
ks
(m2 s 2  k ) x2  kx1 
Then,
x2  (m1s 2  k )( x1  m1 g / ks) / k , and
2
mg
T  m2 g


(m2 s 2  k ) (m1s 2  k )( x1  1 ) / k   kx1 
, or
ks
s


(m m s
1
2
4

 k (m1  m2 ) s 2  k 2  k 2 x1  (


s 2 m1m2 s 2  k (m1  m2 ) x1 
m1 g
k (T  m2 g )
)( m2 s 2  k ) 
, or
ks
s
k (T  m2 g ) m1 g
(
)( m2 s 2  k ).
s
ks
Define the natural frequency of this system to be
n2 
k (m1  m2 )
.
m1m2
Then,
( s 2  n2 ) x1 
k (T  m2 g )
g
(
)( m2 s 2  k ) .
3
m1m2 s
m2 ks3
Now, suppose next that the thrust acceleration is very large compared to g. Then,
( s 2  n2 ) x1 
kT
.
m1m2 s 3
Given this,
x1 
kT
1
.
3
2
m1m2 s ( s  n2 )
Now, what we really care about is the acceleration, A1. Then,
A1 
kT
1
.
2
m1m2 s ( s  n2 )
Next, rewrite this in terms of a partial fraction expansion:
m1m2 A1
1
s
 2  2 2
kT
n s n (s  n2 )
The inverse Laplace transform can be found adding the contributions from both terms:
m1m2n2
A1 (t )  1  cos(nt ) .
kT
3
Finally, the maximum value for A1 is
A1 max 
kT
m1m2
2T
*
*2 
m1m2 k (m1  m2 )
(m1  m2 )
Thus, the effect of a finite stiffness is to increase the steady axial load on m1 by a factor
of 2. Note that this result does not depend on the spring stiffness or the relative
magnitude of the two masses.
Three Mass Model of Ignition Shock
To understand this problem at its most fundamental level, consider a bare-bones
model having only three massive elements, m1 ,m2 and m3 , each pair connected by a
spring whose stiffness is k . At t = 0 the rocket is ignited and thrust T , acting as a step
function is applied to m3 . The mass displacements are measured from the launch pad
upward. Gravity is neglected since it is usually much smaller than thrust.
m1
k
m2
x
k
m3
T
The three coupled equations of motion are:
m1x1  k ( x1  x2 )
m2 x2  k ( x1  x2 )  k ( x2  x3 )
m3 x3  k ( x2  x3 )  T
Next, take the Laplace transform, assuming the stack starts from rest with no initial
displacements.
4
m1s 2 x1  kx1  kx2
m2 s 2 x2  2kx2  k ( x1  x3 )
m3 s 2 x3  kx3  kx2 
T
s
Eliminate x2 :
x2  (m1s 2  k ) x1 / k , and
(m2 s 2  2k )( m1s 2  k ) x1  k 2 ( x1  x3 )
(m3s 2  k ) x3 
T
 (m1s 2  k ) x1
s
Then, eliminate x3 :
T
 (m1s 2  k ) x1
, and
x3  s
(m3 s 2  k )
T
 (m1s 2  k ) x1
(m2 s 2  2k )( m1s 2  k ) x1  k 2 x1  k 2 s
(m3 s 2  k )
m1s 2  k
k 2T
(m1m2 s  k (2m1  m2 ) s  k ) x1  k
x1 
m3 s 2  k
s(m3s 2  k )
4
2
2
2
At this point it becomes expedient to aggregate the stack in a way makes all three masses
the same. Then,
s 2 (m2 s 2  3km) x1 
k 2T
s(ms2  k )
However, what we really want is the acceleration of the topmost (first) mass, A1 :
A1 
k 2T
s(ms2  k )(m2 s 2  3km)
Before inversion, rewrite this as
m3 A1 mA1 1
1
1
, where

 * 2
* 2
2
2
kT
T
s s  n s  3n2
5
n2 
k
.
m
Expand this into partial fractions:
mA1 1/ 3
s/2
s/6
.

 2
 2
2
T
s
s  n s  3n2
Next, invert the Laplace transform term by term:
mA1 (t ) 1 1
1
  cos(nt )  cos(3nt ) .
T
3 2
6
The steady state acceleration is
Ass 
T
3m
Then,
A1 (t )
3
1
 1  cos(nt )  cos(3nt )
Ass
2
2
This is plotted below as the Amplification Factor. It shows a maximum value of about
2.4142.
Top Mass Amplification Factor
Amplification Factor
3
2.5
2
1.5
1
0.5
0
-0.5 0
1
2
3
4
5
6
7
8
9
10
11
12
-1
Omegan * t
6
Next, find the acceleration of the second mass:
ms2  k
kT
n2T
.
A2 
A1  2 2

k
m s( s  3n2 ) ms( s 2  3n2 )
The partial fraction expansion is
mA2 1/ 3 (1/ 3)s

 2
T
s
s  3n2
This can be inverted to give
mA2 (t ) 1 1
  cos 3nt
T
3 3
Normalizing by the steady state acceleration,
A2
 (1  cos 3nt )
Ass
We see, by inspection, that the maximum value of this ratio is
A2
 2.
Ass
Finally, seek the maximum acceleration of the base element, x3 :
T
x3 
 x1 , or
s(ms2  k )
A3 
sT
T
s
 A1  * 2
 A1
2
ms  k
m s  n2
Then, since
T  3mAss ,
A3  3 Ass
s
 A1
s  n2
2
The inversion is easy:
A3
3
1
3
1
 3 cos(nt )  1  cos(nt )  cos(3nt )  1  cos(nt )  cos(3nt )
Ass
2
2
2
2
As shown in the plot below, the maximum Amplification Factor is 3.
7
Bottom Mass Amplification Factor
Amplification Factor
3
2.5
2
1.5
1
0.5
0
-0.5 0
1
2
3
4
5
6
7
8
9
10
11
12
-1
Omegan * t
Then, the amplification factors do not depend on element or total mass, nor do they
depend on the stiffnesses between elements. In summary
Element
1 (top)
2 (middle)
3 (bottom)
Amplification Factor
2.4142
2
3
8