# The Simulation of an Automotive Air Spring
## 2 The Air Spring Stiffness
### The Pneumatic Stiffness  
In this step, two models are used to approach the mechanical/pneumatic behavior of the spring:  
1. The first model assumes that only the compressed air participates as an elastic body in the evaluation of the spring stiffness. The shell torus enclosing the air does not contribute to the suspension stiffness;
2. The second model consists in superimposing the structural stiffness of the shell torus to the one of the air under compression, as two springs working in parallel. In step (a), the spring stiffness due only to the air under compression results from the concept of differential volume contributing to the internal pressure work. The internal energy stored in a pressurized gas enclosed in a volume \(Vol\) is given by the integral expression:  
\[U=\int\limits_{Vol}PdV \tag{5}\]  
which is calculated over the volume of the container. For a small variation \(\partial a\) of parameter \(a\), the internal pressure \(P\) practically remains unchanged. For an axial compression of the toroidal bellows, the variation of the internal energy \(U\) is, from Equation (4), given by:  
\[\partial U=P\partial(\Delta Vol)=\underbrace{2\pi P\left[-\frac{4}{9}r^{2} \partial a+2a\partial a\left(\frac{28r}{90}+\frac{R\pi}{4}\right)\right]}_{Torus}+ \underbrace{P\pi R^{2}\times 2\partial a}_{\textit{Cylindrical core}} \tag{6}\]  
By the theorem of virtual work, the force associated with the internal energy rate due to a virtual \(\partial a\) is obtained as follows:  
\[F_{a}=P\frac{\partial\Delta Vol}{\partial(2a)}=P\pi\left[-\frac{4}{9}r^{2}+2a \left(\frac{28r}{90}+\frac{R\pi}{4}\right)\right]+P\pi R^{2} \tag{7}\](It is noted that there is a pair of symmetrical forces applied to the flanges of the spring; hence, the differentiation in order to the double value \(\partial(2a)\) of the amplitude increment). The calculation of the tangent stiffness associated with the variation of the spring compression force in Equation (7) is also obtained by considering the double variation \(\partial(2a)\) of the relative displacement of the flanges:  
\[K_{a}=\frac{\partial F_{a}}{\partial(2a)}=\frac{\partial^{2}U}{\partial(2a)^{ 2}}=P\frac{\partial^{2}\Delta Vol}{\partial(2a)^{2}}+\frac{\partial P}{ \partial(\Delta Vol)}\bigg{(}\frac{\partial\Delta Vol}{\partial(2a)}\bigg{)}^{2} \tag{8}\]  
From the expression in Equation (7), the tangent stiffness of the pneumatic spring is:  
\[K_{a}=\pi P\bigg{(}\frac{28r}{90}+\frac{R\pi}{4}\bigg{)}+\bigg{\{}\pi\bigg{[}- \frac{4}{9}r^{2}+2a\bigg{(}\frac{28r}{90}+\frac{R\pi}{4}\bigg{)}\bigg{]}+\pi R ^{2}\bigg{\}}^{2}\frac{\partial P}{\partial Vol} \tag{9}\]  
where \(Vol\) is the _current_ total shell volume as in Equation (1) using the expressions in Equation (3). The calculation of the variation rate of the current pressure \(P\) with the volume \(Vol\) due to the displacement \(a\) depends on the thermodynamics law of the air; as the spring is generally subjected to high speed volume changes, then it is coherent to assume an adiabatic law for the pressure/volume variation of the air enclosed in the bellows shell:  
\[PV^{\gamma}=Const=P_{0}V_{0}^{\gamma} \tag{10}\]  
where \(\gamma\) is the adiabatic exponent, and variables with subscript 0 refer to the initial values of the pressure and the volume; then:  
\[\frac{\partial P}{\partial Vol}=\frac{-\gamma P_{0}V_{0}^{\gamma}}{V^{\gamma+1}} \tag{11}\]  
The minus sign in the previous expression is due to the antagonist variations of the pressure and the volume, either in isothermal or adiabatic transformations. Here, an absolute value was assumed for the purpose of obtaining a normalized value for the pneumatic stiffness of the spring exclusively in compression. Generally, \(\gamma\) ranges from 1.2 to 1.4 (perfect gases); in this work, an intermediate value of \(\gamma=1.3\) was considered.