This section will examine how various types of tubes carry torsional loads.  Round versus square cross sections will first be considered, followed by a more general comparison of open and closed cross sections.  Next, adding springs in series within the same structure will be considered, such as is the case in a car chassis where deflections occur in the springs and in structural members.  Finally, charts will be presented to aid in understanding the interaction of the compliant elements in a car chassis.

4.1 General Tube Types

To begin the discussion, observe the following plot of the relative torsional stiffness for different types and thickness tubes.

We can see that larger diameter tubes result in a higher J/A value, or torsional stiffness efficiency.  Additionally, square tubes appear to be more efficient for a given size.  This is because they distribute the material farther away from the center than a round tube.  These observations will be used in the next section.

4.2 Choosing a (Closed) Cross Section for Carrying Torsional Loads

When choosing what type of tube to use in a space frame there are several decision to be made.  These are cross section type (round, square, I-beam, etc.) and dimensions.  For a rounds or square hollow tube the dimensions to be specified are the diameter and thickness.  In some cases the cross section type is dictated by other factors.  For example, Cornell suspension bays use square cross-section tubes with milled slots to take the loads from a bolted rod-end.  Round tubes could be used if a different method of inputting the loads was chosen.  We will examine the case where the cross section is not predetermined by external loading.  It can be shown that for a given area, round tubes have a higher stiffness for a given area than square tubes.  A chart showing a comparison of the polar moment of inertia per area for round and square tubes is shown below.

FIGURE 1 - COMPARISON OF ROUND AND SQUARE TUBE LOAD CARRYING EFFICIENCIES

Each of the results is plotted for a constant area.  This chart uses a 1" square tube for all calculations.  For a given area, a hollow round tube can be determined by choosing either the diameter or the thickness (or interchangeably the outer diameter and the inner diameter).  From the chart we see that for a given area, it is best to choose the wall thickness of the tube, and then calculate the diameter to achieve the most efficient use of the material.  Round tubes of a given area and fixed diameter are less efficient than their square equivalents.  This is due to the moment of inertia (or polar moment of inertia) scaling with the diameter to the fourth power.  Thus, if size is ever a constraint, you should use the largest possible square tube necessary.  If size is not a limitation then the largest possible round tube should be used.  In almost all cases on the FSAE frame then, it is preferable to use large diameter thin walled round tube.  Exceptions would be in the suspension bays.  Welding and manufacturing often dictate thicker walled tubes, in which case the outside diameter should be scaled accordingly to provide the necessary stiffness, without sacrificing weight.

4.3 Multiple Torsional Members

In many situations multiple members are used, either in series or parallel.  For example, two tubes may be welded together and cantilevered from a wall.  A picture of which is shown below

Here the tubes are shown in series.  The deflection that occurs at the end of the assembly has a component from each of the two tubes.  The stiffness, then, is also a function of the stiffness of each tube.  Springs, in this case torsional springs, in series add together like electrical capacitors in series.  If we use d to represent the flexibility of each tube then the flexibility of the system is just

The stiffness is the inverse of the flexibility, which for the entire two-tube system can be found from

Which is the generic equation of stiffness for springs in series.  If we had additional springs they would simply be taken into account by another term at the end of the equation.  Another useful expressions would be one to find the equivalent torsional stiffness for a linear spring at the end of a bar.  A diagram is shown below.

The diagram depicts a bar, pinned at one end, and connected to a linear spring at the other.  The spring is fixed to ground at one end.  From this information we wish to find the equivalent torsional spring constant for the system.  For this calculation we need to find the torque the linear force is producing about the joint, and the angle the bar is moved through.  While the diagram shows the force, F, and the displacement, d, we in fact know the spring constant, K_L.  Knowing either K_L or F and d the other quantities can be calculated.  The following equations use a unit length for the variable d, and assign the numerical value of K_t to the force F.

Since the spring constant is usually measured in force per unit length, we can use the following equation to calculate the angle

Here the units of L must coincide with the length units of the spring constant.  Next, we can calculate the torque as follows

Since we will likely be working with units of inches for L, but we express T in terms of ft-lbs/degree, we need to divide L by 12 in the above expressions before using the torque.  The torsional stiffness constant, K_T, can then be found by combining the above expressions

If we make the small angle approximation, add the conversion to determine K_T in terms of ft-lbs/degree while expressing L in inches and convert radians to degrees we find

This is a convenient expression for determining the torsional spring constant of a linear spring system, such as arises in a car suspension.  In that case we use the wheel rate as our linear spring constant and the length is half of either the front or rear track, depending on which stiffness we are considering.  As mentioned earlier the variable F takes on the numerical value of the spring constant, since the above equation is worked out for a unit displacement.  Thus, if our spring constant is 100 lbs/in and the distance from the front contact patch to centerline of the car is 22" our equivalent torsional stiffness for that corner of the car is

In simpler terms, if we express K_T, the torsional spring stiffness, in units of in-lbs/radian then the equivalent linear spring stiffness, expressed in lbs/in and approximated using the small angle approximation is:

It is also possible to convert from torsional to linear spring stiffness in a similar manner.  Performing the analysis we would find the general equation is

And including the appropriate conversions to express L in units of inches and K_T in ft-lbs/degree the resulting expression for K_L in lbs/in is

Now that we can model both torsion and linear springs in the same system, it is possible to build a model of all the compliant members in an automotive chassis.  Depending on the desired complexity, different elements can be included or ignored in the model.  The simplest model we will consider is to calculate the chassis stiffness for a rigid frame and compliant springs.  In this model we assume the frame and suspension members are all infinitely stiff, and only the actual suspension springs themselves allow for any deflection.  A picture of this model is shown below.

The load is applied at the front left wheel (positive x and y direction).  The other wheels are all constrained from motion in the vertical direction.  We are neglecting forces and movement other than in the vertical direction, though the actual constraints are shown above.  If we draw a free-body diagram of the model and solve using the sum of forces and moments we can determine that the forces at all four wheels are equal.  The back right wheel is of the same direction as the applied load, while the other two wheels have their forces acting in the opposite direction, or trying to hold the car down.  If we apply a force greater than the weight on those two wheels we would lift our car frame off the ground.  For the purposes of this example, and in real world testing, we can assume that we have added weight to those corners to limit wheel lift.  The forces and deflections we are considering are all in addition to the pre-existing forces/deflections that result from the car supporting its own weight.

Since the force applied at each wheel is equal, call it F, the deflection of the spring at that wheel can be calculated if we know the spring constant, by the simple expression F=Kx.  If we assume that each spring has the same rate, then the deflection of each spring will be equal. Note in the figure the node numbers given.  We constrained vertically three nodes, 1, 3 and 4.  The four springs representing the suspension at the four corners of the car are all acting in series to resist the motion of the left wheel, or node 2.  Therefore, the total response of the wheel, reacting against some applied load can be found by the following expression:

Here the subscripts simply denote each of the four springs.  If each spring were the same, say 100 lbs/in, then the total stiffness of the chassis would be 100/4=25 lbs/in.  The deflection at the wheel is simply the deflection calculated by using the overall spring rate and the applied load and the standard spring expression, F=Kx.  Above each of the three contact points, except the one to which the load is applied, the only compliant member is one spring connecting the node to ground.  Therefore at each of these three points, in our above figures occurring at nodes 5, 7 and 8, the vertical deflection is simply the value of spring deflection mentioned in the last paragraph, or x=F/K.  So, with 100 lbs/in springs and an applied load of 100 lbs, node 2 deflects 4” while nodes 5, 7 and 8, deflect 1”.  Different spring rates will produce different deflections for a constant force, but the relationship of 1:4 will hold true in all cases.  If the springs take on different values, front/rear or even side to side, the above method will still yield accurate results, but the relative motion of the nodes will change.

The next model to consider is to represent the torsional compliance of the frame alone.  A diagram of which is shown below.

In the above model a force applied at node 2, the contact patch, causes a torsional deflection in the frame.  Since the other suspension elements are fixed, no other deflections occur.  All other nodes remain at there initial position.  Node 6 moves through a vertical deflection corresponding to the equivalent linear rate of the frame torsion spring.  If the frame stiffness measured in ft-lbs/degree is equivalent to 100 lbs/in, then from a 100lb load node 2 deflects 1”.  It should be noted that the angle of the bar connecting nodes 5 and 6 will change during this condition, but we are considering only vertical deflections at this time. 

Now we can use the principal of superposition to show that the considering deflections from both the translational suspension springs and the frame torsion spring produces a deflection which is the sum of deflections occurring in each element.  The spring-model considering the suspension and frame springs is shown below.

The simple expression describing this behavior is as follows:

Note that K₅ is simply the spring constant of the torsion spring.  In order to use this equation we must use consistent values of spring constants – either all translational spring values or all torsion spring values.  We can convert back and forth by knowing the wheelbase and using the expression developed earlier in this section.

Finally, we may wish to consider the fact that the suspension members also contribute compliance to the overall chassis system.  This could be shown graphically as another torsion spring in series with the frame.  Also, we need to use the installed spring rate for each suspension spring.  This installed spring rate will be the spring rate divided by the motion ratio squared.  The squared term arises because the motion ratio effects both the force transmitted and the displacement the spring moves through.  Conservation of energy is one way to show the motion ratio must be squared.  A mathematical description of the complete system, using more description variable names is given below:

The variable r in the above expression is the motion ratio of the corresponding spring.  Again, the units of spring stiffness must be consistently measured in equivalent stiffness for either a linear spring or rotary spring.

One additional use of the above expression for chassis stiffness is for the calculation of the motion ratio from a chassis mathematical model.  For example, to calculate the front motion ratio, we model the springs with their actual stiffness but set all other structural members to rigid bodies, perhaps by setting their stiffness values to several orders of magnitude higher than the spring stiffness.  The resulting chassis stiffness can then be used to back-calculate the motion ratio.  This method allows Ansys to be used to determine suspension rates, instead of using Matlab, Mitchell or some other software program.  The obvious advantage is that since the frame is already modeled in Ansys, and the suspension should be as well for proper design of the frame, it can be easier to also configure the suspension points directly in Ansys.  The Excel spreadsheet detailed later in this paper allows for rapid changes to be made to the Ansys model, so iteration becomes more rapid.

In Section 4.2 expressions were developed which allow the overall chassis stiffness to be calculated by knowing the spring rates of the frame, suspension structure and the actual installed suspension spring, or wheel rate.  Selecting values of these elements for an actual design is one of compromise and tradeoff.  To assist in this process three graphs are presented below to aid in the initial decision making process.  These graphs show the three changing levels of the spring rates involved: wheel rate, frame rate and suspension structure rate.  The first step is to consider the suspension stiffness resulting from the installed spring rate of the suspension springs, as well as the built in compliance of the suspension structure.  Likely, the spring rate will be much lower and the suspension structure will only reduce the rate by a small fraction.  Once this value of suspension stiffness is obtained it can then be cross-referenced by the frame stiffness on the second chart to obtain a measure of the overall chassis stiffness.  The third graph is to be used if the spring is being neglected and you wish to know the chassis stiffness due only to the compliances of the frame and suspension structure.  The three graphs are presented below.  It is important to note that all values are calculated using a wheelbase of 44”.  A different wheelbase will influence the results.