# Questions related to SAE paper 970960

The following are our responses to questions we have received on SAE paper 970960 "Effects of Restitution in the Application of Crush Coefficients"

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• Question#1: If one performs a revised SMAC calculation, all springs would initially have the (same?) input value of stiffness, K1. When each spring starts to unload, the maximum dynamic crush is known for that spring. At that time, K2, rho and gamma can be calculated for each spring using the formulas you present. But re-calculating K1 would make no sense, since K1 is not used for the rest of the time-history. So how does Equation 8 in SAE paper 970960 get implemented to define K1?

Answer #1: The four fitted constants are calculated only once, before the start of an application run. When a given radial spring starts to unload, the residual crush for that spring is calculated from equations (17) and (24), or the equivalent equation (9). Equation (8) was presented only for the purpose of showing that the value of K1 is independent of the extent of restored energy and, thereby, of the restitution coefficient.

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• Question#2: Equation 8 of SAE paper 970960, specifying the loading stiffness, is identical to Equation 10 of SAE paper 910119 (except for notation). Of course, this follows because both demand a consistency between the loading energy in SMAC and the loading energy in CRASH.
1. Answer #2: The derivation of equation (8) is presented in appendix 2 in the form of equation (26). While the manner of derivation is different, the result is equivalent to equation (10) in SAE #910119.

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• Question#3: Figure 6a of SAE paper 970960 shows an original SMAC specification of A,B and KV. Since original SMAC does not use A and B, I assume that A and B were specified as target values used by the sophisticated analyst to select the KV value. Likewise, on Figure 8 was KV selected based on A, B and the calculated deflections?

Answer #3: In figure 6A, the listed values of A, B, (/)1, ()1, and ()1 were used to calculate the value for Kv (by application of either equations (2) through (5) or equation (8)). In Figures 7, 8 and 9, the same A and B values as in Figure 6A were applied with different inputs for the restitution properties:

 Fig.6A Fig. 7, 8 & 9 A 317 317 LB/IN B 56 56 LB/IN2 (m)1 30 30 IN ()1 0.2000 0.1500 - (/)1 0.8000 0.8775 - K1 54.74 63.7 LB/IN2 K2 54.74 95.5 LB/IN2

Note that a major point in the paper is the fact that a given set of values for A and B can serve to represent a wide range of restitution behavior.

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• Question#4: Clarification #7 of Appendix C of SAE paper 970960 states that the elastic range (full dimensional recovery) is equal to A/B. This implies disturbingly large elastic deformations. For example, the first Figure A3, A/B is 357/13=27.5 inches. Do you see this as a serious problem?
1. Answer#4: Clarification #7 refers specifically to the CRASH (EDCRASH) computer program in which the implied elastic deformation range, in terms of full dimensional recovery is equal to A/B. In prior publications (e.g., SAE papers 910119, 920607, etc.) "A" has been referred to as the "zero residual crush force" or the "stiffness coefficient which represents the maximum force per unit width of the contact area which produces no crush." Therefore, the quantity A/B must be recognized as effectively being the elastic deformation range in terms of full dimensional recovery.

With the revised form of restitution modeling presented in SAE #970960, the corresponding elastic deformation range, in terms of full dimensional recovery, is given by equation (26) with set to zero. In that case, the maximum deformation for elastic behavior, in terms of full dimensional recovery, is defined as:

For the A, B values of the first Figure A3, which were fitted by Monk and Guenther (Monk, M.W., Guenther, D.A., "Update of CRASH II Computer Model Damage Tables, Vol. I", DOT HS-806446, March 1983), the following results can be obtained:

A = 357 LB/IN
B = 13 LB/IN2
()1 = 20 INCHES
()1 = 0.225
(/)1 = 0.750
K1 = 58.6 LB/IN2
K2 = 47.5 LB/IN2

While 12.93 inches of elastic deformation, in terms of full dimensional recovery, also seems large, it is certainly more believable than the 27.5 inches of elastic deformation in terms of full dimensional recovery in CRASH (EDCRASH) for the same (questionable) fitted values of A and B.

The purpose for inclusion of Figure A3 was to demonstrate that (1) unusual A, B fits such as Category 4, Rear, can be accommodated with crush properties that appear to be reasonable and (2) force equilibrium is possible in collisions between vehicles with widely different values for A and B.

As indicated in the paper, progress towards a rigorous and complete validation is data-limited at the present time.

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• Question#5: If you extrapolate your V line for the revised SMAC in Figure 8 of the paper SAE 97-0960 to zero residual crush you come up with a V of approximately 15 MPH. Am I misinterpreting this graph or using it incorrectly?

Answer#5: In Figure 8 of SAE #970960 (revised SMAC) the maximum deformation corresponding to a zero residual deformation, from equation (9):

For = 5.308 Inches, the coefficient of restitution, from equation (1):

Since the calculated value for is > 1.00, is set equal to the maximum value of 1.0.

Thus, extension of the slope of the V line in Figure 8 will include a change in the slope of the V line at that value of residual deflection where the calculated equals 1.0.

From equation (1):

=6.410 Inches

From equation (9) the corresponding value of residual deflection is:

The intercept is then obtained using equation (19) and the coefficient of restitution (1.0):

= (2.00)(6.74) = 13.48 MPH

This means that, for the given fitted properties, the residual deformation would be zero in a 6.74 MPH SAE barrier crash. The corresponding V would be 13.48 MPH.

Another item to note is that because of the selected form of restitution control in the original NHTSA SMAC (EDSMAC), the V line has an intercept through (0,0). In particular, in original SMAC the residual deflection approaches ~95% of the maximum deflection at small deflections (see Figure 6 of the paper). In real-life, the residual deflection should become a very small portion of the maximum deflection at small deflections (i.e, the ratio of should approach 0.0).