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Modelling and analysis of suspension systems 225 Equation 4 3 6x 9 6y 436 6z 0 (4.250) The four equations can now be set up in matrix form ready for solution: ⎡ 3 0 436 9 ⎤ ⎡ A s ⎤ ⎡ 669.564 ⎤ ⎢ 9 436 0 3 ⎥ ⎢ ⎥ ⎢ 28 592.484 ⎥ ⎢ ⎥ ⎢ 6x ⎥ ⎢ ⎥ (4.251) ⎢436 9 3 0 ⎥ ⎢6y ⎥ ⎢ 3838.594⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 3 9 436⎦ ⎣6z ⎦ ⎣ 0 ⎦ Solving equation (4.251) yields the following answers for the four unknowns: A s 7.457 s 2 6x 65.730 rad/s 2 6y 1.147 rad/s 2 6z 0.483 rad/s 2 This gives us the last two angular acceleration vectors for the upper and lower damper bodies: T 6 1 T 7 1 { } [ 65.730 1.147 0.483 ] rad/s { } [ 65.730 1.147 0.483 ] rad/s From equation (4.240) we now have {A C6C7 } 1 2{ 6 } 1 {V C6C7 } 1 A s {R CI } 1 (4.252) ⎡879.162⎤ ⎡ 3 ⎤ ⎡ 856.791 ⎤ { A C C } 1 ⎢ 3626.702 ⎥ 7.457 ⎢ 9 ⎥ ⎢ 3559.589 ⎥ mm/s ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ 80.457 ⎦⎥ ⎣⎢ 436⎦⎥ ⎣⎢ 3331.709⎦⎥ 6 7 2 (4.253) A comparison of the angular accelerations found from the preceding calculations and those found using an equivalent MSC.ADAMS model is shown in Table 4.12. 2 2 Table 4.12 Comparison of angular acceleration vectors computed by theory and MSC.ADAMS Body Angular acceleration vectors Theory MSC.ADAMS x (rad/s 2 ) y (rad/s 2 ) z (rad/s 2 ) x (rad/s 2 ) y (rad/s 2 ) z (rad/s 2 ) 2 11.790 0.0 0.0 11.791 0.0 0.0 3 18.819 0.0 1.145 18.823 0.0 1.146 4 29.386 3.737 20.847 29.395 3.737 20.840 5 23.361 0.065 1.843 23.384 0.157 2.053 6 65.730 1.147 0.483 57.204 2.663 0.449 7 65.730 1.147 0.483 57.204 2.663 0.449

226 Multibody Systems Approach to Vehicle Dynamics Table 4.13 Comparison of translational acceleration vectors computed by theory and MSC.ADAMS Point Translational acceleration vectors Theory MSC.ADAMS A x (mm/s 2 ) A y (mm/s 2 ) A z (mm/s 2 ) A x (mm/s 2 ) A y (mm/s 2 ) A z (mm/s 2 ) C 210.614 28 476.707 3455.690 210.385 28 478.60 3456.320 D 177.625 47 432.261 2920.973 177.846 47 433.70 2921.760 G 0.0 41 481.177 1888.432 0.0 41 480.60 1888.430 H 422.619 45 953.343 3744.647 420.215 45 933.00 3722.660 P 558.211 36 163.671 0.0 557.780 36 161.60 0.0 C 6 C 7 856.791 3559.589 3331.709 957.170 3 544.09 2566.940 A comparison of the translational accelerations found at points within the suspension system from the preceding calculations and those found using an equivalent MSC.ADAMS model is shown in Table 4.13. 4.10.4 Static analysis As discussed in this chapter a starting point for suspension component loading studies is to use equivalent static forces to represent the loads acting through the road wheel, associated with real world driving conditions. In this example the vector analysis method is used to carry out a static analysis where a vertical load of 10 000 N is applied at the tyre contact patch, this being representative in magnitude of the loads used for a 3G bump case on a typical vehicle of this size. In this analysis we are ignoring gravity and the self-weight of the suspension components as this contribution tends to be minor compared with overall vehicle loads reacted at the tyre contact patch and diffused into the suspension system. For completeness the effects of self-weight will be included in a follow-on demonstration of a dynamic analysis. Before attempting any vector analysis to determine the distribution of forces it is necessary to prepare a free-body diagram and label the bodies and forces in an appropriate manner as shown in Figure 4.73. For the action–reaction forces shown acting between the bodies in Figure 4.73 Newton’s third law would apply. The interaction, for example, at point D between Body 3 and Body 4 requires {F D43 } 1 and {F D34 } 1 to be equal and opposite equal. Thus instead of including the six unknowns F D43x , F D43y , F D43z , F D34x , F D34y and F D34z we can reduce this to three unknowns F D43x , F D43y , F D43z . In a similar manner looking at the connections at points G and H we can see for all connections to Body 4 that the following applies: {F D43 } 1 {F D34 } 1 (4.254) {F G42 } 1 {F G24 } 1 (4.255) {F H45 } 1 {F H54 } 1 (4.256)

226 Multibody Systems Approach to Vehicle Dynamics<br />

Table 4.13 Comparison of translational acceleration vectors computed by theory and<br />

MSC.ADAMS<br />

Point<br />

Translational acceleration vectors<br />

Theory<br />

MSC.ADAMS<br />

A x (mm/s 2 ) A y (mm/s 2 ) A z (mm/s 2 ) A x (mm/s 2 ) A y (mm/s 2 ) A z (mm/s 2 )<br />

C 210.614 28 476.707 3455.690 210.385 28 478.60 3456.320<br />

D 177.625 47 432.261 2920.973 177.846 47 433.70 2921.760<br />

G 0.0 41 481.177 1888.432 0.0 41 480.60 1888.430<br />

H 422.619 45 953.343 3744.647 420.215 45 933.00 3722.660<br />

P 558.211 36 163.671 0.0 557.780 36 161.60 0.0<br />

C 6 C 7 856.791 3559.589 3331.709 957.170 3 544.09 2566.940<br />

A comparison of the translational accelerations found at points within the<br />

suspension system from the preceding calculations and those found using<br />

an equivalent MSC.ADAMS model is shown in Table 4.13.<br />

4.10.4 Static analysis<br />

As discussed in this chapter a starting point for suspension component<br />

loading studies is to use equivalent static forces to represent the loads acting<br />

through the road wheel, associated with real world driving conditions.<br />

In this example the vector analysis method is used to carry out a static<br />

analysis where a vertical load of 10 000 N is applied at the tyre contact<br />

patch, this being representative in magnitude of the loads used for a 3G<br />

bump case on a typical vehicle of this size.<br />

In this analysis we are ignoring gravity and the self-weight of the suspension<br />

components as this contribution tends to be minor compared with<br />

overall vehicle loads reacted at the tyre contact patch and diffused into the<br />

suspension system. For completeness the effects of self-weight will be<br />

included in a follow-on demonstration of a dynamic analysis.<br />

Before attempting any vector analysis to determine the distribution of<br />

forces it is necessary to prepare a free-body diagram and label the bodies<br />

and forces in an appropriate manner as shown in Figure 4.73.<br />

For the action–reaction forces shown acting between the bodies in Figure<br />

4.73 Newton’s third law would apply. The interaction, for example, at point<br />

D between Body 3 and Body 4 requires {F D43 } 1 and {F D34 } 1 to be equal<br />

and opposite equal. Thus instead of including the six unknowns F D43x ,<br />

F D43y , F D43z , F D34x , F D34y and F D34z we can reduce this to three unknowns<br />

F D43x , F D43y , F D43z . In a similar manner looking at the connections at points<br />

G and H we can see for all connections to Body 4 that the following<br />

applies:<br />

{F D43 } 1 {F D34 } 1 (4.254)<br />

{F G42 } 1 {F G24 } 1 (4.255)<br />

{F H45 } 1 {F H54 } 1 (4.256)

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