1 rubber composite support deformation and load relationship
The rubber composite support consists of an upper rubber pad, a lower rubber pad, a screw, a metal sleeve, an upper metal gasket, and a lower metal gasket (see 2). The metal sleeve is between the hole in the rubber gasket and the screw. a is the height of the upper rubber nut when the nut is tightened, and b is the height of the lower rubber mat when the nut is tightened, satisfying l = a + b + t, where l is the length of the metal sleeve and t is the thickness of the frame steel plate. With respect to the rigidity of the rubber mat, the metal sleeve and the frame steel plate can be regarded as a rigid body.
When the rubber composite support is tightened, both ends of the metal sleeve are in contact with the upper metal gasket and the lower metal gasket, respectively. When the rubber assembly is pre-supported and tightened, the total deformation Δl of the upper and lower rubber mats is (see 2) Δl = l - (a 0 + b 0 + t) = (a + b) - (a 0 + b 0) (1) The sum of the axial deformations of the two rubber mats should be equal to Δl, where a 0 is the initial height of the upper rubber mat and b0 is the initial height of the lower rubber mat.
The rubber mat in the rubber composite support is generally hard, and in the case of small deformation, the axial deformation and load of the rubber mat approximately linear relationship, and the slope of the straight line is the rigidity of the rubber mat along the axis direction.
Using F0 (3) in combination with equations (2) and (3), the compressive strain Δa 0 and Δb 0 Δa 0 = F 0 ka = kbka + k bΔl, Δb 0 = F resulting from the pre-tightening of the rubber mat and the lower rubber mat are obtained. 0 ka = kaka + k bΔl (4) where Δl = l - (a 0 + b 0 + t) is given by (1).
If the vehicle body and its load are acting on the rubber composite support with pressure P, if the screw transmits the pressure on the upper metal gasket and the lower metal gasket with F, using the types (2) and (5) and the deformation coordination condition, there is Δl = Δa + Δb = F ka + kbkakb + P ka = Δa 0 + Δb 0 (6) Solving the equation (6), obtain F = kakbka + kb (Δl - P ka) (7) FP > k a Δl. Use Δx = Δa - Δa 0 to indicate the displacement at the metal shims on the rubber composite support under pressure P (Note: This assumes that the body moves down Δx is negative and the body moves up Δx is positive), combined with (4) and (7) And (5), get Δx = P ka + kb,0 > P > k aΔl - k bΔl ka + kb + P ka, P ≤ k aΔl (8) When the rubber support is under pressure, if the pressure 0 car is uneven Driving on the road surface, the body rebounds and the rubber composite supports the overall tensile force. The body tension P > 0 is transmitted from the screw to the lower metal gasket, so that the lower metal gasket is under pressure. Assume that the rubber pad is under pressure F - k bΔl and the elastic stiffness is kb.
Set the body and its load so that when the rubber composite support is under pressure, P 0, comprehensive (8) and (12), the relationship between the deformation of the rubber composite support and the external force is Δx = - k bΔl ka + kb + P ka, P Comparison of 组åˆk aΔl P ka + kb, k aΔl 2 rubber composite support test and theoretical results
Three different groups of rubber composite supports were selected for the test, namely 1 support and 2 respectively.
Support, 3
Supports, these rubber composite supports have the same dimensions in the axial direction, l = 70 mm, a 0 + b 0 + t = 80 mm, Δl = l - (a 0 + b 0 + t) = - 10 mm (15 Where l is the length of the sleeve and a 0 and b 0 are the initial heights of the upper and lower rubber pads, respectively. Measured by the device i
The elastic stiffness k(i)a and k(i)b(i=1,2,3) of the rubber mat and the lower rubber mat on the rubber composite support are as follows k(1)a=322.
9 N/mm, k(1)b= 70.95 N/mm k(2)
a= 143.86 N/mm, k(2)
b = 60.20 N/mm k(3)
a=143.86 N/mm, k(3)
b=171.02 N/mm(16)
Substituting (15) and (16) into (3) gives i
The preload of the rubber combined support F(i)0(i=1,2,3)F(1)0=k(1)ak(1)bk(1)ak(1)b?l=-581.
69 N,
F(2)0 = - 424.
40 N, F(3)0 = - 781.
34 N (17)
Substituting equation (16) into equation (14) yields the relationship between deformation and external force of i-rubber composite support P( i)P( 1)- 581.69 +322.91Δx,Δx≤- 8.2 mm 393.86Δx, - 8.2 mm 8 mm (18 a) P(2)
- 424.40+ 143.86Δx, Δx≤- 8.05 mm 204.06Δx, - 7.05 mm Note: The stiffness of the rubber composite support is the slope of the straight line
Using a rubber composite support test device, you can measure i
The rubber composition supports the elastic stiffness coefficients k(i)k(1) = 406 N/mm, k(2) = 211 N/mm, and k(3) = 324 N/mm(19), respectively, under small deformation conditions. The experimental results of the formula are basically consistent with the theoretical results of the small deformation in (18). The theoretical results are 393.86, 204.06, and 314.88 N/mm.
3 conclusions
The rubber composite support can relieve the vibration between the vehicle body and the frame. The damping of the rubber composite support can consume the excitation of the vehicle body caused by road fluctuations when the SUV vehicle is running. The rubber composite support plays an important role in the SUV vehicle, and the structure of the rubber composite support makes It can ease the up and down vibration of the body. The relation (14) between the external load P and Δx displacement of the rubber composite support is deduced, which can be used to guide the design of the rubber composite support. The theoretical calculation result (18) is basically consistent with the test result (19).
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