Calculation of generated force-displacement curve by an automotive switch

1. Arash Vakily For Presentation Only
Confidential Customer
Calculations for lockout button tactile forces
Introduction:
The lockout button for ____ program is a rocker type which activates lockout feature for the rear windows when pushed down.
Ram
Lockout tactile is achieved by passage of lockout button ram through the hump of the lockout spring, located on the case.
Hump
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1. Initial force and kinematic analysis:
Activation point
Occupant finger
d1
7.31
d2
9.16
Contact with lockout
feature
F1
11 N (From CTS)
Pivot
F2
OK F1 = 11 N Max. variance: ± 2 N
OK d1 = 7.31 mm
OK d2 = 9.16 mm
1 M1+M2=0
2 -F1 d1 + F2 d2 = 0
3 F2 = F1 d1 / d2
4 F2 = 8.778 N
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Equiblirium of forces on lockout button ram is shown below:
Forces R1, R2 are result of deformation of the two lockout beams creating tactile feel
Forces f1, f2 are friction forces between the beams and the ram
Note: End float of the button pivot (along the width) allows the ram to self-adjust, creating equal forces on both sides.
Therefore, R1=R2=R and f1=f2=f
f
R Rx
Θ
Θ fx
Ry Θ
Y
F2
X
Θ
Θ
R
f
5 F2 = 2 Rx + 2 f x
6 f=µR
8 2 Ry = 2 Fv Returning force from deflection of the
9 Ry = Fv lockout beams on each of two sides
10 Ry = R Cos θ
R y = Fv
11 R = Fv / Cos θ ( 9 , 10 )
12 f = µ Fv / Cos θ (6)
13 fx = f Cos θ
14 fx = µ Fv Cos θ / Cos θ ( 6 , 13)
15 fx = µ Fv ( 14 )
16 Rx = R Sin θ
17 Rx = Ry Tan θ ( 10 , 16 ) Fv
18 F2 = 2 ( Rx + fx ) (5)
19 F2 = 2 ( Ry Tan θ + µ Fv) ( 15 , 17 )
20 F2 = 2 Fv ( Tan θ + µ ) ( 9 , 19 )
21 Fv = F2 / [ 2 ( Tan θ + µ ) ] ( 20 )
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OK θ= 30 Degrees = 0.523599 Rad
µ= 0.39
22 Fv = 4.537 N ( 4 , 21 )
2. Stress & strain calculations:
Each of the beams have rectangular cross sections and are loaded by a force Fv in the middle and restrained at both ends.
Direction
of force
b
23 Ymax = F l / 192EI
3
at l/2 (F = Fv)
3
24 I = b h / 12 (Moment of inertia for rectangular cross section)
25 σ=My/I h
y
26 M = F (4x - l ) / 8 (F = Fv)
27 σ = F y ( l / 4 - x) / 2 I ( 25 , 26 ) (F = Fv)
28 σmax = F y l / 8 I at x = 0 ( 27 ) (F = Fv)
29 y=h/2
Formulas 23 and 28 calculate Maximum strain and Maximum stress in the beam (which happen in different locations).
n (Safety Factor) = 2
30 σmax = Sy / n
The relationship between σmax, Ymax, and h is calculated as below:
31 σmax / Ymax = 24 y E / l2 ( 23 , 28 )
32 h = σmax l / 12 E Ymax
2
( 29 , 31 )
Then, b can be calculated as follows:
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33 b = F l 3 / 16 E Ymax h 3 ( 23 , 24 ) (F = Fv)
For the selected material (PC+ABS blend) the yield stress and modulus of elasticity are:
Sy = 96.5 MPa
E= 2960 MPa
34 σmax = 48.25 MPa ( 30 )
Based on the desired tolerances of the features, minimum desired amount of Ymax is defined as below.
Even in the lower limit of tolerances, the amount of deflection should be significantly larger than tolerances themselves.
Also, Ymax should be kept to minimum possible to avoid developing of the cracks.
Therefore:
OK 1.4 +/-0.05 spring hump & 2.175 +/-0.05 ram -->> Y max = 2.225-1.35 = 0.875
OK Ymax = 0.875 mm
l
Length of the lockout beams ( ) should be kept at maximum possible.
Packaging information limit this dimension, based on the math data received from the customer.
Therefore:
OK l= 8.2 mm
h is therefore calculated:
35 h= 0.104 mm ( 32 , 34 )
Subsequently, b is calculated:
36 b= 53.075 mm ( 22 , 33 , 35 )
3. Cross section of the lockout feature beams:
From equations 35 & 36 above, the cross section of the beams are rounded to nearest preferred sizes, and defined as:
37 h= 0.1 mm
38 b= 53.1 mm
4. Design for tolerance insensitivity:
In order to assure optimal performance of the lockout feature, the extreme ends of the tolerance zones are considered
and the safety factor based on these dimensions is re-calculated as below.
This is to ensure the design and calculated dimensions are not over sensitive to tolerances of the features.
General tolerance for dimensions h, b:
Tol. = ± 0.1 mm
Note:
Changes in dimension h could result in changes to Ymax due to dependancy of geometry.
Worst case scenario is considered in the following table for all calculations.
39 σmax = 12 h E Ymax / l 2 ( 32 )
40 n = Sy / σmax ( 30 )
The following table shows the effect of tolerances on calculated values above.
Where:
Nominal: All dimensions considered at their nominal
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Case 1: h = h max , b = b max
Case 2: h = h min , b = b min
Case 3: h = h max , b = b min
Case 4: h = h min, b = b max
Note: Change in Ymax is a function in change in h:
41 ∆Ymax = - ∆h
Also, variation in applied force relative to the nominal activation force is calculated based on these tolerances.
Nominal activation force: 11 N (From page 2)
Max acceptable variation in activation force: ± 2 N (From page 2)
42 Fv = F = 16 E Ymax b h3 / l 3 ( 33 )
F2 = 2 Fv ( Tan θ + µ ) ( Equation 20 )
43 F1 = F2 d2 / d1 (2)
Parameter Units Nominal Case 1 Case 2 Case 3 Case 4 Equation
h mm 0.1 0.2 0.0 0.2 0.0
b mm 53.1 53.2 53.0 53.0 53.2
∆h mm - 0.1 -0.1 0.1 -0.1
∆ Ymax mm - -0.1 0.1 -0.1 0.1 ( 41 )
Ymax mm 0.875 0.8 1.0 0.8 1.0
σmax MPa 46.22249 81.87983 0 81.87983 0 ( 39 )
n - 2.09 1.18 N/A 1.18 N/A ( 40 )
Fv N 3.991 28.332 0.000 28.225 0.000 ( 42 )
F2 N 7.721 54.813 0.000 54.607 0.000 ( 20 )
F1 N 9.675 68.686 0.000 68.427 0.000 ( 43 )
∆ F1 N -1.325 57.686 -11.000 57.427 -11.000
Table 1
Conclusions for structural calculations:
1. Safety factor:
From the above table, the safety factor is close to the defined amount ( n = 2 ) in all cases, which shows the design is
not sensitive to the defined tolerances
2. Variation in activation force:
From the above table, variation in the force is always below the acceptable amount which also shows insensitivity
of the design to tolerances.
3. The above calculations are in ambient temperature. Hot / Cold calculations in the following pages.
Please note that the product duty cycle is very low ( 1500 cycles in the life time of vehicle)
5. Temperature effects:
In order to make sure the parts will work in extreme temperature, the stress calculations are made with modifications:
Modified material properties obtained from resin supplier:
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Sy' = 102 MPa at cold Sy' = 87 MPa at hot
E' = 3300 MPa at cold E' = 2218 MPa at hot
44 σmax = 12 h E' Ymax / l 2 ( 39 )
45 n = Sy' / σmax ( 40 )
For the selected material:
CTE = 0.000072 (1/ °C)
Parameter Units Ambient Hot Cold Equation
h mm 0.1000 0.1000 0.1000
b mm 53.1000 53.1038 53.0962
Ymax mm 0.8750 0.8751 0.8749
l mm 8.2000 8.2006 8.1994
σmax MPa 46.22249 34.63563 51.53183 ( 44 )
n - 2.09 2.51 1.98 ( 45 )
Table 2
As observed in the above table, effect of temperature on safety factor for stress is acceptable.
Safety factor is still close to desired amount ( 2 ) and therefore acceptable.
6. Tactile curve for occupant:
Total user displacement: 1.7 mm
Total user displacement is defined in CTS: 1.7mm
Note:
Relationship between angle of rotation and user displacement is cauculated in the kinematics analysis of buttons
The following table shows calculated relationship between force and user displacement:
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Parameter Units Values
Displ. mm 0.00 0.85 1.70 1.95
F Nom N 0.000 9.675 0.000 30.000
F Min N 0.000 0.000 0.000 30.000
F Max N 0.000 68.686 0.000 30.000
Table 3
Lockout Force/Travel Nom Max Min
Curves (ambient)
80.000
70.000
60.000
50.000
Force, N
40.000
30.000
20.000
10.000
0.000
-10.000
0.00 0.85 1.70 1.95
Displacement, mm
Notes:
Location of the tip of the hump is selected at 0.85 to achieve desired tactile curve
Forces at the above table / chart are from table 1 (F1)
Tactile curve above is consistent with customer requirements
General Notes for this document:
Values in yellow boxes are the inputs of the design.
Values with blue font in yellow boxes are non-changeable values (such as material properties, customer requirements, or
values that are dictated by the packaging constraints)
Values with red font in yellow boxes are the ones that are optimized, or considered for optimization, in order
to achieve the design goal.
Results presented in this summary report are after optimization.
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Appendix A: Calculation of the hump width:
d3
14
d2
9.16
d3 = 14 mm
d2 = 9.16 mm (from page 2)
Please refer to the tolerance stack-up analysis for calculation of max / min rotation angle.
Rotation angle of the knob:
Nominal 10.997 degrees 0.191934 rad
Max 11.635 degrees 0.203069 rad
Min 10.359 degrees 0.180799 rad
d3
Φ
y1
d3
d2
y2
Φ
x2 x1
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y1 = d3 Sin Φ
y2 = d2 - y1
x2 = y2 Tan Φ
Therefore:
x2 = y2 Tan Φ = (d2 - y1) Tan Φ
A1 x2 = (d2 - d3 Sin Φ) Tan Φ
x1 = d3 - d3 cos Φ
A2 x1 = d3 (1 - cos Φ)
A3 x = x1 + x2 Total horizontal movement
Parameter Units Nominal Max Min
Angle Rad 0.191934 0.203069 0.180799
x1 mm 0.26 0.29 0.23
x2 mm 1.26 1.30 1.21
x mm 1.52 1.59 1.44
Therefore, the width of the hump is selected:
Width = 1.65 ± 0.05
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11. Arash Vakily For Presentation Only
Appendix B: Effect of friction force
Due to vertical movement of the ram, there is a vertical component for the friction force, in z direction:
At the tip of the hump, the maximum friction force in z direction is:
B1 fz = µ Fv
B2 fz = 1.8 N
Amount of this force is insignificant compared to the force that deflects the beams:
B3 fz / Fv = µ
B4 fz / Fv = 0.39
Fv
ß
fz
Fv'
This force will result in a slight deviation in the amount of force and the angle of which the force is applied:
B5 Fv' = √ Fv2 + fz2
B6 Fv' = 4.870 N
B7 Fv' / Fv = 1.073
This means a 7.3 % increase in force which is negligible considering the safety factor selected
B8 ß = Tan-1 ( Fz / Fv )
B9 ß= 0.371856 rad = 21.31 degrees
As observed above, the deviation in angle and force is smaller than the amount of error resulted from manufacturing,
tolerances, or non-homogenity of the material.
Furthermore, the results of this calculation will be confirmed by FEA for a more accurate stress analysis
Please also note that for a more desireable sound level, the parts are lubricated, which will reduce
coefficient of friction, further reducing the effect of this deviation
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