Working Dimensions Of The Formed Parts Lab Report
4.12.4 Calculation of the Working Dimensions of the Formed Parts
For the calculation of the working size of the molded parts, there are the average shrinkage method and the limit size method. The average shrinkage method is used to calculate the working dimensions of the stress cone cavity. According to literature [55],(3-14): the cavity size of the formula:
(1) The radial size of core and reentrant mold
1) Size calculation
L_M=[(1+S_max ) L_s-∆_1 ]_0^(+δ_z1 )
=[(1+2.2%)×42-0.36]_0^(+0.12)
= 42.564 mm
In the formula above: LM - The radial dimension of the die (mm);
Smax - The maximum shrinkage of plastic (%); Here, Smax = 2.2%
Ls - Radial dimensions of plastic parts (mm); Here, Ls = 42 mm
∆1 - Tolerances of radial dimension of plastic parts …show more content…
2) When repairing mold core shorten
A. Dimensional calculation
H_M=[(1+S_min ) H_s+∆_4 ]_(-δ_z4)^0
=[(1+1.8%)×74+0.54]_(-0.18)^0
=〖75.872〗_(-0.18)^0 mm
In the formula above: The symbols are as defined above.
B. Dimensional check
HM – Smax ∙ Hs – δz4 ≥ Hs
In the formula above: The symbols are as defined above.
Because: 75.872 - 2.2% × 74-0.18 = 74.064 mm ≥ 74 mm
So, to the core depth to meet the requirements, the calculations should be as above.
(5) Center distance size
1) Dimensional calculation
L_M=[(1+S_cp ) L_s ]±δ_z5/2
= [(1+2%) × 62] ± (0.13 / 2)
= 63.24 ± 0.065 mm
In the formula above: LM - The center distance of mold (mm);
Ls - Center distance of plastic parts (mm), Here; Ls = 62 mm δz5 - Mold center distance manufacturing tolerances (mm),
Hence: δz5 = 1/3 ∆5 = 0.13 mm
Scp - Average shrinkage of plastic (%), Here, Scp = 2%
2) Dimensional check
L_M-S_min∙L_s+δ_z5/2-∆_5/2≤L_s
In the formula above: ∆5 - The height tolerance of the plastic part (mm), Here; ∆5 = 0.39 mm
The rest of the calculations are the same as above. 63.24 - 1.8% × 62 + 0.13/2 - 0.39/2 = 61.994 mm ≤ 62 mm
So the center distance from the size to meet the
For the calculation of the working size of the molded parts, there are the average shrinkage method and the limit size method. The average shrinkage method is used to calculate the working dimensions of the stress cone cavity. According to literature [55],(3-14): the cavity size of the formula:
(1) The radial size of core and reentrant mold
1) Size calculation
L_M=[(1+S_max ) L_s-∆_1 ]_0^(+δ_z1 )
=[(1+2.2%)×42-0.36]_0^(+0.12)
= 42.564 mm
In the formula above: LM - The radial dimension of the die (mm);
Smax - The maximum shrinkage of plastic (%); Here, Smax = 2.2%
Ls - Radial dimensions of plastic parts (mm); Here, Ls = 42 mm
∆1 - Tolerances of radial dimension of plastic parts …show more content…
2) When repairing mold core shorten
A. Dimensional calculation
H_M=[(1+S_min ) H_s+∆_4 ]_(-δ_z4)^0
=[(1+1.8%)×74+0.54]_(-0.18)^0
=〖75.872〗_(-0.18)^0 mm
In the formula above: The symbols are as defined above.
B. Dimensional check
HM – Smax ∙ Hs – δz4 ≥ Hs
In the formula above: The symbols are as defined above.
Because: 75.872 - 2.2% × 74-0.18 = 74.064 mm ≥ 74 mm
So, to the core depth to meet the requirements, the calculations should be as above.
(5) Center distance size
1) Dimensional calculation
L_M=[(1+S_cp ) L_s ]±δ_z5/2
= [(1+2%) × 62] ± (0.13 / 2)
= 63.24 ± 0.065 mm
In the formula above: LM - The center distance of mold (mm);
Ls - Center distance of plastic parts (mm), Here; Ls = 62 mm δz5 - Mold center distance manufacturing tolerances (mm),
Hence: δz5 = 1/3 ∆5 = 0.13 mm
Scp - Average shrinkage of plastic (%), Here, Scp = 2%
2) Dimensional check
L_M-S_min∙L_s+δ_z5/2-∆_5/2≤L_s
In the formula above: ∆5 - The height tolerance of the plastic part (mm), Here; ∆5 = 0.39 mm
The rest of the calculations are the same as above. 63.24 - 1.8% × 62 + 0.13/2 - 0.39/2 = 61.994 mm ≤ 62 mm
So the center distance from the size to meet the