An assembly being produced is comprised of two aluminum plates with a hard rubbe

Question

An assembly being produced is comprised of two aluminum plates with a hard rubber insulation in between. The aluminum plates have a nominal thickness value of 0.09375 inch and the insulation is to be 0.0625 inch. thick. Because of the resilience of the hard rubber material, it is deemed necessary that the standard deviation of the process making this material to be one-half that of the process making the aluminum. If the overall assembly has specification of 0.25 + 0.0040 inch., what are the required specifications on the individual parts? Assume a +/- 4 sigma assembly

Explanation / Answer

Let X = thickness of aluminum plate

      Y = thickness of rubber insulation.

      T = thickness of assembly.

Then, T = X + X + Y.

Assuming thickness follow Normal Distribution, say X ~ N(µ1, 12) and

Y ~ N(µ2, 22), thickness of aluminum plates are iid and X and Y are independent,

V(T) = 2V(X) + V(Y)

        = 212 + 22 …………………………………………………………………………(1)

Given ‘it is deemed necessary that standard deviation of rubber making process to be one-half that of aluminum making process’, => 2 = (1/2) ....................…………………………………………(2)

(1) and (2) => V(T) = 212 + (12/4)

Or V(T) = (9/4)12

Or Standard deviation of T = T = (3/2)1 ………………………………………………(3)

Given the specification for T as 0.25 ± 0.0040 and a ± 4sigma assembly,

4T = 0.0040 or

T = 0.0010 ……………………………………….....……………………………………….(4)

(3) and (4) => 1 = 0.0010/(3/2) = 0.00067 ………………………………………………..(5)

(5) and (2) => 2 = 0.00067/2) = 0.00033 ………………………………………………..(6)

Given nominal for aluminum thickness as 0.09375, 1 = 0.00067 and a ± 4sigma,

Specification for aluminum thickness = 0.09375 ± 0.0026 ANSWER 1

Similarly, given nominal for insulation thickness as 0.0625, 2 = 0.00033 and a ± 4sigma,

Specification for aluminum thickness = 0.0625 ± 0.0013 ANSWER 2

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