Modulus Of Resilience Of A Molded Phenolic Plastic
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Modulus Of Resilience Of A Molded Phenolic Plastic

Data from a tension test to determine the elastic properties of a molded phenolic (synthetic resin) plastic are as follows:
Specimen diameter, d = 0.400 + 0.001 in.
Gage length, l = 1 + 0.01 in.
Load at the proportional limit, P = 500 + 20 lb
Elongation due to P, δ = 0.0030 + 0.0001 in.

(a) Determine the modulus of resilience, ur.
(b) Calculate the probable maximum relative error in ur.

The strings: S7P3A31 (Force - Pull).

The math:
Pj Problem of Interest is of type force (pull). Modulus of resilience is strain energy absorbed per unit volume. Energy and Work problems are in general force problems. In this case the force is tension force (pull).

Formulas of interest:
Modulus of resilience ur, is strain energy absorbed per unit volume.
ur = Pδ/2V = σplεpl/2 ---------(1)
Where P is load at the proportional limit. Proportional limit is the maximum stress a material can sustain without deviating from the law of stress-strain proportionality.
δ is elongation due to P
V is volume
σpl is stress at proportional limit
εpl is strain due to σpl

The probable maximum relative error in ur is given by the following equation:
dur/ur = [(dP/P)2 + (dδ/δ)2 + (2dA/A)2 + (dl/l)2]1/2 --------(2)
Where each term in equation (2) represents relative change.
dur/ur = (dP/P) + (dδ/δ) + (2dA/A) + (dl/l) is obtained by taking the logarithm and partial differentiation of ur = Pδ/2V.

(a) ur = Pδ/2V.
So, ur = 500(0.0030)/(2x.04πx1) = 6 lb-in.
So, modulus of resilience = 6 lb-in.

(b) dur/ur = [(dP/P)2 + (dδ/δ)2 + (2dA/A)2 + (dl/l)2]1/2
So, dur/ur = [(20/500)2 + (0.0001/0.0030)2 + 4(0.001/0.400)2 + (0.01/1)2]1/2
= [(0.04)2 + (0.033)2 + 4(0.0025)2 + (0.0001)]1/2 = [0.0028]1/2 = 0.053 = 5.3%
So, probable maximum relative error in ur = 5.3%

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