Modulus Of Resilience Of A Molded Phenolic Plastic
Strings (SiPjAjk) = S7P3A31 Base Sequence = 12735 String Sequence = 12735 - 3 - 31
Data from a tension test to determine the elastic properties of a molded phenolic (sythetic 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.
S7P3A31 (Force - Pull).
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%
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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