MMO A03

import numpy as np import matplotlib.pyplot as plt !pip install uncertainties from uncertainties import ufloat

Oppgave 1

Necking occurs when the applied load ceases to increase, i.e. when the nominal stress-strain curve peaks.

epsilonNom= np.linspace(0,10,1001) sigmaTrue = 30*epsilonNom**0.8 sigmaNom = sigmaTrue/(1+epsilonNom) sigmaNomNeck = max(sigmaNom) #sigmaList = sigmaNom.tolist() #epsilonNeck = sigmaList.index(sigmaNeck) epsilonNomNeckIndex = sigmaNom.tolist().index(sigmaNomNeck) epsilonNomNeck = epsilonNom[epsilonNomNeckIndex] sigmaTrueNeck = sigmaTrue[epsilonNomNeckIndex] plt.plot(epsilonNom,sigmaNom,epsilonNomNeck, sigmaNomNeck, "o") #plt.hlines(sigmaNeck,0,10, 'C2') print("The nominal strain at necking is", epsilonNomNeck) print("The true stress at necking is", np.round(sigmaTrueNeck,2), "Mpa")

Oppgave 2

epsilonNomNeg = np.linspace(-0.99,10,1100) epsilonTrue = np.log(1 + epsilonNomNeg) # (i) plt.plot(epsilonNomNeg, epsilonTrue) plt.xlabel('\u03B5_nom') #epsilon plt.ylabel('\u03B5_t') #epsilon

# (ii) epsilonSmallDeviation = np.zeros(100) counter = 0 deviation = abs((epsilonTrue-epsilonNomNeg)/epsilonNomNeg)*100 for i in range(len(deviation)): if(deviation[i] <= 5.0): #print(deviation[i]) epsilonSmallDeviation[counter] = epsilonNomNeg[i] counter += 1 print("\u03B5_t deviates with more than 5% from \u03B5_nom when:") print("\u03B5_nom <", np.round(epsilonSmallDeviation[0],4), "and \u03B5_nom >", np.round(epsilonSmallDeviation[counter-1],4), "\n\n\n")

epsilonTrueApprox = epsilonNomNeg - epsilonNomNeg**2/2 print(epsilonTrueApprox) plt.plot(epsilonNomNeg, epsilonTrueApprox) plt.xlabel('\u03B5_nom') #epsilon plt.ylabel('\u03B5_t_approx') #epsilon

# (iii) epsilonSmallDeviation2 = np.zeros(100) counter2 = 0 deviation2 = abs((epsilonTrueApprox-epsilonTrue)/epsilonTrue)*100 for i in range(len(deviation)): if(deviation2[i] <= 5.0): #print(deviation[i]) epsilonSmallDeviation2[counter2] = epsilonNomNeg[i] counter2 += 1 print("\u03B5_t_approx deviates with more than 5% from \u03B5_t when:") print("\u03B5_nom <", np.round(epsilonSmallDeviation2[0],4), "and \u03B5_nom >", np.round(epsilonSmallDeviation2[counter2-1],4), "\n\n\n")

Oppgave 3

The elastic modulus of the composite material is given by:

E = E_c f + E_s (1-f)

# b) (i) E_c = 1000 #Gpa E_s = 1 #Gpa f = 0.5 E = E_c*f + E_s*(1-f) print("b) (i)\n E(f = 0.5) =", E, "Gpa\n") # b) (ii) f = 0.9 E = E_c*f + E_s*(1-f) print("b) (ii)\n E(f = 0.9) =", E, "Gpa\n") # c) (i) E = 73 #Gpa f = (E - E_s)/(E_c - E_s) print("c) (i)\n f(E =", E, "Gpa) =", np.round(f,2), "\n") # c) (ii) E = 235 #Gpa f = (E - E_s)/(E_c - E_s) print("c) (ii)\n f(E =", E, "Gpa) =", np.round(f,2), "\n")

Oppgave 4

TS = TS_inf - (A/Mn)

Solving the system of equations to find TS_inf and A.

# Solving the system of equations to find TS_inf and A (2eq) Mn_1 = 73800 # g/mol TS_1 = 16.7 # Mpa Mn_2 = 51300 # g/mol TS_2 = 14.4 # Mpa A = np.array([[1,-1/Mn_1],[1, -1/Mn_2]]) b = np.array([TS_1, TS_2]) X = np.linalg.solve(A,b) TS_inf = X[0] A = X[1] print("TS_inf =",np.round(TS_inf,2),"\nA=", np.round(A,2), "\n") # a) TS = 18.1 #Mpa Mn = A / (TS_inf - TS) print("a)\n Mn(TS =", TS, "Mpa) =", np.round(Mn,2), "g/mol\n") # b) TS = TS_inf * 0.95 Mn = A / (TS_inf - TS) print("b)\n TS = 95% TS_inf =", np.round(TS,2), "Mpa\n Mn(TS =", np.round(TS,2), "Mpa) =", np.round(Mn,2), "g/mol\n")

Oppgave 5

POM = np.loadtxt('polymer_sample.txt') n = POM[:,0] DP = POM[:,1] # a) total weight n_mono_tot = sum(n*DP) print("a)\n m_tot =", n_mono_tot, "g\n") # Vekt for ett mol monomerer [g/mol] M_c = 12 #g/mol M_h = 1 #g/mol M_o = 16 #g/mol M_pom_mol = M_c + 2*M_h + M_o mol = 6.022*10**23 # Vekt for ett monomer [g] M_pom = M_pom_mol/mol # Total masse mass_tot = M_pom * n_mono_tot

# b) # Weight fraction w_i = n*(M_pom*DP)/mass_tot # Numbers fraction x_i = n/sum(n) # Molecular weight for range M_i = DP*M_pom plt.figure(1) plt.plot(M_i, x_i, 'bo') plt.xlabel('M_i [g]') plt.ylabel('w_i') plt.figure(2) plt.plot(M_i, w_i, 'ro') plt.xlabel('M_i [g]') plt.ylabel('x_i') print("b)\n Se plot\n")

# c) M_w = sum(w_i*M_i) M_n = sum(x_i*M_i) DPI = M_w/M_n # Typical range 1.02 to 50+ print("c)\n M_w =", "%.4g" %M_w,"g\n M_n =", "%.4g" %M_n, "g\n DPI =", np.round(DPI,2)) #print(M_w * mol) #print(M_w/mol)

Oppgave 6

One reason for the lack of interessest in defects in polymers might be that they are less critical due to the macro-scale molecules of the polymer. The point defect concept is different in polymers than in metals due to the chainlike macromolecules and the nature of the crystalline state for polymers.

Other reasons might be that polymers are often used for low stress applications, while metals are preferred for load bearing applications, or that there has been done less research on polymers than metals in general.

Crystal polystyrene is not crystalline, it is completely amorphous.

Oppgave 7

a) See derivation in appendix

# b) p_a = 840 # density amorphous PP [kg/m^3] p_c = 950 # density crystalline PP [kg/m^3] p_s = 880 # density of PP specimen [kg/m^3] # The degree of crystallinity (DC) is given by: DC = (p_c*(p_s - p_a))/(p_s*(p_c - p_a))*100 # (i) Assuming no voids we have: print("b) (i) Assuming no voids we have:\n DC =", np.round(DC,2), "[%]\n") # (ii) Estimating uncertainty p_a = ufloat(840, 5) # p_a = 840+/-5 [kg/m^3] p_c = ufloat(950, 5) # p_a = 950+/-5 [kg/m^3] p_s = ufloat(880, 5) # p_s = 880+/-5 [kg/m^3] DC = (p_c*(p_s - p_a))/(p_s*(p_c - p_a))*100 print("b) (ii) With uncertainty we get:\n DC =",DC, "[%]\n") std_DC = DC.std_dev COV_DC = DC.std_dev/DC.nominal_value print(" Standard deviation =", np.round(std_DC,2), "[%]\n COV =",np.round(COV_DC,2), "%")

Oppgave 8

Size teaspoon = 4.93 mL +/- 5%

Size acre = 4046.86 m +/- 20%

where the ucertainties are guesses

volumeTeaspoon = 4.93 # mL COV_teaspoon = 0.05 # 5% stdTeaspoon = COV_teaspoon*volumeTeaspoon # using uncertainty +/-5% areaAcre = 4046.86 # m COV_acre = 0.20 # 10% stdAcre = COV_acre*areaAcre # using uncertainty +/-10% volumeTeaspoon = ufloat(volumeTeaspoon, stdTeaspoon) # mL areaAcre = ufloat(areaAcre, stdAcre) # m area = areaAcre/2 # m volume = volumeTeaspoon*10**(-6) # L thickness = volume/area print("Thickness =", thickness, "m\n") COV_thickness = thickness.std_dev/thickness.nominal_value print("Standard deviation thickness =","%.2g" %thickness.std_dev, "m") print("COV thickness =",np.round(COV_thickness,4)*100, "%")

The order of magnitude of the results are belivable. However it is difficult it is difficult to determine the uncertenties of the measured volume and area, as unconventional units are used. Therefore quite large uncertenties are used, making the uncertenty of the result noteble as well.

Oppgave 9

Se appendix