Answer Happy

Step by step solving the following equation manually. What is
important to me is how do I graph the first and determine the
points of the X axis because it is very small, the second question
with drawing them as well, please, the second graph how do I do it,
the third question how do I get values from the graph
[[The ultimate tensile stress = 745.54 MPa The fracture stress =
641.511 MPa Elongation percentage% = 23 % , From Plotting Area
(1): The modulus of elasticity = 𝝈𝒑𝒍 𝜺𝒑𝒍 = 𝟑𝟖𝟏.𝟒𝟑𝟗 𝟎.𝟎𝟎𝟏𝟖 = 217.965
GPa The yield stress = 409.1803 MPa Modulus of resilience = Ur =
0.5*𝝈𝒑𝒍*𝜺𝒑𝒍 = 0.5*381.439*0.0018 = 0.333759 MJ/m]]]
Load (KN) Elongation (mm) 0 0 6.672 0.0127 20.462 0.0381 35.586 0.0635 48.930 0.0889 52.489 0.127 52.489 0.2032 53.378 0.508 73.840 1.016 88.964 2.54 95.636 7.112 86.740 10.16 82.292 11.684 Solution: Cross-sectional area A = (nt/4)ť = (1/4)(12.78)2 = 128.278 mm2 Stress: 0= F/A = 6.672x103/128.278 = 52 MPa Strain & = AL/L0 = 0.0127/50.8 = 0.00025 = 0.0003 mm/mm Plotting area 1: 800 Stress (MPa) 0 52.01445 159.511 277.4104 381.4393 409.1803 409.1803 416.1156 575.6265 693.5259 745.5404 676.1878 641.5115 Strain (mm/mm) 0 0.0003 0.0008 0.0013 0.0018 0.0025 0.0040 0.0100 0.0200 0.0500 0.1400 0.2000 0.2300 = = 700 Stress (MPa) 600 500 2 400 300 200 100 0 0.0000 0.0500 0.1000 0.1500 0.2000 0.2500 Strain (mm/mm)
From Plotting Area (1): The ultimate tensile stress = 745.54 MPa 1.5 The fracture stress = 641.511 MPa Elongation percentage% = 23 % Plotting area 2: 450 400 Stress (MPa) 350 300 250 200 150 100 50 0 0.0000 0.0020 0.0040 0.0060 0.0120 0.0080 0.0100 Strain (mm/mm) From Plotting Area (1): pl The modulus of elasticity 381.439 = 217.965 GPa 0.0018 Epl 1.5 The yield stress = 409.1803 MPa Modulus of resilience = Ur = 0.5* O pl*Epl = 0.5*381.439*0.0018 = 0.333759 MJ/mº