Question

Please read the questions very well and plot using excel. Don’t waste my question if you’re not gonna do what I asked for. Thank you

Problem 1 A tensile test specimen having a diameter of 0.505 in. and a gage length of 2 in. was tested to fracture. Load and deformation data obtained during the test were as follows. Plot the conventional stress-strain curve for the steel and determine modulus of elasticity, the proportional limit, ultimate stress, yield stress at 0.2% offset, fracture stress and the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.425 in. Note use Excel to plot the curve with your original data and show your work directly on the curve. Load (b) Elongation (in.) Load(lb) Elongation (in.) Load (lb) Elongation (in.) 0 2200 4300 6400 8200 8600 8800 9200 0 0.0008 0.0016 0.0024 0.0032 0.0040 0.0048 0.0064 9500 9600 10600 11800 12600 13200 13900 14300 0.0080 0.0096 0.02 0.04 0.06 0.08 0.12 0.16 14500 14600 14500 14400 14300 13800 13000 0.2 0.24 0.28 0.32 0.36 0.4 Fracture

Answer 1

-
Area of speimen = pi /4 (0.805)^2 = 0.20019/m^2 Stress: Load/Area 1 strain = clongation/original length Proportional limtt = stress at the end of straight portion Modulus of elesticity = 41,000/0.0016 = 25.625 times 10 lb/in^2 Ultimate stress = Ultimate load/Area = 14500/0.2 = 72500 lb/in^2 Fracture stress correspoly to initial Diameter = 13,000/0.2 = 65,000 lb/in^2 True fracture stress = Fracture load/True area True area = pi /4(0.425)^2 = 0.14179 in^2 Fracture stress = 13,000/0.4179 = 91684.48 lb/in^2 Yield stress corresponding to 0.2% offsct slope of initial straight portion for 47000 stress time = tan 89.999771 degree > 47000/stress strain > 47,000/tan 89.999771 greaterthanorequalto 0.188 times 10^-2 adding '0.002' offsct > 0.00188 to 0.002 = 0.00388 Strain at 47,500 lb/in^2 is '0.004'. Therefore, the linear variation will correspond to 47,000 lb/in^2