Question

MFG 333: Strength of Materials Lab 4b: Stress-strain Diagram Given the data set in the table below from a tensile test experiment, generate the strain and stress columns and create the stress-strain diagram for the material in Microsoft Excel. From the graph estimate: a. Proportional limit stress b. Proportional limit strain c. Elastic modulus d. Yield strength (0.2% offset) e. Tensile strength L f. Fracture strain Gauge length Initial diameter 50 mm 12.5 mm Length (mm) 50.00 50.05 50.08 50.12 50.18 50.27 50.55 51.12 51.59 $2.37 53.16 $3.92 54.71 55.5 56.29 57.05 57.83 58.62 61.95 68.78 71.12 7152 7231 7264 Load (kN) 0.00 8.00 12.00 20.00 28.00 32.00 37.00 40.20 42.05 47.43 51.32 54.80 57.59 59.98 62.28 63.77 64.86 66.16 69.35 70.74 70.55 68.95 58.69 48.33

Answer 1

Refer below to determine the above mentioned parameters from tension test;

From the given parameter, determine the area to determine stress.

Gauge length (Lo) = 50 mm Initial diameter (d) = 12.5 mm Area (A) d/4 = 122.73 mm2 122.73X 106 m2

Now determine stress; Stress = load/area

Strain = change in length/original length (determine; change in length = length - gauge length)

Strain &-AL/Lo Length (mm) Load P (kN)Stress o = P/A (kPa) Change in length AL (mm 50 0.00 0 0 65183.74 50.05 8 0.05 0.001 50.08 12 97775.60 0.08 0.0016 50.12 20 162959.34 0.12 0.0024 50.18 28 228143.08 0.18 0.0036 260734.95 50.27 32 0.27 0.0054 50.55 37 301474.78 0.55 0.011 51.12 40.2 327548.28 1.12 0.0224 51.59 42.05 342622.02 1.59 0.0318 52.37 47.43 386458.08 2.37 0.0474 53.16 51.32 418153.67 3.16 0.0632 53.92 54.8 446508.60 3.92 0.0784 54.71 57.59 469241.42 4.71 0.0942 55.5 59.98 488715.07 5.5 0.11 56.29 62.28 507455.39 6.29 0.1258 519595.86 57.05 63.77 7.05 0.141 57.83 64.86 528477.14 7.83 0.1566 58.62 66.16 539069.50 8.62 0.1724 61.95 69.35 565061.52 11.95 0.239 68.78 70.74 576387.19 18.78 0.3756 71.12 70.55 574839.08 21.12 0.4224 71.52 68.95 561802.33 21.52 0.4304 72.31 58.69 478204.19 22.31 0.4462 48.33 72.64 393791.25 22.64 0.4528

Plot stress vs strain curve;

From the graph: (a) Proportional limit stress = 260734.95 kPa

700000.00 600000.00 500000.00 400000.00 300000.00 Proportional limit stress 260734.95 200000.00 Elastic modulus E = 48284249.4 kPa 100000.00 Proportional limit starin = 0.0054 0.00 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Strain Stress (kPa) SC

(Proportional limit is the point, below which no permanent deformation occurs. Stress corresponding to that point is called proportional limit stress)

(b) Proportional limit strain = 0.0054

(Strain corresponding to proportional limit stress is called as proportional limit stain)

(C) Elastic modulus E = 48284249.4 kPa

(Its the slope of strain strain curve within proportional limit. It is calculated as; E=stress/strain at proportional limit)

(d) Yield strength at 0.2% offset = 301474.78 kPa

It is obtained by drawing line parallel to proportional limit region from 0.2% strain offset = 0.002

700000.00 Tensile strength= 576387.19 kPa. 600000.00 500000.00 400000.00 300000.00 Yield strength at 0.2% Offset 301474.78 kPa 200000.00 100000.00 Fracture strain 0.453 starin at 0.2% offset 0.002 mm/mm = 0.00 0.15 0 0.05 0.1 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Stress (kPa)

(e) Tensile strength = 576387.19 kPa

Tensile strength is the resistance of material breaking under tension. Nothing but maximum load or the maximum stress carried by material.

(f) Failure strain = 0.453