Lab3 Data

Steel Grad Nominal Ca Fatigue 9254 0.54 1070 0.7 8620 0.8 8695 0.95 8695 0.95 6150 0.5 1552 0.52 845 533 458 401 509 307 257 Yield 2270 1950 920 780 847 1878 933 Ultimate Hardness 584 613 583 539 539 555 601 2450 2069 1202 1496 979 2343 933

You have been given the following set of data from a soil-geotextile friction test:   Normal stress (kPa) Peak shear strength (kPa) Residual shear strength (kPa) 17 20.2 13 35 31 22 70 52 39.5 140 94 74.5   (a) Plot the Mohr-Coulomb failure envelope (x-axis=normal stress; y-axis=shear stress) (b) Determine the peak friction angle,  and the residual friction angle, . (c) Determine the peak adhesion and the residual adhesion. (d) Calculate the fabric efficiency based on a soil friction angle of  = 35-degrees using the peak friction angle.   **HINT: the equation for efficiency of soil friction angle mobilization is:

Graph the following data on a stress-strain curve. Identify/Determine the following: Elastic Limit, Plastic Limit, Proportional Limit, Upper Yield Point, Lower Yield Point, Ultimate Strength, Rupture Strength, and Modulus of Elasticity. Engr. True Stress Strain o, MPa 102.3 0.00050 204.7 0.00102 293.5 0.00146 272.9 0.00230 264.0 0.00310 268.9 0.00500 267.6 0.00700 264.0 0.01000 318.6 0.0490 372.0 0.1250 394.4 0.2180 395.0 0.2340 384.2 0.3060 360.4 0.3300 327.5 0.3480 290.0 0.3600 266.8 0.3660

Can Young’s modulus have a negative value? What about the bulk modulus?

Calculate: 1. stress 2. strain 3. diameter of wire corrected to zero error

Considering Figure 20-8 (textbook), the torsion geometry parameter, Jw, for case 2 when b=6 and d=9 is most likely:     400.5     562.5     283.5     60.8     279.0     392.6     386.0     572.6     184.1

2) Calculate the following: A) Estimate the modulus of resilience. It uses the expanded scale. B) Estimate the modulus of toughness. It uses the regular scale with each block being 0.04 in width. 8 σ (ksi) 80 70 60 50 40 30 20 10 0 0 0 Stress vs. Strain Uses Regular Scale Uses Expanded Scale 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 € (in./in.)

1.5-6) A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data listed in the accompanying table (see next page). Plot the stress-strain curve and determine the proportional limit, modulus of elasticity (which is the slope of the initial part of the stress-strain curve), and the yield stress at 0.2% offset. Is the material ductile or brittle? PROBLEM 1.5-6

Can anyone help me with this question

The following data were collected from a 12 mm diameter test specimen of Magnesium. LOAD (N) GAUEGE LENGTH (mm) 0 5000 10000 15000 20000 25000 26500 27000 26500 30.000 25000 30.0296 30.0592 30.0888 30.15 30.51 30.90 31.50 (maximum load) 32.10 32.79 (fracture) After the fracture, the gauge length is 32.61 mm and the diameter is 11.74 mm. a) What is the elastic modulus? b) Percent elongation at fracture? c) Percent elongation after fracture? d) What is the Poisson's ratio? e)Draw the engineering stress-strain diagram corresponding to the values in the table. Call this plot I. Now consider this experiment is repeated at a higher temperature with an identical sample. Draw the new engineering stress-strain diagram, call it plot II and highlight the differences (on the same graph) between I and II.

Please show handwritten solution

. Data was collected from the tensile test demonstration for Aluminum 6061-T6. Below are the resulting stress-strain diagrams based on the tensile test we performed. o (MPa) Al 6061-T6 Stress-Strain Curve o (MPa) 350 320 300 250 200 150 100 50 0 Estimate 0 0.05 0.1 0.15 ε (mm/mm) 300 270 250 220 200 150 140 100 a) The modulus of Elasticity, E b) The yield stress, d, using the 0.2% offset method c) The strain at failure, & failure d) The modulus of resilience (approximately!) 50 0 0 Al6061-T6 Stress-Strain Curve, Magnified N 0.0032 0.002 0.004 0.006 0.008 € (mm/mm) The right diagram shows the magnified elastic region of stress-strain diagram to enhance the details.