During stress-strain test, the unit deformation at a stress of 35 MPa was observed to be 167 x 10^-6 m/m and at a stress of 140 MPa it was 667 x 10^-6. If the proportional limit was 200 MPa, what is the modulus of Elasticity (GPa)? What is the strain corresponding to a stress of 80 MPa?
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- Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight (units of in ) that can be stored in each or the materials listed in the accompanying table, assuming that the material is stressed to the proportional limit. DATA FOR PROBLEM 2.7-5 Material Weight Density (lb/in3) Modulus of Elasticity (ksi) Proportional Limit (psi) Mild sleel 0.284 30,000 36,000 Tool steel 0.284 30,000 75,000 Aluminum 0.0984 10,500 60,000 Rubber (soft) 0.0405 0.300 300Tensile test specimens are extracted from the "X" and "y" directions of a rolled sheet of metal. "x" is the rolling direction, "y" is transverse to the rolling direction, and "z" is in the thickness direction. Both specimens were pulled to a longitudinal strain = 0.15 strain. For the sample in the x-direction, the width strain was measured to be ew= -0.0923 at that instant. For the sample in the y-direction, the width strain was measured to be gw=-0.1000 at that instant. The yield strength of the x-direction specimen was 50 kpsi and the yield strength of the y-direction specimen was 52 kpsi. Determine the strain ratio for the x direction tensile test specimen. Determine the strain ratio for the y-direction tensile test specimen. Determine the expected yield strength in the z-direction. Give your answer in units of kpsi (just the number). If the sheet is plastically deformed in equal biaxial tension (a, = 0, to the point where & = 0.15, calculate the strain, 6, that would be expected.The following stress-strain curve was prepared based on a tensile test of a specimen that had a circular cross-section. The gage diameter of the specimen was 0.25 inches and the gage length was 4 inches. The stress scale of the stress-strain diagram is given with the factor a = 10 ksi. Estimate: (a) The modulus of elasticity. (b) The ultimate strength. (c) The yield strength (0.2% offset). (d) The percent elongation at fracture. 2013 Michael Swanbom STRESS VS. STRAIN BY NC SA 7a bat Sat 2at at 0.05 STRAIN 0.01 0.04 0.06 0.08 0.02 0.03 0.07 0.09 STRESS
- 1.7C1 1.4C1 ci 0.9C1 Uniform plastic elongation Necking Elastie 0 4 % 15% 35% Strain in % Stress-Strain diagram Stress in MPaA tensile test is conducted on a mild steel bar. The following data was obtained from the test : diameter of the steel bar= 3cm, Gauge Length of the bar = 20cm, Load at Elastic limit= 250kN, Extension at a load of 150kN= 0.21 mm, Maximum Load = 380kN, Total Extension= 60mm, Diameter of the rod at failure=2.25cm, determine young's modulus?By application of tensile force, the cross-sectional area of bar Pis first reduced by 30% and then by an additional 20% Another bar Q of the same material is reduced in cross-sectional area by 50% in a single step by applying tensile force After deformation, the true strain in bar Pand bar Q will, respectively be (a) 0.5 and 05 (b) 0.58 and 0.69 (c) 0.69 and 0.69 (d) 0.78 and 1.00
- O 0.6% 19% If an isotropic material has a Young's modulus of 85 Gpa and a Poisson's ratio of 0.25, calculate its shear modulus. Select one: O G = 29 Gpa O G = 34 Gpa G = 25 Gpa O G= 77 Gpa O G= 46 Gpa If a rubber material is deformed as shown in the following figure, determine the normal strain along diagonal BD. C 2 mm 4 mm EN DO OStress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…A tension test was performed on a specimen having an original diameter of 12.5 mm and a gage length of 50mm. The data are listed in the table below: Complete the following: Plot the stress-strain curve. Label the y-axis every 50 MPa, and the x-axis every 0.05 mm/mm. Plot the linear portion of the stress-strain curve (first 5 points). Label the y-axis every 50 MPa, and the x-axis every 0.001 mm/mm. Determine the approximate Modulus of Elasticity Determine the approximate Ultimate Stress Determine the approximate Fracture Stress Determine the approximate Modulus of Resilience Determine the approximate Modulus of Toughness Other Requirements: Provide an example hand-written calculation showing how you calculated one point on the curve. Remember to properly label your plots and provide axis labels with units. Hand sketched plots will not be accepted. Use Excel or similar software.
- A tensile test was conducted on a mild steel& the following data was obtained as follows. Diameter of the steel bar = 3 cm,Gauge length of the bar = 20 cm. Load at elastic limit = 250 kN . Extension at a load of 150kN = 0.21 mm . Maximum load = 380 kN . Total extension = 60 mm . Diameter of the rod at failure = 2. 25 cm Determine: (a) Young’s modulus (b) stress at elastic limit (c)the percentage of elongation (d) Percentage decrease in area.You are given a square rod of 6061-T6 aluminum (cross-section = 4mm X 4mm, length = 2.3m). The material has a Young’s modulus of 72 GPa. If a compressive load of 3 kN is applied parallel to the 2.3m dimension, what is the resulting engineering normal strain?The relationship between true strain (ɛ-) and engineering strain (ɛ) in a uniaxial tension test is given as (a) & = In (1 + &7) (c) &7= In (1+ E) (b) ɛg = In (1- E) (d)= in (1- E) %3D %3D