Figure 1: P = 400 kN 30° 1000 mm Area 30 mm 20 mm In Figure 1, if the Young modulus is 200 Gpa, the strain percentage in the direction of the normal force is الاختيارات 0.48% 0.28% 0.18% 0.38% O
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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 0.300 300- 7.2-26 The strains on the surface of an experiment al device made of pure aluminum (E = 70 GPa. v = 0.33) and tested in a space shuttle were measured by means of strain gages. The gages were oriented as shown in the figure. and the measured strains were = 1100 X 106, h = 1496 X 10.6, and = 39.44 X l0_. What is the stress o in the x direction?During a test of an airplane wing, the strain gage readings from a 45° rosette (see figure) are as follows: gage A, 520 × l0-6; gage B. 360 × l0-6; and gage C,-80 × 10-6. Determine the principal strains and maximum shear strains, and show them on sketches of properly oriented elements.
- On the free surface of a component, a strain rosette was used to obtain the following normal strain data: €a = 300µɛ, ɛp = 400µs, and Ec = 200µe. Calculate the normal and shear strains in the x-y plane.A polypropylene has the following tensile creep compliance measured at 40 °C: D(t) = 1.2 x t0.1 GPa-1 where t is expressed in seconds. It is subjected to the following tensile stress sequence at 40 °C. Find the tensile strain at 3200 sec. Stress (MPa) 1.5 1.0 0.5 1000 2000 3000 Time (s)In our previous topic regarding strain, repeat the problem on the strain using thefollowing parameters:F = 150000 +/- 150 lbfSide length = 1.2 +/- 0.025 inModulus of Elasticity = 25000 +/- 11.5 ksiDetermine the relative error. Is it above or below the accepted 5% error?
- 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 OPROPAGATION OF ERRORS Determine the range of values to which the strain will be calculated.F = 150000 +/- 150 lbfSide length = 1.2 +/- 0.025 inModulus of Elasticity = 25000 +/- 11.5 ksiDetermine the relative error. Is it above or below the accepted 5% relative error?Tensile 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.
- Problem 8. Re-write the strain compatibility condition €22,33 + €33,22 in the engineering notation. = 2 e23,233. For a steel alloy, the stress-strain behaviour is shown below. Determine the modulus of elasticity, proportional limit, yield strength at 0.002 and maximum allowable load. 600 500E 400- 500 300 400- 300아 200 200- 100- 100E 0.000 0.002 0.004 0.006 Strain 0.00 0.04 0.08 0.12 0.16 0.20 Strain Stress (MPa) Stress (MPa)● Example 3.2 A 60mm diameter solid shaft has a strain gauge mounted at 65° to the axis of the shaft. In service a torque is applied to the shaft and the strain gauge reads 200 x 10-6. Calculate the value of the torque if the shaft is made from steel with E = 207 GPa and v = 0.3. (BC&A question 11.18) Strain Transformations Mechanical Engineering Mechanics of Materials 2 Solution 3.2 Applied torque alone will produce only shear stress/strain in the directions parallel and perpendicular to the torque. The strain gauge is mounted at 65° to the axis of the shaft. Strain Transformations Mechanical Engineering Mechanics of Materials 2 Now Strain gauge From the strain circle 65⁰ Strain Transformations Note that for circular shaft 4 D J = ² (2) ²* 33 Dr PJ Martin Lecture Notes 34 Dr PJ Martin Lecture Notes 35