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- (d) Figure Q2 (d) shows a typical aluminum stress-strain curve. A mechanical engineer has to carry out a tensile test for a rod of aluminum that can withstand an applied force of 20kN. To ensure safety, the maximum allowable stress on the rod is limited to 400 MPa, which is below the yield strength of the aluminum. The rod must be at least 380 cm long and must deform elastically not more than 0.8 cm when the force is applied. Design the appropriate rod.Mild steel 1 Young;s modulus 1219.5 Yield strain and stress (0.4101,500.08) Failure stress and strain :not able to find because the given data shows the experiment did not reach the failure point. if the material stress and strain does not reach a failure point ,what dose it means , does it means that the material is more stronger?Draw the tensile curves of the following materials, taking into account the appropriate elongation at break, modulus of elasticity and tensile stress, on the side stress-strain graph. a)Low carbon steel (Ecelik=200Gpa),(Syield=300MPa),(Stensile=400MPa) b)High carbon steel (Esteel=200GPa),(Stensile=700MPa) c)Aluminum (Ealuminum=70GPa), (Syield=200MPa),(Stensile=300MPa) d)Cast iron (Ecast iron=120GPa),(Stensile=200MPa)
- If a plate has to be manufactured by punching, with diameter d of 25.4 mm, thickness t 10 mm, and material's shear strength Ss = 115 MPa. How much minimum SHEARING force P in kilogram force must be applied TOWARDS THE PLATE to punch a hole? (Note: a. stress area is the area of failure, b. material is punched halfway down) Answer asap. Thanks . I will rate helpfulThe following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 150 kN. - Length of the specimen is 24 mm. - The yield strength is 78 kN/mm2. - The percentage of elongation is 42 %. Determine the following iii) Stress under an elastic load of 14 kN, iv) Young's Modulus if the elongation is 1.7 mm at 14 kN and (v) Final diameter if the percentage of reduction in area is 26 %. I wont answer for this three questions only: 1)Final Area of the Specimen at Fracture (in mm) 2)Final Diameter of the Specimen after Fracture (in mm) 3)Initial Cross-sectional Area (in mm2)And **The modulus of elasticity** (MPa) I need a clear step by step answer please :)
- A) An engineer works for a company that has many gray cast iron rods of different diameters and would like to design an equivalent rod that could handle a force of P= 10.0 kipkip without fracturing. What is the minimum diameter that this new rod must have to support P? B)A tensile test is being conducted on a steel-rod specimen with a gauge length of L0 = 4.0 in and an initial diameter of d0 = 0.75 in. If the final length of the rod at fracture is Lf= 5.53 in , find the percent elongation of the rod at fracture.1. What are the elastic modulus (E) and the Poisson's ratio () used to indicate? 2. Illustrate the differences between actual stress and engineered stress with strain, and also describe their underlying physical concepts. 3. If the engineering strain is 2% for a specific state of uniaxial stress, what is the real strain? Please solve for all in full detail and step by stepThe following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 157 kN. - Length of the specimen is 23 mm. - The yield strength is 89 kN/mm2. - The percentage of elongation is 45 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 18 kN, iv) Young's Modulus if the elongation is 1.3 mm at 18 kN and (v) Final diameter if the percentage of reduction in area is 25 %. Solution: Initial Cross-sectional Area (in mm2) The Diameter of the Specimen (in mm) Final Length of the Specimen (in mm) Stress at the elastic load (in N/mm2) Young's Modulus of the Specimen (in N/mm2) Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm)
- The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 142 kN. - Length of the specimen is 23 mm. - The yield strength is 82 kN/mm2. - The percentage of elongation is 48 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 15 kN, iv) Young's Modulus if the elongation is 1.6 mm at 15 kN and (v) Final diameter if the percentage of reduction in area is 20 %. Solution Initial Cross-sectional Area (in mm2) Answer for part 1 The Diameter of the Specimen (in mm) Answer for part 2 Final Length of the Specimen (in mm) Answer for part 3 Stress at the elastic load (in N/mm2) Answer for part 4 Young's Modulus of the Specimen (in N/mm2) Answer for part 5 Final Area of the Specimen at Fracture (in mm) Answer for part 6 Final Diameter of the Specimen after Fracture (in mm)The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 151 kN. - Length of the specimen is 23 mm. - The yield strength is 79 kN/mm2. - The percentage of elongation is 45 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 17 kN, iv) Young's Modulus if the elongation is 2.4 mm at 17 kN and (v) Final diameter if the percentage of reduction in area is 24 %. Find: 1)Stress at the elastic load (in N/mm2) 2)Young's Modulus of the Specimen (in N/mm2) 3)Final Area of the Specimen at Fracture (in mm) 4)Final Diameter of the Specimen after Fracture (in mm)The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 151 kN. - Length of the specimen is 23 mm. - The yield strength is 79 kN/mm2. - The percentage of elongation is 45 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 17 kN, iv) Young's Modulus if the elongation is 2.4 mm at 17 kN and (v) Final diameter if the percentage of reduction in area is 24 %. THE QUESTION IS : FIND THE Final Diameter of the Specimen after Fracture (in mm)????