If a steel has a value of oo = 150 MN/m² and k-0.70 MN/m³2, what are the values of yield stress for the k steel with the grain sizes of 100 µm and 1 um? Recall that oy = 0 + √d
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- A compound bar is subjected to a compressive force of P=556 kN by means of a rigid endplate at the top. If the length of the bar is 876mm, determine the stress in the alurninum core. Answer in MPa. Round off your final answer to 2 decimal places. 25 mm Brass core E = 105 GPa a = 20.9 x 10-6/°C Aluminum shell E = 70 GPa Q = 23.6 x 10-6/°C 60 mmProblem- A steel bar 2 m long, 20 mm wide and 15 mm thick is subjected to a tensile load of 30 kN. If Poisson's ratio is 0.25 and Young's modulus is 200 GPa, an increases in volume will be..A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find . The stress at elastic limit. • Young's modulus. . Percentage elongation. Percentage reduction in area. Ultimate tensile stress. .
- For a point on a steel specimen, the principal stresses are known to be 01 = 360 MPa and O2 = 60 MPa. Calculate the minimum yield stress of the material according to the Tresca criterion. Give your answer in MPa to 3 significant figures.A hot-rolled steel bar has a diameter of 12 mm was subjected to a tension test as shown in the figure below. Determine the following. a. If the recorded yield stress was 380 MPa, what is the value of the yielding force (P) in kN? b. If the value of the (P) ultimate was 66 kN, what is the ultimate stress in MPa? c. If the modulus of elasticity (E) of the steel equal to 2 x 105 MPa and stress at the proportional limit equal to 276 MPa, what is the corresponding elastic strain? And what is the unit of this strain? d. If the recorded fracture stress was 490 MPa and the fracture force was 40 kN, what is the diameter of the bar at fracture? e. What can be obtained from the 0.2% offset method? D, = 12 mm10/10 FIND STRESS IN Aluminum. IF E st= .200000 MPa, Ealu.=70000 MPa 100 kN RALU. 30 MM Aluminum Steel 1m RST 20 MM
- A load applied to a machine component results in the state of plane stress ?x=80 MPa, ?y=100 MPa, ?xy=60 MPa. The component is made of a brittle high-strength steel that follows the maximum normal stress criterion with ?u=200 MPa. If increasing the load increases each stress component proportionally, determine the percentage increase that can be applied before the component fails.PROBLEM 1) An aluminum bar carries the axial loads at the positions shown. If E=70GPA, compute the total deformation of the bar. Assume that the bar is suitably braced to prevent buckling. 0.4m D 10KN 0.8m 0.4m 5KN 0.6m A 20KN 3) What is the deformation &cn in mm? A=800 mm? A=1,200 mm²A steel with E = 29 000 ksi with a rectangular cross-section is bent over a rigid mandrel with R = 15 in as shown in the figure. If the maximum flexural stress in the bar is not to exceed the yield strength of 36 ksi, determine the allowable thickness h of the bar.
- In the figure shown below, determine: 1) The final temperature if the normal stress at aluminium is Og = -90 MPa and the initial temperature %3D 20°C. 2) The final length of the aluminium member. Aluminum A=1800mm? Bronze A=1500mm? E=73GPA E=105GPA a=23.2x10-6/°C a=21.6x10-6/°C Gap=0.5mm 0.35m 0.45mSample: Malleable Steel (AISI 4145) Original Diameter: 6.14mm Gauge Length: 55 mm Final Length: 68.12 mm Final Diameter: 3.54mm Load (kN) Deformation (mm) Stress (MPa) 12.95 Strain (%) 0.3838 0.54 0.7841 0.95 26.48 1.3899 1.63 1.9485 2.07 3.3090 3.10 4.5821 4.22 5.9359 5.03 7.0340 5.51 8.2413 6.22 10.9446 7.22 13.1951 8.18 12.8228 8.77 12.2583 9.11 12.5915 9.61 13.2536 10.33 13.6636 10.71 13.9772 11.35 14.5433 12.63 15.1155 13.88 15.4970 15.10 15.6484 17.09 15.4031 17.79 14.7655 18.73 13.6721 19.28 10.4617 19.883. The distribution of stress in an aluminum machine component is given (in megapascals) by Ox = y + z? Oy = x + z Oz = 3x + y Txy = 3z2 Tyz = x Txz = %3D Calculate the state of strain at a point positioned at (1,2,4). Use E=70 GPa and v = 0.3