Select the appropriate Stress -Strain plot which has the highest ultimate tensile strength from the following graph A В A. A В. В С.С D. D
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Q: Given data: Longitudinal strain 0.35 Lateral strain3-0.25 Determine Poisson's ratio
A: Given data : Lateral strain=εLt Longitudinal strain=εL To find : Poisson ratio
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- 1. We can visualize the factor of safety for an arbitrary stress using a surface in principal stress space. For a ductile material that yields according to a von Mises criterion with a yield stress σy, sketch the von Mises surface in σ₁ - 02 space and sketch the stress surface that corresponds to a factor of safety FoS = 2. For a brittle material that yields according to a max normal (Rankine) criterion with a tensile strength Gyt and a compressive strength σvc = 20yt, sketch the yield surface and the surface that corresponds to a factor of safety FoS = 2.diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the fracture 2.00 in. The data is listed in the table. Plot the stress-strain 8-1. A tension test was performed on a steel specimen n original diameter of 0.503 in. and gage length of PROBLEMS *84. origi the f having an for t and stress. Use a scale of 1 in. Dodraw the elastic region, using the same stress scale but a 20 ksi and 1 in. = 0.05 in./in. strain scale of 1 in.= 0.001 in./in. Load (kip) Elongation (in.) 0. 0. 0.0005 0.0015 1.50 4.60 8.00 11.00 0.0025 0.0035 0.0050 11.80 11.80 0.0080 0.0200 12.00 16.60 0.0400 0.1000 0.2800 20.00 21.50 19.50 18.50 0.4000 0.4600 Prob. 8-1A steel pipe with an outside diameter of 115 mm and an inside diameter of 105 mm supports the loadings shown. Assume a = 510 mm, b = 230 mm, Px = 20 kN, P, = 16 kN, and P, = 11 kN. (a) Determine the normal and shear stresses on the top surface of the pipe at point H. (b)Determine the principal stresses and maximum in-plane shear stress magnitude at point H and show the orientation of these stresses on an appropriate sketch. a |P, y P: P H K
- Question: The stress components at point O of a part made of steel material (E = 210 GPa and v= 0.3) are given below. 25 40 - 20 40 30 35| MPа - 20 35 -10 a) Calculate the strain components (Exx, Eyy, Ezz, Yxy, Yxz and Yyz). b) Calculate the principal strain components (E,, E2, and E3) and the maximum shear stress (max). c) Draw the 3-D Mohr circle for the strain components.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.You measure the strains ɛrx = 30E – 6 and ɛyy = 10E – 6 on the surface of a thin part (plane stress). If the part material has a elastic modulus of 100 GPa and a Poisson's ratio of - 0.35. What are the normal stresses in the x and y directions? y-direction normal stress = 2.34 MPа x-direction normal stress = 3.82 MPa O y-direction normal stress = -4.34 MPa x-direction normal stress = -8.27 MPa
- 1. A bicycle wheel is assembled with each spoke pre-tensioned to a tensile force of 95 lbf. The diameter of each spoke is d = 0.0787 in. After the wheel is assembled, a strain gauge is applied to a representative spoke so that the gauge reads zero strain in the pre-tensioned condition. The figure below shows one cycle of strain readings as the wheel rotates through one revolution in use. Zero degrees corresponds to the spoke being between the hub and the ground (i.e., at the six o'clock position). 50 -50 -100 -150 -200 -250 -300 -150 -100 -50 50 100 150 Spoke position in degrees He, microstrainThe figure below shows a hollow circular beam with an outside diameter of 100 mm and a wall thickness of 40 mm loaded in bending with force F = 200 kN. F Ra Rb 1m 1 m Young's modulus of the beam is 80 GPa. Calculate the second moment of area, I, in m* in the form a x 10 0 where the number a is correct to two decimal places. I: x10 m4 Calculate the maximum bending moment, M, in kilonewtonmetres (kNm): M: kNm Calculate the maximum bending stress, o, in megapascals (MPa) correct to two decimal places. MPaIn a volume element with dimensions dx, dy, dz the negatíve y-face has a 2 MPa stress in the positivex-direction, a 3 MPa stress in the negative z-direction, and a 4 MPa stress in the negatíve y-direction. Which of the following describes the stresses per elasticity convention? Oyy = 4 MPa Oyr = -2 MPa 3 MPa A MPa
- Find the strain proportions (ɛ1:E2:E3) that existed in a forming process where O3= 0. Also specify two conditions that can be used the stresses are 01 02=01/3 , to describe the deformation process.Determine the resulting maximum value of the normal stress. Specify the orientation of the plane on which these maximum values occur. **The answer is tensile stress is 0 ksi at 90 degrees. **The answer is compressive stress is 7 ksi at 0 degrees. Can you explain how that is? This was my thought process: I know that tensile would be zero because the force P is actually going inwards and not outwards. I know that means that there would be a compressive force. I am confused on the angles, how is a tensile force going 90 degress if there technically is no force in the tensile direction. And how is there a compressive force at 90 degrees if there is a stress? thank you!Stress 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…