A tensile stress is to be applied along the axis of a cylindrical brass rod with diameter of 200 mm. What is the load magnitude required to produce 2.8 x 10^-2 change in diameter if the deformation is entirely elastic?
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A tensile stress is to be applied along the axis of a cylindrical brass rod with diameter of 200 mm. What is the load magnitude required to produce 2.8 x 10^-2 change in diameter if the deformation is entirely elastic?
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- (a) Solve part (a) of the preceding problem if the pressure is 8.5 psi, the diameter is 10 in., the wall thickness is 0,05 in., the modulus of elasticity is 200 psi, and Poisson's ratio is 0.48. (b) If the strain must be limited to 1.01, find the maximum acceptable inflation pressureThe Highest load sustained druing an uniaxial tensile testing experiment is 7,500lb. If the original cross section has a diamter of 0.25in, what is the ultimate tensile strength? (please also make a drawing)Derive the following equation for computing shear stress in a solid bar along an oblique plane with angle B with respect to the vertical plane. The bar is subjected to centric axial loading. sin 0 cos 0 T = Ao
- 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 tensile load of 3 kN is applied to the ends of the rod (parallel to the 2.3m dimension), what is the applied engineering normal stress?Example: Convert the change in length data in Table 3-2 to engineering stress and strain and plot a stress-strain curve Homework- help Table 3-2 The results of a tensile test of a 0.505 in. diameter aluminum alloy test bar, initial length (1o) = 2 in. Calculated LTO Load (Ib) Change in Length (in.) Stress (psi) Strain (in./in.) 0.000 1000 0.001 0.0005 4,993 14,978 24,963 34,948 37,445 39,442 39,941 39,691 37,944 3000 0.003 0.0015 5000 0.005 0.0025 7000 0.007 0.0035 7500 0.030 0.0150 7900 0.080 0.0400 8000 (maximum load) 0.120 0.0600 7950 0.160 0.0800 7600 (fracture) 0.205 0.1025(Suppose you need to design a tension test machine capable of testing specimens that have nominal ultimate stresses as high as σu = 100 ksi . How much force must the machine be capable of generating? Assume the testing specimen has the ASTM shape shown. Answer for this is 19.6 kip) (If the maximum nominal strain is ϵf = 0.7 just before the test specimen fractures and the test machine operates by moving only one grip, how far must that grip be designed to travel? The total length of the deforming part of the specimen is 3 in. Answer for this is 2.10 in) Do not know if this info is needed but this was the other 2 parts
- a)Draw the stress-strain curve of a ductile material such as that of a ductile steel and discuss the various regions of the curve. Explain the difference between a ductile and brittle material and the concept of ductility and its measurement. b)A machine part 10 mm thick, having the dimensions shown in the figure, is to be subjected to cyclic loading. If the maximum stress is limited to 60 MPa, determine the allowable force P. Approximate the stress concentration factors form the graph given. Where might a potential fracture occur?A. Calculate the axial deformation of a steel rod 20 cm long, 10 cm diameter. If an axial loading 300 MPa is applied at the end. B. If the cross-sectional is square with t=10 mm, is the deformation will be higher or lower. (use E=200 GPa) * Vour onouIorIn the attached picture there is a sketch of a socket wrench. Assume the wrench is held at a fixed point “A”. The yield stress of the material is known to be 500 MPa. Answer the questions below Describe the stresses at point “A” and their causes and calculate the stresses. Determine the factor of safety against yield assuming the Tresca yield criteria. Determine the factor of safety against yield assuming the von Mises yield criteria using both principal stresses and “Cartesian” stresses. Do your values match or not, and is this expected? Explain. Do the calculated values make sense with the respect to the Tresca value? Explain, why or why not?
- A sample test rod made of a ferrous material of 0.5 in. diameter was tested for tensile strength with the gauge length of 2.5 in. The following observations were recorded: Final length = 3.15 in; Final diameter = 0.275 in; Yield load = 7650 lb. and Ultimate load = 10375 lb. (a) Calculate: 1. yield stress, 2. ultimate tensile stress, 3. percentage reduction in area, and 4. percentage elongation. (b) What specific material, if any, is similar to this tested rod? Refer to AT7 and AT8. (c) This same material is then used as a tension member 55 inches long and is subjected to a maximum load of 6000 lb, repeated but not reversed. (d) Solve part (c) if the total elongation is not to exceed 0.025 in.? 1. %3DA stress-strain diagram for a tension test of an alloy steel sample is shown in the diagram below. The initial sample diameter is 0.502”, the diameter of the fractured sample is 0.412”, the original gage length is 2.00” and the final length after fracture is 2.78”. Determine/ calculate the following: (a) The stress at the proportional limit. (b) The modulus of elasticity (E). (c) The yield stress using the 0.2% offset method. (d) Theultimatetensilestrength. (e) The rupture stress. (f) The percent reduction in area. (g) The percent elongation. (h) Is the material considered ductile or brittle?The data shown in the accompanying table are From a tensile test of high-strength steel. The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for Prob. 1.5-3). At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (the slope of the initial part of thestress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in., and percent reduction in area. TENSILE-TEST DATA FOR PROB. L.5-7 Laid (lb) Elongation (in) 1000 0.0002 2000 0.0006 6000 0.0019 10,000 0.0033 12,000 0.0039 12,900 0.0041 13,400 0.0047 13,600 0.0054 13,800 0.0063 14,000 0.0090 14,400 0.0102 15,200 0.0130 16,800 0.0230 18,400 0.0336 20,000 0.0507 22,400 0.1108 22,600 Fracture