S= 1 4 F nd² = 4F πα?
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Why did it became 4F/A from F/A, in (a)
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- Solve the preceding problem if the diameter is 480 mm, the pressure is 20 MPa, the yield stress in tension is 975 MPa, the yield stress in shear is 460 MPa, the factor of safety is 2,75, the modulus of elasticity is 210 GPa, Poissorfs ratio is 0.28, and the normal strain must not exceed 1190 x 10" . For part (b), assume that the tank thickness is 8 mm and the measured normal strain is 990 x 10~ .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 0.1025You are in charge of designing a new fixture for a “universal testing machine” that will attach a tensile specimen to the machine using a clevis—a U-shaped piece with holes drilled through the two arms—and a cylindrical pin that passes through the clevis and the specimen. If the maximum load exerted by the machine is 30 kN and the pin is to be made of some sort of steel, what is the minimum pin diameter needed to ensure that the shear stress in the pin does not exceed 600 MPa? Assume that the steel has similar elastic properties to pure Fe.
- A 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?A part is loaded with a combination of bending, axial, and torsion such that the following stresses are created at a particular location: Bending - Completely reversed, with a maximum stress of 60 MPa Axial - Constant stress of 20 MPa Torsion - Repeated load, varying from 0 MPa to 50 MPa Assume the varying stresses are in phase with each other. The part contains a notch such that Kibending = 1.4, Kaxial = 1.1, and K. = 2.0. The material properties are Sy = 300 MPa and S, = 400 MPa. The completely adjusted endurance limit is found to be Se= 200 MPa. Find the factor of safety for fatigue based on infinite life. If the life is not infinite, estimate the number of cycles. Be sure to check for yielding. f,torsionM P 1: Y A :s.google.com Q1 * Find the number of bolts M8 1 which made from SAE Grade 4 steel that should be used to provide a clamping force of 60 KN between two components of a machine. Also, specify the required tightening torque. (Hint: assume that cach bolt is limited to be stressed to 65% of its proof strength). 1 Add file Q2 * A flat plate with a central hole is subjected to an axial tensile force F of 9.8 KN, as shown in the figure below. Calculate the minimum thickness t of the plate. Given that the material of the plate can support maximum tensile stress of 75 N.mm, w = 10 cm and d = 3 cm. Thickness =t F 1 Add file Submit Clear form II
- 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. %3DWrite a short explanation for the following terms: 1. Strain 2. Young's Modulus of Elasticity 3. Design Limit Load & Design Ultimate load 4. Safe-life 5. Fail-safe structure 6. Fatigue 7. Loads on fuselageA part is loaded with a combination of bending, axial, and torsion such that the following stresses are created at a particular location:Bending: Completely reversed, with a maximum stress of 60 MPaAxial: Constant stress of 20 MPaTorsion: Repeated load, varying from 0 MPa to 70 MPaAssume the varying stresses are in phase with each other. The part contains a notch such that Kf,bending = 1.4, Kf,axial = 1.1, and Kf,torsion = 2.0. The material properties are Sy = 300 MPa and Su = 400 MPa. The completely adjusted endurance limit is found to be Se = 160 MPa. Find the factor of safety for fatigue based on infinite life, using the Goodman criteria. If the life is not infinite, estimate the number of cycles, using the Walker criterion to find the equivalent completely reversed stress. Be sure to check for yielding.
- Q6/ A material has a true stress -strain curve given by o=Kɛ", derive and calculate the true and engineering ultimate tensile strength of this material. (use : k=689.47kpa , n=0.5)A tianium disk (roe 107 CPa, Poisson's ratio = 0,34)precisely 8 mm thick by 30 mm diameter is usedas a cover plate in a mechanical loading device. Ifa 20-kN load is applied to the disk, calculate theresulting dimensions.A solid truncated cone is subject to an axial force P as shown. The exact elongation is d = PL/ (2pc²E). Replace the cone by n equal thickness circular cylinders and derive the expression as a function of n, P, L, C, E. draw a graph showing variation in elongation d with respect to n values. Take n from 1 to 100. Given that the P = 1 kip, L = 12 inch, C= 4 inch, E = 30x10° psi. B 2c