specimen Yield strength based on actual Yield strength based on 12mm based on actual diameter (MPa) diameter (MPa) Ultimate strength Ultimate strength STRAIN based on 12mm diameter (MPa) diameter (MPa)
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- A high-strength steel bar used in a large crane has a diameter d = 2.00 in. (sec figure). The steel has a modulus of elasticity E = 29 × 10 psi and Poisson’s ratio is v = 0.29. Because of clearance requirements, the diameter of the bar is limited to 2.001 in. when it is compressed by axial forces. What is the largest compressive load Pmaxthat is permitted?The rails of a railroad track are welded together at their ends (to form continuous rails and thus eliminate the clacking sound of the wheels) when the temperature is 60°F. What compressive stress ?? =6.5×10-6 /? is produced in the rails when they are heated by the sun to 120"F if the coefficient of thermal expansion a = the modulus of elasticity E = 30 × 106 psi?An aluminum bar has length L = 6 ft and diameter d = 1.375 in. The stress-strain curse for the aluminum is shown in Fig. 1.34. The initial straight, line part of the curve has a slope (modulus of elasticity) of 10.6 × 106 psi. The bar is loaded by tensile forces P = 44.6 k and then unloaded. (a) That is the permanent set of the bar? (b) If the bar is reloaded. what is the proportional limit? hint: Use the concepts illustrated in Figs. l.39b and 1.40.
- (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 pressure6 cans containing soda made out of aluminum ( E = 69 GPa, σY = 240 MPa ) material are shaken. Shaking them creates an internal pressure of 0.14 MPa in each can. A uniform load of W is applied to the cans through a plate which has a negligible weight. The inner diameter of each can is 2R = 62 mm, the wall thickness is t = 0.25 mm and the length is L = 112 mm. According to the von - Mises yield criterion, determine the maximum magnitude of the applied load W. Note: The weight of the cans will be neglected in calculations. Accept that the liquid in the cans does not provide any support to the cans and no buckling occurs in the cans.3. Two brass rods (one square, one circular) are extruded; initial and final dimensions are provided below. Which part is expected to have the greater tensile strength? Which part is expected to have the greater ductility? Part Initial Dimensions Final Dimensions Cylindrical D=2.5 in D = 2.35 in. Square 4 × 4 in. 3.9 × 3.9 in.
- A Brass specimen is subjected to a Tensile test in a laboratory. The following data obtained are: Length of Specimen = 367 mm; Diameter of Specimen = 19 mm; Yield Stress = 124 N/mm?: Ultimate Stress = 217 N/mm?: Fracture Stress = 183 N/mm?: % of Elongation = 50 % % of Reduction in area = 27 %. Determine the elongation at 103 kN, Load at yielding. Maximum load, Load at fracture, Final Length & Final diameter. Take Young's Modulus as 104 GPa.6. Discuss the different types of volume change in concrete at early and long-term ages. 8. A center-point loading flexural test was performed on a concrete beam having a size of100×100×500 mm. The beam is sitting on two supports located at 25 mm from each edge ofbeam. If the failure load was 17.5 kN, find the flexural strength.9. Considering the two tests of flexure, 3-points and 4-points, which one is more suitable forbending test and why?10. The compression strength test was performed on several samples of concrete molded in fourtypes of molds cubes of 0.100 m×0.100 m ×0.100 m and of 0.150 m×0.150 m × 0.150 m, andcylinders having a cross section of d=100 mm by H=200 mm and d=150 mm by H=300 mm,respectively. If the weight of the samples and the loads at failure are as given in the Table. Bothcubes and cylinder samples are cast from the same concrete batch under the same conditionsand subjected to the same curing method.11. Calculate the density (in kg/m3) and the compressive strength…A 1.64 m long rectangular AISI 1020 cold rolled steel bar (10 mm x 25 mm) is subject to a torsional load of 76.6 N·m. What is the safety factor against yielding? (Use Shigley for material properties.)
- 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.A tensile test specimen has a starting gage length 50 mm and a cross-sectional area 200 mm2. During the test, the specimen yields under a load of 30,000 N (this is the 0.2% offset) at a gage length of 52 mm.The maximum load of 63,000 N is reached at a gage length of 57 mm just before necking begins. Final fracture occurs at a gage length of 63.5 mm. Determine (a) yield strength, (b) modulus of elasticity, (c) tensile strength, (d) engineering strain at maximum load, and (e) percent elongation.Experiments were conducted in a tension testing machine to evaluate the material properties of two metallic rods both having diameter equal to 10 mm. First rod having modulus of rigidity of the material equal to 400 tonnes/cm2 when subjected to an axial tensile force of 6 kN, the change in its diameter was observed to be 0.0018 cm. The value for the bulk modulus of the second metallic rod is same as that of the first one but modulus of elasticity 18% more. Based on the material properties calculate and compare the values of different moduli as well as Poisson’s ratio for both metallic rods. Justify your answer with reasons.