Calculation -compression Stress (6c) = A - compression strain (c) = Lo-Lf -young's Modulus (E) = Ec hight = 6-5mm. d = 5.9 mm. Write the full table. 15 fields
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- The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its weight. For materials in tension, use a characteristic tensile stress obtained from a stress-strain curve as a measure of strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus, the strength-to-weight ratio RS/Wfor a material in tension is defined as Rs/w= in which a is the characteristic stress and 7 is the weight density. Note that the ratio has units of length. Using the ultimate stress Uas the strength parameter, calculate the strength-to-weight ratio (in units of meters) for each of the following materials: aluminum alloy 606I-T6, Douglas fir (in bending}, nylon. structural steel ASTM-A57.2, and a titanium alloy. Obtain the material properties from Tables [-1 and 1-3 of Appendix I. When a range of values is given in a table, use the average value.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?A wine of length L = 4 ft and diameter d = 0.125 in. is stretched by tensile forces P = 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by =18,0001+30000.03(=ksi) in which is nondimensional and has units of kips per square inch (ksi). (a) Construct a stress-strain diagram for the material. (bj Determine the elongation, of the wire due to the Forces P. (c) IF the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the proportional limit?
- Solve the preceding problem if the thickness of the steel plate is. t = 12 mm. the gage readings are x = 530 × 10-6 (elongation) and y = -210 -× l0-6 (shortening), the modulus is E = 200 GPa, and Poisson’s ratio is v = 0.30.Derive the relationship between critical resolved shear stress and disloca- tion density given below Terss = To + AVPDThe rod has a diameter of 40 mm. (Figure 1) It is subjected to the force system F = 1390 N, Fy = 510 N, F₂ = 835 N. Figure 5 300 mm 100 N-m F₂ 100 mm B 1 of 1 Determine the normal stress that acts at point B. Express your answer to three significant figures and include appropriate units. OB = Submit Part B (TB)xy 0 = μÅ Value Request Answer Determine the shear stress Try that acts at point B. Express your answer to three significant figures and include appropriate units. 0 μᾶ Units Value ? Units ?
- Quiz ch-2 Strain rosette readings are made at a critical point on the free surface in a structural steel, member. The 60° rosette contains three wire gages positioned at 0°, 60°, and 120° The readings are: Ea=190 ustrain, Eh=200 ustrain, &c=300 µstrain E-200GPa & v=0.3 Determine the principal strains & principal stresses only?Activate WinsExample: 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 0.1025What is the maximum shear stress RESULTING FROM THE COMBINATION OF THE FOLLOWING stresses: Tesile stresses of SX-200 MPA, SY-100 MPA AND SHEAR STRESS of Ss=50 MPa?
- 7:Y. O %Y. O 36 ll ll く ロ quiz 2 elearning.uowa.edu.iq Marked out of 1.00 P Flag question quiz 2: flexural stress which fluctuates between + 250 MN/m2 and - 115 MN/m² applied on a machine component. Calculate the ultimate strength value according to Goodman's relation. Take yield strength = 0.55 Ultimate %3D strength; Endurance strength = 0.5 Ultimate strength; and factor of safety = 1.5. A- В I D II1. 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.M 1:-A O photo_2021-02-1.. -> 200 stress 150 (MPa) L00 50 0.1 0-2 strain From the tensile stress-strain behavior fer the brass specimen the foibuding:- O the diameter is 12-8 mm Calcalate 1- The Yield stress. 2- Modulus of elasticity. 3 - The Tensile strength . II