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Contraction 1. Yield stress at an offset strain of 0. Yield stress at an extension strain of 0. Secant modulus at a stress of 62 ksi. Tangent modulus at a stress of 65 ksi. Proportional limit b.
Ultimate strength d. Modulus of resilience e. Toughness f. Material A 51 ksi 63 ksi Material B 40 ksi 52 ksi ksi 73 ksi 0. Assume that the stress is within the linear elastic range. For a strain of 0. See Sections 1. The stresses and strains can be calculated as follows: Time min. Stress versus time plot for the asphalt concrete sample 14 12 Stress, psi 10 8 6 4 2 0 0 50 Time, minutes Strain versus time plot for the asphalt concrete sample 0.
The phenomenon of the change of specimen height during static loading is called creep while the phenomenon of the change of specimen height during unloading called is called recovery.
See Figure 1. The change in length can be calculated using Equation 1.
The tension load needed to return the length to the original value of 4 meters can be calculated as follows: If the bar was fixed at one end and free at the other end, the bar would have contracted and no stresses would have developed.
In that case, the change in length can be calculated using Equation 1. Since the bar is fixed at both ends, the length of the bar will not change. Therefore, a tensile stress will develop in the bar as follows. The control chart is shown below. The target value is any value above the specification limit of 5, psi. The plant production is meeting the specification requirement.
The control chart shows that most of the samples have asphalt content within the specification limits. Only few samples are outside the limits. The plot shows no specific trend, but large variability especially in the last several samples. No information is given about accuracy. Accuracy can be improved by calibration.
See Section 2. For the BCC lattice structure: For the FCC lattice structure: From Table 2. See Figure 2. Spreading salt reduces the melting temperature of ice. When temperature increases more ice will melt. At a temperature of -5oC, all ice will melt. STEEL 3. At a temperature just higher than C all the austenite will have a carbon content of 0. The ferrite will remain as primary ferrite.
The proportions can be determined from using the lever rule. The ferrite will have 0. See Section 3. A wide-flange shape that is nominally 36 in. Cold forming will almost double the yield strength to 66 ksi. See Figures 3.
See Figure 3. The proportional limit is at a stress of MPa and a strain of 0. Yield strength at 0. No deformation because the applied stress is below the proportional limit and therefore below the elastic limit.
Yes, because the applied stress is much below the yield strength. The proportional limit is at a stress of The true strain at failure is larger than the engineering strain at failure since the increase in length at the vicinity of the neck is much larger than the increase in length outside of the neck.
The specimen experiences the largest deformation contraction of the cross-sectional area and increase in length at the regions closest to the neck due to the nonuniform distribution of the deformation.
The large increase in length at the neck increases the true strain to a large extent for the following reason. The definition of true strain utilizes a ratio of the change in length in an infinitesimal gauge length. By decreasing the gauge length toward an infinitesimal size and increasing the length due to localization in the neck, the numerator of an expression is increased while the denominator stays small resulting in a significant increase in the ratio of the two numbers.
Note that when calculating the true strain, a small gauge length should be used at the neck since the properties of the material such as the cross section at the neck represent the true material properties. The linear portion of the stress-strain relationship is shown below.
No, because the applied stress would result in permanent deformation in the structure. Easiest solution is to "google" the shape S The area is 0. This is conservative since yield strength will increase by strain hardening. No, in compression buckling would control for a thin member.
The proportional limit is at a stress of Mpa and a strain of 0. Using S. Using U. The toughness versus temperature relation is shown below. Therefore, the steel member has adequate Chary V-notch fracture toughness.
See the introduction section of Chapter 4. The modulus of aluminum is lower. Therefore, the aluminum section must be larger. The material property that controls the tension is the yield strength. The yield s of the steel is lower. Therefore, steel would require a larger cross section.
Stress, ksi 4. The stress-strain relationship is shown below.
The proportional limit is at a stress of 58 ksi and a strain of 0. The proportional limit is at a stress of MPa and a strain of 5. See Table 4. The proportional limit is at a stress of 50, psi.
Stress, ksi b. The proportional limit is at a stress of 45, psi and a strain of 0. Observation 0. The yield strength at a strain offset of 0.
Drawing a line parallel to the original part of the curve shows that the final plastic strain is about 0.