Rockwell hardness . figure 6.19 Rockwell hardness Tensile strength = 0 MPA Tensile strength = 0 ksi 60 70 80 90 100 HRB Brinell hardness i HB 20 30 40 50 HRC Rockwell hardness i HRB 250 1500 200 Steels 1000 150 (b) The steel alloy for which the stress-strain behavior is shown in Animated Figure 6.22. You might need to use Animated 100 Figure 6.19. 500 Brass Cast iron (nodular) 50 Brinell hardness i HB Rockwell hardness i HRB 400 500 Brinell hardness number Tensile strength (MPa) () abuaas osua

Rockwell hardness . figure 6.19 Rockwell hardness Tensile strength = 0 MPA Tensile strength = 0 ksi 60 70 80 90 100 HRB Brinell hardness i HB 20 30 40 50 HRC Rockwell hardness i HRB 250 1500 200 Steels 1000 150 (b) The steel alloy for which the stress-strain behavior is shown in Animated Figure 6.22. You might need to use Animated 100 Figure 6.19. 500 Brass Cast iron (nodular) 50 Brinell hardness i HB Rockwell hardness i HRB 400 500 Brinell hardness number Tensile strength (MPa) () abuaas osua

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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The fluctuating stresses listed in the table are found at a critical location of a component made of steel with Se = 40 ksi, Sy = 80 ksi, Sut = 110 ksi and f= 0.87. These stresses are applied on the part within 10 s. What is the accumulative damage of this part? What is the life of the part in hours if this stress pattern continues to repeat for the remained of the part's life? Use Goodman criterion and Miner's rule in your solution. Loading order |Omin o max |Number of cycles -20 30 -10 50 1 3. -30 30 1
Sketch a representative stress-strain digrama for a ductile material showing the following on it (label the axis and provid units). 1- YIELD STRENGTH. 2- TENSILE STRENGTH. 3- FRACTURE POINT. 4- HARDENING PART. 5- SOFTENING PART. 6- A METHOD TO CALCULATE THE STIFFNESS. 7- A METHOD TO CALCULATE THE DUCTILITY.
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
The figure below shows the tensile engineering stress-strain behavior for a steel alloy. (a) What is the modulus of elasticity? (b) What is the yield strength at a strain offset of 0.002? (c) What is the tensile strength? Stress (MPa) 600 500 400 300 200 100 T I 0.00 Stress (MPa) 0.04 500 400 300 200 100 0.000 0.08 0.002 Strain 0.004 Strain 0.12 I 0.006 0.16 0.20 In the previous problem, A load of 85,000 N (19,100 lbf) is applied to a cylindrical specimen of the steel alloy that has a cross-sectional diameter of 15 mm (0.59 in.). (a) Will the specimen experience elastic and/or plastic deformation? Why? (b) If the original specimen length is 250 mm (10 in.), how much will it increase in length when this load is applied?
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Sketch a tensile member with (a) a rectangular crosssection, (b) a solid circular cross section, and (c) a circulartubecross section, and label the dimensions symbolically (e.g.,label the radius for the solid circular case). For each member, write out the definition of engineering stress in terms of the actual dimensions of the component. If therectangular member has dimensions of width and thicknessequal to 1 cm 0.3 cm, what would be the radius of a solidcircular member such that the stress is equal for an equaltensile load? If a tube has an outer radius equal to that ofthis same solid cylinder, what is the maximum inner radiussuch that the stress does not exceed 200% of the stress inthe solid cylinder?
A tensile test was conducted on a mild steel& the following data was obtained as follows. Diameter of the steel bar = 3 cm,Gauge length of the bar = 20 cm. Load at elastic limit = 250 kN . Extension at a load of 150kN = 0.21 mm . Maximum load = 380 kN . Total extension = 60 mm . Diameter of the rod at failure = 2. 25 cm Determine: (a) Young’s modulus (b) stress at elastic limit (c)the percentage of elongation (d) Percentage decrease in area.
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. From the tensile stress-strain behaviour for a casted iron as shown in the following curve. Please answer the following questions (1) Estimate the modulus of elasticity, (2) Calculate the change in length of a specimen with original length of 100mm that is subjected to a tensile stress of 300MPa. (Hint: Using the Hooke's law to get the strain.) 600 500 400 - 300 - 200 - 100 - 2 4 6. 8 10 12 14 16 Strain Strength (MPa)
The following stress-strain curve was prepared based on a tensile test of a specimen that had a circular cross-section. The gage diameter of the specimen was 0.25 inches and the gage length was 4 inches. The stress scale of the stress-strain diagram is given with the factor a = 10 ksi. Estimate: (a) The modulus of elasticity. (b) The ultimate strength. (c) The yield strength (0.2% offset). (d) The percent elongation at fracture. 2013 Michael Swanbom STRESS VS. STRAIN BY NC SA 7a bat Sat 2at at 0.05 STRAIN 0.01 0.04 0.06 0.08 0.02 0.03 0.07 0.09 STRESS