QUESTION 7 Which governing equations are required to solve a plane elasticity problem? Boundary conditions Equilibrium condition and boundary conditions Equilibrium conditions, boundary conditions, and the compatibility equation QUESTION 8 Which of the following conditions must a structure meet to produce a state of plane strain parallel to the x-y plane? Thin-walled Uniform in z-direction Dimension in z-direction that is much smaller than x and y QUESTION 9 What is the major advantage of using the Airy stress function for plane isotropic solids? Reduces the governing equations to only the boundary conditions and compatibility equation Reduces governing equations to a single function with a single unknown QUESTION 10 Airy stress functions are applicable for solving all elasticity problems. Trun

QUESTION 7 Which governing equations are required to solve a plane elasticity problem? Boundary conditions Equilibrium condition and boundary conditions Equilibrium conditions, boundary conditions, and the compatibility equation QUESTION 8 Which of the following conditions must a structure meet to produce a state of plane strain parallel to the x-y plane? Thin-walled Uniform in z-direction Dimension in z-direction that is much smaller than x and y QUESTION 9 What is the major advantage of using the Airy stress function for plane isotropic solids? Reduces the governing equations to only the boundary conditions and compatibility equation Reduces governing equations to a single function with a single unknown QUESTION 10 Airy stress functions are applicable for solving all elasticity problems. Trun

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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Draw a typical stress vs strain tensile test curve for the following materials (two seperate graphs) and label the axis. A ductile metallic test specimen that is stretched to failure displaying a characteristic yield point and show the following parts on the curve. 1- Yield point 2- Ultimate Tensile Strength 3- Breaking point 4- Elastic Region 5- Plastic Region 6- Necking region
Extension Compression Determine the engineering strain "e" and the true strain "e" for each of the following situations of extending a bar of length L or compressing a billet of height h: Extension from L to 1.002 L Compression from h to 0.998 h Extension from L to 1.9L Compression from h to 0.1h Compression from h to zero height
Stress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…
For a given homogeneous, isotropic, linearly elastic material, E = 15e6 psi and v = 0.3. Solve for the shear modulus. 2.1.1 Homogeneous, isotropic, linearly elastic materials For specimens undergoing small deformations, the stress-strain diagram often ex- hibits a linear behavior. Although this is a very crude approximation to the behavior of actual materials, it is a convenient assumption that is often used for preliminary evaluation. A linear relationship between stress and strain can be expressed as 01 = E €1, (2.1) where the coefficient of proportionality, E, is called Young's modulus or modulus of elasticity. Since strains are non-dimensional quantities, this coefficient has the same units as stress quantities, i.e., Pa. This linear relationship is known as Hooke's law. The elongation of a bar in the direction of the applied stress is accompanied by a lateral contraction that is also proportional to the applied stress. The resulting defor- mations for this uniaxial state of stress…
1. A 0.015 meter diameter bar of steel is stress tested and the following stress-strain curve is produced Stress (psix 10 60 50 8 20 10 1 MPa 300 100 psi (x 10%) 50 200-30 OL 40- 20- 10 0.05 0.005 1 L 0.010 0.10 Strain a. What is the ultimate tensile strength (MPa)? d. What is the yield strength(MPa)? 0.15 c. What is the percent plastic elongation at the time of fracture? e. What is the Modulus of Elasticity (GPa)? 400 300 200 Stress (MPa) b. What is the greatest load that can be place on this bar before failure(N)? 100 10 0.20
A tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the ultimate strength. Stress (ksi) 60 50- 40- 30- 20 10- 0 0 0 Upper scale O 66 ksi 28 ksi O 72 ksi O 53 ksi O 62 ksi Lower scale. 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Strain (in./in.)
True stress-strain Engineering stress-strain strain Calculate the engineering ultimate tensile strength of a material whose strength coefficient is 535 MPa and of a tensile-test specimen that necks at a true strain of 0.55. 200 MPa O 267 MPa O 244 MPa O 222 MPa stress
5. Consider a cylindrical specimen of a steel alloy (Figure 6.22) 10 mm in diameter and 75 mm long that is pulled in tension. Determine its elongation when a load of 20,000 N is applied. Stress (MPa) 600 500 400 300 200 100 0 0.00 0.04 Stress (MPa) 500 400 300 200 100 0.000 0.002 0.004 0.006 0.08 Strain Strain 0.12 0.16 0.20 Figure 6.22 Tensile stress-strain behavior for a steel alloy.
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?
An initially rectangular element of a material is deformed into the shape shown in the figure. Deformed Undeformed 16.6 0.1992 mm 15.7 0.2 mm Normal strain in the x- direction * -4.0 x 10^-3 4.0 x 10^-3 10 x 10^-3 -10 x 10^-3 Normal strain in the y- direction O -4.0 x 10^-3 4.0 x 10^-3 O 10 x 10^-3 O -10 x 10^-3 0.1515 mm 0.15 mm