8-22. The elastic portion of the stress-strain diagram for analuminum alloy is shown in the figure. The specimen fromwhich it was obtained has an original diameter of 12.7 mmand a gage length of 50.8 mm. If a load of P - 60 kN isapplied to the specimen, determine its new diameter andlength. Take v-0.35.a (MPa)490(mm/mm)0.007

Young’s modulus can be determined from graph as

Answered: 8-22. The elastic portion of the…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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Determine the nodal displacements and the element stresses. E = 105 GPa, v = 0.3, and t = 5 mm. You don't have to calculate everything manually, but can write a code to get some input from you and find the stiffness matrices for instance. 400 mm 3 5 400 mm 30 kN 2
The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gage length of 2 in. and a diameter of 0.5 in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is Gal = 3.811032 ksi.
The two bars shown in the figure are used to support a load P. When unloaded, joint B has coordinates (0, 0). After load P is applied. joint B moves to the coordinate position (0.40 in., -0.55 in.). Assume a - 10 ft, b-7 ft, and h-8 ft. Determine the normal strain in each bar. Answers: EAB Egc i i B P b με με
Problem 7 The shear stress-strain diagram for an alloy is shown in the figure. If a bolt having a diameter of 0.25 in. is made of this material and used in the lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield. Take v=0.3. 7 (ksi) Ty= 50 0.004 y (rad)
Determine the elastic tensile load "F" that acts on mild steel specimen of Diameter 5 mm & Modulus of Elasticity 207 GPa. It is found that the specimen has undergone an extension of 1.2 mm due to the elastic load "F". Also, determine the length "L" of the specimen, if the strain-induced due to the load "F" on the specimen is 3 x 10-³. Solution Cross-sectional Area, A (in mm²) = Tensile Stress, (in N/mm²) = Tensile Load, (in N) = Length of the Specimen (in mm)
The vertical steel rod has a constant diameter of 32 mm and a length of 400 mm. is hanged from a rigid fixing. A mass is gently placed onto a collar which is attached to the lower end of the rod. The strain energy stored in the steel is 0.400 x 10-³. If Esteet = 209 GPa, 1.1 Determine the gradually applied load and magnitude of the mass. 1.2 Calculate the maximum stress in the rod when gradually applied. 1.3 Determine the magnitude of the mass when the mass is suddenly applied. 1.4 Calculate the extension of the rod when suddenly applied.
8-21. The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. When the applied load on the specimen is 50 kN, the diameter is 12.67494 mm. Determine Poisson's ratio for the material. 8-22. The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. If a load of P = 60 kN is applied to the specimen, determine its new diameter and length. Take v = 0.35. σ (MPa) 490 0.007 Probs. 8-21/22 € (mm/mm)
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The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure below (Figure 1). The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.5 mm. Take v=0.35 Figure σ (MPa) 490 0.007 € (mm/mm) Part A If a load of P = 40 kN is applied to the specimen, determine its new diameter. Express your answer to four significant figures and include appropriate units. d= Submit Part B L= O Value Submit μA Request Answer Determine its length. Express your answer to four significant figures and include appropriate units. Provide Feedback Value Units Request Answer ? Units ?
The normal strain in a suspended bar of material of varying cross section due to its own weight is given by the expression yy/3E where y = 3.0 lb/in.³ is the specific weight of the material, y = 6.3 in. is the distance from the free (i.e., bottom) end of the bar, L = 21 in. is the length of the bar, and E = 27000 ksi is a material constant. Determine, (a) the change in length of the bar due to its own weight. (b) the average normal strain over the length L of the bar (c) the maximum normal strain in the bar. Answer: (a) d = i (b) avg (c) Emax i x10-6 in. με με
Determine the normal stress in each segment of the steel rod shown in the figure. E-200 GPa. A = 600 mm² A B 300 KN + 0.3 m A 200 mm² 200 KN 0.3 m 0.4m D 100 kN
The normal strain in a suspended bar of material of varying cross section due to its own weight is given by the expression yy/3E where y = 2.1 lb/in.³ is the specific weight of the material, y = 2.0 in. is the distance from the free (i.e., bottom) end of the bar, L = 5 in. is the length of the bar, and E = 27000 ksi is a material constant. Determine, (a) the change in length of the bar due to its own weight. (b) the average normal strain over the length L of the bar (c) the maximum normal strain in the bar. Answer: (a) 8 = i (b) avg (c) Emax = i x10 in. με με

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