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 = 2.0 lb/in.³ is the specific weight of the material, y = 2.8 in. is the distance from the free (i.e., bottom) end of the bar, L = 14 in. is the length of the bar, and E = 20000 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) Eavg= i (c) Emax = i x10-6 in. με με

Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter7: Analysis Of Stress And Strain
Section: Chapter Questions
Problem 7.6.13P: A solid spherical ball of magnesium alloy (E = 6.5 × l0-6 psi, v = 0.35) is lowered into the ocean...
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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 = 2.0 lb/in.³ is the specific weight of the material, y = 2.8 in. is the distance from the free (i.e., bottom) end of the bar, L = 14 in. is the
length of the bar, and E = 20000 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) Eavg
(c) Emax =
i
x10-6 in.
με
με
Transcribed Image Text: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 = 2.0 lb/in.³ is the specific weight of the material, y = 2.8 in. is the distance from the free (i.e., bottom) end of the bar, L = 14 in. is the length of the bar, and E = 20000 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) Eavg (c) Emax = i x10-6 in. με με
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