3:10 Odocs.google.com/forms1010 cm x 10 cmAluminium bar38 cmYour answerThe bar shown is subjected to atensile load of 160kN. If the stress inthe middle portion is limited to 150Newton per millimeter squared,determine the diameter of the middleportion. Find also the length of themiddle portion if the total elongationof the bar is O.2mm. Young's modulusis given as equal to 2.1x10^5 Newtonper millimeter squared.160 kN160 KN6 cm DIA6 cm DIA40 cmYour answerII 3:10 Osubmit this form.Not mtomol@cats.edu.ph? SwitchaccountA member formed by connecting asteel bar into an aluminum bar.Assuming that the bars are preventedfrom buckling sideways, calculate themagnitude of force P that will causethe total length of the member todecrease by 0.25mm. The values forelastic modulus for steel andaluminum 2.1x10^5 Newton permillimeter squared and 7x10^4Newton per millimeter squaredrespectively.5 cm x 5 cmŠteel bar30 cmt0 cm x 10 cmAluminium bar38 cmYour answer

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The bar shown is subjected to a tensile load of 160KN. If the stress in the middle portion is limited to 150 Newton per millimeter squared, determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is 0.2mm. Young's modulus is given as equal to 2.1x10^5 Newton

The bar shown is subjected to a tensile load of 160KN. If the stress in the middle portion is limited to 150 Newton per millimeter squared, determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is 0.2mm. Young's modulus is given as equal to 2.1x10^5 Newton

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter2: Axially Loaded Members

Section: Chapter Questions

Problem 2.5.1P: The rails of a railroad track are welded together at their ends (to form continuous rails and thus...

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