Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

To determine the original length of a slender rod given the normal strain and de

ID: 2124418 • Letter: T

Question

To determine the original length of a slender rod given the normal strain and deformation experienced by the rod due to an applied axial load.


if image can't display plz open the link


https://twitter.com/vxiongv/status/368731306927996930/photo/1


To determine the original length of a slender rod given the normal strain and deformation experienced by the rod due to an applied axial load. As shown, rod AB is subjected to the loading of two blocks, each of weight W (and where we neglect the mass of the pulley). The loading creates a normal strain in the rod given by . where z is measured in centimeters. This strain corresponds to the total deformation in the rod given by Delta B= 5.20cm, Determine the initial length of the rod, L. measured before the loading is applied. Express the initial length in centimeters to three significant figures. Determine the average normal strain in the rod. Express the average normal strain in centimeters per centimeter to three significant figures. By doubling the magnitude of each of the weights of the blocks to 2 W. the total deformation of the rod is double its original value. Determine the ratio of the new initial length of the rod to the original initial length of the rod. Express the ratio of lengths to three significant figures.

Explanation / Answer

soln. Ez = Change in length/original length;

Ez = delta L / delta z ;

delta z is the length of a small section;

delta l is the change in length of the section delta z;

Ez = 0.015 z^0.5;

0.015 z^0.5 = dL / dz;

integral z=0 to z=L ( 0.015 z^0.5) dz= integra dL;

[0.015 * z ^(3/2)*2/3] = delta L;

delta L = 5.2 cm;

puttimg value ;

0.015 * L ^ 1.5 * 2/3 = 5.2 ;

so we get L = 64.66 cm

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote