Irodov Problem 1.298

Consider a section of the rod XB that is x units long from the lowest end of the rod. Let the density of the rod be p and its cross-section area be S. There are two forces acting on this part of the rod, i) the force of gravity pxSg pulling it down and ii) the force due to stress in the rod at point X given by . Since the section of the rod is not falling down we have,



The strain experienced by the section XB of the rod is thus given by,



Suppose that the change in length for the section XB of the rod is , then we have,












The net strain in the rod is thus given by,




(b) In addition to the longitudinal strain that elongates the rod, the rod also experiences a lateral strain (dictated by Poisson's ratio) that leads to a reduction in the rod's radius given by . In other words if the radius of the rod is R, then the change in the radius is given by,




The change in volume of the rod is given by,

Irodov Problem 1.297

Let the radius of the cylinder be R then the pressure seen by the cylinder is given by . The longitudinal strain seen by the cylinder is . In other words . The pressure also causes a lateral strain dictated by the Poisson's ratio making the cylinder to widen (radius increases) given by . In other words,. We can now calculate the change in volume as,

Irodov Problem 1.299

Liquids exert equal pressure in all directions, thus the body experiences equal pressure p from all directions. Suppose that the dimensions of the body are a,b, and c as shown in the figure.

Each side experiences a compression due to the pressure along its direction (dictated by Young's modulus) and expansion due to lateral strain (dictated by the Poisson's ratio).

Hence, we have,

















The change in volume is given by,











From he above expression it seems as though for a value of Poisson's ratio that is greater than 0.5, the volume of the body will actually increase. Since the body is being compressed from all directions, its volume cannot possibly increase. Thus, we conclude that the value of Poisson's ratio cannot be greater than 0.5.

Irodov Problem 1.296

Consider an infinitesimally small piece of the rod at a distance r (point X) from the axis of rotation of length dr. If the density of the material is p and surface area of cross-section of the rod is S, then the mass of the infinitesimally small piece of rod will be pSdr. The centrifugal force experienced by this infinitesimally small piece will be .

Let the force with which the part of the rod XB pulls the part AX with a force f(r) given by,








The stress is thus given by,



The strain is thus given by,





Suppose that the part XB was stretched by due to the rotation, then we have,

Irodov Problem 1.295

Suppose that the mass of the plank AB is m, its length is l and that is experiences an acceleration of a. Hence, we have,



Now consider the motion of a section of the plank XB of length x. Since the plank is of uniform density, the mass of the section will be . Let force exerted by the plank section AX at the cross section of XB be f(x). This force f(x) is responsible for accelerating the plank section XB at rate a. Hence we have,








The stress induced in the plank cross-section at point X is given by,
























Suppose that prior to the application of the force F, the plank was A'B, and the point X was located at X'. After the force, the plank gets compressed so that A moves to A' and X moves to X'. Let the change in length of section XB, XX' be . The strain experienced by the section XB of the plank is given by ,




But we already know that,