fix the end of a flexible rod. Place weight W on the other end. What shape does the rod assume and why?

Answer 1

How flexible is the rod and how much is the weight? If you attach a 1 ounce weight to the end of a piece piece of rebarb then it takes the form of a straight line. If you attach a 500lb weight to the end it takes the shape of a parabola.

Answer 2

Assuming your material behaves linearly (obeys Hookes Law), and you do not plastically deform (permanently deform) and are talking about small displacements, then the shape will actually be governed approximately by a cubic polynomial. This is because it is in a state of constant shear across the length of the rod, which is related to the rod displacement by the 3th derivative.

w''' ~ W/EI (you have to include the stiffness which is a function both of the material of the rod and the area moment of inertia of the cross sectional area (EI)).

Also, if you were really going to include the weight of the rod, you would not model it as a point load in the middle. This is statically equivalent, however in terms of the displacement of the beam, you must model it as a distributed load. In that case, the beam will take the shape of a 4th order polynomial.

The beam must deform because it will store the energy equivalent to the work done in the displacement of the weight at the tip. Conservation of energy says it must go somewhere...so it goes into the rod and as a result it deforms. When you take the weight back off, the rod will return to its original shape (assuming you didnt put too much weight on it, then material dissapation of the energy can occur).

Answer 3

This is what is called a cantilever suspension. The beam is supposed to be of negligible weight. If the beam weight is also considered, the beam's weight is supposed to act at the centre of the rod. Then the resultant weight will be at a point decided by the ratios of the rod weight w and the other weight W.

In any case, the shape is cantilever.