Elasticity/Sample final6

Sample Final Exam Problem 6

Two long cylinders are in contact as shown in the figure below. Both cylinders are made of the same material which has a Young's modulus of 10 GPa and a Poisson's ratio 0.20. The smaller cylinder has a radius of 4 cm while the outer one has a radius of 10 cm. What is the width of the region of contact of the two cylinders under the action of a force of 1 kN per unit length of the cylinders.

File:Two cylinder contact.png Contact between two cylinders

Solution

The area of contact per unit length is given by

$a=\sqrt{\frac{P{r}_1{r}_2}{\pi (r_1+r_2)}\left(\frac{\kappa +1}{\mu }\right)}$

For plane strain

$\kappa =3-4\nu$

Therefore, for the material of the cylinders

$\kappa =3-(4)(0.20)=2.2;\mu =\frac{E}{2(1+\nu )}=\frac{10}{2(1+0.20)}=4.2\text{GPa}$

Since the outer cylinder contains the inner one, the radius of curvature can be considered to be negative. Therefore,

$r_1=4\text{cm}=0.04\text{m};r_2=-10\text{cm}=-0.1\text{m};$

The area of contact per unit length of the cylinders is

$a=\sqrt{\frac{(1)(10^3)(0.04)(-0.1)}{\pi (0.04-0.1)}\left(\frac{2.2+1}{(4.2)(10^9)}\right)}=0.13\text{mm}$

The width of the region of contact is

$a=0.13\text{mm}$