Sample Final Exam Problem 6
edit
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.
Contact between two cylinders
Solution
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The area of contact per unit length is given by
a
=
P
r
1
r
2
π
(
r
1
+
r
2
)
(
κ
+
1
μ
)
{\displaystyle a={\sqrt {{\frac {Pr_{1}r_{2}}{\pi (r_{1}+r_{2})}}\left({\frac {\kappa +1}{\mu }}\right)}}}
For plane strain
κ
=
3
−
4
ν
{\displaystyle \kappa =3-4\nu }
Therefore, for the material of the cylinders
κ
=
3
−
(
4
)
(
0.20
)
=
2.2
;
μ
=
E
2
(
1
+
ν
)
=
10
2
(
1
+
0.20
)
=
4.2
GPa
{\displaystyle \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
cm
=
0.04
m
;
r
2
=
−
10
cm
=
−
0.1
m
;
{\displaystyle 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
=
(
1
)
(
10
3
)
(
0.04
)
(
−
0.1
)
π
(
0.04
−
0.1
)
(
2.2
+
1
(
4.2
)
(
10
9
)
)
=
0.13
mm
{\displaystyle 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
mm
{\displaystyle {a=0.13~{\text{mm}}}}
